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98,821 |
I have the following question: How likely it is that an author carefully read through a paper cited by him? Not everyone reads through everything that they have cited. Sometimes, if one wants to use a theorem that is not in a standard textbook, one typically finds another paper which cites the desired result and copies that citation thereby passing the responsibility of ensuring correctness to someone else. This saves a lot of time, but seems to propagate inaccurate citations and poor understanding of the work being cited. The question is thus about what should authors' citing policy be, and to what extent authors should verify results they are citing rather than using them as black boxes.
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I normally do. Right now I'm facing a tough choice though: to read David-Semmes book in honest or to write something like "We prove that A implies B. The reader can juxtapose that with the claim on page ... in [] that B implies C" instead of "We prove that A implies C" in the introduction. Being as lazy as I am, I am inclined to go for the second option but that will certainly make the paper less "sexy", so my co-authors do not feel very happy about it. However this shows how you can avoid both reading the papers you refer to and the uncertainty about whether what you declare proved is actually proved: separate the part you prove from the part you refer to in a crystal clear way and take credit for the reduction only rather than for the full statement (which, frankly speaking, is as much credit as you can really claim anyway). The correction mechanism Nik mentioned works primarily in the way that most things just go unnoticed because nobody reads those papers or uses them in any way. When something is really important, it gets a lot of attention and somebody finally straightens things out. However, that doesn't happen fast and I have learned it hard way. My 2002 Duke paper joint with Treil and Volberg on the system Tb theorem has an error in the proof. It had been cited a lot of times before the error was finally spotted and corrected by Tuomas Hytonen around 2010. This also shows that an erratic argument isn't always useless or fatally flawed. Sometimes it is just an "incomplete proof". To my shame, I should also mention that it was one of the cases when I didn't read the final draft carefully and relied on my co-authors to do that. Apparently, they had a similar attitude...
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98,822 |
What I will ask, more than a solution, is better mathematical definition of my problem and directions to find the solution. I have a set of linear equations, e.g.: \begin{align}
d_1 &= L_1 - 9\,m_1 - 9\,m_2 \\
d_2 &= x_1 + 3\,m_1 + 3\,m_2 \\
d_3 &= y_1 \\
d_4 &= L_2 - 4\,m_2 \\
d_5 &= x_2 + 2\,m_2 \\
d_6 &= y_2
\end{align} where $d_1,d_2,...,d_6$ are linear combinations of $L_1,x_1,y_1,m_1,L_2,x_2,y_2,m_2 \in \mathbb{R}$. $L_1,x_1,y_1,m_1,L_2,x_2,y_2,m_2$ are physical parameters which have physical nonlinear constraints, in the form of polynomial inequalities, e.g.: \begin{align}
m_1 &>0\\
L_1m_1−x^2_1−y^2_1 &>0\\
m_2 &>0\\
L_2m_2−x^2_2−y^2_2 &>0
\end{align} I would like to rewrite constraints in terms of $d_1,d_2,...,d_6$ only. I.e., I would like to find constrains over $d_1,d_2,...,d_6$ parameters (only), so that when a numerical set of $d_1,d_2,...,d_6$ verifies the new mapped constraints that would mean that there is at least one $L_1,x_1,y_1,m_1,L_2,x_2,y_2,m_2$ solution (it doesn't matter what) which verify the former constraints. If the new constraints are not verified it must mean that no $L_1,x_1,y_1,m_1,L_2,x_2,y_2,m_2$ solution exists. Here I presented a particularly small example, for it I was already able to find the constraints doing manual equation manipulation ($d_1 + 6 d_2 > 0$ and $- 9 d_{5}^{2} - 9 d_{6}^{2} + \left(d_{1} + 6 d_{2}\right) \left(d_{4} + 4 d_{5}\right) >0$)
However, I have problems with up to 70 linear equations and 30 higher order polynomial inequalities constraints. What I need is a systematic method to write the constrains over $d_1,d_2,...,d_6$. I think that in geometrical thinking this is like projecting the intersection of a set of non-linear volumes/regions of $n$-dimensional space ($n>>1$) onto a subspace of it, where the coefficients of the linear system are the basis of such subspace. Now the questions: What kind of problem do I have? Which mathematical fields shall I study, and which directions must I follow? My background is in engineering so my mathematical writing is not very formal. Thanks.
|
I normally do. Right now I'm facing a tough choice though: to read David-Semmes book in honest or to write something like "We prove that A implies B. The reader can juxtapose that with the claim on page ... in [] that B implies C" instead of "We prove that A implies C" in the introduction. Being as lazy as I am, I am inclined to go for the second option but that will certainly make the paper less "sexy", so my co-authors do not feel very happy about it. However this shows how you can avoid both reading the papers you refer to and the uncertainty about whether what you declare proved is actually proved: separate the part you prove from the part you refer to in a crystal clear way and take credit for the reduction only rather than for the full statement (which, frankly speaking, is as much credit as you can really claim anyway). The correction mechanism Nik mentioned works primarily in the way that most things just go unnoticed because nobody reads those papers or uses them in any way. When something is really important, it gets a lot of attention and somebody finally straightens things out. However, that doesn't happen fast and I have learned it hard way. My 2002 Duke paper joint with Treil and Volberg on the system Tb theorem has an error in the proof. It had been cited a lot of times before the error was finally spotted and corrected by Tuomas Hytonen around 2010. This also shows that an erratic argument isn't always useless or fatally flawed. Sometimes it is just an "incomplete proof". To my shame, I should also mention that it was one of the cases when I didn't read the final draft carefully and relied on my co-authors to do that. Apparently, they had a similar attitude...
|
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99,230 |
Does someone has a reference to a modern proof of the Baker-Campbell-Hausdorff formula? All proofs I have ever seen are related only to matrix Lie groups / Lie algebras and
are not at all geometric (i.e. depend on indices, bases ect.) By a 'modern' proof I'm thinking of a proof entirely in terms of differential geometry, i.e.
in terms of the tangent bundle on the Lie group manifold or even better in terms of jets. I'll keep the formulation vague on purpose, to higher my chances to get a good reference.
I think the question is pretty clear anyway. Beyond the plain BCH-equation I would like to get a deeper understanding WHY the commutator
(and the linear structure of the Lie algebra) is enough to define the group product locally
and what is geometrically going on.
|
In the Lie Theory notes on my website based on the 2008 class by Mark Haiman, section 3.3, we give the following discussion. I should mention that I think of BCH and breaking into two parts. The first is purely algebraic: Let $U$ be a bialgebra over field $\mathbb k$, and $s$ a formal (commuting) variable. Then we can form the $s$-adic completion $U[\![ s ]\!]$ of $U \otimes_{\mathbb k} \mathbb k[s]$. It is an algebra, of course. It is not a bialgebra in the algebraic sense. On the other hand, we can extend the comultiplication $\Delta : U \to U\otimes U$ by linearity (and continuity – it is $s$-adic-continuous) to $U[\![ s ]\!] \to (U\otimes U)[\![ s ]\!]$, and the map $U[\![ s ]\!] \otimes_{\mathbb k[\![ s ]\!]} U[\![ s ]\!] \to (U\otimes U)[\![ s ]\!]$ realizes the latter as the $s$-adic-completion of the former. So $U[\![ s ]\!]$ is a bialgebra for this $s$-adic-completed tensor product, which I will denote by $\hat\otimes$. Denote this completed comultiplication by $\hat\Delta$. Note that $\hat\Delta$ is continuous in the $s$-adic topology. Choose $\psi \in U[\![ s ]\!]$ such that $\psi(s=0) = 0$. Then $\psi$ is primitive (i.e. $\hat\Delta(\psi) = \psi \otimes 1 + 1\otimes \psi$ iff $e^\psi$ is grouplike (i.e. $\hat\Delta(e^\psi) = e^\psi \hat\otimes e^\psi$). Note that $e^\psi = \sum_n \frac1{n!}\psi^n$ converges in the $s$-adic topology since $\psi(s=0) = 0$. Proof: By continuity, $\hat\Delta(e^\psi) = e^{\hat\Delta(\psi)}$. As a remark, note that the power series $\psi \in U[\![ s ]\!]$ is primitive iff it is primitive term-by-term. Everything above continues to work if we use two commuting formal variables $s$ and $t$. Let $\mathfrak f$ denote the free Lie algebra on two generators $x$ and $y$. Then its universal enveloping algebra is the free associative algebra $T$ on the same two generators. (Proof: "free" functors are left adjoint to "forget to vector spaces" and "universal enveloping" is left adjoint to "forget from associative to Lie", and the composition of left adjoints is left adjoint to the corresponding composition of forgetful functors.) We will work in the (completed) bialgebra $T[\![s,t]\!]$. Define the formal series $b(sx,ty)$ by the formula
$$ e^{b(sx,ty)} = e^{sx}e^{ty}. $$
Since $x$ and $y$ are primitive and the product of grouplikes is grouplike, it follows from 2. that $b(sx,ty)$ is primitive. Indeed, the coefficient on $s^mt^n$ in $b(sx,ty)$ is primitive. The primitives in a universal enveloping algebra (in this case, $T$) are precisely the original Lie algebra (in this case, $\mathfrak f$). Thus the coefficient on $s^mt^n$ in $b(sx,ty)$ is a Lie polynomial (composition of brackets) in the noncommuting variables $x,y$. By degree counting, it is homogeneous of degree $m$ in $x$ and homogeneous of degree $n$ in $y$. More generally, $b(sx,ty)$ is a Lie series. If you work to define the correct topology so that you can talk about power series in noncommuting variables, then it makes sense to talk about the series $b(x,y)$. The second part is manifold-theoretic: Let $G$ denote a Lie group (whose underlying manifold is given analytic structure, and such that multiplication is an analytic map), and $\mathfrak g$ its Lie algebra. Identify the universal enveloping algebra $U\mathfrak g$ with the left-invariant differential operators on $G$. Consider the stalk $\mathcal C(G)_e$ of analytic functions defined in a neighborhood of the identity. Then $u\in U\mathfrak g = 0$ iff $u$ annihilates $\mathcal C(G)_e$. Recall that we have a map $\exp: \mathfrak g \to G$ (defined, for example, by flowing along the left-invariant vector on $G$ field defined by $x\in \mathfrak g$). It is an analytic isomorphism near the identity, and so near the identity we have an inverse map $\log : G \to \mathfrak g$. By passing to smaller neighborhoods of the identity as needed, we can define $\beta(x,y) = \log(\exp x\exp y)$. It is a $\mathfrak g$-valued analytic function on a neighborhood of $(0,0)\in \mathfrak g \times \mathfrak g$. Choose $f\in \mathcal C(G)_e$ and $x,y\in \mathfrak g$. Then $e^{sx} \in U\mathfrak g[\![s]\!]$ makes is $\mathbb k[\![s]\!]$-valued differential operator on $\mathcal C(G)_e$, and $e^{sx} f$ is the Taylor expansion in $s$ of $f(\exp sx)$. Similarly, $e^{sx}e^{ty}f$ computes the two-variable Taylor expansion of $f(\exp sx \exp ty)$. Let $\tilde \beta$ denote the Taylor expansion of $\beta(sx,ty)$. Then
$$ e^{\tilde \beta(sx,ty)}f = f(\exp(\tilde\beta(sx,ty))) = f(\exp sx \exp ty) = e^{sx}e^{ty} f = e^{b(sx,ty)} f. $$
By 1., we must have $\tilde\beta(sx,ty) = b(sx,ty)$ as formal power series. But $\tilde\beta$ is the Taylor expansion of the analytic function $\beta$. Thus in a small enough neighborhood of the origin, it converges. By definition, $\tilde \beta = b $ is the Baker–Campbell–Hausdorff series.
