source_id
int64 1
4.64M
| question
stringlengths 0
28.4k
| response
stringlengths 0
28.8k
| metadata
dict |
|---|---|---|---|
8,147 |
I´m looking for homomorphisms between exterior powers of a free module M of rank m Λ m R M → Λ m-1 R M Exactly, I´m looking for an explicit isomorphism M → Hom R (Λ m R M , Λ m-1 R M) I compare the ranks and the things go, but I can not imagine a concrete expression. Suggestions are welcome
|
Yes, there are. See "Locked and Unlocked Polygonal Chains in 3D", T. Biedl, E. Demaine, M. Demaine, S. Lazard, A. Lubiw, J. O'Rourke, M. Overmars, S. Robbins, I. Streinu, G. Toussaint, S. Whitesides, arXiv:cs.CG/9910009 , figure 6.
|
{
"source": [
"https://mathoverflow.net/questions/8147",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2040/"
]
}
|
8,244 |
Is there a name for those categories where objects posses a given structure and every bijective morphism determines an isomorphism between the corresponding objects? Examples of categories of that type abound: Gr , Set , ... An specific example of a category where the constraint doesn't hold is given by Top : a morphism there is a continuous function between topological spaces. Now, it is easy to give here a concrete example of a bijective morphism between [0,1) and $\mathbb{S}^{1}$ that fails to be an isomorphism of topological spaces (in point of fact, much more is known in this case, right?).
|
The comments on the question point out that it's not really well-posed: the property "bijective" isn't defined for morphisms of an arbitrary category. However, for maps between sets, "bijective" means "injective and surjective". A common way to interpret "injective" in an arbitrary category is "monic", and a common way to interpret "surjective" in an arbitrary category is "epic". So we might interpret "bijective" as "monic and epic". Then JHS's question becomes: is there a name for categories in which every morphism that is both monic and epic is an isomorphism? And the answer is yes: balanced . It's not a particularly inspired choice of name, nor does it seem to be a particularly important concept. But the terminology is quite old and well-established, in its own small way. Incidentally, you don't have to interpret "injective" and "surjective" in the ways suggested. You could, for instance, interpret "surjective" as "regular epic", and indeed that's often a sensible thing to do. But then the question becomes trivial, since any morphism that's both monic and regular epic is automatically an isomorphism.
|
{
"source": [
"https://mathoverflow.net/questions/8244",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1593/"
]
}
|
8,258 |
Before you close for "homework problem", please note the tags. Last week, I gave my calculus 1 class the assignment to calculate the $n$-volume of the $n$-ball. They had finished up talking about finding volume by integrating the area of the cross-sections. I asked them to calculate a formula for $4$ and $5$, and take the limit of the general formula to get 0. Tomorrow I would like to give them a more geometric idea of why the volume goes to zero. Anyone have any ideas? :) Comm wiki in case people want to add/modify this a bit.
|
The ultimate reason is, of course, that the typical coordinate of a point in the unit ball is of size $\frac{1}{\sqrt{n}}\ll 1$. This can be turned into a simple geometric argument (as suggested by fedja) using the fact that an $n$-element set has $2^n$ subsets: At least $n/2$ of the coordinates of a point in the unit ball are at most $\sqrt{\frac{2}{n}}$ in absolute value, and the rest are at most $1$ in absolute value. Thus, the unit ball can be covered by at most $2^n$ bricks (right-angled parallelepipeds) of volume $$\left(2\sqrt{\frac{2}{n}}\right)^{n/2},$$ each corresponding to a subset for the small coordinates. Hence, the volume of the unit ball is at most $$2^n \cdot \left(2\sqrt{\frac{2}{n}}\right)^{n/2} = \left(\frac{128}{n}\right)^{n/4}\rightarrow0.$$ In fact, the argument shows that the volume of the unit ball decreases faster than any exponential, so the volume of the ball of any fixed radius also goes to $0$.
|
{
"source": [
"https://mathoverflow.net/questions/8258",
"https://mathoverflow.net",
"https://mathoverflow.net/users/348/"
]
}
|
8,324 |
All definitions I've seen for the statement "$E,F$ are linearly disjoint extensions of $k$" are only meaningful when $E,F$ are given as subfields of a larger field , say $K$. I am happy with the equivalence of the various definitions I've seen in this case. Lang's Algebra VIII.3-4 and (thanks to Pete) Zariski & Samuel's Commutative Algebra 1 II.15-16 have good coverage of this. "Ambient" definitions of linear disjointness: Wikipedia says it means the map $E\otimes_k F\to E.F$ is injective , where $E.F$ denotes their compositum in $K$, the smallest subfield of $K$ containing them both. An equivalent (and asymmetric) condition is that any subset of $E$ which is linearly independent over $k$ is also linearly independent over $F$ (hence the name); this all happens inside $K$. However, I often see the term used for field extensions which are NOT subfields of a larger one , even when the field extensions are not algebraic (so there is no tacit assumption that they live in the algebraic closure). Some examples of these situations are given below. Question: What is the definition of "linearly disjoint" for field extensions which are not specified inside a larger field? ANSWER: (After reading the helpful responses of Pete L. Clark, Hagen Knaf, Greg Kuperberg, and JS Milne -- thanks guys! -- I now have a satisfying and fairly exhaustive analysis of the situation.) There are two possible notions of abstract linear disjointness for two field extensions $E,F$ of $k$ ( proofs below ): (1) " Somewhere linearly disjoint", meaning " There exists an extension $K$ with maps $E,F\to K$, the images of $E,F$ are linearly disjoint in $K$." This is equivalent to the tensor product $E\otimes_k F$ being a domain . (2) " Everywhere linearly disjoint", meaning " For any extension $K$ with maps $E,F\to K$, the images of $E,F$ are linearly disjoint in $K$." This is equivalent to the tensor product $E\otimes_k F$ being a field. Results: (A) If either $E$ or $F$ is algebraic, then (1) and (2) are equivalent. (B) If neither $E$ nor $F$ is algebraic, then (2) is impossible. Depending on when theorems would read correctly, I'm not sure which of these should be the "right" definition... (1) applies in more situations, but (2) is a good hypothesis for implicitly ruling out pairs of transcendental extensions. So I'm just going to remember both of them :) PROOFS: (for future frustratees of linear disjointness!) (1) Linear disjointness in some field $K$, by the Wikipedia defintion above, means the tensor product injects to $K$, making it a domain. Conversely, if the tensor product is a domain, then $E,F$ are linearly disjoint in its field of fractions. (2) If the tensor product $E\otimes_k F$ is a field, since any map from a field is injective, by the Wikipedia definition above, $E,F$ are linearly disjoint in any $K$. Conversely, if $E \otimes_k F$ is not a field, then it has a non-trivial maximal ideal $m$, with quotient field say $K$, and then since $E\otimes_k F\to K$ has non-trivial kernel $m$, by definition $E,F$ are not linearly disjoint in $K$. (A) Any two field extensions have some common extension (take a quotient of their tensor product by any maximal ideal), so (2) always implies (1). Now let us first show that (1) implies (2) supposing $E/k$ is a finite extension. By hypothesis the tensor product $E\otimes_k F$ is a domain, and finite-dimensional as a $F$-vector space, and a finite dimensional domain over a field is automatically a field: multiplication by an element is injective, hence surjective by finite dimensionality over $F$, so it has an inverse map, and the image of $1$ under this map is an inverse for the element. Hence (1) implies (2) when $E/k$ is finite. Finally, supposing (1) and only that $E/k$ is algebraic, we can write $E$ as a union of its finite sub-extensions $E_\lambda/k$. Since tensoring with fields is exact, $E_\lambda\otimes_k F$ naturally includes in $E\otimes_k F$, making it a domain and hence a field by the previous argument. Then $E\otimes_k F$ is a union of fields, making it a field, proving (1) implies (2). (B) Now this is easy. Let $t_1\in E$, $t_2\in F$ be transcendental elements. Identify $k(t)=k(t_1)=k(t_2)$ by $t\mapsto t_1 \mapsto t_2$, making $E$,$F$ extensions of $k(t)$. Let $K$ be a common extension of $E,F$ over $k(s)$ (any quotient of $E\otimes_{k(s)} F$ by a maximal ideal will do). Then $E,F$ are not linearly disjoint in $K$ because their intersection is not $k$: for example the set { $1,t$ }$\subseteq E$ is linearly independent over $k$ but not over $F$, so they are not linearly disjoint by the equivalent definition at the top. Examples in literature of linear disjointness referring to abstract field extensions: Eisenbud, Commutative Algebra , Theorem A.13 (p.564 in my edition) says, in characteristic $p$, "$K$ is separable over $k$ iff $k^{1/p^{\infty}}$ is linearly disjoint from $K$." Liu, Algebraic Geometry and Arithmetic Curves , Corollary 2.3 (c) (p. 91) says, for an integral algebraic variety $X$ over a field $k$ with function field $K(X)$, "$X$ is geometrically integral iff $K(X)$ and $\overline{k}$ are linearly disjoint over $k$. ( Follow-up: Since in both these situations, one extension is algebraic, the two definitions summarized in the answer above are equivalent, so everything is fantastic.) Old edit: My first guess was (and still is) to say that the tensor product is a domain...
|
The reasonable meaning following example (1) seems to be that $E \otimes_k F$ is a field. If so, then it is isomorphic to every compositum. If not, then there exists a compositum within which they are not linearly disjoint. I am not (yet) getting voter support, but I stand my ground! :-) First, clearly if $E \otimes_k F$ is a field, then it is isomorphic to every compositum. Second, if $E \otimes_k F$ is not a field, then there exists a compositum in which $E$ and $F$ are not linearly disjoint. It has a non-trivial quotient field, and that field can serve as a compositum. As Pete Clark points out, there is a difference between the case that $E \otimes_k F$ is an integral domain and the case that it has zero divisors. (And Pete is right that I forgot about this distinction.) In the former case, there exists a compositum in which they are linearly disjoint, namely the fraction field of $E \otimes_k F$ . In the latter case, $E$ and $F$ are not linearly disjoint in any compositum. If $E$ and $F$ are both transcendental extensions, then there are two different criteria: Weakly linearly disjoint, when $E \otimes_k F$ is an integral domain, and strongly linearly disjoint, when it is a field. Which you think is the more important condition is up to you. In Andrew's examples, $E$ and $F$ aren't both transcendental, so the distinction is moot. (I needed to think about this issue in Finite, connected, semisimple, rigid tensor categories are linear .) Actually, the previous isn't the whole story. If $E$ and $F$ are both transcendental, then they are extensions of purely transcendental extensions $E'$ and $F'$ . $E'$ and $F'$ are only weakly linearly disjoint, and therefore $E$ and $F$ are too. So the distinction is always moot. Pete and Andrew's intuition was more correct all along. The correct statement is that when $E$ and $F$ are both transcendental, linearly disjoint extensions have different behavior.
|
{
"source": [
"https://mathoverflow.net/questions/8324",
"https://mathoverflow.net",
"https://mathoverflow.net/users/84526/"
]
}
|
8,407 |
Is there a generalized method to find the projective closure of an affine curve? For example, I read that the projective closure of $y^2 = x^3−x+1$ in $\mathbb{P}^2$ is $y^2z = x^3−xz^2+z^3$. If I want to find the the closure of another affine curve, what method should I employ? I can't seem to find an adequate description in any text. Thanks
|
When the curve is a plane curve of degree $d$, the formula is simple (and in fact, this works for any hypersurface) you take $f(x,y)=0$ and replace it by $z^d f(x/z,y/z)=0$. This will be homogeneous of degree $d$, and when $z\neq 0$, you recover your curve. Now, if you have a more general affine variety, given by $\langle f_1,\ldots,f_k\rangle$ in $k[x_1,\ldots,x_n]$, then you first compute a Groebner Basis , which you can learn how to do using Cox, Little and O'Shea's "Ideals, Varieties and Algorithms" (assumes no background whatsoever) and then you do this same trick with each polynomial in the Groebner basis. Why can't you just use any basis? Look at the twisted cubic. $t\mapsto (t,t^2,t^3)$. This is cut out, in affine space, by $y-x^2$ and $z-x^3$, though $y-x^2$ and $xy-z$ are better to use. But still, $yw-x^2$ and $xy-zw$ don't give the twisted cubic! Bezout tells us that they give something of degree four containing the cubic, and it's not a hypersurface, so you get the twisted cubic plus a line. To get the cubic itself, you have to use a third quadratic polynomial that you get in the ideal, $y^2-xz$, which is not in the ideal given by the homogenized generators. These three, however, form a Groebner basis for the ideal, and then homogenization gives the homogeneous ideal of the projective twisted cubic.
|
{
"source": [
"https://mathoverflow.net/questions/8407",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2473/"
]
}
|
8,415 |
The number of conjugacy classes in $S_n$ is given by the number of partitions of $n$. Do other families of finite groups have a highly combinatorial structure to their number of conjugacy classes? For example, how much is known about conjugacy classes in $A_n$?
|
Since $A_n$ has index two in $S_n$, every conjugacy class in $S_n$ either is a conjugacy class in $A_n$, or it splits into two conjugacy classes, or it misses $A_n$ if it is an odd permutation. Which happens when is a nice undergraduate exercise in group theory. (And you are a nice undergraduate. :-) ) The pair $A_n \subset S_n$ is typical for this question in finite group theory. You want the conjugacy classes of a finite simple group $G$, but the answer is a little simpler for a slightly larger group $G'$ that involves $G$. Another example is $\text{GL}(n,q)$. It involves the finite simple group $L(n,q)$, but the conjugacy classes are easier to describe in $\text{GL}(n,q)$ . They are described by their Jordan canonical form, with the twist that you may have to pass to a field extension of $\mathbb{F}_q$ to obtain the eigenvalues. The group $\text{GL}(n,q)$ is even more typical. It is a Chevallay group, which means a finite group analogue of a Lie group. All of the infinite sequences of finite simple groups other than $A_n$ and $C_p$ are Chevallay groups. You expect a canonical form that looks something like Jordan canonical form, although it can be rather more complicated. If $G$ is far from simple, i.e., if it has some interesting composition series, then one approach to its conjugacy classes is to chase them down from the conjugacy classes of its composition factors, together with the structure of the extensions. The answer doesn't have to be very tidy. I suppose that finite Coxeter groups give you some exceptions where you do get a tidier answer, just because they all resemble $S_n$ to varying degrees. But I don't know a crisp answer to all cases of this side question. The infinite sequences of finite Coxeter groups consist only of permutation groups, signed permutation groups, and dihedral groups. (And Cartesian products of these.) In the case of signed permutation groups, the answer looks just like $S_n$, except that cycles can also have odd or even total sign. There is also the type $D_n$ Coxeter group of signed permutation matrices with an even number of minus signs; the answer is just slightly different from all of the signed permutation matrices, which is type $B_n$. The crisp answer that I don't have would be a uniform description that includes the exceptional finite Coxeter groups, such as $E_8$.
|
{
"source": [
"https://mathoverflow.net/questions/8415",
"https://mathoverflow.net",
"https://mathoverflow.net/users/960/"
]
}
|
8,443 |
I'm not a functional analyst (though I like to pretend that I am from time to time) but I use it and I think it's a great subject. But whenever I read about locally convex topological vector spaces, I often get bamboozled by all the different types that there are. At one point, I made a little summary of the properties of my favourite space (smooth functions from the circle to Euclidean space) and found that it was: metrisable, barrelled, bornological, Mackey, infrabarrelled, Montel, reflexive, separable, Schwartz, convenient, semi-reflexive, reflexive, $c^\infty$ -top is LCS top, quasi-complete, complete, Baire, nuclear Its dual space (with the strong topology) is reflexive, semi-reflexive, barrelled, infrabarrelled, quasi-complete, complete, bornological, nuclear, Mackey, convenient, $c^\infty$ top = LCS top, Schwartz, Montel, separable, DF space I get the impression that many of these properties (and there are more!) are not "front line" properties but rather are conditions that guarantee that certain Big Theorems (like uniform boundedness, or open mapping theorem) hold. But as an outsider of functional analysis, it's not always clear to me which are "front line" and which are "supporters". So that's my question: which of these properties (and others that I haven't specified) are main properties and which have more of a supporting role? I realise that there's a little vagueness there as to exactly where the division lies - but that's part of the point of the question! If pressed, I would refine it to "Which of these properties would you expect to find used outside functional analysis, and which are more part of the internal machine?".
|
First, there is a great survey of locally convex topological spaces in section 424 of the Encyclopedic Dictionary of Mathematics. (The EDM, if you have not seen it, is a fabulous reference for all kinds of information. Even though we all have Wikipedia, the EDM is still great.) At the end it has a chart of many of these properties of topological vector spaces, indicating dependencies, although not all of them. This chart helped me a lot with your question. On that note, I am not a functional analyst, but I can play one on MO . Maybe a serious functional analyst can give you a better answer. (Or a worse answer? I wonder if to some analysts, everything is the machine.) Every important topological vector space that I have seen in mathematics (of those that are over $\mathbb{R}$ or $\mathbb{C}$) is at least a Banach space , or an important generalization known as a Frechet space , or is derived from one of the two. We all learn what a Banach space is; a Frechet space is the same thing except with a countable family of seminorms instead of one norm. Many of the properties that you list, for instance metrizable and bornological, hold for all Frechet spaces. In emphasizing Banach and Frechet spaces, the completeness property is implicitly important. Since a normed linear space is a metric space, you might as well take its completion, which makes it Banach. A Frechet space is a generalization of a metric space known as a uniform space and you might as well do the same thing. Also the discussion is not complete without mentioning Hilbert spaces. You can think of a Hilbert space either as a construction of a type of Banach space, or a Banach space that satisfies the parallelogram law; of course it's important. Of the properties that do not hold for all Frechet spaces, I can think of four that actually matter: reflexive, nuclear, separable, and unconditional. In addition, a Schwartz space isn't really a space with a property but a specific (and useful) construction of a Frechet space. ( Edit: It seems that "Schwartz" means two things, the Schwartz space of rapidly decreasing smooth functions, and the Schwartz property of a locally convex space.) A discussion of the properties that I think are worth knowing, and why: reflexive . This means a space whose dual is also its pre-dual. If a Banach space has a pre-dual, then its unit ball is compact in the weak-* topology by the Banach-Alaoglu theorem. In particular, the set of Borel probability measures on a compact space is compact. This is important in geometry, for sure. Famously, Hilbert spaces are reflexive. Note also that there is a second important topology, the weak-* topology when a pre-dual exists, which you'd also call the weak topology in the reflexive case. (I am not sure what good the weak topology is when they are different.) separable . As in topology, has a countable dense subset. How much do you use manifolds that do not have a countable dense subset? Inseparable topological vectors are generally not that useful either, with the major exception of the dual of a non-reflexive, separable Banach space. For instance $B(H)$, the bounded operators on a Hilbert space, is inseparable, but it is the dual of the Banach space of trace-class operators $B_1(H)$, which is separable. unconditional . It is nice for a Banach space to have a basis, and the reasonable kind is a topological basis, a.k.a. a Schauder basis. The structure does not resemble familiar linear algebra nearly as much if linear combinations are only conditionally convergent. An unconditional basis is an unordered topological basis, and an unconditional space is a Banach space that has one. There is a wonderful structure theorem that says that, up to a constant factor that can be sent to 1, the norm in an unconditional space is a convex function of the norms of the basis coordinates. All unconditional Banach spaces resemble $\ell^p$ in this sense. Note also that there is a non-commutative moral generalization for operators, namely that the norm be spectral , or invariant under the available unitary group. nuclear . Many of the favorable properties of the smooth functions $C^\infty(M)$ on a compact manifold come from or are related to the fact that it is a nuclear Frechet space. For instance, in defining it as a Frechet space (using norms on the derivatives), you notice that the precise norms don't matter much. This doesn't necessarily mean that you should learn the theory of nuclear spaces (since even most analysts don't). But in drawing a line between constructions and properties, Opinion seems to be split as to whether the theory of nuclear spaces is tangential or fundamental. Either way, my impression is that the main favorable properties of $C^\infty(M)$ are that it is Frechet, reflexive, and nuclear. On the other hand, infinite-dimensional Banach spaces are never nuclear. Prompted by Andrew's question, I compiled some data on relationships between types of locally convex topological spaces. In particular, I made a Hasse diagram (drawn with Graphviz ) of properties included below. It began as a simplification of the one in EDM, but then I added more conditions. The rule for the Hasse diagram is that I only allow "single-name" properties, not things like Frechet and nuclear, although adverbs are allowed. The graph makes the topic of locally convex spaces look laughably complicated, but that's a little unfair. You could argue that there are more than enough defined properties in the field, but of course mathematicians are always entitled to make new questions. Moreover, if you look carefully, relatively few properties toward the top of the diagram imply most of the others, which is part of the point of my answer. If a topological vector space is Banach or even Frechet, then it automatically has a dozen other listed properties. I have the feeling that the Hasse diagram has missing edges even for the nodes listed. If someone wants to add comments about that, that would be great. (Or it could make a future MO question.) The harder question is combinations of properties. I envision a computer-assisted survey to compare all possible combinations of important properties, together with citations to counterexamples and open status as appropriate. I started a similar computer-assisted survey of complexity classes a few years ago.
|
{
"source": [
"https://mathoverflow.net/questions/8443",
"https://mathoverflow.net",
"https://mathoverflow.net/users/45/"
]
}
|
8,445 |
EDIT (Harry): Since this question in its original form was poorly stated (asked about topology rather than graph theory), but we have a list of Topology books in the answers, I guess you should go ahead and post with regard to that topic, rather than graph theory, which the questioner can ask again in another topic. EDIT (David): The original question was asking for places to learn topology with an eye towards applying it to computer science (artificial neural networks in particular)
|
Since the discussion has broadened from the original question to include a wider range of topology books, let me add one more. This is an algebraic topology book by Tammo tom Dieck published just a year ago with the canonical title "Algebraic Topology". Its viewpoint is fairly homotopy-theoretic, as in May's book, and it has a similar density coefficient that some commenters here seem to like. What really impressed me about the book is that in the last few chapters the author manages to give the first ever non-spectral-sequence proofs of some deep and fundamental theorems like Serre's theorem that the homotopy groups of spheres are finitely generated, and Serre's calculation of all the non-torsion. Another is the Hirzebruch signature theorem, the very last theorem in the book. These results are 50 years old, yet apparently no one had previously seen how to prove them without spectral sequences. Of course, spectral sequences are important things that serious topologists should know about, and their use cannot always be avoided, but it's illuminating to see when they are needed and when they are not. Whenever I get around to a second edition of my book I'll have to include tom Dieck's new approach, and I think one can go even further and develop the basic framework of rational homotopy theory without spectral sequences. It's too bad that math books aren't like Google Maps where one can zoom in or out to get the level of detail and density one wants, or switch between satellite and map views to include or omit things like examples and informal discussions of ideas and motivation. Maybe someday this sort of thing will be possible with electronic books.
|
{
"source": [
"https://mathoverflow.net/questions/8445",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2482/"
]
}
|
8,521 |
As a student, I was taught that the Jordan curve theorem is a great example of an intuitively clear statement which has no simple proof. What is the simplest known proof today? Is there an intuitive reason why a very simple proof is not possible?
|
There's a short proof (less than three pages) that uses Brouwer's fixed point theorem, available here: The Jordan Curve Theorem via the Brouwer Fixed Point Theorem The goal of the proof is to take Moise's "intuitive" proof and make it simpler/shorter. Not sure whether you'd consider it "nice," though.
|
{
"source": [
"https://mathoverflow.net/questions/8521",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2498/"
]
}
|
8,560 |
Background: a field is formally real if -1 is not a sum of squares of elements in that field. An ordering on a field is a linear ordering which is (in exactly the sense that you would guess if you haven't seen this before) compatible with the field operations. It is immediate to see that a field which can be ordered is formally real. The converse is a famous result of Artin-Schreier. (For a graceful exposition, see Jacobson's Basic Algebra. For a not particularly graceful exposition which is freely available online, see http://alpha.math.uga.edu/~pete/realspectrum.pdf. ) The proof is neither long nor difficult, but it appeals to Zorn's Lemma. One suspects that the reliance on the Axiom of Choice is crucial, because a field which is formally real can have many different orderings (loc. cit. gives a brief introduction to the real spectrum of a field, the set of all orderings endowed with a certain topology making it a compact, totally disconnected topological space). Can someone give a reference or an argument that AC is required in the technical sense (i.e., there are models of ZF in which it is false)? Does assuming that formally real fields can be ordered recover some weak version of AC, e.g. that any Boolean algebra has a prime ideal? (Or, what seems less likely to me, is it equivalent to AC?)
|
This is equivalent (in ZF) to the Boolean Prime Ideal Theorem (which is equivalent to the Ultrafilter Lemma). Reference:
R. Berr, F. Delon, J. Schmid, Ordered fields and the ultrafilter theorem, Fund Math 159 (1999), 231-241. online
|
{
"source": [
"https://mathoverflow.net/questions/8560",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1149/"
]
}
|
8,579 |
I remember learning some years ago of a theorem to the effect that if a polynomial $p(x_1, ... x_n)$ with real coefficients is non-negative on $\mathbb{R}^n$, then it is a sum of squares of polynomials in the variables $x_i$. Unfortunately, I'm not sure if I'm remembering correctly. (The context in which I saw this theorem was someone asking whether there was a sum-of-squares proof of the AM-GM inequality in $n$ variables, so I'm not 100% certain if the quoted theorem was specific to that case.) So: does anyone know a reference for the correct statement of this theorem, if in fact something like it is true? (Feel free to retag if you don't think it's appropriate, by the way.)
|
One interpretation of the question is Hilbert's seventeenth problem , to characterize the polynomials on $\mathbb{R}^n$ that take non-negative values. The problem is motivated by the nice result, which is not very hard, that a non-negative polynomial in $\mathbb{R}[x]$ (one variable) is a sum of two squares. What is fun about this result is that it establishes an analogy between $\mathbb{C}[x]$, viewed as a quadratic extension by $i$ of the Euclidean domain $\mathbb{R}[x]$; and $\mathbb{Z}[i]$ (the Gaussian integers), viewed as a quadratic extension by $i$ of the Euclidean domain $\mathbb{Z}$. In this analogy, a real linear polynomial is like a prime that is 3 mod 4 that remains a Gaussian prime, while a quadratic irreducible polynomial is like a prime that is not 3 mod 4, which is then not a Gaussian prime. A non-zero integer $n \in \mathbb{Z}$ is a sum of two squares if and only if it is positive and each prime that is 3 mod 4 occurs evenly. Analogously, a polynomial $p \in \mathbb{R}[x]$ is a sum of two squares if and only if some value is positive and each real linear factor occurs evenly. And that is a way of saying that $p$ takes non-negative values. In dimension 2 and higher, the result does not hold for sums of squares of polynomials. But as the Wikipedia page says, Artin showed that a non-negative polynomial (or rational function) in any number of variables is at least a sum of squares of rational functions. In general, if $R[i]$ and $R$ are both unique factorization domains, then some of the primes in $R$ have two conjugate (or conjugate and associate) factors in $R[i]$, while other primes in $R$ are still primes in $R[i]$. This always leads to a characterization of elements of $R$ that are sums of two squares. This part actually does apply to the multivariate polynomial ring $R = \mathbb{R}[\vec{x}]$. What no longer holds is the inference that if $p \in R$ has non-negative values, then the non-splitting factors occur evenly. For instance, $x^2+y^2+1$ is a positive polynomial that remains irreducible over $\mathbb{C}$. It is a sum of 3 squares rather than 2 squares; of course you have to work harder to find a polynomial that is not a sum of squares at all.
|
{
"source": [
"https://mathoverflow.net/questions/8579",
"https://mathoverflow.net",
"https://mathoverflow.net/users/290/"
]
}
|
8,609 |
Christmas is almost here, so imagine you want to buy a good popular math book for your aunt (or whoever you want). Which book would you buy or recommend? It would be nice if you could answer in the following way: Title: The Poincaré Conjecture: In Search of the Shape of the Universe Author: Donal O'Shea Short description: The history of the Poincaré Conjecture. (Perhaps something like "difficulty level": + (no prior knowledge of math, as the book mentioned above), ++ (some prior knowledge of math is helpful), +++ (Roger Penrose: Road to Reality (?)) I hope this is appropriate for MO, since I think is of interest to mathematicians (at least for those who want to buy a popular math book for some aunt :-) ).
|
Title: Godel, Escher, Bach: an Eternal Golden Braid Author: Douglas Hofstadter Short Description: It's mildly debatable whether this is in fact a book about mathematics, but any mathematician who has read this book will understand why I recommend it and any who has not should. Probably best for those with either a philosophical or musical bent.
|
{
"source": [
"https://mathoverflow.net/questions/8609",
"https://mathoverflow.net",
"https://mathoverflow.net/users/675/"
]
}
|
8,663 |
Edit & Note: I'm declaring a convention here because I don't feel like trying to fix this in a bunch of spots: If I said model category and it doesn't make sense, I meant a model-category "model" of an (infinity,1)-category. Also, "model" in quotes means the English word model, whereas without quotes it has do do with model categories. At the very beginning of Lurie's higher topos theory, he mentions that there is a theory of $(\infty,1)$-categories that can be directly constructed by using model categories. What I'd like to know is: Where can I find related papers (Lurie mentions two books that are not available for download)? How dependent on quasicategories is the theory developed in HTT? Can the important results be proven for these $(\infty,1)$-model-categories by proving some sort of equivalence (not equivalence of categories, but some weaker kind of equivalence) to the theory of quasicategories? When would we want to use quasicategories rather than these more abstract model categories? And also, conversely, when would we want to look at model categories rather than quasicategories? Does one subsume the other? Are there disadvantages to the model category construction just because it requires you to have all of the machinery of model categories? Are quasicategories better in every way? The only "models" of infinity categories that I'm familiar with are the ones presented in HTT.
|
You can't produce every ($\infty$,1)-category from a model category. The slogan is that every presentable ($\infty$,1)-category comes from a model category, and every adjoint pair between such comes from a Quillen pair of functors between model categories. The paper by Dugger on Universal model categories works out this formalism from the point of view of model categories. (A companion paper shows that a large class of model categories (the combinatorial ones) are "presentable" in this sense.) (I say it's a slogan, but I'm sure it's a theorem; I just don't have a reference for you.) Presentable ($\infty$,1)-categories are special among all ($\infty$,1)-categories; in particular, they are complete and cocomplete. For instance, you can define the notion of ($\infty$,1)-topos in terms of model categories, since ($\infty$,1)-topoi are presentable, and morphisms between such are certain kinds of adjoint functor pairs.
|
{
"source": [
"https://mathoverflow.net/questions/8663",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1353/"
]
}
|
8,684 |
I have a good grasp of ordinary pullbacks and pushouts; in particular, there are many categorical constructions that can be seen as special cases: e.g., equalizers/coequalizers, kernerls/cokernels, binary products/coproducts, preimages,... I know the (a?) definition of homotopy pullbacks/pushouts, but I am lacking two things: examples and intuition. So here are my questions: What are the canonical examples of homotopy pullbacks/pushouts? E.g., in the category of pointed topological spaces the loop space $\Omega X$ is a homotopy pullback of the map $\ast \to X$ along itself. How should I think about homotopy pullbacks/pushouts? What is the intuition behind the concept?
|
You can think of the pushout of two maps f : A → B, g : A → C in Set as computing the disjoint union of B and C with an identification f(a) = g(a) for each element a of A. We could imagine forming this as either the quotient by an equivalence relation, or by gluing in a segment joining f(a) to g(a) for each a, and taking π 0 of the resulting space. If two elements a, a' of A satisfy f(a) = f(a') and g(a) = g(a'), the pushout is unaffected by removing a' from A. The homotopy pushout is formed by gluing in a segment joining f(a) to g(a) for each a and not forgetting the number of ways in which two elements of B ∐ C are identified; instead we take the entire space as the result. It is the "derived" version of the pushout. In general you can think of the homotopy pushout of A → B, A → C as the "free" thing generated by B and C with "relations" coming from A. But it's important that the "relations" are imposed exactly once, since in the homotopical/derived setting we keep track of such things (and have "relations between relations" etc.) Another, possibly more familiar example: In a derived category, the mapping cone of a morphism f : A → B is the homotopy pushout of f and the zero map A → 0. This certainly depends on A, even when B is the zero object: it is the suspension of A.
|
{
"source": [
"https://mathoverflow.net/questions/8684",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1797/"
]
}
|
8,731 |
Can there be a foundations of mathematics using only category theory, i.e. no set theory? More precisely, the definition of a category is a class/set of objects and a class/set of arrows, satisfying some axioms that make commuting diagrams possible. So although in question 7627, where psihodelia asked for alternative foundations for mathematics without set theory, there Steven Gubkin said that Lawvere and McCarthy did some work in reformulation the set theoretic axioms ZFC as the axioms of elementary topoi, this manner of foundations is still not complete since a category is still ultimately a set! J Williams in his answer below noted that via metacategories, we can have a first order axiomatization of categories. However, this does not provide a foundations of mathematics using only category theory, since set theory permeates the formulation of first order logic. In first order logic, structures are sets together with constants, functions and relations. Here constants, functions and relations are also sets. So even if we say that categories are first order axiomatizable, at the very end, categories are still defined in terms of sets. I admit in wanting foundations totally in terms of categories, then there will be some kind of recursiveness. However, this recursiveness should not be seen as a problem since as described above, first order axiomatization of sets like ZFC, are written in the language of first order logic which (at least in a meta-level) are sets themselves. In fact, this recursiveness is very much a feature of symbolic logic and is partially responsible for the successful that a single primitve concept of set/set-membership can describe so much (or all?) of mathematics. I'm aware also in certain proofs of equivalence of categories in mainstream math, like GAGA theorems by Serre, there is a need to use categories where the objects are of classes of different levels, like the NBG set theory. In the end, the reasons provided for why the argument of using classes can be pushed down to essentially small category, this in the end invokes NBG set theory.
|
On the subject of categorical versus set-theoretic foundations there
is too much complicated discussion about structure that misses the
essential point about whether "collections" are necessary. It doesn't matter exactly what your personal list of mathematical
requirements may be -- rings, the category of them, fibrations,
2-categories or whatever -- developing the appropriate foundational
system for it is just a matter of "programming", once you understand
the general setting. The crucial issue is whether you are taken in by the Great Set-Theoretic
Swindle that mathematics depends on collections (completed infinities).
(I am sorry that it is necessary to use strong language here in order to
flag the fact that I reject a widely held but mistaken opinion.) Set theory as a purported foundation for mathematics does not and cannot
turn collections into objects. It just axiomatises some of the intuitions
about how we would like to handle collections, based on the relationship
called "inhabits" (eg "Paul inhabits London", "3 inhabits N"). This
binary relation, written $\epsilon$, is formalised using first order
predicate calculus, usually with just one sort, the universe of sets.
The familiar axioms of (whichever) set theory are formulae in first order
predicate calculus together with $\epsilon$. (There are better and more modern ways of capturing the intuitions about
collections, based on the whole of the 20th century's experience of algebra
and other subjects, for example using pretoposes and arithmetic universes,
but they would be a technical distraction from the main foundational issue.) Lawvere's "Elementary Theory of the Category of Sets" axiomatises some
of the intuitions about the category of sets, using the same methodology.
Now there are two sorts (the members of one are called "objects" or "sets"
and of the other "morphisms" or "functions"). The axioms of a category
or of an elementary topos are formulae in first order predicate calculus
together with domain, codomain, identity and composition. Set theorists claim that this use of category theory for foundations
depends on prior use of set theory, on the grounds that you need to start
with "the collection of objects" and "the collection of morphisms".
