source_id
int64 1
4.64M
| question
stringlengths 0
28.4k
| response
stringlengths 0
28.8k
| metadata
dict |
|---|---|---|---|
14,613 |
Let's say I start with the polynomial ring in $n$ variables $R = \mathbb{Z}[x_1,...,x_n]$ (in the case at hand I had $\mathbb{C}$ in place of $\mathbb{Z}$).
Now the symmetric group $\mathfrak{S}_n$ acts by permutation on the indeterminates.
The subring of invariant polynomials $R^{\mathfrak{S}_n}$ has a nice description (by generators and relations) in terms of symmetric functions. What if I only consider the action of the cyclic group $Z_n$?
Does anyone know if the ring $R^{Z_n}$ admits a nice presentation?
(in the case at hand I had $\mathbb{C}[x_1,x_2,x_3]$ and the action of the cyclic group $Z_3$. Maybe in this case we can use some formula (assuming there is one) for groups splitting as semidirect products?)
|
The actions of $S_n$ and $\mathbb Z_n$ differ in the sense that in the first case the quotient is smooth (it is again $\mathbb C^n$ ) while in the second case it is singular. This is why in the fist case we have a nice presentation, but in the second not really. For example, the number of generators of the quotient can not be less than the dimension of Zariski tangent space to the singularity at zero of $\mathbb C^n/\mathbb Z_n$ . Still in principle the presentation can be provided by toric geometry ( https://dacox.people.amherst.edu/toric.html ) because the quotient is the toric singularity. For example, in your case of $\mathbb C^3/\mathbb Z_3$ let us change the coordinates so that $\mathbb Z_3$ is acting as $w_0\to w_0$ , $w_1\to \mu w_1$ , $w_2\to \mu^2 w_2$ (here $\mu^3=1$ ). Then you can write the minimal set of four generators: $w_0, w_1^3, w_2^3, w_1w_2$ , and one obvious relation $(w_1^3w_2^3)=(w_1w_2)^3$ The case $\mathbb C^n/\mathbb Z_n$ for $n>3$ will be more involved, but the idea is the same roughly. First you chose the coordinates on $\mathbb C^n$ $w for which the action is diagonal. Then pick the minimal set of monomials (in these new coordinates) that are invariant under the action, and generate the whole set of invariant monomials (of positive degree). Consider one more case $n=4$ , and chose the coordinates $w_i$ , so that $Z_4$ is acting as $w_i\to \mu^iw_i$ , $\mu^4=1$ . The number of generators is $7$ this time: $w_0, w_1^4, w_3^4, w_2^2, w_1w_3, w_1^2w_2, w_3^2w_2$
|
{
"source": [
"https://mathoverflow.net/questions/14613",
"https://mathoverflow.net",
"https://mathoverflow.net/users/3701/"
]
}
|
14,627 |
I was reading Dieudonne's "On the history of the Weil conjectures" and found two things that surprised me. Dieudonne makes some assertions about the work of Artin and Schmidt which are no doubt correct, but he doesn't give references, and the thought of ploughing through Artin's collected works seems a bit daunting to me, so I thought I'd ask here first. Background. If $V$ is a smooth (affine or projective) curve over a finite field $k$ of size $q$, then $k$ has (up to isomorphism) a unique extension $k_n$ of degree $n$ over $k$ (so $k_n$ has size $q^n$) and one can define $N_n$ to be the size of $V(k_n)$. Completely concretely, one can perhaps imagine the case where $V$ is defined by one equation in affine or projective 2-space, for example $y^2=x^3+1$ (note that this equation will give a smooth curve in affine 2-space for $p$, the characteristic of $k$, sufficiently large), and simply count the number of solutions to this equation with $x,y\in k_n\ $to get $N_n$. Let $F_V(u)=\sum_{n\geq1}N_nu^n$ denote the formal power series associated to this counting function. Now it turns out from the "formalism of zeta functions" that this isn't the most ideal way to package the information of the $N_n$, one really wants to be doing a product over closed points of your variety. If $C_d$ is the number of closed points of $V$ of degree $d$, that is, the number of closed points $v$ of (the topological space underlying the scheme) $V$ such that $k(v)$ is isomorphic to $k_d$, then one really wants to define
$$Z_V(u)=\prod_{d\geq1}(1-u^d)^{-C_d}.$$
If one sets $u=q^{-s}$ then this is an analogue of the Riemann zeta function, which is a product over closed points of $Spec(\mathbf{Z})$ of an analogous thing. Now the (easy to check) relation between the $C$s and the $N$s is that $N_n=\sum_{d|n}dC_d$, and this translates into a relation between $F_V$ and $Z_V$ of the form
$$uZ_V'(u)/Z_V(u)=F_V(u).$$ This relation also means one can compute $Z$ given $F$: one divides $F$ by $u$, integrates formally, and then exponentiates formally; this works because $f'/f=(\log(f))'$. The reason I'm saying all of this is just to stress that this part of the theory is completely elementary. The Weil conjectures in this setting. The Weil conjectures imply that for $V$ as above, the power series $Z_V(u)$ is actually a rational function of $u$, and make various concrete statements about its explicit form (and in particular the location of zeros and poles). Note that they are usually stated for smooth projective varieties, but in the affine curve case one can take the smooth projective model for $V$ and then just throw away the finitely many extra points showing up to see that $Z_V(u)$ is a rational function in this case too. How to prove special cases in 1923? OK so here's the question. It's 1923, we are considering completely explicit affine or projective curves over explicit finite fields, and we want to check that this power series $Z_V(u)$ is a rational function. Dieudonne states that Artin manages to do this for curves of the form $y^2=P(x)$ for "many polynomials $P$ of low degree". How might we do this? For $P$ of degree 1 or 2, the curve is birational to projective 1-space and the story is easy. For $V$ equals projective 1-space, we have
$$F_V(u)=(1+q)u+(1+q^2)u^2+(1+q^3)u^3+\ldots=u/(1-u)+qu/(1-qu)$$
from which it follows easily from the above discussion that
$$Z_V(u)=1/[(1-u)(1-qu)].$$
For polynomials $P$ of degree 3 or 4, the curve has genus 1 and again I can envisage how Artin could have approached the problem. The curve will be birational to an elliptic curve, and it will lift to a characteristic zero curve with complex multiplication. The traces of Frobenius will be controlled by the corresponding Hecke character, a fact which surely will not have escaped Artin, and I can believe that he was now smart enough to put everything together. For polynomials of degree 5 or more, given that it's 1923, the problem looks formidable. Q1) When Dieudonne says that Artin verified (some piece of) the Weil conjectures for "many polynomials of low degree", does he mean "of degree at most 4", or did Artin really move into genus 2? How much further can we get in 1931? Now this one really surprised me. Dieudonne claims that in 1931 F. K. Schmidt proved rationality of $Z_V(u)$, plus the functional equation, plus the fact that $Z_V(u)=P(u)/(1-u)(1-qu)$, for $V$ an arbitrary smooth projective curve, and that he showed $P(u)$ was a polynomial of degree $2g$, with $g$ the genus of $V$. This is already a huge chunk of the Weil conjectures. We're missing the statement that $P(u)$ has all its rots of size $q^{-1/2}$ (the "Riemann hypothesis") but this is understandable: one needs a fair amount of machinery to prove this. What startled me (in my naivity) was that I had assumed that all this was due to Weil in the 1940s and I am obviously wrong: "all Weil did" was to prove RH. So I have a very basic history question: Q2) However did Schmidt do this? EDIT: brief summary of answers below (and what I learned from chasing up the references): A1) Artin didn't do anything like what I suggested. He could explicitly compute the zeta function of an arbitrary given hyperelliptic curve over a given finite field by an elegant application of quadratic reciprocity. See e.g. the first of Roquette's three papers below. The method in theory works for all genera although the computations quickly get tiresome. A2) Riemann-Roch. Express the product defining $Z$ as an infinite sum and then use your head.
|
Peter Roquette has written four beautiful papers on the history of the zeta-function in characteristic $p$. The Riemann hypothesis in characteristic p, its origin and development. Part 1. The formation of the zeta functions of Artin and F.K. Schmidt. The Riemann hypothesis in characteristic p, its origin and development. Part 2. The first steps by Davenport and Hasse. The Riemann hypothesis in characteristic p, its origin and development. Part 3: The elliptic case. The Riemann hypothesis in characteristic p, its origin and development. Part 4: Davenport-Hasse fields. Relevant to your questions is part 1. From the abstract "This Part 1 is dealing with the development before Hasse's contributions to the Riemann hypothesis. We are trying to explain what he could build upon. The time interval covered will be roughly between 1921 and 1931. We start with Artin's thesis of 1921 where the Riemann hypothesis for function fields was spelled out and discussed for the first time, namely in the case of quadratic function fields. We will describe the activities following Artin's thesis until F.K.Schmidt's classical paper 1931 on the Riemann-Roch theorem and the zeta function of an arbitrary function field. Finally we will review Hasse's paper in 1934 where he gives a summary about all what was known at that time about zeta functions of function fields. "
|
{
"source": [
"https://mathoverflow.net/questions/14627",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1384/"
]
}
|
14,667 |
In complex projective geometry, we have a specified Kähler class $\omega$ and we have a Lefschetz operator $L:H^i(X,\mathbb{C})\to H^{i+2}(X,\mathbb{C})$ given by $L(\eta)=\omega\wedge \eta$. We then define primitive cohomology $P^{n-k}(X,\mathbb{C})=\ker(L^{k+1}:H^{n-k}(X)\to H^{n+k+2}(X))$, and we even have a nice theorem, the Lefschetz decomposition, that says $H^m(X,\mathbb{C})=\oplus_k L^kP^{n-2k}$. Often, in papers, people just prove their result for primitive classes, as they seem to be easier to work with. So, what exactly ARE primitive classes? Sure, they're things that some power of $\omega$ kills, but what's the intuition? Why are they an interesting distinguished class? Is there a good reason to expect this decomposition?
|
The primitive classes are the highest weight vectors. Hard Lefschetz says that the operator $L$ (which algebraic geometers know as intersecting with a hyperplane) is the "lowering operator" $\rho(F)$ in a representation $\rho
\colon \mathfrak{sl}_2(\mathbb{C})\to End (H^\ast(X;\mathbb{C}))$. The raising operator $\rho(E)$ is $\Lambda$, the restriction to the harmonic forms of the the formal adjoint of $\omega \wedge \cdot$ acting on forms. The weight operator $\rho(H)$ has $H^{n-k}(X;\mathbb{C})$ as an eigenspace (= weight space), with eigenvalue (=weight) $k$. The usual picture of an irreducible representation of $\mathfrak{sl}_2(\mathbb{C})$ is of a string of beads (weight spaces) with $\rho(F)$ moving you down the string and decreasing the weight by 2, and $\rho(E)$ going in the opposite direction. The highest weight is an integer $k$, the lowest weight $-k$. From this picture, it's clear that the space of highest weight vectors in a (reducible) representation is $\ker \rho(E)$. It's also clear that, of the vectors of weight $k$, those which are highest weights are the ones in $\ker \rho(F)^{k+1}$. So the highest weight vectors in $H^{n-k}(X; \mathbb{C})$ are those in $\ker L^{k+1}$. Of course, all this ignores the rather subtle question of how to explain in an invariant way what this $\mathfrak{sl}_2(\mathbb{C})$, or its corresponding Lie group, really is. Added , slipping Mariano an envelope. But here's what that group is. Algebraic geometers, brace yourselves. Fix $x\in X$, and let $O_x = O(T_x X\otimes \mathbb{C})\cong O(4n,\mathbb{C})$. Then $O_x$ acts projectively on $\Lambda^\bullet (T_x X\otimes \mathbb{C})$ via the spinor representation (which lives inside the Clifford action). The holonomy group $Hol_x\cong U(n)$ also acts on complex forms at $x$, and the "Lefschetz group" $\mathcal{L}$ is the centralizer of $Hol_x$ in $O_x$. One proves that $\mathcal{L}\cong GL(\mathbb{C}\oplus \mathbb{C})$. Not only is this the right group, but its Lie algebra comes with a standard basis, coming from the splitting $T_x X \otimes\mathbb{C} = T^{1,0} \oplus T^{0,1}$. Now, $\mathcal{L}$ acts on complex forms on $X$, by parallel transporting them from $y$ to $x$, acting, and transporting back to $y$. Check next that the action commutes with $d$ and $*$, hence with the Laplacian, and so descends to harmonic forms = cohomology. Finally, check that the action of $\mathcal{L}$ exponentiates the standard action of $\mathfrak{gl}_2$ where the centre acts by scaling. (This explanation is Graeme Segal's, via Ivan Smith.)
|
{
"source": [
"https://mathoverflow.net/questions/14667",
"https://mathoverflow.net",
"https://mathoverflow.net/users/622/"
]
}
|
14,714 |
I know the following facts. (Don't assume I know much more than the following facts.) The Atiyah-Singer index theorem generalizes both the Riemann-Roch theorem and the Gauss-Bonnet theorem. The Atiyah-Singer index theorem can be proven using heat kernels. This implies that both Riemann-Roch and Gauss-Bonnet can be proven using heat kernels. Now, I don't think I have the background necessary to understand the details of the proofs, but I would really appreciate it if someone briefly outlined for me an extremely high-level summary of how the above two proofs might go. Mostly what I'm looking for is physical intuition: when does one know that heat kernel methods are relevant to a mathematical problem? Is the mathematical problem recast as a physical problem to do so, and how? (Also, does one get Riemann-Roch for Riemann surfaces only or can we also prove the version for more general algebraic curves?) Edit: Sorry, the original question was a little unclear. While I appreciate the answers so far concerning how one gets from heat kernels to the index theorem to the two theorems I mentioned, I'm wondering what one can say about going from heat kernels directly to the two theorems I mentioned. As Deane mentions in this comments, my hope is that this reduces the amount of formalism necessary to the point where the physical ideas are clear to someone without a lot of background.
|
Here is how the heat kernel proof of Atiyah-Singer goes at a high level. Let $(\partial_t - \Delta)u = 0$ and define the heat kernel (HK) or Green function via $\exp(-t\Delta):u(0,\cdot) \rightarrow u(t,\cdot)$. The HK derives from the solution of the heat equation on the circle: $u(t,\theta) = \sum_n a_n(t) \exp(in\theta) \implies a_n(t) = a_n(0)\cdot \exp(-tn^2)$ For a sufficiently nice case the solution of the heat equation is $u(t,\cdot) = \exp(-t\Delta) * u(0,\cdot)$. The hard part is building the HK: we have to compute the eigenstuff of $\Delta$ (this is the Hodge theorem). But once we do that, a miracle occurs and we get the Atiyah-Singer Theorem: The supertrace
of the HK on forms is constant: viz. $\mathrm{Tr}_s \exp(-t\Delta) = \sum_k (-1)^k \,\mathrm{Tr} \exp(-t\Delta^k) = \mathrm{const}.$ For $t$ large, this can be evaluated topologically; for small $t$, it can be evaluated analytically as an integral of a characteristic class. Edit per Qiaochu's clarification This article of Kotake (really in here as the books seem to be mixed up) proves Riemann-Roch directly using the heat kernel.
|
{
"source": [
"https://mathoverflow.net/questions/14714",
"https://mathoverflow.net",
"https://mathoverflow.net/users/290/"
]
}
|
14,735 |
There is, of course, a complete classification for simple complex Lie algebras. Is there a good reference which lists the group of outer automorphisms for each?
|
Proposition D.40 of Fulton and Harris' Representation Theory states Emerton's comment: the group of outer automorphisms of a simple Lie algebra are precisely the group of graph automorphisms of the associated Dynkin diagram. There is also some discussion of this in Section 16.5 of Humphrey's Introduction to Lie Algebras and Representation Theory . So for $A_n$ and $n>1$, there is an order 2 automorphism which for $sl_{n+1}$ amounts to negative transpose. Type B and C have no outer automorphisms. For $D_n$, there is an order two automorphism swapping the two endpoints, and this corresponds to interchanging the two spin representations. On $so_{2n}$ this is obtained (modulo the inner automorphisms) by conjugating by an orthogonal matrix in $O(2n)$ which has determinant $-1$. For $n=4$, there is also an order 3 automorphism. This is triality, and is discussed in Section 20.3 of Fulton and Harris. For $E_6$, there is an order 2 automorphism, though I don't know enough about the exceptional Lie algebras to say anything useful about it. But you can find a discussion of the automorphism group in Section 7 of Jacobson's Exceptional Lie Algebras where it is described using Jordan algebras. For the other 4 exceptional Lie algebras there are no outer automorphisms.
|
{
"source": [
"https://mathoverflow.net/questions/14735",
"https://mathoverflow.net",
"https://mathoverflow.net/users/3513/"
]
}
|
14,739 |
Given a commutative ring $R$, there is a category whose objects are epimorphisms surjective ring homomorphisms $R \to S$ and whose morphisms are commutative triangles making two such epimorphisms surjections compatible, and the skeleton of this category is a partial order that can be identified with the lattice of ideals of $R$. Now, I have always been under the impression that anything one can say about ideals one can phrase in this purely arrow-theoretic language: most importantly, the intersection of ideals is the product in this category and the sum of ideals is the coproduct. (Since we're working in a partial order, product and coproduct are fancy ways to say supremum and infimum. The direction of the implied ordering on ideals may differ here from the one you're used to, but that's not important.) However, Harry's made some comments recently that made me realize I don't know how to define the product of two ideals purely in terms of this category, that is, via a universal construction like the above. It would be really surprising to me if this were not possible, so maybe I'm missing something obvious. Does anyone know how to do this?
|
Nice question! The answer is that it's not possible! Let $R=\mathbb{F}_3[x,y]/(x^2,y^2)$. The lattice of ideals consists of the eight ideals $(1)$ $(x,y)$ $(x)$ $(y)$ $(x+y)$ $(x-y)$ $(xy)$ $(0)$, in which each ideal contains all ideals at lower levels. In the middle level, some of the ideals have square zero, and some don't, but you can't tell which ones just from looking at the (unlabeled) lattice.
|
{
"source": [
"https://mathoverflow.net/questions/14739",
"https://mathoverflow.net",
"https://mathoverflow.net/users/290/"
]
}
|
14,752 |
It's possible I just haven't thought hard enough about this, but I've been working at it off and on for a day or two and getting nowhere. You can define a notion of " covering graph " in graph theory, analogous to covering spaces. (Actually I think there's some sense -- maybe in differential topology -- in which the notions agree exactly, but that's not the question.) Anyway, it behaves like a covering space -- it lifts paths and so on. There's also a "universal cover," which I think satisfies the same universal property as topological universal covers but I'm not sure. Universal covers are acyclic (simply connected) in graph theory, so they're trees, usually infinite. The universal cover doesn't determine the graph; for instance, any two k-regular graphs (k > 1) have the same universal cover. You can construct plenty of other pairs, too. I'm interested in necessary and sufficient conditions for two graphs $G, H$ to have the same universal cover. One such condition (I'm pretty sure!) is whether you can give a 1-1 correspondence between trails in $G$ and trails in $H$ that preserves degree sequences. Unfortunately this doesn't help me much, since this is still basically an infinite condition. Is there some less-obvious but more easily checkable condition? In particular is it possible to determine if two (finite) graphs have the same universal cover in polynomial time?
|
Two finite graphs have the same universal cover iff they have a common finite cover. This surprising fact was first proved by Tom Leighton here: Frank Thomson Leighton, Finite common coverings of graphs. 231-238 1982 33 J. Comb. Theory, Ser. B I'm quite sure the paper also presents an algorithm for determining if this is the case for two given graphs; essentially you develop a refined "degree" sequence for the graphs, starting from "# of vertices of degree k" and refining to "# of vertices of degree k with so-and-so neighbors of degree l" etc. As an aside, the reason this result is so surprising is that it says something highly non-trivial about groups acting on trees (any two subgroups of Aut(T) with a finite quotient are commensurable, up to conjugation), and proving this result directly via group-theoretic methods is surprisingly difficult (and interesting). There's a paper of Bass and Kulkarni which pretty much does just that. Edit: I just ran a quick search and found this sweet overview: " On Leighton's Graph Covering Theorem ".
|
{
"source": [
"https://mathoverflow.net/questions/14752",
"https://mathoverflow.net",
"https://mathoverflow.net/users/382/"
]
}
|
14,756 |
Suppose we want to recover an input vector $f \in \textbf{R}^n$ from some measurements $y = Af + \varepsilon$. Now $A$ is an $m \times n$ matrix and $\varepsilon$ are some unknown errors. Is this equivalent to finding some function $g: \textbf{R}^m \to \textbf{R}^m$ such that $g(Af+ \varepsilon) = g(Af) + g(\varepsilon) = g(\varepsilon)$? In other words, we are considering the kernel of some unknown function $g(Af)$ so that we can identify the error term? Is syndrome decoding an efficient way of recovering $f$?
|
Two finite graphs have the same universal cover iff they have a common finite cover. This surprising fact was first proved by Tom Leighton here: Frank Thomson Leighton, Finite common coverings of graphs. 231-238 1982 33 J. Comb. Theory, Ser. B I'm quite sure the paper also presents an algorithm for determining if this is the case for two given graphs; essentially you develop a refined "degree" sequence for the graphs, starting from "# of vertices of degree k" and refining to "# of vertices of degree k with so-and-so neighbors of degree l" etc. As an aside, the reason this result is so surprising is that it says something highly non-trivial about groups acting on trees (any two subgroups of Aut(T) with a finite quotient are commensurable, up to conjugation), and proving this result directly via group-theoretic methods is surprisingly difficult (and interesting). There's a paper of Bass and Kulkarni which pretty much does just that. Edit: I just ran a quick search and found this sweet overview: " On Leighton's Graph Covering Theorem ".
|
{
"source": [
"https://mathoverflow.net/questions/14756",
"https://mathoverflow.net",
"https://mathoverflow.net/users/3846/"
]
}
|
14,763 |
Let me stress that I am only interested in $p$-adic fields in this question, for reasons that will become clear later. Let me also stress that in some sense I am basically assuming that the reader knows what the "1970s version of the local Langlands conjectures" are when writing this question---there are plenty of references that will get us this far (I give one below that works in the generality I'm interested in). So let $F$ be a finite extension of $\mathbf{Q}_p$, let $G$ be a connected reductive group over $F$, let $\widehat{G}$ denote the complex dual group of $G$ (a connected complex Lie group) and let ${}^LG$ denote the $L$-group of $G$, the semi-direct product of the dual group and the Weil group of $F$ (formed using a fixed algebraic closure $\overline{F}$ of $F$). Here is the "standard", or possibly "standard in the 1970s", way of formulating what local Langlands should say (for more details see Borel's paper "Automorphic $L$-functions", available online (thanks AMS) here at the AMS website . One defines sets $\Phi(G)$ ($\widehat{G}$-conjugacy classes of admissible Weil-Deligne representations from the Weil-Deligne group to the $L$-group [noting that "admissible" includes assertions about images only landing in so-called "relevant parabolics" in the general case and is quite a subtle notion]) and $\Pi(G)$ (isomorphism classes of smooth irreducible admissible representations of $G(F)$), and one conjectures: LOCAL LANGLANDS CONJECTURE (naive form): There is a canonical surjection $\Pi(G)\to\Phi(G)$ with finite fibres, satisfying (insert list of properties here). See section 10 of Borel's article for the properties required of the map. Now in recent weeks I have had two conversations with geometric Langlands type people both of whom have mocked me when I have suggested that this is what the local Langlands conjecture should look like. They point out that studying some set of representations up to isomorphism is a very "coarse" idea nowadays, and one should reformulate things category-theoretically, considering Tannakian categories of representations, and relating them to...aah, well there's the catch. Looking back at what both of them said, they both at a crucial point slipped in the line "well, now for simplicity let's assume we're in the function field/geometric setting. Now..." and off they went with their perverse sheaves. The happy upshot of all of this is that now one has a much better formulation of local Langlands, because one can demand much more than a canonical surjection with finite fibres, one can ask whether two categories are equivalent. But I have been hoodwinked here, because I am interested in $p$-adic fields. So yes yes yes I'm sure it's all wonderful in the function field/geometric setting, and things have been generalised beyond all recognition. My question is simply: Q) Can we do better than the naive form of Local Langlands (i.e. is there a stronger statement about two categories being equivalent) when $F$ is a p-adic field ? The answer appears to be "yes" in other cases but I am unclear about whether the answer is yes in the $p$-adic case. Even if someone were to be able to explain some generalisation in the case where $G$ is split, I am sure I would learn a lot. To be honest, I think I'd learn a lot if someone could explain how to turn the surjection into a more bijective kind of object even in the case of $SL(2)$. Even in the unramified case! That's how far behind I am! As far as I can see, the Satake isomorphism gives only a surjection in general, because there is more than one equivalence class of hyperspecial maximal compact in general.
|
Now that our paper Geometrization of the local Langlands correspondence with Fargues is finally out (ooufff!!), it may be worth giving an update to Ben-Zvi's answer above. In brief: we give a formulation of Local Langlands over a $p$ -adic field $F$ so that it is finally an actual conjecture, in the sense that it asks for properties of a given construction, not for a construction ; of a form as in geometric Langlands, in particular about an equivalence of categories , not merely a bijection of irreducibles. First, I should say that in the notation of the OP, we construct a canonical map $\Pi(G)\to \Phi(G)$ , and prove some properties about it. However, we are not able to say anything yet about its fibres (not even finiteness). Moreover, we give a formulation of local Langlands as an equivalence of categories, and (essentially) construct a functor in one direction that one expects to realize the equivalence. In particular, this nails down what the local Langlands correspondence should be, it "merely" remains to establish all the desired properties of it. Let me briefly state the main result here. Let $\mathrm{Bun}_G$ be the stack of $G$ -bundles on the Fargues--Fontaine curve. We define an ( $\infty$ -)category $\mathcal D(\mathrm{Bun}_G,\overline{\mathbb Q}_\ell)$ of $\ell$ -adic sheaves on $\mathrm{Bun}_G$ . The stack $\mathrm{Bun}_G$ is stratified into countably many strata enumerated by $b\in B(G)$ , and on each stratum, the category $\mathcal D(\mathrm{Bun}_G^b,\overline{\mathbb Q}_\ell)$ is the derived ( $\infty$ -)category of smooth representations of the group $G_b(F)$ . In particular, for $b=1$ , one gets smooth representations of $G(F)$ . Moreover, there is an Artin stack $Z^1(W_F,\hat{G})/\hat{G}$ of $L$ -parameters over $\overline{\mathbb Q}_\ell$ . Our main result is the construction of the "spectral action": There is a canonical action of the $\infty$ -category of perfect complexes on $Z^1(W_F,\hat{G})/\hat{G}$ on $\mathcal D(\mathrm{Bun}_G,\overline{\mathbb Q}_\ell)$ . The main conjecture is basically that this makes $\mathcal D(\mathrm{Bun}_G,\overline{\mathbb Q}_\ell)^\omega$ a "free module of rank $1$ over $\mathrm{Perf}(Z^1(W_F,\hat{G})/\hat{G})$ ", at least if $G$ is quasisplit (or more generally, has connected center). More precisely, assume that $G$ is quasisplit and fix a Borel $B\subset G$ and a generic character $\psi$ of $U(F)$ , where $U\subset B$ is the unipotent radical, giving the Whittaker representation $c\text-\mathrm{Ind}_{U(F)}^{G(F)}\psi$ , thus a sheaf on $[\ast/G(F)]$ , which is the open substack of $\mathrm{Bun}_G$ of geometrically fibrewise trivial $G$ -bundles; extending by $0$ thus gives a sheaf $\mathcal W_\psi\in \mathcal D(\mathrm{Bun}_G,\overline{\mathbb Q}_\ell)$ , called the Whittaker sheaf. Conjecture. The functor $$ \mathrm{Perf}(Z^1(W_F,\hat{G})/\hat{G})\to \mathcal D(\mathrm{Bun}_G,\overline{\mathbb Q}_\ell)$$ given by acting on $\mathcal W_\psi$ is fully faithful, and extends to an equivalence $$\mathcal D^{b,\mathrm{qc}}_{\mathrm{coh}}(Z^1(W_F,\hat{G})/\hat{G})\cong \mathcal D(\mathrm{Bun}_G,\overline{\mathbb Q}_\ell)^{\omega}.$$ Here the superscript $\mathrm{qc}$ means quasicompact support, and $\omega$ means compact objects. As $Z^1(W_F,\hat{G})$ is not smooth (merely a local complete intersection), there is a difference between perfect complexes and $\mathcal D^b_{\mathrm{coh}}$ , and there is still a minor ambiguity about how to extend from perfect complexes to all complexes of coherent sheaves. Generically over the stack of $L$ -parameters, there is however no difference. It takes a little bit of unraveling to see how this implies more classical forms of the correspondence, like the expected internal parametrization of $L$ -packets; in the case of elliptic $L$ -parameters, everything is very clean, see Section X.2 of our paper. (There are related conjectures and results by Ben-Zvi--Chen--Helm--Nadler , Hellmann and Zhu ; see also the work of Genestier--Lafforgue in the function field case. And this work is heavily inspired by previous work in geometric Langlands, notably the conjectures of Arinkin--Gaitsgory , and the work of Nadler--Yun and Gaitsgory--Kazhdan--Rozenblyum--Varshavsky on spectral actions.) PS: It may be worth pointing out that this conjecture is, at least a priori, of a quite different nature than Vogan's conjecture, mentioned in the other answers, which is based on perverse sheaves on the stack of $L$ -parameters; here, we use coherent sheaves.
|
{
"source": [
"https://mathoverflow.net/questions/14763",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1384/"
]
}
|
14,842 |
Given a formal power series $f \in k[[X]]$, where $k$ is a commutative field, is there any good way to tell whether or not $f\in k(X)$? Edit: To clarify, "good way to tell" means "computable algorithm to tell". Edit 2: I really screwed up this question, so I am recusing myself from accepting an answer. I will accept an answer after a sufficient number of votes have been cast for the best answer.
|
Continued fractions! To motivate this answer, first recall the continued fraction algorithm for testing whether a real number is rational. Namely, given a real number $r$, subtract its floor $\lfloor r \rfloor$, take the reciprocal, and repeat. The number $r$ is rational if and only if at some point subtracting the floor gives $0$. Of course, an infinite precision real number is not something that a Turing machine can examine fully in finite time. In practice, the input would be only an approximation to a real number, say specified by giving the first 100 digits after the decimal point. There is no longer enough information given to determine whether the number is rational, but it still makes sense to ask whether up to the given precision it is a rational number of small height , i.e., with numerator and denominator small relative to the amount of precision given. If the number is rational of small height, one will notice this when computing its continued fraction numerically, because subtracting the floor during one of the first few steps (before errors compound to the point that they dominate the results) will give a number that is extremely small relative to the precision; replacing this remainder by $0$ in the continued fraction built up so far gives the small height rational number. What is the power series analogue? Instead of the field of real numbers, work with the field of formal Laurent series $k((x))$, whose elements are series with at most finitely many terms with negative powers of $x$: think of $x$ as being small. For $f = \sum a_n x^n \in k((x))$, define $\lfloor f \rfloor = \sum_{n \le 0} a_n x^n$; this is a sum with only finitely many nonzero terms. Starting with $f$, compute $f - \lfloor f \rfloor$, take the reciprocal, and repeat. The series $f$ is a rational function (in $k(x)$) if and only if at some point subtracting the floor gives $0$. The same caveats as before apply. In practice, the model is that one has exact arithmetic for elements of $k$ (the coefficients), but a series will be specified only partially: maybe one is given only the first 100 terms of $f$, say. The only question you can hope to answer is whether $f$ is, up to the given precision, equal to a rational function of low height (i.e., with numerator and denominator of low degree). The answer will become apparent when the continued fraction algorithm is applied: check whether subtracting the floor during one of the first few steps gives a series that starts with a high positive power of $x$. Bonus: Just as periodic continued fractions in the classical case correspond to quadratic irrational real numbers, periodic continued fractions in the Laurent series case correspond to series belonging to a quadratic extension of $k(x)$, i.e., to the function field of a hyperelliptic curve over $k$. Abel in 1826 exploited this idea as an ingredient in a method for determining which hyperelliptic integrals could be computed in elementary terms!
|
{
"source": [
"https://mathoverflow.net/questions/14842",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1353/"
]
}
|
14,849 |
At the very end of this 2006 interview (rm) , Kontsevich says "...many great theorems are originally proven but I think the proofs are not, kind of, "morally right." There should be better proofs...I think the Index Theorem by Atiyah and Singer...its original proof, I think it's ugly in a sense and up to now, we don't have "the right proof." Or Deligne's proof of the Weil conjectures, it's a morally wrong proof. There are three proofs now, but still not the right one." I'm trying to understand what Kontsevich means by a proof not being "morally right." I've read this article by Eugenia Cheng on morality in the context of mathematics, but I'm not completely clear on what it means with respect to an explicit example. The general idea seems to be that a "moral proof" would be one that is well-motivated by the theory and in which each step is justified by a guiding principle, as opposed to an "immoral" one that is mathematically correct but relatively ad hoc. To narrow the scope of this question and (hopefully) make it easier to understand for myself, I would like to focus on the second part of the comment. Why would Kontsevich says that Deligne's proof is not "morally right"? More importantly, what would a "moral proof" of the Weil Conjectures entail? Would a morally proof have to use motivic ideas, like Grothendieck hoped for in his attempts at proving the Weil Conjectures? Have there been any attempts at "moralizing" Deligne's proof? How do do the other proofs of the Weil Conjectures measure up with respect to mathematical morality?
|
I would guess that Grothendieck's envisaged proof, via the standard conjectures, would be "morally right" in Kontsevich's sense. (Although there is the question of how the standard
conjectures would be proved; since they remain conjectures, this question is open for now!) The objection to Deligne's proof is that it relies on various techniques (passing to symmetric powers and Rankin--Selberg inspired ideas, analytic arguments related to the positivity of the coefficients of the zeta-function, and other such things) that don't seem to be naturally related to the question at hand. I believe that Grothendieck had a similar objection to
Deligne's argment. As a number-theorist, I think Deligne's proof is fantastic. One of the appeals (at least to me) of number theory is that none of the proofs are "morally right" in Kontsevich's sense. Obviously, this is a very personal feeling. (Of course, a proof of the standard conjectures --- any proof, to my mind --- would also be fantastic!) [Edit, for clarification; this is purely an aside, though:] Some arguments in number theory, for example the primitive root theorem discussed in the comments, are pure algebra when viewed appropriately, and here there are very natural and direct arguments. (For example,
in the case of primitive roots, there is basic field theory combined with Hensel's lemma/Newton approximation; this style of argument extends, in some form, to the very general setting of complete local rings.) When I wrote that none of the proof in number theory are "morally right", I had in mind largely the proofs in modern algebraic number theory, such as the modularity of elliptic curves, Serre's conjecture, Sato--Tate,
and so on. The proofs use (almost) everything under the sun, and follow no dogma. Tate wrote of abelian class field theory that "it is true because it could not be otherwise"
(if I remember the quote correctly), which I took to mean (given the context) that the proofs in the end are unenlightening as to the real reason it is true; they are simply logically correct proofs. This seems to be even more the case with the proofs of results in non-abelian class field theory such as those mentioned above. Despite this, I personally find the arguments wonderful; it is one of the appeals of the subject for me.
|
{
"source": [
"https://mathoverflow.net/questions/14849",
"https://mathoverflow.net",
"https://mathoverflow.net/users/239/"
]
}
|
14,877 |
We may define a topological manifold to be a second-countable Hausdorff space such that every point has an open neighborhood homeomorphic to an open subset of $\mathbb{R}^n$. We can further define a smooth manifold to be a topological manifold equipped with a structure sheaf of rings of smooth functions by transport of structure from $\mathbb{R}^n$, since $\mathbb{R}^n$ has a canonical sheaf of differentiable functions $\mathbb{R}^n\to \mathbb{R}$, with a canonical restriction sheaf to any open subset. This gives a manifold as a locally ringed space. (Of course this definition generalizes to all sorts of other kinds of manifolds with minor adjustments). Then the questions:
If we totally ignore the definition using atlases, will we at some point hit a wall? Can we fully develop differential geometry without ever resorting to atlases? Regardless of the above answer, are there any books that develop differential geometry primarily from a "locally ringed space" viewpoint, dropping into the language of atlases only when necessary? I looked at Kashiwara & Schapira's "Sheaves on Manifolds", but that's much more focused on sheaves of abelian groups and (co)homology. Edit: To clarify (Since Pete and Kevin misunderstood): It's easy to show that the approaches are equivalent, but proofs using charts don't always translate easily to proofs using sheaves.
|
There is the book by Ramanan "Global Calculus" which develops differential geometry relying heavily on sheaf theory (you should see his definition of connection algebra...).
He avoids the magic words "locally ringed space" by requiring the structure sheaf to be a subsheaf of the sheaf of continuous functions (hence maximal ideal of stalks = vanishing functions).
|
{
"source": [
"https://mathoverflow.net/questions/14877",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1353/"
]
}
|
14,918 |
Are there examples of sets of natural numbers that are proven to be decidable but by non-constructive proofs only?
|
When I teach computability, I usually use the following example to illustrate the point. Let $f(n)=1$, if there are $n$ consecutive $1$s somewhere in the decimal expansion of $\pi$, and $f(n)=0$ otherwise. Is this a computable function? Some students might try naively to compute it like this: on input $n$, start to enumerate the digits of $\pi$, and look for $n$ consecutive $1$s. If found, then output $1$. But then they realize: what if on a particular input, you have searched for 10 years, and still not found the instance? You don't seem justified in outputting $0$ quite yet, since perhaps you might find the consecutive $1$s by searching a bit more. Nevertheless, we can prove that the function is computable as follows. Either there are arbitrarily long strings of $1$ in $\pi$ or there is a longest string of $1$s of some length $N$. In the former case, the function $f$ is the constant $1$ function, which is definitely computable. In the latter case, $f$ is the function with value $1$ for all input $n\lt N$ and value $0$ for $n\geq N$, which for any fixed $N$ is also a computable function. So we have proved that $f$ is computable in effect by providing an infinite list of programs and proving that one of them computes $f$, but we don't know which one exactly. Indeed, I believe it is an open question of number theory which case is the right one. In this sense, this example has a resemblence to Gerhard's examples.
|
{
"source": [
"https://mathoverflow.net/questions/14918",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2672/"
]
}
|
15,003 |
Why is it that the vanishing of the integral first Chern class of a compact Kahler manifold is equivalent to the canonical bundle being trivial? I can see that it implies that the canonical bundle must be topologically trivial, but not necessarily holomorphically trivial. Does proving the equivalence require Yau's theorem, in order to produce a flat connection on the canonical bundle, or is there a more elementary proof?
|
I have looked for a while for a proof
which does not use the Calabi-Yau theorem
and nobody seems to know it. Also, there are plenty of non-Kaehler
manifolds with canonical bundle trivial
topologically and non-trivial as a holomorphic
bundle (the Hopf surface is an easiest
example). The argument actually uses the Calabi-Yau
theorem, Bochner's vanishing, Berger's classification
of holonomy and Bogomolov's decomposition theorem. From Calabi-Yau theorem you infer that
there exists a Ricci-flat Kaehler metric.