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99,488 |
The only reasonable way to interpret "$ds$" as a functional on tangent vectors has to be that it takes a tangent vector and spits out its length, but this is not linear. So $ds$ is not a 1-form. It still seems like a nice sort of object to think about integrating. Does $ds$ fit into a larger class of gadgets generalizing differential forms? Or it there some compelling reason that I shouldn't care about $ds$?
|
It is not a 1-form, it is a 1-density : a function that is continuous and homogeneous of degree 1 on the tangent space of the manifold. It also happens to be convex and positive in the complement of the zero section (actually, its restriction to each tangent space is a Euclidean norm). If the norm is not Euclidean, you have the arc-length element of a Finsler metric. The convexity is basically necessary and sufficient for the lower semi-continuity of the length functional (Busemann-Mayer theorem). See my answer to this question for more on densities.
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99,506 |
By a blackbox theorem I mean a theorem that is often applied but whose proof is understood in detail by relatively few of those who use it. A prototypical example is the Classification of Finite Simple Groups (assuming the proof is complete). I think very few people really know the nuts and bolts of the proof but it is widely applied in many areas of mathematics. I would prefer not to include as a blackbox theorem exotic counterexamples because they are not usually applied in the same sense as the Classification of Finite Simple Groups. I am curious to compile a list of such blackbox theorems with the usual CW rules of one example per answer. Obviously this is not connected to my research directly so I can understand if this gets closed.
|
The existence of resolution of singularities in characteristic zero is certainly used by many more people than those who know the details of its proofs, especially the original one.
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99,547 |
There has been a great deal of excitement among topologists about the proof of the Virtual Haken Theorem , and in fact of the Virtual Fibering Theorem (for closed hyperbolic 3-manifolds, but I'm guessing they will soon be proven in full generality). The proof is lucidly discussed in Danny Calegari's blog . The theorems state that every compact orientable irreducible 3-manifold with infinite fundamental group has a finite cover which is Haken or a surface bundle over a circle, correspondingly. This implies various good things for a 3-manifold with fundamental group π, including: π is large, meaning that π has a finite index subgroup which maps onto a free group with at least 2 generators. In particular the Betti numbers of finite covers can become arbitrarily large. π is linear over $\mathbb{Z}$, i.e. π admits a faithful representation $\pi \to \mathrm{GL}(n,\mathbb{Z})$ for some $n$. (Thurston conjectured that $n\leq 4$ is sufficient). π is virtually biorderable. Stefan Friedl, from whose comment the above list is an excerpt, summarizes the situation as follows: It seems like every nice property of fundamental groups which one can possibly ask for either holds for π or a finite index subgroup of π. All well and good. But how could you `sell' that to somebody who isn't a classically-oriented 3-dimensional topologist?
An elevator pitch is defined by Wikipedia as follows: An elevator pitch is a short summary used to quickly and simply define a product, service, or organization and its value proposition. The name "elevator pitch" reflects the idea that it should be possible to deliver the summary in the time span of an elevator ride, or approximately thirty seconds to two minutes. In The Perfect Elevator Speech , Aileen Pincus states that an elevator speech should "sum up unique aspects of your service or product in a way that excites others." The Virtual Fibering Conjecture (or the Virtual Haken Conjecture) was the grand conjecture in 3-manifold topology following Geometrization, and thus must have/ should have/ ought to have (I believe) a compelling elevator pitch. For contrast, Geometrization is easy to `sell' because it directly applies to the Homeomorphism Problem in 3-manifold topology: Given two 3-manifolds, determine whether or not they are homeomorphic. Geometrization allows you to canonically decompose both manifolds into submanifolds with geometric structure, and then to compare geometric invariants. In terms of "The Goals of Mathematical Research" as given in the introduction to The Princeton Companion to Mathematics , this corresponds to the goal of Classifying. Question : What is a good elevator pitch for Virtual Fibering (or for Virtual Haken), explaining the utility of these results in terms of "the fundamental goals of mathematical research" (Solving Equations, Classifying, Generalizing, Discovering Patterns, Explaining Patterns and Coincidences, Counting and Measuring, and Finding Explicit Algorithms). The target would be mathematicians who are not 3-dimensional topologists. Everyone in the approximate vicinity of the field instinctively feels that these are historic results, but I'd like to be able to justify that feeling (in the abovementioned sense) to myself and to others.
|
OK, I will also give it a shot. First of all, I don't like to sell Geometrization because it helps with the homeomorphism problem. The Geometrization Theorem is an object of stunning beauty ("most 3-manifolds are hyperbolic" should be an exciting statement for anybody in an elevator who has seen the art of M.C.Escher), and beauty in mathematics is usually a sign that we are on the right track. And indeed, the beauty of Geometrization begets all kinds of results. Regarding the new results of Agol, Wise et al. one should perhaps not jump right to "virtual Haken" or "virtually fibered" but one should look at the "real theorem",
the Virtually Compact Special Theorem which goes as follows: If $N$ is a finite volume hyperbolic 3-manifold, then $\pi_1(N)$ is virtually compact special, i.e. $\pi_1(N)$ is virtually a quasi-convex subgroup of a Right Angled Artin Group (RAAG). One can explain a RAAG to anybody who has seen group theory in an elevator between about 3 floors. The fact that "simple" objects like RAAGs contain all hyperbolic 3-manifold groups (up to going to a finite index subgroup) is stunning and beautiful. All the goodies, e.g. largeness, linear over $\mathbb{Z}$, virtual fibering, LERF, virtually biorderable etc come from that statement (well, together with Agol's fibering theorem, tameness etc.).
This can be seen clearly by looking at Diagram 4 in a recent survey paper on 3-manifold groups by authors whose names escape me at the moment. It is really stunning how the Virtually Compact Special Theorem answers all open questions at once.
It is one of the great achievements of Dani Wise to have found the "right statement". (Note that largeness, linear over $\mathbb{Z}$, biorderable do NOT follow from virtual fibering or virtual Haken alone.) Back to the elevator: The results make me think that hyperbolic 3-manifolds are like Jack in the Box.
If you take a hyperbolic integral homology sphere you look at a tiny manifold, but when you press a button (i.e. go to an appropriate finite cover), the 3-manifold suddenly becomes a grand object of beauty (e.g. has as many fibered faces in the Thurston norm ball as you could wish). (This analogy also works with tiny seed, a bit of water, blooming flower etc. for the botanically minded elevator companion) So to conclude, I think the Geometrization Theorem and the Virtually Compact Special Theorem of Agol-Wise are stunningly beautiful results. The fact that the statements are so beautiful made it highly plausible that they were right, even before they were proved (I can't imagine that any serious person doubted the Poincare conjecture after Thurston stated the Geometrization conjecture). And ideally it's this beauty which I would like to communicate.
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99,643 |
I do not think it is possible really believe or experimentally check (now), but all modern physical doctrines suggest that out world is NOT 4-dimensional, but higher.
The least sophisticated candidate - bosonic string theory says that out world is 26 dimensional (it is not realistic due to presence of tachion, and so there are super strings with 10 dimensions, M-theory with 11, F-theory with 12). Let us do not care about physical realities and ask: what mathematics stands behind the fact that 26 is the only dimension where bosonic string theory can live ?
Definitely there is some mathematics e.g. 26 in that MO question is surely related. Let me recall the bosonic string theory background.
Our real world is some Riemannian manifold M which is called TS (target space).
We consider the space of all maps from the circle to M, actually we need to consider how
the circle is moving inside M, so we get maps from $S^1\times [0~ T]$ to M ( here $S^1\times [0~ T]$ is called WS - world sheet);
we identify the maps which differs by parametrization (that is how Virasoro comes into game
and hence relation with Leonid's question ). That was pretty mathematical, but now ill-defined physics begin - we need integrate
over this infinite-dimensional space of maps/parametrizations with measure
corresponding to exp( i/h volume_{2d}(image(WS))).
This measure is known NOT to exist mathematically, but somehow this does not stop
physists they do what they call regularization or renormalization or something like
that and 26 appears...
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In addition to Chris Gerig's operator-language approach, let me also show how this magical number appears in the path integral approach. Let $\Sigma$ be a compact surface (worldsheet) and $M$ a Riemannian manifold (spacetime). The string partition function looks like
$$Z_{string}=\int_{g\in Met(\Sigma)}dg\int_{\sigma\in Map(\Sigma,M)}d\sigma\exp(iS(g,\sigma)).$$
Here $Met(\Sigma)$ is the space of Riemannian metrics on $\Sigma$ and $S(g,\sigma)$ is the standard $\sigma$-model action $S(g,\sigma)=\int_{\Sigma} dvol_\Sigma \langle d\sigma,d\sigma\rangle$. In particular, $S$ is quadratic in $\sigma$, so the second integral $Z_{matter}$ does not pose any difficulty and one can write it in terms of the determinant of the Laplace operator on $\Sigma$. Note that the determinant of the Laplace operator is a section of the determinant line bundle $L_{det}\rightarrow Met(\Sigma)$. The measure $dg$ is a 'section' of the bundle of top forms $L_g\rightarrow Met(\Sigma)$. Both line bundles carry natural connections. However, the space $Met(\Sigma)$ is enormous: for example, it has a free action by the group of rescalings $Weyl(\Sigma)$ ($g\mapsto \phi g$ for $\phi\in Weyl(\Sigma)$ a positive function). It also carries an action of the diffeomorphism group. The quotient $\mathcal{M}$ of $Met(\Sigma)$ by the action of both groups is finite-dimensional, it is the moduli space of conformal (or complex) structures, so you would like to rewrite $Z_{string}$ as an integral over $\mathcal{M}$. Everything in sight is diffeomorphism-invariant, so the only question is how does the integrand change under $Weyl(\Sigma)$. To descend the integral from $Met(\Sigma)$ to $Met(\Sigma)/Weyl(\Sigma)$ you need to trivialize the bundle $L_{det}\otimes L_g$ along the orbits of $Weyl(\Sigma)$. This is where the critical dimension comes in: the curvature of the natural connection on $L_{det}\otimes L_g$ (local anomaly) vanishes precisely when $d=26$. After that one also needs to check that the connection is actually flat along the orbits, so that you can indeed trivialize it. Two references for this approach are D'Hoker's lectures on string theory in "Quantum Fields and Strings" and Freed's "Determinants, Torsion, and Strings".
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99,736 |
I'm curious about the beautiful descriptions of exceptional simple complex Lie groups and algebras (and maybe their compact forms). By beautiful I mean: simple (not complicated - it means that we need not so many words to describe this). For $G_2$ we know the automorphisms of octonions and the rolling distribution (and also the intersection of three $Spin_7$-s in $Spin_8$). For $F_4$ we know the automorphisms of Jordan algebra $H_3(\mathbb O)$ and Lie algebra of commutators of right multiplications in this algebra (see Chevalley-Schafer's paper for details). For $E_6$ we know the automorphisms of determinant in $H_3(\mathbb O)$ and Lie algebra linearly spanned by right multiplications and $\mathfrak f_4$. For $\mathfrak f_4$, $\mathfrak e_6$, $\mathfrak e_7$, $\mathfrak e_8$ we know the Vinberg-Freudenthal Magic Square. What do we know (expressing in a simple form) about $E_7$ and $E_8$?