Curiously, they think that their own approach is immune to the same
criticism. I would like to make it clear that I do NOT share this view of Lawvere's. Prior to 1870 completed infinities were considered to be nonsense. When you learned arithmetic at primary school, you learned some rules that
said that, when you had certain symbols on the page in front of you,
such as "5+7", you could add certain other symbols, in this case "=12".
If you followed the rules correctly, the teacher gave you a gold star,
but if you broke them you were told off. Maybe you learned another set of rules about how you could add lines and
circles to a geometrical figure ("Euclidean geometry"). Or another one
involving "integration by parts". And so on. NEVER was there a "completed
infinity". Whilst the mainstream of pure mathematics allowed itself to be seduced
by completed infinities in set theory, symbolic logic continued and
continues to formulate systems of rules that permit certain additions
to be made to arrays of characters written on a page. There are many
different systems -- the point of my opening paragraph is that you can
design your own system to meet your own mathematical requirements --
but a certain degree of uniformity has been achieved in the way that they
are presented. We need an inexhaustible supply of VARIABLES for which we may substitute. There are FUNCTION SYMBOLS that form terms from variables and other terms. There are BASE TYPES such as 0 and N, and CONSTRUCTORS for forming new
types, such as $\times$, $+$, $/$, $\to$, .... There are TRUTH VALUES ($\bot$ and $\top$), RELATION SYMBOLS ($=$)
and CONNECTIVES and QUANTIFIERS for forming new predicates. Each variable has a type, formation of terms and predicates must respect
certain typing rules, and each formation, equality or assertion of a
predicate is made in the CONTEXT of certain type-assignments and
assumptions. There are RULES for asserting equations, predicates, etc. We can, for example, formulate ZERMELO TYPE THEORY in this style. It has
type-constructors called powerset and {x:X|p(x)} and a relation-symbol
called $\epsilon$. Obviously I am not going to write out all of the details
here, but it is not difficult to make this agree with what ordinary
mathematicians call "set theory" and is adequate for most of their
requirements Alternatively, one can formulate the theory of an elementary topos is this
style, or any other categorical structure that you require. Then a "ring"
is a type together with some morphisms for which certain equations are
provable. If you want to talk about "the category of sets" or "the category of rings"
WITHIN your tpe theory then this can be done by adding types known as
"universes", terms that give names to objects in the internal category
of sets and a dependent type that provides a way of externalising
the internal sets. So, although the methodology is the one that is practised by type theorists,
it can equally well be used for category theory and the traditional purposes
of pure mathematics. (In fact, it is better to formalise a type theory
such as my "Zermelo type theory" and then use a uniform construction to
turn it into a category such as a topos. This is easier because the
associativity of composition is awkward to handle in a recursive setting.
However, this is a technical footnote.) A lot of these ideas are covered in my book "Practical Foundations of
Mathematics" (CUP 1999), http://www.PaulTaylor.EU/Practical-Foundations Since writing the book I have written things in a more type-theoretic
than categorical style, but they are equivalent. My programme called
"Abstract Stone Duality", http://www.PaulTaylor.EU/ASD is an example of the
methodology above, but far more radical than the context of this question
in its rejection of set theory, ie I see toposes as being just as bad.
|
{
"source": [
"https://mathoverflow.net/questions/8731",
"https://mathoverflow.net",
"https://mathoverflow.net/users/-1/"
]
}
|
8,741 |
Here is a topic in the vein of Describe a topic in one sentence and Fundamental examples : imagine that you are trying to explain and justify a mathematical theory T to a skeptical mathematician who thinks T is just some sort of abstract nonsense for its own sake. The ideal solution consists
of a problem P which can be stated and understood without knowing anything about T, but which is difficult (or impossible, even better) to solve without T, and easier (or almost-trivial, even better) to solve with the help of T. What should be avoided is an example where T is "superimposed", e.g. when T is a model for some physical phenomenon, because there is always something arbitrary about the choice of a specific model. A classical example is Galois theory for solving polynomial equations. Any examples for homological algebra ? For Fourier analysis ? For category theory ?
|
[In front of a blackboard, in an office at Real College] Skeptic: And why should I care about holomorphic functions? Holomorphic enthusiast:$\;$ Can you compute $\quad$ $\sum_{n={-\infty}}^{\infty} \frac{1}{(a+n)^2}$ ? Here $a$ is one of your cherished real numbers, but not an integer. Skeptic: Well, hm... Holomorphic enthusiast, nonchalantly: Oh, you just get $$\sum_{n={-\infty}}^{\infty} \frac{1}{(a+n)^2}=\pi^2 cosec^2 \pi a $$ It's easy using residues. Skeptic: Well, maybe I should have a look at these "residues". Holomorphic enthusiast (generously): Let me lend you this introduction to Complex Analysis by Remmert, this one by Lang and this oldie by Titchmarsh. As Hadamard said: "Le plus court chemin entre deux vérités dans le domaine réel passe par le domaine complexe".You can look for a translation at Mathoverflow. They have a nice list of mathematical quotations, following a question there. Skeptic: Mathoverflow ?? Holomorphic enthusiast (looking a bit depressed) : I think we should have a nice long walk together now.
[Exeunt]
|
{
"source": [
"https://mathoverflow.net/questions/8741",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2389/"
]
}
|
8,756 |
So there are easy examples for algebraic closures that have index two and infinite index: $\mathbb{C}$ over $\mathbb{R}$ and the algebraic numbers over $\mathbb{Q}$. What about the other indices? EDIT: Of course $\overline{\mathbb{Q}} \neq \mathbb{C}$. I don't know what I was thinking.
|
Theorem (Artin-Schreier, 1927): Let K be an algebraically closed field and F a proper subfield of K with $[K:F] < \infty$. Then F is real-closed and $K = F(\sqrt{-1})$. See e.g. Jacobson, Basic Algebra II, Theorem 11.14.
|
{
"source": [
"https://mathoverflow.net/questions/8756",
"https://mathoverflow.net",
"https://mathoverflow.net/users/96/"
]
}
|
8,789 |
Let $M$ be a (real) manifold. Recall that an analytic structure on $M$ is an atlas such that all transition maps are real-analytic (and maximal with respect to this property). (There's also a sheafy definition.) So in particular being analytic is a structure, not a property. Q1: Is it true that any topological manifold can be equipped with an analytic structure? Q2: Can any $C^\infty$ manifold (replace "analytic" by "$C^\infty$" in the first paragraph) be equipped with an analytic structure (consistent with the smooth structure)?
|
(similar to Mariano's post) Q1: no. There are topological manifolds that don't admit triangulations, let alone smooth structures. All smooth manifolds admit triangulations, this is a theorem of Whitehead's. The lowest-dimensional examples of topological manifolds that don't admit triangulations are in dimension 4, the obstruction is called the Kirby-Siebenmann smoothing obstruction. Q2: $C^1$ manifolds all admit compatible $C^\infty$ and analytic ( $C^\omega$ ) structures. This is a theorem of Hassler Whitney's, in his early papers on manifold theory, where he proves they embed in euclidean space. The basic idea is that your manifold is locally cut out of euclidean space by $C^1$ -functions so you apply a smoothing operator to the function and then argue that the level-set does not change (up to $C^1$ -diffeomorphism), provided your smoothing approximation is small enough in the $C^1$ -sense. I'm not sure who gets the original credit but you can go much further -- compact boundaryless smooth manifolds are all realizable as components of real affine algebraic varieties, planar linkages in particular. There's a Millson and Kapovich paper, Universality theorems for configuration spaces of planar linkages on the topic available. It seems people give a lot of credit to Bill Thurston. edit: Some time ago Riccardo Benedetti sent me some comments to append to my answer. They appear below, with some minor MO-formatting on my part. In the famous paper "Real algebraic manifolds" (Annals of Math 56, 3, 1952), John Nash just proved that: "Every compact closed smooth manifold M embedded in some $R^N$ , with $N$ big enough (as usually $N=2Dim(M)+1$ suffices), can be smoothly approximated by a union of components, say $M_a$ , of the non-singular locus of a real algebraic subset $X$ of $R^N$ ." In the same paper he stated also some conjectures/questions, in particular whether one can get $M_a = X$ (so that $X$ is a non-singular real algebraic model of $M$ ), or whether one can even get such an algebraic model which is rational. A.H. Wallace (1957) solved positively the first question under the assumption that $M$ is a boundary. Finally a complete positive answer was given by A. Tognoli (1973) by using, among other things, a so called "Wallace trick" and the fact (due to Milnor) that the smooth/un-oriented bordism group is generated by real algebraic projective non singular varieties. Starting from this Nash-Tognoli theorem, mostly in the 80's-90's of the last century, a huge activity has been developed about the existence of real algebraic models for several instances of smooth or polyhedral structures, with major contributions by S. Akbulut and H. King (in particular they proved that if M is embedded in $R^n$ , an algebraic model can be realized in $R^{n+1}$ ; to my knowledge it is open if we can stay in the given $R^n$ ). If I am not wrong, the realization of real algebraic varieties via planar linkages (with related credit to Thurston) does not provide an alternative proof of Nash-Tognoli theorem. The "Nash rationality conjecture" is more intriguing and has been basically "solved" in dimension less or equal to 3. This is mentioned for instance in some answers to the questions: What's the difference between a real manifold and a smooth variety? "You might also be interested in some of the articles by Kolla'r on the Nash conjecture contrasting real varieties and real manifolds. such as
"What are the simplest varieties?", Bulletin, vol 38. I like the pair of theorems 54, 51, subtitled respectively: "the Nash conjecture is true in dim 3",
and "The Nash conjecture is false in dim 3". What is known about the MMP over non-algebraically closed fields "Another issue is the rational connectivity and its relation to Mori fiber spaces ...... To illustrate the difficulties, here is a conjecture of Nash (yes, that Nash): Let $Z$ be a smooth real algebraic variety. Then $Z$ can be realized as the real points of a rational complex algebraic variety. This, actually, turns out to be false. Kollár calls it the shortest lived conjecture as it was stated in 1954 and disproved by Comessatti around 1914 (I don't remember the exact year). However, even if the statement is false, just the fact that it was made and no one realized for 50 years that it was false should show that these questions are by no means easy. (Comessatti's paper was in Italian and I have no idea how Kollár found it.) Kollár showed more systematically the possible topology types of manifolds that can satisfy this statement. In particular, Kollár shows that any closed connected
3-manifold occurs as a possibly non-projective real variety birationally equivalent to $P^3$ . In other words the way Nash's conjecture fails is on the verge of the difference between projective and proper again showing that these questions are not easy." (Sandor Kovacs) As I have been even more concerned with, let me add a few comments. Comessatti's result was certainly "well known" at least to some Italian people and also to the real algebraic Russian guys of Arnol'd's and mostly Rokhlin's school. Moreover (before Kollar's work) it had been rediscovered (via modern tools)
for instance by R. Silhol. This allows the following 2D solution of the Nash conjecture: " (1) Every non orientable compact closed surface admits a rational projective non singular model which can be explicitly given by (algebraically) blowing-up $\mathbb RP^2$ at some points; (2) $S^2$ and $T^2$ are the only orientable surfaces that admit non singular projective rational models (Comessatti); (3) Every surface $S$ admits a projective rational model possibly having one singular point (to get it, first smoothly blow-up $S$ at one point $p$ getting a
non orientable $S'$ containing a smooth exceptional curve $C$ over $p$ . Take a projective non singular rational model $S'_r$ of $S$ .
Finally one can prove that $C$ can be approximated in $S'_r$ by a non singular algebraic curve $C_a$ and that we can perform a "singular" blow-down of $C_a$ producing the required singular rational model of $S$ .)" Inspired by this 2D discussion, in the paper (with A. Marin) Dechirures de varietes de dimension trois et la conjecture de Nash de rationalite' en dimension trois Comment. Math. Helv. 67 (4) (1992), 514-545 (a PDF file is available in http://www.dm.unipi.it/~benedett/research.html ) we got a formally similar 3D solution. More precisely we provide at first a complete classification of compact closed smooth 3-manifolds $M$ up to "flip/flop" along smooth centres (see below - these are the "Dechirures" - perhaps we were wrong to write the paper in French ... ). Summarizing (and roughly) the results are: There is a complete invariant $I(M)$ for this equivalence relation. Depending only on $I(M)$ , we explicitly produce a real projective non singular
rational 3-fold $Z$ such that $I(M)=I(Z)$ . There is a smooth link $L$ in $M$ and a non singular real algebraic link $L_a$ in $Z$ such that by smoothly (algebraically) blowing up $M (Z)$ along $L (L_a)$ we get the same manifold $Z'$ and furthermore the disjoint real algebraic exceptional tori over $L_a$ coincide with the exceptional tori over $L$ (thinking all within the smooth category, basically by definition this means that $Z$ and $M$ are equivalent up to flip/flop along smooth centres). Clearly $Z'$ is projective rational non singular and algebraically dominates $Z$ . It also smoothly dominates M. Finally we can convert the smooth
blow-down onto $M$ to a singular algebraic blow-down producing a projective rational model $M_r$ of $M$ , possibly singular at a non singular
real algebraic copy of the link $L$ in $M_r$ . The invariant $I(M)$ is easy when $M$ is orientable; this is just the dimension of $H_2(M;\mathbb Z/2)$ . In this case $Z$ is obtained by algebraically blowing up $S^3$ at some points. $I(M)$ is much more complicated if M is non orientable and involves, among other things, certain quadratic forms on $Ker(i_*:H_1(S;\mathbb Z/2) \to H_1(M;\mathbb Z/2))$ , $S$ being any characteristic surface of $M$ . A combination of our work with Kollar's one roughly gives: (3D "a la" Comessatti) In the projective framework, in general our singular rational models cannot be improved (singularity cannot be avoided, and in a sense we provided models with very mild singularities); in other words those blowing-down $Z'\to M_r$ were intrinsically singular. (Non projective non singular rational models) Starting from our real projective rational non singular 3-fold $Z'$ , as above, Kollar proves that one can realize a non singular blow-down $Z'\to M'_r$ , provided that we leave the projective framework. Finally it is intriguing to note another important occurrence of the opposition projective singular vs non singular (related to the existence of intrinsically singular blow-down). Going back to the original Nash-Tognoli kind of problems, for a while it was conjectured and very "desidered" (for several reasons also related to the general question of characterizing say the compact polyhedron which admit possibly singular real algebraic models) that every M as above admits a "totally algebraic model" M_a, i.e. such that $H_*(M_a;\mathbb Z/2)$ is generated by the $\mathbb Z/2$ -fundamental class of algebraic sub-varieties of $M_a$ . On the contrary we constructed counterexamples in: (with M. Dedo') Counterexamples to representing homology classes by real algebraic subvarieties up to homeomorphism, Compositio Math. 53 (2) (1984), 143-151 (idem) This contrasts with a result by Akbulut-King that M admits singular totally algebraic models.
|
{
"source": [
"https://mathoverflow.net/questions/8789",
"https://mathoverflow.net",
"https://mathoverflow.net/users/78/"
]
}
|
8,800 |
K-theory sits in an intersection of a whole bunch of different fields, which has resulted in a huge variety of proof techniques for its basic results. For instance, here's a scattering of proofs of the Bott periodicity theorem for topological complex K-theory that I've found in the literature: Bott's original proof used Morse theory, which reappeared in Milnor's book Morse Theory in a much less condensed form. Pressley and Segal managed to produce the homotopy inverse of the usual Bott map as a corollary in their book Loop Groups . Behrens recently produced a novel proof based on Aguilar and Prieto, which shows that various relevant maps are quasifibrations, therefore inducing the right maps on homotopy and resulting in Bott periodicity. Snaith showed that $BU$ is homotopy equivalent to $CP^\infty$ once you adjoin an invertible element. (He and Gepner also recently showed that this works in the motivic setting too, though this other proof relies on the reader having already seen Bott periodicity for motivic complex K-theory.) Atiyah, Bott, and Shapiro in their seminal paper titled Clifford Modules produced an algebraic proof of the periodicity theorem. EDIT: Whoops x2! They proved the periodicity of the Grothendieck group of Clifford modules, as cdouglas points out, then used topological periodicity to connect back up with $BU$. Wood later gave a more general discussion of this in Banach algebras and Bott periodicity . Atiyah and Bott produced a proof using elementary methods, which boils down to thinking hard about matrix arithmetic and clutching functions. Variations on this have been reproduced in lots of books, e.g., Switzer's Algebraic Topology: Homotopy and Homology . A proof of the periodicity theorem also appears in Atiyah's book K-Theory , which makes use of some basic facts about Fredholm operators. A differently flavored proof that also rests on Fredholm operators appears in Atiyah's paper Algebraic topology and operations on Hilbert space . Atiyah wrote a paper titled Bott Periodicity and the Index of Elliptic Operators that uses his index theorem; this one is particularly nice, since it additionally specifies a fairly minimal set of conditions for a map to be the inverse of the Bott map. Seminaire Cartan in the winter of '59-'60 produced a proof of the periodicity theorem using "only standard techniques from homotopy theory," which I haven't looked into too deeply, but I know it's around. Now, for my question: the proofs of the periodicity theorem that make use of index theory are in some vague sense appealing to the existence of various Thom isomorphisms. It seems reasonable to expect that one could produce a proof of Bott periodicity that explicitly makes use of the facts that: The Thom space of the tautological line bundle over $CP^n$ is homeomorphic to $CP^{n+1}$. Taking a colimit, the Thom space of the tautological line bundle over $CP^\infty$ (call it $L$) is homeomorphic to $CP^\infty$. The Thom space of the difference bundle $(L - 1)$ over $CP^\infty$ is, stably, $\Sigma^{-2} CP^\infty$. This seems to me like a route to producing a representative of the Bott map. Ideally, it would even have good enough properties to produce another proof of the periodicity theorem. But I can't find anything about this in the literature. Any ideas on how to squeeze a proof out of this -- or, better yet, any ideas about where I can find someone who's already done the squeezing? Hope this isn't less of this is nonsense! -- edit -- Given the positive response but lack of answers, I thought I ought to broaden the question a bit to start discussion. What I was originally looking for was a moral proof of the periodicity theorem -- something short that I could show to someone with a little knowledge of stable homotopy as why we should expect the whole thing to be true. The proofs labeled as elementary contained too much matrix algebra to fit into parlor talk, while the proofs with Fredholm operators didn't seem -- uh -- homotopy-y enough. While this business with Thom spaces over $CP^\infty$ seemed like a good place to look, I knew it probably wasn't the only place. In light of Lawson's response, now I'm sure it isn't the only place! So: does anyone have a good Bott periodicity punchline, aimed at a homotopy theorist? (Note: I'll probably reserve the accepted answer flag for something addressing the original question.)
|
Here is my attempt to address Eric's actual question. Given a real $n$-dimensional vector bundle $E$ on a space $X$, there is an associated Thom space that can be understood as a twisted $n$-fold suspension $\Sigma^E X$. (If $E$ is trivial then it is a usual $n$-fold suspension $\Sigma^n X$.) In particular, if $E=L$ is a complex line bundle, it is a twisted double suspension. In particular, if $X = \mathbb{C}P^\infty$, the twisted double suspension of the tautological line bundle $L$ satisfies the equation
$$\Sigma^L\mathbb{C}P^\infty = \mathbb{C}P^\infty.$$
As I understand it, Eric wants to know whether this periodicity can be interpreted as a Bott map, maybe after some modification, and then used to prove Bott periodicity. What I am saying matches Eric's steps 1 and 2. Step 3 is a modification to make the map look more like Bott periodicity. I think that the answer is a qualified no. On the face of it, Eric's map does not carry the same information as the Bott map. Bott periodicity is a theorem about unitary groups and their classifying spaces. What Eric has in mind, as I understand now, is a result of Snaith that constructs a spectrum equivalent to the Bott spectrum for complex K-theory by modifying $\mathbb{C}P^\infty$. Snaith's model has been called "Snaith periodicity", but the existing arguments that it is the same are a use and not a proof of Bott periodicity. (In that sense, Snaith's model is stone soup , although that metaphor is not really fair to his good paper.) For context, here is a quick definition of Bott's beautiful map as Bott constructed it in the Annals is beautiful. In my opinion, it doesn't particularly need simplification. The map generalizes the suspension relation $\Sigma S^n = S^{n+1}$. You do not need Morse theory to define it; Morse theory is used only to prove homotopy equivalence. Bott's definition: Suppose that $M$ is a compact symmetric space with two points $p$ and $q$ that are connected by many shortest geodesics in the same homotopy class. Then the set of these geodesics is another symmetric space $M'$, and there is an obvious map $\Sigma M' \to M$ that takes the suspension points to $p$ and $q$ and interpolates linearly. For example, if $p$ and $q$ are antipodal points of a round sphere $M = S^{n+1}$, the map is $\Sigma(S^n) \to S^{n+1}$. For complex K-theory, Bott uses $M = U(2n)$, $p = q = I_{2n}$, and geodesics equivalent to the geodesic $\gamma(t) = I_n \oplus \exp(i t) I_n$, with $0 \le t \le 2\pi$. The map is then
$$\Sigma (U(2n)/U(n)^2) \to U(2n).$$
The argument of the left side approximates the classifying space $BU(n)$. Bott show that this map is a homotopy equivalence up to degree $2n$. Of course, you get the nicest result if you take $n \to \infty$. Also, to complete Bott periodicity, you need a clutching function map $\Sigma(U(n)) \to BU(n)$, which exists for any compact group. (If you apply the general setup to $M = G$ for a simply connected, compact Lie group, Bott's structure theorem shows that $\pi_2(G)$ is trivial; c.f. this related MO question .) At first glance, Eric's twisted suspension is very different. It exists for $\mathbb{C}P^\infty = BU(1)$, and of course $\mathbb{C}P^\infty$ is a $K(\mathbb{Z},2)$ space with a totally different homotopy structure from $BU(\infty)$. Moreover, twisted suspensions aren't adjoint to ordinary delooping. Instead, the space of maps $\Sigma^L X \to Y$ is adjoint to sections of a bundle over $X$ with fiber $\mathcal{L}^2 Y$. The homotopy structure of the twisted suspension depends on the choice of $L$. For instance, if $X = S^2$ and $L$ is trivial, then $\Sigma^L S^2 = S^4$ is the usual suspension. But if $L$ has Chern number 1, then $\Sigma^L S^2 = \mathbb{C}P^2$, as Eric computed. However, in Snaith's paper all of that gets washed away by taking infinitely many suspensions to form $\Sigma_+^{\infty}\mathbb{C}P^\infty$, and then as Eric says adjoining an inverse to a Bott element $\beta$. (I think that the "+" subscript just denotes adding a disjoint base point.) You can see what is coming just from the rational homotopy groups of $\Sigma^\infty \mathbb{C}P^\infty$. Serre proved that the stable homotopy of a CW complex $K$ are just the rational homology $H_*(K,\mathbb{Q})$. (This is related to the theorem that stable homotopy groups of spheres are finite.) Moreover, in stable, rational homotopy, twisted and untwisted suspension become the same. So Snaith's model is built from the fact that the homology of $\mathbb{C}P^\infty$ equals the homotopy of $BU(\infty)$. Moreover, there is an important determinant map
$$\det:BU(\infty) \to BU(1) = \mathbb{C}P^\infty$$
that takes the direct sum operation for bundles to tensor multiplication of line bundles. Snaith makes a moral inverse to this map (and not just in rational homology). Still, searching for a purely homotopy-theoretic proof of Bott periodicity is like searching for a purely algebraic proof of the fundamental theorem of algebra . The fundamental theorem of algebra is not a purely algebraic statement! It is an analytic theorem with an algebraic conclusion, since the complex numbers are defined analytically. The best you can do is a mostly algebraic proof, using some minimal analytic information such as that $\mathbb{R}$ is real-closed using the intermediate value theorem. Likewise, Bott periodicity is not a purely homotopy-theoretic theorem; it is a Lie-theoretic theorem with a homotopy-theoretic conclusion. Likewise, the best you can do is a mostly homotopy-theoretic proof that carefully uses as little Lie theory as possible. The proof by Bruno Harris fits this description. Maybe you could also prove it by reversing Snaith's theorem, but you would still need to explain what facts you use about the unitary groups. (The answer is significantly revised now that I know more about Snaith's result.)
|
{
"source": [
"https://mathoverflow.net/questions/8800",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1094/"
]
}
|
8,829 |
Real projective spaces $\mathbb{R}P^n$ have $\mathbb{Z}/2$ cohomology rings $\mathbb{Z}/2[x]/(x^{n+1})$ and total Stiefel-Whitney class $(1+x)^{n+1}$ which is $1$ when $n$ is odd, so it follows that odd dimensional ones are boundaries of compact $(n+1)$-manifolds. My question is: are there any especially nice constructions of these $(n+1)$-manifolds? I'm especially interested in the case $n=3$. I believe we can get an explicit example of a 4-manifold bound by $\mathbb{R}P^3$ using Rokhlin/Lickorish-Wallace but it doesn't look like that would generalize to higher dimensions at all easily. Are there a lot of different 4-manifolds with this property?
|
$\mathbb RP^3$ double-covers the lens space $L_{4,1}$, so it's the boundary of the mapping cylinder of that covering map. In general $\mathbb RP^n$ for $n$ odd double-covers such a lens space. So in general $\mathbb RP^n$ is the boundary of a pretty standard $I$-bundle over the appropriate lens space. To be specific, define the general $L_{4,1}$ as $S^{2n-1} / \mathbb Z_4$ where $Z_4 \subset S^1$ are the 4-th roots of unity, and we're using the standard action of the unit complex numbers on an odd dimensional sphere $S^{2n-1} \subset \mathbb C^n$. Edit: generalizing Tim's construction, you have the fiber bundle $S^1 \to S^{2n-1} \to \mathbb CP^{n-1}$. This allows you to think of $S^{2n-1}$ as the boundary of the tautological $D^2$-bundle over $\mathbb CP^{n-1}$. You can mod out the whole bundle by the antipodal map and you get $\mathbb RP^{2n-1}$ as the boundary of the disc bundle over $\mathbb CP^{n-1}$ with Euler class $2$. So this gives you an orientable manifold bounding $\mathbb RP^{2n-1}$ while my previous example was non-orientable.
|
{
"source": [
"https://mathoverflow.net/questions/8829",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1772/"
]
}
|
8,846 |
Can you give examples of proofs without words? In particular, can you give examples of proofs without words for non-trivial results? (One could ask if this is of interest to mathematicians , and I would say yes, in so far as the kind of little gems that usually fall under the title of 'proofs without words' is quite capable of providing the aesthetic rush we all so professionally appreciate. That is why we will sometimes stubbornly stare at one of these mathematical autostereograms with determination until we joyously see it.) (I'll provide an answer as an example of what I have in mind in a second)
|
A proof of the identity $$1+2+\cdots + (n-1) = \binom{n}{2}$$ (Adapted from an entry I saw at Wolfram Demonstrations , see also the original faster animation ) This proof was discovered by Loren Larson, professor emeritus at St. Olaf College. He included it along with a number of other, more standard, proofs, in "A Discrete Look at 1+2+...+n," published in 1985 in The College Mathematics Journal (vol. 16, no. 5, pp. 369-382, DOI: 10.1080/07468342.1985.11972910 , JSTOR ).
|
{
"source": [
"https://mathoverflow.net/questions/8846",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1409/"
]
}
|
8,938 |
The Torelli map $\tau\colon M_g \to A_g$ sends a curve C to its Jacobian (along with the canonical principal polarization associated to C); see this question for a description which works for families. Theorem (Torelli): If $\tau(C) \cong \tau(C')$, then $C \cong C'$. If one prefers to work with coarse spaces (instead of stacks) it is okay to just say that $\tau$ is injective. Question: Is $\tau$ an immersion? (One remark: $\tau$ isn't a closed immersion -- the closure of its image consists of products of Jacobians!)
|
Respectfully, I disagree with Tony's answer. The infinitesimal Torelli problem fails for $g>2$ at the points of $M_g$ corresponding to the hyperelliptic curves. And in general the situation is trickier than one would expect. The tangent space to the deformation space of a curve $C$ is $H^1(T_C)$, and the tangent space to the deformation space of its Jacobian is $Sym^2(H^1(\mathcal O_C))$. The infinitesimal Torelli map is an immersion iff the map of these tangent spaces $$ H^1(T_C) \to Sym^2( H^1(\mathcal O_C) )$$ is an injection. Dually, the following map should be a surjection: $$ Sym^2 ( H^0(K_C) ) \to H^0( 2K_C ), $$ where $K_C$ denotes the canonical class of the curve $C$. This is a surjection iff $g=1,2$ or $g=3$ and $C$ is not hyperelliptic; by a result of Max Noether. Therefore, for $g\ge 3$ the Torelli map OF STACKS $\tau:M_g\to A_g$ is not an immersion. It is an immersion outside of the hyperelliptic locus $H_g$. Also, the restriction $\tau_{H_g}:H_g\to A_g$ is an immersion. On the other hand, the Torelli map between the coarse moduli spaces IS an immersion in char 0. This is a result of Oort and Steenbrink "The local Torelli problem for algebraic curves" (1979). F. Catanese gave a nice overview of the various flavors of Torelli maps (infinitesimal, local, global, generic) in "Infinitesimal Torelli problems and counterexamples to Torelli problems" (chapter 8 in "Topics in transcendental algebraic geometry" edited by Griffiths). P.S. "Stacks" can be replaced everywhere by the "moduli spaces with level structure of level $l\ge3$ (which are fine moduli spaces). P.P.S. The space of the first-order deformations of an abelian variety $A$ is $H^1(T_A)$. Since $T_A$ is a trivial vector bundle of rank $g$, and the cotangent space at the origin is $H^0(\Omega^1_A)$, this space equals $H^1(\mathcal O_A) \otimes H^0(\Omega^1_A)^{\vee}$ and has dimension $g^2$. A polarization is a homomorphism $\lambda:A\to A^t$ from $A$ to the dual abelian variety $A^t$. It induces an isomorphism (in char 0, or for a principal polarization) from the tangent space at the origin $T_{A,0}=H^0(\Omega_A^1)^{\vee}$ to the tangent space at the origin $T_{A^t,0}=H^1(\mathcal O_A)$. This gives an isomorphism
$$ H^1(\mathcal O_A) \otimes H^0(\Omega^1_A)^{\vee} \to
H^1(\mathcal O_A) \otimes H^1(\mathcal O_A). $$ The subspace of first-order deformations which preserve the polarization $\lambda$ can be identified with the tensors mapping to zero in $\wedge^2 H^1(\mathcal O_A)$, and so is isomorphic to $Sym^2 H^1(\mathcal O_A)$, outside of characteristic 2.
|
{
"source": [
"https://mathoverflow.net/questions/8938",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2/"
]
}
|
8,957 |
Several times I've heard the claim that any Lie group $G$ has trivial second fundamental group $\pi_2(G)$, but I have never actually come across a proof of this fact. Is there a nice argument, perhaps like a more clever version of the proof that $\pi_1(G)$ must be abelian?
|
I don't know of anything as bare hands as the proof that $\pi_1(G)$ must be abelian, but here's a sketch proof I know (which can be found in Milnor's Morse Theory book. Plus, as an added bonus, one learns that $\pi_3(G)$ has no torsion!): First, (big theorem): Every (connected) Lie group deformation retracts onto its maximal compact subgroup (which is, I believe, unique up to conjugacy). Hence, we may as well focus on compact Lie groups. Let $PG = \{ f:[0,1]\rightarrow G | f(0) = e\}$ (I'm assuming everything is continuous.). Note that $PG$ is contractible (the picture is that of sucking spaghetti into one's mouth). The projection map $\pi:PG\rightarrow G$ given by $\pi(f) = f(1)$ has homotopy inverse $\Omega G = $ Loop space of G = $\{f\in PG | f(1) = e \}$ . Thus, one gets a fibration $\Omega G\rightarrow PG\rightarrow G$ with $PG$ contractible. From the long exact sequence of homotopy groups associated to a fibration, it follows that $\pi_k(G) = \pi_{k-1}\Omega G$ Hence, we need only show that $\pi_{1}(\Omega G)$ is trivial. This is where the Morse theory comes in. Equip $G$ with a biinvariant metric (which exists since $G$ is compact). Then, following Milnor, we can approximate the space $\Omega G$ by a nice (open) subset $S$ of $G\times\cdots \times G$ by approximating paths by broken geodesics. Short enough geodesics are uniquely defined by their end points, so the ends points of the broken geodesics correspond to the points in $S$ . It is a fact that computing low (all?...I forget)* $\pi_k(\Omega G)$ is the same as computing those of $S$ . Now, consider the energy functional $E$ on $S$ defined by integrating $|\gamma|^2$ along the entire curve $\gamma$ . This is a Morse function and the critical points are precisely the geodesics**. The index of E at a geodesic $\gamma$ is, by the Morse Index Lemma, the same as the index of $\gamma$ as a geodesic in $G$ . Now, the kicker is that geodesics on a Lie group are very easy to work with - it's pretty straight forward to show that the conjugate points of any geodesic have even index. But this implies that the index at all critical points is even. And now THIS implies that $S$ has the homotopy type of a CW complex with only even cells involved. It follows immediately that $\pi_1(S) = 0$ and that $H_2(S)$ is free ( $H_2(S) = \mathbb{Z}^t$ for some $t$ ). Quoting the Hurewicz theorem, this implies $\pi_2(S)$ is $\mathbb{Z}^t$ . By the above comments, this gives us both $\pi_1(\Omega G) = 0$ and $\pi_2(\Omega G) = \mathbb{Z}^t$ , from which it follows that $\pi_2(G) = 0$ and $\pi_3(G) = \mathbb{Z}^t$ . Incidentally, the number $t$ can be computed as follows. The universal cover $\tilde{G}$ of $G$ is a Lie group in a natural way. It is isomorphic (as a manifold) to a product $H\times \mathbb{R}^n$ where $H$ is a compact simply connected group. H splits isomorphically as a product into pieces (all of which have been classified). The number of such pieces is $t$ . (edits) *- it's only the low ones, not "all", but one can take better and better approximations to get as many "low" k as one wishes. **- I mean CLOSED geodesics here
|
{
"source": [
"https://mathoverflow.net/questions/8957",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2510/"
]
}
|
8,970 |
I've heard that the problem of counting topologies is hard, but I couldn't really find anything about it on the rest of the internet. Has this problem been solved? If not, is there some feature that makes it pretty much intractable?
|
It's wiiiiiiide open to compute it exactly. As far as I know the "feature that makes it intractable" is that there's no real feature that makes it tractable. Very broadly speaking, if you want to count the ways that a generic type of structure can be put on an n-element set, there's no efficient way to do this -- you basically have to enumerate the structures one by one. This is essentially because "given a description of a structure type and $n$, count the number of structures on an n-element set" is a ridiculously broad problem which ends up reducing to lots and lots of different counting and decision problems. Alternatively I think you can argue via Kolmogorov complexity and all that, but that's not my style? So in any case, the burden of proof is on the person who claims that an efficient counting algorithm should exist. (If you believe some crazy things about complexity theory, like P = PSPACE, this starts to become less true since the structure types hard to enumerate will usually be hard to describe. But if you believe that you're a lost cause in any case :P) It's still reasonable to ask for further justification, though. I'd attempt to give some, but I've been awake for like 30 hours and it would be even more handwavey than the above. The short version: If you do enough enumerative combinatorics, you start to see that nice formulas for enumerating structures arise from one of a few situations. Some really big ones are: The ability to derive a sufficiently nice recurrence relation. This is a rather nebulous property, and some surprising structure types have cute recurrence relations. I can't really tell you a good solid reason why the number of finite topologies doesn't admit a nice recurrence relation; if you work with it a while, it just doesn't feel like it does. A classification theorem such that the structures in each class have nice formulas. Sometimes these are deep, sometimes they aren't. One non-deep example is in the usual (non-linear-algebra) proof of Cayley's formula on the number of trees. Point-set topology is weird , and it's pretty weird even in the finite case. This is out of the question. If the set of structures on a set of n elements is very rigid, there may be an "algebraic" way of counting them. Again, point-set topology is too weird for this to kick in. So my answer boils down to: It's intractable 'cause it is. Not a particularly satisfying reason, but sometimes that's the way combinatorics works. Sorry if you read that whole post -- I meant for it to be shorter and have more content, but it ended up like most tales told by idiots. But hopefully you learned something, or at least had fun with it?