Since the Ricci curvature is a curvature
of the canonical bundle, this implies
that the canonical bundle admits a flat
connection. Of course, this does not mean that
it is trivial holomorphically; in fact,
the canonical bundle is flat on Hopf surface
and on the Enriques surface, which are not Calabi-Yau. For Calabi-Yau manifolds, however,
it is known that the Albanese map
is a locally trivial fibration and
and has Calabi-Yau fibers with trivial
first Betti number. This is shown using the
Bochner's vanishing theorem which implies
that all holomorphic 1-forms are parallel. Now, by adjunction formula, you prove that
the canonical bundle of the total space is
trivial, if it is trivial for the base and the fiber.
The base is a torus, and the fiber is a Calabi-Yau
with $H^1(M)=0$. For the later, triviality
of canonical bundle follows from Bogomolov's
decomposition theorem, because such a
Calabi-Yau manifold is a finite quotient
of a product of simple Calabi-Yau manifolds
and hyperkaehler manifolds having holonomy
$SU(n)$ and $Sp(n)$. Bogomolov's decomposition
is itself a non-trivial result, and (in this generality)
I think it can be only deduced from the Berger's
classification. The original proof of Bogomolov was
elementary, but he assumed holomorphic triviality
of a canonical bundle, which we are trying to prove. This argument is extremely complicated; also,
it is manifestly useless in non-Kaehler situation
(and in many other interesting situations).
I would be very interested in any attempt
to simplify it. Update: Just as I was writing the reply,
Dmitri has posted a link to Bogomolov's article, where
he proves that some power of a canonical bundle is
always trivial, without using the Calabi-Yau theorem.
|
{
"source": [
"https://mathoverflow.net/questions/15003",
"https://mathoverflow.net",
"https://mathoverflow.net/users/3952/"
]
}
|
15,075 |
Is there any 'general' topological invariant to tell the difference between $M$ and $N$, where $M$ has homotopy type of a closed manifold and $N$ has homotopy type of a manifold with boundary. I meant something like homotopy group/ homology group can 'detect' the obstruction of being homotopic to a closed manifold.
|
If $M$ is a closed manifold, then you can always cross it with an interval to get a
manifold with non-empty boundary that is homotopic to $M$. So the distinction between the two possibilities (closed or non-closed) is not so well-defined in the homotopy category. On the other hand, suppose that we fix the dimension $n$ of the manifolds we want to consider.
Then,
the most basic difference between closed manifolds and those with boundary is the top dimensional homology, namely: a closed orientable connected manifold of dimension $n$ has $H_n(M) = {\mathbb Z}$,
while on the other hand, if $N$ has non-empty boundary, then $H_n(N) = 0$.
|
{
"source": [
"https://mathoverflow.net/questions/15075",
"https://mathoverflow.net",
"https://mathoverflow.net/users/3922/"
]
}
|
15,087 |
Are there any general methods for computing fundamental group or singular cohomology (including the ring structure, hopefully) of a projective variety (over C of course), if given the equations defining the variety? I seem to recall that, if the variety is smooth, we can compute the H^{p,q}'s by computer -- and thus the H^n's by Hodge decomposition -- is this correct? However this won't work if the variety is not smooth -- are there any techniques that work even for non-smooth things? Also I seem to recall some argument that, at least if we restrict our attention to smooth things only, all varieties defined by polynomials of the same degrees will be homotopy equivalent. The homotopy should be gotten by slowly changing the coefficients of the polynomials. Is something like this true? Does some kind of argument like this work?
|
This is an interesting question. To repeat some of the earlier answers, one should be able to get one's hands on a triangulation algorithmically using real algebro-geometric methods, and thereby compute singular cohomology and (a presentation for) the fundamental group. But this should probably be a last resort in practice. For smooth projective varieties, as people have noted, one can compute the Hodge numbers by writing down a presentation for the sheaf p-forms and then apply standard Groebner basis techniques to compute sheaf cohomology. This does work pretty well on a computer. For specific classes, there are better methods. For smooth complete intersections, there is a generating function for Hodge numbers due to Hirzebruch (SGA 7, exp XI), which is extremely efficient to use. As for the fundamental group, if I had to compute it for a general smooth projective variety, I would probably use a Lefschetz pencil to write down a presentation. For singular varieties, one can still define Hodge numbers using the mixed Hodge structure on cohomology. The sum of these numbers are still the Betti numbers. I expect these Hodge numbers are still computable, but it would somewhat unpleasant to write down a general algorithm. The first step is to build a simplicial resolution using resolution of singularities. My colleagues who know about resolutions assure me that this can be done algorithmically now days. (This is my first reply in this forum. Hopefully it'll go through.)
|
{
"source": [
"https://mathoverflow.net/questions/15087",
"https://mathoverflow.net",
"https://mathoverflow.net/users/83/"
]
}
|
15,115 |
Hi,
I'm looking for examples of groups that are both Hopfian and Co-Hopfian. I have a non trivial (and beautiful, at least to me) example: $\mathrm{SL}(n,\mathbb{Z})$ (with $n>2$). Do you know others (non trivial)? Thank you.
|
Mapping class groups of closed surfaces are both Hopfian and co-Hopfian (the former follows from residual finiteness, and the latter is due to Ivanov-McCarthy). Out(F_n) also has both properties (residual finiteness and a theorem of Farb-Handel).
|
{
"source": [
"https://mathoverflow.net/questions/15115",
"https://mathoverflow.net",
"https://mathoverflow.net/users/3958/"
]
}
|
15,181 |
Can you divide one square paper into five equal squares?
You have a scissor and glue. You can measure and cut and then attach as well. Only condition is You can't waste any paper.
|
The Wallace-Bolyai-Gerwien Theorem theorem says: Any two simple polygons of equal area are equidecomposable (where simple means no self intersections and equidecomposable means finitely cut and glued). For your problem you can take the first polygon to be a unit square and the second to be a sqrt(5) by 1/sqrt(5) rectangle and apply this theorem. Then perform the remaining four cuts. Also, the generalisation of your question is the 2d analogue of Hilbert's 3rd Problem which asks whether given any two polyhedra with equal volume can one be finitely cut and glued into the other. The answer here, unlike in the 2d case, is "no" which was proved by Dehn using Dehn invariants in 1900.
|
{
"source": [
"https://mathoverflow.net/questions/15181",
"https://mathoverflow.net",
"https://mathoverflow.net/users/3056/"
]
}
|
15,202 |
I'm taking my first steps in the language of stacks, and would like something cleared up. The intuitive idea of moduli spaces is that each point corresponds to an object of what we're trying to classify (smooth curves of genus g over ℂ, for example). Fine moduli spaces are defined to be the objects that represent the functor that takes an object and gives you the [set, for schemes; groupoid, as I understand it, for stacks] of ways that that object parametrizes families of the object we want to classify. Now, for schemes - this makes sense in the following way: Let that functor be F, and let it be represented by M. Then F(Spec ℂ) are the families of (desired) objects parametrized by Spec ℂ (a point), so it corresponds to all desired objects (the ones we want to classify). But F(Spec ℂ) is also Hom(Spec ℂ, M), and so corresponds to the closed points of M. Thus M really does, intuitively, have as points the objects it wishes to classify. Does this idea go through to moduli stacks? Of course, it probably does, and this is all probably trivial - but I feel like I need someone to assure me that I'm not crazy. So let me put the question like this: Can you formulate how to think of a fine moduli stack (as an object that represents an F as above; also: how would you define this F in the category of stacks?) in a way that makes it clear that it parametrizes the desired objects?
|
I'll assure you that you're not crazy. Not only does the idea go through for stacks, but it's impossible (or at least very hard) to make sense of stacks without that idea. If you're trying to parameterize wigits, you can build a functor F(T)={flat families of wigits over T}. If there is a space M that deserves to be called the moduli space of wigits, it should represent F. It's not just that the points of M must correspond to isomorphism classes of wigits, so we must have F(Spec ℂ)=Hom(Spec ℂ,M). The points are also connected up in the right way. For example, a family of widits over a curve should correspond to a choice of wigit for every point in the curve in a continuous way, so it should correspond to a morphism from the curve to M. It happens that if wigits have automorphisms, there's no hope of finding a geometric object M so that maps to M are the same thing as flat (read "continuous") families of wigits. The reason is that any geometric object should have the property that maps to it can be determined locally. That is, if U and V cover T, specifying a map from T to M is the same as specifying maps from U and V to M which agree on U∩V. The jargon for this is "representable functors are sheaves." If a wigit X has an automorphism, then you can imagine a family of wigits over a circle so that all the fibers are X, but as you move around the circle, it gets "twisted" by the automorphism (if you want to think purely algebro-geometrically, use a circular chain of ℙ 1 s instead of a circle). Locally, you have a trivial family of wigits, so the map should correspond to a constant map to the moduli space M, but that would correspond to the trivial family globally, which this isn't. Oh dear! Instead of giving up hope entirely, the trick is to replace the functor F by a "groupoid-valued functor" (fibered category), so the automorphisms of objects are recorded. Now of course there won't be a space representing F, since any space represents a set-valued functor, but it turns out that this sometimes revives the hope that F is represented by some mythical "geometric object" M in the sense that objects in F(T) (which should correspond to maps to M) can be determined locally. If this is true, we say that "F is a stack " or that "M is a stack ." Part of what makes your question tricky is that as things get stacky, the line between M and F becomes more blurred. M isn't really anything other than F. We just call it M and treat it as a geometric object because it satisfies this gluing condition. We usually want M to be more geometric than that; we want it to have a cover (in some precise sense) by an honest space. If it does, then we say "M (or F) is an algebraic stack" and it turns out you can do real geometry on it.
|
{
"source": [
"https://mathoverflow.net/questions/15202",
"https://mathoverflow.net",
"https://mathoverflow.net/users/3238/"
]
}
|
15,220 |
Let $K$ be a Galois extension of the rationals with degree $n$ . The Chebotarev Density Theorem guarantees that the rational primes that split completely in $K$ have density $1/n$ and thus there are infinitely many such primes. As Kevin Buzzard pointed out to me in a comment , there is a simpler way to see that there are infinitely many rational primes that split completely in $K$ , namely that the Dedekind zeta-function $\zeta_K(s)$ has a simple pole at $s = 1$ . While this result is certainly much easier to prove than Chebotarev's Theorem, it is still not an elementary proof. Is there a known elementary proof of the fact that there are infinitely many rational primes that split completely in $K$ ? Selberg's elementary proof of Dirichlet's Theorem for primes in arithmetic progressions handles the case where $\text{Gal}(K/\mathbb{Q})$ is Abelian. I don't know anything about the general case. Since Dirichlet's Theorem is stronger than required, it is possible that an simpler proof exists even in the Abelian case. Remarks on the meaning of elementary . I am aware that there is no uniformly recognized definition of "elementary proof" in number theory. While I am not opposed to alternate definitions, my personal definition is a proof which can be carried out in first-order arithmetic, i.e. without quantification over real numbers or higher-type objects. Obviously, I don't require it to be explicitly formulated in that way — even logicians don't do that! Odds are that whatever you believe is elementary is also elementary in my sense. Kurt Gödel observed that proofs of (first-order) arithmetical facts can be much, much shorter in second-order arithmetic than in first-order arithmetic. This observation explains some of the effectiveness of analytic number theory, which is implicitly second-order. In view of Gödel's observation, it is possible that we have encountered arithmetical facts with a reasonably short second-order proof (i.e. could be found in an analytic number theory textbook) but no reasonable first-order proof (i.e. the production of any such proof would necessarily exhaust all of our natural resources). The above is unlikely to be such, but it is interesting to know that beasts of this type could exist...
|
By the primitive element theorem, $K=\mathbb{Q}(\alpha)$ for some nonzero $\alpha \in K$, and we may assume that the minimal polynomial $f(x)$ of $\alpha$ has integer coefficients. Let $\Delta$ be the discriminant of $f$. Since $K/\mathbb{Q}$ is Galois, a prime $p \nmid \Delta$ splits completely in $K$ if and only if there is a degree $1$ prime above $p$, which is if and only if $p | f(n)$ for some $n \in \mathbb{Z}$. Suppose that the set $P$ of such primes is finite. Enlarge $P$ to include the primes dividing $\Delta$. Let $t$ be a positive integer such that $\operatorname{ord}_p t> \operatorname{ord}_p f(0)$ for all $p \in P$. For any integer $m$, we have $f(mt) \equiv f(0) \;(\bmod \; t)$, so $\operatorname{ord}_p f(mt) = \operatorname{ord}_p f(0)$ for all $p \in P$. But $f(mt) \to \infty$ as $m \to \infty$, so eventually it must have a prime factor outside $P$, contradicting the definition of $P$.
|
{
"source": [
"https://mathoverflow.net/questions/15220",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2000/"
]
}
|
15,292 |
Lawrence Evans wrote in discussing the work of Lions fils that there is in truth no central core
theory of nonlinear partial
differential equations, nor can there
be. The sources of partial
differential equations are so many -
physical, probabilistic, geometric etc.
- that the subject is a confederation of diverse subareas, each studying
different phenomena for different
nonlinear partial differential
equation by utterly different methods. To me the second part of Evans' quote does not necessarily imply the first. So my question is: why can't there be a core theory of nonlinear PDE? More specifically it is not clear to me is why there cannot be a mechanical procedure (I am reminded here by [very] loose analogy of the Risch algorithm ) for producing estimates or good numerical schemes or algorithmically determining existence and uniqueness results for "most" PDE. (Perhaps the h-principle has something to say about a general theory of nonlinear PDE, but I don't understand it.) I realize this question is more vague than typically considered appropriate for MO, so I have made it CW in the hope that it will be speedily improved. Given the paucity of PDE questions on MO I would like to think that this can be forgiven in the meantime.
|
I find Tim Gowers' " two cultures " distinction to be relevant here. PDE does not have a general theory , but it does have a general set of principles and methods (e.g. continuity arguments, energy arguments, variational principles, etc.). Sergiu Klainerman's " PDE as a unified subject " discusses this topic fairly exhaustively. Any given system of PDE tends to have a combination of ingredients interacting with each other, such as dispersion, dissipation, ellipticity, nonlinearity, transport, surface tension, incompressibility, etc. Each one of these phenomena has a very different character. Often the main goal in analysing such a PDE is to see which of the phenomena "dominates", as this tends to determine the qualitative behaviour (e.g. blowup versus regularity, stability versus instability, integrability versus chaos, etc.) But the sheer number of ways one could combine all these different phenomena together seems to preclude any simple theory to describe it all. This is in contrast with the more rigid structures one sees in the more algebraic sides of mathematics, where there is so much symmetry in place that the range of possible behaviour is much more limited. (The theory of completely integrable systems is perhaps the one place where something analogous occurs in PDE, but the completely integrable systems are a very, very small subset of the set of all possible PDE.) p.s. The remark Qiaochu was referring to was Remark 16 of this blog post .
|
{
"source": [
"https://mathoverflow.net/questions/15292",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1847/"
]
}
|
15,309 |
Qiaochu Yuan in his answer to this question recalls a blog post (specifically, comment 16 therein) by Terry Tao: For instance, one cannot hope to find an algorithm to determine the existence of smooth solutions to arbitrary nonlinear partial differential equations, because it is possible to simulate a Turing machine using the laws of classical physics, which in turn can be modeled using (a moderately complicated system of) nonlinear PDE Is this "it is possible" an application of a Newton thesis , much as the usual Church-Turing thesis , going along the lines of «if something is imaginably doable in real life, one can simulate it with the usual equations of physics», or has the simulation actually been actually carried out? Does one need PDEs to simulate Turing machines, or are ODEs good enough?
|
ODEs are good enough. Your comment got me started digging.
|
{
"source": [
"https://mathoverflow.net/questions/15309",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1409/"
]
}
|
15,327 |
There are already a lot of discussion about the motivation for prime spectrum of commutative ring. In my perspective(highly non original), there are following reasons for the importance of prime spectrum. $\text{Spec}(R)$ is the minimal spectrum containing $\text{Spec}_{\rm max}(R)$ which has good functoriality which means the preimage of a prime ideal is still a prime ideal. if $p\in \text{Spec}(R)$, then $S_{p} = R-p$ is multiplicative set. Then one can localize. $S_{p}^{-1}R$ for $p\in \text{Spec}(R)$ is a local ring (has unique maximal ideal which is equivalent to have unique isomorphism class of simple modules). Local ring is easy to deal with and the maximal ideal can be described in explicitly, i.e $m=S_p^{-1}p$ (My advisor told me P.Cartier pushed Grothendieck to built commutative algebraic geometry machinery based on prime spectrum and the reasons mentioned above are the reasons they used prime spectrum) Addtional reason: one can have good definitions of topological space and a structure sheaf on it so that one can recover this commutative ring back as its global section Now, my question is for the people coming from commutative world, what other properties do you expect the spectrum of a noncommutative ring should have? I am aware that people are coming from different branches, there might be various kinds of noncommutative ring arising in your study. Therefore, the question for people coming from different branches of mathematics is that which kind of noncommutative ring do you meet and what properties do you feel that the spectrum of noncommutative ring should have to satisfy your need? The main motivation for me to ask this question is I am learning noncommutative algebraic geometry. In the existence work by Rosenberg, there are several kinds of spectrum(at least six different spectrum) for different purposes and they satisfy the analogue properties(noncommutative version)I mentioned above and coincide with prime spectrum when one impose the condition of commutativity. I wonder check whether these spectrum satisfied the other reasonable demand
|
I know almost nothing about noncommutative rings, but I have thought a bit about what the general concept of spectra might or should be, so I'll venture an answer. One other property you might ask for is that it has a good categorical description. I'll explain what I mean. The spectrum of a commutative ring can be described as follows. (I'll just describe its underlying set, not its topology or structure sheaf.) We have the category CRing of commutative rings, and the full subcategory Field of fields. Given a commutative ring $A$, we get a new category $A/$ Field : an object is a field $k$ together with a homomorphism $A \to k$, and a morphism is a commutative triangle. The set of connected-components of this category $A/$ Field is $\mathrm{Spec} A$. There's a conceptual story here. Suppose we think instead about algebraic topology. Topologists (except "general" or "point-set" topologists) are keen on looking at spaces from the point of view of Euclidean space. For example, a basic thought of homotopy theory is that you probe a space by looking at the paths in it, i.e. the maps from $[0, 1]$ to it. We have the category Top of all topological spaces, and the subcategory Δ consisting of the standard topological simplices $\Delta^n$ and the various face and degeneracy maps between them. For each topological space $A$ we get a new category Δ $/A$, in which an object is a simplex in $A$ (that is, an object $\Delta^n$ of Δ together with a map $\Delta^n \to A$) and a morphism is a commutative triangle. This new category is basically the singular simplicial set of $A$, lightly disguised. There are some differences between the two situations: the directions have been reversed (for the usual algebra/geometry duality reasons), and in the topological case, taking the set of connected-components of the category wouldn't be a vastly interesting thing to do. But the point is this: in the topological case, the category Δ $/A$ encapsulates how $A$ looks from the point of view of simplices. In the algebraic case, the category $A/$ Field encapsulates how $A$ looks from the point of view of fields. $\mathrm{Spec} A$ is the set of connected-components of this category, and so gives partial information about how $A$ looks from the point of view of fields.
|
{
"source": [
"https://mathoverflow.net/questions/15327",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1851/"
]
}
|
15,366 |
I wonder if anyone else has noticed that the market for expository papers in mathematics is very narrow (more so than it used to be, perhaps). Are there any journals which publish expository work, especially at the "intermediate" level? By intermediate, I mean neither (i) aimed at an audience of students, especially undergraduate students (e.g. Mathematics Magazine) nor (ii) surveys of entire fields of mathematics and/or descriptions of spectacular new results written by veteran experts in the field (e.g. the Bulletin, the Notices). Let me give some examples from my own writing, mostly just to fix ideas. (I do not mean to complain.) About six years ago I submitted an expository paper "On the discrete geometry of Chicken McNuggets" to the American Mathematical Monthly. The point of the paper was to illustrate the utility of simple reasoning about lattices in Euclidean space to give a proof of Schur's Theorem on the number of representations of an integer by a linear form in positive integers. The paper was rejected; one reviewer said something like (I paraphrase) "I have the feeling that this would be a rather routine result for someone versed in the geometry of numbers." This shows that the paper was not being viewed as expository -- i.e., a work whose goal is the presentation of a known result in a way which will make it accessible and appealing to a broader audience. I shared the news with my officemate at the time, Dr. Gil Alon, and he found the topic interesting. Together we "researchized" the paper by working a little harder and proving some (apparently) new exact formulas for representation numbers. This new version was accepted by the Journal of Integer Sequences: https://cs.uwaterloo.ca/journals/JIS/VOL8/Clark/clark80.html This is not a sad story for me overall because I learned more about the problem ("The Diophantine Problem of Frobenius") in writing the second version with Gil. But still, something is lost: the first version was a writeup of a talk that I have given to advanced undergraduate / basic graduate audiences at several places. For a long time, this was my "general audience" talk, and it worked at getting people involved and interested: people always came up to me afterward with further questions and suggested improvements, much more so than any arithmetic geometry talk I have ever given. The main result in our JIS paper is unfortunately a little technical [not deep, not sophisticated; just technical: lots of gcd's and inverses modulo various things] to state, and although I have recommended to several students to read this paper, so far nothing has come of it. A few years ago I managed to reprove a theorem of Luther Claborn (every abelian group is isomorphic to the class group of some Dedekind domain) by using elliptic curves along the lines of a suggestion by Michael Rosen (who reproved the result in the countable case). I asked around and was advised to submit the paper to L'Enseignement Mathematique . In my writeup, I made the conscious decision to write the paper in an expository way: that is, I included a lot of background material and explained connections between the various results, even things which were not directly related to the theorem in question. The paper was accepted; but the referee made it clear that s/he would have preferred a more streamlined, research oriented approach. Thus EM , despite its name ("Mathematical Education"), seems to be primarily a research journal (which likes papers taking new looks at old or easily stated problems: it's certainly a good journal and I'm proud to be published in it), and I was able to smuggle in some exposition under the cover of a new research result. I have an expository paper on factorization in integral domains: http://alpha.math.uga.edu/~pete/factorization.pdf [ Added : And a newer version: http://alpha.math.uga.edu/~pete/factorization2010.pdf . ] It is not finished and not completely polished, but it has been circulating around the internet for about a year now. Again, this completely expository paper has attracted more attention than most of my research papers. Sometimes people talk about it as though it were a preprint or an actual paper, but it isn't: I do not know of any journal that would publish a 30 page paper giving an intermediate-level exposition of the theory of factorization in integral domains. Is there such a journal? Added : In my factorization paper, I build on similar expositions by the leading algebraists P. Samuel and P.M. Cohn. I think these two papers, published in 1968 and 1973, are both excellent examples of the sort of "intermediate exposition" I have in mind (closer to the high end of the range, but still intermediate: one of the main results Samuel discusses, Nagata's Theorem, was published in 1957 so was not exactly hot off the presses when Samuel wrote his article). Both articles were published by the American Mathematical Monthly ! I don't think the Monthly would publish either of them nowadays. Added : I have recently submitted a paper to the Monthly: http://alpha.math.uga.edu/~pete/coveringnumbersv2.pdf (By another coincidence, this paper is a mildly souped up answer to MO question #26. But I did the "research" on this paper in the lonely pre-MO days of 2008.) Looking at this paper helps me to see that the line between research and exposition can be blurry. I think it is primarily an expository paper -- in that the emphasis is on the presentation of the results rather than the results themselves -- but I didn't have the guts to submit it anywhere without claiming some small research novelty: "The computation of the irredundant linear covering number appears to be new." I'll let you know what happens to it. ( Added : it was accepted by the Monthly.)
|
I'm not too familiar with Expositiones Mathematicae , but have you given them a look? EDIT: The article I happened to have seen, which made me think that Expo Math might be along the lines Pete Clark was looking for, is this paper of T. Bühler - it modestly claims to no originality save for assembling disparate parts of the literature and writing down what's old news to connoisseurs (I'm paraphrasing here!) but of course this is, in a sense, precisely its originality & worth.
|
{
"source": [
"https://mathoverflow.net/questions/15366",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1149/"
]
}
|
15,438 |
On one of my exams last year, we were given a problem (we chose five or six out of eight problems) on an exam, the goal of which was to prove the Bruhat decomposition for $GL_n(k)$. I was one of the two people to choose said problem. I gave a very long convoluted argument which although correct was really inelegant. I proved it more than once because I wasn't satisfied with my proof, and I figured out a somewhat slick contradiction argument based on maximizing leading zeroes of rows (number of zeroes before the pivot), but the proof was still a real mess. Statement of the problem: Let $G:=GL(V)$ for $V$ a finite dimensional $k$ vector space. Let $B$ be the stabilizer of the standard flag (these will be invertible upper triangular matrices), and let $W$ be the subgroup of permutation matrices. Show that $G=\coprod_{w\in W} BwB$, where the $BwB$ are double cosets. That is, show that $G=BWB$. Question:
Is there a slick proof of this fact maybe using "more machinery"? In particular, is there any sort of "coordinate-free" proof (Can we even define the Borel and Weyl subgroups without coordinates?)?
|
You are asking to compute the double quotient $B\backslash G /B.$ This is the same as computing $G\backslash (G/B \times G/B)$. A point in $G/B$ is a full flag on $k^n$. So you
are trying to compute the set of pairs $(F_1,F_2)$ of flags, modulo the simultaneous action
of $G$. Another way to think about $G/B$ is that it is the space of Borel subgroups (the coset of
$g$ corresponds to the conjugate $g B g^{-1}$, where $B$ is the upper triangular Borel
that was fixed in the statement of the question).
The passage from flags to Borels is given by mapping $F$ to its stablizer in $G$. So you can also think that you're trying to describe pairs of Borels $(B_1,B_2)$,
modulo simultaneous conjugation by $G$. Now recall that a torus $T$ in $G$ is a conjugate of the diagonal subgroup. Choosing
a torus in $G$ is the same as choosing a decomposition of $k^n$ as a direct sum of
1-dimensional subspaces (or lines, for short). (These will be the various eigenspaces of the torus acting
on $k^n$.) The diagonal torus corresponds to the standard decomposition of $k^n$
as $n$ copies of $k$. Now a torus $T$ is contained in a Borel $B$ (let me temporarily use $B$ to
denote any Borel, not just the upper triangular one) if and only if the corresponding
decomposition of $k^n$ into a sum of lines is compatible with the flag that $B$ fixes, i.e. if the flag is given by taking first one line, than the sum of that one with a second, then the sum of those two
with a third, and so on. In particular, choosing a torus $T$ contained in a Borel $B$
determines a "labelled decomposition" of $k^n$, i.e. we may write
$k^n = L_1 \oplus \ldots \oplus L_n$, where $L_i$ is the $i$th line;
just to be clear, the labelling is chosen so that the corresponding flag is just
$L_1 \subset L_1\oplus L_2 \subset \cdots.$ (Again, to be completely clear,
if $T$ is the conjugate by $g \in G$ of the diagonal torus, then $L_i$ is the
translate by $g$ of the line spanned by the $i$th standard basis vector.) Note that this labelled decomposition depends not just on $T$ (which only gives an
unlabelled decomposition) but on the Borel $B$ containing $T$ as well. (In more Lie
theoretic language, this is a reflection of that fact that a torus determines
a collection of weights in any representation of $G$,
while a choice of a Borel containing the torus lets you order the weights as well,
by determining a set of positive roots.) Of course, $B$ will contain more than one torus; or more geometrically,
$k^n$ will admit more than one decomposition into lines adapted to the filtration
$F$ of which $B$ is the stabilizer. But if one thinks about the different possible
lines, you see that $L_1$ is uniquely determined (it must be the first step in
the flag), $L_2$ is uniquely determined modulo $L_1$ (since together with $L_1$
it spans the second step in the flag), and so on, which shows that any two tori
$T$ in $B$ are necessarily conjugate by an element of $B$, and the same sort
of reasoning shows that the normalizer of $T$ in $B$ is just $T$ (because
if $g \in G$ is going to preserve both the flag and the collection of lines, which
is the same as preserving the ordered collection of lines, all it can do
is act by a scalar on each line, which is to say, it must be an element of $T$). Now a key fact is that any two Borels, $B_1$ and $B_2$, contain a common torus.
In other words, given two filtrations, we can always choose an (unordered)
decomposition of $k^n$ into a direct sum of lines which is adapted to both filtrations. (This is an easy exercise.) Of course the ordering of the
lines will depend which of the two filtrations we use. In other words, we get a set
of $n$ lines in $k^n$ which are ordered one way according to the filtration
$F_1$ given by $B_1$, and in a second way according to the filtration
given $F_2$ by $B_2$. If we let $w \in S_n$ be the permutation which takes the first
ordering to the second, then we see that the pair $B_1$ and $B_2$ determines
an element $w \in S_n$. This is the Bruhat decomposition. It wouldn't be hard to continue with this point of view to completely prove
the claimed decomposition, but it will be easier for me (at least notationally)
to switch back to the $B\backslash G/B$ picture. Thus consider the coset $gB$ in $G/B$, corresponding to the Borel $g B g^{-1}$.
Let me use slightly nonstandard notation, and write $D$ for the diagonal torus;
of course $D \subset B$. We may also find a torus $T \subset B \cap g B g^{-1}$.
Now there is an element $b \in B$, determined modulo $D$, such that $T = b D b^{-1}$.
(This follows from the discussion above about conjugacy properties of tori in
Borel subgroups.) We also have $g D g^{-1} \subset g B g^{-1}$, and there exists
$g b'g^{-1}\in g B g^{-1},$ well defined modulo $g D g^{-1}$, such that
$T = (g b'g^{-1}) g D g^{-1} (g b' g^{-1})^{-1} = g b' D (b')^{-1} g^{-1}.$ We thus find that $b^{-1} g b' \in N(D)/D$, and thus that $g \in B w B$ for
some $w$ in the Weyl group $N(D)/D$. Note that since $b$ and $b'$ are
well defined modulo $D$, the map from $T$ to $w$ is well-defined. Thus certainly $G$ is the union of the $B w B$. If you consider what I've already
written carefully, you will also see that the different double cosets are disjoint. We can also prove this directly as follows: given $B$ and $g B g^{-1}$, the map
$T \mapsto w$ constructed above is a map from the set of $T$ contained in $B \cap
g B g^{-1}$ to the set $N(D)/D$. Now any two such $T$ are in fact conjugate
by an element of $B \cap g B g^{-1}$. The latter group is connected,
and hence the space of such $T$ is connected.
(These assertions are perhaps most easily seen by thinking in terms of
filtrations and decompositions of $k^n$ into sums
of lines, as above). Since $N(D)/D$ is discrete, we see that $T \mapsto w$ must
in fact be constant, and so $w$ is uniquely determined just by $g B g^{-1}$ alone.
In other words, the various double cosets $B w B$ are disjoint. The preceding discussion is a litte long, since I've tried to explain (in the
particular special cases under consideration) some general facts about conjugacy
of maximal tori in algebraic groups, using the translation of group theoretic
facts about $G$, $B$, etc., into linear algebraic statements about $k^n$. Nevertheless, I believe that this is the standard proof of the Bruhat decomposition,
and explains why it is true: the relative position of two flags is described by an element of the Weyl group.
|
{
"source": [
"https://mathoverflow.net/questions/15438",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1353/"
]
}
|
15,440 |
There is the following theorem: "A space $X$ is the inverse limit of a system of discrete finite spaces, if and only if $X$ is totally disconnected, compact and Hausdorff." A finite discrete space is totally disconnected, compact and Hausdorff and all those properties pass to inverse limits. I guess the other direction might be proved by taking the system of all decompositions of $X$ into disjoint clopen sets. The inverse limit should give $X$ back. So what happens, if I dismiss the finiteness condition. As mentioned above every inverse limit of discrete spaces is totally disconnected, Hausdorff. So the question is: "Which totally disconnected Hausdorff spaces are inverse limits of discrete spaces?" For example I think it is impossible to write $\mathbb{Q}$ as an inverse limit of discrete spaces, but I don't have a proof.
|
These are the completely ultrametrizable spaces. Recall that a d:E 2 →[0,∞) is an ultrametric if d(x,y) = 0 ↔ x = y d(x,y) = d(y,x) d(x,z) ≤ max(d(x,y),d(y,z)) As usual, (E,d) is a complete ultrametric space if every Cauchy sequence converges. Suppose E ∞ is the inverse limit of the sequence E n of discrete spaces, with f n :E ∞ →E n being the limit maps. Then d ∞ (x,y) = inf { 2 -n : f n (x) = f n (y) } is a complete ultrametric on E ∞ , which is compatible with the inverse limit topology. Conversely, given a complete ultrametric space (E,d), the relation x ∼ n y defined by d(x,y) ≤ 2 -n is an equivalence relation. Let E n be the quotient E/∼ n , with the discrete topology. These spaces have obvious commuting maps between them, let E ∞ be the inverse limit of this system. The map which sends each point of E to the sequence of its ∼ n equivalence classes is a continuous map f:E→E ∞ . Because E is complete, this map f is a bijection. Moreover, a simple computation shows that this bijection is in fact a homeomorphism. Indeed, with d ∞ defined as above, we have d ∞ (f(x),f(y)) ≥ d(x,y) ≥ d ∞ (f(x),f(y))/2. As Pete Clark pointed out in the comments, the above is an incomplete answer since the question does not assume that the inverse system is countable. However, the general case does admit a similar characterization in terms of uniformities . For the purposes of this answer, let us say that an ultrauniformity is a unformity with a fundamental system of entourages which consists of open (hence clopen) equivalence relations. The spaces in question are precisely the complete Hausdorff ultrauniform spaces. Suppose E is the inverse limit of the discrete spaces E i with limit maps f i :E→E i . Without loss of generality, this is a directed system. Then the sets U i = {(x,y): f i (x) = f i (y)} form a fundamental system of entourages for the topology on E, each of which is a clopen equivalence relation on E. The universal property of inverse limits guarantees that E is complete and Hausdorff. Indeed, every Cauchy filter on E defines a compatible sequence of points in the spaces E i , which is the unique limit of this filter. Conversely, suppose E is a complete Hausdorff ultrauniform space. If U is a fundamental entourage (so U is a clopen equivalence relation on E) then the quotient space E/U is a discrete space since the diagonal is clopen. In fact, E is the inverse limit of this directed system of quotients. It is a good exercise (for Pete's students) to show that completeness and Hausdorffness of E ensure that E satisfies the universal property of inverse limits.
|
{
"source": [
"https://mathoverflow.net/questions/15440",
"https://mathoverflow.net",
"https://mathoverflow.net/users/3969/"
]
}
|
15,444 |
Define an "eventual counterexample" to be $P(a) = T $ for $a < n$ $P(n) = F$ $n$ is sufficiently large for $P(a) = T\ \ \forall a \in \mathbb{N}$ to be a 'reasonable' conjecture to make. where 'reasonable' is open to interpretation, and similar statements for rational, real, or more abstractly ordered sets for $n$ to belong to are acceptable answers. What are some examples of eventual counterexamples, famous or otherwise, and do different eventual counterexamples share any common features? Could we build an 'early warning system' set of heuristics for seemingly plausible theorems? edit: The Polya conjecture is a good example of what I was trying to get at, but answers are not restricted to number theory or any one area.
|
It was once conjectured that factors of $x^n-1$ over the rationals had no coefficient exceeding 1 in absolute value. The first counterexample comes at $n=105$.
|
{
"source": [
"https://mathoverflow.net/questions/15444",
"https://mathoverflow.net",
"https://mathoverflow.net/users/3623/"
]
}
|
15,493 |
Is a closed morphism with proper fibres proper?
|
The answer is no. Consider an integral nodal curve $Y$ over an algebraically closed field, normalize the node and remove one of the two points lying over the node. Then you get a morphisme $f : X\to Y$ which is bijective (hence homeomorphic), separated and of finite type, and the fibers are just (even reduced) points. But $f$ is not proper (otherwise it would be finite and birational hence coincides with the normalization map). In the positive direction, you can look at EGA, IV.15.7.10. [ Add ] There is an elementary way to see that $f$ is not proper just using the definition. Let $Y'\to Y$ be the normalization of $Y$. So $X$ is $Y'$ minus one closed point $y_0$. It is enough to show that the base change of $f$ to $X\times Y' \to Y \times Y'$ is not closed. Consider the closed subset
$$\Delta=\left\lbrace (x, x) \mid x\in X \right\rbrace \subset X\times Y'.$$
Its image by $f_{Y'} : X\times Y' \to Y\times Y'$ is $\left\lbrace (f(x), x) \mid x\in X\right\rbrace$ which is the graph of $Y'\to Y$ minus one point $(f(y_0), y_0)$. So $f$ is not universally closed, thus not proper.
|
{
"source": [
"https://mathoverflow.net/questions/15493",
"https://mathoverflow.net",
"https://mathoverflow.net/users/-1/"
]
}
|
15,569 |
How does one show that if $U \subseteq \mathbb{C}^n$ is nonempty and Zariski open then $U$ is also dense in the analytic topology on $\mathbb{C}^n$?
|
It is enough to show that the complement of $U$ has empty interior. Also, that complement is contained in the zero set $Z$ of a non-constant polynomial $f$, so it is enough to show that $Z$ does not contain open sets. If $z\in Z$ is a point in the interior of $Z$, then the Taylor series of $f$ at $z$ is of course zero. Since $f$ is an entire function, this is absurd.
|
{
"source": [
"https://mathoverflow.net/questions/15569",
"https://mathoverflow.net",
"https://mathoverflow.net/users/4048/"
]
}
|
15,611 |
In his answer to a question about simple proofs of the
Nullstellensatz
( Elementary / Interesting proofs of the Nullstellensatz ),
Qiaochu Yuan referred to a really simple proof for the case of an
uncountable algebraically closed field. Googling, I found this construction also in Exercise 10 of a 2008 homework
assignment from a course of J. Bernstein (see the last page of http://www.math.tau.ac.il/~bernstei/courses/2008%20spring/D-Modules_and_applications/pr/pr2.pdf ).