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It is not always clear what one means by 'the simplest description' of one of the exceptional Lie groups. In the examples you've given above, you quote descriptions of these groups as automorphisms of algebraic structures, and that's certainly a good way to do it, but that's not the only way, and one can argue that they are not the simplest in terms of a very natural criterion, which I'll now describe: Say that you want to describe a subgroup $G\subset \text{GL}(V)$ where $V$ is a vector space (let's not worry too much about the ground field, but, if you like, take it to be $\mathbb{R}$ or $\mathbb{C}$ for the purposes of this discussion). One would like to be able to describe $G$ as the stabilizer of some element $\Phi\in\text{T}(V{\oplus}V^\ast)$, where $\mathsf{T}(W)$ is the tensor algebra of $W$. The tensor algebra $\mathsf{T}(V{\oplus}V^\ast)$ is reducible under $\text{GL}(V)$, of course, and, ideally, one would like to be able to chose a 'simple' defining $\Phi$, i.e., one that lies in some $\text{GL}(V)$-irreducible submodule $\mathsf{S}(V)\subset\mathsf{T}(V{\oplus}V^\ast)$. Now, all of the classical groups are defined in this way, and, in some sense, these descriptions are as simple as possible. For example, if $V$ with $\dim V = 2m$ has a symplectic structure $\omega\in \Lambda^2(V^\ast)$, then the classical group $\text{Sp}(\omega)\subset\text{GL}(V)$ has codimension $m(2m{-}1)$ in $\text{GL}(V)$, which is exactly the dimension of the space $\Lambda^2(V^\ast)$. Thus, the condition of stabilizing $\omega$ provides exactly the number of equations one needs to cut out $\text{Sp}(\omega)$ in $\text{GL}(V)$. Similarly, the standard definitions of the other classical groups as subgroups of linear transformations that stabilize an element in a $\text{GL}(V)$-irreducible subspace of $\mathsf{T}(V{\oplus}V^\ast)$ are as 'efficient' as possible. In another direction, if $V$ has the structure of an algebra, one can regard the multiplication as an element $\mu\in \text{Hom}\bigl(V\otimes V,V\bigr)= V^\ast\otimes V^\ast \otimes V$, and the automorphisms of the algebra $A = (V,\mu)$ are, by definition, the elements of $\text{GL}(V)$ whose extensions to $V^\ast\otimes V^\ast \otimes V$ fix the element $\mu$. Sometimes, if one knows that the multiplication is symmetric or skew-symmetric and/or traceless, one can regard $\mu$ as an element of a smaller vector space, such as $\Lambda^2(V^\ast)\otimes V$ or even the $\text{GL}(V)$-irreducible module $\bigl[\Lambda^2(V^\ast)\otimes V\bigr]_0$, i.e., the kernel of the natural contraction mapping $\Lambda^2(V^\ast)\otimes V\to V^\ast$. This is the now-traditional definition of $G_2$, the simple Lie group of dimension $14$: One takes $V = \text{Im}\mathbb{O}\simeq \mathbb{R}^7$ and defines $G_2\subset \text{GL}(V)$ as the stabilizer of the vector cross-product $\mu\in \bigl[\Lambda^2(V^\ast)\otimes V\bigr]_0\simeq \mathbb{R}^{140}$. Note that the condition of stabilizing $\mu$ is essentially $140$ equations on elements of $\text{GL}(V)$ (which has dimension $49$), so this is many more equations than one would really need. (If you don't throw away the subspace defined by the identity element in $\mathbb{O}$, the excess of equations needed to define $G_2$ as a subgroup of $\text{GL}(\mathbb{O})$ is even greater.) However, as was discovered by Engel and Reichel more than 100 years ago, one can define $G_2$ over $\mathbb{R}$ much more efficiently: Taking $V$ to have dimension $7$, there is an element $\phi\in \Lambda^3(V^\ast)$ such that $G_2$ is the stabilizer of $\phi$. In fact, since $G_2$ has codimension $35$ in $\text{GL}(V)$, which is exactly the dimension of $\Lambda^3(V^\ast)$, one sees that this definition of $G_2$ is the most efficient that it can possibly be. (Over $\mathbb{C}$, the stabilizer of the generic element of $\Lambda^3(V^\ast)$ turns out to be $G_2$ crossed with the cube roots of unity, so the identity component is still the right group, you just have to require in addition that it fix a volumme form on $V$, so that you wind up with $36$ equations to define the subgroup of codimension $35$.) For the other exceptional groups, there are similarly more efficient descriptions than as automorphisms of algebras. Cartan himself described $F_4$, $E_6$, and $E_7$ in their representations of minimal dimsension as stabilizers of homogeneous polynomials (which he wrote down explicitly) on vector spaces of dimension $26$, $27$, and $56$ of degrees $3$, $3$, and $4$, respectively. There is no doubt that, in the case of $F_4$, this is much more efficient (in the above sense) than the traditional definition as automorphisms of the exceptional Jordan algebra. In the $E_6$ case, this is the standard definition. I think that, even in the $E_7$ case, it's better than the one provided by the 'magic square' construction. In the case of $E_8\subset\text{GL}(248)$, it turns out that $E_8$ is the stabilizer of a certain element $\mu\in \Lambda^3\bigl((\mathbb{R}^{248})^\ast\bigr)$, which is essentially the Cartan $3$-form on on the Lie algebra of $E_8$. I have a feeling that this is the most 'efficient' description of $E_8$ there is (in the above sense). This last remark is a special case of a more general phenomenon that seems to have been observed by many different people, but I don't know where it is explicitly written down in the literature: If $G$ is a simple Lie group of dimension bigger than $3$, then $G\subset\text{GL}({\frak{g}})$ is the identity component of the stabilizer of the Cartan $3$-form $\mu_{\frak{g}}\in\Lambda^3({\frak{g}}^\ast)$. Thus, you can recover the Lie algebra of $G$ from knowledge of its Cartan $3$-form alone. On 'rolling distributions': You mentioned the description of $G_2$ in terms of 'rolling distributions', which is, of course, the very first description (1894), by Cartan and Engel (independently), of this group. They show that the Lie algebra of vector fields in dimension $5$ whose flows preserve the $2$-plane field defined by
$$
dx_1 - x_2\ dx_0 = dx_2 - x_3\ dx_0 = dx_4 - {x_3}^2\ dx_0 = 0
$$
is a $14$-dimensional Lie algebra of type $G_2$. (If the coefficients are $\mathbb{R}$, this is the split $G_2$.) It is hard to imagine a simpler definition than this. However, I'm inclined not to regard it as all that 'simple', just because it's not so easy to get the defining equations from this and, moreover, the vector fields aren't complete. In order to get complete vector fields, you have to take this $5$-dimensional affine space as a chart on a $5$-dimensional compact manifold. (Cartan actually did this step in 1894, as well, but that would take a bit more description.) Since $G_2$ does not have any homogeneous spaces of dimension less than $5$, there is, in some sense, no 'simpler' way for $G_2$ to appear. What doesn't seem to be often mentioned is that Cartan also described the other exceptional groups as automorphisms of plane fields in this way as well. For example, he shows that the Lie algebra of $F_4$ is realized as the vector fields whose flows preserve a certain 8-plane field in 15-dimensional space. There are corresponding descriptions of the other exceptional algebras as stabilizers of plane fields in other dimensions. K. Yamaguchi has classified these examples and, in each case, writing down explicit formulae turns out to be not difficult at all. Certainly, in each case, writing down the defining equations in this way takes less time and space than any of the algebraic methods known. Further remark: Just so this won't seem too mysterious, let me describe how this goes in general: Let $G$ be a simple Lie group, and let $P\subset G$ be a parabolic subgroup. Let $M = G/P$. Then the action of $P$ on the tangent space of $M$ at $[e] = eP\in M$ will generally preserve a filtration
$$
(0) = V_0 \subset V_1\subset V_2\subset \cdots \subset V_{k-1} \subset V_k = T_{[e]}M
$$
such that each of the quotients $V_{i+1}/V_i$ is an irreducible representation of $P$. Corresponding to this will be a set of $G$-invariant plane fields $D_i\subset TM$ with the property that $D_i\bigl([e]\bigr) = V_i$. What Yamaguchi shows is that, in many cases (he determines the exact conditions, which I won't write down here), the group of diffeomorphisms of $M$ that preserve $D_1$ is $G$ or else has $G$ as its identity component. What Cartan does is choose $P$ carefully so that the dimension of $G/P$ is minimal among those that satisfy these conditions to have a nontrivial $D_1$. He then takes a nilpotent subgroup $N\subset G$ such that $T_eG = T_eP \oplus T_eN$ and uses the natural immersion $N\to G/P$ to pull back the plane field $D_1$ to be a left-invariant plane field on $N$ that can be described very simply in terms of the multiplication in the nilpotent group $N$ (which is diffeomorphic to some $\mathbb{R}^n$). Then he verifies that the Lie algebra of vector fields on $N$ that preserve this left-invariant plane field is isomorphic to the Lie algebra of $G$. This plane field on $N$ is bracket generating, i.e., 'non-holonomic' in the classical terminology. This is why it gets called a 'rolling distribution' in some literature. In the case of the exceptional groups $G_2$ and $F_4$, the parabolic $P$ is of maximal dimension, but this is not so in the case of $E_6$, $E_7$, and $E_8$, if I remember correctly.
|
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|
99,750 |
Let $G$ be a reductive group, $F$ a Frobenius morphism, $B$ a Borel subgroup $F$-stable and consider the finite groups $G^F$ and $U^F$ where $U$ is the radical unipotent of $B=UT$ ($T$ torus). I would like a reference for the description of the algebra $End_{G^F}( \mathbb{C}[G^F/U^F] )$. More precisely, I'd like to relate it with a structure of Hecke algebra, which is usually defined as $End_{G^F}( \mathbb{C}[G^F/B^F] ) := End_{G^F} ( Ind_{B^F}^{G^F} 1 )$. I hope to find that the endomorphism algebra is isomorphic to some kind of extension of the Hecke algebra by the torus $T$. Thank you!