|
{
"source": [
"https://mathoverflow.net/questions/8970",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1353/"
]
}
|
8,972 |
I have been thinking about which kind of wild non-measurable functions you can define. This led me to the question: Is it possible to prove in ZFC, that if a ( Edit : measurabel) set $A\subset \mathbb{R}$ has positive Lebesgue-measure, it has the same cardinality as $\mathbb{R}$? It is obvious if you assume CH, but can you prove it without CH?
|
I found the answer in the paper "Measure and cardinality" by Briggs and Schaffter . In short: not if I interpret positive measure to mean positive outer measure. A proof is given that every measurable subset with cardinality less than that of $\mathbb{R}$ has Lebesgue measure zero. However, they then survey results of Solovay that show that there are models of ZFC in which CH fails and every subset of cardinality less than that of $\mathbb{R}$ is measurable, and that there are models of ZFC in which CH fails and there are subsets of cardinality $\aleph_1$ that are nonmeasurable. So it is undecidable in ZFC. If it was intended that our sets are assumed to be measurable, then the answer would be yes by the first part above. Edit : In light of the comment by Konrad I added a couple of lines to clarify.
|
{
"source": [
"https://mathoverflow.net/questions/8972",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2097/"
]
}
|
9,000 |
Wikipedia says that the intermediate value theorem “depends on (and is actually equivalent to) the completeness of the real numbers.” It then offers a simple counterexample to the analogous proposition on ℚ and a proof of the theorem in terms of the completeness property. Does an analogous result hold for the computable reals (perhaps assuming that the function in question is computable)? If not, is there a nice counterexample?
|
Let me assume that you are speaking about computable reals and functions in the sense of computable analysis , which is one of the most successful approaches to the topic. (One must be careful, since there are several incompatible notions of computability on the reals.) In computable analysis, the computable real numbers are those that can be computed to within any desired precision by a finite, terminating algorithm (e.g. using Turing machines). One should imagine receiving rational approximations to the given real. In this subject, functions on the reals are said to be computable, if there is an algorithm that can compute, for any desired degree of accuracy, the value of the function, for any algorithm that produces approximations to the input with sufficient accuracy. That is, if we want to know f(x) to within epsilon, then the algorithm is allowed to ask for x to within any delta it cares to. The Computable Intermediate Value Theorem would be the assertion that if f is a computable continuous function and f(a)< c<f(b) for computable reals a, b, c, then there is a computable real d with f(d)=c. The book Computable analysis: an introduction by Klaus Weihrauch discusses exactly this question in Example 6.3.6. The basic situation is as follows. The answer is Yes . If f happens to be increasing, then the usual bisection proof of existence turns out to be effective. For other f, however, one can use a trisection proof. Theorem 6.3.8 says that if f is computable and f(x)*f(z)<0, then f has a computable zero. This implies the Computable Intermediate Value theorem above. In contrast, the same theorem also says that there is a non-negative computable continuous function f on [0,1], such that the sets of zeros of f has Lebesgue measure greater than 1/2, but f has no computable zero. In summary, if the function crosses the line, you can compute a crossing point, but if it stays on one side, then you might not be able to compute a kissing point, even if it is kissing on a large measure set.
|
{
"source": [
"https://mathoverflow.net/questions/9000",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2599/"
]
}
|
9,007 |
Let X be a topological space, let $\mathcal{U} = \{U_i\}$ be a cover of X, and let $\mathcal{F}$ be a sheaf of abelian groups on X. If X is separated, each $U_i$ is affine, and $\mathcal{F}$ is quasi-coherent, then Cech cohomology computes derived functor cohomology; in general one only gets a spectral sequence
$$
H^p(\mathcal{U},\underline{H}^q(\mathcal{F})) \Rightarrow H^{p+q}(X,\mathcal{F})
$$
where $\underline{H}^q(\mathcal{F})$ is the presheaf $U \mapsto H^q(U,\mathcal{F}|_U)$. Question : For q > 0, $\underline{H}^q(\mathcal{F})$ sheafifies to 0. For a quasi-coherent sheaf $\mathcal{F}$ this is clear because cohomology vanishes on affines. Is this really true in general? Brian Conrad states this in the introduction to his notes on cohomological descent.
|
Yes, this is true in general. It suffices to show the stalks vanish. Pick $x \in X$ and take an injective resolution $0 \to {\cal F} \to I^0 \to \cdots$. For any open $U$ containing $x$, we get a chain complex $$0 \to I^0(U) \to I^1(U) \to \cdots$$ whose cohomology groups are $H^p(U,{\cal F}|_U)$. Taking direct limits of these sections gives the chain complex $$0 \to I^0_x \to I^1_x \to \cdots$$ of stalks, which has zero cohomology in positive degrees because the original complex was a resolution. However, direct limits are exact and so we find $$0 = {\rm colim}_{x \in U} H^p(U,{\cal F}|_U) = {\underline H}^p({\cal F})_x$$ as desired. Generally, cohomology tells you the obstructions to patching local solutions into global solutions, and this says that locally those obstructions vanish.
|
{
"source": [
"https://mathoverflow.net/questions/9007",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2/"
]
}
|
9,037 |
My apologies if this is too elementary, but it's been years since I heard of this paradox and I've never heard a satisfactory explanation. I've already tried it on my fair share of math Ph.D.'s, and some of them postulate that something deep is going on. The Problem: You are on a game show. The host has chosen two (integral and distinct) numbers and has hidden them behind doors A and B. He allows you to open one of the doors, thus revealing one of the numbers. Then, he asks you: is the number behind the other door bigger or smaller than the number you have revealed? Your task is to answer this question correctly with probability strictly greater than one half. The Solution: Before opening any doors, you choose a number $r$ at random using any continuous probability distribution of your choice. To simplify the analysis, you repeat until $r$ is non-integral. Then you open either door (choosing uniformly at random) to reveal a number $x$. If $r < x$, then you guess that the hidden number $y$ is also smaller than $x$; otherwise you guess that $y$ is greater than $x$. Why is this a winning strategy? There are three cases: 1) $r$ is less than $x$ and $y$. In this case, you guess "smaller" and win the game if $x > y$. Because variables $x$ and $y$ were assigned to the hidden numbers uniformly at random, $P(x > y) = 1/2$. Thus, in this case you win with probability one half. 2) $r$ is greater than $x$ and $y$. By a symmetric argument to (1), you guess "larger" and win with probability one half. 3) $r$ is between $x$ and $y$. In this case, you guess "larger" if $x < y$ and "smaller" if $x > y$ -- that is, you always win the game. Case 3 occurs with a finite non-zero probability $\epsilon$, equivalent to the integral of your probability distribution between $x$ and $y$. Averaging over all the cases, your chance of winning is $(1+\epsilon)/2$, which is strictly greater than half. The Paradox: Given that the original numbers were chosen "arbitrarily" (i.e., without using any given distribution), it seems impossible to know anything about the relation between one number and the other. Yet, the proof seems sound. I have some thoughts as to the culprit, but nothing completely satisfying. Insightful members, could you please help me out with this one? Thanks!
|
After Bill's latest clarifications in the commentary on Critch's answer, I think the question is interesting again. My take: One thing that always seemed to fall through the cracks when I learned about probability theory is that probability is intricately tied to information, and probabilities are only defined in the context of information. Probabilities aren't absolute; two people who have different information about an event may well disagree on its probability, even if both are perfectly rational. Similarly, if you get new information relevant to a certain event, then you should probably reevaluate what you think is the probability that it will occur. Your particular problem is interesting because the new information you get isn't enough for you to revise that probability by purely mathematical considerations, but I'll get to that in good time. With the previous paragraph in mind, let's compare two games: G1. You are given two closed doors, A and B, with two numbers behind them, and your goal is to choose the door with the higher number. You are given no information about the doors or numbers. G2. You are given two closed doors, A and B, with two numbers behind them, and your goal is to choose the door with the higher number. You are allowed to look behind one of the doors and then make your choice. For the first game, by symmetry, you clearly can't do better than choosing a door randomly, which gives you a success probability of exactly 1/2. However, the second game has a chance of being better. You are playing for the same goal with strictly more information, so you might expect to be able to do somewhat better. [I had originally said that it was obviously better, but now I'm not so sure that it's obvious.] The tricky thing is quantifying how much better, since it's not clear how to reason about the relationship between two numbers if you know one of the numbers and have no information about the other one. Indeed, it isn't even possible to quantify it mathematically. "But how can that be?" you may ask. "This is a mathematical problem, so how can the solution not be mathematically definable?" There's the rub: part of the issue is that the problem isn't formulated in a mathematically rigorous way. That can be fixed in multiple ways, and any way we choose will make the paradox evaporate. The problem is that we're asked to reason about "the probability of answering the question correctly," but it's not clear what context that probability should be computed in. (Remember: probabilities aren't absolute.) In common probability theory problems and puzzles, this isn't an issue because there is usually an unambiguous "most general applicable context": we should obviously assume exactly what's given in the problem and nothing else. We can't do that here because the most general context, in which we assume nothing about how the numbers $x$ and $y$ are chosen, does not define a probability space at all and thus the "probability of answering the question correctly" is not a meaningful concept. Here's a simpler ostensible probability question that exhibits the same fallacy: "what's the probability that a positive integer is greater than 1,000,000?" In order to answer that, we have to pick a probability distribution on the positive integers; the question is meaningless without specifying that. As I said, there are multiple ways to fix this. Here are a couple: I1. (Tyler's interpretation.) We really want the probability of answering the question correctly given a particular $x$ and $y$ to be greater than 1/2. (The exact probability will of course depend on the two numbers.) I2. (Critch's interpretation.) More generally, we want the probability of answering correctly given a particular probability distribution for $(x,y)$ to be greater than 1/2. (The exact probability will of course depend on the distribution.) (Those two are actually equivalent mathematically.) Clearly, if we knew what that distribution was, we could cook up strategies to get a success probability strictly above 1/2. That's pretty much obvious. It is not nearly as obvious that a single strategy (such as the one in the statement of the question) can work for all distributions of $(x,y)$ , but it's true, as Bill's proof shows. It's an interesting fact, but hardly paradoxical now. Let me summarize by giving proper mathematical interpretations of the informal statement "there is a strategy that answers the question correctly with probability strictly greater than 1/2," with quantifiers in place: (1a) $\exists \text{ strategy } S: \forall x, y: \exists \delta > 0$ : $S$ answers correctly on $x$ , $y$ with probability at least $1/2 + \delta$ . (1b) $\exists \text{ strategy } S: \forall \text{ probability distributions } P \text{ on } \mathbb{N}^2: \exists \delta > 0$ : $S$ answers correctly, when $x$ , $y$ are chosen according to $P$ , with probability at least $1/2 + \delta$ . I think with the proper quantifiers and the dependence on $x$ and $y$ explicit, it becomes a cool mathematical result rather than a paradox. Actually, based on my arguments at the beginning, it's not even that surprising: we should expect to do better than random guessing, since we are given information. However, simply knowing one number doesn't seem very useful in determining whether the other number is bigger, and that's reflected in the fact that we can't improve our probability by any fixed positive amount without more context. Edit: It occurred to me that the last part of my discussion above has a nonstandard-analytical flavor. In fact, using the first version of the formula for simplicity (the two being equivalent), and the principle of idealisation, I think we immediately obtain: (2) $\exists \text{ strategy } S: \exists \delta > 0: \forall \text{ standard }x, y:$ $S$ answers correctly on $x$ , $y$ with probability at least $1/2 + \delta$ . (Please correct me if I'm wrong.)
The number $\delta$ is not necessarily standard, and a basic argument shows that it must actually be smaller than all standard positive reals, i. e., infinitesimal. Thus, we can say that being able to look behind one door gives us an unquantifiably small, infinitesimal edge over random guessing. That actually meshes pretty well with my intuition! (It might still a nontrivial observation that the strategy $S$ can be taken standard; I'm not sure about that...)
|
{
"source": [
"https://mathoverflow.net/questions/9037",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2614/"
]
}
|
9,066 |
Is that true that there is no rational curves contained in an Abelian variety? If it's true, is that because abelian varieties are not uniruled? How do I know whether an abelian variety is not uniruled?
|
There are no rational curves in an abelian variety, this is much stronger than not being uniruled. If there is a map $P^1 \to A$, $A$ abelian, the map would factor through the Albanese variety of $P^1$, by definition. However, for curves, the Albanese is the Jacobian (from general theory of the Jacobian) and the Jacobian of $P^1$ is a point.
|
{
"source": [
"https://mathoverflow.net/questions/9066",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2348/"
]
}
|
9,073 |
Does anyone have examples of when an object is positive, then it has (or does not have) a square root? Or more generally, can be written as a sum of squares? Example. A positive integer does not have a square root, but is the sum of at most 4 squares. (Lagrange Theorem). However, a real positive number has a square root. Another Example. A real quadratic form that is postive definite (or semi-definite) is, after a change of coordinates, a sum of squares. How about rational or integral quadratic forms? Last Example. A positive definite (or semidefinite) real or complex matrix has a square root. How about rational or integral matrices? Do you have other examples?
|
For many examples of this kind, see Olga Taussky, "Sums of squares", Amer. Math. Monthly 77 (1970) 805-830.
|
{
"source": [
"https://mathoverflow.net/questions/9073",
"https://mathoverflow.net",
"https://mathoverflow.net/users/-1/"
]
}
|
9,125 |
The use of the term "spectrum" to denote the prime ideals of a ring originates from the case that the ring is, say, $\mathbb{C}[T]$ where $T$ is a linear operator on a finite-dimensional vector space; then the prime spectrum (which is equal to the maximal spectrum) is precisely the set of eigenvalues of $T$. The use of the term "spectrum" in the operator sense, in turn, seems to have originated with Hilbert, and was apparently not inspired by the connection to atomic spectra. (This appears to have been a coincidence.) A cursory Google search indicates that Hilbert may have been inspired by the significance of the eigenvalues of Laplacians, but I don't understand what this has to do with non-mathematical uses of the word "spectrum." Does anyone know the full story here?
|
Hilbert, in fact, got the term from Wilhelm Wirtinger (the first one to propose it according to, say http://www.mathphysics.com/opthy/OpHistory.html ) the paper of Wirtinger is "Beiträge zu Riemann’s Integrationsmethode für hyperbolische Differentialgleichungen, und deren Anwendungen auf Schwingungsprobleme" (1897). In http://jeff560.tripod.com/s.html it says "...Wirtinger drew upon the similarity with the optical spectra of molecules when he used the term "Bandenspectrum" with reference to Hill’s (differential) equation." I haven't read Wirtinger's paper, nor do I know how reliable these sources are :)
|
{
"source": [
"https://mathoverflow.net/questions/9125",
"https://mathoverflow.net",
"https://mathoverflow.net/users/290/"
]
}
|
9,134 |
Thinking of arbitrary tensor products of rings, $A=\otimes_i A_i$ ($i\in I$, an arbitrary index set), I have recently realized that $Spec(A)$ should be the product of the schemes $Spec(A_i)$, a priori in the category of affine schemes, but actually in the category of schemes, thanks to the string of equalities (where $X$ is a not necessarily affine scheme) $$ Hom_{Schemes} (X, Spec(A))= Hom_{Rings}(A,\Gamma(X,\mathcal O))=\prod_ {i\in I}Hom_{Rings}(A_i,\Gamma(X,\mathcal O)) $$ $$ =\prod_ {i\in I}Hom_{Schemes}(X,Spec(A_i))$$ Since this looks a little too easy, I was not quite convinced it was correct but a very reliable colleague of mine reassured me by explaining that the correct categorical interpretation of the more down to earth formula above is that the the category of affine schemes is a reflexive subcategory of the category of schemes.
(Naturally the incredibly category-savvy readers here know that perfectly well, but I didn't at all.) And now I am stumped: I had always assumed that infinite products of schemes don't exist and I realize I have no idea why I thought so! Since I am neither a psychologist nor a sociologist, arguments like "it would be mentioned in EGA if they always existed " don't particularly appeal to me and I would be very grateful if some reader could explain to me what is known about these infinite products.
|
Let me rephrase the question (and Ilya's answer). Given an arbitrary collection $X_i$ of schemes, is the functor (on affine schemes, say) $Y \mapsto \prod_i Hom(Y, X_i)$ representable by a scheme? If the $X_i$ are all affine, the answer is yes, as explained in the statement of the question. More generally, any filtered inverse system of schemes with essentially affine transition maps has an inverse limit in the category of schemes (this is in EGA IV.8). The topology in that case is the inverse limit topology, by the way. It is easy to come up with examples of infinite products of non-separated schemes that are not representable by schemes. This is because any scheme has a locally closed diagonal. In other words, if $Y \rightrightarrows Z$ is a pair of maps of schemes then the locus in $Y$ where the two maps coincide is locally closed in $Y$. Suppose $Z$ is the affine line with a doubled origin. Every distinguished open subset of an affine scheme $Y$ occurs as the locus where two maps $Y \rightrightarrows Z$ agree. Let $X = \prod_{i = 1}^\infty Z$. Every countable intersection of distinguished open subsets of $Y$ occurs as the locus where two maps $Y \rightarrow X$ agree. Not every countable intersection of open subsets is locally closed, however, so $X$ cannot be a scheme. Since the diagonal of an infinite product of separated schemes is closed, a more interesting question is whether an infinite product of separated schemes can be representable by a scheme. Ilya's example demonstrates that the answer is no. Let $Z = \mathbf{A}^2 - 0$. This represents the functor that sends $Spec A$ to the set of pairs $(x,y) \in A^2$ generating the unit ideal. The infinite product $X = \prod_{i = 1}^\infty Z$ represents the functor sending $A$ to the set of infinite collections of pairs $(x_i, y_i)$ generating the unit ideal. Let $B$ be the ring $\mathbf{Z}[x_i, y_i, a_i, b_i]_{i = 1}^\infty / (a_i x_i + b_i y_i = 1)$. There is an obvious map $Spec B \rightarrow X$. Any (nonempty) open subfunctor $U$ of $X$ determines an open subfunctor of $Spec B$, and this must contain a distinguished open subset defined by the invertibility of some $f \in B$. Since $f$ can involve at most finitely many of the variables, the open subset determined by $f$ must contain the pre-image of some open subset $U'$ in $\prod_{i \in I} Z$ for some finite set $I$. Let $I'$ be the complement of $I$. If we choose a closed point $t$ of $U'$ then $U$ contains the pre-image of $t$ as a closed subfunctor. Since the pre-image of $t$ is $\prod_{i \in I'} Z \cong X$ this shows that any open subfunctor of $X$ contains $X$ as a closed subfunctor. In particular, if $X$ is a scheme, any non-empty open affine contains a scheme isomorphic to $X$ as a closed subscheme. A closed subscheme of an affine scheme is affine, so if $X$ is a scheme it is affine. Now we just have to show $X$ is not an affine scheme. It is a subfunctor of $W = \prod_{i = 1}^\infty \mathbf{A}^2$, so if $X$ is an affine scheme, it is locally closed in $W$. Since $X$ is not contained in any closed subset of $W$ except $W$ itself, this means that $X$ is open in $W$. But then $X$ can be defined in $W$ using only finitely many of the variables, which is impossible. Edit: Laurent Moret-Bailly pointed out in the comments below that my argument above for this last point doesn't make sense. Here is a revision: Suppose to the contrary that $X$ is an affine scheme. Then the morphism $p : X \rightarrow X$ that projects off a single factor is an affine morphism. If we restrict this map to a closed fiber then we recover the projection from $Z$ to a point, which is certainly not affine. Therefore $X$ could not have been affine in the first place.
|
{
"source": [
"https://mathoverflow.net/questions/9134",
"https://mathoverflow.net",
"https://mathoverflow.net/users/450/"
]
}
|
9,185 |
How do I feasibly generate a random sample from an $n$-dimensional $\ell_p$ ball? Specifically, I'm interested in $p=1$ and large $n$. I'm looking for descriptions analogous to the statement for $p=2$: Take $n$ standard gaussian random variables and normalize.
|
For arbitrary p, this paper does exactly what you want. Specifically, pick $X_1,\ldots,X_n$ independently with density proportional to $\exp(-|x|^p)$, and $Y$ an independent exponential random variable with mean 1. Then the random vector
$$\frac{(X_1,\ldots,X_n)}{(Y+\sum |X_i|^p)^{1/p}}$$
is uniformly distributed in the unit ball of $\ell_p^n$. The paper also shows how to generate certain other distributions on the $\ell_p^n$ ball by modifying the distribution of $Y$.
|
{
"source": [
"https://mathoverflow.net/questions/9185",
"https://mathoverflow.net",
"https://mathoverflow.net/users/825/"
]
}
|
9,218 |
What are some interesting examples of probabilistic reasoning to establish results that would traditionally be considered analysis? What I mean by "probabilistic reasoning" is that the approach should be motivated by the sort of intuition one gains from a study of probability, e.g. games, information, behavior of random walks and other processes. This is very vague, but hopefully some of you will know what I mean (and perhaps have a better description for what this intuition is). I'll give one example that comes to mind, which I found quite inspiring when I worked through the details. Every Lipschitz function (in this case, $[0,1] \to \mathbb{R}$) is absolutely continuous, and thus is differentiable almost everywhere. We can use a probabilistic argument to actually construct a version of its derivative. One begins by considering the standard dyadic decompositions of [0,1), which gives us for each natural n a partition of [0,1) into $2^{n-1}$ half-open intervals of width $1/{2^{n-1}}$. We define a filtration by letting $\mathcal{F}_n$ be the sigma-algebra generated by the disjoint sets in our nth dyadic decomposition. So e.g. $\mathcal{F}_2$ is generated by $\{[0,1/2), [1/2,1)\}$. We can then define a sequence of random variables $Y_n(x) = 2^n (f(r_n(x)) - f(l_n(x))$ where $l_n(x)$ and $r_n(x)$ are defined to be the left and right endpoints of whatever interval contains x in our nth dyadic decomposition (for $x \in [0,1)$). So basically we are approximating the derivative. The sequence $Y_n$ is in fact a martingale with respect to $\mathcal{F}_n$, and the Lipschitz condition on $f$ makes this a bounded martingale. So the martingale convergence theorem applies and we have that $Y_n$ converges almost everywhere to some $Y$. Straightforward computations yield that we indeed have $f(b) - f(a) = \int_a^b Y$. What I really like about this is that once you get the idea, the rest sort of works itself out. When I came across the result it was the first time I had thought of dyadic decompositions as generating a filtration, but it seems like a really natural idea. It seems much more structured than just the vague idea of "approximation", since e.g. the martingale condition controls the sort of refinement the next approximating term must yield over its predecessor. And although we could have achieved the same result easily by a traditional argument, I find it interesting to see from multiple points of view. So that's really my goal here.
|
One nice example is Bernstein's proof of the Weierstrass theorem. This proof analyses a simple game: Let $f$ be a continuous function on $[0,1]$, and run $n$ independent yes/no experiments in which the “yes” probability is $x$. Pay the gambler $f(m/n)$ if the answer “yes” comes up $m$ times. The gambler's expected gain from this is, of course, $$p_n(x)=\sum_{k=0}^n f(k/n)\binom{n}{k}x^k(1-x)^{n-k}$$ (known as the Bernstein polynomial). The analysis shows that $p_n(x)\to f(x)$ uniformly. S. N. Bernstein, A demonstration of the Weierstrass theorem based on the theory of probability , first published (in French) in 1912. It has been reprinted in Math. Scientist 29 (2004) 127–128 ( MR2102260 ).
|
{
"source": [
"https://mathoverflow.net/questions/9218",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1228/"
]
}
|
9,220 |
Let $x$ be a formal (or small, since the function is analytic) variable, and consider the power series
$$ A(x) = \frac{x}{1 - e^{-x}} = \sum_{m=0}^\infty \left( -\sum_{n=1}^\infty \frac{(-x)^n}{(n+1)!} \right)^m = 1 + \frac12 x + \frac1{12}x^2 + 0x^3 - \frac1{720}x^4 + \dots $$
where I might have made an arithmetic error in expanding it out. Are all the coefficients egyptian , in the sense that they are given by $A^{(n)}(0)/n! = 1/N$ for $N$ an integer? The answer is no, unless I made an error, e.g. the third coefficient. But maybe every non-zero coefficient is egyptian? If all the coefficients were positive eqyptian, then the sequence of denominators might count something — one hopes that the $n$th element of any sequence of nonnegative integers counts the number of ways of putting some type of structure on an $n$-element set. Of course, generating functions really come in two types: ordinary and exponential. The difference is whether you think of the coefficients as $\sum a_n x^n$ or as $\sum A^{(n)} x^n/n!$. If it makes more sense as an exponential generating function, that's cool too. So my question really is: is there a way of computing the $n$th coefficient of $A(x)$, or equivalently of computing $A^{(n)}(0)/n!$, without expanding products of power series the long way? Where you might have seen this series Let $\xi,\psi$ be non-commuting variables over a field of characteristic $0$, and let $B(\xi,\psi) = \log(\exp \xi \exp \psi)$ be the Baker-Campbell-Hausdorff series. Fixing $\xi$ and thinking of this as a power series in $\psi$, it is given by
$$B(\xi,\psi) = \xi + A(\text{ad }\xi)(\psi) + O(\psi^2)$$
where $A$ is the series above, and $\text{ad }\xi$ is the linear operator given by the commutator: $(\text{ad }\xi)(\psi) = [\xi,\psi] = \xi\psi - \psi\xi$. More generally, $B$ can be written entirely in terms of the commutator, and so makes sense as a $\mathfrak g$-valued power series on $\mathfrak g$ for any Lie algebra $\mathfrak g$. It converges in a neighborhood of $0$ when $\mathfrak g$ is finite-dimensional over $\mathbb R$, in which case $\mathfrak g$ is a (generally noncommutative) "partial group". (More generally, you can consider the "formal group" of $\mathfrak g$. Namely, take the commutative ring $\mathcal P(\mathfrak g)$ of formal power series on $\mathfrak g$; then $B$ defines a non-cocommutative comultiplication, making $\mathcal P = \mathcal P(\mathfrak g)$ into a Hopf algebra. Or rather, $B(\mathcal P)$ does not land in the algebraic tensor product $\mathcal P \otimes \mathcal P$. Instead, $\mathcal P$ is cofiltered , in the sense that it is a limit $\dots \to \mathcal P_2 \to \mathcal P_1 \to \mathcal P_0 = 0$, where (over characteristic 0, anyway) $\mathcal P_n = \text{Poly}(\mathfrak g)/(\mathfrak g \text{Poly}(\mathfrak g))^n$, where $\text{Poly}(\mathfrak g)$ is the ring of polynomial functions on $\mathfrak g$, and $\mathfrak g \text{Poly}(\mathfrak g)$ is the ideal of functions vanishing at $0$. Then $B$ lands in the cofiltered tensor product , which is just what it sounds like. (In arbitrary characteristic, $\mathcal P$ is the cofiltered dual of the filtered Hopf algebra $\mathcal S \mathfrak g$, the symmetric algebra of $\mathfrak g$, filtered by degree.)) Why I care When $\mathfrak g$ is finite-dimensional over $\mathbb R$, and $U$ is the open neighborhood of $0$ in which $B$ converges, then $\mathfrak g$ acts as left-invariant derivations on $U$, where by left-invariant I mean under the multiplication $B$. Hence there is a canonical identification of the universal enveloping algebra $\mathcal U\mathfrak g$ with the algebra of left-invariant differential operators on $U$. Since $\mathfrak g$ is in particular a vector space, the "symbol" map gives a canonical identification between the algebra of differential operators on $U$ and the algebra of functions on the cotangent bundle $T^{\ast} U$ that are polynomial (of uniformly bounded degree) in the cotangent directions. Left-invariance then means that the operators are uniquely determined by their restrictions to the fiber $T^{\ast}_0\mathfrak g = \mathfrak g^{\ast}$, and the space of polynomials on $\mathfrak g^{\ast}$ is canonically the symmetric algebra $\mathcal S \mathfrak g$. This gives a canonical PBW map $\mathcal U \mathfrak g \to \mathcal S \mathfrak g$, a fact I learned from J. Baez and J. Dolan. (In the formal group language, the noncocommutative cofiltered Hopf algebra $\mathcal P(\mathfrak g)$ is precisely the cofiltered dual to the filtered algebra $\mathcal U\mathfrak g$, whereas with its cocommutative Hopf structure $\mathcal P(\mathfrak g)$ is dual to $\mathcal S \mathfrak g$. But as algebras these are the same , and unpacking the dualizations gives the PBW map $\mathcal U\mathfrak g \cong \mathcal S \mathfrak g$, and explains why it is actually an isomorphism of coalgebras.) Anyway, in one direction, the isomorphism $\mathcal U\mathfrak g \cong \mathcal S \mathfrak g$ is easy. Namely, the map $\mathcal S \mathfrak g \to \mathcal U \mathfrak g$ is given on monomials by the "symmetrization map" $\xi_1\cdots \xi_n \mapsto \frac1{n!} \sum_{\sigma \in S_n} \prod_{k=1}^n \xi_{\sigma(k)}$, where $S_n$ is the symmetric group on $n$ letters, and the product is ordered. (In this direction, the isomorphism of coalgebras is obvious. In fact, the corresponding symmetrization map into the full tensor algebra is a coalgebra homomorphism.) In the reverse direction, I can explain the map $\mathcal U \mathfrak g \to \mathcal S \mathfrak g$ as follows. On a monomial $\xi_1\cdots \xi_n$, it acts as follows. Draw $n$ dots on a line, and label them $\xi_1,\dots,\xi_n$. Draw arrows between the dots so that each arrow goes to the right (from a lower index to a higher index), and each dot has either 0 or 1 arrow out of it. At each dot, totally order the incoming arrows. Then for each such diagram , evaluate it as follows. What you want to do is collapse each arrow $\psi\to \phi$ into a dot labeled by $[\psi,\phi]$ at the spot that was $\phi$, but never collapse $\psi\to \phi$ unless $\psi$ has no incoming arrows, and if $\phi$ has multiple incoming arrows, collapse them following your chosen total ordering. So at the end of the day, you'll have some dots with no arrows left, each labeled by an element of $\mathfrak g$; multiply these elements together in $\mathcal S\mathfrak g$. Also, multiply each such element by a numerical coefficient as follows: for each dot in your original diagram, let $m$ be the number of incoming arrows, and multiply the final product by the $m$th coefficient of the power series $A(x)$. Sum over all diagrams. Anyway, the previous paragraph is all well and cool, but it would be better if the numerical coefficient could be read more directly off the diagram somehow, without having to really think about the function $A(x)$.
|
Two people have pointed it out already, but somehow I can't resist: your formal power series is precisely the defining power series of the Bernoulli numbers: http://en.wikipedia.org/wiki/Bernoulli_number#Generating_function Accordingly, they are far from Egyptian: as came up recently in response to the question When does the zeta function take on integer values? the odd-numbered terms (except the first) are all zero, whereas the even-numbered terms alternate in sign and grow rapidly in absolute value, so only finitely many are reciprocals of integers. I find it curious that you are looking at this sequence from such a sophisticated perspective and didn't know its classical roots. I feel like there should be a lesson here, but I don't know exactly what it is. Here's a possibility: every young mathematician should learn some elementary number theory regardless of their primary interests. Comments?
|
{
"source": [
"https://mathoverflow.net/questions/9220",
"https://mathoverflow.net",
"https://mathoverflow.net/users/78/"
]
}
|
9,255 |
How do I test whether a given undirected graph is the 1-skeleton of a polytope ? How can I tell the dimension of a given 1-skeleton?
|
A few comments: In general, you can't tell the dimension of a polytope from its graph. For any $n \geq 6$ , the complete graph $K_n$ is the edge graph of both a $4$ -dimensional and a $5$ -dimensional polytope. (Thanks to dan petersen for correcting my typo.) The term for such polytopes is "neighborly". On the other hand, you can say that the dimension is bounded above by the lowest vertex degree occurring anywhere in the graph. A beautiful paper of Gil Kalai shows that, given a $d$ -regular graph, there is at most one way to realize it as the graph of a $d$ -dimensional polytope, and gives an explicit algorithm for reconstructing that polytope. You could try running his algorithm on your graph. (Or a more efficient version recently found by Friedman .) This algorithm will output some face lattice; that is to say, it will tell you which collections of vertices should be $2$ -faces, which should be $3$ -faces and so forth. Unfortunately, going from the face lattice to the polytope is very hard. According to the MathSciNet review, Richter-Gebert has shown that it is NP-hard to, given a lattice of subsets of a finite set, decide whether it is the face lattice of a polytope. Note that this is a lower bound for the difficulty of your problem. Let me be more explicit about the last statement. Richter-Gebert shows that, given a collection $L$ of subsets of $[n]$ , it is NP-hard to determine whether there is a polytope with vertices labeled by $[n]$ whose edges, $2$ -faces and $3$ -faces are the given sets. (Here $[n] = \{ 1,2, \ldots, n \}$ .) Suppose we had an algorithm to decide whether a graph could be the edge graph of a polytope. Take our collection $L$ and look at the two-element sets within it. These form a graph with vertex set $[n]$ . Run the algorithm on it. If the output is NO, then the answer to Richter-Gebert's problem is also no. If the answer is YES, then we have the problem that our algorithm might have found a polytope whose $2$ -faces and $3$ -faces differ from those prescribed by $L$ . If our graph is $4$ -regular, this problem doesn't come up by Kalai's result. But, not having read Richter-Gebert myself, I don't know whether the problem is still NP-hard when we restrict to $4$ -regular graphs. However, even if Richter-Gebert's result doesn't apply directly, I find it difficult to imagine that there could be an efficient algorithm to solve the graph realization problem, since there isn't one to solve the face lattice problem.
|
{
"source": [
"https://mathoverflow.net/questions/9255",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2672/"
]
}
|
9,269 |
In Lawvere, F. W., 1966, “The Category of
Categories as a Foundation for
Mathematics”, Proceedings of the
Conference on Categorical Algebra, La
Jolla, New York: Springer-Verlag,
1–21. Lawvere proposed an elementary theory of the category of categories which can serve as a foundation for mathematics. So far I have heard from several sources that there are some flaws with this theory so that it does not completely work as proposed. So my question is whether there is currently any (accepted) elementary theory of the category of categories that is rich enough so that one can formulate, say, the following things in the theory: The category of sets. Basic notions of category theory (functor categories, adjoints, Kan extensions, etc.). Other important categories (like the category of rings or the category of schemes). The elementary theory I am looking for should allow me to identify what should be called a category of commutative rings (at best I would like to see this category defined by a universal 2-categorical property) or how to work with this category. I am not interested in defining groups, rings, etc. as special categories as this seems to be better done in an elementary theory of sets. P.S.: The same question has an analogue one level higher. Assume that we have constructed an object in the category of categories (=: CAT) which can serve as a, say, category C of spaces. Classically, we can associate to each space X in C the sheaf topos over it. In the picture I have in mind, one should ask whether there is a similar elementary theory of the category of 2-categories (=: 2-CAT). Then one should be able to lift the object C from CAT to 2-CAT (as one is able to form the discrete category from a set), define an object T in 2-CAT that serves as the 2-category of toposes, and a functor C -> T in 2-CAT.