Interestingly, this exercise ends with the
following (asterisked, hard) question: (*) Reduce the case of arbitrary field $k$ to the case of
an uncountable field. After some tries to prove it myself, I gave up and returned to googling.
I found several references to the proof provided by Qiaochu Yuan, but no
answer to exercise (*) above. So, my question is: To prove the Nullstellensatz, how can the general case
of an arbitrary algebraically closed field be reduced to the
easily-proved case of an uncountable algebraically closed field? The exercise is from a course of Bernstein called 'D-modules and their
applications.' One possibility is that the answer arises somehow when
learning D-modules, but unfortunately I know nothing of
D-modules. Hence, proofs avoiding D-modules would be particularly
helpful.
|
These logic/ZFC/model theory arguments seem out of proportion to the task at hand. Let $k$ be a field and $A$ a finitely generated $k$-algebra over a field $k$. We want to prove that there is a $k$-algebra map from $A$ to a finite extension of $k$. Pick an algebraically closed extension field $k'/k$ (e.g., algebraic closure of a massive transcendental extension, or whatever), and we want to show that if the result is known in general over $k'$ then it holds over $k$. We just need some very basic commutative algebra, as follows. Proof: We may replace $k$ with its algebraic closure $\overline{k}$ in $k'$ and $A$ with a quotient $\overline{A}$ of $A \otimes_k \overline{k}$ by a maximal ideal (since if the latter equals $\overline{k}$ then $A$ maps to an algebraic extension of $k$, with the image in a finite extension of $k$ since $A$ is finitely generated over $k$). All that matters is that now $k$ is perfect and infinite. By the hypothesis over $k'$, there is a $k'$-algebra homomorphism
$$A' := k' \otimes_k A \rightarrow k',$$
or equivalently a $k$-algebra homomorphism $A \rightarrow k'$. By expressing $k'$ as a direct limit of finitely generated extension fields of $k$ such an algebra homomorphism lands in such a field (since $A$ is finitely generated over $k$). That is, there is a finitely generated extension field $k'/k$ such that the above kind of map exists. Now since $k$ is perfect, there is a separating transcendence basis $x_1, \dots, x_n$, so
$k' = K[t]/(f)$ for a rational function field $K/k$ (in several variables) and a monic (separable) $f \in K[t]$ with positive degree. Considering coefficients of $f$ in $K$ as rational functions over $k$, there is a localization
$$R = k[x_1,\dots,x_n][1/h]$$
so that $f \in R[t]$. By expressing $k'$ as the limit of such $R$ we get such an $R$ so that there is a $k$-algebra map
$$A \rightarrow R[t]/(f).$$
But $k$ is infinite, so there are many $c \in k^n$ such that $h(c) \ne 0$. Pass to the quotient by $x_i \mapsto c_i$.
QED I think the main point is twofold: (i) the principle of proving a result over a field by reduction to the case of an extension field with more properties (e.g., algebraically closed), and (ii) spreading out (descending through direct limits) and specialization are very useful for carrying out (i).
|
{
"source": [
"https://mathoverflow.net/questions/15611",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2734/"
]
}
|
15,658 |
I have heard the claim that the derived category of an abelian category is in general additive but not abelian. If this is true there should be some toy example of a (co)kernel that should be there but isn't, or something to that effect (for that matter, I could ask the same question just about the homotopy category). Unless I'm mistaken, the derived category of a semisimple category is just a ℤ-graded version of the original category, which should still be abelian. So even though I have no reason to doubt that this is a really special case, it would still be nice to have an illustrative counterexample for, say, abelian groups.
|
The following nicely does the trick I think... Lemma Every monomorphism in a triangulated category splits. Proof: Let $T$ be a triangulated category and suppose that $f\colon x\to y$ is a monomorphism. Complete this to a triangle $x \stackrel{f}{\to} y \stackrel{g}{\to} z \stackrel{h}{\to} \Sigma x$ then $f\circ \Sigma^{-1}h = 0$ as we can rotate backward and maps in triangles compose to zero. Since $f$ is a monomorphism we deduce that $\Sigma^{-1}h$ and hence $h$ are zero. But this implies that $y\cong x\oplus z$ (a proof of this can be found in the first part of my answer here so that $f$ is a split monomorphism. █ Since every kernel is a monomorphism we get the following counterexample. The map $\mathbb{Z}/p^2\mathbb{Z} \to \mathbb{Z}/p\mathbb{Z}$
does not have a kernel in $D(Ab)$ by virtue of the fact that $\mathbb{Z}/p^2\mathbb{Z}$ is indecomposable. Of course the same thing works in the homotopy category.
|
{
"source": [
"https://mathoverflow.net/questions/15658",
"https://mathoverflow.net",
"https://mathoverflow.net/users/361/"
]
}
|
15,664 |
Here is the criteria for a "perfect" graph editor: it should be able to perform an automated, but controllable layout one is able to make "manual" enforcements to nodes and edges locations when you need it (or at least such fine automated layout so you don't need "manual" enforcements) one could add some math symbols and formulae on a graph Common vector graphics editors could do the trick, but there is a lot of overhead efforts to draw every node, every edge, every label. Graphviz is good enough, but sometimes you cannot get needed layout (even if you use several tricks like additional invisible nodes etc) and you should use ladot or dot2tex for math formulae yEd has nice layouts, but there is a problem with a math text. This is probably not a math question, but it is common to draw graphs in articles i think. Result graph (Update: 27.12.2010)
Here is another candidate for the best editor in TeX - Asymptote (asymptote.sf.net).
The very powerful tool at first glance. (Update: 28.04.2014)
The very tasty semiautomated tool to use with PGF/TiKZ is TiKZEdt. You can extend its palette with your own tools and make the process of diagram creation very simple!
|
I already had about 30 pages of graphs typeset with xymatrix for my thesis before discovering tikz; but was so impressed by it that I was happy to rewrite them all. As well as (imho) looking better, it gave me cross-platform compatibility - xypic seems to need pstricks, so on the mac with TeXshop (which uses pdflatex, I assume) the old graphs couldn't even be rendered. Its ability to construct graphs iteratively can also be a massive timesaver- for instance, I wanted a bunch of otherwise identical rectangles at various positions, so with tikz could just loop over a list of their first coordinate rather than having to tediously cut,paste and modify an appropriate number of copies of the command for a rectangle. Particularly handy when I then decided they all needed to be slightly wider! There's a gallery of tikz examples here , to give you some idea of what it's capable of (and with the relevant source code- I did find the manual a bit hard to understand and learnt mostly by examples or trial and error). The vector graphics package inkscape (which I used to use for drawing more complicated graphs for inclusion as eps images) also apparently has a plugin to export as tikz, although I haven't tried that out.
|
{
"source": [
"https://mathoverflow.net/questions/15664",
"https://mathoverflow.net",
"https://mathoverflow.net/users/3315/"
]
}
|
15,685 |
From Model Theory article from wikipedia : "A theory is satisfiable if it has a model $ M\models T$ i.e. a structure (of the appropriate signature) which satisfies all the sentences in the set $T$". In structure definition there is also requirement for "container of a structure" to be set. As we assume then, every model have to lay inside of some set-container . This obviously give us in serious trouble, as for set theory, there is no model of this type, and even maybe cannot be. One of possible explanations why set theory cannot be closed inside set ( which will lead us to some well known paradoxes) is opinion that " there can be no end to the process of set formation " so we have "structure" which cannot be closed inside itself which is obviously rather well state. As we know that not every theory may have a model (see set-theory) then some question arises: What are the coses (other than pure practical - if they are set we know how to work with them) of putting so strong
requirements for model to be set? Is there any way to weaken this
requirement? Are there any "explorations" of
possible extension of model theory
with fundamental objects other than
sets? I presume that some categorical point of view may be useful here, but is there any? I am aware about questions asked before, specially here 8731 , but it was asked in context of category theory which is of course valid point of view but somehow too fine. I would like to ask in general. And last one, philosophical question: is then justified, that condition for a theory to have model in set universe is some kind of anthropomorphic point of view - just because we cannot build any other structures in effective way we build what is accessible for us way but it has no objective nor universal meaning? Is true that model theory is only a "universal algebra+logic" in universe of the set, or it justifications may be extended to some broader point of view? If yes: which one? I have hope that this question is good enough for mathoverflow: at least please try to deal wit as kindly request for references. Remark: Well formulated point from n-CathegoryCafe discussion : "In the centre of Model Theory there is " fundamental existence theorem says that the syntactic analysis of a theory [the existence or non-existence of a contradiction] is equivalent to the semantic analysis of a theory [the existence or non-existence of a model]." In fact the most important point is: may it be extended on non "set container" universes? I would like to thank everyone who put here some comments or answers. In is the most interesting that in a light of answer of Joel David Hamkins it seems that for first order theories (FOT) consistency is equivalent to having set model. It is nontrivial, because it is no matter of somehow arbitrary definition of "having model" but it is related to constructive proof of Completeness Theorem of Gödel. From ontological point of view it then states that for FOT there is no weaker type of consistency than arising from model theory based on sets, and in some way it is maximal form of consistency simultaneously. Then there is no way to extend for FOT equivalence to non-set containers, which is nontrivial part - the only theories which are consistent in FOT are those which has a set models and this statement is proved not using set theoretical constructions in nonconstructive ways. So it was important to me, and I learn a lot from this even if for specialist it is somehow maybe obvious. I have hope that I understand it right;-) @Tran Chieu Minh: thank You for pointing to interesting discussion, I will try to understand the meaning of Your remarks here and there.
|
You seem to believe that it is somehow contradictory to have a set model of ZFC inside another model of ZFC. But this belief is mistaken. As Gerald Edgar correctly points out, the Completeness Theorem of first order logic asserts that if a theory is consistent (i.e. proves no contradiction), then it has a countable model. To be sure, the proof of the Completeness Theorem is fairly constructive, for the model is built directly out of the syntactic objects (Henkin constants) in an expanded language. To re-iterate, since you have mentioned several times that you find something problematic with it: Completeness Theorem. (Goedel 1929) If a theory is consistent, then it has a model that is a set. The converse is much easier, so in fact we know that a theory is consistent if and only if it has a set model. This is the answer to your question. More generally, if a theory is consistent, then the upward Lowenheim-Skolem theorem shows that it has models of every larger cardinality, all of which are sets. In particular, since the language of set theory is countable, it follows that if ZFC is consistent, then it has models of any given (set) cardinality. The heart of your confusion appears to be the mistaken belief that somehow there cannot be a model M of ZFC inside another structure V satisfying ZFC. Perhaps you believe that if M is a model of ZFC, then it must be closed under all the set-building operations that exist in V. For example, consider a set X inside M. For M to satisfy the Power Set axiom, perhaps you might think that M must have the full power set P(X). But this is not so. All that is required is that M have a set P, which contains as members all the subsets of X that exist in M. Thus, M's version of the power set of X may be much smaller than the power set of X as computed in V. In other words, the concept of being the power set of X is not absolute between M and V. Different models of set theory can disagree about the power set of a set they have in common, and about many other things, such as whether a given set is countable, whether the Continuum Hypothesis holds, and so on. Some of the most interesting arguments in set theory work by analyzing and taking advantage of such absoluteness and non-absoluteness phenomenon. Perhaps your confusion about models-in-models arises from the belief that if there is a model of ZFC inside another model of ZFC, then this would somehow mean that we've proved that ZFC is consistent. But this also is not right. Perhaps some models of ZFC have models of ZFC inside them, and others think that there is no model of ZFC. If ZFC is consistent, then these latter type of models definitely exist. Incompleteness Theorem. (Goedel 1931) If a (representable) theory T is consistent (and sufficiently strong to interpret basic arithmetic), then T does not prove the assertion "T is consistent". Thus, there is a model of T which believes T is inconsistent. In particular, if ZFC is consistent, then there will be models M of ZFC that believe that there is no model of ZFC. In the case that ZFC+Con(ZFC) is consistent, then some models of ZFC will have models of ZFC inside them, and some will believe that there are no such models. A final subtle point, which I hesitate to mention because it can be confusing even to experts, is that it is a theorem that every model M of ZFC has an object m inside it which M believes to be a first order structure in the language of set theory, and if we look at m from outside M, and view it as a structure of its own, then m is a model of full ZFC. This was recently observed by Brice Halmi, but related observations are quite classical. The proof is that if M is an ω-model, then it will have the same ZFC as we do in the meta-theory and the same proofs, and so it will think ZFC is consistent (since we do), and so it will have a model. If M is not an ω-model, then we may look at the fragments of the (nonstandard) ZFC inside M that are true in some V α of M. Every standard fragment is true in some such set in M by the Reflection Theorem. Thus, by overspill (since M cannot see the standard cut of its ω) there must be some V α in M that satisfies a nonstandard fragment of its ZFC. Such a model m = V α M will therefore satisfy all of the standard ZFC. But M may not look upon it as a model of ZFC, since M has nonstandard axioms which it thinks may fail in m.
|
{
"source": [
"https://mathoverflow.net/questions/15685",
"https://mathoverflow.net",
"https://mathoverflow.net/users/3811/"
]
}
|
15,687 |
In Toen's and Vezzosi's article From HAG to DAG: derived moduli stacks a kind of definition of DAG is given. I am not an expert and can't see what's the relation between DAG and the motivic cohomology ideas of Voevodsky (particularly his category $DG$). It would be nice if somebody could explain that to me. I addition I am very keen on seeing how these ideas can be used in an explicit example. If I should explain someone why motivic cohomology is a good thing, I would certainly mention the proof of the Milnor conjecture. But can one see the use of derived ideas in a more explicit and down-to-earth example?
|
They are very different. Both involve a mixing of algebraic geometry and homotopy theory, but not at all in the same way: DAG is when you use more homotopyish rings for your basic affines, whereas in Voevodsky/Morel's work you're thinking of a variety as a kind of space and trying to capture the homotopy type of that space in a universal way. Even more loosely, in DAG you're injecting the heroin of homotopy theory right into the foundations of algebraic geometry, keeping the same form; but in motivic homotopy theory you're trying to push varieties into that decadent realm, loosen them up a little, and you fundamentally change their form.
|
{
"source": [
"https://mathoverflow.net/questions/15687",
"https://mathoverflow.net",
"https://mathoverflow.net/users/4011/"
]
}
|
15,703 |
What did Newton himself do, so that the "Newton polygon" method is named after him?
|
The Newton polygon and Newton's method are closely related. The following theorem was first proven by Puiseux: if $K$ is an algebraically closed field of characteristic zero, then the field of Puiseux series over $K$ is the algebraic closure of the field of formal Laurent series over $K$ However according to Wikipedia This property was implicit in Newton's use of the Newton polygon as early as 1671 and therefore known either as Puiseux's theorem or as the Newton–Puiseux theorem. A place where this is illustrated in more detail is "A history of algorithms: from the pebble to the microchip" By Jean-Luc Chabert, Évelyne Barbin, page 191. I will quote the first paragraph Immediately following his description of his numerical method for solving equations, Newton used the same principle to show how to obtain algebraic solutions of equations. He explains how the method of successive linear approximations can be adapted by using a ruler and "small parallelograms", the first version of what is called Newton's polygon. The method was applied in a more general case later, by Puiseux in 1850, both in considering multiple branches and in considering functions of a complex variable Then it explains Newton's approach in detail. You can follow the references given there. And then we know the story that this nice tool is now used for the understanding of polynomials over local fields even tough originally the local field was the field of formal Laurent series.
|
{
"source": [
"https://mathoverflow.net/questions/15703",
"https://mathoverflow.net",
"https://mathoverflow.net/users/454/"
]
}
|
15,731 |
I am curious to collect examples of equivalent axiomatizations of mathematical structures. The two examples that I have in mind are Topological Spaces. These can be defined in terms of open sets, closed sets, neighbourhoods, the Kuratowski closure axioms, etc. Matroids . These can be defined via independent sets, bases, circuits, rank functions, etc. Are there are other good examples? Secondly, what are some advantages of multiple axiomatizations? Obviously, one advantage is that one can work with the most convenient definition depending on the task at hand. Another is that they allow different generalizations of the object in question. For example, infinite matroids can be axiomatized by adapting the independent set axioms, but it is unknown how to axiomatize them via the circuit axioms. An acceptable answer to the second question would be an example of a proof in one axiom system that doesn't translate easily (not sure how to make this precise) into another axiom system.
|
The phenomenon that I think you have in mind has a name: cryptomorphism . I learned the name from the writings of Gian-Carlo Rota; Rota's favorite example was indeed matroids. Gerald Edgar informs me that the name is due to Garrett Birkhoff. I think modern mathematics is replete with cryptomorphisms. In my class today, I presented the "Omnibus Hensel's Lemma". Part a) was: the following five conditions on a valued field are all equivalent. Part b) was: complete fields satisfy these equivalent properties. There are lots more equivalent conditions than the five I listed: see An unfamiliar (to me) form of Hensel's Lemma and especially Franz Lemmermeyer's answer for further characterizations. I would say that the existence of cryptomorphisms is a sign of the richness and naturality of a mathematical concept -- it means that it has an existence which is independent of any particular way of thinking about it -- but that on the other hand the existence of not obviously equivalent cryptomorphisms tends to make things more complicated, not easier: you have to learn several different languages at once. For instance, the origin of the question I cited above was the fact that in Tuesday's class I st p dly chose the wrong form of Hensel's Lemma to use to try to deduce yet another version of Hensel's Lemma: it didn't work! Since we are finite, temporal beings, we often settle for learning only some of the languages, and this can make it harder for us to understand each other and also steer us away from problems that are more naturally phrased and attacked via the languages in which we are not fluent. Some further examples: I think that the first (i.e., most elementary) serious instance of cryptomorphism is the determinant. Even the Laplace expansion definition of the determinant gives you something like $n$ double factorial different ways to compute it; the fact that these different computations are not obviously equivalent is certainly a source of consternation for linear algebra students. To say nothing of the various different ways we want students to think about determinants. It is "just" the signed change of volume of a linear transformation in Euclidean space (and the determinant over a general commutative ring can be reduced to this case). And it is "just" the induced scaling factor on the top exterior power. And it is "just" the unique scalar $\alpha(A)$ which makes the adjugate equation $A*\operatorname{adj}(A) = \alpha(A) I_n$ hold. And so forth. You have to be fairly mathematically sophisticated to understand all these things. Other examples: Nets versus filters for convergence of topological spaces. Most standard texts choose one and briefly allude to the other. As G. Laison has pointed out, this is a disservice to students: if you want to do functional analysis (or read works by American mathematicians), you had better know about nets. If you want to do topological algebra and/or logic (or read works by European mathematicians), you had better know about filters. There are (at least) three axiomatizations of the concept of uniform space : (i) entourages, (ii) uniform covers, (iii) families of pseudometrics. One could develop the full theory using just one, but at various points all three have their advantages. Is there anyone who doesn't wish that there were just one definition that would work equally well in all cases?
|
{
"source": [
"https://mathoverflow.net/questions/15731",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2233/"
]
}
|
15,841 |
This question is about the space of all topologies on a
fixed set X. We may order the topologies by refinement, so
that τ ≤ σ just in case every τ open set is open in σ.
Equivalently, we say in this case that τ is coarser than σ, that σ is finer than τ or that
σ refines τ. (See wikipedia on comparison of
topologies .)
The least element in this order is the indiscrete topology and the largest topology is the discrete topology. One can show that the collection of all topologies on a fixed set is a complete lattice. In the downward direction, for example, the intersection of any collection of
topologies on X remains a topology on X, and this intersection
is the largest topology contained in them all. Similarly,
the union of any number of topologies generates a smallest
topology containing all of them (by closing under finite
intersections and arbitrary unions). Thus, the collection of all topologies on X is a complete
lattice. Note that the compact topologies are closed downward in
this lattice, since if a topology τ has fewer open sets than
σ and σ is compact, then τ is compact.
Similarly, the Hausdorff topologies are closed upward,
since if τ is Hausdorff and contained in σ, then
σ is Hausdorff. Thus, the compact topologies inhabit
the bottom of the lattice and the Hausdorff topologies the
top. These two collections kiss each other in the compact
Hausdorff topologies. Furthermore, these kissing points,
the compact Hausdorff topologies, form an antichain in the
lattice: no two of them are comparable. To see this,
suppose that τ subset σ are both compact
Hausdorff. If U is open with respect to σ, then the
complement C = X - U is closed with respect to σ and
hence compact with respect to σ in the subspace
topology. Thus C is also compact with respect to τ in
the subspace topology. Since τ is Hausdorff, this
implies (an elementary exercise) that C is closed with respect to τ, and so U is
in τ. So τ = σ. Thus, no two distinct compact Hausdorff topologies are comparable, and so these topologies are spread out sideways, forming an antichain of the lattice. My first question is, do the compact Hausdorff topologies
form a maximal antichain? Equivalently, is every topology
comparable with a compact Hausdorff topology? [Edit: François points out an easy counterexample in the comments below.] A weaker version of the question asks merely whether every
compact topology is refined by a compact Hausdorff
topology, and similarly, whether every Hausdorff topology
refines a compact Hausdorff topology. Under what
circumstances is a compact topology refined by a unique
compact Hausdorff topology? Under what circumstances does a
Hausdorff topology refine a unique compact Hausdorff
topology? What other topological features besides compactness and
Hausdorffness have illuminating interaction with this
lattice? Finally, what kind of lattice properties does the lattice
of topologies exhibit? For example, the lattice has atoms,
since we can form the almost-indiscrete topology having
just one nontrivial open set (and any nontrivial subset
will do). It follows that every topology is the least upper
bound of the atoms below it. The lattice of topologies is
complemented .
But the lattice is not distributive (when X has at least
two points), since it embeds N 5 by the
topologies involving {x}, {y} and the topology generated by
{{x},{x,y}}.
|
This is a community wiki of the answers in the comments. The compact Hausdorff topologies do not generally form a maximal antichain. If X is infinite, split X into two infinite halves and put the discrete topology on one half and the indiscrete topology on the other half. (Comment by François G. Dorais) Addendum: Without sufficient Choice, the infinite set $X$ may be amorphous . Amorphous sets are precisely the infinite sets for which this approach doesn't work. Very little Choice is needed to ensure that no such beast exists. (Edit by Cameron Buie) There is a maximal compact topology on a countable space which is not Hausdorff. See Steen & Seebach 99 . (Comment by Gerald Edgar) There is a minimal Hausdorff topology on a countable space which is not compact. See Steen & Seebach 100 . (Comment by François G. Dorais) Those examples can be lifted to any cardinality space, simply by using the disjoint sum with any given compact Hausdorff space. (Comment by Gerald Edgar) Every set admits a compact Hausdorff topology, by topologizing it as the one-point compactification of the discrete space structure on the complement of any point. (Answer below by Cameron Buie) ( Feel free to edit and expand )
|
{
"source": [
"https://mathoverflow.net/questions/15841",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1946/"
]
}
|
15,901 |
I'm thinking about running a graduate student seminar in the summer. Having both organized and participated in such seminars in the past, I have witnessed first-hand that, contrary to what one might expect, they can be rather successful. However I haven't been able to quite put my finger on what makes a good seminar good. Certainly there are obvious necessary conditions for success, such as having sufficiently many (dedicated) participants and at least some semblance of a goal. But in my experience these conditions aren’t at all sufficient. And on the other hand there are clear pitfalls that should be avoided, such as going too fast or not going fast enough, or scheduling the seminar at 8 in the morning. But there are also more subtle pitfalls that aren't as easily avoided: for example, having consecutive speakers of a certain style that might put off or discourage other participants. (Of course a plausible solution to this specific problem is to not have such people speak one after another, but often this is infeasible.) So I turn to the collective wisdom of MO: In your experience, what has made a specific learning seminar feel successful to you? Feel free to interpret the word "successful" any way you want. Anecdotes and horror stories welcome. (Aside: The seminar I'm planning is a "classics in geometry and topology" type of deal. By this I mean, each participant will select a classic paper at the beginning of the summer and then briefly discuss its contents sometime during the course of the seminar. I would consider this seminar successful if, at the end, each participant walks away with a set of their own notes on each paper, explaining why it's important, and containing a sketch of its main ideas and how it fits in the grand scheme of things; the hope is that such a set of notes might prove useful if one were to take a closer look at the paper down the road. If anyone has any experience about running a seminar of this type, then I'd be especially interested in hearing their comments!)
|
One big key is for the organizers/leaders of the seminar to be willing to set the tone by asking frequent and sometimes stupid questions. If the audience gets into a low-question equilibrium rather than a high-question one then you're in trouble. You want people in the audience to feel like they can and should understand everything that happens, and should hassle the speaker until they do understand everything. So make sure that there are enough people in the audience who are willing to do this.
|
{
"source": [
"https://mathoverflow.net/questions/15901",
"https://mathoverflow.net",
"https://mathoverflow.net/users/430/"
]
}
|
15,951 |
I've been reading up a bit on the fundamentals of formal logic, and have accumulated a few questions along the way. I am pretty much a complete beginner to the field, so I would very much appreciate if anyone could clarify some of these points. A complete (and consitent) propositional logic can be defined in a number of ways, as I understand, which are all equivalent. I have heard it can be defined with one axiom and multiple rules of inferences or multiple axioms and a single rule of inference (e.g. Modus Ponens) - or somewhere inbetween. Are there any advantage/disvantages to either? Which is more conventional? Propositional (zeroth-order) logic is simply capable of making and verifying logical statements. First-order (and higher order) logics can represent proofs (or increasing hierarchial complexity) - true/false, and why? What exactly is the relationship between an nth-order logic and an (n+1)th-order logic, in general. An explanation mathematical notation would be desirable here, as long as it's not too advanced. Any formal logic above (or perhaps including?) first-order is sufficiently powerful to be rendered inconsistent or incomplete by Godel's Incompleteness Theorem - true/false? What are the advantages/disadvantages of using lower/higher-order formal logics? Is there a lower bound on the order of logic required to prove all known mathematics today, or would you in theory have to use an arbitrarily high-order logic? What is the role type theory plays in formal logic? Is it simply a way of describing nth-order logic in a consolidated theory (but orthogonal to formal logic itself), or is it some generalisation of formal logic that explains everything by itself? Hopefully I've phrased these questions in some vaguely meaningful/understandable way, but apologies if not! If anyone could provide me with some details on the various points without assuming too much prior knowledge of the fields, that would be great. (I am an undergraduate Physics student, with a background largely in mathematical methods and the fundamentals of mathematical analysis, if that helps.)
|
This is a long list of questions! These are all related to a certain extent, but you might consider breaking it up into separate questions next time. Proof theorists tend to prefer systems with many rules and few axioms such as natural deduction systems and Gentzen systems. The reason is that these are much easier to manipulate and visualize. (Look at the statement and proof of the Cut Elimination Theorem for an illustrative example.) Model theorists tend to prefer lots of axioms and just modus ponens, like Hilbert systems. The reason is that the semantics of such systems are easy to handle, and semantics is what model theorists really care about. The real difference between propositional and first-order logic is quantification. Quantifiers are naturally more expressive than logical connectives. There are many, many flavors of higher-order logic. The two main classifications are set-based systems and function-based systems. There are variants which don't really fit this division and there are variants which mix both. With strong enough internal combinatorics, these are all equivalent. The set-based system have a set A of type 0 objects, which are considered atomic. Type 1 objects are elements of the powersets ℘(A n ). When the base theory has an internal pairing function (like arithmetic and set theory), the exponent n can be dropped. Then, type 2 objects are elements of the second powerset ℘(℘(A)), and similarly for higher types. Set-based higher types are somewhat inconvenient without an internal pairing function. Function-based systems are similar, with functions A n →A as type 1. Again, with an internal pairing function, the higher types streamline to A → A, (A → A) → A, ((A → A) → A) → A, etc. However, it is common to use more complex types in these systems. Be careful how you read Gödel's Incompleteness Theorem. For first-order logic, the hypotheses state that the theory in question must be recusively enumerable and that it must be powerful enough to interpret a reasonable amount of arithmetic. Those are important hypotheses. (And they explain why no such theorem exists for propositional logic.) There are many variants for higher-order systems, but you need to be even more careful when stating them. You can appreciate the difference between the different levels of higher-order logic by reading about Gödel's Speed-Up Theorems . Most of mathematics today is based on set theory, at least in a theoretical sense. In set theory, higher types are interpreted internally using powersets and sets of functions. They also extend transfinitely and this transfinite hierarchy of higher types was shown necessary to prove theorems very low in the hierarchy, such as Martin's Borel Determinacy Theorem . You may want to read on categorical logic .
|
{
"source": [
"https://mathoverflow.net/questions/15951",
"https://mathoverflow.net",
"https://mathoverflow.net/users/602/"
]
}
|
16,026 |
Books can be written about the finite subgroups of $\mathrm{SL}(2,\mathbb C)$ (and their immediate family, like the polyhedral groups...) I am about to start writing notes for a short course about them and I would like to include references to as much useful and interesting information about them as possible. Since they show up in quite different contexts, and can be looked upon from many different points of view, I am sure the very varied MO audience knows lots of things about them I don't. So, despite this being more or less canonically too broad/vague a question for MO according to the FAQ : Can you tell me (or at least point me to) all about the finite subgroups of $\mathrm{SL}(2,\mathbb C)$? LATER: Thanks to everyone who answered. So far, the information is essentially of algebraic and geometric nature. I wonder now about combinatorics and such beasts. For example, it is a theorem of Whitney (or maybe it just follows easily from a theorem of Whitney) that a 3-connected simple planar graph with $e$ edges has an automorphism group of order at most $4e$, and that the order is $4e$ precisely when the graph comes from a polyhedron, so that the group is a polyhedral group. Do you know of similar results?
|
This question is heavily related to one of my favorite relations between geometry and representation theory. Consider simple Lie algebras of the following types: $A_n$ $D_n$ $E_6$ $E_7$ $E_8$ Then these Dynkin diagrams correspond to all possible finite subgroups of $SL_2.$ The relation is given by special classes of isolated surface singularities known as Kleinian or Du Val singularities. These arise as follows. Given a finite subgroup $G \subset SL_2,$ we have an action of $G$ on $\mathbb{C}^2$ with no fixed points other than the origin. If we then look at the geometric quotient $\mathbb{C}^2/G$ corresponding to $G$-invariant polynomials in $\mathbb{C}[x,y],$ it is generated by three homogeneous polynomials $f_1, f_2, f_3$ which are related by a weighted homogeneous polynomial $g$ of degree 3 such that $g(f_1, f_2, f_3) = 0.$ We can then identify $\mathbb{C}^2/G$ with the hypersurface $\{ g = 0 \} \subset \mathbb{C}^3.$ The resulting hypersurfaces have the following equations (with corresponding subgroup): $A_n: x^{n+1} + y^2 + z^2$ (cyclic) $D_n: x^{n-1} + xy^2 + z^2$ (dihedral) $E_6: x^4 + y^3 + z^2$ (tetrahedral) $E_7: x^3y + y^3 + z^2$ (octahedral) $E_8: x^5 + y^3 + z^2$ (icosahedral) The Dynkin diagram enter as follows. Each of these surfaces can be resolved through a finite number of blow-ups, and the exceptional fiber in the resolution will consist of a copy of $\mathbb{P}^1$ for each node of the Dynkin diagram, each of which is joined to the point of another $\mathbb{P}^1$ if there is a corresponding edge in the Dynkin diagram connecting the two nodes (so in the cyclic case, it's just a chain of $\mathbb{P}^1$'s). Lastly, there is a neat connection between this and Springer theory that goes as follows. Let $\mathcal{N}$ denote the nilpotent cone of a Lie algebra of one of the types listed above, and let $\mathcal{O}$ denote the subregular orbit. Then $\mathcal{O}$ has codimension two in $\mathcal{N}$ and hence the corresponding Kostant/Slodowy slice is a surface in $\mathcal{N}.$ It then turns out that this surface is one of the surface singularities listed above, and that the corresponding Springer fiber of a subregular element is isomorphic to the exceptional fiber in the resolution of the surface mentioned above. So the Springer resolution encodes the information of the successive blow-ups of these surfaces. A few good references: Milnor, Singular Points of Complex Hypersurfaces Dimca, Singularities and Topology of Hypersurfaces Slodowy, Simple Singularities and Simple Algebraic Groups
|
{
"source": [
"https://mathoverflow.net/questions/16026",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1409/"
]
}
|
16,048 |
I found myself "naturally" dealing with an object of this form: X is a complex vector space, with a "product" (a,b) → {aba} which is quadratic in the first variable, linear in the second, and satisfies some associativity conditions. These conditions are actually complicated, but more or less say that {aba} looks like the product (aba) in an alternative algebra Y containing X as a subspace. For example, the main "associativity condition" I am interested in is:
{a{b{aca}b}a}={{aba}c{aba}} Examples Symmetric matrices Octonions, or indeed any alternative algebra Let J belong to GL(n,ℂ), with t J=-J and J²=-Id, and W={w∈M(n×n,ℂ)|J t wJ=-w} all with the standard product {aba}=aba. All of these examples are Jordan algebras, with respect to the symmetrized product a∘b=½(ab+ba), but I cannot see any direct link between the Jordan product and my product.
|
In a Jordan algebra with product $\cdot$, a triple product is defined by $$\{abc\}=(a\cdot b)\cdot c+(b\cdot c)\cdot a-(a\cdot c)\cdot b.$$
In a special Jordan algebra (constructed by symmetrising an associative product) one has $\{aba\}=aba$, and it is easy to show that in such algebras one always has the identity $$\{\{aba\}c\{aba\}\}=\{a\{b\{aca\}b\}a\}.$$ Now, there is an amazing general theorem of Macdonald's that states that any identity in three variables which is of degree at most one in one of them and which is valid in special Jordan algebras actually holds in all Jordan algebras. This is proved in Jacobson's breath-taking Structure and representations of Jordan algebras . So your identity holds in all Jordan algebras. As a consequence, from the information you give it is more or less impossible to distinguish your structure from Jordan algebras, as far as I can see. By the way, in his book, Jacobson notes that McCrimmon has developed the theory of Jordan algebras based exclusively on the composition $(a,b)\mapsto aba$, and gives [McCrimmon, Kevin. A general theory of Jordan rings. Proc. Nat. Acad. Sci. U.S.A. 56 1966 1072--1079. MR0202783 (34 #2643)] as reference. I do not have access to the paper, though. The paper can be gotten from this link Andrea provided in a comment below.
|
{
"source": [
"https://mathoverflow.net/questions/16048",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1626/"
]
}
|
16,074 |
This is maybe not an entirely mathematical question, but consider it a pedagogical question about representation theory if you want to avoid physics-y questions on MO. I've been reading Singer's Linearity, Symmetry, and Prediction in the Hydrogen Atom and am trying to come to terms with the main physical (as opposed to mathematical) argument of the text. The argument posits, if I understand it correctly, that a quantum system described by a Hilbert space $H$ on which a group $G$ of symmetries acts by unitary transformations should have the property that its "elementary states" "are" irreducible subrepresentations of the representation of $G$ on $H$. She begins this argument in section 5.1: Invariant subspaces are the only physically natural subspaces. Recall from Section 4.5 that in a quantum system with symmetry, there is a natural representation $(G, V, \rho)$. Any physically natural object must appear the same to all observers. In particular, if a subspace has physical significance, all equivalent observers must agree on the question of a particular state's membership in that subspace. and continues it in section 6.3: We know from numerous experiments that every quantum system has *elementary states*. An elementary state of a quantum system should be **observer-independent**. In other words, any observer should be able (in theory) to recognize that state experimentally, and the observations should all agree. Secondly, an elementary state should be indivisible. That is, one should not be able to think of the elementary state as a superposition of two or more "more elementary" states. If we accept the model that every recognizable state corresponds to a vector subspace of the state space of the system, then we can conclude that elementary states correspond to irreducible representations. The independence of the choice of observer compels the subspace to be invariant under the representation. The indivisible nature of the subspace requires the subspace to be irreducible. So elementary states correspond to irreducible representations. More specifically, if a vector $w$ represents an elementary state, then $w$ should lie in an *irreducible* invariant subspace $W$, that is, a subspace whose only invariant subspaces are itself and $0$. In fact, every vector in $W$ represents a state "indistinguishable" from $w$, as a consequence of Exercise 6.6. (For people who actually know their quantum, Singer is ignoring the distinction between representations and projective representations until later in the book.) My first problem with this argument is that Singer never gives a precise definition of "elementary state." My second problem is that I'm not sure what physical principle is at work when she posits that physically natural subspaces and elementary states should be observer-independent (i.e. invariant under the action of $G$). What underlying assumption of quantum mechanics, or whatever, is at work here? Why should a mathematician without significant training in physics find this reasonable? (I have the same question about the identification of elementary particles with irreducible representations of the "symmetry group of the universe," so any comments about this physical argument are also welcome.) Singer goes on to use this assumption to deduce the number of electrons that fill various electron orbitals, and I won't be able to convince myself that this makes sense until I understand the physical assumption that allows us to use irreducible representations to do this.
|
Invariant states are not the only meaningful ones. Even in classical mechanics, a baseball traveling 90 mph toward my head is quite meaningful to me, even though it is of no consequence to my fellow mathematician a mile away. The focus on invariant subspaces comes not from an assumption, but from the way physicists do their work. They want to predict behavior by making calculations. They want to find laws that are universal. They want equations and calculation rules that will be invariant under a change of observers. Any particular calculation might require a choice of coordinates, but the rules must be invariant under that choice. Once we're talking about one particular baseball trajectory, that trajectory will look different in different coordinate systems; the rules governing baseball flight, however, must look the same in all equivalent coordinate systems. The natural features of baseballs arise from the equivalence classes of trajectories of baseballs -- equivalence under the group action. Here, if we pretend the earth is flat, gravity is vertical, and air does not resist the baseball, the group is generated by translations and rotations of the plane. Any physically natural, intrinsic property of the baseball itself (such as its mass) or its trajectory (such as the speed of the baseball) must be invariant under the group action. If you don't know a priori what these properties will be, a good way to find them is to pass from individual instances (the baseball heading toward me at 90mph) to the equivalence class generated by individual instances under the group action (the set of all conceivable baseballs traveling at 90mph). Note that the equivalence class is invariant under the group action, and it is exactly this invariance that makes the equivalence class a useful object of the physicists' study. More generally, if you are studying a physical system with symmetry, it's a good bet that the invariant objects will lead to physically relevant, important quantities. It's more a philosophy than an axiom, but it has worked for centuries.