|
It is not always clear what one means by 'the simplest description' of one of the exceptional Lie groups. In the examples you've given above, you quote descriptions of these groups as automorphisms of algebraic structures, and that's certainly a good way to do it, but that's not the only way, and one can argue that they are not the simplest in terms of a very natural criterion, which I'll now describe: Say that you want to describe a subgroup $G\subset \text{GL}(V)$ where $V$ is a vector space (let's not worry too much about the ground field, but, if you like, take it to be $\mathbb{R}$ or $\mathbb{C}$ for the purposes of this discussion). One would like to be able to describe $G$ as the stabilizer of some element $\Phi\in\text{T}(V{\oplus}V^\ast)$, where $\mathsf{T}(W)$ is the tensor algebra of $W$. The tensor algebra $\mathsf{T}(V{\oplus}V^\ast)$ is reducible under $\text{GL}(V)$, of course, and, ideally, one would like to be able to chose a 'simple' defining $\Phi$, i.e., one that lies in some $\text{GL}(V)$-irreducible submodule $\mathsf{S}(V)\subset\mathsf{T}(V{\oplus}V^\ast)$. Now, all of the classical groups are defined in this way, and, in some sense, these descriptions are as simple as possible. For example, if $V$ with $\dim V = 2m$ has a symplectic structure $\omega\in \Lambda^2(V^\ast)$, then the classical group $\text{Sp}(\omega)\subset\text{GL}(V)$ has codimension $m(2m{-}1)$ in $\text{GL}(V)$, which is exactly the dimension of the space $\Lambda^2(V^\ast)$. Thus, the condition of stabilizing $\omega$ provides exactly the number of equations one needs to cut out $\text{Sp}(\omega)$ in $\text{GL}(V)$. Similarly, the standard definitions of the other classical groups as subgroups of linear transformations that stabilize an element in a $\text{GL}(V)$-irreducible subspace of $\mathsf{T}(V{\oplus}V^\ast)$ are as 'efficient' as possible. In another direction, if $V$ has the structure of an algebra, one can regard the multiplication as an element $\mu\in \text{Hom}\bigl(V\otimes V,V\bigr)= V^\ast\otimes V^\ast \otimes V$, and the automorphisms of the algebra $A = (V,\mu)$ are, by definition, the elements of $\text{GL}(V)$ whose extensions to $V^\ast\otimes V^\ast \otimes V$ fix the element $\mu$. Sometimes, if one knows that the multiplication is symmetric or skew-symmetric and/or traceless, one can regard $\mu$ as an element of a smaller vector space, such as $\Lambda^2(V^\ast)\otimes V$ or even the $\text{GL}(V)$-irreducible module $\bigl[\Lambda^2(V^\ast)\otimes V\bigr]_0$, i.e., the kernel of the natural contraction mapping $\Lambda^2(V^\ast)\otimes V\to V^\ast$. This is the now-traditional definition of $G_2$, the simple Lie group of dimension $14$: One takes $V = \text{Im}\mathbb{O}\simeq \mathbb{R}^7$ and defines $G_2\subset \text{GL}(V)$ as the stabilizer of the vector cross-product $\mu\in \bigl[\Lambda^2(V^\ast)\otimes V\bigr]_0\simeq \mathbb{R}^{140}$. Note that the condition of stabilizing $\mu$ is essentially $140$ equations on elements of $\text{GL}(V)$ (which has dimension $49$), so this is many more equations than one would really need. (If you don't throw away the subspace defined by the identity element in $\mathbb{O}$, the excess of equations needed to define $G_2$ as a subgroup of $\text{GL}(\mathbb{O})$ is even greater.) However, as was discovered by Engel and Reichel more than 100 years ago, one can define $G_2$ over $\mathbb{R}$ much more efficiently: Taking $V$ to have dimension $7$, there is an element $\phi\in \Lambda^3(V^\ast)$ such that $G_2$ is the stabilizer of $\phi$. In fact, since $G_2$ has codimension $35$ in $\text{GL}(V)$, which is exactly the dimension of $\Lambda^3(V^\ast)$, one sees that this definition of $G_2$ is the most efficient that it can possibly be. (Over $\mathbb{C}$, the stabilizer of the generic element of $\Lambda^3(V^\ast)$ turns out to be $G_2$ crossed with the cube roots of unity, so the identity component is still the right group, you just have to require in addition that it fix a volumme form on $V$, so that you wind up with $36$ equations to define the subgroup of codimension $35$.) For the other exceptional groups, there are similarly more efficient descriptions than as automorphisms of algebras. Cartan himself described $F_4$, $E_6$, and $E_7$ in their representations of minimal dimsension as stabilizers of homogeneous polynomials (which he wrote down explicitly) on vector spaces of dimension $26$, $27$, and $56$ of degrees $3$, $3$, and $4$, respectively. There is no doubt that, in the case of $F_4$, this is much more efficient (in the above sense) than the traditional definition as automorphisms of the exceptional Jordan algebra. In the $E_6$ case, this is the standard definition. I think that, even in the $E_7$ case, it's better than the one provided by the 'magic square' construction. In the case of $E_8\subset\text{GL}(248)$, it turns out that $E_8$ is the stabilizer of a certain element $\mu\in \Lambda^3\bigl((\mathbb{R}^{248})^\ast\bigr)$, which is essentially the Cartan $3$-form on on the Lie algebra of $E_8$. I have a feeling that this is the most 'efficient' description of $E_8$ there is (in the above sense). This last remark is a special case of a more general phenomenon that seems to have been observed by many different people, but I don't know where it is explicitly written down in the literature: If $G$ is a simple Lie group of dimension bigger than $3$, then $G\subset\text{GL}({\frak{g}})$ is the identity component of the stabilizer of the Cartan $3$-form $\mu_{\frak{g}}\in\Lambda^3({\frak{g}}^\ast)$. Thus, you can recover the Lie algebra of $G$ from knowledge of its Cartan $3$-form alone. On 'rolling distributions': You mentioned the description of $G_2$ in terms of 'rolling distributions', which is, of course, the very first description (1894), by Cartan and Engel (independently), of this group. They show that the Lie algebra of vector fields in dimension $5$ whose flows preserve the $2$-plane field defined by
$$
dx_1 - x_2\ dx_0 = dx_2 - x_3\ dx_0 = dx_4 - {x_3}^2\ dx_0 = 0
$$
is a $14$-dimensional Lie algebra of type $G_2$. (If the coefficients are $\mathbb{R}$, this is the split $G_2$.) It is hard to imagine a simpler definition than this. However, I'm inclined not to regard it as all that 'simple', just because it's not so easy to get the defining equations from this and, moreover, the vector fields aren't complete. In order to get complete vector fields, you have to take this $5$-dimensional affine space as a chart on a $5$-dimensional compact manifold. (Cartan actually did this step in 1894, as well, but that would take a bit more description.) Since $G_2$ does not have any homogeneous spaces of dimension less than $5$, there is, in some sense, no 'simpler' way for $G_2$ to appear. What doesn't seem to be often mentioned is that Cartan also described the other exceptional groups as automorphisms of plane fields in this way as well. For example, he shows that the Lie algebra of $F_4$ is realized as the vector fields whose flows preserve a certain 8-plane field in 15-dimensional space. There are corresponding descriptions of the other exceptional algebras as stabilizers of plane fields in other dimensions. K. Yamaguchi has classified these examples and, in each case, writing down explicit formulae turns out to be not difficult at all. Certainly, in each case, writing down the defining equations in this way takes less time and space than any of the algebraic methods known. Further remark: Just so this won't seem too mysterious, let me describe how this goes in general: Let $G$ be a simple Lie group, and let $P\subset G$ be a parabolic subgroup. Let $M = G/P$. Then the action of $P$ on the tangent space of $M$ at $[e] = eP\in M$ will generally preserve a filtration
$$
(0) = V_0 \subset V_1\subset V_2\subset \cdots \subset V_{k-1} \subset V_k = T_{[e]}M
$$
such that each of the quotients $V_{i+1}/V_i$ is an irreducible representation of $P$. Corresponding to this will be a set of $G$-invariant plane fields $D_i\subset TM$ with the property that $D_i\bigl([e]\bigr) = V_i$. What Yamaguchi shows is that, in many cases (he determines the exact conditions, which I won't write down here), the group of diffeomorphisms of $M$ that preserve $D_1$ is $G$ or else has $G$ as its identity component. What Cartan does is choose $P$ carefully so that the dimension of $G/P$ is minimal among those that satisfy these conditions to have a nontrivial $D_1$. He then takes a nilpotent subgroup $N\subset G$ such that $T_eG = T_eP \oplus T_eN$ and uses the natural immersion $N\to G/P$ to pull back the plane field $D_1$ to be a left-invariant plane field on $N$ that can be described very simply in terms of the multiplication in the nilpotent group $N$ (which is diffeomorphic to some $\mathbb{R}^n$). Then he verifies that the Lie algebra of vector fields on $N$ that preserve this left-invariant plane field is isomorphic to the Lie algebra of $G$. This plane field on $N$ is bracket generating, i.e., 'non-holonomic' in the classical terminology. This is why it gets called a 'rolling distribution' in some literature. In the case of the exceptional groups $G_2$ and $F_4$, the parabolic $P$ is of maximal dimension, but this is not so in the case of $E_6$, $E_7$, and $E_8$, if I remember correctly.
|
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|
100,033 |
I am looking for mathematical documentaries, both technical and non-technical. They should be "interesting" in that they present either actual mathematics, mathematicians or history of mathematics. I am in charge of nourishing our departmental math library (PUCV) and I would like to add this kind of material in order to attract undergraduates toward mathematics. For this reason, I am not looking for videos of conferences or seminar talks, but rather for introductory or "wide public" material. Here are some good examples. "Dimensions", by Leys, Ghys & Alvarez, http://www.dimensions-math.org/ which explains actual maths and is excellent. "Julia Robinson and Hilbert's tenth problem", https://www.vismath.eu/en/films/julia-robinson , about the life of some great mathematicians. BBC documentary on "Fermat's last theorem" (by the way, any information about how to purchase it would be welcome, it does not seem to be possible to do it from the BBC site http://www.bbc.co.uk/iplayer/episode/b0074rxx/horizon-19951996-fermats-last-theorem . Maybe http://vimeo.com/18216532 ?). Are there more examples? Thanks, Ricardo.
|
$N$ Is a Number: A Portrait of Paul Erdős . ( Official site ; Wikipedia link )
|
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|
100,265 |
Question: I'm asking for a big list of not especially famous, long open problems that anyone can understand. Community wiki, so one problem per answer, please. Motivation: I plan to use this list in my teaching, to motivate general education undergraduates, and early year majors, suggesting to them an idea of what research mathematicians do. Meaning of "not too famous" Examples of problems that are too famous might be the Goldbach conjecture, the $3x+1$ -problem, the twin-prime conjecture, or the chromatic number of the unit-distance graph on ${\Bbb R}^2$ . Roughly, if there exists a whole monograph already dedicated to the problem (or narrow circle of problems), no need to mention it again here. I'm looking for problems that, with high probability, a mathematician working outside the particular area has never encountered. Meaning of: anyone can understand The statement (in some appropriate, but reasonably terse formulation) shouldn't involve concepts beyond high school (American K-12) mathematics. For example, if it weren't already too famous, I would say that the conjecture that "finite projective planes have prime power order" does have barely acceptable articulations. Meaning of: long open The problem should occur in the literature or have a solid history as folklore. So I do not mean to call here for the invention of new problems or to collect everybody's laundry list of private-research-impeding unproved elementary technical lemmas. There should already exist at least of small community of mathematicians who will care if one of these problems gets solved. I hope I have reduced subjectivity to a minimum, but I can't eliminate all fuzziness -- so if in doubt please don't hesitate to post! To get started, here's a problem that I only learned of recently and that I've actually enjoyed describing to general education students. https://en.wikipedia.org/wiki/Union-closed_sets_conjecture Edit: I'm primarily interested in conjectures - yes-no questions, rather than classification problems, quests for algorithms, etc.