|
My personal opinion is that one should consider the 2-category of categories, rather than the 1-category of categories. I think the axioms one wants for such an "ET2CC" will be something like: Firstly, some exactness axioms amounting to its being a "2-pretopos" in the sense I described here: http://ncatlab.org/michaelshulman/show/2-categorical+logic . This gives you an "internal logic" like that of an ordinary (pre)topos. Secondly, the existence of certain exponentials (this is optional). Thirdly, the existence of a "classifying discrete opfibration" $el\to set$ in the sense introduced by Mark Weber ("Yoneda structures from 2-toposes") which serves as "the category of sets," and internally satisfies some suitable axioms. Finally, a "well-pointedness" axiom saying that the terminal object is a generator, as is the case one level down with in ETCS. This is what says you have a 2-category of categories, rather than (for instance) a 2-category of stacks. Once you have all this, you can use finite 2-categorical limits and the "internal logic" to construct all the usual concrete categories out of the object "set". For instance, "set" has finite products internally, which means that the morphisms $set \to 1$ and $set \to set \times set$ have right adjoints in our 2-category Cat (i.e. "set" is a "cartesian object" in Cat). The composite $set \to set\times set \to set$ of the diagonal with the "binary products" morphism is the "functor" which, intuitively, takes a set $A$ to the set $A\times A$. Now the 2-categorical limit called an "inserter" applied to this composite and the identity of "set" can be considered "the category of sets $A$ equipped with a function $A\times A\to A$," i.e. the category of magmas. Now we have a forgetful functor $magma \to set$, and also a functor $magma \to set$ which takes a magma to the triple product $A\times A \times A$, and there are two 2-cells relating these constructed from two different composites of the inserter 2-cell defining the category of magmas. The "equifier" (another 2-categorical limit) of these 2-cells it makes sense to call "the category of semigroups" (sets with an associative binary operation). Proceeding in this way we can construct the categories of monoids, groups, abelian groups, and eventually rings. A more direct way to describe the category of rings with a universal property is as follows. Since $set$ is a cartesian object, each hom-category $Cat(X,set)$ has finite products, so we can define the category $ring(Cat(X,set))$ of rings internal to it. Then the category $ring$ is equipped with a forgetful functor $ring \to set$ which has the structure of a ring in $Cat(ring,set)$, and which is universal in the sense that we have a natural equivalence $ring(Cat(X,set)) \simeq Cat(X,ring)$. The above construction then just shows that such a representing object exists whenever Cat has suitable finitary structure. One can hope for a similar elementary theory of the 3-category of 2-categories, and so on up the ladder, but it's not as clear to me yet what the appropriate exactness properties will be.
|
{
"source": [
"https://mathoverflow.net/questions/9269",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1841/"
]
}
|
9,309 |
Recall the two following fundamental theorems of mathematical logic: Completeness Theorem: A theory T is syntactically consistent -- i.e., for no statement P can the statement "P and (not P)" be formally deduced from T -- if and only if it is semantically consistent: i.e., there exists a model of T. Compactness Theorem: A theory T is semantically consistent iff every finite subset of T is semantically consistent. It is well-known that the Compactness Theorem is an almost immediate consequence of the
Completeness Theorem: assuming completeness, if T is inconsistent, then one can deduce
"P and (not P)" in a finite number of steps, hence using only finitely many sentences of T. The traditional proof of the completeness theorem is rather long and tedious: for instance, the book Models and Ultraproducts by Bell and Slomson takes two chapters to establish it, and Marker's Model Theory: An Introduction omits the proof entirely. There is a quicker proof due to Henkin (it appears e.g. on Terry Tao's blog), but it is still relatively involved. On the other hand, there is a short and elegant proof of the compactness theorem using ultraproducts (again given in Bell and Slomson). So I wonder: can one deduce completeness from compactness by some argument which is easier than Henkin's proof of completeness? As a remark, I believe that these two theorems are equivalent in a formal sense: i.e., they are each equivalent in ZF to the Boolean Prime Ideal Theorem. I am asking about a more informal notion of equivalence. UPDATE: I accepted Joel David Hamkins' answer because it was interesting and informative. Nevertheless, I remain open to the possibility that (some reasonable particular version of) the completeness theorem can be easily deduced from compactness.
|
There are indeed many proofs of the Compactness theorem. Leo Harrington once told me that he used a different method of proof every time he taught the introductory graduate logic course at UC Berkeley. There is, of course, the proof via the Completeness Theorem, as well as proofs using ultrapowers, reduced products, Boolean-valued models and so on. (In my day, he used Boolean valued models, but that was some time ago, and I'm not sure if he was able to keep this up since then!) Most model theorists today appear to regard the Compactness theorem as the significant theorem, since the focus is on the models---on what is true---rather than on what is provable in some syntactic system. (Proof-theorists, in contast, may focus on the Completeness theorem.) So it is not because Completness is too hard that Marker omits it, but rather just that Compactness is the important fact. Surely it is the Compactness theorem that has deep applications (or at least pervasive applications) in model theory. I don't think formal deductions appear in Marker's book at all. But let's get to your question. Since the exact statement of the Completeness theorem depends on which syntactic proof system you set up---and there are a huge variety of such systems---any proof of the Completeness theorem will have to depend on those details. For example, you must specify which logical axioms are formally allowed, which deduction rules, and so on. The truth of the Completness Theorem depends very much on the details of how you set up your proof system, since if you omit an important rule or axiom, then your formal system will not be complete. But the Compactness theorem has nothing to do with these formal details. Thus, there cannot be hands-off proof of Completeness using Compactness, that does not engage in the details of the formal syntactic proof system. Any proof must establish some formal properties of the formal system, and once you are doing this, then the Henkin proof is not difficult (surely it fits on one or two pages). When I prove Completeness in my logic courses, I often remark to my students that the fact of the theorem is a foregone conclusion, because at any step of the proof, if we need our formal system to be able to make a certain kind of deduction or have a certain axiom, then we will simply add it if it isn't there already, in order to make the proof go through. Nevertheless, Compactness can be viewed as an abstract Completness theorem. Namely, Compactness is precisely the assertion that if a theory is not satisfiable, then it is because of a finite obstacle in the theory that is not satisfiable. If we were to regard these finite obstacles as abstract formal "proofs of contradiction", then it would be true that if a theory has no proofs of contradiction, then it is satisfiable. The difference between this abstract understanding and the actual Completness theorem, is that all the usual deduction systems are highly effective in the sense of being computable. That is, we can computably enumerate all the finite inconsistent theories by searching for formal syntactic proofs of contradiction. This is the new part of Completness that the abstract version from Compactness does not provide. But it is important, for example, in the subject of Computable Model Theory, where they prove computable analogues of the Completeness Theorem. For example, any consistent decidable theory (in a computable language) has a decidable model, since the usual Henkin proof of Completeness is effective when the theory is decidable. Edit: I found in Arnold Miller's lecture
notes an entertaining account of an easy proof of (a fake version of) Completeness from Compactenss (see page 58). His system amounts to the abstract formal system I describe above. Namely, he introduces the MM proof system (for Mickey Mouse),
where the axioms are all logical validities, and the only
rule of inference is Modus Ponens. In this system, one can
prove Completeness from Compactness easily as follows: We
want to show that T proves φ if and only if every model
of T is a model of φ. The forward direction is
Soundness, which is easy. Conversely, suppose that every
model of T is a model of φ. Thus, T+¬φ has no
models. By Compactness, there are finitely many axioms
φ 0 , ..., φ n in T such that
there is no model of them plus ¬φ. Thus,
(φ 0 ∧...∧φ n implies
φ) is a logical validity. And from this, one can easily
make a proof of φ from T in MM. QED! But of course, it is a joke proof system, since the
collection of validities is not computable, and Miller uses this example to illustrate the point as follows: The poor MM system went to the Wizard of OZ and said, “I
want to be more like all the other proof systems.” And the
Wizard replied, “You’ve got just about everything any other
proof system has and more. The completeness theorem is easy
to prove in your system. You have very few logical rules
and logical axioms. You lack only one thing. It is too hard
for mere mortals to gaze at a proof in your system and tell
whether it really is a proof. The difficulty comes from
taking all logical validities as your logical axioms.” The
Wizard went on to give MM a subset Val of logical
validities that is recursive and has the property that
every logical validity can be proved using only Modus
Ponens from Val. And he then goes on to describe how one might construct
Val, and give what amounts to a traditional proof of
Completeness.
|
{
"source": [
"https://mathoverflow.net/questions/9309",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1149/"
]
}
|
9,418 |
Why does a space with finite homotopy groups [for every n] have finite homology groups? How can I proof this [not only for connected spaces with trivial fundamental group]? The converse is false. $\mathbb{R}P^2$ is a counterexample. Do finitely generated homotopy groups imply finitely generated homology groups? I can proof this only for connected spaces with trivial fundamental group. The converse is false. $S^1\vee S^2$ is a counterexample.
|
(This answer has been edited to give more details.) Finitely generated homotopy groups do not imply finitely generated homology groups. Stallings gave an example of a finitely presented group $G$ such that $H_3(G;Z)$ is not finitely generated. A $K(G,1)$ space then has finitely generated homotopy groups but not finitely generated homology groups. Stallings' paper appeared in Amer. J. Math 83 (1963), 541-543. Footnote in small print: Stallings' example can also be found in my algebraic topology book, pp.423-426, as part of a more general family of examples due to Bestvina and Brady. As Stallings noted, it follows that any finite complex $K$ with $\pi_1(K)=G$ has $\pi_2(K)$ nonfinitely generated, even as a module over $\pi_1(K)$. This is in contrast to the example of $S^1 \vee S^2$. Finite homotopy groups do imply finite homology groups, however. In the simply-connected case this is a consequence of Serre's mod C theory, but for the nonsimply-connected case I don't know a reference in the literature. I asked about this on Don Davis' algebraic topology listserv in 2001 and got answers from Bill Browder and Tom Goodwillie. Here's the link to their answers: http://www.lehigh.edu/~dmd1/tg39.txt The argument goes as follows. First consider the special case that the given space $X$ is $BG$ for a finite group $G$. The standard model for $BG$ has finite skeleta when $G$ is finite so the homology is finitely generated. A standard transfer argument using the contractible universal cover shows that the homology is annihilated by $|G|$, so it must then be finite. For a general $X$ with finite homotopy groups one uses the fibration $E \to X \to BG$ where $G=\pi_1(X)$ and $E$ is the universal cover of $X$. The Serre spectral sequence for this fibration has $E^2_{pq}=H_p(BG;H_q(E))$ where the coefficients may be twisted, so a little care is needed. From the simply-connected case we know that $H_q(E)$ is finite for $q>0$. Since $BG$ has finite skeleta this implies $E_{pq}^2$ is finite for $q>0$, even with twisted coefficients. To see this one could for example go back to the $E^1$ page where $E_{pq}^1=C_p(BG;H_q(E))$, the cellular chain group, a finite abelian group when $q>0$, which implies finiteness of $E_{pq}^2$ for $q>0$. When $q=0$ we have $E_{p0}^2=H_p(BG;Z)$ with untwisted coefficients, so this is finite for $p>0$ by the earlier special case. Now we have $E_{pq}^2$ finite for $p+q>0$, so the same must be true for $E^\infty$ and hence $H_n(X)$ is finite for $n>0$. Sorry for the length of this answer and for the multiple edits, but it seemed worthwhile to get this argument on record.
|
{
"source": [
"https://mathoverflow.net/questions/9418",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2625/"
]
}
|
9,465 |
Gabor Toth's Glimpses of Algebra and Geometry contains the following beautiful proof (perhaps I should say "interpretation") of the formula $\displaystyle \frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} \mp ...$, which I don't think I've ever seen before. Given a non-negative integer $r$, let $N(r)$ be the number of ordered pairs $(a, b) \in \mathbb{Z}^2$ such that $a^2 + b^2 \le r^2$, i.e. the number of lattice points in the ball of radius $r$. Then if $r_2(n)$ is the number of ordered pairs $(a, b) \in \mathbb{Z}^2$ such that $a^2 + b^2 = n$, it follows that $N(r^2) = 1 + r_2(1) + ... + r_2(r^2)$. On the other hand, once one has characterized the primes which are a sums of squares, it's not hard to show that $r_2(n) = 4(d_1(n) - d_3(n))$ where $d_i(n)$ is the number of divisors of $n$ congruent to $i \bmod 4$. So we want to count the number of divisors of numbers less than or equal to $r^2$ congruent to $i \bmod 4$ for $i = 1, 3$ and take the difference. This gives $\displaystyle \frac{N(r^2) - 1}{4} = \left\lfloor r^2 \right\rfloor - \left\lfloor \frac{r^2}{3} \right\rfloor + \left\lfloor \frac{r^2}{5} \right\rfloor \mp ...$ and now the desired result follows by dividing by $r^2$ and taking the limit. Question: Does a similar proof exist of the formula $\displaystyle \frac{\pi^2}{6} = 1 + \frac{1}{2^2} + \frac{1}{3^2} + ...$? By "similar" I mean one first establishes a finitary result with a clear number-theoretic or combinatorial meaning and then takes a limit.
|
I think that the 14th and last proof in Robin Chapman's collection is just that. It relies on the formula for the number of representations of an integer as a sum of four squares, which is kind of overkill, but anyway.
|
{
"source": [
"https://mathoverflow.net/questions/9465",
"https://mathoverflow.net",
"https://mathoverflow.net/users/290/"
]
}
|
9,474 |
Reading Serre's letter to Gray , I wonder if now modern expositions of the themes in Klein's book exist. Do you know any?
|
"Geometry of the Quintic" is available for free at my website. Jerry Shurman
|
{
"source": [
"https://mathoverflow.net/questions/9474",
"https://mathoverflow.net",
"https://mathoverflow.net/users/451/"
]
}
|
9,525 |
Where can I find a concrete description of mapping class group of surfaces? I know the mapping class group of the torus is $SL(2, \mathbb{Z})$. Perhaps, there is a simple description for the sphere with punctures or the torus with punctures. Also, I would appreciate any literature reference for an arbitrary surface of genus g with n punctures. Mapping class groups come up in my reading about billiards and the geodesic flow on flat surfaces. I wonder: the moduli space of complex structures on the torus is $\mathbb{H}/SL(2, \mathbb{Z})$, is it a coincidence the mapping class group appears here?
|
1) Let me start by dealing with punctures and higher genus mapping class groups. Aside from a few low-genus cases, there is no easy description of the mapping class group. As you said, the mapping class group of a torus is $SL_2(\mathbb{Z})$, and adding one puncture to a torus does not change its mapping class group (adding a boundary component, however, turns it into the 3-strand braid group). In general (ie for $(g,n)$ not equal to the degenerate cases of $(1,1)$ or $(0,k)$ with $k$ at most $3$), you can relate the mapping class group $\Gamma(g,n)$ of a genus $g$ surface with $n \geq 1$ punctures to the mapping class group of a surface with fewer punctures via the Birman exact sequence. Two forms of it are: $$1 \longrightarrow \pi_1(S_{g,n-1}) \longrightarrow \Gamma(g,n) \longrightarrow \Gamma(g,n-1) \longrightarrow 1$$ and $$1 \longrightarrow B(g,n) \longrightarrow \Gamma(g,n) \longrightarrow \Gamma(g,0) \longrightarrow 1$$ Here $B(g,n)$ is the $n$-strand braid group on a genus $g$ surface. The map to the cokernel comes from "forgetting" punctures, and the kernel comes from "dragging" punctures around the surface. A good reference for this material is the book "A Primer on mapping class groups" by Farb and Margalit, which is available here . 2) As far as moduli space goes, the moduli space of complex structures on a genus $g$ surface with $n$ punctures (as long as $g$ and $n$ are not too small) is isomorphic to the quotient of Teichmuller space by the mapping class group. It is thus no accident that the mapping class group appears in the description of moduli space. Again, Farb and Margalit's book is a nice source.
|
{
"source": [
"https://mathoverflow.net/questions/9525",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1358/"
]
}
|
9,628 |
Here are two questions about finitely generated and finitely presented groups (FP): Is there an example of an FP group that does not admit a homomorphism to $\operatorname{GL}(n,C)$ with trivial kernel for any n? The second question is modified according to the sujestion of Greg below. For which $n$ given two subgroups of $\operatorname{GL}(n,C)$ generated by explicit lists of matrices, together with finite lists of relations and the promise that they are sufficient, is there an algorithm to determine if they are isomorphic as groups?" In both cases we don't impose any conidtion on the group (apart from been FP), in particular it need not be discrete in $\operatorname{GL}(n,C)$ .
|
Here is a more complete picture to go with David's and Richard's answers. It is a theorem of Malcev that a finitely presented group $G$ is residually linear if and only if it is residually finite. The proof is very intuitive: The equations for a matrix representation of $G$ are algebraic, so there is an algebraic solution if there is any solution. Then you can reduce the field of the solution to a finite field, as long as you avoid all primes that occur in the denominators of the matrices. The same proof shows that $G$ has no non-trivial linear representations if and only if it has no subgroups of finite index. So Higman's group has this property. A refined question is to find a finitely presented group which is residually finite, but nonetheless isn't "linear" in the sense of having a single faithful finite-dimensional representation. It seems that the automorphism group of a finitely generated free group, $\text{Aut}(F_n)$, is an example. Nielsen found a finite presentation for this group, it is also known to be residually finite , yet Formanek and Procesi showed that it is not linear when $n \ge 3$. More recently, Drutu and Sapir found an example with two generators and one relator.
|
{
"source": [
"https://mathoverflow.net/questions/9628",
"https://mathoverflow.net",
"https://mathoverflow.net/users/943/"
]
}
|
9,641 |
A while ago I heard of a nice characterization of compactness but I have never seen a written source of it, so I'm starting to doubt it. I'm looking for a reference, or counterexample, for the following . Let $X$ be a Hausdorff topological space. Then, $X$ is compact if and only if $X^{\kappa}$ is Lindelöf for any cardinal $\kappa$. If the above is indeed a fact, can one restrict the class of $\kappa$'s for which the characterization is still valid? Note: Here I'm thinking under ZFC.
|
The answer is Yes . Theorem. The following are equivalent for any Hausdorff
space $X$. $X$ is compact. $X^\kappa$ is Lindelöf for any cardinal
$\kappa$. $X^{\omega_1}$ is Lindelöf. Proof. The forward implications are easy, using Tychonoff
for 1 implies 2, since if $X$ is compact, then
$X^\kappa$ is compact and hence Lindelöf. So suppose that we have a space $X$ that is not compact, but
$X^{\omega_1}$ is Lindelöf. It
follows that $X$ is Lindelöf. Thus, there is a countable
cover having no finite subcover. From this, we may
construct a strictly increasing sequence of open sets
$U_0 \subset U_1 \subset \dots U_n \dots$
with the union $\bigcup\lbrace U_n \; | \; n \in \omega \rbrace = X$. For each $J \subset \omega_1$ of size $n$, let $U_J$ be
the set $\lbrace s \in X^{\omega_1} \; | \; s(\alpha) \in U_n$ for each $\alpha \in J \rbrace$. As the size of $J$ increases, the set $U_J$ allows more freedom on the
coordinates in $J$, but restricts more coordinates. If $J$ has
size $n$, let us call $U_J$ an open $n$-box, since it
restricts the sequences on $n$ coordinates. Let $F$ be the
family of all such $U_J$ for all finite $J \subset \omega_1$ This $F$ is a cover of $X^{\omega_1}$. To
see this, consider any point $s \in X^{\omega_1}$. For each $\alpha \in
\omega_1$, there is some $n$ with $s(\alpha) \in
U_n$. Since $\omega_1$ is uncountable,
there must be some value of $n$ that is repeated unboundedly
often, in particular, some $n$ occurs at least $n$ times. Let $J$
be the coordinates where this $n$ appears. Thus, $s$ is in
$U_J$. So $F$ is a cover. Since $X^{\omega_1}$ is Lindelöf,
there must be a countable subcover $F_0$. Let $J^*$ be
the union of all the finite $J$ that appear in the
$U_J$ in this subcover. So $J^*$ is a countable subset
of $\omega_1$. Note that $J^*$ cannot be finite,
since then the sizes of the $J$ appearing in $F_0$
would be bounded and it could not cover
$X^{\omega_1}$. We may rearrange indices
and assume without loss of generality that $J^*=\omega$ is
the first $\omega$ many coordinates. So $F_0$ is
really a cover of $X^\omega$, by ignoring the
other coordinates. But this is impossible. Define a sequence $s \in
X^{\omega_1}$ by choosing $s(n)$ to be
outside $U_{n+1}$, and otherwise arbitrary. Note that
$s$ is in $U_n$ in fewer than $n$ coordinates below
$\omega$, and so $s$ is not in any $n$-box with $J \subset \omega$, since any such box has $n$ values in $U_n$.
Thus, $s$ is not in any set in $F_0$, so it is not a
cover. QED In particular, to answer the question at the end, it suffices to take any uncountable $\kappa$.
|
{
"source": [
"https://mathoverflow.net/questions/9641",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2089/"
]
}
|
9,661 |
The "Motivation" section is a cute story, and may be skipped; the "Definitions" section establishes notation and background results; my question is in "My Question", and in brief in the title. Some of my statements go wrong in non-zero characteristic, but I don't know that story well enough, so you are welcome to point them out, but this is my characteristic-zero disclaimer. Motivation In his 1972 talk "Missed Opportunities" (MR0522147), F. Dyson tells the following story of how mathematicians could have invented special and much of general relativity long before the physicists did. Following the physicists, I will talk about Lie groups, but I really mean Lie algebras, or maybe connected Lie groups, ... The physics of Galileo and Newton is invariant under the action of the Galileo Group (indeed, this is the largest group leaving classical physics invariant and fixing a point), which is the group $G_\infty = \text{SO}(3) \ltimes \mathbb R^3 \subseteq \text{GL}(4)$ , where $\text{SO}(3)$ acts as rotations of space, fixing the time axis, and $\mathbb R^3$ are the nonrelativistic boosts $\vec x \mapsto t\vec u$ , $t\mapsto t$ . This group is not semisimple. But it is the limit as $c\to \infty$ of the Lorentz Group $G_c = \text{SO}(3,1)$ , generated by the same $\text{SO}(3)$ part but the boosts are now $$ t \mapsto \frac{t + c^{-2}\vec u \cdot \vec x}{\sqrt{1 - c^{-2}u^2}} \quad \quad \vec x \mapsto \frac{\vec x + t\vec u}{\sqrt{1 - c^{-2}u^2}} $$ Since semisimple groups are easier to deal with than nonsemisimple ones, by Occam's Razor we should prefer the Lorentz group. Fortunately, Maxwell's equations are not invariant under $G_\infty$ , but rather under $G_c$ , where $c^{-2}$ is the product of the electric permititivity and magnetic permeability of free space (each of which is directly measurable, giving the first accurate measurement of the speed of light). Actually, from this perspective, $c^{-2}$ is the fundamental number, and we should really think of the Galileo Group as a limit to $0$ , not $\infty$ , of something. But of course from this perspective we should go further. Actual physics is invariant under more than just the Lorentz group, which is the group that physics physics and a point. So Special Relativity is invariant under the Poincare Group $P = G_c \ltimes \mathbb R^{3+1}$ . Again this is not semisimple. It is the limit as $r \to \infty$ of the DeSitter Group $D_r$ , which in modern language "is the invariance group of an empty expanding universe whose radius of curvature $R$ is a linear function of time" (Dyson), so that $R = rt$ in absolute time units. Hubble measured the expansion of the universe in the first half of the twentieth century. Anyway, I'm curious to know if it's true that every Lie group is a limit of a semisimple one: how typical are these physics examples? To make this more precise, I'll switch to Lie algebras. Definitions Let $V$ be an $n$ -dimensional vector space. If you like, pick a basis $e_1,\dots,e_n$ of $V$ , and adopt Einstein's repeated index notation, so that given $n$ -tuples $a^1,\dots,a^n$ and $b_1,\dots,b_n$ , then $a^ib_i = \sum_{i=1}^n a^ib_i$ , and if $v \in V$ , we define the numbers $v^i$ by $v = v^ie_i$ ; better, work in Penrose's index notation. Anyway, a Lie algebra structure on $V$ is a map $\Gamma: V \otimes V \to V$ satisfying two conditions, one homogeneous linear in (the matrix coefficients) of $\Gamma$ and the other homogeneous quadratic: $$ \Gamma^k_{ij} + \Gamma^k_{ji} = 0 \quad\text{and}\quad \Gamma^l_{im}\Gamma^m_{jk} + \Gamma^l_{jm}\Gamma^m_{ki} + \Gamma^l_{km}\Gamma^m_{ij} = 0 $$ Thus the space of Lie algebra structures on $V$ is an algebraic variety in $V \otimes V^{*} \otimes V^{*}$ , where $V^{*}$ is the dual space to $V$ . If $\Gamma$ is a Lie algebra structure on $V$ , the corresponding Killing form $\beta$ is the symmetric bilinear pairing $\beta_{ij} = \Gamma^k_{im}\Gamma^m_{jk}$ . Then $\Gamma$ is semisimple if and only if $\beta$ is nondegenerate. Nondegeneracy is a Zariski-open condition on bilinear forms, since $\beta$ is degenerate if and only if a certain homogeneous-of-degree- $n$ expression in $\beta$ vanishes (namely $\sum_{\sigma \in S_n} (-1)^\sigma \prod_{k=1}^n \beta_{i_k,j_{\sigma(k)}} = 0$ as a map out of $V^{\otimes n} \otimes V^{\otimes n}$ , where $S_n$ is the symmetric group on $n$ objects and $(-1)^\sigma$ is the "sign" character of $S_n$ ). Since $\beta$ is expressed algebraically in terms of $\Gamma$ , semisimplicity is a Zariski-open condition on the variety of Lie algebra structures on a given vector space. Incidentally, Cartan classified all semisimple Lie algebras (at least over $\mathbb C$ and $\mathbb R$ ) up to isomorphism, and the classification is discrete. So any two semisimple Lie algebras in the same connected component of the space of semisimple structures are isomorphic. My Question Is the space of semisimple Lie algebra structures on $V$ dense in the space of all Lie algebra structures on $V$ ? (I.e. if $\Gamma$ is a Lie algebra structure on $V$ and $U \ni \Gamma$ is an open set of Lie algebra structures, does it necessarily contain a semisimple one?) This is really two questions. One is whether it is Zariski-dense. But we can also work over other fields, e.g. $\mathbb R$ or $\mathbb C$ , which have topologies of their own. Is the space of semisimple Lie algebra structures on a real vector space $V$ dense with respect to the usual real topology? (The answer is no when $\dim V = 1$ , as then the only Lie algebra structure is the abelian one $\Gamma = 0$ , which is not semisimple, and it is no when $\dim V = 2$ , as there are nontrivial two-dimensional Lie algebras but no semisimple ones. So I should ask my question for higher-dimensional things.) Edit: I have posted the rest of these as this follow-up question . If the answer is no in general, is it possible to (nicely) characterize the Lie algebra structures that are in the closure of the semisimple part? A related question is whether given a nonsemisimple Lie algebra structure, are all its nearby semisimple neighbors isomorphic? Of course the answer is no: the abelian Lie algebra structure $\Gamma = 0$ is near every Lie algebra but in general there are nonisomoprhic semisimple Lie algebras of the same dimension, and more generally we could always split $V = V_1 \oplus V_2$ , and put a semisimple Lie algebra structure on $G_1$ and a trivial one on $V_2$ . So the converse question: are there any nonsemisimple Lie algebras so that all their semisimple deformations are isomorphic? Yes, e.g. anyone on the three-dimensional vector space over $\mathbb C$ . So: is there a (computationally useful) characterization of those that are? If the answer to all my questions are "yes", then it's probably been done somewhere, so a complete response could consist of a good link. The further-further question is to what extent one can deform representations, but that's probably pushing it.
|
The answer to the question in the title is "no". Semisimplicity is an open condition; however, it is not a dense open condition. Indeed, the variety of Lie algebras is reducible. There is one equation which nonsemisimple and only nonsemisimple Lie algebra structures satisfy, namely, that the Killing form Tr(ad(x)ad(y)) is degenerate, just as stated in the question. But there is also a system of equations which all semisimple Lie algebra structures satisfy, as do also all reductive and nilpotent Lie algebra structures, but solvable Lie algebra structures in general don't. These are the unimodularity equations Tr(ad(x))=0 for all x in the Lie algebra. These mean that the top exterior power of the adjoint representation is a trivial representation of the Lie algebra, which is obvious for any Lie algebra that coincides with its commutator subalgebra. But the nonabelian 2-dimensional Lie algebra is not unimodular. Hence in any dimension n, the direct sum of the nonabelian 2-dimensional Lie algebra with the abelian (n-2)-dimensional Lie algebra does not belong to the Zariski closure of semisimple Lie algebras.
|
{
"source": [
"https://mathoverflow.net/questions/9661",
"https://mathoverflow.net",
"https://mathoverflow.net/users/78/"
]
}
|
9,667 |
I am preparing to teach a short course on "applied model theory" at UGA this summer. To draw people in, I am looking to create a BIG LIST of results in mathematics that have nice proofs using model theory. (I do not require that model theory be the first or only proof of the result in question.) I will begin with some examples of my own (the attribution is for the model-theoretic proof, not the result itself). An injective regular map from a complex variety to itself is surjective (Ax). The projection of a constructible set is constructible (Tarski). Solution of Hilbert's 17th problem (Tarski?). p-adic fields are "almost C_2" (Ax-Kochen). "Almost" every rationally connected variety over Q_p^{unr} has a rational point (Duesler-Knecht). Mordell-Lang in positive characteristic (Hrushovski). Nonstandard analysis (Robinson). [But better would be: a particular result in analysis which has a snappy nonstandard proof.] Added : The course was given in July of 2010. So far as I am concerned, it went well. If you are interested, the notes are available at http://alpha.math.uga.edu/~pete/MATH8900.html Thanks to everyone who answered the question. I enjoyed and learned from all of the answers, even though (unsurprisingly) many of them could not be included in this introductory half-course. I am still interested in hearing about snappy applications of model theory, so further answers are most welcome.
|
Plane geometry is decidable. That is, we have a computable algorithm that will tell us the truth or falsity of any geometrical statement in the cartesian plane. This is a consequence of Tarski's theorem showing that the theory of real closed fields admits elimination of quantifiers. The elimination algorithm is effective and so the theory is decidable. Thus, we have a computable procedure to determine the truth of any first order statement in the structure (R,+,.,0,1,<). The point is that all the classical concepts of plane geometry, in any finite dimension, are expressible in this language. Personally, I find the fact that plane geometry has been proven decidable to be a profound human achievement. After all, for millennia mathematicians have struggled with geometry, and we now have developed a computable algorithm that will in principle answer any question. I admit that I have been guilty, however, of grandiose over-statement of the situation---when I taught my first logic course at UC Berkeley, after I explained the theorem some of my students proceeded to their next class, a geometry class with Charles Pugh, and a little while later he came knocking on my door, asking what I meant by telling the students "geometry is finished!". So I was embarrassed. Of course, the algorithm is not feasible--its double exponential time. Nevertheless, the fact that there is an algorithm at all seems amazing to me. To be sure, I am even more surprised that geometers so often seem unaware of the fact that they are studying a decidable theory.
|
{
"source": [
"https://mathoverflow.net/questions/9667",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1149/"
]
}
|
9,708 |
Has the solution of the Poincaré Conjecture helped science to figure out the shape of the universe?
|
In Einstein's theory of General Relativity, the universe is a 4-manifold that might well be fibered by 3-dimensional time slices. If a particular spacetime that doesn't have such a fibration, then it is difficult to construct a causal model of the laws of physics within it. (Even if you don't see an a priori argument for causality, without it, it is difficult to construct enough solutions to make meaningful predictions.) There isn't usually a geometrically distinguished fibration, but if you have enough symmetry or even local symmetry, the symmetry can select one. An approximate symmetry can also be enough for an approximately canonical fibration. Once you have all of that, the topology of spacelike slices of the universe is not at all a naive or risible question, at least not until you see more physics that might demote the question. The narrower question of whether the Poincaré Conjecture is relevant is more wishful and you could call it naive, but let's take the question of relating 3-manifold topology in general to cosmology. The cosmic microwave background , discovered in the 1964 by Penzias and Wilson, shows that the universe is very nearly isotropic at our location. (The deviation is of order $10^{-5}$ and it was only announced in 1992 after 2 years of data from the COBE telescope .) If you accept the Copernican principle that Earth isn't at a special point in space, it means that there is an approximately canonical fibration by time slices, and that the universe, at least approximately and locally, has one of the three isotropic Thurston geometries, $E^3$, $S^3$, or $H^3$. The Penzias-Wilson result makes it a really good question to ask whether the universe is a 3-manifold with some isotropic geometry and some fundamental group. I have heard of the early discussion of this question was so naive that some astronomers only talked about a 3-torus. They figured that if there were other choices from topology, they could think about them later. Notice that already, the Poincaré conjecture would have been more relevant to cosmology if it had been false! The topologist who has done the most work on the question is Jeff Weeks. He coauthored a respected paper in cosmology and wrote an interesting article in the AMS Notices that promoted the Poincaré dodecahedral space as a possible topology for the universe. But after he wrote that article... There indeed is other physics that does demote the 3-manifold question, and that is inflationary cosmology . The inflation theory posits that the truthful quantum field theory has a vaguely stable high-energy phase, which has such high energy density that the solution to the GR equations looks completely different. In the inflationary solution, hot regions of the universe expand by a factor of $e$ in something like $10^{-36}$ seconds. The different variations of the model posit anywhere from 60 to thousands of factors of $e$, or "$e$-folds". Patches of the hot universe also cool down, including the one that we live in. In fact every spot is constantly cooling down, but cooling is still overwhelmed by expansion. Instead of tacitly accepting certain observed features of the visible universe, for instance that it is approximately isotropic, inflation explains them. It also predicts that the visible universe is approximately flat and non-repeating, because macroscopic curvature and topology have been stretched into oblivion, and that observable anisotropies are stretch marks from the expansion. The stretch marks would have certain characteristic statistics in order to fit inflation. On the other hand, in the inflationary hot soup that we would never see directly, the rationale for canonical time slices is gone, and the universe would be some 4-manifold or even some fractal or quantum generalization of a 4-manifold. The number of $e$-folds is not known and even the inflaton field (the sector of quantum field theory that governed inflation) is not known, but most or all models of inflation predict the same basic features. And the news from the successor to COBE, called WMAP , is that the visible universe is flat to 2% or so, and the anistropy statistically matches stretch marks. There is not enough to distinguish most of the models of inflation. There is not enough to establish inflation in the same sense that the germ theory of disease or the the heliocentric theory are established. What is true is that inflation has made experimental predictions that have been confirmed. After all that news, the old idea that the universe is a visibly periodic 3-manifold is considered a long shot. WMAP didn't see any obvious periodicity, even though Weeks et al were optimistic based on its first year of data. But I was told by a cosmologist that periodicity should still be taken seriously as an alternative cosmological model, if possibly as a devil's advocate. A theory is incomplete science if it is both hard to prove, and if every alternative is laughed out of the room. In arguing for inflation, cosmologists would also like to have something to argue against. In the opinion of the cosmologist that I talked to some years ago, the model of a 3-manifold with a fundamental group, developed by Weeks et al, is as good at that as any proposal. José makes the important point that, in testing whether the universe has a visible fundamental group, you wouldn't necessarily look for direct periodicity represented by non-contractible geodesics. Instead, you could use harmonic analysis, using a suitable available Laplace operator, and this is what used by Luminet, Weeks, Riazuelo, Lehoucq and Uzan. I also that I have not heard of any direct use of homotopy of paths in astronomy, but actually the direct geometry of geodesics does sometimes play an important role. For instance, look closely at this photograph of galaxy cluster Abell 1689 . You can see that there is a strong gravitational lens just left of the center, between the telescope and the dimmer, slivered galaxies. Maybe no analysis of the cosmic microwave background would be geometry-only, but geometry would modify the apparent texture of the background, and I think that that is part of the argument from the data that the visible universe is approximately flat. Who is to say whether a hypothetical periodicity would be seen with geodesics, harmonic expansion, or in some other way. Part of Gromov's point seems fair. I think it is true that you can always expand the scale of proposed periodicity to say that you haven't yet seen it, or that the data only just starts to show it. Before they saw anisotropy with COBE, that kept getting pushed back too. The deeper problem is that the 3-manifold topology of the universe does not address as many issues in cosmology, either theoretical or experimental, as inflation theory does.