|
{
"source": [
"https://mathoverflow.net/questions/16074",
"https://mathoverflow.net",
"https://mathoverflow.net/users/290/"
]
}
|
16,121 |
Can someone give a really concrete example of such a sequence? I am looking at several notes related with such things, but haven't seen any well-calculated example. And I'm really confused at this point. Besides asking for a good example, I am also wondering about the following two things: There is an exact sequence for elliptic curves defined over a local field $K$, $0 \rightarrow \hat E(m) \rightarrow E(K) \rightarrow \tilde E(k) \rightarrow 0$, where $\hat E(m)$ is the formal group associated to $E$ and $\tilde E(k)$ is the reduction. (See Silverman AEC I, page 118), is this sequence related with connected-etale sequence? 2.Take the p-torsion kernel $E[p]$ of $[p]: E \rightarrow E$ for $E$ defined over $K$ a local field.Is $E[p]$ a finite flat group scheme over $R$ the valuation ring? And if so, what is its connected-etale sequence? (maybe I should change $p$ to an $n$, but I'm also curious what will happen if $p$ is the characteristic of $k$?) Thank you.
|
For concepts related to algebraic geometry when the base is not a field, it can be difficult for a beginner to reconcile the approach in Silverman with the approach via schemes. I wasted a lot of time as a student trying to relate the "3 points through a line" definition of the group law over fields with the concept of "reduction mod $p$" on points. Likewise, the approach with formal groups tends to make things confusing, despite their apparent "concreteness". This sort of stuff drove me crazy when I was a student, until I realized that the best way to understand such topics is to give up working over fields and with equations, and to work over the valuation ring and with functorial viewpoints (only translating into field language at the very end). The relevant schemes in the question are really torsion schemes over the valuation ring, not torsion in the separate fibers (so you mean to assume $E$ has good reduction in both questions). I address this below. The following answer is way too long, since I do not know of a suitable reference not involving EGA/SGA. Tate's article on finite flat group schemes probably explains some aspects, but I doubt it addresses the link with the concrete stuff for elliptic curves. If $R$ is a local ring, then an "elliptic curve over $R$" can be defined in two ways: the concrete way is as a Weierstrass plane cubic with unit discriminant, and the right way is as a smooth proper $R$-scheme with geometrically connected fibers of dimension 1 and genus 1 and a distinguished section. As usual, the concrete way is hard to use to actually prove anything interesting (and it is the "wrong" notion when the base is non-local and especially has non-trivial line bundles; e.g., over suitable number fields with class number $> 1$ one can make CM elliptic curves with "everywhere good reduction" which do not admit a global planar model with unit discriminant). How to prove there is a unique $R$-group structure with the indicated section as the identity? A real nightmare with the concrete definition, and elegantly explained in Chapter 2 of Katz-Mazur with the right definition. Likewise, that $E$ is functorial in its generic fiber when $R$ is a discrete valuation ring is a mess to prove by hand (which affine opens to use?), but has an elegant proof when approached through the "smooth and proper" viewpoint. Of course it is important and interesting that these concrete and abstract notions agree, and that is explained in Katz-Mazur Chapter 2. That being said, if $E$ is an elliptic curve over any noetherian (say) scheme $S$ and $[n]_E:E \rightarrow E$ is multiplication by a positive integer $n$, then on geometric fibers this is a finite flat map, so $[n]$ is quasi-finite. Now proper and quasi-finite maps are finite (by Zariski's Main Theorem), so $[n]_E$ is a finite map, and the fibral flatness criterion implies that it is also flat. Being a finite flat map between noetherian schemes, it has a "degree" which is locally constant on the target and yet is $n^2$ on fibers over $S$. Hence, we conclude that $E[n] := {\rm{ker}}([n]_E)$ is a finite flat commutative $S$-group with constant fiber rank $n^2$. Honestly, I do not know any way to prove this which avoids the serious results that I just cited. But that's why the theorems are useful: because we can use them to make our intuition over fields carry over to cases when the base is not a field. (The noetherian condition can be dropped if we are more careful with the phrase "finite flat". I won't dwell on it here.) This answers the first part of the 2nd question (taking the base to be spectrum of the valuation ring there). It the notation there, the $p$-torsion of the elliptic curve over $K$ is not a finite $R$-scheme, and in general it may extend to a finite flat $R$-group in many ways. But the elliptic curve over $K$ uniquely extends to one over $R$ by the theory of Neron models, and its torsion levels provide the "right" finite flat groups you want to use over the valuation ring. OK, now assume $R$ is a complete local noetherian ring (e.g., a complete discrete valuation ring). Could even assume it is a henselian local ring, but the complete case is easier to deal with and covers the case in the question. Let $G$ be a finite flat $R$-group, a case of interest being $E[n]$ for an elliptic curve $E$ over $R$. Let $k$ be the residue field, and consider $G_k$. Being a finite $k$-scheme, it has an open and closed identity component $G_k^0$ which is cut out by an idempotent. By 8.15 (or thereabouts) in Matsumura's Commutative Ring Theory, every idempotent in the special fiber of a finite $R$-algebra uniquely lifts. In particular, if $X$ is a finite $R$-scheme then its connected component decomposition uniquely lifts that of $X_k$. If $X$ is $R$-flat then so is each of its connected components. This is all compatible with products, so if $X$ has a structure of $R$-group then the open and closed connected component $X^0$ containing the identity section is an $R$-subgroup. Returning to our friend $G$, we get the so-called "relative identity component" $G^0$, an open and closed (hence finite flat) $R$-subgroup. Remark: The formation of $G^0$ commutes with any flat local extension on $R$, as follows from the uniqueness! It doesn't usually commute with non-local extension, such as inclusion of a complete dvr into its fraction field. Example: $G = E[n]$. Suppose $R$ is a complete discrete valuation ring with fraction field $K$, and $n \in K^{\times}$. What is $(G^0)_K$? Well, each "point" occurs over a finite extension $K'/K$, say with valuation ring $R'$, and $G(K') = G(R')$ by elementary integrality considerations (or in fancy terms, valuative criterion, which is killing a fly with a sledgehammer). Since the spectrum of $R'$ is connected , a point in $G(R')$ lies in $G^0(R')$ if and only if its specialization into $G_k(k')$ vanishes ($k'$ the residue field of $R'$). In other words, $(G^0)(\overline{K})$ consists of the $n$-torsion geometric points of $E_K$ whose specialization into geometric points of $E_k$ by valuative criterion for the $R$-proper $E$ ($E_K(K') = E(R') \rightarrow E(k') = E_k(k')$!) is 0. Now we need to explain the "etale quotient" in concrete terms. This is best understood as a generalization of the following procedure over a field. Example: Let $k$ be a field and $A_0$ a finite $k$-algebra. There is a unique maximal \'etale $k$-subalgebra $A_0'$ in $A_0$: concretely, in each local factor ring of $A_0$ uniquely lift the separable closure of $k$ in the residue field up into the local factor ring via Hensel's Lemma and the primitive element theorem. Since it is uniquely characterized by lifting separable closures of $k$ in the residue fields of the factor rings, it is a good exercise to check the following crucial thing: if $B_0$ is another finite $k$-algebra then $(A_0 \otimes_k B_0)' = A_0' \otimes_k B_0'$, and $A_0'$ is functorial in $A_0$. Observe that $A_0' \rightarrow A_0$ is faithfully flat since at the level of factor rings of $A_0$ it is an inclusion of a field into a nonzero ring. Also observe that any \'etale $k$-algebra equipped with a map to $A_0$ uniquely factors through $A_0'$. Exercise: The formation of $A_0'$ commutes with any field extension on $k$. (Hint: use Galois descent to reduce to the separate cases of separable algebraic extensions and the easy case $k = k_s$.) In geometric terms, for a finite $k$-scheme $X_0$, the preceding Example constructs a finite \'etale $k$-scheme $X_0'$ and a faithfully flat $k$-map $f_0:X_0 \rightarrow X_0'$ which is initial among all $k$-maps from $X_0$ to finite \'etale $k$-schemes, and its formation is functorial in $X_0$ and commutes with products in $X_0$ and with any extension on $k$. In particular, if $X_0$ is a $k$-group then $X_0'$ has a unique $k$-group structure making $f_0:X_0 \rightarrow X_0'$ a $k$-homomorphism. Example: Now let $R$ be a complete discrete valuation ring with residue field $k$, and let $X$ be a finite flat $R$-scheme. (Can relax the hypothesis on $R$ if familiar with finite \'etale maps in general.) In this setting, "finite 'etale" over $R$ just means "product of finitely many unramified finite extensions". By using Hensel's Lemma in finite local $R$-algebras, to give a map from a finite \'etale $R$-algebra $A$ to a finite $R$-algebra $B$ is the same as to give a map $A_0 \rightarrow B_0$ between their special fibers. In particular, finite \'etale $k$-algebras uniquely and functorially lift to finite \'etale $R$-algebras, and so $X_0'$ uniquely lifts to a finite \'etale $R$-scheme $X'$ and there is a unique $R$-map $f:X \rightarrow X'$ lifting $f_0:X_0 \rightarrow X_0'$. By fibral flatness (using $X$ is $R$-flat!), $f$ is faithfully flat since $f_0$ is. By uniqueness of everything in sight, the formation of $f$ commutes with products and local extension on $R$ and is also functorial in $X$. In particular, if $G$ is a finite flat $R$-group then $G'$ admits a unique $R$-group structure making $f$ an $R$-homomorphism. We call $G'$ the maximal \'etale quotient of $G$. Now we can put it all together and obtain the connected-etale sequence: Proposition: Let $G$ be a finite flat group scheme over a complete discrete valuation ring $R$. (Even ok for complete local noetherian $R$, or even henselian local $R$.) The faithfully flat $R$-homomorphism $f:G \rightarrow G'$ to the maximal \'etale quotient has scheme-theoretic kernel $G^0$. Proof: The kernel $H = \ker f$ is a finite flat $R$-group. To show it contains $G^0$ we have to check that the composite map $G^0 \rightarrow G \rightarrow G'$ vanishes. Being a map from a finite $R$-scheme to a finite \'etale $R$-scheme, the map is determined by what it does on the special fiber, so it suffices to show that $G_k^0 \rightarrow G_0'$ vanishes. This is a map from a finite infinitsimal $k$-scheme to a finite \'etale $k$-scheme which carries the unique $k$-point to the identity point. Thus, it factors through the identity section of $G_0'$, which is open and closed since $G_0'$ is finite etale over $k$. Now that $H$ contains $G^0$, to prove the resulting closed immersion $G^0 \hookrightarrow H$ between finite flat $R$-schemes is an isomorphism it suffices to do so on special fibers. But that reduces us to the variant of our problem over the residue field. We can increase it to be algebraically closed, and so the problem is to show that if $G$ is a finite flat group scheme over an algebraically closed field $k$ then $G \rightarrow G'$ has kernel exactly $G^0$. But $G'$ is a constant $k$-scheme since it is etale and $k$ is algebraically closed, so by construction $G'$ is just the disjoint union of the $k$-points of the connected components of $G$. It is then physically obvious that the kernel is $G^0$. QED Remark: If $X$ is any finite flat $R$-scheme, with $X \rightarrow X'$ the initial map to
a finite \'etale $R$-scheme, then the induced map on $\overline{k}$-points is bijective. Indeed, we can pass to geometric special fibers and connected components to reduce to the case when $X$ is local finite over an algebraically closed field (in place of $R$), in which case the assertion is clear. By this Remark, the geometric points of the $n$-torsion in $E_k$ are identified with the geometric points of the special fiber of the maximal etale quotient $E[n]'$. In particular, if $n$ is not divisible by the characteristic of $K$ and if $K'/K$ is a sufficiently big finite separable extension which splits $E_K[n]$ then the finite etale $R'$-scheme $E[n]'_{R'}$ is constant (as it may be checked on $K'$-fiber), so the map
$$E _K[n] (\overline{K}) = E _K[n] (K') = E[n] (R') \rightarrow E[n]' (k') \hookrightarrow
E _k[n]' (\overline{k}) = E[n]'(R') = E[n]' (\overline{K})$$
is identified with the naive map in question 1. In other words, that step computes the "quotient" part of the connected-etale sequence of $E[n]$ after passing to $\overline{K}$-points! Example: If $E$ has supersingular reduction then $E[p] = E[p]^0$ and the etale part of the sequence for $E[p]$ vanishes. Example: If $E$ has ordinary reduction then working over an algebraic closure of the residue field shows that $E[p]^0$ and $E[p]'$ each have rank $p$ as finite flat $R$-groups. Finally, it remains to relate $E[n]^0$ to $n$-torsion in the so-called "formal group" of $E$ ( not the formal group of $E_K$, which loses contact with the integral structure and for ${\rm{char}} (K) = 0$ is actually the formal additive group which has no nontrivial $n$-torsion!). A moment's reflection on the definition of the formal group in Silverman shows that its $R'$-points for any finite local valuation ring extension of $R$ are precisely the local $R'$-points of the complete local ring $\widehat{\mathcal{O}}_{E,0_k}$ at the origin of the special fiber (or the completion along the identity section, comes to the same since $R$ is complete). By the universal properties of local rings on schemes and completions of local noetherian rings, such $R'$-points of the latter type are simply points in $E(R')$ specializing to $0_k$ in $E_k (k')$. But we saw earlier that $E[n]^0 (R')$ is exactly the set of points in $E[n] (R')$ specializing to $0_k$ on $E_k$. So indeed $E[n]^0 (R')$ inside of $E[n] (R') = E_K[n] (K')$ is exactly the $n$-torsion in the $K'$-points of the "formal group" in the sense of Silverman's book. Voila, so that answers the questions. The arguments used are designed to apply equally well to abelian varieties.
|
{
"source": [
"https://mathoverflow.net/questions/16121",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1238/"
]
}
|
16,125 |
Given a set of points ( 1-Dimensional), Can we get a measurement of the entropy of the data without calculating the underlying distribution. I mean , Is there a closed form for measuring entropy given a set of data-points. The closed form need not give me exact entropy..but any measure of information content would help.
|
For concepts related to algebraic geometry when the base is not a field, it can be difficult for a beginner to reconcile the approach in Silverman with the approach via schemes. I wasted a lot of time as a student trying to relate the "3 points through a line" definition of the group law over fields with the concept of "reduction mod $p$" on points. Likewise, the approach with formal groups tends to make things confusing, despite their apparent "concreteness". This sort of stuff drove me crazy when I was a student, until I realized that the best way to understand such topics is to give up working over fields and with equations, and to work over the valuation ring and with functorial viewpoints (only translating into field language at the very end). The relevant schemes in the question are really torsion schemes over the valuation ring, not torsion in the separate fibers (so you mean to assume $E$ has good reduction in both questions). I address this below. The following answer is way too long, since I do not know of a suitable reference not involving EGA/SGA. Tate's article on finite flat group schemes probably explains some aspects, but I doubt it addresses the link with the concrete stuff for elliptic curves. If $R$ is a local ring, then an "elliptic curve over $R$" can be defined in two ways: the concrete way is as a Weierstrass plane cubic with unit discriminant, and the right way is as a smooth proper $R$-scheme with geometrically connected fibers of dimension 1 and genus 1 and a distinguished section. As usual, the concrete way is hard to use to actually prove anything interesting (and it is the "wrong" notion when the base is non-local and especially has non-trivial line bundles; e.g., over suitable number fields with class number $> 1$ one can make CM elliptic curves with "everywhere good reduction" which do not admit a global planar model with unit discriminant). How to prove there is a unique $R$-group structure with the indicated section as the identity? A real nightmare with the concrete definition, and elegantly explained in Chapter 2 of Katz-Mazur with the right definition. Likewise, that $E$ is functorial in its generic fiber when $R$ is a discrete valuation ring is a mess to prove by hand (which affine opens to use?), but has an elegant proof when approached through the "smooth and proper" viewpoint. Of course it is important and interesting that these concrete and abstract notions agree, and that is explained in Katz-Mazur Chapter 2. That being said, if $E$ is an elliptic curve over any noetherian (say) scheme $S$ and $[n]_E:E \rightarrow E$ is multiplication by a positive integer $n$, then on geometric fibers this is a finite flat map, so $[n]$ is quasi-finite. Now proper and quasi-finite maps are finite (by Zariski's Main Theorem), so $[n]_E$ is a finite map, and the fibral flatness criterion implies that it is also flat. Being a finite flat map between noetherian schemes, it has a "degree" which is locally constant on the target and yet is $n^2$ on fibers over $S$. Hence, we conclude that $E[n] := {\rm{ker}}([n]_E)$ is a finite flat commutative $S$-group with constant fiber rank $n^2$. Honestly, I do not know any way to prove this which avoids the serious results that I just cited. But that's why the theorems are useful: because we can use them to make our intuition over fields carry over to cases when the base is not a field. (The noetherian condition can be dropped if we are more careful with the phrase "finite flat". I won't dwell on it here.) This answers the first part of the 2nd question (taking the base to be spectrum of the valuation ring there). It the notation there, the $p$-torsion of the elliptic curve over $K$ is not a finite $R$-scheme, and in general it may extend to a finite flat $R$-group in many ways. But the elliptic curve over $K$ uniquely extends to one over $R$ by the theory of Neron models, and its torsion levels provide the "right" finite flat groups you want to use over the valuation ring. OK, now assume $R$ is a complete local noetherian ring (e.g., a complete discrete valuation ring). Could even assume it is a henselian local ring, but the complete case is easier to deal with and covers the case in the question. Let $G$ be a finite flat $R$-group, a case of interest being $E[n]$ for an elliptic curve $E$ over $R$. Let $k$ be the residue field, and consider $G_k$. Being a finite $k$-scheme, it has an open and closed identity component $G_k^0$ which is cut out by an idempotent. By 8.15 (or thereabouts) in Matsumura's Commutative Ring Theory, every idempotent in the special fiber of a finite $R$-algebra uniquely lifts. In particular, if $X$ is a finite $R$-scheme then its connected component decomposition uniquely lifts that of $X_k$. If $X$ is $R$-flat then so is each of its connected components. This is all compatible with products, so if $X$ has a structure of $R$-group then the open and closed connected component $X^0$ containing the identity section is an $R$-subgroup. Returning to our friend $G$, we get the so-called "relative identity component" $G^0$, an open and closed (hence finite flat) $R$-subgroup. Remark: The formation of $G^0$ commutes with any flat local extension on $R$, as follows from the uniqueness! It doesn't usually commute with non-local extension, such as inclusion of a complete dvr into its fraction field. Example: $G = E[n]$. Suppose $R$ is a complete discrete valuation ring with fraction field $K$, and $n \in K^{\times}$. What is $(G^0)_K$? Well, each "point" occurs over a finite extension $K'/K$, say with valuation ring $R'$, and $G(K') = G(R')$ by elementary integrality considerations (or in fancy terms, valuative criterion, which is killing a fly with a sledgehammer). Since the spectrum of $R'$ is connected , a point in $G(R')$ lies in $G^0(R')$ if and only if its specialization into $G_k(k')$ vanishes ($k'$ the residue field of $R'$). In other words, $(G^0)(\overline{K})$ consists of the $n$-torsion geometric points of $E_K$ whose specialization into geometric points of $E_k$ by valuative criterion for the $R$-proper $E$ ($E_K(K') = E(R') \rightarrow E(k') = E_k(k')$!) is 0. Now we need to explain the "etale quotient" in concrete terms. This is best understood as a generalization of the following procedure over a field. Example: Let $k$ be a field and $A_0$ a finite $k$-algebra. There is a unique maximal \'etale $k$-subalgebra $A_0'$ in $A_0$: concretely, in each local factor ring of $A_0$ uniquely lift the separable closure of $k$ in the residue field up into the local factor ring via Hensel's Lemma and the primitive element theorem. Since it is uniquely characterized by lifting separable closures of $k$ in the residue fields of the factor rings, it is a good exercise to check the following crucial thing: if $B_0$ is another finite $k$-algebra then $(A_0 \otimes_k B_0)' = A_0' \otimes_k B_0'$, and $A_0'$ is functorial in $A_0$. Observe that $A_0' \rightarrow A_0$ is faithfully flat since at the level of factor rings of $A_0$ it is an inclusion of a field into a nonzero ring. Also observe that any \'etale $k$-algebra equipped with a map to $A_0$ uniquely factors through $A_0'$. Exercise: The formation of $A_0'$ commutes with any field extension on $k$. (Hint: use Galois descent to reduce to the separate cases of separable algebraic extensions and the easy case $k = k_s$.) In geometric terms, for a finite $k$-scheme $X_0$, the preceding Example constructs a finite \'etale $k$-scheme $X_0'$ and a faithfully flat $k$-map $f_0:X_0 \rightarrow X_0'$ which is initial among all $k$-maps from $X_0$ to finite \'etale $k$-schemes, and its formation is functorial in $X_0$ and commutes with products in $X_0$ and with any extension on $k$. In particular, if $X_0$ is a $k$-group then $X_0'$ has a unique $k$-group structure making $f_0:X_0 \rightarrow X_0'$ a $k$-homomorphism. Example: Now let $R$ be a complete discrete valuation ring with residue field $k$, and let $X$ be a finite flat $R$-scheme. (Can relax the hypothesis on $R$ if familiar with finite \'etale maps in general.) In this setting, "finite 'etale" over $R$ just means "product of finitely many unramified finite extensions". By using Hensel's Lemma in finite local $R$-algebras, to give a map from a finite \'etale $R$-algebra $A$ to a finite $R$-algebra $B$ is the same as to give a map $A_0 \rightarrow B_0$ between their special fibers. In particular, finite \'etale $k$-algebras uniquely and functorially lift to finite \'etale $R$-algebras, and so $X_0'$ uniquely lifts to a finite \'etale $R$-scheme $X'$ and there is a unique $R$-map $f:X \rightarrow X'$ lifting $f_0:X_0 \rightarrow X_0'$. By fibral flatness (using $X$ is $R$-flat!), $f$ is faithfully flat since $f_0$ is. By uniqueness of everything in sight, the formation of $f$ commutes with products and local extension on $R$ and is also functorial in $X$. In particular, if $G$ is a finite flat $R$-group then $G'$ admits a unique $R$-group structure making $f$ an $R$-homomorphism. We call $G'$ the maximal \'etale quotient of $G$. Now we can put it all together and obtain the connected-etale sequence: Proposition: Let $G$ be a finite flat group scheme over a complete discrete valuation ring $R$. (Even ok for complete local noetherian $R$, or even henselian local $R$.) The faithfully flat $R$-homomorphism $f:G \rightarrow G'$ to the maximal \'etale quotient has scheme-theoretic kernel $G^0$. Proof: The kernel $H = \ker f$ is a finite flat $R$-group. To show it contains $G^0$ we have to check that the composite map $G^0 \rightarrow G \rightarrow G'$ vanishes. Being a map from a finite $R$-scheme to a finite \'etale $R$-scheme, the map is determined by what it does on the special fiber, so it suffices to show that $G_k^0 \rightarrow G_0'$ vanishes. This is a map from a finite infinitsimal $k$-scheme to a finite \'etale $k$-scheme which carries the unique $k$-point to the identity point. Thus, it factors through the identity section of $G_0'$, which is open and closed since $G_0'$ is finite etale over $k$. Now that $H$ contains $G^0$, to prove the resulting closed immersion $G^0 \hookrightarrow H$ between finite flat $R$-schemes is an isomorphism it suffices to do so on special fibers. But that reduces us to the variant of our problem over the residue field. We can increase it to be algebraically closed, and so the problem is to show that if $G$ is a finite flat group scheme over an algebraically closed field $k$ then $G \rightarrow G'$ has kernel exactly $G^0$. But $G'$ is a constant $k$-scheme since it is etale and $k$ is algebraically closed, so by construction $G'$ is just the disjoint union of the $k$-points of the connected components of $G$. It is then physically obvious that the kernel is $G^0$. QED Remark: If $X$ is any finite flat $R$-scheme, with $X \rightarrow X'$ the initial map to
a finite \'etale $R$-scheme, then the induced map on $\overline{k}$-points is bijective. Indeed, we can pass to geometric special fibers and connected components to reduce to the case when $X$ is local finite over an algebraically closed field (in place of $R$), in which case the assertion is clear. By this Remark, the geometric points of the $n$-torsion in $E_k$ are identified with the geometric points of the special fiber of the maximal etale quotient $E[n]'$. In particular, if $n$ is not divisible by the characteristic of $K$ and if $K'/K$ is a sufficiently big finite separable extension which splits $E_K[n]$ then the finite etale $R'$-scheme $E[n]'_{R'}$ is constant (as it may be checked on $K'$-fiber), so the map
$$E _K[n] (\overline{K}) = E _K[n] (K') = E[n] (R') \rightarrow E[n]' (k') \hookrightarrow
E _k[n]' (\overline{k}) = E[n]'(R') = E[n]' (\overline{K})$$
is identified with the naive map in question 1. In other words, that step computes the "quotient" part of the connected-etale sequence of $E[n]$ after passing to $\overline{K}$-points! Example: If $E$ has supersingular reduction then $E[p] = E[p]^0$ and the etale part of the sequence for $E[p]$ vanishes. Example: If $E$ has ordinary reduction then working over an algebraic closure of the residue field shows that $E[p]^0$ and $E[p]'$ each have rank $p$ as finite flat $R$-groups. Finally, it remains to relate $E[n]^0$ to $n$-torsion in the so-called "formal group" of $E$ ( not the formal group of $E_K$, which loses contact with the integral structure and for ${\rm{char}} (K) = 0$ is actually the formal additive group which has no nontrivial $n$-torsion!). A moment's reflection on the definition of the formal group in Silverman shows that its $R'$-points for any finite local valuation ring extension of $R$ are precisely the local $R'$-points of the complete local ring $\widehat{\mathcal{O}}_{E,0_k}$ at the origin of the special fiber (or the completion along the identity section, comes to the same since $R$ is complete). By the universal properties of local rings on schemes and completions of local noetherian rings, such $R'$-points of the latter type are simply points in $E(R')$ specializing to $0_k$ in $E_k (k')$. But we saw earlier that $E[n]^0 (R')$ is exactly the set of points in $E[n] (R')$ specializing to $0_k$ on $E_k$. So indeed $E[n]^0 (R')$ inside of $E[n] (R') = E_K[n] (K')$ is exactly the $n$-torsion in the $K'$-points of the "formal group" in the sense of Silverman's book. Voila, so that answers the questions. The arguments used are designed to apply equally well to abelian varieties.
|
{
"source": [
"https://mathoverflow.net/questions/16125",
"https://mathoverflow.net",
"https://mathoverflow.net/users/4005/"
]
}
|
16,141 |
I was looking at Wilson's theorem: If $P$ is a prime then $(P-1)!\equiv -1\pmod P$. I realized this
implies that for primes $P\equiv 3\pmod 4$, that $\left(\frac{P-1}{2}\right)!\equiv \pm1 \pmod P$. Question: For which primes $P$ is $\left(\frac{P-1}{2}\right)!\equiv 1\pmod P$? After convincing myself that it's not a congruence condition for $P,$ I found this sequence in OEIS . I'd appreciate any comments that shed light on the nature of such primes (for example, they appear to be of density 1/2 in all primes that are $3\bmod 4$). Thanks, Jacob
|
I am a newcomer here. If p >3 is congruent to 3 mod 4, there is an answer which involves only $p\pmod 8$ and $h\pmod 4$, where $h$ is the class number of $Q(\sqrt{-p})$ .
Namely one has $(\frac{p-1}{2})!\equiv 1 \pmod p$ if an only if either (i) $p\equiv 3 \pmod 8$ and $h\equiv 1 \pmod 4$ or (ii) $p\equiv 7\pmod 8$ and $h\equiv 3\pmod 4$. The proof may not be original: since $p\equiv 3 \pmod 4$, one has to determine the Legendre symbol $${{(\frac{p-1}{2})!}\overwithdelims (){p}}
=\prod_{x=1}^{(p-1)/2}{x\overwithdelims (){p}}=\prod_{x=1}^{(p-1)/2}(({x\overwithdelims (){p}}-1)+1).$$ It is enough to know this modulo 4 since it is 1 or -1. By developping, one gets $(p+1)/2+S \pmod 4$, where
$$S=\sum_{x=1}^{(p-1)/2}\Bigl({x\over p}\Bigr).$$ By the class number formula, one has $(2-(2/p))h=S$ (I just looked up Borevich-Shafarevich, Number Theory), hence the result, since $\Bigl({2\over p}\Bigr)$ depends only on $p \pmod 8$. Edit: For the correct answer see KConrad's post or Mordell's article.
|
{
"source": [
"https://mathoverflow.net/questions/16141",
"https://mathoverflow.net",
"https://mathoverflow.net/users/4181/"
]
}
|
16,145 |
Let $K$ be a field and $G:=SL_2(K)$, then $G$ is a $K-$split reductive group (to use some big words). These groups are classified by a based root datum $(X,D,X',D')$. Let $G'$ be group associated to $(X',D',X,D)$, the so called dual group.
Is it correct that $G'=PGL_2(K)$? I"m wondering how $PSL_2(K)$ fits into this picture.
I'm aware of the fact that if C is algebraically closed, then $PSL_2(C) \cong PGL_2(C)$ as abstract groups; can this be made into an isomorphism of algebraic groups, i.e. is $PSL$ a $K-$form of $PGL$?
|
Yes, the dual of $SL_2$ is $PGL_2$. But you're not going down the right track with $PSL_2$. The problem with $PSL_2$ is that it's not a variety at all! You can quotient out the variety $SL_2$ by the subgroup $\pm1$ but the quotient is the variety $PGL_2$ (recall that quotients in the category of sheaves (for these are really fppf sheaves) don't have to be surjective on global sections, so the statement that there's a surjection $SL_2\to PGL_2$ does not imply that the induced map $SL_2(\mathbf{Q})\to PGL_2(\mathbf{Q})$ is a surjection). The problem with $PSL_2$ is that it is a functor from, say, rings to groups, but it's not a representable one, so in particular it's not an algebraic group. If you like, you can imagine $PSL_2$ as a presheaf quotient, and $PGL_2$ as the associated (representable) sheaf. Edit: I tried to think of a way to make this observation more enlightening. Let's for example try and build an affine variety over $\mathbf{Q}$ representing the $PSL_2$ functor. Well we can certainly build an affine variety over $\mathbf{Q}$ representing the $SL_2$ functor: it's just $A:=\mathbf{Q}[a,b,c,d]/(ad-bc-1)$. Now let's see what happens if we try and quotient out by the group $\pm1$. The quotient is affine, and is represented by the invariants of the action, that is, the subring of $A$ consisting of polynomials in $a$, $b$, $c$, $d$ with the property that every monomial mentioned in the polynomial has total degree even. Now here's the problem: I can see a $\mathbf{Q}$-point of this subring (that is, a map from this subring to $\mathbf{Q}$) that corresponds to the matrix $(s,0;0,1/s)$ for $s=\sqrt{p}$, $p$ a prime number! It's the point that sends $a^2$ to $p$, $ab$ to $0$, and so on and so on, and finally $d^2$ goes to $1/p$ and $ad$ goes to $1$, $bc$ goes to $0$, so $ad-bc=1$. The map from the subring to $\mathbf{Q}$ doesn't extend to a map from the whole ring to $\mathbf{Q}$, so our putative construction has failed because the $\mathbf{Q}$-points of this ring are a group that canonically but strictly contains $PSL_2(\mathbf{Q})$ (this subgroup being the $\mathbf{Q}$-points that extend to $\mathbf{Q}$-points of $A$).
|
{
"source": [
"https://mathoverflow.net/questions/16145",
"https://mathoverflow.net",
"https://mathoverflow.net/users/3380/"
]
}
|
16,193 |
I've been thinking about the value of writing "of course" in mathematical papers (or its variants such as "obviously" etc.). In particular, my current train of thought is, if something is obvious, then it is obvious that it is obvious (so why include it at all?). The example that inspired this post is: If d divides a and d divides b, then of course d also divides a+b. Are there examples in the mathematical literature where the term "of course" is of value? More precisely, I'm after an example (or a few), ideally by a well-known author, where "of course" or "obviously" or similar actually adds tangible value to a sentence (rather than just implying: (a) it's obvious to me, I'm so smart or (b) I can't actually be bothered working out the details)
|
I don't agree that if something is obvious, then it is obvious that it is obvious. When an author declares in a mathematical exposition that a fact is obvious, or says "of course" or something with a similar meaning, then it is a signal that the reader should be able to find a very easy reason justifying the statement, rather than a complex one. This is useful information for the author to signal, and I for one as a reader have often been grateful for it. The truly aggravating uses of these phrases occur when the author says that something is "obvious" or "clear", but it isn't really. Surely many of us have been in situations reading a paper where the author says "clearly", but after a lot of thought, it still isn't so clear. I believe that these phrases are often used because the author didn't want to work out all the details, a kind of laziness, which can also be a dangerous source of mathematical error. I have seen papers where the author states, "It is obvious that X, and let me explain why..."
|
{
"source": [
"https://mathoverflow.net/questions/16193",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2264/"
]
}
|
16,214 |
Recall that a function $f\colon X\times X \to \mathbb{R}_{\ge 0}$ is a metric if it satisfies: definiteness: $f(x,y) = 0$ iff $x=y$, symmetry: $f(x,y)=f(y,x)$, and the triangle inequality: $f(x,y) \le f(x,z) + f(z,y)$. A function $f\colon X\times X \to X$ is associative if it satisfies: associativity: $f(x,f(y,z)) = f(f(x,y),z)$. If $X=\mathbb{R}_{\ge 0}$, then it might be possible for the same function to be a metric and associative. Is there an associative metric on the non-negative reals? Note that these demands actually make $X$ into a group. The element $0$ is the identity because $f(f(0,x),x) = f(0,f(x,x)) = f(0,0) = 0$ by associativity and definiteness, so again by definiteness $f(0,x) = x$. Every element is its own inverse because $f(x,x)=0$. In fact, the following question is equivalent. Is there an abelian group on the non-negative reals such that the group operation satisfies the triangle inequality? Note also that the answer is yes if $X=\mathbb{N}$, the non-negative numbers! Click here for a spoiler . "Also known as nim-sum." The question is originally due to John H. Conway . To my knowledge, the question is unsolved even for $X = \mathbb{Q}_{\ge 0}$, but he does not seem to care about that case. The spoiler above does extend to the non-negative dyadic rationals $\mathbb{N}[\frac 12]$, but apparently not to $\mathbb{N}[\frac 13]$.
|
Seems that this is possible. Here is a (non-constructive) proof.
Suggestions are welcome. The proof is inspired by Mazurkiewicz's argument . This is second version
of the proof: it includes improvements in
the set-theoretic argument suggested by Joel David Hamkins, and also
hopefully clarifies some issues
raised in comments. Thanks for the comments! Goal: Construct a commutative group structure $\star$ on non-negative
reals
${\mathbb R}_{\ge 0}$ such that $x\star y\le x+y$ and $x\star x=0$. Remark: Note that $0$ is automatically a neutral element, and that such a
commutative group is in fact
a vector space over $ {\mathbb F}_2 $. Also, we automatically have the
triangle inequality:
$$x\star z=x\star y\star y\star z\le x\star y+y\star z.$$ Step 1: Let us order ${\mathbb R}_{\ge 0}$ in order type $c$ (continuum). Equivalently,
we choose a bijection $\iota:[0,c)\to{\mathbb R}_{\ge 0}$, where $[0,c)$ is the set
of ordinals smaller than $c$. Note that for any $ \alpha < c $, we have
$$|\iota([0,\alpha))| < c.$$ We may choose $\iota$ so that $\iota(0)=0$, although
it is not strictly necessary. Plan: For every $\alpha\le c$, we will construct a subset
$S_\alpha\subset {\mathbb R}_{\ge 0}$ and a group operation
$\star:S_\alpha\times S_\alpha\to S_\alpha$. The group operation will have the
required properties: $S_\alpha$ is
a vector space over $F_2$ with $0$ being the neutral element,
and $x\star y\le x+y$. Besides
it will also have the additional property that $S_\alpha$ is generated as
a group by $\iota([0,\alpha))$
(in particular, the image is contained in $S_\alpha$). Moreover, if
$\beta\prec\alpha$, $S_\beta$ is a subgroup
of $S_\alpha$. In particular, we get a group structure with required properties on $S_c={\mathbb R}_{\ge 0}$,
as claimed. Step 2: The construction proceeds by transfinite recursion. The base is
$S_0=\lbrace 0\rbrace$ (generated by the empty set). Step 3. Let us now define $S_\alpha$ assuming that $S_\beta$ is already defined for
$\beta<\alpha$. If $\alpha$ is a limit ordinal, take
$$S_\alpha=\bigcup_{\beta<\alpha}S_\beta.$$
Therefore, let us assume $\alpha=\beta+1$. If $\iota(\alpha)\in S_\beta$, take $S_\alpha=S_\beta$. Step 4. It remains to consider the case when $\alpha=\beta+1$ but $\iota(\alpha)\not\in S_\beta$.