|
One problem which I think is mentioned in Guy's book is the integer block problem: does there exist a cuboid (aka "brick") where the width, height, breadth, length of diagonals on each face, and the length of the main diagonal are all integers? update 2012-07-12 Since the question has returned to the front page, I'm taking the liberty to add some links that I found after Scott Carnahan's comments. (Scott deserves the credit, really, but I thought the links belonged in the answer rather than in the comments.) On perfect cuboids , by Ronald van Luijk, master thesis, 2000. The surface parametrizing cuboids , by Michael Stoll and Damiano Testa, arXiv.org:1009.0388.
|
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|
100,281 |
Hello, I ask myself, whether the curvature determines the metric. Concretely: Given a compact Riemannian manifold $M$, are there two metrics $g_1$ and $g_2$, which are not everywhere flat, such that they are
not isometric to one another, but that there is a diffeomorphism which preserves the curvature? If the answer is yes: Can we chose $M$ to be a compact 2-manifold? Can we classify manifolds, where the curvature determines the metric? What happens, if we also allow semi-riemannian manifolds for the above questions? Thank you for help.
|
For concrete 2-dimensional counter-example see page 328 of Kulkarni's paper "Curvature and metric", Annals of Math, 1970. Weinstein's argument there shows that every Riemannian surface provides a counter-example (using flow orthogonal to the gradient of the curvature). On the positive side, if $M$ is compact of dimension $\ge 3$ and has nowhere constant sectional curvature, then combination of results of Kulkarni and Yau show that
a diffeomorphism preserving sectional curvature is necessarily an isometry. Concerning 2-dimensional counter-examples: First of all, every surface which admits an open subset where curvature is (nonzero) constant would obviously yield a counter-example. Thus, I will assume now that curvature is nowhere constant. Kulkarni refers to Kreyszig's "Introduction to Differential Geometry and Riemannian Geometry", p. 164, for a counter-example attributed to Stackel and Wangerin. You probably can get the book through interlibrary loan if you are in the US. Here is the Weinstein's argument. Consider a Riemannian surface $(S,g)$ and let $K$ denote the curvature function. Let $X$ be a nonzero vector field on $S$ orthogonal to the gradient field of $K$. (Such $X$ always exists.) Now, consider the flow $F_t$ along $X$. Then $F_t$ will preserve the curvature but will not be (for most metrics $g$ and vector fields $X$) isometric. For instance, $F_t$ cannot be isometric if genus of $S$ is at least $2$ or if $X$ has more than 2 zeros.
|
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|
100,493 |
There is a (very) long essay by Grothendieck with the ominous title La Longue Marche à travers la théorie de Galois (The Long March through Galois Theory). As usual, Grothendieck knew what he was talking about: Galois Theory, far from being confined to its primary example, namely field extensions, is pervasive throughout mathematics, and still to be fully understood. Ever since I first heard of forcing I was struck by its compelling analogy with field extensions: the ground model (read Q), the generic G (read a new element, say $\sqrt{2}$), the new M[G] (read
Q[$\sqrt{2}$]), etc. I quote Joel Hamkins 's words here, in his sparkling paper on the multiverse : In effect, the forcing extension has
adjoined the “ideal” object G to V ,
in much the same way that one might
build a field extension such as
Q[$\sqrt{2}$] Of course, matters are a bit more complicated in set theory, you have to make sure the extension satisfies the axioms, that it does not "bother" the ordinals, and so on. Yet, one cannot really think that the analogy stops here. And it does not: in Galois theory, the main thing is the central theorem, establishing the Galois Connection between subfields of the extension and the subgroups of the Galois group, ie the group of the automorphisms of the extension leaving fixed the underlying field. No Galois connection, no Galois Theory. But wait, first we need the group. So where is it? A hint is in the great classic result by Jech-Sochor on showing the independence of AC: by considering a group of automorphisms of P, the ordered set of forcing conditions, one can obtain a new model which is (essentially) the set of fixed points of the induced automorphisms. This is even clearer when one looks at it from the point of view of boolean valued models: each automorphism of the boolean algebra induces an automorphism of the extended universe. Now my question: is there some
systematic work on classifying forcing
extensions by their Galois group? Can one develop a full machinery which will apply to relative extensions? NOTE: I think this is no idle brooding: someone for instance has asked here on MO about the 2-category of the multiverse. That is a tough question, and my sense is that before giving it a satisfactory answer some preliminary work needs to be done. Which work? Well, one needs to re-think the classical set constructions from a structural standpoint , leaving behind its gory technical details. Now, forcing is a huge part of the multiverse, and understanding the structural algebra underpinning it would be, I trust, a huge step toward an algebraic understanding of the multiverse.
|
Yes and No... There are strong parallels between forcing and symmetric extensions and field extensions and this way of thinking has been fruitful. However, like in the case of general ring extensions and group extensions and similar problems, this analogy is not perfect and pushing the similarity too far may actually obscure what is really going on. That said, symmetric extensions do indeed show a great deal of similarity with Galois theory. Some work, notably that of Serge Grigorieff [ Intermediate Submodels and Generic Extensions in Set Theory , Annals of Mathematics 101 (1975), 447–490] shows that there is indeed a way to understand intermediate forcing extensions in a manner extremely similar to the way understand field extensions through Galois theory. Some have even pushed this analogy so far as to study some problems roughly analogous to the Inverse Galois Problem in this context, for example [Groszek and Laver, Finite groups of OD-conjugates , Period. Math. Hungar. 18 (1987), 87–97]. There are some very algebraic ways of understanding forcing and symmetric models in a more global sense. For example, forcing extensions correspond in a well-understood way to the category of complete Boolean algebras under complete embeddings. Moreover, the automorphism groups of these algebras plays a crucial role in our understanding of the inner model structure of forcing extensions. An even farther reaching approach comes from transposing the sheaf constructions from topos theory into the set-theoretic context [Blass and Scedrov, Freyd's models for the independence of the axiom of choice , Mem. Amer. Math. Soc. 79 (1989), no. 404]. One could argue that this suggests a stronger analogy with topology rather than algebra, but there are plenty of very deep analogies between Galois theory and topology. The above is not a complete survey of these types of connections, it is just to demonstrate that connections do exist and that they have been looked at and useful for a long time. Because of the relative sparsity of the literature, one could argue that these aspects are underdeveloped but that is hasty judgement. The truth is that there appear to be disappointingly few practical aspects to this kind of approach, perhaps because they are not relevant to most current questions in set theory or perhaps for deeper reasons. For example, the inner model structure of the first (and largely regarded as the simplest) forcing extension, namely the simple Cohen extension, is incredibly rich and complex [Abraham and Shore, The degrees of constructibility of Cohen reals , Proc. London Math. Soc. 53 (1986), 193–208] and there does not appear to be a reasonable higher-level approach that may help us sort through this quagmire in a similar way that Galois theory can help us sort through the complex structure of $\overline{\mathbb{Q}}$.
|
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|
100,565 |
I admit freely that the following question is a bit of a fishing expedition inspired by this lovely "definition" of a module as found on Wikipedia : A module is a ring action on an abelian group. It takes a while for the novice to unzip this definition into the usual list of identities, but for someone with a basic understanding of (group) actions who needs a quick shorthand to remember what being a module entails, it seems hard to beat this definition. Now, volumes have been written about group actions. This makes sense, because set theory underlies modern mathematics and for any set $X$ , the automorphisms $\operatorname{Aut}(X)$ naturally have a group structure. And of course, a group $G$ acting on $X$ is just a group homomorphism $G \to \operatorname{Aut}X$ . On the other hand — aside from this definition of a module — it is hard to come across a general theory of ring actions. In order to find interesting actions of a ring $R$ , one analogously needs the set of endomorphisms $\operatorname{End}X$ to have the structure of a ring so that one may search for ring morphisms $R \to \operatorname{End}X$ . What are the most general objects $O$ for which $\operatorname{End}O$ canonically has the structure of a ring? Certainly, $O$ at least contains abelian groups, but is there anything else?
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First, I am not really sure what you mean by "it is hard to come across a general theory of ring actions." This is precisely module theory! If $f : R \to S$ is any ring homomorphism whatsoever, then composition with the natural map $S \to \text{End}(S)$ (where $S$ is regarded as an abelian group and $S$ acts on this abelian group by left multiplication; this proves "Cayley's theorem for rings") gives $S$ the structure of a left $R$ -module. Anyway, the objects you're looking for are the objects in categories such that Hom-sets canonically have the structure of an abelian group. These are the categories enriched over $(\text{Ab}, \otimes)$ , or $\text{Ab}$ -enriched categories. As Will Sawin says, a more traditional name is preadditive category, but I don't like this name because I don't think I should have to remember the distinction between preadditive, additive, and abelian categories to refer to something as fundamental as $\text{Ab}$ -enrichment. $\text{Ab}$ -enriched categories are abundant. In fact, any ordinary category $C$ has a free $\text{Ab}$ -enriched category $\mathbb{Z}[C]$ equipped with the universal functor from $C$ to an $\text{Ab}$ -enriched category. This functor $\mathbb{Z}[-]$ is left adjoint to the forgetful functor from $\text{Ab}$ -enriched categories to ordinary categories and can be explicitly described by taking free abelian groups on hom-sets. This is a slight generalization of the point Scott Carnahan makes in the comments, as this functor is monoidal, so sends monoids to their monoid algebras and sends monoid actions to ring actions. More generally, let $V$ be any monoidal category . Then one can define $V$ -enriched categories, and a $V$ -enriched category with one object (more precisely the monoid of endomorphisms of such a category) is precisely a monoid object in $V$ , and one gets various generalizations of monoids and rings this way. For example: A $\text{Set}$ -enriched category with one object is a monoid. A $\text{Top}$ -enriched category with one object is a topological monoid. An $\text{Ab}$ -enriched category with one object is a ring. A $\text{Vect}$ -enriched category with one object is a (unital, associative) algebra. A $\text{Ban}$ -enriched category with one object is a (unital) Banach algebra. A $\text{Ch}$ -enriched category with one object is a dg-algebra. And so forth. (Note that $V$ itself need not be $V$ -enriched; I believe this condition is equivalent to $V$ being closed monoidal ). It is also possible to give a uniform definition of what it means for a $V$ -monoid $A$ to act on an element $M$ of $V$ in terms of a map $A \otimes M \to M$ satisfying the usual axioms (this circumvents the need for $V$ to be $V$ -enriched) which reproduces the notion of action of a monoid, module of a ring, representation of an algebra, etc. Just as it is fruitful to generalize groups to groupoids by allowing multiple objects, one can generalize monoids to monoidoids (that is, categories!) in any of the above settings by allowing multiple objects. From this perspective, an $\text{Ab}$ -enriched category is just a ringoid : a ring with many objects! Thinking in this way makes certain aspects of ring theory look more natural, I think. For example, for certain nice $V$ there is a notion of Cauchy completion of a $V$ -enriched category $C$ generalizing both the notion of Karoubi envelope and the notion of Cauchy completion of a metric space. The Cauchy completion of an $\text{Ab}$ -enriched category is obtained by formally adjoining finite biproducts and then splitting idempotents. Now: The Cauchy completion of a ring $R$ (as an $\text{Ab}$ -enriched category) is the category of finitely-generated projective right $R$ -modules. This result gives a very conceptual approach to Morita equivalence : as it turns out, two rings are Morita equivalent if and only if their Cauchy completions are equivalent!