|
{
"source": [
"https://mathoverflow.net/questions/9708",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1172/"
]
}
|
9,721 |
The number ${n \choose k}$ of $k$-element subsets of an $n$-element set and the number $\left( {n \choose k} \right)$ of $k$-element multisets of an $n$-element set satisfy the reciprocity formula $\displaystyle {-n \choose k} = (-1)^k \left( {n \choose k} \right)$ when extended to negative integer indices, for example by applying the usual recurrence relations to all integers. There's an interesting way to think about the "negative cardinalities" involved here using Euler characteristic, which is due to Schanuel; see, for example, this paper of Jim Propp . Another (related?) way to think about this relationship is in terms of the symmetric and exterior algebras; see, for example, this blog post . The number $S(n, k)$ of $k$-block partitions of a set with $n$ elements and the number $c(n, k)$ of permutations of a set with $n$ elements with $k$ cycles satisfy a well-known inverse matrix relationship, but they also satisfy the reciprocity formula $c(n, k) = S(-k, -n)$ when extended to negative integer indices, again by applying the usual recurrence relations. Question: Are there any known highbrow interpretations of this reciprocity formula?
|
Supplementary Exercise 3.2(d,e) on page 313 of my book Enumerative Combinatorics , vol. 1, second printing, shows that this Stirling number reciprocity is a special case of the reciprocity theorem for order polynomials (Exercise 3.61(a)). Thus it is related to a lot of "highbrow" math, such as the reciprocity between a Cohen-Macaulay ring and its canonical module. (For the basic properties of canonical modules, see Section I.12 of my other book Combinatorics and Commutative Algebra, second ed.)
|
{
"source": [
"https://mathoverflow.net/questions/9721",
"https://mathoverflow.net",
"https://mathoverflow.net/users/290/"
]
}
|
9,731 |
Lagrange proved that every nonnegative integer is a sum of 4 squares. Gauss proved that every nonnegative integer is a sum of 3 triangular numbers. Is there a 2-variable polynomial $f(x,y) \in \mathbf{Q}[x,y]$ such that $f(\mathbf{Z} \times \mathbf{Z})=\mathbf{N}$?
|
This is a cute problem! I toyed with it and didn't really get anywhere - I got the strong impression that it requires fields of mathematics that I am not expert in. Indeed, given that the problem seems related to that of counting integer solutions to the equation $f(x,y) = c$, one may need to use arithmetic geometry tools (e.g. Faltings' theorem). In particular if we could reduce to the case when the genus is just 0 or 1 then presumably one could kill off the problem. (One appealing feature of this approach is that arithmetic geometry quantities such as the genus are automatically invariant (I think) with respect to invertible polynomial changes of variable such as $(x,y) \mapsto (x,y+P(x))$ or $(x,y) \mapsto (x+Q(y),y)$ and so seem to be well adapted to the problem at hand, whereas arguments based on the raw degree of the polynomial might not be.) Of course, Faltings' theorem is ineffective, and so might not be directly usable, but perhaps some variant of it (particularly concerning the dependence on c) could be helpful. [Also, it is overkill - it controls rational solutions, and we only care here about integer ones.] This is far outside of my own area of expertise, though... The other thing that occurred to me is that for fixed c and large x, y, one can invert the equation $f(x,y) = c$ to obtain a Puiseux series expansion for y in terms of x or vice versa (this seems related to resolution of singularities at infinity, though again I am not an expert on that topic; certainly Newton polytopes seem to be involved). In some cases (if the exponents in this series expansion are favourable) one could then use Archimedean counting arguments to show that f cannot cover all the natural numbers (this is a generalisation of the easy counting argument that shows that a 1D polynomial of degree 2 or more cannot cover a positive density set of integers), but this does not seem to work in all cases, and one may also have to use some p-adic machinery to handle the other cases. One argument against this approach though is that it does not seem to behave well with respect to invertible polynomial changes of variable, unless one works a lot with geometrical invariants. Anyway, to summarise, it seems to me that one has to break out the arithmetic geometry and algebraic geometry tools. (Real algebraic geometry may also be needed, in order to fully exploit the positivity, though it is also possible that positivity is largely a red herring, needed to finish off the low genus case, but not necessary for high genus, except perhaps to ensure that certain key exponents are even.) EDIT: It occurred to me that the polynomial $f(x,y)-c$ might not be irreducible, so there may be multiple components to the associated algebraic curve, each with a different genus, but presumably this is something one can deal with. Also, the geometry of this curve may degenerate for special c, but is presumably stable for "generic" c (or maybe even all but finitely many c). It also occurs to me that one use of real algebraic geometry here is to try to express f as something like a sum of squares. If there are at least two nontrivial squares in such a representation, then f is only small when both of the square factors are small, which is a 0-dimensional set and so one may then be able to use counting arguments to conclude that one does not have enough space to cover all the natural numbers (provided that the factors are sufficiently "nonlinear"; if for instance $f(x,y)=x^2+y^2$ then the counting arguments barely fail to provide an obstruction, one has to use mod p arguments or something to finish it off...) EDIT, FOUR YEARS LATER: OK, now I know a bit more arithmetic geometry and can add to some of my previous statements. Firstly, it's not Faltings' theorem that is the most relevant, but rather Siegel's theorem on integer points on curves - the enemy appears to be those points $(x,y)$ where $x,y$ are far larger than $f(x,y)$, and Siegel's theorem is one of the few tools available to exclude this case. The known proofs of this theorem are based on two families of results in Diophantine geometry: one is the Thue-Siegel-Roth theorem and its variants (particularly the subspace theorem), and the other is the Mordell-Weil theorem and its variants (particularly the Chevalley-Weil theorem). A big problem here is that all of these theorems have a lot of ineffectivity in them. Even for the very concrete case of Hall's conjecture on lower bounding $|x^2-y^3|$ for integers $x,y$ with $x^2 \neq y^3$, Siegel's theorem implies that this bound goes to infinity as $x,y \to \infty$, but provides no rate; as I understand it, the only known lower bounds are logarithmic and come from variants of Baker's method. As such, a polynomial such as $f(x,y) = (x^2 - y^3 - y)^4 - y + C$ for some large constant C already looks very tough to analyse. (I've shifted $y^3$ here by $y$ to avoid the degenerate solutions to $x^2=y^3$, and to avoid some cheap way to deal with this polynomial from the abc conjecture or something.) The analogue of Hall's conjecture for $|x^2-y^3-y|$ suggests that $f(x,y)$ goes to $+\infty$ as $x,y \to \infty$ (restricting $x,y$ to be integers), but we have no known growth rate here due to all the ineffectivity. As such, we can't unconditionally rule out the possibility of an infinite number of very large pairs $(x,y)$ for which $x^2-y^3-y$ happens to be so close to $y^{1/4}$ that we manage to hit every positive integer value in $f(x,y)$ without hitting any negative ones. However, one may be able to get a conditional result assuming some sufficiently strong variant of the abc conjecture. One should also be able to exclude large classes of polynomials $f$ from working; for instance, if the curve $f(x,y)=0$ meets the line at infinity at a lot of points in a transverse manner, then it seems that the subspace theorem may be able to get polynomial bounds on solutions $(x,y)$ to $f(x,y)=c$ in terms of $c$, at which point a lot of other tools (e.g. equidistribution theory) become available. Another minor addendum to my previous remarks: the generic irreducibility of $f(x,y)-c$ follows from Bertini's second theorem, as one may easily reduce to the case when $f$ is non-composite (not the composition of two polynomials of lower degree).
|
{
"source": [
"https://mathoverflow.net/questions/9731",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2757/"
]
}
|
9,733 |
I have heard during some seminar talks that there are applications of the theory of
matrix factorizations in string theory. A quick search shows mostly papers written by physicists. Are there any survey type papers aimed at an algebraic audience on this topic, especially with current state/open questions motivated by physics? Background: A matrix factorization of an element $x$ in a ring $R$ is a pair of square matrices $A,B$ of size $n$ such that $AB=BA=xId_n$. For more see, for example, Section 3 of this paper .
|
Indeed matrix factorizations come up in string theory. I don't know if there are good survey articles on this stuff, but here is what I can say about it. There might be an outline in the big Mirror Symmetry book by Hori-Katz-Klemm-etc., but I am not sure. When we are considering the B-model of a manifold, for example a compact Calabi-Yau, the D-branes (boundary states of open strings) are given by coherent sheaves on the manifold (or to be more precise, objects of the derived category of coherent sheaves). Matrix factorizations come up in a different situation, namely, they are the D-branes in the B-model of a Landau-Ginzburg model . Mathematically, a Landau-Ginzburg model is just a manifold (or variety) $X$, typically non-compact, plus the data of a holomorphic function $W: X \to \mathbb{C}$ called the superpotential. In this general situation, a matrix factorization is defined to be a pair of coherent sheaves $P_0, P_1$ with maps $d : P_0 \to P_1$, $d : P_1 \to P_0$ such that $d^2 = W$. I guess you could call this a "twisted (or maybe it's 'curved'? I forget the terminology) 2-periodic complex of coherent sheaves". When $X = \text{Spec}R$ is affine, and when the coherent sheaves are free $R$-modules, this is the same as the definition that you gave. The relationship between matrix factorization categories and derived categories of coherent sheaves was worked out by Orlov: http://arxiv.org/abs/math/0503630 http://arxiv.org/abs/math/0503632 http://arxiv.org/abs/math/0302304 I believe that the suggestion to look at matrix factorizations was first proposed by Kontsevich. I think the first paper that explained Kontsevich's proposal was this paper by Kapustin-Li: http://arXiv.org/abs/hep-th/0210296v2 There are some interesting recent papers regarding the relationship between the open-string B-model of a Landau Ginzburg model (which is, again, mathematically given by the matrix factorizations category) and the closed-string B-model, which I haven't described, but an important ingredient is the Hochschild (co)homology of the matrix factorizations category. Take a look at Katzarkov-Kontsevich-Pantev http://arxiv.org/abs/0806.0107 section 3.2. There is a paper of Tobias Dyckerhoff http://arxiv.org/abs/0904.4713 and a paper of Ed Segal http://arxiv.org/abs/0904.1339 which work out in particular the Hochschild (co)homology of some matrix factorization categories. The answer is it's the Jacobian ring of the superpotential. This is the correct answer in terms of physics: the Jacobian ring is the closed state space of the theory. Katzarkov-Kontsevich-Pantev also has some interesting stuff about viewing matrix factorization categories as "non-commutative spaces" or "non-commutative schemes". Edit 1: I forgot to mention: Kontsevich's original homological mirror symmetry conjecture stated that the Fukaya category of a Calabi-Yau is equivalent to the derived category of coherent sheaves of the mirror Calabi-Yau. Homological mirror symmetry has since been generalized to non-Calabi-Yaus. The rough expectation is that given any compact symplectic manifold, there is a mirror Landau-Ginzburg model such that, among other things, the Fukaya category of the symplectic manifold should be equivalent to the matrix factorizations category of the Landau-Ginzburg model. For example, if your symplectic manifold is $\mathbb{CP}^n$, the mirror Landau-Ginzburg model is given by the function $x_1+\cdots+x_n + \frac{1}{x_1\cdots x_n}$ on $(\mathbb{C}^\ast)^n$. This is sometimes referred to as the Hori-Vafa mirror http://arxiv.org/abs/hep-th/0002222 I think that various experts probably know how to prove this form of homological mirror symmetry, at least when the symplectic manifold is, for example, a toric manifold or toric Fano manifold, but it seems that very little of this has been published. There may be some hints in this direction in Fukaya-Ohta-Oh-Ono http://arxiv.org/abs/0802.1703 http://arxiv.org/abs/0810.5654 , but I'm not sure. There is an exposition of the case of $\mathbb{CP}^1$ in this paper of Matthew Ballard http://arxiv.org/abs/0801.2014 -- this case is already non-trivial and very interesting, and the answer is very nice: the categories in this case are equivalent to the derived category of modules over a Clifford algebra. I quite like Ballard's paper; you might be interested in taking a look at it anyway. Edit 2: Seidel also has a proof of this form of homological mirror symmetry for the case of the genus two curve. Here is the paper http://arxiv.org/abs/0812.1171 and here is a video http://www.maths.ed.ac.uk/~aar/atiyah80.htm of a talk he gave on this stuff at the Atiyah 80 conference.
|
{
"source": [
"https://mathoverflow.net/questions/9733",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2083/"
]
}
|
9,754 |
I am interested in magic tricks whose explanation requires deep mathematics. The trick should be one that would actually appeal to a layman. An example is the following: the magician asks Alice to choose two integers between 1 and 50 and add them. Then add the largest two of the three integers at hand. Then add the largest two again. Repeat this around ten times. Alice tells the magician her final number $n$. The magician then tells Alice the next number. This is done by computing $(1.61803398\cdots) n$ and rounding to the nearest integer. The explanation is beyond the comprehension of a random mathematical layperson, but for a mathematician it is not very deep. Can anyone do better?
|
" The best card trick ", an article by Michael Kleber. Here is the opening paragraph: "You, my friend, are about to witness
the best card trick there is.
Here, take this ordinary deck of cards,
and draw a hand of five cards from
it. Choose them deliberately or randomly,
whichever you prefer--but do
not show them to me! Show them instead
to my lovely assistant, who will
now give me four of them: the 7 of spades,
then the Q of hearts, the 8 of clubs, the 3 of diamonds.
There is one card left in your hand, known
only to you and my assistant. And the
hidden card, my friend, is the K of spades."
|
{
"source": [
"https://mathoverflow.net/questions/9754",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2807/"
]
}
|
9,764 |
How do sheaves arise in studying solutions to ordinary differential equations? EDIT: Is it possible to construct non-isomorphic sheaves on a domain $D \subset \mathbb{R}^n$ using solution sets to differential equations? EDIT: Is the sheaf of vector spaces arising from the solution set of a linear ODE necessarily a vector bundle?
|
Let $U$ be an open subset of $\mathbb R^n$, and let $X$ be a vector field on $U$. You can construct a sheaf $\mathcal F$ of solutions of the ODE $Xf=0$ by letting $\mathcal F(U)$, for each open subset $V\subseteq U$, be the vector space of all $C^\infty$ functions $f$ on $V$ such that $Xf=0$. By changing the field $X$ you can certainly change the isomorphism clas of $\mathcal F$. Let $U=\mathbb R^2\setminus\{(0,0)\}$, define fields $X_1(x,y)=\Bigl((\frac1r-1)\frac xr-y,(\frac1r-1)\frac yr+x\Bigr)$ and $X_2(x,y)=(y,-x)$ and consider the corresponding sheaves $\mathcal F_1$ and $\mathcal F_2$. It is not difficult show show that $\mathcal F_1(U)$ is one-dimensional as a real vector space, while $\mathcal F_2(U)$ is infinite dimensional. It follows that $\mathcal F_1\not\cong\mathcal F_2$. Notice that $\mathcal F_1$ and $\mathcal F_2$ are locally isomorphic. This follows easily from the fact that the fields $X_1$ and $X_2$ are non-zero on their domain.
|
{
"source": [
"https://mathoverflow.net/questions/9764",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1358/"
]
}
|
9,778 |
There are two great first examples of complete discrete valuation ring with residue field $\mathbb{F}_p = \mathbb{Z}/p$: The $p$-adic integers $\mathbb{Z}_p$, and the ring of formal power series $(\mathbb{Z}/p)[[x]]$. Any complete DVR over $\mathbb{Z}/p$ is a ring structure on left-infinite strings of digits in $\mathbb{Z}/p$. The difference between these two examples is that in the $p$-adic integers, you add the strings of digits with carries. (In any such ring, you can say that a $1$ in the $j$th place times a 1 in the $k$th place is a 1 in the $(j+k)$th place.) At one point I realized that these two examples are not everything: You can also add with carries but move the carry $k$ places to the left instead of one place to the left. The ring that you get can be described as $\mathbb{Z}_p[p^{1/k}]$, or as the $x$-adic completion of $\mathbb{Z}[x]/(x^k - p)$. This sequence has the interesting feature that the terms are made from $\mathbb{Z}_p$ and have characteristic $0$, but the ring structure converges topologically to $(\mathbb{Z}/p)[[x]]$, which has characteristic $p$. ( Edit: Per Mariano's answer, I have in mind a discrete valuation in the old-fashioned sense of taking values in $\mathbb{Z}$, not $\mathbb{Z}^n$.) I learned from Jonathan Wise in a question on mathoverflow that these examples are still not everything. If $p$ is odd and $\lambda$ is a non-quadratic residue, then the $x$-adic completion of $\mathbb{Z}[x]/(x^2-\lambda p)$ is a different example. You can also call it $\mathbb{Z}_p[\sqrt{\lambda p}]$. So my question is, is there is a classification or a reasonable moduli space of complete DVRs with residue field $\mathbb{Z}/p$? Or whose residue field is any given finite field? Or if not a classification, an indexed family that includes every example at least once? I suppose that the question must be related to the Galois theory of $\mathbb{Q}_p$; maybe the best answer would be a relevant sketch of that theory. But part of my interest is in continuous families of DVR structures on the Cantor set of strings of digits in base $p$. As question authors often say in mathoverflow, I learned stuff from many of the answers and it felt incomplete to only accept one of them. I upvoted several others, though. Following Kevin, the set of pairs $(R,\pi)$, where $R$ is a CDVR and $\pi$ is a uniformizer, is an explicit indexed family that includes every example (of $R$) at least once. The mixed characteristic cases are parametrized by Eisenstein polynomials, and there is only one same-characteristic choice of $R$. One side issue that was not addressed as much is continuous families of CDVRs. To explain that concern a little better, all CDVRs with finite residue field are homeomorphic (to a Cantor set). For any given finite residue field, there are many homeomorphisms that commute with the valuation. So you could ask what a continuous family of CDVRs can look like, where both the addition and multiplication laws can vary, but the topology and the valuation stay the same. Now that I have been told what the CDVRs are, I suppose that not all that much can happen. It seems key to look at $\nu(p)$, the valuation of the integer element $p$, in a sequence of such structures (say). If $\nu(p) \to \infty$, then the CDVRs have to converge to $k[[x]]$. Otherwise $\nu(p)$ is eventually constant; it eventually equals some $n$. Then it looks like you can convert the convergent sequence of CDVRs to a convergent sequence of Eisenstein polynomials of degree $n$.
|
Greg, I want to say some basic things, but people are giving quite "high-brow" answers and what I want to say is a bit too big to fit into a comment. So let me leave an "answer" which is not really an answer but which is basically background on some other answers. So firstly there is this amazing construction of Witt vectors, which takes as input a finite field $k$ of characteristic p (well, it can take a lot more than that but let me stick with finite fields of characteristic p) and spits out a canonical complete DVR $W(k)$ with residue field $k$ and uniformiser $p$. If you feed in $Z/pZ$ it spits out $Z_p$ and if you feed in, say, the field $Z/5Z[\sqrt{2}]$ with 25 elements it spits out something isomorphic to $Z_5[\sqrt{2}]$, and so on. Witt vectors are just a way of formalising the "carry" business---it gets a bit trickier when the residue field isn't Z/pZ. Turns out that any complete DVR $R$ with fraction field of characteristic 0 and residue field $k$ is canonically and uniquely a $W(k)$-algebra. So that's pretty cool. Furthermore, $R$ will be finite and free over $W(k)$, and $R$ is the "ring of integers" (this makes sense) in its fraction field $K$, which is a finite extension of the fraction field of $W(k)$, something which in turn will be finite over $Q_p$ of degree $d$ if $[k:Z/pZ]=d$. [Note in particular that "fraction field of Witt vectors" gives us a whole bunch of extension of $Q_p$---the so-called "unramified" ones---one for each finite extension of $Z/pZ$.] Conversely, given an arbitrary finite field extension $K$ of $Q_p$, the ring of integers of $K$ (that is, the elements in $K$ satisfying a monic polynomial with coefficients in $Z_p$) will be a complete DVR with residue field some finite field $k$, and then the field of fractions of $W(k)$ embeds into $K$. This subfield of $K$---the "maximal unramified subextension of $K$" is canonical and intrinsic. An extension $K$ of $Q_p$ is "totally ramified" if the residue field of $R$ is $Z/pZ$. A FABULOUS place to read about this stuff is Serre's book "local fields". Everything there, with complete canonical proofs. So one aspect of your question is whether there is a moduli space of totally ramified extensions of $Q_p$. I wouldn't rule such a gadget out but I've not seen one. This might be just a question in algebra. If $K$ is a finite totally ramified extension of $Q_p$ then $K=Q_p(\pi)$ with $\pi$ a uniformiser of the integers of $K$, and $\pi$ will satisfy a degree $d$ equation if $d=[K:Q_p]$. Furthermore this degree $d$ equation will be monic, with coefficients in $Z_p$, and the constant term will be $p$ times a unit. Conversely any such equation will give a totally ramified extension of $Q_p$ (they will all be irreducible by Eisenstein's criterion). So now we have a list of all totally ramified extensions of $Q_p$, and hence all complete DVRs with residue characteristic $Z/pZ$ but generic characteristic zero, because we just list all polynomials of this form. Unfortunately each $R$ is in our list infinitely often. So to make the $R$s the $Z/pZ$-points of a moduli space we need to quotient out the set of degree $d$ Eisenstein polynomials by the relation "the extension of $Q_p$ generated by these polynomials are the same". A map $Q_p(\pi_1)\to Q_p(\pi_2)$ is just another polynomial so it looks to me like there is hope that this can be done, but I've not done it. Finally, everything I said above has a natural generalisation to any finite field, not just $Z/pZ$. When people talk about class field theory or Lubin-Tate groups above, what they're saying is that there are certain totally ramified extensions of $Q_p$, namely those which are Galois over $Q_p$ with abelian Galois group, which can be constructed explicitly using other techniques (like formal groups or Artin maps or whatever), and these constructions generalise to give all abelian extensions of an arbitrary finite extension of $Q_p$. However if you're looking for general moduli spaces then it seems to me that these notions might not be of too much use to you because they don't give all extensions, just abelian ones. There. So really that was just a comment but it was visibly too large. Hopefully someone will now quotient out the Eisenstein polynomials by an equivalence relation thus giving you your moduli space, because that seems to me to be the heart of your question.
|
{
"source": [
"https://mathoverflow.net/questions/9778",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1450/"
]
}
|
9,799 |
Hi Everyone, Famous anecdotes of G.H. Hardy relay that his work habits consisted of working no more than four hours a day in the morning and then reserving the rest of the day for cricket and tennis. Apparently his best ideas came to him when he wasn't "doing work." Poincare also said that he solved problems after working on them intensely, getting stuck and then letting his subconscious digest the problem. This is communicated in another anecdote where right as he stepped on a bus he had a profound insight in hyperbolic geometry. I am less interested in hearing more of these anecdotes, but rather I am interested in what people consider an appropriate amount of time to spend on doing mathematics in a given day if one has career ambitions of eventually being a tenured mathematician at a university. I imagine everyone has different work habits, but I'd like to hear them and in particular I'd like to hear how the number of hours per day spent doing mathematics changes during different times in a person's career: undergrad, grad school, post doc and finally while climbing the faculty ladder. "Work" is meant to include working on problems, reading papers, math books, etcetera (I'll leave the question of whether or not answering questions on MO counts as work to you). Also, since teaching is considered an integral part of most mathematicians' careers, it might be good to track, but I am interested in primarily hours spent on learning the preliminaries for and directly doing research. I ask this question in part because I have many colleagues and friends in computer science and physics, where pulling late nights or all-nighters is commonplace among grad students and even faculty. I wonder if the nature of mathematics is such that putting in such long hours is neither necessary nor sufficient for being "successful" or getting a post-doc/faculty job at a good university. In particular, does Malcom Gladwell's 10,000 hour rule apply to mathematicians? Happy Holidays!
|
I agree that hard work and stubbornness are very important (I think we should all take after Wiles and Perelman as much as we can). But it is also important how you spend the many hours you dedicate to mathematics. For instance, choice of problems is quite important: it is important to make sure that when you work on something, you spend your time usefully, i.e. you not only make progress on this particular problem, but also learn something new about mathematics in general. It is also important not to get hyperfocused on a fruitless attempt to solve a problem; after some time and effort spent on it, it becomes addictive. In such a situation, it is sometimes better to stop and ask for help/read something or switch to another problem for a while. Often, you'll wake up one morning a month or a year later and see that the insurmountable obstacle has magically disappeared! Or maybe this "Aha!" moment will come during a discussion with another mathematician, or while listening to a talk. For many people it is also helpful to have many simultaneous projects, so that when you get stuck on one, you can work on another. To summarize, I think that not only the number of hours matters, but also how efficiently you spend them, not only in terms of publishable results,
but also in terms of your personal growth as a mathematician.
|
{
"source": [
"https://mathoverflow.net/questions/9799",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1622/"
]
}
|
9,834 |
What is the heuristic idea behind the Fourier-Mukai transform? What is the connection to the classical Fourier transform? Moreover, could someone recommend a concise introduction to the subject?
|
First, recall the classical Fourier transform. It's something like this: Take a function $f(x)$, and then the Fourier transform is the function $g(y) := \int f(x)e^{2\pi i xy} dx$. I really know almost nothing about the classical Fourier transform, but one of the main points is that the Fourier transform is supposed to be an invertible operation. The Fourier-Mukai transform in algebraic geometry gets its name because it at least superficially resembles the classical Fourier transform. (And of course because it was studied by Mukai.) Let me give a rough picture of the Fourier-Mukai transform and how it resembles the classical situation. Take two varieties $X$ and $Y$, and a sheaf $\mathcal{P}$ on $X \times Y$. The sheaf $\mathcal{P}$ is sometimes called the "integral kernel". Take a sheaf $\mathcal{F}$ on $X$. Think of $\mathcal{F}$ as being analogous to the function $f(x)$ in the classical situation. Think of $\mathcal{P}$ as being analogous to, in the classical situation, some function of $x$ and $y$. Now pull the sheaf back along the projection $p_1 : X \times Y \to X$. Think of the pullback $p_1^\ast \mathcal{F}$ as being analogous to the function $F(x,y) := f(x)$. Think of $\mathcal{P}$ as being analogous to the function $e^{2\pi i xy}$ (but maybe not exactly, see below). Next, take the tensor product $p_1^\ast \mathcal{F} \otimes \mathcal{P}$. This is analogous to the function $F(x,y) e^{2\pi i xy}$ $=$ $f(x)e^{2\pi i xy}$. Finally, push $p_1^\ast\mathcal{F} \otimes \mathcal{P}$ down along the projection $p_2: X \times Y \to Y$. The result is the Fourier-Mukai transform of $\mathcal{F}$ --- it is $p_{2,\ast} (p_1^\ast \mathcal{F} \otimes \mathcal{P})$. This last pushforward step can be thought of as "integration along the fiber" --- here the fiber direction is the $X$ direction. So the analogous thing in the classical situation is $g(y) = \int f(x)e^{2\pi i xy}dx$ --- the Fourier transform of $f(x)$! But to make all of this actually work out, we have to actually use the derived pushforward, not just the pushforward. And so we have to work with the derived categories. When $X$ is an abelian variety, $Y$ is the dual abelian variety, and $\mathcal{P}$ is the so-called Poincare line bundle on $X \times Y$, then the Fourier-Mukai transform gives an equivalence of the derived category of coherent sheaves on $X$ with the derived category of coherent sheaves on $Y$. I think this was proven by Mukai. I think this is supposed to be analogous to the statement I made about the classical Fourier transform being invertible. In other words I think the Poincare line bundle is really supposed to be analogous to the function $e^{2\pi i xy}$. A more general choice of $\mathcal{P}$ corresponds to, in the classical situation, so-called integral transforms, which have been previously discussed here . This is probably why $\mathcal{P}$ is called the integral kernel. You may also be interested in reading about Pontryagin duality , which is a version of the Fourier transform for locally compact abelian topological groups --- this is obviously quite similar, at least superficially, to Mukai's result about abelian varieties. However I don't know enough to say anything more than that. There are some cool theorems of Orlov, I forget the precise statements (but you can probably easily find them in any of the books suggested so far), which say that in certain cases any derived equivalence is induced by a Fourier-Mukai transform. Note that the converse is not true: some random Fourier-Mukai transform (i.e. some random choice of the sheaf $\mathcal{P}$) is probably not a derived equivalence. I think Huybrechts' book "Fourier-Mukai transforms in algebraic geometry" is a good book to look at. Edit: I hope this gives you a better idea of what is going on, though I have to admit that I don't know of any good heuristic idea behind, e.g., Mukai's result --- it is analogous to the Fourier transform and to Pontryagin duality, and thus I suppose we can apply whatever heuristic ideas we have about the Fourier transform to the Fourier-Mukai transform --- but I don't know of any heuristic ideas that explain the Fourier-Mukai transform in a direct way, without appealing to any analogies to things that are outside of algebraic geometry proper. Hopefully somebody else can say something about that. But --- there is certainly something deep going on. Just as CommRing behaves a lot like Set op , I think there is probably some kind of general phenomenon that sheaves (or vector bundles) behave a lot like functions, which is what's happening here. Pullback of sheaves behave a lot like pullback of functions... Pushforward of sheaves behave a lot like integration of functions... Tensor product of sheaves behave a lot like multiplication of functions...
|
{
"source": [
"https://mathoverflow.net/questions/9834",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1547/"
]
}
|
9,864 |
Presburger arithmetic apparently proves its own consistency. Does anyone have a reference to an exposition of this? It's not clear to me how to encode the statement "Presburger arithmetic is consistent" in Presburger arithmetic. In Peano arithmetic this is possible since recursive functions are representable, so a recursive method of assigning Godel numbers to formulas and proofs means that Peano arithmetic can represent its own provability relation (of course, showing all that requires a lot of work). In particular, we can write a Peano arithmetic sentence which says "there is no natural number which encodes a proof of $\bot$". On the other hand, Presburger arithmetic can't represent all recursive functions. It can't even represent all the primitive recursive ones, so this same trick doesn't work. If it did, the first incompleteness theorem would apply.
|
Presburger arithmetic does NOT prove its own consistency. Its only function symbols are addition and successor, which are not sufficient to represent Godel encodings of propositions. However, consistent self-verifying axiom systems do exist -- see the work of Dan Willard ( "Self-Verifying Axiom Systems, the Incompleteness Theorem and Related Reflection Principles" ). The basic idea is to include enough arithmetic to make Godel codings work, but not enough to make the incompleteness theorem go through. In particular, you remove the addition and multiplication function symbols, and replace them with subtraction and division. This is enough to permit representing the theory arithmetically, but the totality of multiplication (which is essential for the proof of the incompleteness theorem) is not provable, which lets you consistently add a reflection principle to the logic.
|
{
"source": [
"https://mathoverflow.net/questions/9864",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2693/"
]
}
|
9,898 |
What's the most common way of writing the all-ones vector, that is, the vector, when projected onto each standard basis vector of a given vector space, having length one? The zero vector is frequently written $\vec{0}$, so I'm partial to writing the all-ones vector as $\vec{1}$, but I don't know how popular this is, and I don't know if a reader might confuse it with the identity matrix. I'm writing for a graph theory audience, if that helps pick a notation.
|
I have used the notation $\vec{1}$ in a paper. I think that it's a good choice if you help the reader by defining it. I did a Google Scholar such of "vector of all ones", and I found a lot of so-so notation such as $e$, $u$, $\mathbf{e}$, $\mathbf{1}$, and even just plain $1$. I don't think that the literature is loyal to any particular choice. Confusing $\vec{1}$ with a matrix would be a little strange, because a matrix is suggested by a two-headed arrow, or $\stackrel{\leftrightarrow}{1}$.
|
{
"source": [
"https://mathoverflow.net/questions/9898",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2836/"
]
}
|
9,901 |
Which are the rigid suborders of the real line? If A is any set of reals, then it can be viewed as an order structure itself under the induced order (A,<). The question is, when is this structure rigid? That is, for which sets A does the structure (A,<) have no nontrivial order automorphisms? For example, the positive integers are rigid under the usual order. More generally, any well-ordered subset of R is rigid. Similarly, any anti-well-ordered set, such as the negative integers, is also rigid. It is also true that the order sum of any well-order plus an anti-well-order is rigid. For example, a sequence converging upward to 0 plus a sequence converging downward to 0 will have order type ω+ω*, which is rigid. (Whereas it is easy to see that the sum of an infinite anti-well-order and an infinite well-order will not be rigid, since it has a copy of Z in the center.) A more elaborate example will be a well-ordered sum of anti-well-orders, such as the set consisting of k+1/n for any positive integers k and n. All these examples are countable; are there uncountable examples? Perhaps there will be some ZFC independence for certain types of examples? I am primarily interested in the situation under ZFC. In ZF without the Axiom of Choice, there can be weird anomalies of uncoutable sets that are Dedekind finite. All such sets are rigid, as I explained in this question . But if someone can provide a ZF characterization, that would also be interesting.
|
Here is a simple example of size continuum. Do the ordinary middle-third construction of the Cantor set, except that whenever you delete the $n$-th (numbered by level and then left to right, say) middle-third interval leave in exactly $n$ points from that interval. Let's call the resulting set $X$. Any automorphism of $X$ must map (resp. left-, right-) isolated points to (resp. left-, right-) isolated points. Since we left a different number of points in each middle-third interval, the automorphism must fix every inner point of each middle-third interval as well as its two endpoints. By density, it follows that the automorphism fixes every point of $X$. All of your examples are scattered, I checked the Rosenstein's Linear Orderings to see if he had anything good to say about which scattered linear orders are rigid. To my dismay, this is what I found: "These considerations seem to make impossible an inductive argument (on $F$-rank or $VD$-rank) to determine which scattered types are rigid." (p. 133) However, he does cite a result of Anne Morel ( Ordering relations admitting automorphisms , Fund. Math. 54 (1964), 279-284.) which says that a linear order $A$ is not rigid if and only if $A \cong A_1 + A_2\times\mathbb{Z} + A_3$ for some linear orderings $A_1,A_2,A_3$, with $A_2$ nonempty. Dense examples of rigid subsets of $\mathbb{R}$ would be interesting to see. This is probably not too difficult to construct under CH. But a ZFC example might have to deal with important barriers such as Baumgartner's result that All ${\aleph_1}$-dense subsets of $\mathbb{R}$ can be isomorphic , Fund. Math. 79 (1973), 101-106. Maybe there are some examples in this classic paper of Sierpinski Sur les types d'ordre des ensembles linéaires , Fund. Math. 37 (1950), 253-264. All three papers can be found here . Addendum (after sdcvvc's comment): For the sake of completeness, I'm including a simplification of the Dushnik-Miller argument that produces a dense subset $X$ of $\mathbb{R}$ which is rigid (though not the stronger result that $X$ has no self-embeddings). To ensure density, the set $X$ will contain all rational numbers. Note that an automorphism $f$ of $X$ is then completely determined by its restriction to $\mathbb{Q}$. Indeed, since $f[\mathbb{Q}]$ must be dense (in $X$ and) in $\mathbb{R}$, we always have $f(x) = \sup\{f(q):q \in (-\infty,x)\cap\mathbb{Q}\} = \inf\{f(q):q \in (x,\infty)\cap\mathbb{Q}\}.$ There are only $c = 2^{\aleph_0}$ increasing maps $f:\mathbb{Q}\to\mathbb{R}$ with dense range. Let $\langle f_\alpha:\alpha<c \rangle$ enumerate all such maps, except for the identity on $\mathbb{Q}$. We will define by induction a sequence $\langle (x_\alpha,y_\alpha) : \alpha<c \rangle$ of pairs of irrational numbers. The $x_\alpha$ will be points of $X$ while the $y_\alpha$ will be in the complement of $X$. For each $\alpha$, we will have $f_\alpha(x_\alpha) = y_\alpha$ (in the sense of the inf/sup definition above). Suppose we have defined $(x_\beta,y_\beta)$ for $\beta<\alpha$. Since $f_\alpha$ is not the identity, there is a rational $q$ such that $f_\alpha(q) \neq q$. Let's suppose that $f_\alpha(q) > q$ (the case $f_\alpha(q) < q$ is symmetric). Since the real interval $(q,f_\alpha(q))$ has size $c$ and the extension of $f_\alpha$ to all of $\mathbb{R}$ is injective, we can always pick $x_\alpha \in (q,f_\alpha(q)) \setminus(\mathbb{Q}\cup\{y_\beta:\beta<\alpha\})$ such that $y_\alpha = f_\alpha(x_\alpha) \notin \mathbb{Q}\cup\{x_\beta:\beta<\alpha\}$. Note that $x_\alpha < f_\alpha(q) < y_\alpha$ so $x_\alpha \neq y_\alpha$. In the end, we will have $\{x_\alpha: \alpha < c\} \cap \{y_\alpha : \alpha<c\} = \varnothing$ and any set $X$ such that $\mathbb{Q}\cup\{x_\alpha:\alpha<c\} \subseteq X \subseteq \mathbb{R}\setminus\{y_\alpha:\alpha<c\}$ is necessarily rigid since $f_\alpha(x_\alpha) = y_\alpha \notin X$ for each $\alpha<c$.
|
{
"source": [
"https://mathoverflow.net/questions/9901",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1946/"
]
}
|
9,924 |
I thought that the order of the Tate-Shafarevich group should always be a square (it's also supposed to be finite, but for the purposes of this question let's assume we know this) but I don't seem to find a good explanation; Wikipedia is silent on the matter. While I know it may be an open problem, is there a good argument pro or contra this?
|
The first example of an abelian variety with nonsquare Sha was discovered in a computation by Michael Stoll in 1996. He emailed it to me and Ed Schaefer, because his calculation depended on a paper that Ed and I had written. At first none of us believed that it was what it was: instead we thought it must be due to either an error in Stoll's calculations or an error in the Poonen-Schaefer paper. Stoll and I worked together over the next few weeks to develop a theory that explained the phenomenon, and this led to the paper http://math.mit.edu/~poonen/papers/sha.ps - that paper contains a detailed answer to your question. To summarize a few of the key points: If the abelian variety over a global field $k$ has a principal polarization coming from a $k$-rational divisor (as is the case for every elliptic curve), then the order of Sha is a square (if finite), because it carries an alternating pairing - this is what Tate proved, generalizing Cassels' result for elliptic curves. For principally polarized abelian varieties in general, the pairing satisfies the skew-symmetry condition $\langle x,y \rangle = - \langle y,x \rangle$ but not necessarily the stronger, alternating condition $\langle x,x \rangle=0$, so all one can say is that the order of Sha is either a square or twice a square (if finite). Stoll and I gave an explicit example of a genus 2 curve over $\mathbf{Q}$ whose Jacobian had Sha isomorphic to $\mathbf{Z}/2\mathbf{Z}$ unconditionally (in particular, finiteness could be proved in this example). If the polarization on the abelian variety is not a principal polarization, then the corresponding pairing need not be even skew-symmetric, so there is no reason to expect Sha to be even within a factor of $2$ of a square. And indeed, William Stein eventually found explicit examples and published them in the 2004 paper cited by Simon. A final remark: Ironically, my result with Stoll quantifying the failure of Sha to be a square is used by Liu-Lorenzini-Raynaud to prove that the Brauer group $\operatorname{Br}(X)$ of a surface over a finite field is a square (if finite)!
|
{
"source": [
"https://mathoverflow.net/questions/9924",
"https://mathoverflow.net",
"https://mathoverflow.net/users/65/"
]
}
|
9,944 |
let $G$ be a group such that $\mathrm{Aut}(G)$ is abelian. is then $G$ abelian? This is a sort of generalization of the well-known exercise, that $G$ is abelian when $\mathrm{Aut}(G)$ is cyclic, but I have no idea how to answer it in general. At least, the finitely generated abelian groups $G$ such that $\mathrm{Aut}(G)$ is abelian can be classified.
|
From MathReviews: MR0367059 (51 #3301)
Jonah, D.; Konvisser, M.