Since $I=\iota([0,\beta))$ generates $S_\beta$, the cardinality of $S_\beta$ is at most the cardinality of the set
of finite subsets of $I$. Therefore, $|S_\beta| < c$. Fix a number $k$ between $0$ and $1$, to be chosen later. Define a
function $f:{\mathbb R}_{\ge 0}\to{\mathbb R}_{\ge 0}$ by
$$f(x)=\cases{\iota(\alpha)+k x, \ x \le \iota(\alpha)\cr x+k \iota(\alpha), \ x > \iota(\alpha)}.$$
Now choose $k$ so that $f(S_\beta)\cap S_\beta=\emptyset$. This is possible because
for every $x,y\in S_\beta$, the equation $f(x)=y$ has at most one solution in
$k$, so the set of prohibited values of $k$ has cardinality at most
$|S_\beta\times S_\beta|$. (We can use $\iota$ to well-order the interval $(0,1)$;
we can then choose $k$ to be the minimal acceptable value, so as to remove arbitrary choice.) Step 5. Now define $S_\alpha=S_\beta\cup f(S_\beta)$ and set $\iota(\alpha)\star x=f(x)$ for
$x\in S_\beta$. The product naturally extends to all of $S_\alpha$:
$$f(x)\star f(y)=x\star y\qquad f(x)\star y=y\star f(x)=f(x\star y).$$
It is not hard to see that it has the required properties. First of all, $S_\alpha$ is an isomorphic image of $S_\beta\times({\mathbb Z}/2{\mathbb Z})$;
this takes care of group-theoretic requirement. It remains to
check two inequalities: Step 5a: $$f(x)\star f(y)\le f(x)+f(y)\quad(x,y\in S_\beta),$$
which is true because $f(x)\ge x$, so
$$f(x)\star f(y)=x\star y\le x+y\le f(x)+f(y).$$ Step 5b: $$f(x)\star y\le f(x)+y\quad(x,y\in S_\beta),$$
which is true because $f$ is increasing and $f(x+t)\le f(x)+t$, so
$$f(x)\star y=f(x\star y)\le f(x+y)\le f(x)+y.$$ That's it.
|
{
"source": [
"https://mathoverflow.net/questions/16214",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1079/"
]
}
|
16,216 |
I am writing an undergraduate thesis on local and global class field theory from a classical (i.e., non-cohomological) approach and am hoping to obtain copies of the early groundbreaking publications in the field. I am primarily interested in finding English translations of articles from Weber, Hasse, Hilbert, Kronecker, and Takagi from 1850 to around 1935 as possible. A fairly comprehensive list of them is found in Hasse's "History of Class Field Theory" in Cassels & Frohlich's ANT, and I can provide a list if needed. Any recommendations for sites that provide translated back issues of Mathematische Annalen and/or Gottinger Nachrichten from these time periods would be of great utility. Edit: Thank you all for responding! You're answers will definitely help me develop the historical portion of my thesis. In response to several comments, I realized when posting that it was probably naive to assume that English translations of all these works were available but presented the question in the above manner for the sake of brevity. @KConrad: I actually found your cfthistory.pdf file while googling the subject early in my research and have benefited from it greatly! Small world! :) I would be fine with library copies of the works, of course, but my university would have to order copies from other institutions. As such, I thought I'd first make sure there wasn't some large repository of translated articles from Hilbert et al. online somewhere. I can't thank you enough for this list of sources, too! I will definitely attempt to get my hands on as many as possible. Thanks again for helping and posting your history of CFT online! @Ben Lenowitz: Thank you for pointing me to the Gottingen database. I haven't been in awhile and not while searching for these articles. I will give it a shot.
|
First, there's no need to focus on online copies, as asked for in the question. We used to have things called libraries which contain journal articles in them. :) Try looking there. More seriously, I think your task is to a large extent hopeless. Most of those works were never translated into English. But there are numerous English language sources which describe some aspect of how class field theory was originally developed and you should start there. Here are some: G. Frei, Heinrich Weber and the Emergence of Class Field Theory, in ``The History of Modern Mathematics, vol. 1: Ideas and their Reception,'' (J. McCleary and D. E. Rowe, ed.) Academic Press, Boston, 1989, 424--450. H. Hasse, ``Class Field Theory,'' Lecture Notes # 11, Dept. Math. Univ. Laval, Quebec, 1973. [This is basically adapted from his paper in Cassels and Frohlich, but has some nuggets that were not in C&F.] K. Iwasawa, On papers of Takagi in Number Theory, in ``Teiji Takagi Collected Papers,'' 2nd ed., Springer-Verlag, Tokyo, 1990, 342--351. S. Iyanaga, ``The Theory of Numbers,'' North-Holland, Amsterdam, 1975. [The end of the book has a nice exposition of how alg. number theory developed up to class field theory.] S. Iyanaga, On the life and works of Teiji Takagi, in ``Teiji Takagi Collected Papers,'' 2nd ed., Springer-Verlag, Tokyo, 1990, 354--371. S. Iyanaga, Travaux de Claude Chevalley sur la th\'eorie du corps de classes:
Introduction, Japan. J. Math. 1 (2006), 25--85. [Are you OK with French?] M. Katsuya, The Establishment of the Takagi--Artin Class Field Theory, in
``The Intersection of History and Mathematics,'' (C. Sasaki, M. Sugiura, J. W. Dauben ed.),
Birkhauser, Boston, 1995, 109--128. T. Masahito, Three Aspects of the Theory of Complex Multiplication, ``The Intersection of History and Mathematics,'' (C. Sasaki, M. Sugiura, J. W. Dauben ed.),
Birkhauser, Boston, 1995, 91--108. K. Miyake, Teiji Takagi, Founder of the Japanese School of Modern Mathematics,
Japan. J. Math. 2 (2007), 151--164. P. Roquette, Class Field Theory in Characteristic $p$, its Origin and Development,
in ``Class Field Theory -- its Centenary and Prospect,'' Math. Soc. Japan, Tokyo, 2001,
549--631. H. Weyl, David Hilbert and His Mathematical Work, Bull. Amer. Math. Soc. {\bf 50} (1944), 612--654. [Hilbert's obituary] I did write up something myself about a year or so ago on the history of class field theory just to put in one place what I was able to cobble together from these kinds of sources. See http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/cfthistory.pdf which contains the above references as the bulk of the bibliography (I did not just type all those articles references above by hand!) The main thing which had baffled me at first was how they originally defined the local norm residue symbol at ramified primes. I give some examples of how this was determined in the original language of central simple algebras.
|
{
"source": [
"https://mathoverflow.net/questions/16216",
"https://mathoverflow.net",
"https://mathoverflow.net/users/4204/"
]
}
|
16,257 |
What I am talking about are reconstruction theorems for commutative scheme and group from category. Let me elaborate a bit. (I am not an expert, if I made mistake, feel free to correct me) Reconstruction of commutative schemes Given a quasi compact and quasi separated commutative scheme $(X,O_{X})$ (actually, quasi compact is not necessary), we can reconstruct the scheme from $Qcoh(X)$ as category of quasi coherent sheaves on $(X,O_{X})$ . This is Gabriel-Rosenberg theorem. Let me sketch the statement of this theorem which led to the question I want to ask: The reconstruction can be taken as geometric realization of $Qcoh(X)$ (abelian category). Let $C_X$ = $Qcoh(X)$ . We define spectrum of $C_X$ and denote it by $Spec(X)$ (For example, if $C_X=R-mod$ .where $R$ is commutative ring, then $Spec(X)$ coincides with prime spectrum $Spec(R)$ ). We can define Zariski topology on $Spec(X)$ , we have open sets respect to Zariski topology. Then we have contravariant pseudo functor from category of Zariski open sets of the spectrum $Spec(X)$ to $Cat$ , $U\rightarrow C_{X}/S_{U}$ ,where $U$ is Zariski open set of $Spec(X)$ and $S_U=\bigcap_{Q\in U}^{ }\hat{Q}$ (Note: $Spec(X)$ is a set of subcategories of $C_{X}$ satisfying some conditions, so here, $Q$ is subcategory belongs to open set $U$ and $\hat{Q}$ =union of all topologizing subcategories of $C_{X}$ which do not contain $Q$ .) For each embedding: $V\rightarrow U$ , we have correspondence localization functor: $C_{X}/S_{U}\rightarrow C_{X}/S_{V}$ . Then we have fibered category over the Zariski topology of $Spec(X)$ , so we have given a geometric realization of $Qcoh(X)$ as a stack of local category which means the fiber(stalk)at each point $Q$ , $colim_{Q\in U} C_{X}/S_{U}$ = $C_{X}/\hat{Q}$ is a local category. Zariski geometric center Define a functor $O_{X}$ : $Open(Spec(X))\rightarrow CRings$ $U|\rightarrow End(Id_{C_{X}/S_{U}})$ It is easy to show that $O_{X}$ is a presheaf of commutative rings on $Spec(X)$ , then Zariski geometric center is defined as $(Spec(X),\hat{O_{X}})$ ,where $\hat{O_{X}}$ is associated sheaf of $O_{X}$ . Theorem : Given X = $(\mathfrak{X},O_{\mathfrak{X}})$ a quasi compact and quasi separated commutative scheme. Then the scheme X is isomorphic to Zariski geometric center of $C_X=Qcoh(X)$ . If we denote the fibered category mentioned above by $\mathfrak{F}_{X}$ . then center of each fiber (stalk) of $\mathfrak{F}_{X}$ recover the presheaf of commutative ring defined above and hence Zariski geometric center. Moreover, Catersion section of this fibered category is equivalent to $Qcoh(X)$ when $X$ is commutative scheme Question It is well known that for a compact group $G$ , we can use Tanaka formalism to reconstruct this group from category of its representation $(Rep(G),\bigotimes _{k},Id)$ . Is this reconstruction Morally the same as Gabriel-Rosenberg pattern reconstruction theorem? From my perspective, I think $Qcoh(X)$ and category of group representations are very similar because $Qcoh(X)$ can be taken as "category of representation of scheme". On the otherhand, group scheme is a scheme compatible with group operations. Therefore, I think it should have united formalism to reconstruct both of them. Then I have following questions: 1 Is there a Tanaka formalism for reconstruction of general scheme $(X,O_{X})$ from $Qcoh(X)$ ? 2.Is there a Gabriel-Rosenberg pattern reconstruction (geometric realization of category of category of representations of group) to recover a group scheme? I think this formalism is natural (because the geometric realization of a category is just a stack of local categories and the center(defined as endomorphism of identical functor)of the fiber of this stack can recover the original scheme) and has good generality (can be extend to more general settings) Maybe one can argue that these two reconstruction theorems live in different nature because reconstruction of group schemes require one to reconstruct the group operations which force one to go to monoidal categories (reconstruct co-algebra structure) while reconstruction of
non-group scheme doesn't (just need to reconstruct algebra structure). However, If we stick to the commutative case. $Qcoh(X)$ has natural symmetric monoidal structure. Then, in this case, I think this argue goes away. More Concern I just look at P.Balmer's paper on reconstruction from derived category , it seems that he also used Gabriel-Rosenberg pattern in triangulated categories: He defined spectrum of triangulated category as direct imitation of prime ideals of commutative ring. Then, he used triangulated version of geometric realization (geometric realization of triangulated category as a stack of local triangulated category) hence, a fibered category arose, then take the center of the fiber at open sets of spectrum of triangulated categories to recove the presheaf (hence sheaf) of commutative rings. This triangulated version of geometric realization is also mentioned in Rosenberg's lecture notes Topics in Noncommutative algebraic geometry,homologial algebra and K-theory page 43-44, it also discuss the relation with Balmer's construction. But in Balmer's consideration, there is tensor structure in his derived category Therefore, the further question is: Is there a triangulated version of Tannaka Formalism which can recover P.Balmer's reconstruction theorem. ? I heard from some experts in derived algebraic geometry that Jacob Lurie developed derived version of Tannaka formalism. I know there are many experts in DAG on this site. I wonder whether one can answer this question. In fact, I still have some concern about Bondal-Orlov reconstruction theorem but I think it seems that I should not ask too many questions at one time.So, I stop here.All the related and unrelated comments are welcome Thanks in advance! Reconstruction theorem in nLab reconstruction theorem EDIT : A nice answer provided by Ben-Zvi for related questions
|
The Tannakian reconstruction for schemes and more generally, geometric stacks from QCoh(X) with its tensor structure (due to Jacob Lurie) is explained in my answer to Tannakian Formalism . One can (and one has) try to extend this to the derived setting, where one ought to get a statement much stronger than Balmer's reconstruction theorem. First of all one needs to replace triangulated categories, which are too coarse to work with effectively, with symmetric monoidal $(\infty,1)$-categories (or symm. monoidal dg categories say in characteristic zero). There's an obvious functor from such objects C to derived stacks Spec C given by Tannakian reconstruction -- it's defined just as in the abelian case (see my answer above). Namely we build Spec C's functor of points, as a functor from derived rings to spaces or simplicial sets: Spec C(R) = the $\infty$-groupoid of tensor functors from C to R-modules.
In the case that C is the infinity-categorical form of QCoh (X) for X a scheme this recovers X, giving a result
refining Balmer's (this was worked out by Nadler, Francis and myself, and many others I think -- in particular Toen and Lurie understood this long ago). I feel strongly this
is a better point of view than trying to define a locally ringed space from a triangulated category, and has more of a chance to extend to stacks. In particular C tautologically sheafifies (as a sheaf of symmetric monoidal $\infty$-categories) over X. Things get much more interesting though in the case that C=Qcoh X for X a stack (with affine diagonal, or everything fails immediately) --
eg for C=Reps(G)=QCoh (BG), the setting of usual Tannakian formalism.
There one quickly learns that the above naive picture is false. Namely take G to be just the multiplicative group $G_m$. Then there are a lot of interesting fiber functors
on the derived Rep G = complexes of graded vector spaces that are not given by
points of BG_m (ie are not the usual "forgetful" fiber functor) --- namely we can send the defining one-dim rep of $G_m$ not to the one-dim vector space in degree zero,
but in any degree we want (ie apply shift) - since this is a generator it determines the rest of the fiber functor. This proves that a naive Tannakian theorem fails in the derived setting, without imposing some "connectivity" hypotheses (ie without having t-structures around). I believe a student of Toen's is writing a thesis on this subject, but I don't know the precise statements (and so won't "out" his results). Of course it's a little disappointing to need BOTH a tensor structure and a t-structure, since morally either one should be enough for reconstruction (having a t-structure means we can recover the abelian category of quasicoherent sheaves, which by Rosenberg recovers X already in the case of schemes without need for tensor structure). But for general stacks I don't know of a better answer.
|
{
"source": [
"https://mathoverflow.net/questions/16257",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1851/"
]
}
|
16,312 |
So, I can understand how non-standard analysis is better than standard analysis in that some proofs become simplified, and infinitesimals are somehow more intuitive to grasp than epsilon-delta arguments (both these points are debatable). However, although many theorems have been proven by non-standard analysis and transferred via the transfer principle , as far as I know all of these results were already known to be true. So, my question is: Is there an example of a result that was first proved using non-standard analysis? To wit, is non-standard analysis actually useful for proving new theorems? Edit : Due to overwhelming support of François' comment , I've changed the title of the question accordingly.
|
From the Wikipedia article : the list of new applications in
mathematics is still very small. One
of these results is the theorem proven
by Abraham Robinson and Allen
Bernstein that every polynomially
compact linear operator on a Hilbert
space has an invariant subspace. Upon
reading a preprint of the
Bernstein-Robinson paper, Paul Halmos
reinterpreted their proof using
standard techniques . Both papers
appeared back-to-back in the same
issue of the Pacific Journal of
Mathematics . Some of the ideas used in
Halmos' proof reappeared many years
later in Halmos' own work on
quasi-triangular operators.
|
{
"source": [
"https://mathoverflow.net/questions/16312",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2233/"
]
}
|
16,393 |
Is there any result about the time complexity of finding a cycle of fixed length $k$ in a general graph?
All I know is that Alon, Yuster and Zwick use a technique called "color-coding",
which has a running time of $O(M(n))$, where $n$ is the number of vertices of the input graph and $M(n)$ is the time required to multiply two $n \times n$
matrices. Is there any better result?
|
Finding a cycle of any even length can be found in $O(n^2)$ time, which is less than any known bound on $O(M(n))$. For example, a cycle of length four can be found in $O(n^2)$ time via the following simple procedure: Assume the vertex set is $\{1,...,n\}$. Prepare an $n$ x $n$ matrix $A$ which is initially all zeroes. For all vertices $i$ and all pairs of vertices $j, k$ which are neighbors to $i$, check $A[j,k]$ for a $1$. If it has a $1$, output four-cycle , otherwise set $A[j,k]$ to be $1$. When this loop finishes, output no four-cycle . The above algorithm runs in at most $O(n^2)$ time, since for every triple $i,j,k$ we either flip a $0$ to $1$ in $A$, or we stop. (We assume the graph is in adjacency list representation, so it is easy to select pairs of neighbors of a vertex.) The general case is treated by Raphy Yuster and Uri Zwick in the paper: Raphael Yuster, Uri Zwick: Finding Even Cycles Even Faster. SIAM J. Discrete Math. 10(2): 209-222 (1997) As for finding cycles of odd length, it's just as David Eppstein says: nothing better is known than $O(M(n))$, including the case where $k=3$. However, if you wished to detect paths of length $k$ instead of cycles, you can indeed get $O(m+n)$ time, where $m$ is the number of edges. I am not sure if the original color-coding paper can provide this time bound, but I do know that the following paper by some random self-citing nerd gets it: Ryan Williams: Finding paths of length k in O*(2^k) time. Inf. Process. Lett. 109(6): 315-318 (2009)
|
{
"source": [
"https://mathoverflow.net/questions/16393",
"https://mathoverflow.net",
"https://mathoverflow.net/users/4248/"
]
}
|
16,416 |
I'm looking for a good book in commutative algebra, so I ask here for some advice. My ideal book should be: -More comprehensive than Atiyah-MacDonald -More readable than Matsumura (maybe better organized?) -Less thick than Eisenbud, and more to the point To put this in context, I'm an algebraic geometer, so I know enough commutative algebra, but I didn't study it systematically (apart from a first course on A-M which I followed as an undergraduate); rather I learned the things I needed from time to time. So I would like to give me an occasion to get a better grasp on the subject. EDIT: I will be more specific about the level. As I said I already had a course on Atiyah-MacDonald, and I know that material well, so I'm not interested in books of a comparable level. But I'm not completely familiar with Cohen-Macaulay rings and the relationship between regular sequences and the Koszul complex for example. And I know very little of Gorenstein rings and duality. So I'm looking for something a little bit more sophisticated than what has been already proposed. Yes, I know Eisenbud does these things but it's easy to get lost in that book. Something more to the point would be nice.
|
For a reference on Cohen-Macaulay and Gorenstein rings, you can try "Cohen-Macaulay rings" by Bruns-Herzog. Also, Huneke's lecture note "Hyman Bass and Ubiquity: Gorenstein Rings" is a great introduction to Gorenstein rings, very easy to read and to the point, I highly recommend it. EDIT: Since this question is already bumped up, I will take this opportunity to make a longer list. There are of course some classic references which are still very useful (I find myself having to look in them quite often despite the new sources available): Bourbaki, EGA IV, Serre's "Local Algebras" (very nice read and culminated in the beautiful Serre intersection formula ). There has been some work done in commutative algebra since the 60s, so here is a more up-to-date list of reference for some currently active topics (Disclaimer: I am not an expert in any of these, the list was formed by randomly looking at my bookself, and put in alphabetical order (-:). This is community-wiki, so feel free to add or edit or suggest things you found missing. Cohen-Macaulay modules, from a representation theory perspective: Yoshino is excellent. Another one is being written . Combinatorial commutative algebra: Miller-Sturmfels . Free resolutions (over non-regular rings): Avramov lecture note Geometry of syzygies: Eisenbud , shorter but free version here . Homological conjectures: Hochster , Roberts (more connections to intersection theory), Hochster notes . Integral closures: Huneke-Swanson , which is available free at the link. Intersection theory done in a purely algebraic way: Flenner-O'Carrol-Vogel (for a very interesting story about this, see Eisenbud beautiful reminiscences , especially page 4) Local Cohomology: Brodmann-Sharp , Huneke's lecture note (very easy to read), 24 hours of local cohomology (I have been told that this one was a pain to write, which is probably a good sign). Tight closure and characteristic $p$ method: Huneke , Karen Smith's lecture note (more geometric, number 24 here ), and of course many well-written introductions available on Hochster website .
|
{
"source": [
"https://mathoverflow.net/questions/16416",
"https://mathoverflow.net",
"https://mathoverflow.net/users/828/"
]
}
|
16,460 |
Two things today motivated this question. First, the professor said that in a lecture Thurston mentioned Any manifold can be seen as the configuration space of some physical system. Clearly we need to be careful here, so the first question is 1) What is a precise formulation and an argument to see why the previous statement is true. Second, the professor went on to say that because of the Poisson Bracket, we see the phase space of a physical system as the Cotangent Bundle of a manifold. I understand that we associate a symplectic form to the cotangent bundle, and that we want to think of Phase Space with a symplectic structure, but my second question is 2) Could you provide an example of a physical system, give the associated "configuration manifold" show the cotangent space, and explain why this is the phase space of the system. I pushed the lecturer quite a bit to get this level of detail, so pushing much further would of probably been considered rude. I should also mention he is speaking about these things because we want to quantize the geometry associated to this manifold. So he looks at the cotangent space which apparently has a symplectic structure, and sends the symplectic form to the Lie Bracket. If any of the things I have said are incorrect, please comment with corrections. I am just learning this material and trying to understand how it fits with my current understanding. One last bonus question(tee hee), If your manifold is a Lie Group, we get a Lie algebra structure on the Tangent Space. Is there a relationship between this Lie Algebra structure and the one you would get by considering the cotangent space and then quantizing in the fashion of above? Thanks in advance!
|
Let's start by answering the first question. Let $M$ be any manifold. Consider a physical system consisting of a point-particle moving on $M$. What are the configurations of this physical system? The points of $M$. Hence $M$ is the configuration space. Typically one takes $M$ to be riemannian and we may add a potential function on $M$ in order to define the dynamics. (More complicated dynamics are certainly possible -- this is just the simplest example.) As an example, let's consider a point particle of mass $m$ moving in $\mathbb{R}^3$ under the influence of a central potential
$$V= k/r,$$
where $r$ is the distance from the origin. The configuration space is $M = \mathbb{R}^3\setminus\lbrace 0\rbrace$. Classical trajectories are curves $x(t)$ in $M$ which satisfy Newton's equation
$$m \frac{d^2 x}{dt^2} = \frac{k}{|x|^2}.$$
To write this equation as a first order equation we introduce the velocity $v(t) = \frac{dx}{dt}$. Geometrically $v$ is a vector field (a section of the tangent bundle $TM$) and hence the classical trajectory $(x(t),v(t))$ defines a curve in $TM$ satisfying a first order ODE:
$$\frac{d}{dt}(x(t),v(t)) = (v(t), \frac{k}{m|x(t)|^2})$$
This equation can be derived from a variational problem associated to a lagrangian function $L: TM \to \mathbb{R}$ given by
$$L(x,v) = \frac12 m v^2 - \frac{k}{|x|}.$$
The fibre derivative of the lagrangian function defines a bundle morphism $TM \to T^*M$:
$$(x,v) \mapsto (x,p)$$
where
$$p(x,v) = \frac{\partial L}{\partial v}.$$ In this example, $p = mv$. The Legendre transform of the lagrangian function $L$ gives a hamiltonian function $H$ on $T^*M$, which in this example is the total energy of the system:
$$H(x,p) = \frac{1}{2m}p^2 + \frac{k}{|x|}.$$ The equations of motion can be recovered as the flow along the hamiltonian vector field associated to $H$ via the standard Poisson brackets in $T^*M$: $$ \frac{dx}{dt} = \lbrace x,H \rbrace \qquad\mathrm{and}\qquad \frac{dp}{dt} = \lbrace p,H \rbrace.$$ Being integral curves of a vector field, there is a unique classical trajectory through any given point in $T^*M$, hence $T^*M$ is a phase space for the system; that is, a space of states of the physical system. Of course $TM$ is also a space of states, but historically one calls $T^*M$ the phase space of the system with configuration space $M$. (I don't know the history well enough to know why. There are brackets in $TM$ as well and one could equally well work there.) Not every space of states is a cotangent bundle, of course. One can obtain examples by hamiltonian reduction from cotangent bundles by symmetries which are induced from diffeomorphisms of the configuration space, for instance. Or you could consider systems whose physical trajectories satisfy an ODE of order higher than 2, in which case the cotangent bundle is not the space of states, since you need to know more than just the position and the velocity at a point in order to determine the physical trajectory. It's late here, so I'll forego answering the bonus question for now.
|
{
"source": [
"https://mathoverflow.net/questions/16460",
"https://mathoverflow.net",
"https://mathoverflow.net/users/348/"
]
}
|
16,532 |
This question is about an issue left unresolved by Chad
Groft's excellent
question and John Stillwell's excellent
answer of
it. Since I find the possibility of an affirmative answer
so tantalizing, I would like to pursue it further here. For background, Rice's Theorem asserts essentially that no nontrivial question about computably enumerable sets is decidable. If W e is the set enumerated by program e, then the theorem states: Rice's Theorem. If A is a collection of computably
enumerable sets and { e |
W e ∈ A } is decidable, then either A is
empty or A contains all computably enumerable sets. In short, one can decide
essentially nothing about a program e, if the answer is to
depend only on what the program computes rather than how it
computes it. The question here is about the extent to which a similar
phenomenon holds for finitely presented groups, using the
analogy between programs and finite group presentations: a program e is like a finite group presentation p the set W e enumerated by e is like the group
⟨p⟩ presented by p. According to this analogy, the analogue of Rice's theorem
would state that any decidable collection of finitely
presented groups (closed under isomorphism) should be either trivial or everything.
John Stillwell pointed out in answer to Chad Groft's
question that this is not true, because from a presentation
p we can easily find a presentation of the abelianization of
⟨p⟩, by insisting that all generators commute,
and many nontrivial questions are decidable about finitely
presented abelian groups. Indeed, since the theory of
abelian groups is a decidable theory, there will be many
interesting questions about finitely presented abelian
groups that are decidable from their presentations. My question is whether this is the only obstacle. Question. Does Rice's theorem hold for finitely
presented groups modulo abelianization? In other words, if
A is a set of finitely presented groups (closed under
isomorphism) and the corresponding set of presentations { p | ⟨p⟩ ∈ A } is
decidable, then does A completely reduce to a question
about the abelianizations of the groups, in the sense that there is a set B of abelian groups such that G ∈ A iff Ab(G) ∈ B? Of course, in this case B consists exactly of the abelian groups in A. The question is equivalently asking whether A respects the equivalence of groups having isomorphic abelianizations. In other words, must it be that G ∈ A iff Ab(G) ∈ A?
The question is asking whether every decidable set of finitely presented groups amounts actually
to a decidable set of abelian groups, extended to all
finitely presented groups just by saturating with respect to abelianization. In particular, the set A should contain either none or all perfect groups. An affirmative answer would seem to provide a thorough
explanation of the pervasive undecidability phenomenon in
group presentations. But perhaps this may simply be too much to hope for... In any event, I suppose that there is an equivalence relation on finite group presentations, saying that p ≡ q just in case ⟨p⟩ and ⟨q⟩ have the same answer with repsect to any decidable question about finitely presented groups. The question above asks whether this equivalence relation is just Ab(⟨p⟩) = Ab(⟨q⟩). If this turns out not to be true, then what can be said about ≡?
|
The question "Is there a nonzero homomorphism from your group to $A_5$?" is decidable. (Just write down all ways of sending the generators of your group to $A_5$, and see whether they satisfy the required relations.) The same is true with $A_5$ replaced by any finite group. I don't see how to reduce this to questions about the abelianization.
|
{
"source": [
"https://mathoverflow.net/questions/16532",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1946/"
]
}
|
16,578 |
What the title said. In a slightly more leisurely fashion:- Let $X$ be a compact, connected subset of $\mathbb{R}^2$ with more than one point, and let $x\in X$. Can $X\smallsetminus\{x\}$ be totally disconnected? Note that the Knaster-Kuratowski fan shows that, in the absence of the compactness hypothesis, the answer can be 'yes'. To give credit where it's due, this question was inspired by one that I was asked by Barry Simon.
|
Being planar has nothing to do with the problem. Suppose a totally disconnects $X$ and choose $b$ different from $a$. By passing to a sub continuum, assume that no proper sub continuum contains both $a$ and $b$. Take non empty disjoint open sets $U$ and $V$ whose union is $ X\sim a$. WLOG $b$ is in $U$, and observe that $U\cup \{a\}$ is closed and connected.
|
{
"source": [
"https://mathoverflow.net/questions/16578",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1463/"
]
}
|
16,600 |
Why do elliptic curves have bad reduction at some point if they are defined over Q, but not necessarily over arbitrary number fields?
|
Here are two answers: (a) If you try to write down an elliptic curve $y^2 = x^3 + a x + b$ with everywhere good reduction, you need to choose $a$ and $b$ such that $4a^3 + 27 b^2 = $ a unit. We can certainly solve this equation over some (lots!) of number fields, say if we set the unit equal to $1$ or $-1$, or a unit in some fixed base number field. But we can't solve it in ${\mathbb Q}$. [Edit: As Bjorn intimates in his comment below, one has to be a little more careful than I am being here to be sure of good reduction mod primes above 2; the details are left to the interested reader (or, I imagine, can be found in Silverman in the section where he discusses the proof that there are no good reduction elliptic curves over $\mathbb Q$).] (b) There are many non-trivial everywhere unramified extensions of number fields (e.g. $\mathbb Q(\sqrt{-5}, i)$ over $\mathbb Q(\sqrt{-5})$), but there are no everywhere unramified extensions of the particular number field $\mathbb Q$. The situation with elliptic curves is completely analogous.
|
{
"source": [
"https://mathoverflow.net/questions/16600",
"https://mathoverflow.net",
"https://mathoverflow.net/users/5730/"
]
}
|
16,632 |
An interesting thing happened the other day. I was computing the Stiefel-Whitney numbers for $\mathbb{C}P^2$ connect sum $\mathbb{C}P^2$ to show that it was a boundary of another manifold. Of course, one can calculate the signature, check that it is non-zero and conclude that it can't be the boundary of an oriented manifold. I decided it might be interesting to calculate the first and only Pontrjagin number to check that it doesn't vanish. I believe Hirzebruch's Signature Theorem can be used to show that it is 6, but I was interested in relating the Stiefel-Whitney classes to the Pontrjagin classes. I believe one relation is $p_i (\mathrm{mod} 2) \equiv w_{2i}^2$ (pg. 181 Milnor-Stasheff) So I went ahead and did a silly thing. I took my first Chern classes of the original connect sum pieces say 3a and 3b, used the fact that the inclusion should restrict my 2nd second "Stiefel-Whitney Class" (scare quotes because we haven't reduced mod 2) on each piece to these two to get $w_2(connect sum)=(3\bar{a},3\bar{b})$. I can use the intersection form to square this and get $3\bar{a}^2+3\bar{b}^2=6c$ since the top dimensional elements in a connect sum are identified. Evaluating this against the fundamental class gives us exactly the first Pontrjagin number! This is false. Of course this is wrong because it should be 9+9=18 as pointed out below. This does away with my supposed miracle example. My Apologies! This brings me to a broader question, namely of defining Stiefel-Whitney Classes over the integers. This was hinted at in Ilya Grigoriev's response to Solbap's question when he says On thing that confuses me: why are the pullbacks of the integer cohomology of the real Grassmanian never called characteristic classes? Of course the natural reason to restrict to $\mathbb{Z}/2$ coefficients is to get around orientability concerns. But it seems like if we restrict our orientation to orientable bundles we could use a construction analogous to those of the Chern classes where Milnor-Stasheff inductively declare the top class to be the Euler class, then look at the orthogonal complement bundle to the total space minus its zero section and continue. I suppose the induction might break down because the complex structure is being used, but I don't see where explicitly. If someone could tell me where the complex structure is being used directly, I'd appreciate it. Note the Euler class on odd dimensional fibers will be 2-torsion so this might produce interesting behavior in this proposed S-W class extension. Another way of extending Stiefel-Whitney classes would be to use Steenrod squares. Bredon does use Steenrod powers with coefficient groups other than $\mathbb{Z}/2$ (generally $\mathbb{Z}/p$ $p\neq 2$), but this creates awkward constraints on the cohomology groups. Is this an obstruction to extending it to $\mathbb{Z}$ coefficients? It would be interesting to see what these two proposed extensions of S-W classes do and how they are related.
|
I'm grateful to Allen Hatcher, who pointed out that this answer was incorrect. My apologies to readers and upvoters. I thought it more helpful to correct it than delete outright, but read critically. If $X$ and $Y$ are cell complexes, finite in each degree, and two maps $f_0$ and $f_1\colon X\to Y$ induce the same map on cohomology with coefficients in $\mathbb{Q}$ and in $\mathbb{Z}/(p^l)$ for all primes $p$ and natural numbers $l$, then they induce the same map on cohomology with $\mathbb{Z}$ coefficients. To see this, write $H^n(Y;\mathbb{Z})$ as a direct sum of $\mathbb{Z}^{r}$ and various primary summands $\mathbb{Z}/(p^k)$, and note that the summand $\mathbb{Z}/(p^k)$ restricts injectively to the mod $p^l$ cohomology when $l\geq k$. One can take only those $p^l$ such that there is $p^l$-torsion in $H^\ast(Y;\mathbb{Z})$. (I previously claimed that one could take $l=1$, which on reflection is pretty implausible, and is indeed wrong.) We can try to apply this to $Y=BG$, for $G$ a compact Lie group. For example, $H^{\ast}(BU(n))$ is torsion-free (and Chern classes generate the integer cohomology), and so rational characteristic classes suffice. In $H^{\ast}(BO(n))$ and $H^{\ast}(BSO(n))$ there's only 2-primary torsion. That leaves the possibility that the mod 4 cohomology contains sharper information than the mod 2 cohomology. It does not, because, as Allen Hatcher has pointed out in this recent answer ,
all the torsion is actually 2-torsion. It's sometimes worthwhile to consider the integral Stiefel-Whitney classes $W_{i+1}=\beta_2(w_i)\in H^{i+1}(X;\mathbb{Z})$, the Bockstein images of the usual ones. These classes are 2-torsion, and measure the obstruction to lifting $w_i$ to an integer class. For instance, an oriented vector bundle has a $\mathrm{Spin}^c$-structure iff $W_3=0$. [I'm sceptical of your example in $2\mathbb{CP}^2$. So far as I can see, $3a+3b$ squares to 18, not 6, and indeed, $p_1$ is not a square.]
|
{
"source": [
"https://mathoverflow.net/questions/16632",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1622/"
]
}
|
16,651 |
I was reading a series of article from the Corvallis volume. There are couple of questions which came to my mind: Why do we need to consider representation of Weil-Deligne group? That is what is an example of irreducible admissible representation of $ Gl(n,F)$ which does not correspond to a representation of $W_F$ of dimension $n$ ? An example for $ n=2 $ will be of great help. In the setting of global Langlands conjecture, why extension of $W_F$ by $G_a$ or products of $W'_{F_v}$ does not work? Thank you.
|
Regarding (1), from the point of view of Galois representations, the point is that continuous Weil group representations on a complex vector space, by their nature,
have finite image on inertia. On the other hand, while a continuous $\ell$-adic Galois representation of $G_{\mathbb Q_p}$ (with $\ell \neq p$ of course) must have finite image on wild inertia, it can have infinite image on tame inertia. The formalism
of Weil--Deligne representations extracts out this possibly infinite image, and encodes it as a nilpotent operator (something that is algebraic, and doesn't refer to the $\ell$-adic topology,
and hence has a chance to be independent of $\ell$). As for (2): Representations of the Weil group are essentially the same thing as representations
of $G_{\mathbb Q}$ which, when restricted to some open subgroup, become abelian. Thus
(as one example) if $E$ is an elliptic curve over $\mathbb Q$ that is not CM, its $\ell$-adic Tate module cannot be explained by a representation of the Weil group (or any simple modification thereof). Thus neither can the weight 2 modular form to which it corresponds. In summary: the difference between the global and local situations is that an $\ell$-adic representation of $G_{\mathbb Q_p}$ (or of $G_E$ for any $p$-adic local field) becomes, after
a finite base-change to kill off the action of wild inertia, a tamely ramified representation,
which can then be described by two matrices, the image of a lift of Frobenius and the image of a generator of tame inertia, satisfying a simple commutation relation. On the other hand, global Galois representations arising from $\ell$-adic cohomology of varieties over number fields are much more profoundly non-abelian. Added: Let me also address the question about a product of $W_{F_v}'$. Again, it is simplest to think in terms of Galois representations (which roughly correspond to motives,
which, one hopes, roughly correspond to automorphic forms). So one can reinterpret the question as asking: is giving a representation of $G_F$ (for a number field $F$) the same as giving representations of each $G_{F_v}$ (as $v$ ranges over the places of $F$). Certainly, by Cebotarev, the restriction of the global representation
to the local Galois groups will determine it; but it will overdetermine it; so giving a collection of local representations, it is unlikely that they will combine into a global one. ($G_F$ is very far from being the free product of the $G_{F_v}$, as Cebotarev shows.) To say something on the automorphic side, imagine writing down a random degree 2 Euler product. You can match this with a formal $q$-expansion, which will be a Hecke eigenform, by taking Mellin transforms,
and with a representation of $GL_2(\mathbb A_F)$, by writing down a corresponding tensor product of unramified representations of the various $G_{F_v}$. But what chance is there
that this object is an automorphic representation? What chance is there that your random formal Hecke eigenform is actually a modular form? What chance is there that your random Euler product is actually an automorphic $L$-function? Basically none. You have left out some vital global glue, the same glue which describes the interrelations of all the $G_{F_v}$ inside $G_F$. Teasing out the nature of this glue is at the heart of proving the conjectured relationship between automorphic forms and motives; its mysterious nature is what makes the theories of automorphic forms, and of Galois representations, so challenging.
|
{
"source": [
"https://mathoverflow.net/questions/16651",
"https://mathoverflow.net",
"https://mathoverflow.net/users/4291/"
]
}
|
16,666 |
How many field automorphisms does $\mathbf{C}$ have? If you assume the axiom of choice, there are tons of them -- $2^{2^{\aleph_0}}$ , I believe. And what if you don't -- how essential is the axiom of choice to constructing "wild" automorphisms of $\mathbf{C}$ ? Specifically, if you assume that $\mathsf{ZF}$ admits a model, does that imply that $\mathsf{ZF}$ admits a model where $\mathbf{C}$ has no wild automorphisms: $\operatorname{Aut}(\mathbf{C})=\mathbf{Z}/2\mathbf{Z}$ ? I suppose if that's true, then the next logical question is to construct models of $\mathsf{ZF}$ where $\operatorname{Aut}(\mathbf{C})$ has cardinality strictly between $2$ and $2^{2^{\aleph_0}}$ --pretty disturbing if you ask me. Which finite groups can you hit?