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|
100,577 |
We know that the intersection of a smooth cubic surface in $\mathbb{P}^3$ with a plane is a cubic, a line+a conic or three disjoint lines. Can this intersection consist of a conic and its tangent line?
|
First, I am not really sure what you mean by "it is hard to come across a general theory of ring actions." This is precisely module theory! If $f : R \to S$ is any ring homomorphism whatsoever, then composition with the natural map $S \to \text{End}(S)$ (where $S$ is regarded as an abelian group and $S$ acts on this abelian group by left multiplication; this proves "Cayley's theorem for rings") gives $S$ the structure of a left $R$ -module. Anyway, the objects you're looking for are the objects in categories such that Hom-sets canonically have the structure of an abelian group. These are the categories enriched over $(\text{Ab}, \otimes)$ , or $\text{Ab}$ -enriched categories. As Will Sawin says, a more traditional name is preadditive category, but I don't like this name because I don't think I should have to remember the distinction between preadditive, additive, and abelian categories to refer to something as fundamental as $\text{Ab}$ -enrichment. $\text{Ab}$ -enriched categories are abundant. In fact, any ordinary category $C$ has a free $\text{Ab}$ -enriched category $\mathbb{Z}[C]$ equipped with the universal functor from $C$ to an $\text{Ab}$ -enriched category. This functor $\mathbb{Z}[-]$ is left adjoint to the forgetful functor from $\text{Ab}$ -enriched categories to ordinary categories and can be explicitly described by taking free abelian groups on hom-sets. This is a slight generalization of the point Scott Carnahan makes in the comments, as this functor is monoidal, so sends monoids to their monoid algebras and sends monoid actions to ring actions. More generally, let $V$ be any monoidal category . Then one can define $V$ -enriched categories, and a $V$ -enriched category with one object (more precisely the monoid of endomorphisms of such a category) is precisely a monoid object in $V$ , and one gets various generalizations of monoids and rings this way. For example: A $\text{Set}$ -enriched category with one object is a monoid. A $\text{Top}$ -enriched category with one object is a topological monoid. An $\text{Ab}$ -enriched category with one object is a ring. A $\text{Vect}$ -enriched category with one object is a (unital, associative) algebra. A $\text{Ban}$ -enriched category with one object is a (unital) Banach algebra. A $\text{Ch}$ -enriched category with one object is a dg-algebra. And so forth. (Note that $V$ itself need not be $V$ -enriched; I believe this condition is equivalent to $V$ being closed monoidal ). It is also possible to give a uniform definition of what it means for a $V$ -monoid $A$ to act on an element $M$ of $V$ in terms of a map $A \otimes M \to M$ satisfying the usual axioms (this circumvents the need for $V$ to be $V$ -enriched) which reproduces the notion of action of a monoid, module of a ring, representation of an algebra, etc. Just as it is fruitful to generalize groups to groupoids by allowing multiple objects, one can generalize monoids to monoidoids (that is, categories!) in any of the above settings by allowing multiple objects. From this perspective, an $\text{Ab}$ -enriched category is just a ringoid : a ring with many objects! Thinking in this way makes certain aspects of ring theory look more natural, I think. For example, for certain nice $V$ there is a notion of Cauchy completion of a $V$ -enriched category $C$ generalizing both the notion of Karoubi envelope and the notion of Cauchy completion of a metric space. The Cauchy completion of an $\text{Ab}$ -enriched category is obtained by formally adjoining finite biproducts and then splitting idempotents. Now: The Cauchy completion of a ring $R$ (as an $\text{Ab}$ -enriched category) is the category of finitely-generated projective right $R$ -modules. This result gives a very conceptual approach to Morita equivalence : as it turns out, two rings are Morita equivalent if and only if their Cauchy completions are equivalent!
|
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|
100,596 |
Possible Duplicate: When to start reviewing I was recently asked to become a MathSciNet reviewer, which, as far as I understand, is something that they ask almost any mathematician at a certain point in their career. Should I accept? What are your motivations in both directions? Is this a service to the community that I should help with, like serving as a referee for journal papers? I'll write my own pros and cons as an answer.
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The question is natural but the answer is highly individual. For what it's worth, I'll comment from my own perspective as a now-retired reviewer who produced almost 500 reviews over many decades. In its early years the print version of Mathematical Reviews was still relatively slim but covered the most widely circulated journals and had as reviewers most of the active mathematicians of the time. It had the great advantage of bringing together both a print database of current literature and (often) helpful commentaries by specialists. For many decades the authors were compensated only by receiving free papers (sometimes books). As mathematics and its offshoots proliferated in the 1960s and later, it became impossible for most people to skim all reviews. But the evolving classification scheme helped, even though it could never meet all needs. Until the Internet (and arXiv) developed far enough, the reviews and database played a mostly constructive role in communication. But managing the flow of papers and editing the submitted reviews required a lot of expensive professionals, as it still does. Some reviews were of course eccentric, such as one which simply quoted verbatim half of a two page note from a widely circulated Springer journal. For me personally it was a way to keep in touch with a wider range of interesting mathematics than I actually worked on at the time. But to do the reviewing task well is time-consuming, since I always felt the need to delve into the related literature. (At least once I discovered an earlier proof in a slightly offbeat journal of a theorem published anew in a mainstream journal by an author who hadn't been aware of the earlier proof.) Sometimes you get correspondence (even arguments) from an author whose work you have reviewed. All very interesting, but optional activity like refereeing. By now MathSciNet functions mainly as an excellent database, still very expensive to maintain, and reviewers are given AMS credits for their use. Fewer papers get full reviews, which is usually the right decision but not always. People rely more for up-do-date stuff on other Internet sources, but the organized and flexibly searchable database is worth the cost for those institutions which can afford it. (Not all can.)
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|
101,014 |
In many areas of mathematics there are fundamental problems that are embarrasingly natural or simple to state, but whose solution seem so out of reach that they are barely mentioned in the literature even though most practitioners know about them. I'm specifically looking for open problems of the sort that when one first hears of them, the first reaction is to say: that's not known ??!! As examples, I'll mention three problems in geometry that I
think fall in this category and I hope that people will pitch in either more problems of this type, or direct me to the literature where these problems are studied . The first two problems are "holy grails" of systolic geometry---the study of inequalities involving the volume of a Riemannian manifold and the length of its shortest periodic geodesic---, the third problem is one of the Busemann-Petty problems and, to my mind, one of the prettiest open problems in affine convex geometry. Systolic geometry of simply-connected manifolds. Does there exist a constant $C > 0$ so that for every Riemannian metric $g$ on the three-sphere, the volume of $(S^3,g)$ is bounded below by the cube of the length of its shortest periodic geodesic times the constant $C$ ? Comments. For the two-sphere this is a theorem of Croke. Another basic test for studying this problem is $S^1 \times S^2$ . In this case the
fundamental group is non-trivial, but in some sense it is small (i.e., the manifold is not essential in the sense of Gromov). There is a very timid hint to this problem in Gromov's Filling Riemannian manifods . Sharp systolic inequality for real projective space. If a Riemannian metric in projective three-space has the same volume as the canonical metric, but is not isometric to it, does it carry a (non-contractible) periodic geodesic of length smaller than $\pi$ ? Comments. For the real projective plane this is Pu's theorem. In his Panoramic view of Riemannian geometry , Berger hesitates in conjecturing that this is the case (he says it is not clear that this is the right way to bet). In a recent preprint with Florent Balacheff, I studied a parametric version of this problem. The results suggest that the formulation above is the right way to bet. Isoperimetry of metric balls. For what three-dimensional normed spaces are metric balls solutions of the isoperimetric inequality? Comments. In two dimensions this problem was studied by Radon. There are plenty of norms on the plane for which metric discs are solutions of the isoperimetric problem. For example, the normed plane for which the disc is a regular hexagon. This is one of the Busemann-Petty problems. The volume and area are defined using the Hausdorff $2$ and $3$ -dimensional measure. I have not seen any partial solution, even of the most modest kind, to this problem. Busemann and Petty gave a beautiful elementary interpretation of this problem: Take a convex body symmetric about the origin and a plane supporting it at some point $x$ . Translate the plane to the origin, intersect it with the body, and consider the solid cone formed by this central section and the point $x$ . The conjecture is that if the
volume of all cones formed in this way is always the same, then the body is an ellipsoid. Additional problem: I had forgotten another beautiful problem from the paper of Busemann and Petty: Problems on convex bodies , Mathematica Scandinavica 4: 88–94. Minimality of flats in normed spaces. Given a closed $k$ -dimensional polyhedron in an $n$ -dimensional normed space with $n > k$ , is it true that the area (taken as $k$ -dimensional Hausdorff measure) of any facet does not exceed the sum of the areas of the remaining facets? Comments. When $n = k + 1$ this is a celebrated theorem of Busemann, which convex geometers are more likely to recognize in the following form: the intersection body of a centrally symmetric convex body is convex. A nice proof and a deep extension of this theorem was given by G. Berck in Convexity of Lp-intersection bodies , Adv. Math. 222 (2009), 920-936. When $k = 2$ this has "just" been proved by D. Burago and S. Ivanov: https://arxiv.org/abs/1204.1543 It is not true that totally geodesic submanifolds of a Finsler space (or a length metric space) are minimal for the Hausdorff measure. Berck and I gave a counter-example in What is wrong with the Hausdorff measure in Finsler spaces , Advances in Mathematics, vol. 204, no. 2, pp. 647-663, 2006.
|
Is every algebraic curve in $\mathbb P^3$ the set-theoretic intersection of two algebraic surfaces ? Not known!
|
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|
101,031 |
Is there any resource that might help non-experts gains some understanding of why
the Kervaire invariant problem remains open now only in dimension $126$? ($126 =2^7-2=2^{j+1}-2$;
whether $\theta_j=\theta_6$ exists in the "$128$-stem"), i.e., why the celebrated Hill-Hopkins-Ravenel
proof technique fails in this last remaining case?
I read a delightful exposition by Erica Klarreich ("Mathematicians solve 45-year-old Kevaire Invariant Puzzle," The Best Writing on Mathematics , 2010, 373ff, Simons Foundation link ), which piqued my interest.
But Michael Hopkins' online presentation slides announcing the result
in 2009 (" Applications of algebra to a problem in topology ") are beyond my ken.
Perhaps this an area too abstruse for all but the experts?
I'd appreciate pointers to expositions. Thanks!
|
I'll give a shot at an answer. The relevant dimensions are of the form $2^j-2$. For
$j\leq 4$, it is easy and classical that we can construct manifolds of Kervaire invariant one. The
problem was ``reduced'' from differential topology to pure stable homotopy
theory by Browder in 1969. Direct calculational methods in
homotopy theory were used by Barratt, Jones, and Mahowald to construct a
cell complex that can be used to solve the homotopy theory problem and prove
that such manifolds also exist in dimensions 30 and 62. I believe a construction
of such a manifold has been worked out in dimension 30, but that has certainly not
been done in dimension 62. Periodicity phenomena play a huge role in modern stable
homotopy theory, and a crucial feature of the Hill, Hopkins, Ravenel proof is a
periodicity of order $2^8 = 256$. That enables them to solve the stable homotopy problem
and prove there is no manifold of Kervaire invariant one for $j\geq 8$. The reasons $j=7$
is so hard are several. Nobody has a really good reason for guessing which way the answer
will go. There is no reason to expect a relevant periodicity of order $2^7$. Direct calculation
of the Adams spectral sequence through dimension $126$ is just plain hard: the
calculations blow up. There is a chance that the methodology of Barratt, Jones,
and Mahowald might extend to prove existence (if that is how the answer turns out!),
but it will probably be much harder to prove nonexistence (if that is the answer).