Some non-abelian $p$ -groups with abelian automorphism groups.
Arch. Math. (Basel) 26 (1975), 131--133. This paper exhibits, for each prime $p$ , $p+1$ nonisomorphic groups of order $p^8$ with elementary abelian automorphism group of order $p^{16}$ . All of these groups have elementary abelian and isomorphic commutator subgroups and commutator quotient groups, and they are nilpotent of class two. All their automorphisms are central. With the methods of the reviewer and Liebeck one could also construct other such groups, but the orders would be much larger. FYI, I found this via a google search. The first to construct such a group (of order $64 = 2^6$ ) was G.A. Miller * in 1913. If you know something about this early American group theorist (he studied groups of order 2, then groups of order 3, then...and he was good at it, and wrote hundreds of papers!), this is not so surprising. I found a nice treatment of "Miller groups" in Section 8 of http://arxiv.org/PS_cache/math/pdf/0602/0602282v3.pdf (*): The wikipedia page seems a little harsh. As the present example shows, he was a very clever guy.
|
{
"source": [
"https://mathoverflow.net/questions/9944",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2841/"
]
}
|
9,951 |
Is it possible to regard limits in analysis (say, of real sequences or more generally nets in topological spaces) as limits in category theory? Is there some formal connection? Edit ('13): Perhaps it is more interesting to ask whether limits in category theory can be seen as special limits of ultrafilters or nets.
|
I have asked this question on math.stackexchange last year, and got satisfying answer.
(So this construction did not come from me.) Let $(X,\mathcal O)$ be a topological space, $\mathcal F(X)$ the partialy ordered set of filters on $X$ with respect to inclusions, considered as a small category in the usual way. Given $x\in X$ and $F\in\mathcal F(X)$ let $\mathcal U_X(x)$ denote the neighbourhood filter of $x$ in $(X,\mathcal O)$ and $\mathcal F_{x,F}(X)$ the full subcategory of $\mathcal F(X)$ generated by $\{G\in\mathcal F(X):F\cup\mathcal U_X(x)\subseteq G\}$, let $E:\mathcal F_{x,F}\hookrightarrow\mathcal F(X)$ be the obvious (embedding) diagram, $\Delta$ the usual diagonal functor and $\lambda:\Delta(F)\rightarrow E$ the natural transformation where $\lambda(G):F\hookrightarrow G$ is the inclusion for each $G\in\mathcal F_{x,F}$. It is not hard to see that $F$ tends to $x$ in $(X,\mathcal O)$ iff $\lambda$ is a limit of $E$.
|
{
"source": [
"https://mathoverflow.net/questions/9951",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2841/"
]
}
|
9,961 |
This is related to another question . I've found many remarks that the category of schemes is not cocomplete. The category of locally ringed spaces is cocomplete, and in some special cases this turns out to be the colimit of schemes, but in other cases not (which is, of course, no evidence that the colimit does not exist). However, I want to understand in detail a counterexample where the colimit does not exist, but I hardly found one. In FGA explained I've found the reference, that Example 3.4.1 in Hartshorne, Appendix B is a smooth proper scheme over $\mathbb{C}$ with a free $\mathbb{Z}/2$-action, but the quotient does not exist (without proof). To be honest, this is too complicated to me. Are there easy examples? You won't help me just giving the example, because there are lots of them, but the hard part is to prove that the colimit really does not exist.
|
Edit: BCnrd gave a proof in the comments that this example works, so I've edited in that proof. A possible proven example I suspect There is no scheme which is "two $\mathbb A^1$'s glued together along their generic points" (or "$\mathbb A^1$ with every closed point doubled"). In other words, the coequalizer of the two inclusions $Spec(k(t))\rightrightarrows \mathbb A^1\sqcup \mathbb A^1$ does not exist in the category of schemes. Intuitively, this coequalizer should be "too non-separated" to be a scheme. I don't have a proof, but I thought other people might have ideas if I posted this here. If a coequalizer $P$ does exist, then no two closed points of $\mathbb A^1\sqcup \mathbb A^1$ map to the same point in $P$. To show this, it is enough to find functions from $\mathbb A^1\sqcup \mathbb A^1$ to other schemes which agree on the generic points but disagree on any other given pair of points. The obvious map $\mathbb A^1\sqcup \mathbb A^1\to \mathbb A^1$ separates most pairs of closed points. To see that a point on one $\mathbb A^1$ is not identified with "the same point on the other $\mathbb A^1$", consider the map from $\mathbb A^1\sqcup \mathbb A^1$ to $\mathbb A^1$ with the given point doubled. On the other hand, let $U$ be an affine open around the image of the generic point in $P$. $U$ has dense open preimages $V$ and $V'$ in both affine lines. Let $W=V\cap V'$ inside the affine line, so we have two maps from $W$ to the affine $U$ which coincide at the generic point of $W$, and hence are equal (as $U$ is affine). In particular, the two maps from affine line to categorical pushout $P$ coincide at each "common pair" of closed points of the two copies of $W$, contradicting the previous paragraph. Edit: The questions below are no longer relevant, but I'd like to leave them there for some reason. Here are some questions that might be helpful to answer: If the coequalizer above does exist, must the map from $\mathbb A^1\sqcup \mathbb A^1$ be surjective? (see the related question Can a coequalizer of schemes fail to be surjective? ) Is the coequalizer of $Spec(k(t))\rightrightarrows \mathbb A^1\sqcup \mathbb A^1$ in the category of separated schemes equal to $\mathbb A^1$? (probably) What are some ways to determine that a functor $Sch\to Set$ is not corepresented by a scheme?
|
{
"source": [
"https://mathoverflow.net/questions/9961",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2841/"
]
}
|
10,014 |
As the title suggests I am interested in CRT applications. Wikipedia article on CRT lists some of the well known applications (e.g. used in the RSA algorithm, used to construct an elegant Gödel numbering for sequences...) Do you know some other (maybe not so well known) applications? Or interesting problems (recreational? or from mathematical competitions like IMO?) which can be solved using CRT. Or any good references or examples in that direction. I hope that with this I will have better understanding of CRT and how to use it in general.
|
Lagrange interpolation is a special case of the Chinese remainder theorem. (Fixing a dead link: https://artofproblemsolving.com/community/c1157h990758_the_chinese_remainder_theorem_and_lagrange_interpolation ) The Jordan normal form can be proven extremely quickly using the Chinese remainder theorem for modules over a commutative ring. This proceeds by first proving the Jordan-Chevalley decomposition, and then the rest is a simple exercise of showing what the Jordan blocks actually look like. The first one is very surprising to people, but if you state Lagrange interpolation correctly, it's easy to see that the idea is not only similar but identical.
|
{
"source": [
"https://mathoverflow.net/questions/10014",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1866/"
]
}
|
10,023 |
I want a ring $R$ of "numbers" such that: For any sequence of congruences $x\equiv a_1 \pmod{n_1}, x\equiv a_2 \pmod{n_2},\dots$ with $a_i\in \mathbb{Z}$ and $n_i\in \mathbb{N}$ such than any finite set of these congruences has a solution $x\in\mathbb{Z}$, there is a $r\in R$ such that $r\equiv a_1 \pmod{n_1}, r\equiv a_2 \pmod{n_2},\dots$ and For any $r\in R$ and $n\in\mathbb{N}$ there is a $a, 0\leq a< n $ such that $r\equiv a \pmod{n}$. I think that $R$ has to be the product set of the p-adic integers over all primes p, but what do you call this ring? (Perhaps there should be a "terminology" tag? Edit: It already exists but it is called "names")
|
The standard notation is $\widehat{\mathbb{Z}}$. The names I know are "the profinite completion of $\mathbb{Z}$" and "$\mathbb{Z}$-hat".
|
{
"source": [
"https://mathoverflow.net/questions/10023",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2097/"
]
}
|
10,033 |
Is there a standard example of two abelian varieties $A$, $B$ over some number field $k$ which are $k_v$-isomorphic for every place $v$ of $k$ but not $k$-isomorphic ?
|
(If you upvote this answer, please consider upvoting the answers by Felipe Voloch and David Speyer too, since this answer builds on their ideas.) The smallest examples are in dimension $2$. Let $E$ be any elliptic curve over $\mathbf{Q}$ without complex multiplication, e.g., $X_0(11)$. We will construct two twists of $E^2$ that are isomorphic over $\mathbf{Q}_p$ for all $p \le \infty$ but not isomorphic over $\mathbf{Q}$. Let $K:=\mathbf{Q}(\sqrt{-1},\sqrt{17})$. Let $G:=\operatorname{Gal}(K/\mathbf{Q}) = (\mathbf{Z}/2\mathbf{Z})^2$. Let $\alpha \colon G \to \operatorname{GL}_2(\mathbf{Z}) = \operatorname{Aut}(E^2)$ be a homomorphism sending the two generators to the reflections in the coordinate axes of $\mathbf{Z}^2$, and let $A$ be the $K/\mathbf{Q}$-twist of $E^2$ given by $\alpha$. Define $\beta$ and $B$ similarly, but with the lines $y=x$ and $y=-x$ in place of the coordinate axes. The representations $\alpha$ and $\beta$ of $G$ on $\mathbf{Z}^2$ are not conjugate: only the former is such that the lattice vectors fixed by nontrivial elements of $G$ generate all of $\mathbf{Z}^2$. Thus $A$ and $B$ are not isomorphic over $\mathbf{Q}$. On the other hand, every decomposition group $D_p$ in $G$ is smaller than $G$ since $-1$ is a square in $\mathbf{Q}_{17}$ and $17$ is a square in $\mathbf{Q}_2$. Also, the restrictions of $\alpha$ and $\beta$ to any proper subgroup of $G$ are conjugate: any single line spanned by a primitive vector in $\mathbf{Z}^2$ can be mapped to any other by an element of $\operatorname{GL}_2(\mathbf{Z})$. Thus $A$ and $B$ become isomorphic after base extension to $\mathbf{Q}_p$ for any $p \le \infty$. $\square$ Remark: The abelian surfaces $A$ and $B$ constructed above are isogenous even over $\mathbf{Q}$, because the $\mathbf{Z}^2$ with one Galois action can be embedded into the $\mathbf{Z}^2$ with the other Galois action: rotate $45^\circ$ and dilate. Remark: The nonexistence of examples in dimension $1$ follows from these two well-known facts: 1) Twists of an elliptic curve over a field $k$ of characteristic $0$
are classified by
$H^1(k,\mu_n)=k^\times/k^{\times n}$ where $n$ is 2, 4, or 6. 2) If $n<8$, the map $k^\times/k^{\times n} \to \prod_v k_v^\times/k_v^{\times n}$ is injective. [ Edit: This answer was edited to simplify the construction and to add those remarks at the end.]
|
{
"source": [
"https://mathoverflow.net/questions/10033",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2821/"
]
}
|
10,066 |
In dimension 2 we know by the Riemann mapping theorem that any simply connected domain ( $\neq \mathbb{R}^{2}$) can be mapped bijectively to the unit disk with a function that preserves angles between curves, ie is conformal. I have read the claim that conformal maps in higher dimensions are pretty boring but does anyone know a proof or even a intuitive argument that conformal maps in higher dimensions are trivial?
|
I think you're looking for Liouville's theorem . This theorem states that for $n >2$, if $V_1,V_2 \subset \mathbb{R}^n$ are open subsets and $f : V_1 \rightarrow V_2$ is a smooth conformal map, then $f$ is the restriction of a higher-dimensional analogue of a Mobius transformation. By the way, observe that there are no assumptions on the topology of the $V_i$ -- they don't have to be simply-connected, etc. EDIT : I'm updating this ancient answer to link to a blog post by Danny Calegari which contains a sketch of a beautifully geometric argument for Liouville's theorem.
|
{
"source": [
"https://mathoverflow.net/questions/10066",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2888/"
]
}
|
10,103 |
When I was a teenager, I was given the book Men of Mathematics by E. T. Bell, and I rather enjoyed it. I know that this book has been criticized for various reasons and I might even agree with some of the criticism, but let's not digress onto that. E. T. Bell made a reasonable list of 34 of the greatest mathematicians from the ancient Greek period to the end of the 19th century. His list isn't perfect — maybe he should have included Klein or skipped Poncelet — but no such list can be perfect anyway. I think that his selection was good. He also made the careers of these mathematicians exciting and he addressed why mathematicians care about them today. It helped me learn what achievements and topics in mathematics are important. (Just to head off discussion, the standard complaints include that the title is sexist and so is the book, that Bell was loose with biographical facts, and that he "chewed the scenery".) After the period covered by Bell, there is a 50-70 year gap, followed by the the Fields Medals. The list of Fields Medalists has its own limitations, but it is another interesting, comparably long list of great mathematicians. Somewhat accidentally, this list also orients and motivates advanced mathematics students today. But who are the great mathematicians in the gap itself, that is, those born between 1850 and 1900, or say between 1860 and 1910? (Or 1920 at the latest; that is what I had before.) There is a good expanded list of mathematicians with biographies at the St. Andrews site . However, it is too long to work as a sequel to Bell's book. (Not that I plan to write one; I'm just asking.) If you were to make a list of 20 to 50 great mathematicians in this period, how would you do it or who would they be? Presumably it would include Hilbert, but who else? (Poincaré was born in 1854 and is the latest born in Bell's list.) For example, I know that there was the IBM poster "Men of Modern Mathematics" that also earned some criticism. But I don't remember who was listed, and my recollection is that the 1850-1900 period was somewhat cramped. I decided based on the earliest responses to convert this list to community wiki. But I am not just asking people to throw out names one by one and then vote them up. If a good reference for this question already exists, or if there is some kind of science to make a list, then that would be ideal. If you would like to post a full list, great. If you would like to list one person who surely should be included and hasn't yet been mentioned, then that's also reasonable. It may be better to add years of birth and death in parentheses, for instance "Hilbert (1862-1943)". Note: To make lists, you should either add two spaces to the end of each line, or " - " (space dash space) to the beginning of each line.
|
The St. Andrews site is an invaluable resource. From that list, I picked (usually) at most one great mathematician born in each year from 1860 to 1910: $\textbf{EDIT: By popular demand, the list now extends from 1849 to 1920.}$ 1849: Felix Klein, Ferdinand Georg Frobenius 1850: Sofia Vasilyevna Kovalevskaya 1851: honorable mention: Schottky 1852: William Burnside 1853: honorable mentions: Maschke, Ricci-Curbastro, Schoenflies 1854: Henri Poincare 1856: Emile Picard (honorable mention: Stieltjes) 1857: honorable mention: Bolza 1858: Giuseppe Peano (honorable mention: Goursat) 1859: Adolf Hurwitz (honorable mention: Holder) 1860: Vito Volterra 1861: honorable mention: Hensel 1862: David Hilbert 1864: Hermann Minkowski 1865: Jacques Hadamard (honorable mention: Castelnuovo) 1868: Felix Hausdorff 1869: Elie Cartan 1871: Emile Borel (honorable mentions: Enriques, Steinitz, Zermelo) 1873: honorable mentions: Caratheodory, Levi-Civita, Young 1874: Leonard Dickson 1875: Henri Lebesgue (honorable mentions: Schur, Takagi) 1877: Godfrey Harold Hardy 1878: Max Dehn 1879: honorable mentions: Hahn, Severi 1880: Frigyes Riesz 1881: Luitzen Egbertus Jan Brouwer 1882: Emmy Amalie Noether (honorable mentions: Sierpinski, Wedderburn) 1884: George Birkhoff, Solomon Lefschetz 1885: Hermann Weyl (honorable mention: Littlewood) 1887: Erich Hecke (honorable mentions: Polya, Ramanujan, Skolem) 1888: Louis Joel Mordell (honorable mention: Alexander) 1891: Ivan Matveevich Vinogradov 1892: Stefan Banach 1894: Norbert Wiener 1895: honorable mention: Bergman 1896: Carl Ludwig Siegel (honorable mention: Kuratowski) 1897: honorable mention: Jesse Douglas 1898: Emil Artin, Helmut Hasse (honorable mentions: Kneser, Urysohn) 1899: Oscar Zariski (honorable mentions: Bochner, Krull, Ore) 1900: Antoni Zygmund 1901: Richard Brauer 1902: Alfred Tarski (honorable mention: Hopf) 1903: John von Neumann (hm's: Hodge, Kolmogorov, de Rham, Segre, Stone, van der Waerden) 1904: Henri Cartan (honorable mentions: Hurewicz, Whitehead) 1905: Abraham Adrian Albert 1906: Kurt Godel, Andre Weil (honorable mentions: Dieudonne, Feller, Leray, Zorn) 1907: Lars Ahlfors, Hassler Whitney (honorable mentions: Coxeter, Deuring) 1908: Lev Pontrjagin 1909: Claude Chevalley, Saunders Mac Lane (honorable mentions: Stiefel, Ulam) 1910: Nathan Jacobson (honorable mention: Steenrod) 1911: Shiing-shen Chern (honorable mentions: Birkhoff, Chow, Kakutani, Witt) 1912: Alan Mathison Turing (honorable mentions: Eichler, Zassenhaus) 1913: Samuel Eilenberg, Paul Erdos, Israil Moiseevich Gelfand (dis/honorable mention: Teichmuller) 1914: honorable mentions: Dantzig, Dilworth, Kac 1915: Kunihiko Kodaira (honorable mentions: Hamming, Linnik, Tukey) 1916: Claude Elwood Shannon (honorable mention: Mackey) 1917: Atle Selberg (honorable mentions: Iwasawa, Kaplansky) 1918: Abraham Robinson 1919: honorable mention: Julia Robinson 1920: Alberto Calderon
|
{
"source": [
"https://mathoverflow.net/questions/10103",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1450/"
]
}
|
10,124 |
Well, $n!$ is for integer $n < 0$ not defined — as yet. So the question is: How could a sensible generalization of the factorial for negative integers look like? Clearly a good generalization should have a clear combinatorial meaning which combines well with the nonnegative case.
|
It's not that it's not defined... Actually it has been defined more than it should have. There are plenty of functions that interpolate the factorials, some of them extend to the negative integers as well. Hadamard's Gamma function is entire, logarithmic single inflected factorial function is another example. But on the other hand, for some mysterious reason, the nice property that we want an extension of the factorial to enjoy is log-convexity. The Bohr-Mollerup-Artin Theorem tells us that the only function which is logarithmically convex on the positive real line and satisfies $f(z)=zf(z-1)$ there (also $f(1)=1$ and $f(z)>0$ ) is the Gamma function . Unfortunately the gamma function doesn't extend to negative integers, and that is why I guess people don't really care that much for defining them as they know that no "good" answer can be found.
|
{
"source": [
"https://mathoverflow.net/questions/10124",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2797/"
]
}
|
10,128 |
This is totally elementary, but I have no idea how to solve it: let $A$ be an abelian group such that $A$ is isomorphic to $A^3$ . is then $A$ isomorphic to $A^2$ ? probably no, but how construct a counterexample? you can also ask this in other categories as well, for example rings. if you restrict to Boolean rings, the question becomes a topological one which makes you think about fractals: let $X$ be Stone space such that $X \cong X + X + X$ , does it follow that $X \cong X + X$ (here $+$ means disjoint union)? Edit : In the answers there are already counterexamples. but you may add others in other categories (with products/coproducts), especially if they are easy to understand :).
|
The answer to the first question is no. That is, there exists an abelian group $A$ isomorphic to $A^3$ but not $A^2$ . This result is due to A.L.S. (Tony) Corner, and is the case $r = 2$ of the theorem described in the following Mathematical Review. MR0169905 Corner, A.L.S., On a conjecture of Pierce concerning direct decomposition of Abelian groups. 1964 Proc. Colloq. Abelian Groups (Tihany, 1963) pp.43--48 Akademiai Kiado, Budapest. It is shown that for any positive integer $r$ there exists a countable torsion-free abelian group $G$ such that the direct sum of $m$ copies of $G$ is isomorphic to the direct sum of $n$ copies of $G$ if and only if $m \equiv n \pmod r$ . This remarkable result is obtained from the author's theorem on the existence of torsion-free groups having a prescribed countable, reduced, torsion-free endomorphism ring by constructing a ring with suitable properties. It should be mentioned that the question of the existence of algebraic systems with the property stated above has been considered by several mathematicians. The author has been too generous in crediting this "conjecture" to the reviewer. Reviewed by R.S. Pierce
|
{
"source": [
"https://mathoverflow.net/questions/10128",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2841/"
]
}
|
10,227 |
So, say we are working with non-CH mathematics. This means, AFAIK, that there is at least one set $S$ in our non-CH mathematics, whose cardinality is intermediate between $|\mathbb{N}|$ (card. of naturals) and $|\mathbb{R}|=2^\mathbb{N}$, the continuum. Question: what kind of objects would we find in this set $S$? Also: is this mathematics radically different from the one where CH holds? Specifically, are there results that are used in everyday math , at a relatively introductory level, which do not hold on our non-CH math.?. What results that we find in everyday math would not hold in our new math? Would there, e.g., still exist non-measurable sets? Maybe more specifically: what results depend on the CH?
|
The question of what happens when CH fails is, of course, intensely studied in set theory. There are entire research areas, such as the area of cardinal characteristics of the continuum , which are devoted to studying what happens with sets of reals when the Continuum Hypothesis fails. The lesson of much of this analysis is that many of the most natural open questions turn out to be themselvesd independent of ZFC, even when one wants ¬CH. For example, the question of whether all sets that are intermediate in size between the natural numbers and the continuum should be Lebesgue measure 0, is independent of ZFC+¬CH. The question of whether only the countable sets have continuum many subsets is independent of ZFC+¬CH. There are a number of cardinal characteristics that I mention here , whose true nature becomes apparant only when CH fails. For example, must every unbounded family of functions from ω to ω have size continuum? It is independent of ZFC+¬CH. Must every dominating family of such functions have size continuum? It is independent of ZFC+¬CH. Those question are relatively simple to state and could easily be considered part of "ordinary" mathematics. However, much of the rest of what you might think of as ordinary mathematics is simply not affected by CH or not CH. In particular, the existence of non-measurable sets that you mentioned is provable in ZFC, whether or not CH holds. (This proof requires the use of the Axiom of Choice, however, unless large cardinals are inconsistent, a result proved by Solovay and Shelah.) Nevertheless, there is a growing body of research on some sophisticated axioms in set theory called forcing axioms, which have powerful consequences, and many of these new axioms imply the failure of CH. This topic began with Martin's Axiom MA ω 1 , and has continued with the Proper Forcing Axiom, Martin's Maximum and now many other variations. Lastly, in your title you asked what are the new sets like. The consistency of the failure of the Continuum Hypothesis was proved by Paul Cohen with the method of forcing . This highly sophisticated and versatile method is now used pervasively in set theory, and is best thought of as a fundamental method of constructing models of set theory, sharing many affinities with construction methods in algebra, such as the construction of algebraic or transcendental field extensions. Cohen built a model of ZFC+ ¬CH by starting with a model V of ZFC+CH, and then using the method of forcing to add ω 2 many new real numbers to construct the forcing extension V[G]. Since V and V[G] have the same cardinals (by a detailed combinatorial argument), it follows that the set of reals in V[G] has size at least (in fact, exactly) ω 2 . In particular, the old set of reals from V, which had size ω 1 , is now one of the sets of reals of intermediate size. Thus, these intermediate sets are not so mysterious after all!
|
{
"source": [
"https://mathoverflow.net/questions/10227",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2915/"
]
}
|
10,237 |
The Convolution Theorem is often exploited to compute the convolution of two sequences efficiently: take the (discrete) Fourier transform of each sequence, multiply them, and then perform the inverse transform on the result. The same thing can be done for convolutions in the quotient ring Z/pZ via the analogous Number Theoretic Transform . Does this procedure generalize to other algebraic structures? Arbitrary rings? Semirings? Fast convolutions over the semiring (min,+) would be particularly useful. I'm led to suspect this because both sorting and FFT can be computed using the same butterfly-like network with simple operations like "min", "max", "+", and "*" at the nodes of the network, and sorting can be thought of as a kind of convolution.
|
In general, it is a major open question in discrete algorithms as to which algebraic structures admit fast convolution algorithms and which do not. (To be concrete, I define the $(\oplus,\otimes)$ convolution of two $n$-vectors $[x_0,\ldots,x_{n-1}]$ and $[y_0,\ldots,y_{n-1}]$, to be the vector $[z_0,\ldots,z_{n-1}]$ with $$z_k = (x_0 \otimes y_k) \oplus (x_1 \otimes y_{k-1}) \oplus \cdots \oplus (x_k \otimes y_0).$$ Here, $\otimes$ and $\oplus$ are the multiplication and addition operations of some underlying semiring.) For any $\otimes$ and $\oplus$, the convolution can be computed trivially in $O(n^2)$ operations. As you note, when $\otimes = \times$, $\oplus = +$, and we work over the integers, this convolution can be done efficiently, in $O(n \log n)$ operations. But for more complex operations, we do not know efficient algorithms, and we do not know good lower bounds. The best algorithm for $(\min,+)$ convolution is $n^2/2^{\Omega (\sqrt{\log n})}$ operations, due to combining my recent APSP paper Ryan Williams: Faster all-pairs shortest paths via circuit complexity. STOC 2014: 664-673 and David Bremner, Timothy M. Chan, Erik D. Demaine, Jeff Erickson, Ferran Hurtado, John Iacono, Stefan Langerman, Perouz Taslakian: Necklaces, Convolutions, and X + Y. ESA 2006: 160-171 A substantially subquadratic algorithm for $(\min,+)$ convolution would (to my knowledge) imply a subcubic algorithm all-pairs shortest paths in general graphs, a longstanding open problem. The above ESA06 reference also gives a $O(n^2 (\log \log n)^2/\log n)$ algorithm for a "(median,+) convolution". The situation is subtle. It's not clear when convolution over a semiring is easy and when it's hard. For instance, the $(\min,\max)$ convolution can be computed in subquadratic time: I believe that $O(n^{3/2} \log n)$ operations suffice. This can be obtained from adapting the $(\min,\max)$ matrix multiplication algorithm in my work with Vassilevska and Yuster on all-pairs bottleneck paths. Basically you reduce the problem to computing $\sqrt{n}$ instances of a $(+,\times)$ ring convolution.
|
{
"source": [
"https://mathoverflow.net/questions/10237",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2361/"
]
}
|
10,239 |
Is it true that, in the category of $\mathbb{Z}$-modules, $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}[x],\mathbb{Z})\cong\mathbb{Z}[[x]]$ and $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}[[x]],\mathbb{Z})\cong\mathbb{Z}[x]$? The first isomorphism is easy since any such homomorphism assigns an integer to $x^i, \forall\ {i>0}$ which defines a power series. For the second one might think similarly that if $S$ is the set of all power series with non-zero constant term then $\mathbb{Z}[[x]]=0\oplus{S}\oplus{x}S\oplus{x^2}S\dots$, but it doesn't quite work since it is not clear how to map $S$.
|
Yes: this is an old chestnut. Let me write $\oplus_n\mathbf{Z}$ for what you call $\mathbf{Z}[x]$ and $\prod_n\mathbf{Z}$ for what you call $\mathbf{Z}[[x]]$ (all products and sums being over the set {$0,1,2,\ldots$}). Clearly the homs from the product to $\mathbf{Z}$ contain the sum; the issue is checking that equality holds. So say I have $f:\prod_n\mathbf{Z}\to\mathbf{Z}$ and let me prove $f$ is in the sum. Let $e_n$ ($n\geq0$) be the $n$th basis element in the product (so, what you called $x^n$).