|
The use of inaccessible cardinals is not necessary here, the Baire property works just as well as Lebesgue measure. Shelah ( Can you take Solovay's inaccessible away , Isr. J. Math. 48, 1984, 1-47) shows that $\mathsf{ZF}$ + $\mathsf{DC}$ + "every subset of $\mathbf{R}$ has the Baire property " is relatively consistent with $\mathsf{ZF}$ . (This is also the paper where Shelah also shows that the inaccessible cardinal is necessary for Solovay's result.) The connection is an old theorem of Banach and Pettis which says that any Baire measurable homomorphism between Polish groups is automatically continuous. This result is provable in $\mathsf{ZF}$ + $\mathsf{DC}$ . Since $\mathbf{C}$ is a Polish group under addition, it follows that every additive endomorphism of $\mathbf{C}$ is continuous in Shelah's model. Since the continuous additive endomorphisms of $\mathbf{C}$ are precisely the $\mathbf{R}$ -vector space endomorphisms, it follows that the only field automorphisms of $\mathbf{C}$ in Shelah's model are the identity and conjugation. As pointed out by Pete Clark in the comments, the Artin-Schreier Theorem goes through using only the Boolean Prime Ideal Theorem ( $\mathsf{PIT}$ ), which is significantly weaker than full $\mathsf{AC}$ . This shows that $\mathsf{AC}$ is not completely necessary to show that there is a unique conjugacy class of elements of order $2$ in $\mathrm{Aut}(\mathbf{C})$ and that these correspond precisely to the finite subgroups of $\mathrm{Aut}(\mathbf{C})$ . Looking at Pete Clark's Field Theory Notes , specifically at Steps 4 and 5 of his proof of the Grand Artin-Schreier Theorem on pages 62-63, I think that it is a theorem of $\mathsf{ZF}$ that the only possible order for a nontrivial finite subgroup of $\mathrm{Aut}(\mathbf{C})$ is $2$ .
|
{
"source": [
"https://mathoverflow.net/questions/16666",
"https://mathoverflow.net",
"https://mathoverflow.net/users/271/"
]
}
|
16,668 |
I wonder how strong the power of Tannaka philosophy is, and if we accept that a tensor category is a generalized bialgebra, what difficulties we will come up against ? Edit: Whether most tensor categories are representable, or whether for every "good enough" tensor category there exist a bialgebra with its module category isomorphic to the given category?
|
I'd like to explain Bruce's answer a bit more. The fusion categories Bruce mentioned have non-integer Frobenius-Perron dimensions, so it is very easy to see that they are not categories of finite dimensional modules over a bialgebra. E.g. one of the simplest of them, the so called Yang-Lie category, has simple objects $1,X$ with $X^2=X+1$. So if $X$ were a finite dimensional representation of a bialgebra, then the dimension of $X$ would be the golden ratio, which is absurd. This, however, can be fixed if we allow weak bialgebras and weak Hopf algebras. In fact, any fusion category is the category of modules over a finite dimensional weak Hopf algebra, see arXiv.math/0203060. As to Akhil's example (Deligne's categories), it is also true that they cannot be realized as categories of finite dimensional representations of a bialgebra (or even a weak bialgebra), but for a different reason. Namely, if X is a finite dimensional representation of a bialgebra, then the length of the object $X^{\otimes n}$ is at most ${\rm dim}(X)^n$, where ${\rm dim}$ means the vector space dimension. But in Deligne's categories, the length of $X^{\otimes n}$ grows faster as $n\to \infty$. Actually, in another paper, Deligne shows that if in a symmetric rigid tensor category over an algebraically closed field of characteristic zero, the length of $X^{\otimes n}$ grows at most exponentially, then this is the category of representations of a proalgebraic supergroup, where some fixed central order 2 element acts by parity
(so essentially this IS the category of (co)modules over a bialgebra). This is, however, violently false in characteristic $p$, since if the root of unity $q$ is of order $p$, where $p$ is a prime, the the fusion categories for $U_q({\mathfrak g})$ mentioned by Bruce admit good reduction to characteristic $p$, which are semisimple symmetric rigid tensor categories with finitely many simple objects and non-integer Frobenius-Perron dimensions. A third very simple example of a tensor category not coming from a bialgebra is the category of vector spaces graded by a finite group $G$ with associator defined by a nontrivial $3$-cocycle. This category, however, is the category of representatins of a quasibialgebra (and also of a weak bialgebra, as mentioned above). So the conclusion is as in the previous two answers: tensor categories are more general than bialgebras. More precisely, the existence of a bialgebra for a tensor category is equivalent to the existence of a fiber functor to vector spaces, which is an additional structure that does not always exist. And if it exists, it is often not unique, so you may have many different bialgebras giving rise to the same tensor category.
|
{
"source": [
"https://mathoverflow.net/questions/16668",
"https://mathoverflow.net",
"https://mathoverflow.net/users/4155/"
]
}
|
16,829 |
Related: question #879, Most interesting mathematics mistake . But the intent of this question is more pedagogical. In many branches of mathematics, it seems to me that a good counterexample can be worth just as much as a good theorem or lemma. The only branch where I think this is explicitly recognized in the literature is topology, where for example Munkres is careful to point out and discuss his favorite counterexamples in his book, and Counterexamples in Topology is quite famous. The art of coming up with counterexamples, especially minimal counterexamples, is in my mind an important one to cultivate, and perhaps it is not emphasized enough these days. So: what are your favorite examples of counterexamples that really illuminate something about some aspect of a subject? Bonus points if the counterexample is minimal in some sense, bonus points if you can make this sense rigorous, and extra bonus points if the counterexample was important enough to impact yours or someone else's research, especially if it was simple enough to present in an undergraduate textbook. As usual, please limit yourself to one counterexample per answer.
|
A polynomial $p(x) \in \mathbb{Z}[x]$ is irreducible if it is irreducible $\bmod l$ for some prime $l$. This is an important and useful enough sufficient criterion for irreducibility that one might wonder whether it is necessary: in other words, if $p(x)$ is irreducible, is it necessarily irreducible $\bmod l$ for some prime $l$? The answer is no. For example, the polynomial $p(x) = x^4 + 16$ is irreducible in $\mathbb{Z}[x]$, but reducible $\bmod l$ for every prime $l$. This is because for every odd prime $l$, one of $2, -2, -1$ is a quadratic residue. In the first case, $p(x) = (x^2 + 2 \sqrt{2} x + 4)(x^2 - 2 \sqrt{2} x + 4)$. In the second case, $p(x) = (x^2 + 2 \sqrt{-2} x - 4)(x^2 - 2 \sqrt{-2} x - 4)$. In the third case, $p(x) = (x^2 + 4i)(x^2 - 4i)$. This result can be thought of as a failure of a local-global principle, and the counterexample is minimal in the sense that the answer is yes for quadratic and cubic polynomials.
|
{
"source": [
"https://mathoverflow.net/questions/16829",
"https://mathoverflow.net",
"https://mathoverflow.net/users/290/"
]
}
|
16,848 |
Each orientable 3-manifold can be obtained by doing surgery along a framed link in the 3-sphere. Kirby's theorem says that two framed links give homeomorphic manifolds if and only if they are obtained from one another by a sequence of isotopies and Kirby moves. The original proof by R. Kirby (Inv Math 45, 35-56) uses Morse theory on 5-manifolds and is quite involved. There are two simpler proofs that use Wajnryb's presentations of mapping class groups. One is due to N. Lu (Transactions AMS 331, 143-156) and the other to S. Matveev and M. Polyak (Comm Math Phys 160, 537-556). I would like to ask what other proofs of Kirby's theorem are known. In particular, is there a proof that uses only Morse theory/handle decompositions of 3-manifolds?
|
There is Bob Craggs' 1974 proof, which was never published. It relies on Wall's result, that any two 2-handle cobordisms between S^3 and itself are stably homeomorphic if the associated bilinear forms have the same signature and type. It's very much what you are after. I have a hard-copy in my office. I do not fully follow all aspects of the argument and therefore cannot yet vouch for it. I uploaded a scan of his preprint HERE (thanks Ryan for help combining PDF files!). I'm not positive which proof (Cerf theory or MCG) is really "simpler", because both proofs rely on a lot of background "black boxes". For Kirby's proof, later simplified by Fenn and Rourke, and then later by Justin Roberts in by Kirby calculus in manifolds with boundary , the only black box is Cerf's theorem. Surely the proof of Cerf's theorem could be tremendously simplified, and after this is eventually done by somebody, my money would be on the Cerf theory proof to be slicker. The MCG proof uses presentations of the mapping class group, which use simple connectedness of the Hatcher-Thurston complex (itself not so easy to prove), and results on buildings (a result of Brown) in order to construct presentations of the group from its action on a simply connected simplicial complex. This is actually a lot of machinery, if you think about it. Again, you can simplify the proof by using Gervais's presentation from A finite presentation of the mapping class group of an oriented surface , proven directly by Silvia Benvenuti ( Finite presentations for the mapping class group via the ordered complex of curves ) using an ordered complex of curves, or by Susumu Hirose Action of the mapping class group on a complex of curves and a presentation for the mapping class group of a surface using a complex of non-separating curves. When all is said and done, I am personally not satisfied with any of the proofs out there. The Cerf theory proof, while being conceptual, takes you into deep and hard analytic terrain- while there is nothing at all conceptual about the MCG proof, despite it being "easy" (it's definitely much easier than Cerf theory, at least for me). The heart of the proof is to check that it happens to be possible to realize "relations" in your favourite finite presentation of the MCG by Kirby moves on links which generate the Dehn twists. The presentation itself is almost incidental, and the relations don't represent Kirby moves in any obvious way (the proof only goes one way- you could not prove a finite presentation of the MCG from Kirby's theorem).
|
{
"source": [
"https://mathoverflow.net/questions/16848",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2349/"
]
}
|
16,850 |
Here is an updated formulation of the question, which is more precise and I think completely correct: Suppose $M$ is a Riemannian manifold. Pick a point $p$ in $M$ and let $U$ be a neighborhood of the origin in $T_p M$ on which $exp_p$ restricts to a diffeomorphism. Let $X$ and $Y$ be tangent vectors in $T_p M$, and let $V$ be the intersection of $U$ and the plane spanned by $X$ and $Y$. Let $c(t)$ be a piecewise smooth simple closed curve in $V$. I claim that for any vector $Z$ in $T_p M$, $R(X,Y)Z=(P_c(Z)−Z)Area(c)+o(Area(c))$ where $R$ is the Riemannian curvature tensor, $P_c$ is parallel transport around the image of $c$ under $exp_p$, and $Area(c)$ is the area enclosed by the image of $c$ under $exp_p$. Can anyone refer me to a proof of this statement or something similar? I am fairly sure the argument has something to do with integrating the curvature 2-form over the embedded surface obtained by restricting $exp_p$ to the region enclosed by $c$, but I am having trouble with the estimates. Unfortunately I can't find anything in Kobayashi and Nimazu. Thanks in advance! Paul
|
It appears to me that one reason why nobody has proved the formula yet is that the formula is still wrong. First, the formula has to depend on $X$ and $Y$. If you rescale $X$ and $Y$, the left side of the formula scales but the right side stays constant. That can't be. Second, the two sides of the equation do not scale the same under a constant scaling of the metric. Warning: I wrote this up very quickly and did not check for typos and errors. It's possible that my final formula is still not right, but I am confident that my argument can be used to obtain a correct formula. I also did not provide every last detail, so, if you're unfamiliar with an argument like this, you need to do a lot of work making sure that everything really works. The key trick is pulling everything back to the unit square, where elementary calculus can be used. I'm sure this trick can be replaced by Stokes' theorem on the manifold itself, but that's too sophisticated for my taste. Holonomy calculation ADDED: The correct formula, if you assume $|X\wedge Y| = 1$, is $P_\gamma Z - Z = Area(c) R(X,Y)Z$ This scales properly when you rescale the metric by a constant factor. Notice that the left side is invariant under rescaling of the metric. I recommend looking at papers written by Hermann Karcher, especially the one with Jost on almost linear functions, the one with Heintze on a generalized comparison theorem, and the one on the Riemannian center of mass. I haven't looked at this or anything else in a long time, but I have the impression that I learned a lot about how to work with Jacobi fields and Riemann curvature from these papers. Finally, don't worry about citing anything I've said or wrote. Just write up your own proof of whatever you need. If it happens to look very similar to what I wrote, that's OK. I consider all of this "standard stuff" that any good Riemannian geometer knows, even if they would say it differently from me. EVEN MORE: There are similar calculations in my paper with Penny Smith: P. D. Smith and Deane Yang Removing Point Singularities of Riemannian Manifolds , TAMS (333) 203-219, especially in section 7 titled "Radially parallel vector fields". In section 5, we attribute our approach to H. Karcher and cite specific references.
|
{
"source": [
"https://mathoverflow.net/questions/16850",
"https://mathoverflow.net",
"https://mathoverflow.net/users/4362/"
]
}
|
16,857 |
Some time ago I heard this question and tried playing around with it. I've never succeeded to making actual progress. Here it goes: Given a finite (nonempty) set of real numbers, $S=\{a_1,a_2,\dots, a_n\}$, with the property that for each $i$ there exist $j,k$ (not necessarily distinct) so that $a_i=a_j+a_k$ (i.e. every element in $S$ can be written as a sum of two elements in $S$, note that this condition is trivially satisfied if $0\in S$ as then every $x\in S$ can be written as $x+0$). Must there exist $\{i_1,i_2,\dots, i_m\}$ (distinct) so that $a_{i_1}+a_{i_2}+\cdots +a_{i_m}=0$? ETA: A possible reformulation can be made in terms of graphs. We can take the vertex set $\{1,\dots ,n\}$ and for each equation $a_i=a_j+a_k$ in S add an edge $[ij]$ and its "dual" $[ik]$. The idea is to find a cycle in this graph, whose dual is a matching.
|
The answer is in the affirmative; indeed, If $S$ is a finite non-empty subset of any abelian group such that every element of $S$ is a sum of two other (possibly, equal to each other) elements, then $S$ has a non-empty, zero-sum subset. For a complete proof, see this recent preprint by János Nagy, Péter Pach, and myself. The proof is a little too long to be presented here but at least, to indicate the general direction, here is our main lemma. Lemma. Suppose that $M$ is an integer square matrix of order $n$ , representable as $A-I$ where $A$ has all its elements non-negative and all row sums equal to $2$ , and $I$ is the identity matrix. Then there exist nonzero vectors $u,v\in\{0,1\}^n$ such that $M^Tu=v$ ; that is, there exists a system of rows of $M$ such that their sum is a nonzero, zero-one vector. In fact, for our purposes it would suffice to have any nonzero vector $u$ satisfying $M^Tu\in\{0,1\}^n$ ; having $u$ itself to be zero-one is an extra (compare with Bill Thurston's answer above). The proof of the main lemma is completely elementary, by induction on $n$ . Historically, the question seems to originate from a problem contributed by Dan Schwarz to the EGMO 2012 (European Girls’ Mathematical Olympiad): Does there exist a set of integers such that every element of the set
is a sum of two other elements, while the set does not contain a
finite nonempty zero-sum subset? The answer to this question is positive since the set here is allowed to be infinite.
|
{
"source": [
"https://mathoverflow.net/questions/16857",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2384/"
]
}
|
16,858 |
Given a finite group G, I'm interested to know the smallest size of a set X such that G acts faithfully on X. It's easy for abelian groups - decompose into cyclic groups of prime power order and add their sizes. And the non-abelian group of order pq (p, q primes, q = 1 mod p) embeds in the symmetric group of degree q as shown here: www.jstor.org/stable/2306479. How much is known about this problem in general?
|
It is difficult to find this number for arbitrary finite groups, but many families have been solved. A somewhat early paper that has motivated a lot of work in this area is: Johnson, D. L. "Minimal permutation representations of finite groups."
Amer. J. Math. 93 (1971), 857-866. MR 316540 DOI: 10.2307/2373739 . This paper classifies those groups for which the regular permutation representation is the minimal faithful permutation representation (cyclic of prime power order, K4, or generalized quaternion) and some results on nilpotent groups (improved in later papers). The minimal permutation degrees of the finite simple groups are known from: Cooperstein, Bruce N. "Minimal degree for a permutation representation of a classical group."
Israel J. Math. 30 (1978), no. 3, 213-235. MR 506701 DOI: 10.1007/BF02761072 . This paper only finds the degrees. A fuller description of the permutation representations are given in: Grechkoseeva, M. A. "On minimal permutation representations of classical simple groups." Siberian Math. J. 44 (2003), no. 3, 443-462 MR 1984704 DOI: 10.1023/A:1023860730624 . There are a great deal of topics associated with minimal permutation degrees. I'll just briefly sketch them below, let me know if any interest you and I can give citations or longer descriptions: The minimal degree of a subgroup is never larger than the minimal degree of the parent group, but the minimal degree of a quotient group may be much, much larger than that of the original. This poses problems in computational group theory, since quotient groups may be difficult to represent. Some quotients are easy to represent, and this has had a significant impact on CGT in the last 10 years or so. Finding minimal permutation representations of covering groups can be difficult, and here I think the results are much less complete. Basically what you want are large subgroups not containing normal subgroups. In a covering group Z(G) is contained in really quite a few of the "good choices". This is because it is contained in Φ(G), the Frattini subgroup, the intersection of the maximal subgroups. One has to give up using maximal subgroups (at least if Z(G) is cyclic of prime power order), and so the minimal degree can increase dramatically. Minimal degrees of primitive permutation groups is a topic with a different flavor (rather than specific families of groups, it is more of the interactions between group properties), but a great deal is known. Similar techniques are used to describe asymptotic behavior of minimal degrees of arbitrary families of finite groups, and quite powerful results are available there.
|
{
"source": [
"https://mathoverflow.net/questions/16858",
"https://mathoverflow.net",
"https://mathoverflow.net/users/4336/"
]
}
|
16,953 |
Are all submodules of free modules free? I would like a reference to a proof or counterexample please.
|
Вот общий пример: неглавной идеал в кольце $A$. Кольцо $A$ -- свободный $A$-модуль. Идеал в кольце -- подмодуль, а он тоже свободный $A$-модуль только в случае, что он главной идеал: ненулевые элементы $a$ и $b$ в кольце удовлетворяют нетривиальному $A$-линейному соотношению $c_1a + c_2b = 0$, где $c_1 = b$ и $c_2 = -a$, поэтому если существует базис, то мощность является одним.
|
{
"source": [
"https://mathoverflow.net/questions/16953",
"https://mathoverflow.net",
"https://mathoverflow.net/users/3787/"
]
}
|
16,967 |
$\DeclareMathOperator\Ho{Ho}$ Is there a model category $C$ on an additive category such that its homotopy category $\Ho(C)$ is the stable homotopy category of spectra and the additive structure on $\Ho(C)$ is induced from that on $C$ . Basically I want to add and subtract maps in $C$ without going to its homotopy category. I'm not asking for $C$ to be a derived category or anything like that. Just that it should have an additive structure. As John Palmieri pointed out I should really say what structure I want the equivalence (between $\Ho(C)$ and the stable homotopy category) to preserve. Since I do want it to be a triangulated equivalence, Cisinski indicates why this is not possible.
|
The answer is: no there isn't such a thing. Here is a rough argument (a full proof would deserve a little more care). Using the main result of S. Schwede, The stable homotopy category is rigid , Annals of Mathematics 166 (2007), 837-863 your question is equivalent to the following: does there exist a model category $C$ , which is additive, and such that $C$ is Quillen equivalent to the usual model category of spectra? In particular, we might ask: does there exist an additive category $C$ , endowed with a Quillen stable model category structure, such that the corresponding stable $(\infty,1)$ -category is equivalent to the stable $(\infty,1)$ -category of spectra? Replacing $C$ by its full subcategory of cofibrant objects, your question might be reformulated as: does there exist a category of cofibrant objects $C$ (in the sense of Ken Brown ), with small sums (and such that weak equivalences are closed under small sums), and such that the corresponding $(\infty,1)$ -category (obtained by inverting weak equivalence of $C$ in the sense of $(\infty,1)$ -categories) is equivalent to the stable $(\infty,1)$ -category of spectra? If the answer is no, then there will be no additive model category $C$ such that $Ho(C)$ is (equivalent to) the category of spectra (as a triangulated category). So, assume there is an additive category of cofibrant objects $C$ , with small sums, such that $Ho(C)$ is (equivalent to) the category $S$ of spectra (as a triangulated category). Let $C_f$ be the full subcategory of $C$ spanned by the objects which correspond to finite spectra in $S$ . Then $Ho(C_f)\simeq S_f$ , where, by abuse of notations, $Ho(C_f)$ is the $(\infty,1)$ -category obtained from $C_f$ by inverting weak equivalences, while $S_f$ stands for the stable $(\infty,1)$ -category of finite spectra (essentially the Spanier-Whitehead category of finite CW-complexes). Given any (essentially) small additive category $A$ denote by $K(A)$ the "derived $(\infty,1)$ -category of $A$ " (that is the $(\infty,1)$ -category obtained from the category of bounded complexes of $A$ , by inverting the chain homotopy equivalences). Then, the canonical functor $A\to K(A)$ (which sends an object $X$ to itself, seen as a complex concentrated in degree $0$ ), has the following universal property: given a stable $(\infty,1)$ -category $T$ , any functor $A\to T$ which sends split short exact sequences of $A$ to distinguished triangles (aka homotopy cofiber sequences) in $T$ extends uniquely into a finite colimit preserving functor $K(A)\to T$ . In particular, the functor $C_f\to Ho(C_f)\simeq S_f$ extends uniquely to a finite colimit preserving functor $F:K(C_f)\to S_f$ . Let $Ker(F)$ be the full $(\infty,1)$ -subcategory of $K(C_f)$ spanned by objects which are sent to zero in $S_f$ . Then the induced functor $$K(C_f)/Ker(F)\to S_f$$ is an equivalence of (stable) $(\infty,1)$ -categories (to see this, you may use the universal property of $S_f$ : given a stable $(\infty,1)$ -category $T$ , a finite colimit preserving functor $S_f\to T$ is the same as an object of $T$ ; see Corollary 10.16 in DAG I ). This implies that, for any object $X$ of $S_f$ , if $X/n$ denotes the cone of the map $n:X\to X$ (multiplication by an integer $n$ ), then $n.X/n\simeq 0$ (see Proposition 1 in Schwede's paper Algebraic versus topological triangulated categories ). But such a property is known to fail whenever $X$ is a finite spectrum for $n=2$ (see Proposition 2 in loc. cit. ). Hence there isn't such a $C$ ...
|
{
"source": [
"https://mathoverflow.net/questions/16967",
"https://mathoverflow.net",
"https://mathoverflow.net/users/3557/"
]
}
|
16,991 |
I watched a video that said the probability for Gaussian integers to be relatively prime is an expression in $\pi$, and I also know about $\zeta(2) = \pi^2/6$ but I am wondering what are more connections between $\pi$ and prime numbers?
|
Well, first of all, $\pi$ is not just a random real number. Almost every real number is transcendental so how can we make the notion " $\pi$ is special" (in a number-theoretical sense) more precise? Start by noticing that $$\pi=\int_{-\infty}^{\infty}\frac{dx}{1+x^2}$$ This already tells us that $\pi$ has something to do with rational numbers. It can be expressed as "a complex number whose real and imaginary parts are values of absolutely convergent integrals of rational functions with rational coefficients, over domains in $\mathbb{R}^n$ given by polynomial inequalities with rational coefficients." Such numbers are called periods .
Coming back to the identity $$\zeta(2)=\frac{\pi^2}{6}$$ There is a very nice proof of this (that at first seems very unnatural) due to Calabi. It shows that $$\frac{3\zeta(2)}{4}=\int_0^1\int_0^1\frac{dx\,dy}{1-x^2y^2}$$ by expanding the corresponding geometric series, and then evaluates the integral to $\pi^2/8$ . (So yes, $\pi^2$ and all other powers of $\pi$ are periods.) But the story doesn't end here as it is believed that there are truly deep connections between values of zeta functions (or L-functions ) and certain evaluations involving periods, such as $\pi$ . Another famous problem about primes is Sylvester's problem of which primes can be written as a sum of two rational cubes. So one studies the elliptic curve $$E_p: p=x^3+y^3$$ and one wants to know if there is one rational solution, the central value of the corresponding L-function will again involve $\pi$ up to some integer factor and some Gamma factor. Next, periods are also values of multiple zeta functions: $$\zeta(s_1,s_2,\dots,s_k)=\sum_{n_1>n_2>\cdots>n_k\geq 1}\frac{1}{n_1^{s_1}\cdots n_k^{s_k}}$$ And they also appear in other very important conjectures such as the Birch and Swinnerton-Dyer conjecture . But of course all of this is really hard to explain without using appropriate terminology, the language of motives etc. So, though, this answer doesn't mean much, it's trying to show that there is an answer to your question out there, and if you study a lot of modern number theory, it might just be satisfactory :-).
|
{
"source": [
"https://mathoverflow.net/questions/16991",
"https://mathoverflow.net",
"https://mathoverflow.net/users/4361/"
]
}
|
16,992 |
I've read in (abstracts of) papers that there are abelian varieties over fields of positive characteristic that admit no prinicipal polarization. Apparently its not the easiest thing to find an example of, but I was thinking it should be much easier over the complex numbers. All abelian varieties of dimension 1 are elliptic curves which always have a principal polarization. So any example would have to be at least two dimensional. So my question is given an abelian variety with a polarization, is there a good way of telling if there is or isn't a principal polarization? Or if that's in general a difficult question, are there some relatively simple examples where you can really see that there are or aren't any principal polarizations?
|
Here is another construction, followed by some comments on how to solve the existence problem in general. If $A$ is a $g$-dimensional principally polarized abelian variety over $\mathbf{C}$ with $\operatorname{End} A = \mathbf{Z}$, and $G$ is a finite subgroup whose order $n$ is not a $g$-th power, then $B:=A/G$ is an abelian variety that admits no principal polarization. Proof: If $B$ had a principal polarization, its pullback to $A$, given by the composition $A \to B \to B' \to A' \simeq A$ (where $A'$ is the dual of $A$ and so on) would be an endomorphism of degree $n^2$. But this endomorphism is multiplication-by-$m$ for some integer $m$, which has degree $m^{2g}$. So $n$ would have to be a $g$-th power. $\square$ To complete this answer, observe that most abelian varieties $A$ over $\mathbf{C}$ satisfy $\operatorname{End} A=\mathbf{Z}$. An explicit example is the Jacobian of the hyperelliptic curve that is the smooth projective model of the affine curve
$$y^2= a_{2g+1} x^{2g+1} + \cdots + a_1 x + a_0$$
where $a_{2g+1},\ldots,a_0 \in \mathbf{C}$ are algebraically independent over $\mathbf{Q}$. Remarks: 1) Of course there is no reason to restrict to $\mathbf{C}$. For instance, one can find examples over $\mathbf{Q}$ by using the fact that the endomorphism ring injects into the endomorphism ring of the reduction modulo any prime of good reduction, and combining this information for several primes. 2) For an arbitrary abelian variety $A$, if you are given a polarization $A \to A'$, then if there is a principal polarization, following the first by the inverse of the second would give you an endomorphism of $A$. So one way of answering the existence question is to determine the endomorphism ring of $A$ and to study those endomorphisms that factor through your given polarization. (That's not quite sufficient, but it gives you an idea of the complexity of the problem since determining the endomorphism ring can be rather difficult.)
|
{
"source": [
"https://mathoverflow.net/questions/16992",
"https://mathoverflow.net",
"https://mathoverflow.net/users/7/"
]
}
|
16,994 |
Can anyone suggest a relatively gentle linear algebra text that integrates vector spaces and matrix algebra right from the start? I've found in the past that students react in very negative ways to the introduction of abstract vector spaces mid-way through a course. Sometimes it feels as though I've walked into class and said "Forget math. Let's learn ancient Greek instead." Sometimes the students realize that Greek is interesting too, but it can take a lot of convincing! Hence I would really like to let students know, right from the start, what they're getting themselves into. Does anyone know of a text that might help me do this in a not-too-advanced manner? One possibility, I guess, is Linear Algebra Done Right by Axler, but are there others? Axler's book might be too advanced. Or would anyone caution me against trying this, based on past experience?
|
For teaching the type of course that Dan described, I'd like to recommend David Lay's "Linear algebra". It is very thoroughly thought out and well written, with uniform difficulty level, some applications, and several possible routes/courses that he explains in the instructor's edition. Vector spaces are introduced in Chapter 4, following the chapters on linear systems, matrices, and determinants. Due to built-in redundancy, you can get there earlier, but I don't see any advantage to that. The chapter on matrices has a couple of sections that "preview" abstract linear algebra by studying the subspaces of $\mathbb{R}^n$.
|
{
"source": [
"https://mathoverflow.net/questions/16994",
"https://mathoverflow.net",
"https://mathoverflow.net/users/4042/"
]
}
|
17,006 |
Simple linear algebra methods are a surprisingly powerful tool to prove combinatorial results. Some examples of combinatorial theorems with linear algebra proofs are the (weak) perfect graph theorem , the Frankl-Wilson theorem , and Fisher's inequality . Are there other good examples?
|
Some other examples are the Erdos-Moser conjecture (see R. Proctor, Solution of two difficult problems with linear algebra, Amer. Math. Monthly 89 (1992), 721-734), a few results at http://math.mit.edu/~rstan/312/linalg.pdf , and Lovasz's famous result on the Shannon capacity of a 5-cycle and other graphs ( IEEE Trans. Inform. Theory 25 (1979), 1-7). For a preliminary manuscript of Babai and Frankl on this subject (Linear Algebra Methods in Combinatorics), see http://people.cs.uchicago.edu/~laci/CLASS/HANDOUTS-COMB/BaFrNew.pdf .
|
{
"source": [
"https://mathoverflow.net/questions/17006",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2233/"
]
}
|
17,020 |
Today, I heard that people think that if you can prove the Hodge conjecture for abelian varieties, then it should be true in general. Apparently this case is important enough (and hard enough) that Weil wrote up some families of abelian 4-folds that were potential counterexamples to the Hodge conjecture, but I've never heard of another potential counterexample. Anyway, in short: 1) Does the Hodge Conjecture for abelian varieties imply the full Hodge conjecture? 2) If not, is there an intuitive reason why abelian varieties should be the hardest case?
|
I would say the answer to both questions is no. In fact, abelian varieties should be an "easy" case. For example, it is known that for abelian varieties (but not other varieties), the variational Hodge conjecture implies the Hodge conjecture. It is disconcerting that we can't prove the Hodge conjecture even for abelian varieties, even for abelian varieties of CM-type, and we can't even prove that the Hodge classes Weil described are algebraic. So if the Hodge conjecture was proved in one interesting case, e.g., abelian varieties, that would be a big boost. Added: As follow up to Matt Emerton's answer, a proof that the Hodge conjecture for abelian varieties implies the Hodge conjecture for all varieties would (surely) also show that Deligne's theorem (that Hodge classes on abelian varieties are absolutely Hodge) implies the same statement for all varieties. But no such result is known (and would be extremely interesting).
|
{
"source": [
"https://mathoverflow.net/questions/17020",
"https://mathoverflow.net",
"https://mathoverflow.net/users/622/"
]
}
|
17,023 |
Background Let $\mathfrak{A}$, $\mathfrak{B}$, and $\mathfrak{C}$ be subcategories of the category of Banach spaces (over $\mathbb{R}$). Suppose we have a functor $\lambda:\mathfrak{A}^{op}\times\mathfrak{B}\to \mathfrak{C}$. Let $f:E'\to E$ be a morphism belonging to $\mathfrak{A}$, and let $g:F\to F'$ be a morphism belonging to $\mathfrak{B}$. (Note: These are morphisms of topological vector spaces). Then we have a map $$\matrix{Hom(E',E) \times Hom(F,F')\to Hom(\lambda(E,F),\lambda(E',F'))\\
(f,g)\mapsto\lambda(f,g)}$$ We say $\lambda$ is of class $C^p$ if for all manifolds $U$, and any two $C^p$ morphisms $U\to Hom(E',E)$ and $U\to Hom(F,F')$, the composition $$U\to Hom(E',E) \times Hom(F,F')\to Hom(\lambda(E,F),\lambda(E',F'))$$
is also of class $C^p$. (Note: We can replace $\mathfrak{A}$ and $\mathfrak{B}$ with categories of tuples to generalize this to several variables. In fact, this is what we do below.) It is not hard to show that this induces a unique functor $$\lambda_X:VB(X, \mathfrak{A})^{op}\times VB(X,\mathfrak{B})\to VB(X,\mathfrak{C}).$$ on vector bundles taking values in the appropriate vector bundle categories over $X$. We define a tensor bundle of type $\mathbf{\lambda}$ on $X$ to be $\lambda_X(TX)=\lambda_X((TX,\dots,TX),(TX,\dots,TX))$, where $TX$ is the tangent bundle. However, this doesn't agree with the definition given on Wikipedia or anywhere else I've looked. Questions Is this terminology nonstandard? Is the notion itself nonstandard? If the terminology is nonstandard, but the notion is standard, does it have a different name? Is this definition useful? Does this include more vector bundles as tensor bundles than the standard definition?
|
I would say the answer to both questions is no. In fact, abelian varieties should be an "easy" case. For example, it is known that for abelian varieties (but not other varieties), the variational Hodge conjecture implies the Hodge conjecture. It is disconcerting that we can't prove the Hodge conjecture even for abelian varieties, even for abelian varieties of CM-type, and we can't even prove that the Hodge classes Weil described are algebraic. So if the Hodge conjecture was proved in one interesting case, e.g., abelian varieties, that would be a big boost. Added: As follow up to Matt Emerton's answer, a proof that the Hodge conjecture for abelian varieties implies the Hodge conjecture for all varieties would (surely) also show that Deligne's theorem (that Hodge classes on abelian varieties are absolutely Hodge) implies the same statement for all varieties. But no such result is known (and would be extremely interesting).
|
{
"source": [
"https://mathoverflow.net/questions/17023",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1353/"
]
}
|
17,032 |
What is $\mathbb{Q}_p \cap \overline{\mathbb{Q}}$ ? For instance, we know that $\mathbb{Q}_p$ contains the $p-1$st roots of unity, so we might say that $\mathbb{Q}(\zeta) \subset \mathbb{Q}_p \cap \overline{\mathbb{Q}}$, where $\zeta$ is a primitive $p-1$st root. As a more specific example, $x^2 - 6$ has 2 solutions in $\mathbb{Q}_5$, so we could also say that $\mathbb{Q}(\sqrt{6},\sqrt{-1})\subset \mathbb{Q}_p \cap \overline{\mathbb{Q}}$. Edit: I removed the motivation for this question (which I think stands by itself), as it will be better as a separate question once I think it through a bit better.
|
The field $K_p = \mathbb{Q}_p \cap \overline{\mathbb{Q}}$ is a very natural and well-studied one. I can throw some terminology at you, but I'm not sure exactly what you want to know about it. 1) It is often called the field of "$p$-adic algebraic numbers". This comes up in model theory: it is a $p$-adically closed field, which is the $p$-adic analogue of a real-closed field. In particular, it is elementarily equivalent to $\mathbb{Q}_p$. 2) It is the Henselization of $\mathbb{Q}$ with respect to the $p$-adic valuation, or the
fraction field of the Henselization of the ring $\mathbb{Z}_{(p)}$ -- i.e., $\mathbb{Z}$ localized at the prime ideal $p$. The idea is that this field is not complete but is Henselian -- it satisfies the conclusion of Hensel's Lemma. Alternately and somewhat more gracefully, Henselian valued fields are characterized by the fact that the valuation extends uniquely to any algebraic field extension. Roughly speaking, Henselian fields are as good as complete fields for algebraic constructions but are not "big enough" to have the same topological properties. For instance, note that $K_p$ cannot possibly be complete with respect to the $p$-adic valuation, because it is countably infinite and without isolated points: apply the Baire Category Theorem.
|
{
"source": [
"https://mathoverflow.net/questions/17032",
"https://mathoverflow.net",
"https://mathoverflow.net/users/434/"
]
}
|
17,062 |
A student asked me why $\mathcal{O}_K$ is the notation used for the ring of integers in a number field $K$ and why $h$ is the notation for class numbers. I was able to tell him the origin of $\mathcal{O}$ (from Dedekind's use of Ordnung, the German word for order, which was taken from taxonomy in the same way the words class and genus had been stolen for math usage before him), but I was stumped by $h$. Does anyone out there know how $h$ got adopted? I have a copy of Dirichlet's lecture notes on number theory (the ones Dedekind edited with his famous supplements laying out the theory of ideals), and in there he is using $h$. So this convention goes back at least to Dirichlet -- or maybe Dedekind? -- but is that where the notation starts? And even if so, why the letter $h$? I had jokingly suggested to the student that $h$ was for Hilbert, but I then told him right away it made no historical sense (Hilbert being too late chronologically).
|
Gauss, in his Disquisitiones, used ad hoc notation for the class number when he needed it. He did not use h. Dirichlet used h for the class number in 1838 when he proved the class number formula for binary quadratic forms. I somewhat doubt that he was thinking of "Hauptform" in this connection - back then, the group structure was not as omnipresent as it is today, and the result that $Q^h$ is the principal form was known (and written additively), but did not play any role. Kummer, 10 years later, used H for the class number of the field of p-th roots of unity, and h for the class number of a subfield generated by Gaussiam periods (and "proved" that $h \mid H$); in the introduction he quotes Dirichlet's work on forms at length.
|
{
"source": [
"https://mathoverflow.net/questions/17062",
"https://mathoverflow.net",
"https://mathoverflow.net/users/3272/"
]
}
|
17,072 |
This question is inspired by the recent question "The finite subgroups of SL(2,C)" . While reading the answers there I remembered reading once that identifying the finite subgroups of SU(3) is still an open problem. I have tried to check this and it seems it was at least still open in the Eighties. Can anyone confirm or deny that the finite subgroups of SU(3) are not all known? And if this is true, then what is the source of the difficulty? Secondly, what is known of the finite subgroups of SU(n) for n > 3? UPDATE: Thanks to those below who have corrected my ignorance! It seems that I may have been tricked by some particularly sensationalised abstracts (or perhaps just misunderstood them.)
|
There is an algorithm due to Zassenhaus which, in principle, lists all conjugacy classes of finite subgroups of compact Lie groups. I believe that the algorithm was used for $\mathrm{SO}(n)$ for at least $n=6$ if not higher. I believe it is expensive to run, which means that in practice it is only useful for low dimension. Added Now that I'm in my office I have my orbifold folder with me and I can list some relevant links: Zassenhaus's original paper (in German) Über einen Algorithmus zur Bestimmung der Raumgruppen There is a book by RLE Schwarzenberger N-dimensional crystallography with lots of references There are a couple of papers in Acta Cryst. by Neubüser, Wondratschek and Bülow titled On crystallography in higher dimensions There is a sequence of papers in Math. Comp. by Plesken and Pohst titled On maximal finite irreducible subgroups of GL(n,Z) which I remember were relevant. Independent of this algorithm, there is some work on $\mathrm{SU}(n)$ from the physics community motivated by elementary particle physics and more modern considerations of the use of orbifolds in the gauge/gravity correspondence. The case of $\mathrm{SU}(3)$ was done in the mid 1960s and is contained in the paper Finite and Disconnected Subgroups of SU(3) and their Application to the Elementary-Particle Spectrum by Fairbairn, Fulton and Klink. For the case of $\mathrm{SU}(4)$ there is a more recent paper A Monograph on the Classification of the Discrete Subgroups of SU(4) by Hanany and He, and references therein. Further edit The paper Non-abelian finite gauge theories by Hanany and He have the correct list of finite subgroups of SU(3), based on Yau and Yu's paper Gorenstein quotient singularities in dimension three .
|
{
"source": [
"https://mathoverflow.net/questions/17072",
"https://mathoverflow.net",
"https://mathoverflow.net/users/3623/"
]
}
|
17,189 |
The following problem is not from me, yet I find it a big challenge to give a nice (in contrast to 'heavy computation') proof. The motivation for me to post it lies in its concise content. If $a$ and $b$ are nonnegative real numbers such that $a+b=1$, show that $a^{2b} + b^{2a}\le 1$.
|
Fixed now. I spent some time looking for some clever trick but the most unimaginative way turned out to be the best. So, as I said before, the straightforward Taylor series expansion does it in no time. Assume that $a>b$. Put $t=a-b=1-2b$. Step 1:
$$
\begin{aligned}
a^{2b}&=(1-b)^{1-t}=1-b(1-t)-t(1-t)\left[\frac{1}2b^2+\frac{1+t}{3!}b^3+\frac{(1+t)(2+t)}{4!}b^4+\dots\right]
\\
&\le 1-b(1-t)-t(1-t)\left[\frac{b^2}{1\cdot 2}+\frac{b^3}{2\cdot 3}+\frac{b^4}{3\cdot 4}+\dots\right]
\\&
=1-b(1-t)-t(1-t)\left[b\log\frac 1{a}+b-\log\frac {1}a\right]
\\
&=1-b(1-t^2)+(1-b)t(1-t)\log\frac{1}a=1-b\left(1-t^2-t(1+t)\log\frac 1a\right)
\end{aligned}
$$
(in the last line we rewrote $(1-b)(1-t)=(1-b)2b=b(2-2b)=b(1+t)$) Step 2.