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101,043 |
I am interested in seeing if and how Morse Theory can "do everything". Some core things are handle decomposition, Bott periodicity, and Euler characteristic. But what do the normal (co)homology operations look like from Morse Homology? Poincare duality $H_*(M)\cong H^{n-\ast}(M)$ is the symmetry $f\to -f$ , i.e. the reversal of flowlines. Cup product $H^i(M)\otimes H^j(M)\to H^{i+j}(M)$ is given by counting Y-shaped flowlines, using Morse functions along each of the three edges. The cap product is connected to the above two. Kunneth isomorphism $H_\ast(M\times N)\cong H_\ast(M)\otimes H_\ast(N)$ is combining flowlines from $f_1:M\to\mathbb{R}$ and $f_2:N\to \mathbb{R}$ to get flowlines for $f_1+f_2:M\times N\to\mathbb{R}$ . Leray-Serre spectral sequence : pull back a Morse function on the base (flowlines of total space project onto flowlines of base space) and use a filtration by ordering the critical-point indices. Does someone know what goes on for the following? Slant product Alexander duality Steenrod operations (in particular, the Cartan formula ) Massey triple product My guess for (4) is counting X-shaped flowlines, and then I get suspicious about its relation to $A_\infty$ -structures from Lagrangian-Intersection Floer homology. [[Edit]] There was a MathOverflow post for (2), here . Alexander duality $H_\ast(S^n-M)\cong H^{n-1-\ast}(M)$ arises by taking a height function on $S^n$ and perturbing it to become Morse on the subspace $M\subset S^n$ , and then separating the critical points according to its tubular neighborhood and its complement. [[Edit]] Cohen and Schwarz' paper "A Morse Theoretic Description of String Topology" provides the relative cohomology and the Thom isomorphism , as well as homomorphisms arising from proper embeddings of submanifolds.
|
Massey products are discussed in Section 1.3 of Fukaya, Kenji. Morse homotopy,
$A_{\infty}$-category, and Floer
homologies. Proceedings of GARC
Workshop on Geometry and Topology '93
(Seoul, 1993), 1--102, available here (pdf). The Massey products are obtained by counting gradient flow graphs with four external edges and one (finite-, possibly zero-, length) internal edge. Fukaya sketches a construction of an $A_{\infty}$ category whose objects are Morse functions $f$ and with morphisms from $f$ to $g$ given by the Morse chain complex of $f-g$. In particular the Massey products can then be seen as arising from the $A_{\infty}$ structure in a standard formal way. There is a relation to Lagrangian Floer theory: a Morse function $f:M\to \mathbb{R}$ corresponds to a Lagrangian submanifold $graph(df)$ of $T^*M$ and intersections between $graph(df)$ and $graph(dg)$ are in obvious bijection with critical points of $f-g$. There are results, pioneered by Fukaya, Kenji; Oh, Yong-Geun.
Zero-loop open strings in the
cotangent bundle and Morse homotopy.
Asian J. Math. 1 (1997), no. 1,
96--180, which relate the gradient flow graphs appearing in the Morse $A_{\infty}$ operations to the holomorphic curves appearing in the Lagrangian Floer $A_{\infty}$ operations.
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|
101,160 |
In Peano Arithmetic, the induction axiom states that there is no proper subset of the natural numbers that contains 0 and is closed under the successor function. This is intended to rule out the possibility of extra natural numbers beyond the familiar ones. It doesn't accomplish that goal; there remains the possibility that other natural numbers exist and the familiar ones do not form a set. In Internal Set Theory (IST), which is an extension of ZFC that is consistent relative to ZFC, there is a distinction between standard and nonstandard sets, and it can be shown that (1) 0 is standard; (2) if $n \in \mathbb{N}$ is standard, then so is its successor; (3) $\mathbb{N}$ has nonstandard elements. The induction axiom is not violated because the standard natural numbers do not form a set. Is there a way to axiomatize set theory so that no such nonstandard natural numbers can exist? (Note: this question is not about nonstandard models of arithmetic. In IST, $\mathbb{N}$ is a standard set, and within a given model of IST, all models of second-order Peano arithmetic are isomorphic to $\mathbb{N}$.)
|
As long as you axiomatize set theory in first-order logic, the answer to your question is no. The axioms would be consistent with each finite subset of the following set of sentences involving a new constant symbol $c$: "$c$ is a natural number" and "$c\neq n$" for each (standard name of a) natural number $n$. By compactness, there would be a model of the axioms plus all of these sentences, and in that model $c$ would denote a nonstandard natural number. On the other hand, if you're willing to go beyond first-order logic, then the answer to your question is yes. For example, in second-order logic, you can express the induction axiom as a single sentence and be confident that "set" really means arbitrary set (not "internal set" or anything like that). In other words, once you're sure that "set" has its intended meaning, the induction principle guarantees that "natural number" also has its intended meaning. (To me, this doesn't look very helpful, since the intended meaning of "set" seems more complicated than the intended meaning of "natural number".) For another example, if you're willing to use infinitary logic, then you can formulate the axiom "every natural number is equal to 0 or to 1 or to 2 or to 3, or ..."
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|
101,176 |
Many introductory texts on algebraic geometry set up some sort of algebra-geometry dictionary in which radical ideals correspond to varieties, and so on. I am wondering if there is a geometric way to think about a subalgebra in a polynomial ring? For instance for invariant rings one may develop some intuition, but what about infinitely generated subalgebras? A simple example would be the infinitely generated monomial algebra
$k[x,xy, xy^2, xy^3, \dots]$ as a subalgebra in $k[x,y]$ for some field $k$.
|
Because geometry and algebra are connected by the contavariant $\textrm{Spec}$ functor, a subalgebra corresponds to the image of a dominant morphism. So subalgebras of $k[x_1,..,x_n]$ are affine schemes with dominant maps from $\mathbb A^n_k$. Sometimes, like in your example, this is in addition a surjective morphism, so it looks like the quotient of $\mathbb A^n$ by some equivalence relation. But sometimes, the simplest example being $k[x,xy] \subset k[x,y]$, it fails to be surjective. Here there is a map from the plane to the plane whose image is the plane minus a line plus a point. Since the image is always dense, we can geometrically think of it as the quotient of $\mathbb A^n_k$ by an equivalence relation followed by a completion or partial compactification.
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|
101,395 |
Many ODE's and PDE's arising in nature have a variational formulation. An example of what I mean is the following. Classical motions are solutions $q(t)$ to Lagrange's equation
$$
\frac{d}{dt}\frac{\partial L(q,\dot q)}{\partial\dot q}=\frac{\partial L(q,\dot q)}{\partial q},
$$
and these are critical points of the functional $$
I(q)=\int L(q,\dot q)dt.
$$
Of course one needs to be precise with what considers a solution to both equations. This amounts to specifying regularity and a domain of the functional. This example is an ODE, but many PDE examples are possible as well (for example electromagnetism, or more exotic physical theories). Once one knows a variational description of the problem, many more methods are available to solve the problem. Now I do not expect that any PDE or ODE can be viewed (even formally) as a critical point of a suitable action functional. This is because this whole set up reminds me of De Rham cohomology: "which one-forms (the differential equations) are exact (that is, the $d$ of a functional)?". The last sentence is not correct, but the analogy maybe is? Anyway, my question is: Are there any criteria to determine if a given differential equation admits a variational formulation?
|
Others give useful references that discuss what is known about the answer, but no statement of the answer itself. The relevant algebraic setting is the variational bicomplex, which is discussed in the works of Anderson and others. In this setting, there are two differentials, the horizontal differential $d_H$ (representing derivatives with respect to independent variables like $t$) and the vertical differential $d_V$ (representing variational derivatives with respect to dependent variables like $q(t)$). Each of these differentials is "de Rham-like" and they anticommute with each other, which explains the cohomological flavor of the answer. A rough statement of the answer is as follows. A Lagrangian $L$ density gives rise to a set of Euler-Lagrange equations $E_i=0$ as follows:
$$ d_V L = E_i ~ d_V q^i - d_H\theta , $$
that is, the vertical 1-form $E_i ~ d_V q^i$ is vertically exact (up to a horizontally exact term $d_H \theta$). So, it is necessary for $E_i=0$ to be the Euler-Lagrange system of some Lagrangian that $E_i ~ d_V q^i$ is closed up to a horizontally exact term, namely
$$ d_V(E_i~d_V q^i) = d_H \theta' (= -d_H d_V \theta) . $$
In fact, the same condition is also sufficient , up to obstructions related to the global topology of the manifold where the dependent variables $q$ take their values. This condition was formulated already classically by Helmholtz . However, the above statement is restrictive in that it answers the question only when $E_i=0$ is already in Euler-Lagrange form. However, there are many transformations that one can apply to the system $E_i=0$ that gives an equivalent system $F_a=0$. Given only the system $F_a=0$, is it still possible to decide whether it is equivalent to some system $E_i=0$ in Euler-Lagrange form? This is the hard inverse problem (aka the multiplier problem ). The only general result that I'm aware of in that direction is this. If there exists a form $\omega$ of vertical degree 2 and horizontal degree $n-1$, where $n$ is the number of independent variables, such that it is both horizontally and vertically closed modulo the equations $F_a=0$ (namely $d_V \omega = A^a F_a$ and $d_H \omega = B^a F_a$), then there exists (again, up to global topological obstructions) a Lagrangian density $L$ whose Euler-Lagrange equations $E_i=0$ are equivalent to a subsystem of $F_a=0$. To my knowledge, the above observation first appeared in Henneaux ( AnnPhys , 1982) for ODEs and in Bridges, Hydon & Lawson ( MathProcCPS , 2010) for PDEs. The calculation demonstrating this observation is given in a bit more detail on this nLab page . ( Edit: At risk of shameless self-promotion, I'll also note that I collected these observations in a self-contained paper ( arXiv ; JMP , 2013).) It reduces the solution of the hard inverse problem to classifying all such forms $\omega$ (corresponding to the so-called characteristic cohomology of the variational bicomplex restricted to $F_a=0$ in the corresponding degree) and checking that there exists a candidate that gives rise to a Lagrangian density whose Euler-Lagrange system $E_i=0$ is equivalent to the full system $F_a=0$. The calculation of the corresponding characteristic cohomology of the system $F_a=0$ is still non-trivial, but there exist ways of attacking it, which include Vinogradov's $\mathcal{C}$-spectral sequence mentioned in other responses.