First I claim that $f(e_n)=0$ for all $n$ sufficiently large. Let's prove this by contradiction. If it were false then I would have infinitely many $n$ with $f(e_n)\not=0$, so by throwing away the $e_n$ such that $f(e_n)=0$ (this is just for simplicity of notation; otherwise I would have to let this infinite set of $n$ be called $n_0$, $n_1\ldots$ and introduce another subscript) we may as well assume that $f(e_n)=c_n\not=0$ for all $n=0,1,2,\ldots$. Now choose any old integers $d_i$ such that $\tau:=\sum_{i\geq0}2^id_ic_i$, a 2-adic integer, is not in $\mathbf{Z}$ (this can easily be done: infinitely many $d_i$ are "the last to change a binary digit of $\tau$" and hence one can recursively rule out all elements of $\mathbf{Z}$), and consider the integer $t:=f(\sum_{i\geq0}2^id_ie_i)\in\mathbf{Z}$. The point is that $\sum_{i\geq N}2^id_ie_i$ is a multiple of $2^N$ in the product, and hence its image under $f$ must be a multiple of $2^N$ in $\mathbf{Z}$. So one checks easily that $t-\tau$ is congruent to zero mod $2^N$ for all $N\geq1$ and hence $t=\tau$, a contradiction. [Remark: in my first "answer" to this question, I stopped here. Thanks to Qiaochu for pointing out that my answer wasn't yet complete.] We deduce that $f$ agrees with an element $P$ of the sum on the subgroup $\oplus_n\mathbf{Z}$ of $\prod_n\mathbf{Z}$. So now let's consider $f-P$; this is a map from the product to $\mathbf{Z}$ which is zero on the sum, and our job is to show that it is zero. So far I have used the fact that $\mathbf{Z}$ has one prime but now I need to use the fact that it has two. Firstly, any map $(\prod_n\mathbf{Z})/(\oplus_n\mathbf{Z})\to\mathbf{Z}$ is clearly going to vanish on the infinitely $p$-divisible elements of the left hand size for any prime $p$ (because there are no infinitely $p$-divisible elements of $\mathbf{Z}$ other than $0$). In particular it will vanish on elements of $\prod_n\mathbf{Z}$ of the form $(c_0,c_1,c_2,\ldots,c_n,\ldots)$ with the property that $c_n$ tends to zero $p$-adically. Call such a sequence a "$p$-adically convergent sequence". But using the Chinese Remainder Theorem it is trivial to check that every element of $\prod_n\mathbf{Z}$ is the difference of a 2-adically convergent sequence and a 3-adically convergent sequence, and so now we are done. Remark: I might be making a meal of this. My memory of what Kaplansky writes is that he uses the second half of my argument but does something a bit simpler for the first half.
|
{
"source": [
"https://mathoverflow.net/questions/10239",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2533/"
]
}
|
10,255 |
Eric Mazur has a wonderful video describing how physics is taught at many universities and his description applies word for word to the way I learned mathematics and the way it is still being taught, i.e. professors lecture to students and sketch some proofs. Suffice it to say I'm not a fan of the current methods and I don't think it would be too far from the truth to say that I do all the actual learning outside the classroom. Has anyone tried anything different and seen any difference in student understanding and comprehension in graduate or undergraduate courses? Some background motivation: I'm a TA and my current method of doing things is to just write some problems on the board and then go through their solutions. This is fine and it's what the students expect but sometimes I feel guilty because I'm just teaching them problem/solution patterns and reinforcing all the bad stereotypes about what mathematics is instead of showing them the underlying conceptual tapestry and helping them rethink their attitudes toward mathematics. It's kinda like the old saying “Give a man a fish; you have fed him for today. Teach a man to fish; and you have fed him for a lifetime”. So basically I throw a bunch of fish at the students hoping it will feed them for the semester.
|
The topic you touch upon is vast, but I wanted to comment on this phrase: "problem/solution patterns which is very different from showing them the underlying conceptual tapestry". If for some reason you have to use this format (department restrictions or whatnot) choosing your problems well will simultaneously introduce some of the conceptual tapestry. Rather than introducing a mathematical tool and then the problem that goes with it, you introduce the problem first (just out of range of the student ability) and bring it to the point where things get stuck, where something new is needed to go further. Then the motivation is clear for the new tool.
|
{
"source": [
"https://mathoverflow.net/questions/10255",
"https://mathoverflow.net",
"https://mathoverflow.net/users/-1/"
]
}
|
10,334 |
I am a non-mathematician. I'm reading up on set theory. It's fascinating, but I wonder if it's found any 'real-world' applications yet. For instance, in high school when we were learning the properties of i , a lot of the kids wondered what it was used for. The teacher responded that it was used to describe the properties of electricity in circuits. So is there a similar practical app of set theory? Something we wouldn't be able to do or build without set theory? Edit: Actually, I'm asking about the practicality of the knowledge of the properties of infinite sets, and their cardinality. I'm reading Peter Suber's [A Crash Course in the Mathematics Of Infinite Sets][1] ([Wayback Machine](https://web.archive.org/web/20110703003113/https://earlham.edu/~peters/writing/infapp.htm)). The properties of infinite sets seem unintuitive, but of course, the proofs show that they are true. My guess is that whoever came up with the square root of -1 did so many years before it 'escaped' from mathematics and found a practical use. Before then perhaps people thought it was clever, but not necessarily useful or even 'true'. So then, if you need to understand electricity, and you can do it best by using i , then even someone who thinks it's silly to have a square root of negative -1 would have to grudgingly admit that there's some 'reality' to it, despite its unintuitiveness, because electricity behaves as if it 'exists'. Seeing as how there was so much resistance to infinite sets at the beginning, even among mathematicians, I wonder: has the math of infinite sets been 'proven worthwhile' by having a practical application outside of mathematics, so that no one can say it's just some imaginative games?
|
The purpose of set theory is not practical application in the same way that, for example, Fourier analysis has practical applications. To most mathematicians (i.e. those who are not themselves set theorists), the value of set theory is not in any particular theorem but in the language it gives us. Nowadays even computer scientists describe their basic concept - Turing machines - in the language of set theory. This is useful because when you specify an object set-theoretically there is no question what you are talking about and you can unambiguously answer any questions you might have about it. Without precise definitions it is very difficult to do any serious mathematics. I guess another important point here is that it is hard to appreciate the role of set theory in mathematics without knowing some of the history behind the crisis of foundations in mathematics, but I don't know any particularly good references. Your second question is more specific, so I'll give a more specific answer: to thoroughly understand the mathematics behind, say, modern physics does in fact require (among many other things) that you understand the properties of infinite sets because topology has become an important part of this mathematics, and understanding general topology depends heavily on understanding properties of infinite sets. Whether this means that set theory has any bearing on "reality" depends on how much faith you have in topological spaces as a good model for the real world. As a specific example, the mathematics behind general relativity is called differential geometry . I think it's fair to say the development of general relativity would have been impossible without the mathematical language to express it. Differential geometry takes place on special kinds of manifolds, which are special kinds of topological spaces. So to understand differential geometry you need to understand at least some topology. And I don't think I need to justify the usefulness of general relativity!
|
{
"source": [
"https://mathoverflow.net/questions/10334",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2929/"
]
}
|
10,335 |
It seems to me that Hilbert modular varieties (forms) are generalization from Q to totally real fields. While Siegel modular varieties (forms) are generalization from 1 dimensional to higher dimensional abelian varieties. But they should both be some kind of Shimura variety (automorphic forms), right? According to Milne's note of Shimura varieties, Siegel modular varieties are Shimura varieties coming from the Shimura datum (G, X) where G is the symplectic similitude group of a symplectic space (V, \phi). So what is the corresponding Shimura datum for the Hilbert modular variety? Or am I asking a wrong question? Also, in the definition of Shimura varieties G(Q)\G(A_f)X/K, why we only consider Q and its adele group? why not general number fields and their adeles? (again, maybe a wrong question) Thank you.
|
The purpose of set theory is not practical application in the same way that, for example, Fourier analysis has practical applications. To most mathematicians (i.e. those who are not themselves set theorists), the value of set theory is not in any particular theorem but in the language it gives us. Nowadays even computer scientists describe their basic concept - Turing machines - in the language of set theory. This is useful because when you specify an object set-theoretically there is no question what you are talking about and you can unambiguously answer any questions you might have about it. Without precise definitions it is very difficult to do any serious mathematics. I guess another important point here is that it is hard to appreciate the role of set theory in mathematics without knowing some of the history behind the crisis of foundations in mathematics, but I don't know any particularly good references. Your second question is more specific, so I'll give a more specific answer: to thoroughly understand the mathematics behind, say, modern physics does in fact require (among many other things) that you understand the properties of infinite sets because topology has become an important part of this mathematics, and understanding general topology depends heavily on understanding properties of infinite sets. Whether this means that set theory has any bearing on "reality" depends on how much faith you have in topological spaces as a good model for the real world. As a specific example, the mathematics behind general relativity is called differential geometry . I think it's fair to say the development of general relativity would have been impossible without the mathematical language to express it. Differential geometry takes place on special kinds of manifolds, which are special kinds of topological spaces. So to understand differential geometry you need to understand at least some topology. And I don't think I need to justify the usefulness of general relativity!
|
{
"source": [
"https://mathoverflow.net/questions/10335",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1238/"
]
}
|
10,358 |
Just because a problem is NP-complete doesn't mean it can't be usually solved quickly. The best example of this is probably the traveling salesman problem, for which extraordinarily large instances have been optimally solved using advanced heuristics, for instance sophisticated variations of branch-and-bound . The size of problems that can be solved exactly by these heuristics is mind-blowing, in comparison to the size one would naively predict from the fact that the problem is NP. For instance, a tour of all 25000 cities in Sweden has been solved, as has a VLSI of 85900 points (see here for info on both). Now I have a few questions: 1) There special cases of reasonably small size where these heuristics either cannot find the optimal tour at all, or where they are extremely slow to do so? 2) In the average case (of uniformly distributed points, let's say), is it known whether the time to find the optimal tour using these heuristics is asymptotically exponential n, despite success in solving surprisingly large cases? Or is it asymptotically polynomial, or is such an analysis too difficult to perform? 3) Is it correct to say that the existence of an average-case polynomial, worst-case exponential time algorithm to solve NP problems has no importance for P=NP? 4) What can be said about the structure of problems that allow suprisingly large cases to be solved exactly through heuristic methods versus ones that don't?
|
This phenomenon extends beyond the traveling salesman problem, and even beyond NP, for there are even some undecidable problems with the feature that most instances can be solved very quickly. There is an emerging subfield of complexity theory called generic-case complexity , which is concerned with decision problems in the generic case, the problem of solving most or nearly all instances of a given problem. This contrasts with the situtation in the classical complexity theory, where it is in effect the worst-case complexity that drives many complexity classifications. (And even for approximate solutions in NP-hard problems, the worst-case phenomenon is still present.) Particularly interesting is the black-hole phenomenon, the phenomenon by which the difficulty of an infeasible or even undecidable problem is concentrated in a very tiny region, outside of which it is easy. (Here, tiny means tiny with respect to some natural measure, such as asymptotic density.) For example, many of the classical decision problems from combinatorial group theory, such as the word problem and the conjugacy problem are linear time solvable in the generic case. This phenomenon provides a negative answer to analogue of your question 1 for these problems. The fact that the problems are easily solved outside the black hole provides a negative answer to the analogue of question 2. And I think that the fact that these problems are actually undecidable as total problems suggests that this manner of solving almost all cases of a problem will not help us with P vs. NP, in your question 3. For question 4, let me mention that an extreme version of the black-hole phenomenon is provided even by the classical halting problem. Of course, this is the most famous of undecidable problems. Nevertheless, Alexei Miasnikov and I proved that for one of the standard Turing machine models with a one-way infinite tape, there is an algorithm that solves the halting problem on a set of asymptotic measure one. That is, there is a set A of Turing machine programs, such that (1) almost every program is in A, in the sense of asymptotic density, (2) A is linear time decidable, and (3) the halting problem is linear time decidable for programs in A. This result appears in (J. D. Hamkins and A. Miasnikov, The halting problem is decidable on a set of asymptotic probability one, Notre Dame J. Formal Logic 47, 2006. http://arxiv.org/abs/math/0504351 ). Inside the black hole, the complement of A, of course, the problem is intractible. The proof, unfortunately, does not fully generalize to all the other implementations of Turing machines, since for other models one finds a black hole of some measure intermediate between 0 and 1, rather than measure 0.
|
{
"source": [
"https://mathoverflow.net/questions/10358",
"https://mathoverflow.net",
"https://mathoverflow.net/users/942/"
]
}
|
10,409 |
Does anyone know of an introduction and motivation for W-algebras ? Edit: Okay, sorry I try to add some more background. W algebras occur, for example when you study nilpotent orbits: Take a nice algebraic/Lie group. It acts on its Lie-algebra by the adjoint action. Fix a nilpotent element e and make a sl_2 triple out of it. A W algebra is some modification of the universal enveloping, based on this data. A precise definition is for example given here: https://arxiv.org/pdf/0707.3108 . But this definition looks quite complicated and not very natural to me. I don't see what's going on. Therefor I wonder if there exists an easier introduction.
|
W-algebras appear in at least three interrelated contexts. Integrable hierarchies , as in the article by Leonid Dickey that mathphysicist mentions in his/her answer. Integrable PDEs like the KdV equation are bihamiltonian, meaning that the equations of motion can be written in hamiltonian form with respect to two different Poisson structures. One of the Poisson structures is constant, whereas the other (the so-called second Gelfand-Dickey bracket ) defines a so-called classical W-algebra . For the KdV equation it is the Virasoro Lie algebra, but for Boussinesq and higher-order reductions of the KP hierarchy one gets more complicated Poisson algebras. Drinfeld-Sokolov reduction , for which you might wish to take a look at the work of Edward Frenkel in the early 1990s. This gives a homological construction of the classical W-algebras starting from an affine Lie algebra and a nilpotent element. You can also construct so-called finite W-algebras in this way, by starting with a finite-dimensional simple Lie algebra and a nilpotent element. The original paper is this one by de Boer and Tjin . A lot of work is going on right on on finite W-algebras. You might wish to check out the work of Premet. Conformal field theory . This is perhaps the original context and certainly the one that gave them their name. This stems from this paper of Zamolodchikov . In this context, a W-algebra is a kind of vertex operator algebra: the vertex operator algebra generated by the Virasoro vector together with a finite number of primary fields. A review about this aspect of W-algebras can be found in this report by Bouwknegt and Schoutens . There is a lot of literature on W-algebras, of which I know the mathematical physics literature the best. They had their hey-day in Physics around the late 1980s and early 1990s, when they offered a hope to classify rational conformal field theories with arbitrary values of the central charge. The motivation there came from string theory where you would like to have a good understanding of conformal field theories of $c=15$ . The rational conformal field theories without extended symmetry only exist for $c<1$ , whence to overcome this bound one had to introduce extra fields (à la Zamolodchikov). Lots of work on W-algebras (in the sense of 3 ) happened during this time. The emergence of matrix models for string theory around 1989-90 (i.e., applications of random matrix theory to string theory) focussed attention on the integrable hierarchies, whose $\tau$ -functions are intimately related to the partition functions of the matrix model. This gave rise to lots of work on classical W-algebras (in the sense of 1 above) and also to the realisation that they could be constructed à la Drinfeld-Sokolov . The main questions which remained concerned the geometry of W-algebras, by which one means a geometric realisation of W-algebras analogous to the way the Virasoro algebra is (the universal central extension of) the Lie algebra of vector fields on the circle, and the representation theory. I suppose it's this latter question which motivates much of the present-day W-algebraic research in Algebra. Added In case you are wondering, the etymology is pretty prosaic. Zamolodchikov's first example was an operator vertex algebra generated by the Virasoro vector and a primary weight field $W$ of weight 3. People started referring to this as Zamolodchikov's $W_3$ algebra and the rest, as they say, is history. Added later Ben's answer motivates the study of finite W-algebras from geometric representation theory and points out that a finite W-algebra can be viewed as the quantisation of a particular Poisson reduction of the dual of the Lie algebra with the standard Kirillov Poisson structure. The construction I mentioned above is in some sense doing this in the opposite order: you first quantise the Kirillov Poisson structure and then you take BRST cohomology , which is the quantum analogue of Poisson reduction.
|
{
"source": [
"https://mathoverflow.net/questions/10409",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2837/"
]
}
|
10,413 |
Denote Zermelo Fraenkel set theory without choice by ZF. Is the following true: In ZF, every definable non empty class A has a definable member; i.e. for every class $A = \lbrace x : \phi(x)\rbrace$ for which ZF proves "A is non empty", there is a class $a = \lbrace x : \psi(x)\rbrace$ such that ZF proves "a belongs to A"?
|
Update. (June, 2017) François Dorais and I have completed a paper that ultimately grew out of this questions and several follow-up questions. F. G. Dorais and J. D. Hamkins, When does every definable nonempty set have a definable element? ( arχiv:1706.07285 ) Abstract. The assertion that every definable set has a definable element is equivalent over ZF to the principle $V=\newcommand\HOD{\text{HOD}}\HOD$, and indeed, we prove, so is the assertion merely that every $\Pi_2$-definable set has an ordinal-definable element. Meanwhile, every model of ZFC has a forcing extension satisfying $V\neq\HOD$ in which every $\Sigma_2$-definable set has an ordinal-definable element. Similar results hold for $\HOD(\mathbb{R})$ and $\HOD(\text{Ord}^\omega)$ and other natural instances of $\HOD(X)$. Read more at the blog post . Click on the history to see the original answer. Related questions and answers appear at: Can $V\neq\HOD$, if every $\Sigma_2$-definable set has an ordinal-definable element? Is it consistent with ZFC (or ZF) that every definable family of sets has at least one definable member? We also make use of the $\Sigma_2$ conception of Local properties in set theory .
|
{
"source": [
"https://mathoverflow.net/questions/10413",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2689/"
]
}
|
10,419 |
How does a depressed graduate student go about recovering his enthusiasm for the subject and the question at hand? Edit: I am not that grad student; it is a very talented friend of mine. Moderator's update: The discussion about this question should happen at this link on meta . This page is reserved for answers to the question as stated.
|
google suggests that to guard against burnout: Be sure to get regular exercise Eat well. Maintain hobbies and interests
outside of work Take time for social activities
|
{
"source": [
"https://mathoverflow.net/questions/10419",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2938/"
]
}
|
10,481 |
I hope someone can point me to a quick definition of the following terminology. I keep coming across wild and tame in the context of classification problems, often adorned with quotes, leading me to believe that the terms are perhaps not being used in a formal sense. Yet I am sure that there is some formal definition. For example, the classification problem for nilpotent Lie algebras is said to be wild in dimension $\geq 7$. All that happens is that in dimension $7$ and above there are moduli. In what sense is this wild? Thanks in advance!
|
I am not an expert but in the algebra and representation theory the apparently standard definition is as follows (see also Drozd - Tame and wild matrix problems and Belitskii and Sergeichuk - Complexity of matrix problems ): a problem is wild if it contains a subproblem which is equivalent to the problem of simultaneously reducing to canonical form two linear operators on a finite-dimensional space.
|
{
"source": [
"https://mathoverflow.net/questions/10481",
"https://mathoverflow.net",
"https://mathoverflow.net/users/394/"
]
}
|
10,512 |
[I have rewritten this post in a way which I hope will remain faithful to the questioner and make it seem more acceptable to the community. I have also voted to reopen it. -- PLC] There are many ways to approach noncommutative geometry. What are some of the most important currently known approaches? Who are the principal creators of each of these approaches, and where are they coming from? E.g., what more established mathematical fields are they using as jumping off points? What problems are they trying to solve? Two examples: 1) The Connes school, with an approach from C$^*$-algebras/mathematical physics. 2) The Kontsevich school, with an approach from algebraic geometry.
|
In accordance with the suggestion of Yemon Choi, I am going to suggest some further delineation of the approaches to "Non-commutative Algebraic Geometry". I know very little about "Non-commutative Differential Geometry", or what often falls under the heading "à la Connes". This will be completely underrepresented in this summary. For that I trust Yemon's summary to be satisfactory. ( edit by YC : BB is kind to say this, but my attempted summary is woefully incomplete and may be inaccurate in details; I would encourage anyone reading to investigate further, keeping in mind that the NCG philosophy and toolkit in analysis did not originate and does not end with Connes.) Also note that much of what I know about these approaches comes from two sources: The paper by Mahanta My advisor A. Rosenberg. Additionally, much useful discussion took place at Kevin Lin's question (as Ilya stated in his answer). I think a better break down for the NCAG side would be: A. Rosenberg/Gabriel/Kontsevich approach Following the philosophy of Grothendieck: "to do geometry, one needs only the category of quasi-coherent sheaves on the would-be space" ( edit by KL : Where does this quote come from?) In the famous dissertation of Gabriel , he introduced the injective spectrum of an abelian category, and then reconstructed the commutative noetherian scheme, which is a starting point of noncommutative algebraic geometry. Later, A. Rosenberg introduced the left spectrum of a noncommutative ring as an analogue of the prime spectrum in commutative algebraic geometry, and generalized it to any abelian category. He used one of the spectra to reconstruct any quasi-separated (not necessarily quasi-compact), commutative scheme. (Gabriel-Rosenberg reconstruction theorem.) In addition, Rosenberg has described the NC-localization (first observed also by Gabriel) which has been used by him and Kontsevich to build NC analogs of more classical spaces (like the NC Grassmannian) and more generally, noncommutative stacks. Rosenberg has also developed the homological algebra associated to these 'spaces'. Applications of this approach include representation theory (D-module theory in particular), quantum algebra, and physics. References in this area are best found through the MPIM Preprint Series, and a large collection is linked here . Additionally, a book is being written by Rosenberg and Kontsevich furthering the work of their previous paper . Some applications of these methods are used here , here , here , and here . The first two are focusing on representation theory, the second two on non-commutative localization. Kontsevich/Soibelman approach They might refer to their approach as "formal deformation theory", and quoting directly from their book The subject of deformation theory can be defined as the "study of moduli spaces of structures...The subject of this book is formal deformation theory. This means $\mathcal{M}$ will be a formal space(e.g. a formal scheme), and a typical category $\mathcal{W}$ will be the category of affine schemes..." Their approach is related to $A_{\infty}$ algebras and homological mirror symmetry. References that might help are the papers of Soibelman . Also, I think this is related to the question here . (Note: I know hardly anything beyond that this approach exists. If you know more, feel free to edit this answer! Thanks for your understanding!) ( Some comments by KL : I am not sure whether it is appropriate to include Kontsevich-Soibelman's deformation theory here. This kind of deformation theory is a very general thing, which intersects some of the "noncommutative algebraic geometry" described here, but I think that it is neither a subset nor a superset thereof. In any case, I've asked some questions related to this on MO in the past, see [this][22] and [this][23]. However, there is the approach of noncommutative geometry via categories, as elucidated in, for instance, [Katzarkov-Kontsevich-Pantev][24]. Here the idea is to think of a category as a category of sheaves on a (hypothetical) non-commutative space. The basic "non-commutative spaces" that we should have in mind are the "Spec" of a (not necessarily commutative) associative algebra, or dg associative algebra, or A-infinity algebra. Such a "space" is an "affine non-commutative scheme". The appropriate category is then the category of modules over such an algebra. Definitively commutative spaces, for instance quasi-projective schemes, are affine non-commutative schemes in this sense: It is a theorem of van den Bergh and Bondal that the derived category of quasicoherent sheaves on a quasi-projective scheme is equivalent to a category of modules over a dg algebra. (I should note that in my world everything is over the complex field; I have no idea what happens over more general fields.) Lots of other categories are or should be affine non-commutative in this sense: [Matrix factorization categories][25] (see in particular [Dyckerhoff][26]), and probably various kinds of Fukaya categories are conjectured to be so as well. Anyway I have no idea how this kind of "noncommutative algebraic geometry" interacts with the other kinds explained here, and would really like to hear about it if anybody knows.) Lieven Le Bruyn's approach As I know nearly nothing about this approach and the author is a visitor to this site himself, I wouldn't dare attempt to summarize this work. As mentioned in a comment, his website contains a plethora of links related to non-commutative geometry. I recommend you check it out yourself . Approach of Artin, Van den Berg school Artin and Schelter gave a regularity condition on algebras to serve as the algebras of functions on non-commutative schemes. They arise from abstract triples which are understood for commutative algebraic geometry. (Again edits are welcome!) Here is a nice report on Interactions between noncommutative algebra and algebraic geometry . There are several people who are very active in this field: Michel Van den Berg, James Zhang, Paul Smith, Toby Stafford, I. Gordon, A. Yekutieli. There is also a very nice page of Paul Smith: noncommutative geometry and noncommutative algebra , where you can find almost all the people who are currently working in the noncommutative world. References: [This][16] paper introduced the need for the regularity condition and showed the usefulness. Again I defer to [Mahanta][17] for details.
Serre's FAC is the starting point of noncommutative projective geometry. But the real framework is built by Artin and James Zhang in their famous paper [Noncommutative Projective scheme][18]. Non-commutative Deformation Theory by Laudal Olav Laudal has approached NCAG using NC-deformation theory. He also applies his method to invariant theory and moduli theory. (Please edit!) References are on his page [here][19] and [this][20] paper seems to be a introductory article. Apologies Without a doubt, I have made several errors, given bias, offended the authors, and embarrassed myself in this post. Please don't hold this against me, just edit/comment on this post until it is satisfactory. As it was said before, the [nlab][21] article on noncommutative geometry is great, you should defer to it rather than this post. Thanks! [16]: https://books.google.com/books?hl=en&lr=&id=_BnSoQSKnNUC&oi=fnd&pg=PA33&dq=%252522Artin%252522+%252522Some+algebras+associated+to+automorphisms+of+elliptic+curves%252522+&ots=hRXnP7udMW&sig=t77CnWnsYPHhuonQQffrSXedyj0#v=onepage&q="Artin" "Some algebras associated to automorphisms of elliptic curves"&f=false
[17]: https://arxiv.org/abs/math/0501166 [18]: https://web.archive.org/web/20121023193142/http://www.ingentaconnect.com/content/ap/ai/1994/00000109/00000002/art01087 [19]: https://web.archive.org/web/20181103123848/http://folk.uio.no:80/arnfinnl/ [20]: https://web.archive.org/web/20080425144650/http://folk.uio.no/arnfinnl/Noncom.alg.geom.pdf [21]: https://ncatlab.org/nlab/show/noncommutative%20geometry [22]: What is a deformation of a category? [23]: Deformation theory and differential graded Lie algebras [24]: https://arxiv.org/abs/0806.0107 [25]: Matrix factorizations and physics [26]: https://arxiv.org/abs/0904.4713
|
{
"source": [
"https://mathoverflow.net/questions/10512",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2938/"
]
}
|
10,535 |
This seems to be a favorite question everywhere, including Princeton quals. How many ways are there? Please give a new way in each answer, and if possible give reference. I start by giving two: Ahlfors, Complex Analysis, using Liouville's theorem. Courant and Robbins, What is Mathematics?, using elementary topological considerations. I won't be choosing a best answer, because that is not the point.
|
Here is a standard algebraic proof. It suffices to show that if $L/\mathbb{C}$ is a finite extension, then $L=\mathbb{C}$. By passing to a normal closure we assume that $L/\mathbb{R}$ is Galois with Galois group $G$. Let $H$ be the Sylow-2 subgroup of $G$ and $M=L^H$. By the Fundamental Theorem of Galois Theory, $M/\mathbb{R}$ has odd degree. Let $\alpha\in M$ and $f(x)$ be its minimal polynomial. Then $f(x)$ has odd degree and by the Intermediate Value Theorem, a real root. As $f(x)$ is irreducible, it must have degree one. Then $\alpha\in\mathbb{R}$ and $M=\mathbb{R}$. So $G=H$ is a 2-group. Then $G_1:=Gal(L/\mathbb{C})$ is a 2-group as well. Assuming that $G_1$ is not trivial, there must exist a degree 2 subextension $K$ of $\mathbb{C}$. But every quadratic complex polynomial has a root (by the quadratic formula), so we have a contradiction.
|
{
"source": [
"https://mathoverflow.net/questions/10535",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2938/"
]
}
|
10,574 |
One way to define the algebra of differential forms $\Omega(M)$ on a smooth manifold $M$ (as explained by John Baez's week287 ) is as the exterior algebra of the dual of the module of derivations on the algebra $C^{\infty}(M)$ of smooth functions $M \to \mathbb{R}$. Given that derivations are vector fields, 1-forms send vector fields to smooth functions, and some handwaving about area elements suggests that k-forms should be built from 1-forms in an anticommutative fashion, I am almost willing to accept this definition as properly motivated. One can now define the exterior derivative $d : \Omega(M) \to \Omega(M)$ by defining $d(f dg_1\ \dots\ dg_k) = df\ dg_1\ \dots\ dg_k$ and extending by linearity. I am almost willing to accept this definition as properly motivated as well. Now, the exterior derivative (together with the Hodge star and some fiddling) generalizes the three main operators of multivariable calculus: the divergence, the gradient, and the curl. My intuition about the definitions and properties of these operators comes mostly from basic E&M, and when I think about the special cases of Stokes' theorem for div, grad, and curl, I think about the "physicist's proofs." What I'm not sure how to do, though, is to relate this down-to-earth context with the high-concept algebraic context described above. Question: How do I see conceptually that differential forms and the exterior derivative, as defined above, naturally have physical interpretations generalizing the "naive" physical interpretations of the divergence, the gradient, and the curl? (By "conceptually" I mean that it is very unsatisfying just to write down the definitions and compute.) And how do I gain physical intuition for the generalized Stokes' theorem? (An answer in the form of a textbook that pays special attention to the relationship between the abstract stuff and the physical intuition would be fantastic.)
|
Here's a sketch of the relation between div-grad-curl and the de Rham complex, in case you might find it useful. The first thing to realise is that the div-grad-curl story is inextricably linked to calculus in a three-dimensional euclidean space . This is not surprising if you consider that this stuff used to go by the name of "vector calculus" at a time when a physicist's definition of a vector was "a quantity with both magnitude and direction". Hence the inner product is essential part of the baggage as is the three-dimensionality (in the guise of the cross product of vectors). In three-dimensional euclidean space you have the inner product and the cross product and this allows you to write the de Rham sequence in terms of div, grad and curl as follows:
$$ \matrix{ \Omega^0 & \stackrel{d}{\longrightarrow} & \Omega^1 & \stackrel{d}{\longrightarrow} & \Omega^2 & \stackrel{d}{\longrightarrow} & \Omega^3 \cr
\uparrow & & \uparrow & & \uparrow & & \uparrow \cr
\Omega^0 & \stackrel{\mathrm{grad}}{\longrightarrow} & \mathcal{X} & \stackrel{\mathrm{curl}}{\longrightarrow} & \mathcal{X} & \stackrel{\mathrm{div}}{\longrightarrow} & \Omega^0 \cr}$$
where $\mathcal{X}$ stands for vector fields and the vertical maps are, from left to right, the following isomorphisms: the identity: $f \mapsto f$ the musical isomorphism $X \mapsto \langle X, -\rangle$ $X \mapsto \omega$, where $\omega(Y,Z) = \langle X, Y \times Z \rangle$ $f \mapsto f \mathrm{dvol}$, where $\mathrm{dvol}(X,Y,Z) = \langle X, Y \times Z\rangle$ up to perhaps a sign here and there that I'm too lazy to chase. The beauty of this is that, first of all, the two vector calculus identities $\mathrm{div} \circ \mathrm{curl} = 0$ and $\mathrm{curl} \circ \mathrm{grad} = 0$ are now subsumed simply in $d^2 = 0$, and that whereas div, grad, curl are trapped in three-dimensional euclidean space, the de Rham complex exists in any differentiable manifold without any extra structure. We teach the language of differential forms to our undergraduates in Edinburgh in their third year and this is one way to motivate it. As for the integral theorems, I always found Spivak's Calculus on manifolds to be a pretty good book. Another answer mentioned Gravitation by Misner, Thorne and Wheeler. Personally I found their treatment of differential forms very confusing when I was a student. I'm happier with the idea of a dual vector space than I am with the "milk crates" they draw to illustrate differential forms. Wald's book on General Relativity had, to my mind, a much nicer treatment of this subject.
|
{
"source": [
"https://mathoverflow.net/questions/10574",
"https://mathoverflow.net",
"https://mathoverflow.net/users/290/"
]
}
|
10,630 |
Suppose for some reason one would be expecting a formula of the kind $$\mathop{\text{ch}}(f_!\mathcal F)\ =\ f_*(\mathop{\text{ch}}(\mathcal F)\cdot t_f)$$ valid in $H^*(Y)$ where $f:X\to Y$ is a proper morphism with $X$ and $Y$ smooth and quasiprojective, $\mathcal F\in D^b(X)$ is a bounded complex of coherent sheaves on $X$, $f_!: D^b(X)\to D^b(Y)$ is the derived pushforward, $\text{ch}:D^b(-)\to H^*(-)$ denotes the Chern character, and $t_f$ is some cohomology class that depends only on $f$ but not $\mathcal F$. According to the Grothendieck–Hirzebruch–Riemann–Roch theorem (did I get it right?) this formula is true with $t_f$ being the relative Todd class of $f$, defined as the Todd class of relative tangent bundle $T_f$. So, let's play at "guessing" the $t_f$ pretending we didn't know GHRR ($t_f$ are not uniquely defined , so add conditions on $t_f$ in necessary). Question. Expecting the formula of the above kind, how to find out that $t_f = \text{td}\, T_f$? You don't have to show this choice works (that is, prove GHRR), but you have to show no other choice works.