We need the inequality $e^{ku}\ge (1+u)(1+u+\dots+u^{k-1})+\frac k{k+1}u^{k+1}$ for $u\ge 0$.
For $k=1$ it is just $e^u\ge 1+u+\frac{u^2}{2}$. For $k\ge 2$, the Taylor coefficients on the left are $\frac{k^j}{j!}$ and on the right $1,2,2,\dots,2,1$ (up to the order $k$) and then $\frac{k}{k+1}$. Now it remains to note that $\frac{k^0}{0!}=1$, $\frac{k^j}{j!}\ge \frac {k^j}{j^{j-1}}\ge k\ge 2$ for $1\le j\le k$, and $\frac{k^{k+1}}{(k+1)!}\ge \frac{k}{k+1}$. Step 3:
Let $u=\log\frac 1a$. We've seen in Step 1 that $a^{2b}\le 1-b(1-t\mu)$ where $\mu=u+(1+u)t$. In what follows, it'll be important that $\mu\le\frac 1a-1+\frac 1a t=1$ (we just used $\log\frac 1a\le \frac 1a-1$ here. We have $b^{2a}=b(a-t)^t$. Thus, to finish, it'll suffice to show that $(a-t)^t\le 1-t\mu$. Taking negative logarithm of both sides and recalling that $\frac 1a=e^u$, we get the inequality
$$
tu+t\log(1-te^u)^{-1}\ge \log(1-t\mu)^{-1}
$$
to prove.
Now, note that, according to Step 2,
$$
\begin{aligned}
&\frac{e^{uk}}k\ge \frac{(1+u)(1+u+\dots+u^{k-1})}k+\frac{u^{k+1}}{k+1}
\ge\frac{(1+u)(\mu^{k-1}+\mu^{k-2}u+\dots+u^{k-1})}k+\frac{u^{k+1}}{k+1}
\\
&=\frac{\mu^k-u^k}{kt}+\frac{u^{k+1}}{k+1}
\end{aligned}
$$
Multiplying by $t^{k+1}$ and adding up, we get
$$
t\log(1-te^u)^{-1}\ge -ut+\log(1-t\mu)^{-1}
$$
which is exactly what we need. The end. P.S. If somebody is still interested, the bottom line is almost trivial once the top line is known. Assume again that $a>b$, $a+b=1$. Put $t=a-b$. $$
\begin{aligned}
&\left(\frac{a^b}{2^b}+\frac{b^a}{2^a}\right)^2=(a^{2b}+b^{2a})(2^{-2b}+2^{-2a})-\left(\frac{a^b}{2^a}-\frac{b^a}{2^b}\right)^2
\\
&\le 1+\frac 14\{ [\sqrt 2(2^{t/2}-2^{-t/2})]^2-[(1+t)^b-(1-t)^a]^2\}
\end{aligned}
$$
Now it remains to note that $2^{t/2}-2^{-t/2}$ is convex on $[0,1]$, so, interpolating between the endpoints, we get $\sqrt 2(2^{t/2}-2^{-t/2})\le t$. Also, the function $x\mapsto (1+x)^b-(1-x)^a$ is convex on $[0,1]$ (the second derivative is $ab[(1-x)^{b-2}-(1+x)^{a-2}]$, which is clearly non-negative). But the derivative at $0$ is $a+b=1$, so $(1+x)^b-(1-x)^a\ge x$ on $[0,1]$. Plugging in $x=t$ finishes the story.
|
{
"source": [
"https://mathoverflow.net/questions/17189",
"https://mathoverflow.net",
"https://mathoverflow.net/users/3818/"
]
}
|
17,197 |
This is a question, or really more like a cloud of questions, I wanted to ask awhile ago based on this SBS post and this post I wrote inspired by it, except that Math Overflow didn't exist then. As the SBS post describes, the Catalan numbers can be obtained as the moments of the trace of a random element of $SU(2)$ with respect to the Haar measure. This is equivalent to the integral identity
$$\int_{0}^{1} (2 \cos \pi x)^{2k} (2 \sin^2 \pi x) \, dx = C_k.$$ I can prove this identity "combinatorially" as follows: let $A_n$ denote the Dynkin diagram with $n$ vertices and $n-1$ undirected edges connecting those vertices in sequence. The adjacency matrix of $A_n$ has eigenvectors $\mathbf{v}\_i$ with entries $\mathbf{v}\_{i,j} = \sin \frac{\pi ij}{n+1}$ with corresponding eigenvalues $2 \cos \frac{\pi i}{n+1}$. If $k \le n-1$, then a straightforward computation shows that the number of closed walks from one end of $A_n$ to itself of length $2k$ is
$$\frac{1}{n+1} \sum_{i=1}^{n} \left( 2 \cos \frac{\pi i}{n+1} \right)^{2k} 2 \sin^2 \frac{\pi}{n+1} = C_k$$ by the combinatorial definition of the Catalan numbers. Taking the limit as $n \to \infty$ gives the integral identity; in other words, the integral identity is in some sense equivalent to the combinatorial definition of the Catalan numbers in terms of closed walks on the "infinite path graph" $A_{\infty}$. (Is this the correct notation? I mean the infinite path graph with one end.) Now, closed walks of length $2k$ from one end of $A_n$ to itself can be put in bijection with ordered rooted trees of depth at most $n$ and $k$ non-root vertices. (Recall that the Catalan numbers also count ordered rooted trees of arbitrary depth.) The generating function $P_n$ of ordered rooted trees of depth at most $n$ satisfies the recursion
$$P_1(x) = 1, P_n(x) = \frac{1}{1 - x P_{n-1}(x)}.$$ This is because an ordered rooted tree of depth $n$ is the same thing as a sequence of ordered rooted trees of depth $n-1$ together with a new root. (Recall that the generating function of the Catalan numbers satisfies $C(x) = \frac{1}{1 - x C(x)}$. In other words, $C(x)$ has a continued fraction representation, and $P_n$ is its sequence of convergents.) On the other hand, since $P_n(x)$ counts walks on the graph $A_n$, it is possible to write down the generating function $P_n$ explicitly in terms of the characteristic polynomials of the corresponding adjacency matrices, and these polynomials have roots the eigenvalues $2 \cos \frac{\pi i}{n+1}$. This implies that they must be the Chebyshev polynomials of the second kind, i.e. the ones satisfying
$$q_n(2 \cos x) = \frac{\sin (n+1) x}{\sin x}.$$ But the Chebyshev polynomials of the second kind are none other than the characters of the irreducible finite-dimensional representations of $SU(2)$! In particular, they're orthogonal with respect to the Haar measure. The overarching question I have is: how does this sequence of computations generalize, and what conceptual framework ties it together? But I should probably ask more specific sub-questions. Question 1: I remember hearing that the relationship between the Catalan numbers and the Chebyshev polynomials generalizes to some relation between moments, continued fractions, and orthogonal polynomials with respect to some measure. Where can I find a good reference for this? Question 2: I believe that adding another edge and considering the family of cycle graphs gives the sequence ${2k \choose k}$ and the Chebyshev polynomials of the first kind, both of which are related to $SO(2)$. According to the SBS post, this is a "type B" phenomenon, whereas the Catalan numbers are "type A." What exactly does this mean? What would happen if I repeated the above computations for other Dynkin diagrams? Do I get continued fractions for the other infinite families? Question 3: Related to the above, in what sense is it natural to relate walks on the Dynkin diagram $A_n$ to representations of $SU(2)$? This seems to have something to do with question #16026 . How do the eigenvectors of the adjacency matrices fit into the picture? I want to think of the eigenvectors as "discrete harmonics"; does this point of view make sense? Does it generalize? As you can see, I'm very confused, so I would greatly appreciate any clarification.
|
The Catalan numbers enumerate (amongst everything else!) the bases of the Temperley-Lieb algebras. These algebras $TL_n(q)$ are exactly $\operatorname{End}_{U_q \mathfrak{su}_2}(V^{\otimes n})$ where $V$ is the standard representation. If $q$ is a $2k+4$-th root of unity, the semisimplified representation theory of $U_q \mathfrak{su}_2$ becomes '$A_{k+1}$': that is, it has $k+1$ simples, and the principal graph for tensor product with the standart is $A_{k+1}$. The algebra $\operatorname{End}_{U_q \mathfrak{su}_2}(V^{\otimes n})$ is now smaller (some of the summands of the tensor power cancel out when you use the quantum Racah rule), and in fact it is enumerated by loops on $A_{k+1}$ based at the first vertex. You can pick out a subset of the entire Temperley-Lieb that gives a basis: just use your description of paths as trees, and take the dual graph. Since we're looking at a unitary tensor category, the dimension of the standard object is exactly the Frobenius-Perron eigenvalue (largest real eigenvalue of the adjacency matrix) of the principal graph.
|
{
"source": [
"https://mathoverflow.net/questions/17197",
"https://mathoverflow.net",
"https://mathoverflow.net/users/290/"
]
}
|
17,202 |
I am interested in the function $$f(N,k)=\sum_{i=0}^{k} {N \choose i}$$ for fixed $N$ and $0 \leq k \leq N $. Obviously it equals 1 for $k = 0$ and $2^{N}$ for $k = N$, but are there any other notable properties? Any literature references? In particular, does it have a closed form or notable algorithm for computing it efficiently? In case you are curious, this function comes up in information theory as the number of bit-strings of length $N$ with Hamming weight less than or equal to $k$. Edit: I've come across a useful upper bound: $(N+1)^{\underline{k}}$ where the underlined $k$ denotes falling factorial. Combinatorially, this means listing the bits of $N$ which are set (in an arbitrary order) and tacking on a 'done' symbol at the end. Any better bounds?
|
I'm going to give two families of bounds,
one for when $k = N/2 + \alpha \sqrt{N}$ and one for when $k$ is fixed. The sequence of binomial coefficients ${N \choose 0}, {N \choose 1}, \ldots, {N \choose N}$ is symmetric. So you have $\sum_{i=0}^{(N-1)/2} {N \choose i} = {2^N \over 2} = 2^{N-1}$ when $N$ is odd. (When $N$ is even something similar is true
but you have to correct for whether you include the term ${N \choose N/2}$ or not. Also, let $f(N,k) = \sum_{i=0}^k {N \choose i}$. Then you'll have, for real constant $\alpha$, $ \lim_{N \to \infty} {f(N,\lfloor N/2+\alpha \sqrt{N} \rfloor) \over 2^N} = g(\alpha) $ for some function $g$. This is essentially a rewriting of a special case of the central limit theorem. The Hamming weight of a word chosen uniformly at random is a sum of Bernoulli(1/2) random variables. For fixed $k$ and $N \to \infty$, note that
$$ {{N \choose k} + {N \choose k-1} + {N \choose k-2}+\dots \over {N \choose k}}
= {1 + {k \over N-k+1} + {k(k-1) \over (N-k+1)(N-k+2)} + \cdots} $$
and we can bound the right side from above by the geometric series
$$ {1 + {k \over N-k+1} + \left( {k \over N-k+1} \right)^2 + \cdots} $$
which equals ${N-(k-1) \over N - (2k-1)}$. Therefore we have
$$ f(N,k) \le {N \choose k} {N-(k-1) \over N-(2k-1)}.$$
|
{
"source": [
"https://mathoverflow.net/questions/17202",
"https://mathoverflow.net",
"https://mathoverflow.net/users/4405/"
]
}
|
17,205 |
I am supervising an undergraduate for a project in which he's going to talk about the relationship between Galois representations and modular forms. We decided we'd figure out a few examples of weight 1 modular forms and Galois representations and see them matching up. But I realised when working through some examples that computing the conductor of the Galois representation was giving me problems sometimes at small primes. Here's an explicit question. Set $f=x^4 + 2x^2 - 1$ and let $K$ be the splitting field of $f$ over $\mathbf{Q}$. It's Galois over $\mathbf{Q}$ with group $D_8$. Let $\rho$ be the irreducible 2-dimensional representation of $D_8$. What is the conductor of $\rho$? Note that I don't particularly want to know the answer to this particular question, I want to know how to work these things out in general. In fact I think I could perhaps figure out the conductor of $\rho$ by doing calculations on the modular forms side, but I don't want to do that (somehow the point of the project is seeing that calculations done in 2 different ways match up, rather than using known modularity results to do the calculations). Using pari or magma I see that $K$ is unramified outside 2, and the ideal (2) is an 8th power in the integers of $K$. To compute the conductor of $\rho$ the naive approach is to figure out the higher ramification groups at 2 and then just use the usual formula. But the only computer algebra package I know which will compute higher ramification groups is magma, and if I create the splitting field of $f$ over $\mathbf{Q}_2$ (computed using pari's "polcompositum" command) Qx<x>:=PolynomialRing(Rationals());
g:=x^8 + 20*x^6 + 146*x^4 + 460*x^2 + 1681;
L := LocalField(pAdicField(2, 50),g);
DecompositionGroup(L); then I get an instant memory overflow (magma wants 2.5 gigs to do this, apparently), and furthermore the other calculations I would have to do if I were to be following up this idea would be things like RamificationGroup(L, 3); which apparently need 11 gigs of ram to run. Ouch. Note also that if I pull the precision of the $p$-adic field down from 50 then magma complains that the precision isn't large enough to do some arithmetic in $L$ that it wants to do. I think then my question must be: are there any computer algebra resources that will compute higher ramification groups for local fields without needing exorbitant amounts of memory? Or is it a genuinely an "11-gigs" calculation that I want to do?? And perhaps another question is: is there another way of computing the conductor of a (non-abelian finite image) Galois representation without having to compute these higher ramification groups (and without computing any modular forms either)?
|
You can also compute some higher ramification groups in Sage. At the moment it gives lower numbering, not upper numbering, but here it is anyway: sage: Qx.<x> = PolynomialRing(QQ) sage: g=x^8 + 20*x^6 + 146*x^4 + 460*x^2 + 1681 sage: L.<a> = NumberField(g) sage: G = L.galois_group() sage: G.ramification_breaks(L.primes_above(2)[0]) {1, 3, 5} You can also get explicit presentations of G as a permutation group and generators for ramification and decomposition subgroups. The above only takes about half a second on my old laptop -- no 2.5 gigs computations here. (The point is that it is much easier to do computations over a number field, because everything is exact, rather than over a p-adic field which is represented inexactly.)
|
{
"source": [
"https://mathoverflow.net/questions/17205",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1384/"
]
}
|
17,209 |
I assume a number of results have been proven conditionally on the Riemann hypothesis, of course in number theory and maybe in other fields. What are the most relevant you know? It would also be nice to include consequences of the generalized Riemann hypothesis (but specify which one is assumed).
|
I gave a talk on this topic a few months ago, so I assembled a list then which could be appreciated by a general mathematical audience. I'll reproduce it here. (Edit: I have added a few more examples to the end of the list, starting at item m, which are meaningful to number theorists but not necessarily to a general audience.) Let's start with three applications of RH for the Riemann zeta-function only. a) Sharp estimates on the remainder term in the prime number theorem: $\pi(x) = {\text{Li}}(x) + O(\sqrt{x}\log x)$ , where ${\text{Li}}(x)$ is the logarithmic integral (the integral from 2 to $x$ of $1/\log t$ ). b) Comparing $\pi(x)$ and ${\text{Li}}(x)$ . All the numerical data shows $\pi(x)$ < ${\text{Li}}(x)$ , and Gauss thought this was always true, but in 1914 Littlewood used the Riemann hypothesis to show the inequality reverses infinitely often. In 1933, Skewes used RH to show the inequality reverses for some $x$ below 10^10^10^34. In 1955 Skewes showed without using RH that the inequality reverses for some $x$ below 10^10^10^963. Maybe this was the first example where something was proved first assuming RH and later proved without RH. c) Gaps between primes. In 1919, Cramer showed RH implies $p_{k+1} - p_k = O(\sqrt{p_k}\log p_k)$ , where $p_k$ is the $k$ th prime. (A conjecture of Legendre is that there's always a prime between $n^2$ and $(n+1)^2$ -- in fact there should be a lot of them -- and this would imply $p_{k+1} - p_k = O(\sqrt{p_k})$ . This is better than Cramer's result, so it lies deeper than a consequence of RH. Cramer also conjectured that the gap is really $O((\log p_k)^2)$ .) Now let's move on to applications involving more zeta and $L$ -functions than just the Riemann zeta-function. Note that typically we will need to assume GRH for infinitely many such functions to say anything. d) Chebyshev's conjecture. In 1853, Chebyshev tabulated the primes
which are $1 \bmod 4$ and $3 \bmod 4$ and noticed there are always at least as many $3 \bmod 4$ primes up to $x$ as $1 \bmod 4$ primes. He conjectured this was always true and also gave an analytic sense in which there are more $3 \bmod 4$ primes: $$
\lim_{x \rightarrow 1^{-}} \sum_{p \not= 2} (-1)^{(p+1)/2}x^p = \infty.
$$ Here the sum runs over odd primes $p$ . In 1917, Hardy-Littlewood and Landau (independently) showed this second conjecture of Chebyshev's is equivalent to
GRH for the $L$ -function of the nontrivial character mod 4. (In 1994, Rubinstein and Sarnak used simplicity and linear independence hypotheses on zeros of $L$ -functions to say something about Chebyshev's first conjecture, but as the posted question asked only about consequences of RH and GRH, I leave the matter there and move on.) e) The Goldbach conjecture (1742). The "even" version says all even integers $n \geq 4$ are a sum of 2 primes, while the "odd" version says all odd integers $n \geq 7$ are a sum of 3 primes. For most mathematicians, the Goldbach conjecture is understood to mean the even version, and obviously the even version implies the odd version. The odd version turns out to be a consequence of GRH. In 1923, assuming all Dirichlet $L$ -functions are nonzero in a right half-plane ${\text{Re}}(s) \geq 3/4 - \varepsilon$ , where $\varepsilon$ is fixed (independent of the $L$ -function), Hardy and Littlewood showed the odd Goldbach conjecture is true for all sufficiently large odd $n$ . In 1937, Vinogradov proved the same result without needing GRH as a hypothesis. In 1997, Deshouillers, Effinger, te Riele, and Zinoviev showed GRH implies the odd Goldbach conjecture is true for all odd $n \geq 7$ . So GRH completely settles the odd Goldbach conjecture. Update: This is now an obsolete application of GRH since the odd Goldbach Conjecture was proved by Harald Helfgott in 2013 without needing GRH as a hypothesis.
An account of the current status of Helfgott's work is here . f) Polynomial-time primality tests. By results of Ankeny (1952) and Montgomery (1971), GRH for all Dirichlet $L$ -functions implies that the first nonmember of every proper subgroup of the unit group $({\mathbf Z}/m{\mathbf Z})^\times$ is $O((\log m)^2)$ , where the $O$ -constant is independent of $m$ . In 1985, Bach showed GRH for all Dirichlet $L$ -functions implies the constant in that $O$ -estimate can be taken to be 2. That is, GRH for all Dirichlet $L$ -functions implies that each proper subgroup of $({\mathbf Z}/m{\mathbf Z})^\times$ is missing some integer from 1 to $2(\log m)^2$ . Put differently, GRH for all Dirichlet $L$ -functions implies that if a subgroup of $({\mathbf Z}/m{\mathbf Z})^\times$ contains all positive integers below $2(\log m)^2$ then the subgroup is the whole unit group mod $m$ . (To understand one way that GRH has an influence on that upper bound, if all the nontrivial zeros of all Dirichlet $L$ -functions have ${\text{Re}}(s) \leq 1 - \varepsilon$ then the first nonmember of every proper subgroup of $({\mathbf Z}/m{\mathbf Z})^\times$ is $O((\log m)^{1/\varepsilon})$ . Set $\varepsilon = 1/2$ to get the previous result I stated that uses GRH.) In 1976, Gary Miller introduced a deterministic primality test that he could prove runs in polynomial time using GRH for all Dirichlet $L$ -functions. (Part of the test involves deciding if a subgroup of units mod $m$ is proper or not.) Shortly afterwards, Solovay and Strassen described a different primality test using Jacobi symbols. GRH for Dirichlet $L$ -functions implies their test runs in polyomial time, and the subgroups of units mod $m$ occurring for their test contain $-1$ , so the proof that their test runs in polynomial time "only" needs GRH for Dirichlet $L$ -functions of even characters. (Solovay and Strassen described their test as a probabilistic test rather than a deterministic test, so they didn't mention GRH one way or the other.) In 2002 Agrawal, Kayal, and Saxena created a new primality test that they could prove runs in polynomial time without needing GRH. This is a nice example showing how GRH guides mathematicians in the direction of what should be true and then people tried to find a proof of those results by methods not involving GRH. g) Euclidean rings of integers. In 1973, Weinberger showed that GRH for all Dedekind zeta-functions implies that every ring of algebraic integers with an infinite unit group (so ignoring $\mathbf Z$ and the ring of integers of imaginary quadratic fields) is Euclidean if it has class number 1. As a special case, in concrete terms, if $d$ is a positive integer that is not a perfect square then a ring ${\mathbf Z}[\sqrt{d}]$ that is a unique factorization domain must be Euclidean. The same theorem is true for rings of $S$ -integers: an infinite unit group plus class number $1$ plus GRH for zeta-functions of all number fields implies the ring is Euclidean. Progress has been made by Ram Murty and others in the direction of proving class number 1 and infinite unit group implies Euclidean for rings of $S$ -integers without needing GRH, and as a striking special case let's consider ${\mathbf Z}[\sqrt{14}]$ . It has class number 1 (which must have been known to Gauss in the early 19th century, in the language of quadratic forms) and an infinite unit group, so it should be Euclidean. This particular real quadratic ring was first proved to be Euclidean only in 2004 (by M. Harper). So $\mathbf Z[\sqrt{14}]$ is a ring that was known to have unique factorization for over 100 years before it was proved to be Euclidean. h) Artin's primitive root conjecture. In 1927, Artin conjectured that each nonzero integer $a$ that is not $-1$ or a perfect square is a generator of $({\mathbf Z}/p{\mathbf Z})^\times$ for infinitely many primes $p$ , and in fact for a positive proportion of such $p$ .
As a special case, taking $a = 10$ , this says for primes $p$ the unit fraction $1/p$ has decimal period $p-1$ for a positive proportion of $p$ . (For each prime $p$ other than $2$ and $5$ , the decimal period for $1/p$ is a factor of $p-1$ , so this special case is saying the largest possible period is realized infinitely often in a precise sense.) In 1967, Hooley showed GRH for zeta-functions of number fields implies Artin's primitive root conjecture. More precisely, Artin's primitive root conjecture for $a$ follows from GRH for the zeta-functions of all the number fields $\mathbf Q(\sqrt[n]{a},\zeta_n)$ where $n$ runs over the squarefree positive integers. In 1984, R. Murty and Gupta showed Artin's primitive root conjecture is true for infinitely many $a$ without having to use GRH, but their proof couldn't pin down even one specific $a$ for which Artin's primitive root conjecture is true. In 1986, Heath-Brown showed without using GRH that Artin's primitive root conjecture is true for all prime values of $a$ with at most two exceptions (and of course there should not be any exceptions). Without using GRH, no definite $a$ is known for which Artin's conjecture is true. i) First prime in an arithmetic progression. If $\gcd(a,m) = 1$ then there are infinitely many primes $p \equiv a \bmod m$ . When does the first one appear, as a function of $m$ ? In 1934, Chowla showed GRH implies the first prime $p \equiv a \bmod m$ is $O(m^2(\log m)^2)$ . In 1944, Linnik showed without GRH that the bound is $O(m^L)$ for some universal exponent $L$ . The latest choice for $L$ (Xylouris, 2009) without using GRH is $L = 5.2$ . j) Gauss' class number problem. Gauss (1801) conjectured in the language of quadratic forms that there are only finitely many imaginary quadratic fields with class number 1. (He actually conjectured more precisely that the 9 known examples are the only ones, but for what I want to say the weaker finiteness statement is simpler.) In 1913, Gronwall showed this is true if the $L$ -functions of all imaginary quadratic Dirichlet characters have no zeros in some common strip $1- \varepsilon < {\text{Re}}(s) < 1$ . That is weaker than GRH (we only care about $L$ -functions of a restricted collection of characters), but it is a condition like GRH for infinitely many $L$ -functions. In 1933, Deuring and Mordell showed Gauss' conjecture is true if the ordinary RH (for the Riemann zeta-function) is false , and then in 1934 Heilbronn showed Gauss' conjecture is true if GRH is false for some Dirichlet $L$ -function of an imaginary quadratic character. Since Gronwall proved Gauss' conjecture is true when GRH is true for the Riemann zeta-function and the Dirichlet $L$ -functions of all imaginary quadratic Dirichlet characters and Deuring--Mordell--Heilbronn proved Gauss' conjecture is true when GRH is false for at least one of those functions, Gauss' conjecture is true by baby logic. In 1935, Siegel proved Gauss' conjecture is true without using GRH, and in the 1950s and 1960s Baker, Heegner, and Stark gave separate proofs of Gauss' precise "only 9" conjecture without using GRH. k) Missing values of a quadratic form. Lagrange (1772) showed every positive integer is a sum of four squares. However, not every integer is a sum of three squares: $x^2 + y^2 + z^2$ misses all $n \equiv 7 \bmod 8$ . Legendre (1798) showed a positive integer is a sum of three squares iff it is not of the form $4^a(8k+7)$ . This can be phrased as a local-global problem: $x^2 + y^2 + z^2 = n$ is solvable in integers iff the congruence $x^2 + y^2 + z^2 \equiv n \bmod m$ is solvable for all $m$ . More generally, the same local-global
phenomenon applies to the three-variable quadratic form $x^2 + y^2 + cz^2$ for all integers $c$ from 2 to 10 except $c = 7$ and $c = 10$ . What happens for these two special values? Ramanujan looked at $c = 10$ . He found 16 values of $n$ for which there is local solvability (that is, we can solve $x^2 + y^2 + 10z^2 \equiv n \bmod m$ for all $m$ ) but not global solvability (no integral solution for $x^2 + y^2 + 10z^2 = n$ ). Two additional values of $n$ were found later, and in 1990 Duke and Schulze-Pillot showed that local solvability implies global solvability except for (ineffectively) finitely many positive integers $n$ . In 1997, Ono and Soundararajan showed GRH implies the 18 known exceptions are the only ones. l) Euler's convenient numbers. Euler called an integer $n \geq 1$ convenient if each odd integer greater than 1 that has a unique representation as $x^2 + ny^2$ in positive integers $x$ and $y$ , and which moreover has $(x,ny) = 1$ , is a prime number. (These numbers were convenient for Euler to use to prove certain numbers that were large in his day, like $67579 = 229^2 + 2\cdot 87^2$ , are prime.) Euler found 65 convenient numbers below 10000 (the last one being 1848). In 1934, Chowla showed there are finitely many convenient numbers. In 1973, Weinberger showed there is at most one convenient number not in Euler's list, and that GRH for $L$ -functions of all quadratic Dirichlet characters implies that Euler's list of convenient numbers is complete. What he needed from GRH is the lack of real zeros in the interval $(53/54,1)$ . m) Removing a condition in the Brauer-Siegel theorem. In 1947, Brauer proved the Brauer-Siegel theorem for sequences of number fields $K_n$ such that (i) $[K_n:\mathbf Q]/\log |{\rm disc}(K_n)| \to 0$ as $|{\rm disc}(K_n)| \to \infty$ and (ii) $K_n$ is Galois over $\mathbf Q$ . If the zeta-functions $\zeta_{K_n}(s)$ all satisfy GRH (what is actually needed is no real zero in $(1/2,1)$ for those zeta-functions) then we can drop condition (ii). That is, GRH implies the Brauer-Siegel theorem holds for sequences of number fields $K_n$ fitting condition (i). n) Lower bounds on root discriminants. In 1975, Odlyzko showed GRH for zeta-functions of number fields implies a lower bound on root discriminants of number fields: $$
|{\rm disc}(K)|^{1/n} \geq (94.69...)^{r_1/n}(28.76...)^{2r_2/n} + o(1)
$$ as $n = [K:\mathbf Q] \to \infty$ . (The lower bound was improved later to $136^{r_1/n}34.5^{2r_2/n}$ for sufficiently large $n$ .) Building on ideas of Stark, he also showed that GRH for zeta-functions of number fields implies there are only finitely many CM number fields with a given class number (a big generalization of the known fact that only finitely many imaginary quadratic fields have any particular class number). o) Lower bounds on class numbers. In 1990, Louboutin showed GRH for zeta-functions of imaginary quadratic fields (what is really needed is the lack of real zeros in $(1/2,1)$ for these zeta-functions) implies the lower bound $h(\mathbf Q(\sqrt{-d})) \geq (\pi/(3e))\sqrt{d}/\log d$ for imaginary quadratic fields with discirminant $-d$ . The point here is the explicit constant factor $\pi/(3e)$ . (Hecke had shown such a lower bound with a "computable constant" $c$ in place of $\pi/(3e)$ but he did not compute the constant.) Without GRH, lower bounds for $h(\mathbf Q(\sqrt{-d}))$ are on the order of $\log d$ , which is far smaller than $\sqrt{d}/\log d$ . For example, Louboutin's GRH-based lower bound shows if $h(\mathbf Q(\sqrt{-d})) \leq 100$ then $d \leq $ 18,916,898. To put this 8-digit upper bound in perspective,
when Watkins determined all imaginary quadratic fields with class number up to 100 in 2004, he used lower bounds on $h(\mathbf Q(\sqrt{-d}))$ that do not depend on GRH and the upper bound of his search space for $d$ was $e^{298368000}$ , a number with 129,579,576 digits. Just the exponent in that upper bound on $d$ is greater than the upper bound on $d$ coming from GRH. p) Proof of André-Oort conjecture. In 2014, Klingler and Yafaev showed GRH for zeta-functions of CM number fields implies the André-Oort conjecture. Daw and Orr gave another proof also using the same version of GRH. Update: This is now an obsolete application of GRH since the André-Oort conjecture has been proved without GRH. See here . q) Class number calculations. In 2015, J. C. Miller showed that GRH implies the class number $h_p^+$ of the real cyclotomic field $\mathbf Q(\zeta_p)^{+}$ for prime $p$ is $1$ for all $p$ from 157 to 241 except $h_{163}^+ = 4$ , $h_{191}^{+} = 11$ , and $h_{229}^+ = 3$ . r) Elliptic curve rank values. In 2019, Klagsbrun, Sherman, and Weigandt used GRH for $L$ -functions of elliptic curves and for zeta-functions of number fields to prove that the elliptic curve found by Elkies in 2006 with 28 independent rational points has rank equal to 28.
|
{
"source": [
"https://mathoverflow.net/questions/17209",
"https://mathoverflow.net",
"https://mathoverflow.net/users/828/"
]
}
|
17,226 |
Recently, the Department of Mathematics at
our University issued a recommendation encouraging its members
to publish their research in non-specialized, mainstream mathematical journals. For
numerical analysts this will make an additional obstacle for their promotions. But even
for discrete mathematicians this recommendation is causing concerns. For several top mainstream journals I checked with tools offered by MathSciNet what percentage
of discrete mathematical papers they published in recent years. Some statistics indicate that in some journals the number of papers with primary MSC classification, say 05 or 06 decreased significantly in the past 30 years. There are several possible explanations to this fact. The quality of research in DM is dropping. The majority of research in discrete mathematics is so specialized that it is of no interest for the rest of mathematics Some discrete mathematics journals attract even the best work of discrete mathematicians. Some top journals may be biased against discrete math. Maybe discrete math is no longer part of mainstream mathematics and will, like theoretical computer science, eventually develop into an independent body of research. But the key issue is whether discrete math is nowadays perceived as mainstream mathematics.
|
There are two different questions here, one objective and one subjective. I will try to give my view, for what it's worth. Bear with me. First, you are asking what is the publication history of discrete mathematics? (Even if I suspect you know this much better than I do.) Well, originally there was no such thing as DM. If I understand the history correctly, classical papers like A logical expansion in mathematics by Hassler Whitney (on coefficients of chromatic polynomials) were viewed as contributions to "mainstream mathematics". What happened is that starting maybe late 60s there was a rapid growth in the number of papers in mathematics in general, with an even greater growth in discrete mathematics. While the overall growth is relatively easy to explain as a consequence of expansion of graduate programs, the latter is more complicated. Some would argue that CS and other applications spurned the growth, while others would argue that this area was neglected for generations and had many easy pickings, inherent in the nature of the field. Yet others would argue that the growth is a consequence of pioneer works by the "founding fathers", such as Paul Erdős, Don Knuth, G.-C. Rota, M.-P. Schützenberger, and W.T. Tutte, which transformed the field. Whatever the reason, the "mainstream mathematics" felt a bit under siege by numerous new papers, and quickly closed ranks. The result was a dozen new leading journals covering various subfields of combinatorics, graph theory, etc., and few dozen minor ones. Compare this with the number of journals dedicated solely to algebraic geometry to see the difference. Thus, psychologically, it is very easy to explain why journals like Inventiones even now have relatively few DM papers - if the DM papers move in, the "mainstream papers" often have nowhere else to go. Personally, I think this is all for the best, and totally fair. Now, your second question is whether DM is a "mainstream mathematics", or what is it? This is much more difficult to answer since just about everyone has their own take. E.g. miwalin suggests above that number theory is a part of DM, a once prevalent view, but which is probably contrary to the modern consensus in the field. Still, with the growth of "arithmetic combinatorics", part of number theory is definitely a part of DM. While most people would posit that DM is "combinatorics, graph theory + CS and other applications", what exactly are these is more difficult to decide. The split of Journal of Combinatorial Theory into Series A and B happened over this kind of disagreement between Rota and Tutte (still legendary). I suggest combinatorics wikipedia page for a first approximation of the modern consensus, but when it comes to more concrete questions this becomes a contentious issue sometimes of "practical importance". As an editor of Discrete Mathematics , I am routinely forced to decide whether submissions are in scope or not. For example if someone submits a generalization of R–R identities — is that a DM or not? (If you think it is, are you sure you can say what exactly is "discrete" about them?) Or, e.g. is Cauchy theorem a part of DM, or metric geometry, or both? (or neither?) How about "IP = PSPACE" theorem? Is that DM, or logic, or perhaps lies completely outside of mathematics? Anyway, my (obvious) point is that there is no real boundary between the fields. There is a large spectrum of papers in DM which fall somewhere in between "mainstream mathematics" and applications. And that's another reason to have separate "specialized" journals to accommodate these papers, rather than encroach onto journals pre-existing these new subfields. Your department's "encouragement" to use only the "mainstream mathematical journals" for promotion purposes is narrow minded and very unfortunate.
|
{
"source": [
"https://mathoverflow.net/questions/17226",
"https://mathoverflow.net",
"https://mathoverflow.net/users/4400/"
]
}
|
17,230 |
Let $\rho : S_n \rightarrow \text{GL}(n, \mathbb{C})$ be the homomorphism mapping a permutation $g$ to its permutation matrix. Let $\chi(g) = \text{Trace}(\rho(g))$. What is the value of $\langle \chi, \chi \rangle = \displaystyle \frac{1}{n!} \sum_{g \in S_n} \chi(g)^2$ ? Computing this expression for small $n$ yields $2$. Is this always true?
|
You are computing the inner product of $\chi$ with itself. Since $\chi=\mathrm{triv}+\mathrm{std}$ as a $S_n$-module, with $\mathrm{triv}$ being the trivial reprsentation, and $\mathrm{std}$ its orthogonal complement, which is an irreducible $S_n$-module, and since the inner product is, well, an inner product and distinct irreducible characters are orthogonal, your $2$ follows from $$\langle\chi,\chi\rangle=\langle\mathrm{triv}+\mathrm{std},\mathrm{triv}+\mathrm{std}\rangle=\langle\mathrm{triv},\mathrm{triv}\rangle+\langle\mathrm{std},\mathrm{std}\rangle=1+1$$.
|
{
"source": [
"https://mathoverflow.net/questions/17230",
"https://mathoverflow.net",
"https://mathoverflow.net/users/4197/"
]
}
|
17,325 |
You don't need a metric to define the differential of a function,
and the cotangent bundle carries a canonical one-form. But you do need a metric to define the gradient, and the
tangent bundle does not have a canonical vector field. These are not difficult truths, but still... why the preference
toward "co"?
|
If you want to differentiate functions from a manifold to (say) the real line R, then you want to use the cotangent bundle on the manifold. If instead you want to to differentiate functions to the manifold from the real line (i.e. parameterised curves), then you want to use the tangent bundle on the manifold. So the preference comes from whether you want to use the manifold as the domain or as the range of the functions one is differentiating.
|
{
"source": [
"https://mathoverflow.net/questions/17325",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1186/"
]
}
|
17,344 |
In a paper I developed some theory; some of the applications require extensive computations that are not part of the paper. I wrote a Mathematica notebook. I want to publish a PDF and .nb version somewhere to refer to from the paper. arXiv.org seems a good choice, but they won't accept the .nb file. I do not want to put this through peer review. Where to publish it?
|
You should upload the notebook along with the sources of the actual article to the arXiv, of course. The official (but rather hard to find) advice on this from the arXiv is to place your code in a directory called /aux/. (This is problematic for windows users.) You can see an example of this in my recent paper on the extended Haagerup subfactor. A footnote in the text of the article explains that the code is available along with the source download. It would also be appropriate to use the comments field in an arxiv submission to explain that source code is available in the /aux/ directory.
|
{
"source": [
"https://mathoverflow.net/questions/17344",
"https://mathoverflow.net",
"https://mathoverflow.net/users/4438/"
]
}
|
17,501 |
I was asked the following question by a colleague and was embarrassed not to know the answer. Let $f(x), g(x) \in \mathbb{Z}[x]$ with no root in common. Let $I = (f(x),g(x))\cap \mathbb{Z}$, that is, the elements of $\mathbb{Z}$ which are linear combinations of $f(x), g(x)$ with coefficients in $\mathbb{Z}[x]$. Then $I$ is clearly an ideal in $\mathbb{Z}$. Let $D>0$ be a generator of this ideal. The question is: what is $D$? Now, we do have the standard resultant $R$ of $f,g$, which under our hypotheses, is a non-zero integer. We know that $R \in I$ and it's not hard to show that a prime divides $R$ if and only if it divides $D$. I thought $R = \pm D$ but examples show that this is not the case.
|
This quantity $D$ is known as the "congruence number" or "reduced resultant" of the polynomials f and g. I first saw this in a preprint by Wiese and Taixes i Ventosa, http://arxiv.org/abs/0909.2724 . They ascribe the concept to a paper which I don't have a copy of: M. Pohst. A note on index divisors . In Computational number theory (Debrecen, 1989) , 173–
182, de Gruyter, Berlin, 1991.
|
{
"source": [
"https://mathoverflow.net/questions/17501",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2290/"
]
}
|
17,532 |
Given a category $C$ and a commutative ring $R$, denote by $RC$ the $R$-linearization: this is the category enriched over $R$-modules which has the same objects as $C$, but the morphism module between two objects $x$ and $y$ is the free $R$-module on $\operatorname{Hom}_C(x,y)$. Thus in $RC$ we allow arbitrary $R$-linear combinations of morphisms from the original category $C$. Question: if two objects in $x$, $y \in C$ are isomorphic in $RC$, are they already isomorphic in $C$? I do not know the answer to this question for any nontrivial ring $R$, but I'm particularly interested in $R=\mathbb{Z}$ and $R=\mathbb{Z}/2\mathbb{Z}$. What's obviously not true is that every isomorphism in $RC$ comes from an isomorphism in $C$ (take $-id_x$). (Thus the word "isomorphism" in the title refers to a relation on objects rather than to a property of morphisms.) Of course, it is enough to consider categories $C$ with two objects $x$, $y$, but we cannot assume that $C$ is finite. It's fairly elementary to see that if $x$ and $y$ are isomorphic in $RC$ then in $C$, $x$ is a retract of $y$ and vice versa, but the latter does in general not imply that $x \cong y$. A more catchy way of phrasing this problem is: can we always classify objects in a category up to isomorphism by means of functors taking values in $R$-linear categories? (The inclusion $C \to RC$ is the universal such functor.) Edit : A lot of people have posted an "answer" that wasn't, and deleted it, so here's something that will not work, to save others going down the same road. I said that we cannot assume that the category is finite; in fact, it must be infinite. Here is an elementary argument: Since $x$ and $y$ are mutual retracts, there are maps $f,\;f'\colon x \to y$ and $g,\;g'\colon y \to x$ with $fg=\operatorname{id}$ and $g'f'=id$. Consider the powers of $fg' \in \operatorname{End}(y)$. If $\operatorname{End}(y)$ is finite then $(fg')^n = (fg')^m$ for some $m \neq n$; since $fg'$ has a right inverse (viz, $f'g$), we must have that $(fg')^n=\operatorname{id}$ for some $n>0$. So we see that $g'$ has not only a right inverse ($f'$) but also a left inverse: $(fg')^{n-1}f$. So they are the same and $g'$ is already an isomorphism.
|
Hi Tilman. I believe I proved that (in your language) linearization reflects isomorphism. The following is a sketch. I will send you a more detailed version. The general case may be reduced to the case of prime fields $F_p$ and certain categories $C$ with fixed objects $x$ and $y$ and morphisms $f_1,\dots,f_m\colon x\to y$ and $g_1,\dots,g_n\colon y\to x$ subject to relations which correspond to the fact that $u=f_1+\dots+f_m$ and $u^{-1}=g_1+\dots+g_n$ are mutually inverse in the $F_p$-linearization. Apart from trivial cases, we may reindex these generators such that $f_1g_1 = 1_y$ and $g_nf_m=1_x$, while the other summands in the expansion of $uu^{-1}$ and $u^{-1}u$, respectively, fall into equivalence classes whose size is a multiple of $p$. It is then possible to derive a sequence of pairs
$(i_1,j_1),(i_2,j_2),\dots,(i_k,j_k)$
such that $f_{i_r}g_{j_r} = f_{i_{r+1}}g_{j_{r+1}}$ for $r=2,3,\dots,k-1$ and $g_{j_r}f_{i_r}=g_{j_{r+1}}f_{i_{r+1}}$ for $r=1,3,\dots,k-2$. Then
$f_{i_1}g_{j_2}f_{i_3}g_{j_4}\dots f_{i_k}$ and
$g_{j_k}f_{i_{k-1}}g_{j_{k-2}}f_{i_{k-3}}\dots g_{j_1}$
are mutual inverses of $C$.
|
{
"source": [
"https://mathoverflow.net/questions/17532",
"https://mathoverflow.net",
"https://mathoverflow.net/users/4183/"
]
}
|
17,545 |
Because the theory of sheaves is a functorial theory, it has been adopted in algebraic geometry (both using the functor of points approach and the locally ringed space approach) as the "main theory" used to describe geometric data. All sheaf data in the LRS approach can be described by bundles using the éspace étalé construction. It's interesting to notice that the sheafification of a presheaf is the sheaf of sections of the associated éspace étalé. However, in differential geometry, bundles are for some reason preferred. Is there any reason why this is true? Are there some bundle constructions which don't have a realization as a sheaf? Are there advantages to the bundle approach?
|
If $X$ is a manifold, and $E$ is a smooth vector bundle over $X$ (e.g. its tangent bundle),
then $E$ is again a manifold. Thus working with bundles means that one doesn't have to leave
the category of objects (manifolds) under study; one just considers manifold with certain extra structure (the bundle structure). This is a big advantage in the theory; it avoids introducing another class of objects (i.e. sheaves), and allows tools from the theory of manifolds to be applied directly to bundles too. Here is a longer discussion, along somewhat different lines: The historical impetus for using sheaves in algebraic geometry comes from the theory of several complex variables, and in that theory sheaves were introduced, along with cohomological techniques, because many important and non-trivial theorems can be stated as saying that certain sheaves are generated by their global sections, or have vanishing higher cohomology. (I am thinkin of Cartan's Theorem A and B, which have as consequences many
earlier theorems in complex analysis.) If you read Zariski's fantastic report on sheaves in algebraic geometry, from the 50s, you will see a discussion by a master geometer of how sheaves, and especially their cohomology,
can be used as a tool to express, and generalize, earlier theorems in algebraic geometry.
Again, the questions being addressed (e.g. the completeness of the linear systems of hyperplane sections) are about the existence of global sections, and/or vanishing of higher cohomology. (And these two usually go hand in hand; often one establishes existence results about global sections of one sheaf by showing that the higher cohomology of some related sheaf vanishes, and using a long exact cohomology sequence.) These kinds of questions typically don't arise in differential geometry. All the sheaves
that might be under consideration (i.e. sheaves of sections of smooth bundles) have global sections in abundance, due to the existence of partions of unity and related constructions. There are difficult existence problems in differential geometry, to be sure: but these are very often problems in ODE or PDE, and cohomological methods are not what is required to solve them (or so it seems, based on current mathematical pratice). One place where a sheaf theoretic perspective can be useful is in the consideration of flat (i.e. curvature zero) Riemannian manifolds; the fact that the horizontal sections of a bundle with flat connection form a local system, which in turn determines the
bundle with connection, is a useful one, which is well-expressed in sheaf theoretic language. But there are also plenty of ways to discuss this result without sheaf-theoretic language, and in any case, it is a fairly small part of differential geometry, since typically the curvature of a metric doesn't vanish, so that sheaf-theoretic methods don't seem to have much to say. If you like, sheaf-theoretic methods are potentially useful for dealing with problems,
especially linear ones, in which local existence is clear, but the objects are suffiently rigid that there can be global obstructions to patching local solutions. In differential geomtery, it is often the local questions that are hard: they become difficult non-linear PDEs. The difficulties are not of the "patching local solutions"
kind. There are difficult global questions too, e.g. the one solved by the Nash embedding theorem, but again, these are typically global problems of a very different type to those that are typically solved by sheaf-theoretic methods.
|
{
"source": [
"https://mathoverflow.net/questions/17545",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1353/"
]
}
|
17,560 |
I'm fascinated by this open problem (if it is indeed still that) and every few years I try to check up on its status. Some background: Let $x$ be a positive real number. If $n^x$ is an integer for every $n \in \mathbb{N}$ then $x$ must be an integer. This is a fun little puzzle. If $2^x$, $3^x$ and $5^x$ are integers then $x$ must be an integer. This requires fairly sophisticated tools, and can be derived from the results in e.g. Lang, Algebraic values of meromorphic functions. II., Topology 5, 1966. Finally, if all you know is that $2^x$ and $3^x$ are integers, then as far as I know it is not known if $x$ is forced to be an integer (unbelievable, isn't it?). Although of course one can never be certain , I am quite sure this was still the case as recently as 2003. So the question is, is that still an open problem, and is there any sort of relevant progress that may provide some hope?
|
Still open, to the best of my knowledge. The $2^x,3^x,5^x$ result follows from the Six Exponentials Theorem , q.v., and the $2^x,3^x$ would follow from the Four Exponentials Conjecture, q.v.
|
{
"source": [
"https://mathoverflow.net/questions/17560",
"https://mathoverflow.net",
"https://mathoverflow.net/users/25/"
]
}
|
17,578 |
I've had a few undergraduate students ask me for references for the classical fact (due to Rado) that closed topological surfaces can be triangulated. I know two sources for this, namely Ahlfors's book on Riemann surfaces and Moise's book "Geometric topology in dimensions 2 and 3". Both of these strike me as being a bit much for a bright undergraduate. Question : in the 30+ years since Moise's book, has anyone written a more accessible account?
|
[Three years later …] All the published proofs of triangulability of surfaces that I am aware of use the Schoenflies theorem, which is not exactly an easy thing to prove. There is however another line of proof which avoids the Schoenflies theorem and instead uses the Kirby torus trick that underlies Kirby-Siebenmann theory in higher dimensions. There is a 1974 paper by A.J.S.Hamilton that gives much simpler proofs of Moise's theorems on triangulability of 3-manifolds using the torus trick, and the same ideas can be applied even more simply for surfaces. Instead of the Schoenflies theorem one needs a few results about surfaces strictly in the PL (or smooth if one prefers) category. Namely, one needs to know that PL structures are unique up to PL homeomorphism in the following four cases: $S^1\times S^1$ , $S^1\times{\mathbb R}$ , $[0,1]\times{\mathbb R}$ , and $D^2$ . These can be regarded as special cases of the usual classification theorem for compact PL surfaces, extended to include a few noncompact cases. I haven't seen this proof in the literature, so I've written it up as a short expository paper "The Kirby torus trick for surfaces" and posted it on the arXiv here , working in the smooth category rather than the PL category. It's not clear how suitable this proof would be for an undergraduate course. Besides the ingredients mentioned above, a little basic covering space theory is also needed. If one were in the fortunate position of already having covered these things, then this proof might be accessible to undergraduates. On the other hand, it could be of some interest to go through a proof of the often-quoted-but-seldom-proved Schoenflies theorem. (In this connection I might mention a paper by Larry Siebenmann on the Schoenflies theorem in the Russian Math Surveys in 2005, giving history as well as a proof.)
|
{
"source": [
"https://mathoverflow.net/questions/17578",
"https://mathoverflow.net",
"https://mathoverflow.net/users/317/"
]
}
|
17,605 |
I found the following equation on some web page I cannot remember, and found it interesting: $$f(f(x))=\cos(x)$$ Out of curiosity I tried to solve it, but realized that I do not have a clue how to approach such an iterative equation except for trial and error. I also realized that the solution might not be unique, from the solution of a simpler problem $$f(f(x)) = x$$ which has, for example, solutions $f(x) = x$ and $f(x) = \frac{x+1}{x-1}$ . Is there a general solution strategy to equations of this kind? Can you perhaps point me to some literature about these kind of equations? And what is the solution for $f(f(x))=\cos(x)$ ?
|
There are no continuous solutions. Since the cosine has a unique fixed point $x_0$ (such that $\cos x_0=x_0$ ), it should be a fixed point of $f$ . And f should be injective and hence monotone (increasing or decreasing) in a neighborhood of $x_0$ . Then $f(f(x))$ is increasing in a (possibly smaller) neighborhood of $x_0$ while the cosine is not. As for discontinuous ones, there are terribly many of them ( $2^{\mathbb R}$ ) and you probably cannot parametrize them in any reasonable way. You can describe them in terms of orbits of iterations of $\cos x$ , but I doubt this would count as a solution of the equation. UPDATE: Here is how to construct a solution (this is technical and I might overlook something). Let X be an infinite set and $g:X\to X$ is a map, I am looking for a sufficient conditions for the existence of a solution of $f\circ f=g$ . Define the following equivalence relation on $X$ : $x$ and $y$ are equivalent iff $g^n(x)=g^m(y)$ for some positive integers $m$ and $n$ . Equivalence classes will be referred to as orbits (the term is wrong but I don't know what is a correct one). Two orbits are said to be similar is there is a bijection between them commuting with $g$ . If $Y$ and $Z$ are two similar orbits, one can define $f$ on $Y\cup Z$ as follows: on $Y$ , $f$ is that bijection to $Z$ , and on $Z$ , $f$ is the inverse bijection composed with $g$ . So if the orbits can be split into pairs of similar ones, we have a desired $f$ . Now remove from the real line the fixed point of cos and all its roots ( $\pi/2$ and the like). Then, if I am not missing something, in the remaining set $X$ all orbits of $\cos$ are similar, so we can define $f$ as above. Define $f$ so that $0$ has a nonempty pre-image (that is, the orbit containing $0$ should be used as $Z$ and not as $Y$ ). Finally, map the fixed point of $\cos$ to itself, and the roots of $\cos$ to some pre-image of $0$ .
|
{
"source": [
"https://mathoverflow.net/questions/17605",
"https://mathoverflow.net",
"https://mathoverflow.net/users/4503/"
]
}
|
17,608 |
If ZFC (Zermelo-Fraenkel set theory with the Axiom of Choice) is consistent, does it remain consistent
when the following statement is added to it as a new axiom? "There exists a denumerably infinite and ordinal definable set of real numbers, not all of whose elements
are ordinal definable" If the answer to the above question is negative, then it must be provable in ZFC that every denumerably
infinite and ordinal definable set of real numbers is hereditarily ordinal definable. This is because
every real number can be regarded as a set of finite ordinal numbers and every finite ordinal number is
ordinal definable.
Garabed Gulbenkian
|
The original problem solves in the positive: there is a model of ZFC in which there exists a countable OD (well, even lightface $\Pi^1_2$, which is the best possible) set of reals $X$ containing no OD elements. The model (by the way, as conjectured by Ali Enayat at http://cs.nyu.edu/pipermail/fom/2010-July/014944.html ) is a $\mathbf P^{<\omega}$-generic extension of $L$, where $\mathbf P$ is Jensen's minimal $\Pi^1_2$ real singleton forcing and $\mathbf P^{<\omega}$ is the finite-support product of $\omega$ copies of $\mathbf P$. A few details. Jensen's forcing is defined in $L$ so that $\mathbf P =\bigcup_{\xi<\omega_1} \mathbf P_\xi$, where each $\mathbf P_\xi$ is a ctble set of perfect trees in $2^{<\omega}$, generic over the outcome $\mathbf P_{<\xi}=\bigcup_{\eta<\xi}\mathbf P_\eta$ of all earlier steps in such a way that any $\mathbf P_{<\xi}$-name $c$ for a real ($c$ belongs to a minimal countable transitive model of a fragment of ZFC, containing $\mathbf P_{<\xi}$), which $\mathbf P_{<\xi}$ forces to be different from the generic real itself, is pushed by $\mathbf P_{\xi}$ (the next layer) not to belong to any $[T]$ where $T$ is a tree in $\mathbf P_{\xi}$. The effect is that the generic real itself is the only $\mathbf P$-generic real in the extension, while examination of the complexity shows that it is a $\Pi^1_2$ singleton. Now let $\mathbf P^{<\omega}$ be the finite-support product of $\omega$ copies of $\mathbf P$. It adds a ctble sequence of $\mathbf P$-generic reals $x_n$. A version of the argument above shows that still the reals $x_n$ are the only $\mathbf P$-generic reals in the extension and the set $\{x_n:n<\omega\}$ is $\Pi^1_2$. Finally the routine technique of finite-support-product extensions ensures that $x_n$ are not OD in the extension. Addendum. For detailed proofs of the above claims, see this manuscript . Jindra Zapletal informed me that he got a model where a $\mathsf E_0$-equivalence class $X=[x]_{E_0}$ of a certain Silver generic real is OD and contains no OD elements, but in that model $X$ does not seem to be analytically definable, let alone $\Pi^1_2$. The model involves a combination of several forcing notions and some modern ideas in descriptive set theory recently presented in Canonical Ramsey Theory on Polish Spaces . Thus whether a $\mathsf E_0$-class of a non-OD real can be $\Pi^1_2$ is probably still open. Further Kanovei's addendum of Aug 23 .
It looks like a clone of Jensen's forcing on the base of Silver's (or $\mathsf E_0$-large Sacks) forcing instead of the simple Sacks one leads to a lightface $\Pi^1_2$ generic $\mathsf E_0$-class with no OD elements. The advantage of Silver's forcing here is that it seems to produce a Jensen-type forcing closed under the 0-1 flip at any digit, so that the corresponding extension contains a $\mathsf E_0$-class of generic reals instead of a generic singleton. I am working on details, hopefully it pans out. Further Kanovei's addendum of Aug 25 .
Yes it works, so there is a generic extension $L[x]$ of $L$ by a real in which the
$\mathsf E_0$-class $[x]_{\mathsf E_0}$ is a lightface $\Pi^1_2$ (countable) set with no OD elements. I'll send it to Axriv in a few days. Further Kanovei's addendum of Aug 29 . arXiv:1408.6642
|
{
"source": [
"https://mathoverflow.net/questions/17608",
"https://mathoverflow.net",
"https://mathoverflow.net/users/4423/"
]
}
|
17,614 |
This question is of course inspired by the question How to solve f(f(x))=cosx and Joel David Hamkins' answer, which somehow gives a formal trick for solving equations of the form $f(f(x))=g(x)$ on a bounded interval. [EDIT: actually he can do rather better than this, solving the equation away from a bounded interval (with positive measure)]. I've always found such questions ("solve $f(f(x))=g(x)$") rather vague because I always suspect that solutions are highly non-unique, but here are two precise questions which presumably are both very well-known: Q1) Say $g:\mathbf{R}\to\mathbf{R}$ is an arbitrary function. Is there always a function $f:\mathbf{R}\to\mathbf{R}$ such that $f(f(x))=g(x)$ for all $x\in\mathbf{R}$? Q2) If $g$ is as above but also assumed continuous, is there always a continuous $f$ as above? The reason I'm asking is that these questions are surely standard, and perhaps even easy, but I feel like I know essentially nothing about them. Apologies in advance if there is a well-known counterexample to everything. Of course Q1 has nothing to do with the real numbers; there is a version of Q1 for every cardinal and it's really a question in combinatorics. EDIT: Sergei Ivanov has answered both of these questions, and Gabriel Benamy has raised another, which I shall append to this one because I only asked it under an hour ago: Q3) if $g$ is now a continuous function $\mathbf{C}\to\mathbf{C}$, is there always continuous $f$ with $f(f(x))=g(x)$ for all $x\in\mathbf{C}$? EDIT: in the comments under his answer Sergei does this one too, and even gives an example of a continuous $g$ for which no $f$, continuous or not, can exist. Related MO questions: f(f(x))=exp(x) and other functions just in the middle between linear and exponential , and Does the exponential function has a square root .
|
Q1: No. Let $g(0)=1, g(1)=0$ and $g(x)=x$ for all $x\in\mathbb R\setminus\{0,1\}$ .
Assuming $f\circ f=g$ , let $a=f(0)$ , then $f(a)=1$ and $f(1)=g(a)=a$ since $a\notin\{0,1\}$ .
Then $g(1)=f(f(1))=f(a)=1$ , a contradiction. Q2: No. Let $g(x)=-x$ or, in fact, any decreasing function $\mathbb R\to\mathbb R$ . Then $f$ must be injective and hence monotone. Whether $f$ is increasing or decreasing, $f\circ f$ is increasing.
|
{
"source": [
"https://mathoverflow.net/questions/17614",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1384/"
]
}
|
17,617 |
I'm merely a grad student right now, but I don't think an exploration of the sporadic groups is standard fare for graduate algebra, so I'd like to ask the experts on MO. I did a little reading on them and would like some intuition about some things. For example, the order of the monster group is over $8\times 10^{53}$, yet it is simple, so it has no
normal subgroups...how? What is so special about the prime factorization of its order? Why is it $2^{46}$ and not $2^{47}$? Why is it not possible to extend it to obtain that additional power of 2 without creating a normal subgroup? Some of the properties seem really arbitrary, and yet must be very fundamental to the algebra of groups. I don't think I'm the only person curious about this, but I hesitated posting due to my relative inexperience.
|
The question seems to be made of several smaller questions, so I'm afraid my answer may not seem entirely coherent. I have to agree with the other posters who say that the sporadic simple groups are not really so large. For example, we humans can write down the full decimal expansions of their orders, where a priori one might think we'd have to resort to crude upper bounds using highly recursive functions. (In contrast, one could say that almost of the groups in the infinite families are too large for their orders to have a computable description that fits in the universe.) Furthermore, as of 2002 we can load matrix representatives of elements into a computer, even for the monster. Noah pointed out that the monster has a smaller order than $A_{50}$, but I think a more apt comparison is that the monster has a smaller order than even the smallest member of the infinite $E_8$ family. Of course, one could ask why $E_8$ has dimension as large as 248... There was a more explicit question: how is it possible that a group with as many as $8 \times 10^{ 53 }$ elements doesn't have any normal subgroups? I think the answer is that the order of magnitude of a group says very little about its complexity. There are prime numbers very close to the order of the monster, and there are simple cyclic groups of those orders, so you might ask yourself why that fact doesn't seem as conceptually disturbing. Perhaps slightly more challenging is the fact that there aren't any elements of order greater than 119, but again, there is work on the bounded and restricted Burnside problems that shows that you can have groups of very small exponent that are extremely complicated. A second point regarding the large lower bound on order is that there are smaller groups that could be called sporadic, in the sense that they fit into reasonably natural (finite) combinatorial families together with the sporadics, but they aren't designated as sporadic because small-order isomorphisms get in the way. For example, the Mathieu group $M_{10}$ is the symmetry group of a certain Steiner system, much like the simple Mathieu groups, and it is an index 11 subgroup of $M_{11}$. While it isn't simple, it contains $A_6$ as an index 2 subgroup, and no one calls $A_6$ sporadic. Similarly, we describe the 20 "happy family" sporadic subquotients of the monster, but we forget about the subquotients like $A_5$, $L_2(11)$, and so on. Since the order of a nonabelian simple group is bounded below by 60, there isn't much room to maneuver before you get to 7920, a.k.a. "huge" range. The question about the why the 2-Sylow subgroup has a certain size is rather subtle, and I think a good explanation would require delving into the structure of the classification theorem. A short answer is that centralizers of order 2 elements played a pivotal role in the classification after the Odd Order Theorem, and there was a separation into cases by structural features of centralizers. One of the cases involved a centralizer that ended up having the form $2^{1 + 24} . Co1$, which has a 2-Sylow subgroup of order $2^{46}$ (and naturally acts on a double cover of the Leech lattice). This is the case that corresponds to the monster. Regarding the prime factorization of the order of the monster, the primes that appear are exactly the supersingular primes, and this falls into the general realm of "monstrous moonshine". I wrote a longer description of the phenomenon in reply to Ilya's question , but the question of a general conceptual explanation is still open. I'll mention some folklore about the organization of the sporadics. There seems to be a hierarchy given by level 0: subquotients of $M_{24}$ = symmetries of the Golay code level 1: subquotients of $Co1$ = symmetries of the Leech lattice, mod $\{ \pm 1 \}$ level 2: subquotients of the monster = conformal symmetries of the monster vertex algebra where the groups in each level naturally act on (objects similar to) the exceptional object on the right. I don't know what explanatory significance the sequence [codes, lattices, vertex algebras] has, but there are some level-raising constructions that flesh out the analogy a bit. One interesting consequence of the existence of level 2 is that for some finite groups, the most natural (read: easiest to construct) representations are infinite dimensional, and one can reasonably argue using lattice vertex algebras that this holds for some exceptional families as well. John Duncan has some recent work constructing structured vertex superalgebras whose automorphism groups are sporadic simple groups outside the happy family. I think one interesting question that has not been suggested by other responses (and may be too open-ended for MO) is why the monster has no small representations. There are no faithful permutation representations of degree less than $9 \times 10^{ 19 }$ and there are no faithful linear representations of dimension less than 196882. Compare this with the cases of the numerically larger groups $A_{50}$ and $E_8(\mathbb{F}_2)$, where we have linear representations of dimension 49 and 248. This is a different sense of hugeness than in the original question, but one that strongly impacts the computational feasibility of attacking many questions.
|
{
"source": [
"https://mathoverflow.net/questions/17617",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1876/"
]
}
|
17,662 |
Prof. Conrad mentioned in a recent answer that most of the (introductory?) books on reductive groups do not make use of scheme theory. Do any books using scheme theory actually exist? Further, are there any books that use the functor of points approach? Demazure-Gabriel's second book would have covered general group schemes in this way, but it was never written, and it's not clear whether or not it would have covered reductive groups anyway. There is a lot of material in SGA 3 using more modern machinery to study group schemes, but I'm not aware of any significant treatment of reductive groups in that book (although I haven't read very much of it). Correction: Prof. Conrad has noted that SGA 3 does contain a significant treatment of reductive groups using modern machinery.
|
Personally, I find the "classical" books (Borel, Humphreys, Springer) unpleasant to read because they work in the wrong category, namely, that of reduced algebraic group schemes rather than all algebraic group schemes. In that category, the isomorphism theorems in group theory fail, so you never know what is true. For example, the map $H/H\cap N\rightarrow HN/N$ needn't be an isomorphism (take $G=GL_{p}$, $H=SL_{p}$, $N=\mathbb{G}_{m}$ embedded diagonally). Moreover, since the terminology they use goes back to Weil's Foundations, there are strange statements like "the kernel of a homomorphism of algebraic groups defined over $k$ need not be defined over $k$". Also I don't agree with Brian that if you don't know descent theory, EGA, etc. then you don't "know scheme theory well enough to be asking for a scheme-theoretic treatment'. Which explains why I've been working on a book whose goal is to allow people to learn the theory of algebraic group schemes (including the structure of reductive algebraic group schemes) without first reading the classical books and with only the minimum of prerequisites (for what's currently available, see my website under course notes). In a sense, my aim is to complete what Waterhouse started with his book. So my answer to the question is, no, there is no such book, but I'm working on it....
|
{
"source": [
"https://mathoverflow.net/questions/17662",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1353/"
]
}
|
17,736 |
Consider the Sobolev spaces $W^{k,p}(\Omega)$ with a bounded domain $\Omega$ in n-dimensional Euclidean space. When facing the different embedding theorems for the first time, one can certainly feel lost. Are there certain tricks to memorize the (continuous and compact) embeddings between the different $W^{k,p}(\Omega)$ or into $C^{r,\alpha}(\bar{\Omega})$ ?
|
Sobolev norms are trying to measure a combination of three aspects of a function: height (amplitude), width (measure of the support), and frequency (inverse wavelength). Roughly speaking, if a function has amplitude $A$, is supported on a set of volume $V$, and has frequency $N$, then the $W^{k,p}$ norm is going to be about $A N^k V^{1/p}$. The uncertainty principle tells us that if a function has frequency $N$, then it must be spread out on at least a ball of radius comparable to the wavelength $1/N$, and so its support must have measure at least $1/N^d$ or so: $V \gtrsim 1/N^d.$ This relation already encodes most of the content of the Sobolev embedding theorem, except for endpoints. It is also consistent with dimensional analysis, of course, which is another way to derive the conditions of the embedding theorem. More generally, one can classify the integrability and regularity of a function space norm by testing that norm against a bump function of amplitude $A$ on a ball of volume $V$, modulated by a frequency of magnitude $N$. Typically the norm will be of the form $A N^k V^{1/p}$ for some exponents $p$, $k$ (at least in the high frequency regime $V \gtrsim 1/N^d$). One can then plot these exponents $1/p, k$ on a two-dimensional diagram as mentioned by Jitse to get a crude "map" of various function spaces (e.g. Sobolev, Besov, Triebel-Lizorkin, Hardy, Lipschitz, Holder, Lebesgue, BMO, Morrey, ...). The relationship $V \gtrsim 1/N^d$ lets one trade in regularity for integrability (with an exchange rate determined by the ambient dimension - integrability becomes more expensive in high dimensions), but not vice versa. These exponents $1/p, k$ only give a first-order approximation to the nature of a function space, as they only inspect the behaviour at a single frequency scale N. To make finer distinctions (e.g. between Sobolev, Besov, and Triebel-Lizorkin spaces, or between strong L^p and weak L^p) it is not sufficient to experiment with single-scale bump functions, but now must play with functions with a non-trivial presence at multiple scales. This is a more delicate task (which is particularly important for critical or scale-invariant situations, such as endpoint Sobolev embedding) and the embeddings are not easily captured in a simple two-dimensional diagram any more. I discuss some of these issues in my lecture notes http://terrytao.wordpress.com/2009/04/30/245c-notes-4-sobolev-spaces/ EDIT: Another useful checksum with regard to remembering Sobolev embedding is to remember the easy cases: $W^{1,1}({\bf R}) \subset L^\infty({\bf R})$ (fundamental theorem of calculus) $W^{d,1}({\bf R}^d) \subset L^\infty({\bf R}^d)$ (iterated fundamental theorem of calculus + Fubini) $W^{0,p}({\bf R}^d) = L^p({\bf R}^d)$ (trivial) These are the extreme cases of Sobolev embedding; everything else can be viewed as an interpolant between them. EDIT: I decided to go ahead and draw the map of function spaces I mentioned above, at http://terrytao.wordpress.com/2010/03/11/a-type-diagram-for-function-spaces/
|
{
"source": [
"https://mathoverflow.net/questions/17736",
"https://mathoverflow.net",
"https://mathoverflow.net/users/3509/"
]
}
|
17,778 |
I have recently been told of a proposal to produce an English translation
of Landau's Handbuch der Lehre von der Verteilung der Primzahlen , and
this prompts me to ask a more general question: Which foreign-language
books would you most like to see translated into English? These could be classics of historical interest, books you would like your
students to read, books you would like to teach from, or books of use in
your own research.
|
Grothendieck's EGA and SGA.
|
{
"source": [
"https://mathoverflow.net/questions/17778",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1587/"
]
}
|
17,826 |
I hope this question isn't too open-ended for MO --- it's not my favorite type of question, but I do think there could be a good answer. I will happily CW the question if commenters want, but I also want answerers to pick up points for good answers, so... Let $X,Y$ be smooth manifolds. A smooth map $f: Y \to X$ is a bundle if there exists a smooth manifold $F$ and a covering $U_i$ of $X$ such that for each $U_i$, there is a diffeomorphism $\phi_i : F\times U_i \overset\sim\to f^{-1}(U_i)$ that intertwines the projections to $U_i$. This isn't my favorite type of definition, because it demands existence of structure without any uniqueness, but I don't want to define $F,U_i,\phi_i$ as part of the data of the bundle, as then I'd have the wrong notion of morphism of bundles. A definition I'm much happier with is of a submersion $f: Y \to X$, which is a smooth map such that for each $y\in Y$, the differential map ${\rm d}f|_y : {\rm T}_y Y \to {\rm T}_{f(y)}X$ is surjective. I'm under the impression that submersions have all sorts of nice properties. For example, preimages of points are embedded submanifolds (maybe preimages of embedded submanifolds are embedded submanifolds?). So, I know various ways that submersions are nice. Any bundle is in particular a submersion, and the converse is true for proper submersions (a map is proper if the preimage of any compact set is compact), but of course in general there are many submersions that are not bundles (take any open subset of $\mathbb R^n$, for example, and project to a coordinate $\mathbb R^m$ with $m\leq n$). But in the work I've done, I haven't ever really needed more from a bundle than that it be a submersion. Then again, I tend to do very local things, thinking about formal neighborhoods of points and the like. So, I'm wondering for some applications where I really need to use a bundle --- where some important fact is not true for general submersions (or, surjective submersions with connected fibers, say).
|
One would be that a fibre bundle $F \to E \to B$ has a homotopy long exact sequence $$ \cdots \to \pi_{n+1} B \to \pi_n F \to \pi_n E \to \pi_n B \to \pi_{n-1} F \to \cdots $$ This isn't true for a submersion, for one, the fibre in a submersion does not have a consistent homotopy-type as you vary the point in the base space.
|
{
"source": [
"https://mathoverflow.net/questions/17826",
"https://mathoverflow.net",
"https://mathoverflow.net/users/78/"
]
}
|
17,886 |
This question is perhaps somewhat soft, but I'm hoping that someone could provide a useful heuristic. My interest in this question mainly concerns various derived equivalences arising in geometric representation theory. Background For example, Bezrukavnikov, Mirkovic, and Rumynin have proved the following: Let $G$ be a reductive algebraic group over an algebraically closed field of positive characteristic. Then there is an equivalence between the bounded derived category of modules for the sheaf $\cal D$ of crystalline (divided-power) differential operators on the flag variety, and the bounded derived category of modules with certain central character for the enveloping algebra $\cal U$ of Lie($G$). What is interesting is that it is not true that this equivalence holds on the non-derived level: The category of $\cal D$-modules is not equivalent to the category of $\cal U$-modules with the appropriate central character. This is true in characteristic 0 (this is the Beilinson-Bernstein correspondence), but something is broken in positive characteristic: there are certain "bad" sheaves that are $\cal D$-modules which make the correspondence not hold. There are other results in geometric representation theory of this form. For example, Arkhipov, Bezrukavnikov, and Ginzburg have proved that there is an equivalence (in characteristic 0) between the bounded derived category of a certain block of representations for the quantum group associated to $G$, and the bounded derived category of $G \times \mathbb C^*$-equivariant sheaves on the cotangent bundle of the flag variety of $G$. Again, this equivalence does not hold on the non-derived level. In general, there are a number of results in geometric representation theory that hold on the derived level, but not the non-derived level. Question Here's my question: Why would one be led to expect that a derived equivalence holds, when the non-derived equivalence does not? It seems as though the passage to the derived level in some sense fixes something that was broken on the non-derived level; how does it do that?
|
One crude answer is that passing to derived functors fixes one obstruction to being an equivalence. Any equivalence of abelian categories certainly is exact (i.e. it preserves short exact sequences), though lots of exact functors are not equivalences (for example, think about representations of a group and forgetting the G-action). What derived functor does is fix this problem in a canonical way; you have to replace short exact sequences with exact triangles, but you get a functor which is your original "up to zeroth order," exact, and uniquely distinguished by these properties. So, what BMR do is take a functor which is not even exact (and thus obviously not an equivalence), and show that the lack of exactness is "the only problem" for this being an equivalence. EDIT: Let me just add, from a more philosophical perspective, that derived equivalences are just a lot more common. There are just more of them out in the world. Given an algebra A, Morita equivalences to A are classified essentially by projective generating A-modules, whereas derived Morita equivalences of dg-algebras are in bijection with all objects in the derived category of $A-mod$ which generate (in the sense that nothing has trivial Ext with them): you look at the dg-Ext algebra of the object with itself. If you have an interesting algebra (say, a finite dimensional one of wild representation type), there are a lot more of the latter than the former in a very precise sense. Of course, the vast majority of these are completely uncomputable an tell you nothing, but there are enough of them in the mix to make things interesting.
|
{
"source": [
"https://mathoverflow.net/questions/17886",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1528/"
]
}
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.