Also, let's not use Hirzebruch–Riemann–Roch : I'm curious exactly how and where Todd classes will appear.
|
You look at the case when $X=D$ is a Cartier divisor on $Y$ (so that the relative tangent bundle -- as an element of the K-group -- is the normal bundle $\mathcal N_{D/X}=\mathcal O_D(D)$ (conveniently a line bundle, so is its own Chern root), and $\mathcal F=\mathcal O_D$. And the Todd class pops out right away. Indeed from the exact sequence $0\to \mathcal O_Y(-D) \to \mathcal O_Y\to \mathcal O_D\to 0$, you get that
$$ch(f_! \mathcal O_D)=ch(O_Y(D))=
ch(\mathcal O_Y) - ch(\mathcal O_Y(-D)) = 1- e^{-D}.$$
And you need to compare this with the pushforward of $[D]$ in the Chow group, which is $D$. The ratio
$$ \frac{D}{1-e^{-D}} = Td( \mathcal O(D) )$$
is what you are after. Now you have just discovered the Todd class. I suspect that this is how Grothendieck discovered his formula, too -- after seeing that this case fits with Hirzebruch's formula, that the same Todd class appears in both cases.
|
{
"source": [
"https://mathoverflow.net/questions/10630",
"https://mathoverflow.net",
"https://mathoverflow.net/users/65/"
]
}
|
10,631 |
I had this simple question when formulating the Todd class question . Does there exist an example of proper morphism $f:X\to Y$ together with nontrivial homology class $t\in H^*(X)$ such that for all coherent sheaves on $X$ the equality $f_*(\mathop{\text{ch}}(u)\cdot t) = 0$ holds?
|
You look at the case when $X=D$ is a Cartier divisor on $Y$ (so that the relative tangent bundle -- as an element of the K-group -- is the normal bundle $\mathcal N_{D/X}=\mathcal O_D(D)$ (conveniently a line bundle, so is its own Chern root), and $\mathcal F=\mathcal O_D$. And the Todd class pops out right away. Indeed from the exact sequence $0\to \mathcal O_Y(-D) \to \mathcal O_Y\to \mathcal O_D\to 0$, you get that
$$ch(f_! \mathcal O_D)=ch(O_Y(D))=
ch(\mathcal O_Y) - ch(\mathcal O_Y(-D)) = 1- e^{-D}.$$
And you need to compare this with the pushforward of $[D]$ in the Chow group, which is $D$. The ratio
$$ \frac{D}{1-e^{-D}} = Td( \mathcal O(D) )$$
is what you are after. Now you have just discovered the Todd class. I suspect that this is how Grothendieck discovered his formula, too -- after seeing that this case fits with Hirzebruch's formula, that the same Todd class appears in both cases.
|
{
"source": [
"https://mathoverflow.net/questions/10631",
"https://mathoverflow.net",
"https://mathoverflow.net/users/65/"
]
}
|
10,635 |
I remember one of my professors mentioning this fact during a class I took a while back, but when I searched my notes (and my textbook) I couldn't find any mention of it, let alone the proof. My best guess is that it has something to do with Galois theory, since it's enough to prove that the characters are rational - maybe we have to find some way to have the symmetric group act on the Galois group of a representation or something. It would be nice if an idea along these lines worked, because then we could probably generalize to draw conclusions about the field generated by the characters of any group. Is this the case?
|
If $g$ is an element of order $m$ in a group $G$, and $V$ a complex representation of $G$, then $\chi_V(g)$ lies in $F=\mathbb{Q}(\zeta_m)$. Since the Galois group of $F/\mathbb{Q}$ is $(\mathbb{Z}/m)^\times$, for any $k$ relatively prime to $m$ the elements $\chi_V(g)$ and $\chi_V(g^k)$ differ by the action of the appropriate element of the Galois group. If $G$ is a symmetric group and $g$ an element as above, then $g$ and $g^k$ are conjugate: they have the same cycle decomposition. So $\chi_V(g)=\chi_V(g^k)$ whenever $(k,m)=1$, and thus $\chi_V(g)\in \mathbb{Q}$. Now, because $\chi_V(g)$ is an algebraic integer (true for every finite group, every complex character) and a rational number, it is a rational integer, that is, an integer: $\chi_V(g)\in\mathbb{Z}$.
|
{
"source": [
"https://mathoverflow.net/questions/10635",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2363/"
]
}
|
10,667 |
Let $Df$ denote the derivative of a function $f(x)$ and $\bigtriangledown f=f(x)-f(x-1)$ be the discrete derivative. Using the Taylor series expansion for $f(x-1)$, we easily get $\bigtriangledown = 1- e^{-D}$ or, by taking the inverses,
$$ \frac{1}{\bigtriangledown} = \frac{1}{1-e^{-D}} =
\frac{1}{D}\cdot \frac{D}{1-e^{-D}}=
\frac{1}{D} + \frac12+
\sum_{k=1}^{\infty} B_{2k}\frac{D^{2k-1}}{(2k)!}
,$$
where $B_{2k}$ are Bernoulli numbers. ( Edit: I corrected the signs to adhere to the most common conventions.) Here, $(1/D)g$ is the opposite to the derivative, i.e. the integral; adding the limits this becomes a definite integral $\int_0^n g(x)dx$. And $(1/\bigtriangledown)g$ is the opposite to the discrete derivative, i.e. the sum $\sum_{x=1}^n g(x)$. So the above formula, known as Euler-Maclaurin formula, allows one, sometimes, to compute the discrete sum by using the definite integral and some error terms. Usually, there is a nontrivial remainder in this formula. For example, for $g(x)=1/x$, the remainder is Euler's constant $\gamma\simeq 0.57$. Estimating the remainder and analyzing the convergence of the power series is a long story, which is explained for example in the nice book "Concrete Mathematics" by Graham-Knuth-Patashnik. But the power series becomes finite with zero remainder if $g(x)$ is a polynomial. OK, so far I am just reminding elementary combinatorics. Now, for my question. In the (Hirzebruch/Grothendieck)-Riemann-Roch formula one of the main ingredients is the Todd class which is defined as the product, going over Chern roots $\alpha$, of the expression $\frac{\alpha}{1-e^{-\alpha}}$. This looks so similar to the above, and so suggestive (especially because in the Hirzebruch's version
$$\chi(X,F) = h^0(F)-h^1(F)+\dots = \int_X ch(F) Td(T_X)$$
there is also an "integral", at least in the notation) that it makes me wonder: is there a connection? The obvious case to try (which I did) is the case when $X=\mathbb P^n$ and $F=\mathcal O(d)$. But the usual proof in that case is a residue computation which, to my eye, does not look anything like Euler-Maclaurin formula. But is there really a connection? An edit after many answers: Although the connection with Khovanskii-Pukhlikov's paper and the consequent work, pointed out by Dmitri and others, is undeniable, it is still not obvious to me how the usual Riemann-Roch for $X=\mathbb P^n$ and $F=\mathcal O(d)$ follows from them. It appears that one has to prove the following nontrivial Identity: The coefficient of $x^n$ in $Td(x)^{n+1}e^{dx}$ equals
$$\frac{1}{n!}
Td(\partial /\partial h_0) \dots Td(\partial /\partial h_n)
(d+h_0+\dots + h_n)^n |_{h_0=\dots h_n=0}$$ A complete answer to my question would include a proof of this identity or a reference to where this is shown. (I did not find it in the cited papers.) I removed the acceptance to encourage a more complete explanation.
|
As far as I understand this connection was observed (and generalised) by Khovanskii and Puhlikov in the article A. G. Khovanskii and A. V. Pukhlikov, A Riemann-Roch theorem for integrals and sums of quasipolynomials over
virtual polytopes, Algebra and Analysis 4 (1992), 188–216, translation in St. Petersburg Math. J. (1993), no. 4, 789–812. This is related to toric geometry, for which some really well written introduction articles are contained on the page of David Cox http://www3.amherst.edu/~dacox/ Since 1992 many people wrote on this subject, for example EXACT EULER MACLAURIN FORMULAS FOR SIMPLE LATTICE POLYTOPES http://arxiv.org/PS_cache/math/pdf/0507/0507572v2.pdf Or Riemann sums over polytopes http://arxiv.org/PS_cache/math/pdf/0608/0608171v1.pdf
|
{
"source": [
"https://mathoverflow.net/questions/10667",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1784/"
]
}
|
10,842 |
The Hawaiian Earring is usually constructed as the union of circles of radius 1/n centered at (0,1/n): $\bigcup_1^\infty \left[ (0, \frac{1}{n}) + \frac{1}{n}S^1 \right]$. However, nothing stops us from using the sequence of radii $1/n^2$ or any other sequence of numbers $a_n$. I will call a Hawaiian Earring for a sequence (of distinct real numbers), $A = \lbrace a_n \rbrace$, the union of circles of radius $a_n$ centered at $(0, a_n )$. Let the union inherit its topological structure from $\mathbb{R}^2$. Are all of these spaces homeomorphic? If $a_n$ is a monotone decreasing sequence converging to $0$, is its Hawaiian Earring homeomorphic to that of the sequence $\lbrace 1/n \rbrace$?
|
The Hawaian earring is the one-point compactification of a countable union of open intervals (with the coproduct or disjoint sum topology). This description is independent of the radii used to construct it. A beautiful reference about this space is [Cannon, J. W.; Conner, G. R. The combinatorial structure of the Hawaiian earring group. Topology Appl. 106 (2000), no. 3, 225--271 MR1775709 ) If I recall correctly, they prove my claim there.
|
{
"source": [
"https://mathoverflow.net/questions/10842",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1358/"
]
}
|
10,860 |
Motivation I learned about this question from a wonderful article Rational points on curves by Henri Darmon. He gives a list of statements (some are theorems, some conjectures) of the form the set $\{$ objects $\dots$ over field $K$ with good reduction everywhere except set $S$ $\}$ is finite/empty One interesting thing he mentions is about abelian schemes in the most natural case $K = \mathbb Q$ , $S$ empty. I think according to the definition we have a trivial example of relative dimension 0. Question Why is the set of non-trivial abelian schemes over $\mathop{\text{Spec}}\mathbb Z$ empty? Reference This is proven in Il n'y a pas de variété abélienne sur Z by Fontaine, but I'm asking because: (1) Springer requires subscription, (2) there could be new ideas after 25 years, (3) the text is French and could be hard to read (4) this knowledge is worth disseminating.
|
It's a result related in spirit to Minkowski's theorem that $\mathbb Q$ admits no non-trivial unramified extensions. If $A$ is an abelian variety over $\mathbb Q$ with everywhere good reduction, then for any integer $n$ the $n$-torsion scheme $A[n]$ is a finite flat group scheme over $\mathbb Z$. Although this group scheme will be ramified at primes $p$ dividing $n$, Fontaine's theory shows that the ramification is of a rather mild type: so mild, that
a non-trivial such family of $A[n]$ can't exist. In the last 25 years, there has been much research on related questions, including by
Brumer--Kramer, Schoof, and F. Calegari, among others. (One particularly interesting recent variation is a joint paper of F. Calegari and Dunfield in which they use related ideas to construct a tower of closed hyperbolic 3-manifolds that are rational homology spheres, but whose injectivity radii grow without bound.) EDIT: I should add that the case of elliptic curves is older, due to Tate I believe,
and uses a different argument: he considers the equation computing the discriminant of
a cubic polynomial $f(x)$ (corresponding to the elliptic curve $y^2 = f(x)$) and shows
that this solution equation has no integral solutions giving a discriminant of $\pm 1$. This direction of argument generalizes in different ways, but is related to a result of Shafarevic (I think) proving that there are only finitely many elliptic curves with good reduction outside a finite set of primes. (A result which was generalized by Faltings to abelian varieties as part of his proof of Mordell's conjecture.) Finally, one could add that in Faltings's argument, he also relied crucially on ramification results for $p$-divisible groups, due also to Tate, I think, results which Fontaine's theory generalizes. So one sees that the study of ramification of finite flat groups schemes and $p$-divisible groups (and more generally Fontaine's $p$-adic Hodge theory) plays a crucial role in these sorts of Diophantine questions. A colleague describes it as the ``black magic'' that makes all Diophantine arguments (including Wiles' proof of FLT as well) work. P.S. It might be useful to give a toy illustrative example of how finite flat group schemes give rise to mildly ramified extensions: consider all the quadratic extensions of $\mathbb Q$ ramified only at $2$: they are ${\mathbb Q}(\sqrt{-1}),$ ${\mathbb Q}(\sqrt{2})$, and ${\mathbb Q}(\sqrt{-2})$, with discriminants $-4$, $8$, and $-8$ respectively. Thus ${\mathbb Q}(\sqrt{-1})$ is the least ramified, and not coincidentally, it is the splitting field of the finite flat group scheme $\mu_4$ of 4th roots of unity.
|
{
"source": [
"https://mathoverflow.net/questions/10860",
"https://mathoverflow.net",
"https://mathoverflow.net/users/65/"
]
}
|
10,870 |
My question is about the concept of nonstandard metric space that would arise from a use of the nonstandard reals R* in place of the usual R-valued metric. That is, let us define that a topological space X is a nonstandard metric space , if there is a distance function, not into the reals R, but into some nonstandard R* in the sense of nonstandard analysis . That is, there should be a distance function d from X 2 into R*, such that d(x,y)=0 iff x=y, d(x,y)=d(y,x)
and d(x,z) <= d(x,y)+d(y,z). Such a nonstandard metric would give rise to the nonstandard open balls, which would generate a metric-like
topology on X. There are numerous examples of such spaces, beginning with R* itself. Indeed, every metric space Y has a nonstandard analogue Y*, which is a
nonstandard metric space. In addition, there are nonstandard metric spaces that do not arise as Y* for any metric space Y. Most of these examples
will not be metrizable, since we may assume that R* has uncountable cofinality (every countable set is bounded), and this will usually prevent
the existence of a countable local basis. That is, the nested sequence of balls around a given point will include the balls of infinitesimal
radius, and the intersection of any countably many will still be bounded away from 0. For example, R* itself will not be metrizable. The space R* is not connected, since it is the union of the infinitesimal neighborhoods of each point. In fact, one can show it is totally disconnected. Nevertheless, it appears to me that these nonstandard metric spaces are as useful in many ways as their standard counterparts. After all, one can still
reason about open balls, with the triangle inequality and whatnot. It's just that the distances might be nonstandard. What's more, the
nonstandard reals offer some opportunities for new topological constructions: in a nonstandard metric space, one has the standard-part operation,
which would be a kind of open-closure of a set---For any set Y, let Y+ be all points infinitesimally close to a point in Y. This is something
like the closure of Y, except that Y+ is open! But we have Y subset Y+, and Y++=Y+, and + respects unions, etc. In classical topology, we have various metrization theorems , such as Urysohn's theorem that any second-countable regular Hausdorff space is metrizable. Question. Which topological spaces admit a nonstandard metric? Do any of the classical metrization theorems admit an analogue for nonstandard metrizability? How can we tell if a given topological space admits a nonstandard metric? I would also be keen to learn of other interesting aspects of nonstandard metrizability, or to hear of interesting applications. I have many related questions, such as when is a nonstandard metric space actually metrizable? Is there any version of R* itself which is metrizable? (e.g. dropping the uncountable cofinality hypothesis)
|
The uniformity defined by a *R-valued metric is of a special kind. Let $(n_i)_{i<\kappa}$ be a cofinal sequence of positive elements in *R. We may assume that $i < j$ implies that $n_i/n_j$ is infinitesimal. Given a *R-valued metric space $(X,d)$ we can define a family $(d_i)_{i<\kappa}$ of pseudometrics by $d_i(x,y) = st(b(n_id(x,y)))$, where $st$ is the standard part function and $b(z) = z/(1+z)$ to make the metrics bounded by $1$. The topology on $X$ is defined by the family of pseudometrics $(d_i)_{i<\kappa}$. Note that these pseudometrics have a the following special property: (+) If $i < j < \kappa$ and $d_j(x,y) < 1$ then $d_i(x,y) = 0$. Every uniform space can be defined by a family of pseudometrics, but it is rather unusual for the family to have property (+) when $\kappa > \omega$. On the other hand, given a family of pseudometrics $(d_i)_{i<\kappa}$ bounded by $1$ with property (+), then we can recover a *R-valued metric by defining $d(x,y) = d_i(x,y)/n_i$, where $i$ is minimal such that $d_i(x,y) > 0$.
|
{
"source": [
"https://mathoverflow.net/questions/10870",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1946/"
]
}
|
10,879 |
I'm beginning to learn cohomology for cyclic groups in preparation for use in the proofs of global class field theory (using ideal-theoretic arguments). I've seen the proof of the long exact sequence and of basic properties of the Herbrand quotient, and I've started to look through how these are used in the proofs of class field theory. So far, all I can tell is that the cohomology groups are given by some ad hoc modding out process, then we derive some random properties (like the long exact sequence), and then we compute things like $H^2(\mathrm{Gal}(L/K),I_{L})$, where $I_L$ denotes the group of fractional ideals of a number field $L$, and it just happens to be something interesting for the study of class field theory such as $I_K/\mathrm{N}(I_L)$, where $L/K$ is cyclic and $\mathrm{N}$ denotes the ideal norm. We then find that the cohomology groups are useful for streamlining the computations with various orders of indexes of groups. What I don't get is what the intuition is behind the definitions of these cohomology groups. I do know what cohomology is in a geometric setting (so I know examples where taking the kernel modulo the image is interesting), but I don't know why we take these particular kernels modulo these particular images. What is the intuition for why they are defined the way they are? Why should we expect that these cohomology groups so-defined have nice properties and help us with algebraic number theory? Right now, I just see theorem after theorem, I see the algebraic manipulation and diagram chasing that proves it, but I don't see a bigger picture. For context, if $A$ is a $G$-module where $G$ is cyclic and $\sigma$ is a generator of $G$, then we define the endomorphisms $D=1+\sigma+\sigma^2+\cdots+\sigma^{|G|-1}$ and $N=1-\sigma$ of $A$, and then $H^0(G,A)=\mathrm{ker}(N)/\mathrm{im}(D)$ and $H^1(G,A)=\mathrm{ker}(D)/\mathrm{im}(N)$. Note that this is a slight modification of group cohomology, i.e. Tate cohomology, which the cohomology theory primarily used for Class Field Theory. Group cohomology is the same but with $H^0(G,A) = \mathrm{ker}(N)$. The advantage of Tate cohomology is that it is $2$-periodic for $G$ cyclic.
|
Here is a completely elementary example which shows that group cohomology is not empty verbiage, but can solve a problem ("parametrization of rational circle") whose statement has nothing to do with cohomology. Suppose you somehow know that for a finite Galois extension $k\subset K$ with group $G$ the first cohomology group $H^1(G,K^*)$ is zero : this is the homological version of Hilbert's Theorem 90 ( you can look it up in Weibel's book on homological algebra, pages 175-176). If moreover $G $ is cyclic with generator $s$, this implies that an element of $K$
has norm one if and only if it can be written $\frac{a}{s(a)}$ for some $a\in K$. Consider now the quadratic extension $k=\mathbb Q \subset K=\mathbb Q(i)$ with generator $s$ of $Gal(\mathbb Q (i)/\mathbb Q)$ the complex conjugation.The statement above says that $x+iy\in \mathbb Q(i)$ satisfies $x^2+y^2=1$ iff
$x+iy=\frac{u+iv}{s(u+iv)}=\frac{u+iv}{u-iv}=\frac{u^2-v^2}{u^2+v^2}+i\frac{2uv}{u^2+v^2}$ for some $u+iv\in \mathbb Q (i)$ . So we have obtained from group cohomology the well-known parametrization for the rational points of the unit circle $x^2+y^2=1$
$$x=\frac{u^2-v^2}{u^2+v^2}, \quad y=\frac{2uv}{u^2+v^2}$$.
|
{
"source": [
"https://mathoverflow.net/questions/10879",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1355/"
]
}
|
10,885 |
I want to interpret an $n\times n$ matrix $D$ as a set of pairwise distances, and assume that $D$ obeys metric properties. Namely, $D_{ii} = 0$, $D_{ij} \geq 0$, $D_{ij} = D_{ji}$ and $D_{ij} \leq D_{ik} + D_{kj}$ for all $1 \leq i,j,k \leq n$. For convenience, let $\bigtriangleup_n$ denote the set of such matrices. Now, I need to integrate some "simple" functions over this set. The simplest would be an exponential. Namely, I want to compute something like $\int_{\bigtriangleup_n} \exp\left[-\lambda \sum_{i=1}^n \sum_{j=i+1}^n D_{ij}\right] d D$. I've been able to work this out for the simplest nontrivial case: namely $n=3$. But for higher $n$, my brute force way of calculation gets really ugly. The approach I've been taking is to basically first integrate over $D_{11}, D_{12}, D_{13}, ..., D_{1n}$, all of which have no constraints... then integrate over $D_{23}$ which is just a definite integral from $|D_{12} - D_{23}|$ to $D_{12}+D_{23}$ of $\exp[-\lambda D_{13}]$ and then, in the general case, integrating over $D_ij$ becomes the definite integral from $\max_{k \neq i,j} |D_{ik}-D_{jk}|$ to $\min_{k\neq i,j} D_{ik}+D_{jk}$, but this is the point at which I get stuck, because these things becomes nasty quite quickly (even for just $n=4$). At the end of the day, I've love to be able to integrate more complex functions, like a chi-square type function rather than an exponential type function, but the exponential is the most trivial case that is interesting... To be precise, I'm looking for a closed form evaluation of the integral above, preferably with some derivation that will help me work out more complex examples.
|
Here is a completely elementary example which shows that group cohomology is not empty verbiage, but can solve a problem ("parametrization of rational circle") whose statement has nothing to do with cohomology. Suppose you somehow know that for a finite Galois extension $k\subset K$ with group $G$ the first cohomology group $H^1(G,K^*)$ is zero : this is the homological version of Hilbert's Theorem 90 ( you can look it up in Weibel's book on homological algebra, pages 175-176). If moreover $G $ is cyclic with generator $s$, this implies that an element of $K$
has norm one if and only if it can be written $\frac{a}{s(a)}$ for some $a\in K$. Consider now the quadratic extension $k=\mathbb Q \subset K=\mathbb Q(i)$ with generator $s$ of $Gal(\mathbb Q (i)/\mathbb Q)$ the complex conjugation.The statement above says that $x+iy\in \mathbb Q(i)$ satisfies $x^2+y^2=1$ iff
$x+iy=\frac{u+iv}{s(u+iv)}=\frac{u+iv}{u-iv}=\frac{u^2-v^2}{u^2+v^2}+i\frac{2uv}{u^2+v^2}$ for some $u+iv\in \mathbb Q (i)$ . So we have obtained from group cohomology the well-known parametrization for the rational points of the unit circle $x^2+y^2=1$
$$x=\frac{u^2-v^2}{u^2+v^2}, \quad y=\frac{2uv}{u^2+v^2}$$.
|
{
"source": [
"https://mathoverflow.net/questions/10885",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2569/"
]
}
|
10,897 |
There's a famous story about an exercise from Lang's Algebra that says something along the lines of "pick up a homological algebra book and prove all of the theorems yourself" . I cannot find it in the third revised edition, and I'm wondering if it's still in the third revised edition, if it's only in the older editions, or if it's an urban legend.
|
It's real, but only in the first and second editions. (I don't have any electronic proof, but I've seen it in my copy of the second edition and someone else's copy of the first edition.) It's the only exercise in the chapter. The full quote in the second edition is: Take any book on homological algebra,
and prove all the theorems without
looking at the proofs given in that
book. Homological algebra was invented by
Eilenberg-MacLane. General category
theory (i. e. the theory of
arrow-theoretic results) is generally
known as abstract nonsense (the
terminology is due to Steenrod). If I'm not mistaken, the quote is the same in the first edition. First edition: page 105; Second edition: page 175; so you can look in your library to see if I messed up the quote! And I do have an electronic copy of the third edition, which I've searched to confirm it is not there. The historical remarks were expanded, written less dismissively and put at the intro to Part Four. Update : I've scanned the evidence. First , Second .
|
{
"source": [
"https://mathoverflow.net/questions/10897",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1353/"
]
}
|
10,911 |
Kronecker's paper Zwei Sätze über Gleichungen mit ganzzahligen Coefficienten apparently proves the following result that I'd like to reference: Let $f$ be a monic polynomial with integer coefficients in $x$. If all roots of $f$ have absolute value at most 1, then $f$ is a product of cyclotomic polynomials and/or a power of $x$ (that is, all nonzero roots are roots of unity). However, I don't have access to this article, and even if I did my 19th century German skills are lacking; does anyone know a reference in English I could check for details of the proof?
|
I don't know a reference, but here is a quick proof: Let the roots of the polynomial be $\alpha_1$, $\alpha_2$, ..., $\alpha_r$. Let $$f_n(x) = \prod_{i=1}^r (x- \alpha_i^n).$$ All the coefficients of $f_n$ are rational, because they are symmetric functions of the $\alpha$'s, and are algebraic integers, because the $\alpha$'s are, so they are integers.
Also, since $|\alpha_i| \leq 1$, the coefficient of $x^k$ in $f_n$ is at most $\binom{r}{k}$. Combining the above observations, the coefficients of the $f_n$ are integers in a range which is bounded independent of $n$. So, in the infinite sequence $f_i$, only finitely many polynomials occur. In particular, there is some $k$ and $\ell$, with $\ell>0$, such that $f_{2^k} = f_{2^{k + \ell}}$. So raising to the $2^{\ell}$ power permutes the list $(\alpha_1^{2^{k}}, \ldots, \alpha_r^{2^k})$. For some positive $m$, raising to the $2^{\ell}$ power $m$ times will be the trivial permutation. In other words, $$\alpha_i^{2^k} = \alpha_i^{2^{k+\ell m}}$$. Every root of the above equation is $0$ or a root of unity.
|
{
"source": [
"https://mathoverflow.net/questions/10911",
"https://mathoverflow.net",
"https://mathoverflow.net/users/987/"
]
}
|
10,934 |
The statement that the class number measures the failure of the ring of integers to be a ufd is very common in books. ufd iff class number is 1. This inspires the following question: Is there a quantitative statement relating the class number of a number field to the failure of unique factorization in the maximal order - other than $h = 1$ iff $R$ is a ufd? In what sense does a maximal order of class number 3 "fail more" to be a ufd than a maximal order of class number 2? Is it true that an integer in a field of greater class number will have more distinct representations as the product of irreducible elements than an integer in a field with smaller class number?
|
Theorem (Carlitz, 1960): The ring of integers $\mathbb{Z}_F$ of an algebraic number field $F$ has class number at most $2$ iff for all nonzero nonunits $x \in \mathbb{Z}_F$ , any two factorizations of $x$ into irreducibles have the same number of factors. A proof of this (and a 1990 generalization of Valenza) can be found in $\S 22.3$ of my commutative algebra notes . This paper has spawned a lot of research by ring theorists on half-factorial domains : these are rings in which every nonzero nonunit factors into irreducibles and such that the number of irreducible factors is independent of the factorization. To be honest though, I think there are plenty of number theorists who think of the class number as measuring the failure of unique factorization who don't know Carlitz's theorem (or who know it but are not thinking of it when they make that kind of statement). Here is another try [ edit: this is essentially the same as Olivier's response, but said differently; I think it is worthwhile to have both ]: when trying to solve certain Diophantine problems (over the integers), one often gets nice results if the class number of a certain number field is prime to a certain quantity. The most famous example of this is Fermat's Last Theorem, which is easy to prove for an odd prime $p$ for which the class number of $\mathbb{Q}(\zeta_p)$ is prime to $p$ : a so-called "regular" prime. For an application to Mordell equations $y^2 + k = x^3$ , see http://alpha.math.uga.edu/~pete/4400MordellEquation.pdf Especially see Section 4, where the class of rings "of class number prime to 3" is defined axiomatically and applied to the Mordell equation. (N.B.: These notes are written for an advanced undergraduate / first year grad student audience.) The Mordell equation is probably a better example than the Fermat equation because: (i) the argument in the "regular" case is more elementary than FLT in the regular case (the latter is too involved to be done in a first course), and (ii) when the "regularity" hypothesis is dropped, it is not just harder to prove that there are no nontrivial solutions, it is actually very often false!
|
{
"source": [
"https://mathoverflow.net/questions/10934",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2024/"
]
}
|
10,966 |
Let $M$ be a differentiable manifold of dimension $n$. First I give two definitions of Orientability. The first definition should coincide with what is given in most differential topology text books, for instance Warner's book. Orientability using differential forms: There exists a nowhere vanishing differential form $\omega$ of degree $n$ on $M$. The second one is from Greenberg and Harper, "Algebraic Topology". This is the "fundamental class" approach. Let $x$ be a point on $X$, and let $R$ be a commutative ring and in the following the homologies are with coefficients in $R$. Local orientability: A local $R$ orientation of $X$ at $x$ is a choice of a generator of the $R$-module $H_n(X, X-x)$. By a simple application of Excision, it is seen that the above homology module is indeed isomorphic to $R$. We can also so arrange a neighborhood around every point that this local orientation can be "continued to a neighborhood" and is "coherent". Forgive me for being imprecise here; the detailed lemmas are in the reference given above. With this background in mind, we define: A Global $R$-orientation of $X$ consists of: 1. A family $U_i$ of open sets covering of $X$, 2. For each $i$, a local orientation $\alpha_i \in H_n(X, X -U_i)$ of along $X$, such that a "compatibility condition" holds. Here again I am imprecise about the compatibility condition; please check in the reference given above for details. I mean this basically as a question for those who already know both the definitions, as fully writing down the second definition would take 2-3 pages with all the necessary lemmas. Also we define "orientation" to be a such a global choice. Now the question: How do the two definitions, the first one using differential forms, and the second one using homology, match? Of course, to match we have to take $\mathbb{Z}$ to be the base ring for homology. A related question is about the meaning of orientability and orientation when we take a base ring other than $\mathbb{Z}$. It is nice when the base ring is $\mathbb{Z}/2\mathbb{Z}$; every manifold is orientable. But what on earth does it mean to have $4$ possible orientations for the circle or real line for instance, when you take the base ring to be $\mathbb{Z}/5\mathbb{Z}$? Also I ask, are there any additional ways to define orientability/orientation for a differentiable manifold(not just for a vector space)?
|
If $X$ is a differentiable manifold, so that both notions are defined, then they coincide. The ``patching'' of local orientations that you describe can be expressed more formally as follows: there is a locally constant sheaf $\omega_R$ of $R$-modules on $X$ whose stalk at a point is $H^n(X,X\setminus\{x\}; R).$ Of course, $\omega_R = R\otimes_{\mathbb Z} \omega_{\mathbb Z}$. This sheaf is called the orientation sheaf, and appears in the formulation of Poincare duality for not-necessarily orientable manifolds. It is not the case that any section of this sheaf gives an orientation. (For example, we always have the zero section.)
I think the usual definition would be something like a section which generates each stalk. I will now work just with $\mathbb Z$ coefficients, and write $\omega = \omega_{\mathbb Z}$. Since the stalks of $\omega$ are free of rank one over $\mathbb Z$, to patch them together you
end up giving a 1-cocyle with values in $GL_1({\mathbb Z}) = \{\pm 1\}.$ Thus underlying
$\omega$ there is a more elemental sheaf, a locally constant sheaf that is a principal bundle for $\{\pm 1\}$. Equivalently, such a thing is just a degree two (not necessarily connected) covering space
of $X$, and it is precisely the orientation double cover of $X$. Now giving a section of $\omega$ that generates each stalk, i.e. giving an orientation of $X$, is precisely the same as giving a section of the orientation double cover (and so $X$ is orientable, i.e. admits an orientation, precisely when the orientation double cover is disconnected). Instead of cutting down from a locally constant rank 1 sheaf over $\mathbb Z$ to just a double cover, we could also build up to get some bigger sheaves. For example, there is the sheaf $\mathcal{C}_X^{\infty}$ of smooth functions on $X$.
We can form the tensor product $\mathcal{C}_X^{\infty} \otimes_{\mathbb Z} \omega,$
to get a locally free sheaf of rank one over ${\mathcal C}^{\infty}$, or equivalently, the sheaf of sections of a line bundle on $X$. This is precisely the line bundle of top-dimensional forms on $X$. If we give a section of $\omega$ giving rise to an orientation of $X$, call it $\sigma$, then we certainly get a nowhere-zero section
of $\mathcal{C}_X^{\infty} \otimes_{\mathbb Z} \omega$, namely $1\otimes\sigma$. On the other hand, if we have a nowhere zero section of $\mathcal{C}_X^{\infty} \otimes_{\mathbb Z}
\omega$, then locally (say on the the members of some cover $\{U_i\}$ of $X$ by open balls) it has the form $f_i\otimes\sigma_i,$ where $f_i$ is a nowhere zero real-valued function on $U_i$ and $\sigma_i$ is a generator of $\omega_{| U_i}.$ Since $f_i$ is nowhere zero, it is either always positive or always negative; write
$\epsilon_i$ to denote its sign. It is then easy to see that sections $\epsilon_i\sigma_i$
of $\omega$ glue together to give a section $\sigma$ of $X$ that provides an orientation. One also sees that two different nowhere-zero volume forms will give rise to the same orientation if and only if their ratio is an everywhere positive function. This reconciles the two notions.
|
{
"source": [
"https://mathoverflow.net/questions/10966",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2938/"
]
}
|
10,974 |
Is the following true: If two chain complexes of free abelian groups have isomorphic homology modules then they are chain homotopy equivalent.
|
Yes, this is true. Suppose $C_*$ is such a chain complex of free abelian groups. For each $n$, choose a splitting of the boundary map $C_n \to B_{n-1}$, so that $C_n \cong Z_n \oplus B_{n-1}$. (You can do this because $B_{n-1}$, as a subgroup of a free group, is free.) For all $n$, you then have a sub-chain-complex $\cdots \to 0 \to B_n \to Z_n \to 0 \to \cdots$ concentrated in degrees $n$ and $n+1$, and $C_*$ is the direct sum of these chain complexes. Given two such chain complexes $C_*$ and $D_*$, you get a direct sum decomposition of each, and so it suffices to show that any two complexes $\cdots \to 0 \to R_i \to F_i \to 0 \to \cdots$, concentrated in degrees $n$ and $n+1$, which are resolutions of the same module $M$ are chain homotopy equivalent; but this is some variant of the fundamental theorem of homological algebra. This is special to abelian groups and is false for modules over a general ring.
|
{
"source": [
"https://mathoverflow.net/questions/10974",
"https://mathoverflow.net",
"https://mathoverflow.net/users/3046/"
]
}
|
11,084 |
I understand this might be a slightly subjective question, but I am honestly curious what programming languages are used by the mathematics community. I would imagine that there is a group of mathematicians out there that use haskell because it might be more consistent with ideas from mathematics. Is this true? What about APL? Do mathematicians today use APL or is that just a relic of the past?
|
Bryan Birch is credited with once saying that he programmed in a very high-level programming language called "graduate student".
|
{
"source": [
"https://mathoverflow.net/questions/11084",
"https://mathoverflow.net",
"https://mathoverflow.net/users/3063/"
]
}
|
11,192 |
Let $\mathcal{X}$ be a real or complex Banach space.
It is a well known fact that $\mathcal{X}$ is a Hilbert space (i.e. the norm comes from an inner product) if the parallelogram identity holds. Question: Are there other (simple) characterizations for a Banach space to be a Hilbert space?
|
From this article by O. N. Kosukhin: A real Banach space $(X, \|\cdot\|)$ is a Hilbert space if and only if for any three points $A$ , $B$ , $C$ of this space not belonging to a line there are three altitudes in the triangle $ABC$ intersecting at one point. Many other references show when Googling "is a hilbert space if" banach
|
{
"source": [
"https://mathoverflow.net/questions/11192",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2989/"
]
}
|
11,219 |
I am interested in learning about Shimura curves. Unlike most of the people who post reference requests however (see this question for example), my problem is not sorting through an abundance of books but rather dealing with what appears to be an extreme paucity of sources. Anyway, I'm a graduate student and have spent the last year or so thinking about the arithmetic of orders in quaternion algebras (and more generally in central simple algebras). The study of orders in quaternion algebras seems to play an important role in Shimura curves, and I'd like to study these connections more carefully. Unfortunately, it has been very difficult for me to find a good place to start. I only really know of two books that explicitly deal with Shimura curves: Shimura's Introduction to the arithmetic theory of automorphic functions Alsina and Bayer's Quaternion Orders, Quadratic Forms, and Shimura Curves Neither book has been particularly helpful however; the first only mentions them briefly in the final section, and the second has much more of a computational focus then I'd like. Question 1 : Is there a book along the lines of Silverman's The Arithmetic of Elliptic Curves for Shimura curves? I kind of doubt that such a book exists. Thus I've tried to read the introductory sections of a few papers & theses, but have run into a problem. There seem to be various ways of thinking about a Shimura curve, and it has been the case that every time I look at an article I'm confronted with a different one. For example, this talk by Voight and this paper by Milne. By analogy, it seems to be a lot like trying to learn class field theory by switching between articles with ideal-theoretic statements and articles taking an adelic slant without having a definitive source which tells you that both are describing the same theorems. My second question is therefore: Question 2 : Can anyone suggest a 'roadmap' to Shimura curves? Which theses or papers have especially good expository accounts of the basic properties that one needs in order to understand the literature. Clearly I need to say something about my background. As I mentioned above, I'm an algebraic number theorist with a particular interest in quaternion algebras. I don't have the best algebraic geometry background in the world, but have read Mumford's Red Book , the first few chapters of Hartshorne and Qing Liu's Algebraic Geometry and Arithmetic Curves . I've also read Silverman's book The Arithmetic of Elliptic Curves and Diamond and Shurman's A First Course in Modular Forms . Thanks.
|
First of all, Kevin is being quite modest in his comment above: his paper Buzzard, Kevin. Integral models of certain Shimura curves. Duke Math. J. 87 (1997), no. 3, 591--612. contains many basic results on integral models of Shimura curves over totally real fields, and is widely cited by workers in the field: 22 citations on MathSciNet. The most recent is a paper of mine: Clark, Pete L. On the Hasse principle for Shimura curves. Israel J. Math. 171 (2009), 349--365. http://alpha.math.uga.edu/~pete/plclarkarxiv7.pdf Section 3 of this paper spends 2-3 pages summarizing results on the structure of the canonical integral model of a Shimura curve over $\mathbb{Q}$ (with applications to the existence of local points). From the introduction to this paper: "This result [something about local points] follows readily enough from a description of their [certain Shimura curves over Q] integral canonical models. Unfortunately I know of no unique, complete reference for this material. I have myself written first (my 2003 Harvard thesis) and second (notes from a 2005 ISM course in Montreal) approximations of such a work, and in so doing I have come to respect the difficulty of this expository problem." I wrote that about three years ago, and I still feel that way today. Here are the documents: http://alpha.math.uga.edu/~pete/thesis.pdf is my thesis. "Chapter 0" is an exposition on Shimura curves: it is about 50 pages long. For my (incomplete) lecture notes from 2005, go to http://alpha.math.uga.edu/~pete/expositions2012.html and scroll down to "Shimura Curves". There are 12 files there, totalling 106 pages [perhaps I should also compile them into a single file]. On the other hand, the title of the course was Shimura Varieties , and although I don't so much as attempt to give the definition of a general Shimura variety, some of the discussion includes other PEL-type Shimura varieties like Hilbert and Siegel moduli space. These notes do not entirely supercede my thesis: each contains some material that the other omits. When I applied for an NSF grant 3 years ago, I mentioned that if I got the grant, as part of my larger impact I would write a book on Shimura curves. Three years later I have written up some new material (as yet unreleased) but am wishing that I had not said that so directly: I would need at least a full semester off to make real progress (partly, of course, to better understand much of the material). Let me explain the scope of the problem as follows: there does not even exist a single, reasonably comprehensive reference on the arithmetic geometry of the classical modular curves (i.e., $X_0(N)$ and such). This would-be bible of modular curves ought to contain most of the material from Shimura's book (260 pages) and the book of Katz and Mazur Arithmetic Moduli of Elliptic Curves (514 pages). These two books don't mess around and have little overlap, so you get a lower bound of, say, 700 pages that way. Conversely, I claim that there is some reasonable topology on the arithmetic geometry of modular curves whose compactification is the theory of Shimura curves. The reason is that in many cases there are several ways to establish a result about modular curves, and "the right one" generalizes to Shimura curves with little trouble. (For example, to define the rational canonical model for classical modular curves, one could use the theory of Fourier expansions at the cusps -- which won't generalize -- or the theory of moduli spaces -- which generalizes immediately. Better yet is to use Shimura's theory of special points, which nowadays you need to know anyway to study Heegner point constructions.) Most of the remainder concerns quaternion arithmetic, which, while technical, is nowadays well understood and worked out.
|
{
"source": [
"https://mathoverflow.net/questions/11219",
"https://mathoverflow.net",
"https://mathoverflow.net/users/-1/"
]
}
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.