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|
101,452 |
Recall that complex $K$-theory is a cohomology theory on topological spaces, which can be described in several equivalent ways: Given a finite complex $X$, $K^0(X)$ is the Grothendieck group of vector bundles on $X$. $K^*$ is even-periodic, and this determines the entire cohomology theory. Using the tensor product of vector bundles, $K$ becomes a multiplicative cohomology theory. There is a corresponding ring spectrum. The classifying space $BU \times \mathbb{Z}$ for $K^0$ is, by a theorem of Atiyah, the space of Fredholm operators on a countably-dimensional Hilbert space. So we can think of classes in $K^0(X)$ as "families of Fredholm operators" parametrized by $X$: the "index" of such a family should be a virtual vector bundle, which connects to the previous definition. $K$-theory is an even-periodic theory, so it is complex-orientable and corresponds to a formal group on $K^0(\ast) = \mathbb{Z}$. This formal group is the multiplicative one, which turns out to be Landweber-exact. Consequently, one can construct $K$-theory directly from the formal multiplicative group (once one has the spectrum $MU$) via $K_\bullet(X) = MU_\bullet(X) \otimes_{MU_\bullet} K_\bullet$. The spectrum for $K$-theory can be obtained by taking the ring spectrum $\Sigma^\infty \mathbb{CP}^\infty_+$ (which is a ring spectrum as $\mathbb{CP}^\infty$ is a topological abelian monoid) and inverting the natural element in $\pi_2$. (This is a theorem of Snaith.) It's sort of remarkable that $K$-theory can be described both geometrically (via vector bundles or operators) or algebraically (via formal groups or Snaith's theorem). The only explanation that I can think of for this is that the correspondence between (complex-orientable) ring spectra and formal groups is given more or less in terms of Chern classes of vector bundles, so a cohomology theory built directly from vector bundles would have a good chance of furnishing a fairly simple formal group law. (One can use this sort of argument to prove Snaith's theorem, for instance.) A much less formal example of a formal group is that associated to an elliptic curve. If $E/\mathrm{Spec} R$ is an elliptic curve, then under appropriate hypothesis (Landweber exactness, or flatness of the map $\mathrm{Spec} R \to M_{1,1} \to M_{FG}$, or more concretely that $R$ is torsion-free and for each $p$, the Hasse invariant $v_1$ is a nonzerodivisor in $R/pR$) we can construct an "elliptic cohomology" theory $\mathrm{Ell}^*$ which is even-periodic and whose formal group is that of $E/R$.
The associated formal group can have height up to $2$, so we get something much more complicated than $K$-theory. It has been suggested that there should be a geometric interpretation of elliptic cohomology. This seems a lot more difficult, because the formal group law associated to an elliptic curve is less elementary than $\hat{\mathbb{G}_m}$. There are various programs (which start with Segal's survey , I believe), all of which I know nothing about, to interpret elliptic cohomology classes in terms of von Neumann algebras, loop group representations, conformal field theories, ... I can understand why a geometric interpretation of elliptic cohomology would be desirable, but it's mystifying to me why researchers in this area are concentrating on these specific objects. Is there a "high-concept" explanation for this, and motivation (to someone without a background in geometry) for how one might "believe" in these visions? Is there a reason loop groups should be "height two" where the unitary group is "height one"?
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David Roberts mentioned in his comments the relationship K-theory : spin group TMF : string groups Let me recommend the first 6 pages of my unfinished article for a uniform construction of $SO(n)$ , $Spin(n)$ , and $String(n)$ ,
which suggests the existence of a similar uniform construction of $H\mathbb R$ , $KO$ (or $KU$ ), and $TMF$ .
You'll see that von Neumann algebras appear in the construction.
More precisely, von Neumann algebras appear in the definition of conformal nets . The latter are functors from 1-manifolds to von Neumann algberas. For a summary of the conjectural relationship between conformal nets and $TMF$ , have a look at page 8 of
this other paper of mine (joint with Chris Douglas). Loop groups yield non-trivial examples of conformal nets.
Those conformal nets are related (conjecturally) to equivariant $TMF$ .
|
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|
101,531 |
Background: The Strassen Algorithm , described here , has a computational complexity of $\text{O}(n^{2.807})$ for the multiplication of two $n \times n$ matrices (the exponent is $\frac{\log7}{\log2}$ ). However, the constant is so large that this algorithm is in fact slower in practice than naive matrix multiplication for small $n$ . Similarly, the Coppersmith-Winograd algorithm, which has the lowest asymptotic complexity of all known matrix multiplication algorithms, has an exponent of $2.376$ and was discussed here previously. Question: Recently, I made a claim in a submitted paper that the Smith normal form algorithm has super-cubical complexity and a reviewer countered by saying that actually, the complexity has been reduced to matrix multiplication time = $n^{2.37\ldots}$ . I am not an expert on matrix algorithms and would happily change the offending line, but the experience has forced me to wonder, what are the practical implications of saying " X can be done in matrix multiplication time "? More precisely, Does there exist an actual software implementation of Coppersmith Winograd? If not, is there a theoretical obstacle to its existence? By a theoretical obstacle I don't mean something like "Well, it would only be better than existing techniques for $n$ larger than the number of atoms in the universe so what's the point?", but rather something like "the algorithm uses the axiom of choice, or the classification of finite simple groups" etc. PS: Okay, so there is also this paper which apparently reduces the complexity of the Coppersmith-Winograd approach to $2.3737$ from $2.376$ , so I stand corrected about CW being the fastest. The question still stands if we replace CW by the method of V. V. Williams.
|
There are currently no practical implications of any fast matrix multiplication algorithms besides Strassen's. The Coppersmith/Winograd algorithm and its descendants (Stothers, Williams) are very complex, depend on probabilistic constructions, etc. There's no theoretical obstacle to implementing them in the sense you're asking about, and it's something that's humanly possible, but there's little point to it and I don't believe anyone has ever actually done it. It would be complicated and painful, and the only purpose would be really learning how the algorithm works, since the cross-over point for where it would improve on the naive cubic-time algorithm is enormous (so you'll never actually see any improvement). There are other algorithms that would be somewhat easier to implement, at the cost of worse asymptotic performance, but they are also utterly impractical. There's also a deeper issue if you try to use algebraic algorithms in practice. The algebraic complexity model typically used for these problems counts only arithmetic operations and considers memory access to be free. This made sense way back when, since a single floating point operation was comparatively expensive, but nowadays memory management can be the real bottleneck in practice. Algebraic complexity is beautiful and theoretically important, but it ignores important practical issues. If you want to do fast matrix multiplication in practice, it will presumably be on a parallel computer. That introduces further issues of communication complexity; see http://arxiv.org/abs/1202.3173 for an analysis of the Strassen case (both theoretically and in practice).
|
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|
101,725 |
I have never seen any algebraic number theory book discuss the origin of the term "ray class group." Does anyone know where the word "ray" comes from in this context? I always thought it might be a person, but I never see it capitalized. For quick background: the ray class group for a modulus $\mathfrak{m}$ of a number field $K$ is the group of fractional ideals of $K$ prime to $\mathfrak{m}$ modulo the principal ideals generated by elements of $K$ congruent to $1$ modulo $\mathfrak{m}$, where "congruent to $1$" for a real place means "positive."
|
There are many introductions to number theory, but few are as original (my term for what others would call weird) as Fueter's "Synthetische Zahlentheorie" published in 1925.
It starts with elementary number theory, and discusses the arithmetic of cyclotomic
fields up to the Dedekind zeta function and applications to quadratic and cubic reciprocity. In § 5, Fueter defines rays in the field of rational numbers: Definition. If a set of numbers has the property that it contains the product and the quotient of any two of its elements, then this set is called a ray. Actually Fueter does not use the word "set" [Menge] but rather talks about a "domain of numbers" [Bereich von Zahlen]. At the end of this paragraph he makes the following historical remarks: The necessity of considering sets with the property of rays was first realized by Weber. He
called these sets "number groups". Independently, these groups were introduced by R. Fueter
(Der Klassenkörper der quadratischen Körper und die komplexe Multiplikation, Diss. Univ. Göttingen, 1903; see also Crelle 130 (1905), p. 208), who called them rays. We keep this name here since it is similar in nature to the geometric names of fields [in German: Körper, i.e., solid] and ring, and since the word "group" does
not imply commutativity, which is always satisfied by rays. I would have been surprised had there been connections with infinite primes; back in 1903,
Hilbert had already defined infinite primes, but their prominent role in class field theory only became apparent through the work of Furtwängler and Takagi, which took place after the term "ray" had been coined.
|
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|
101,888 |
In 1986 C. C. Hsiung published a paper "Nonexistence of a Complex Structure on the Six-Sphere" and in 1995 he even wrote a monograph "Almost Complex and Complex Structures" to further elaborate on his proof.
Yet answers to the 2009 question on this site all agree that the existence of complex structures on $S^6$ is still an open problem. Some recent preprints answering the question with opposite answers are also cited there.
I would like to know if there are any known mistakes in Hsiung's approach and if so I would appreciate some reference to a paper that points them out.
|
While it's good to have a source, such as Datta's paper that points out the error, I find that his explanation of why the key equation is wrong is not as clear as it could be. In fact, with a little thought (requiring essentially no computation), it's clear why this equation must be wrong and what is wrong with the approach. Since it's relatively short, I thought I'd put it in: On page 263 of Hsuing's monograph "Almost Complex and Complex Structures", he claims the following result, from which, if it were correct, the non-existence of a complex structure on the $6$-sphere would follow immediately (and, in fact, Hsiung 'applies' this result to get exactly this 'conclusion'): Theorem 6.1. Let $J$ be an almost complex structure on a Riemannian $2n$-manifold $M^{2n}$ ($n\ge2$) with a Riemannian metric $g_{ij}$ but without a flat metric or a nonzero constant sectional curvature or both, and let $J_i^j$ and $R_{hijk}$ be respectively the components of the tensor of $J$ and the Riemann curvature tensor of $M^{2n}$ with respect to $g_{ij}$, where all indices take the values $1,2,\ldots,2n$. If $J$ is complex structure on $M^{2n}$, then
$$
J_{i_1}^iJ_{i_2}^jR_{iji_3k}+J_{i_2}^iJ_{i_3}^jR_{iji_1k}+J_{i_3}^iJ_{i_1}^jR_{iji_2k}=0
$$
for all $i_1,i_2,i_3,k$. Now, this result cannot possibly be correct, as you can see from the following observations. First, note that no relation between $g$ and $J$ is supposed. If it weren't for the peculiar assumptions about $M$ not admitting a flat or constant curvature metric (which might have nothing to do with $g$), this would be a purely local statement, but, no matter, let's let $M$ be $\mathbb{CP}^n$ and note that, since $n\ge2$, $M$ cannot carry either kind of metric. Let $J$ be the standard complex structure on $M$. Then the above 'Theorem' would imply that, for any metric $g$ on $M$, its Riemann curvature tensor $R$ would satisfy the above equation. Since any metric in dimension $2n$ can be locally transplanted onto $\mathbb{CP}^n$ and since all complex structures are locally equivalent, it follows easily that the above 'Theorem' implies that the above relation (which is a purely pointwise statement) must hold identically as an algebraic relation for any local pair $J$ and $g$.
(Moreover, since this doesn't involve any derivatives of $J$, the hypothesis that $J$ be integrable is irrelevant.) Second, it's easy to check that this 'identity' does not hold: Just choose a metric $g$ of nonzero constant sectional curvature and any local $J$ that is $g$-orthogonal, and you'll see that this says that the $2$-form $\Omega$ associated to $J$ by $g$ must satisfy $\Omega^2 = 0$, contradicting the fact that $\Omega^n$ cannot vanish because $\Omega$ must be nondegenerate. (This is, in fact, Hsuing's argument as to why $S^6$ can't carry an integrable complex structure, because it has a metric of constant sectional curvature.)
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|
102,138 |
This question has a very general part and a rather concrete part. General: When one wants to prove something in algebraic topology (actually in all parts of mathematics) one obviously needs some good ideas, but first one has to have a good set of tools at hand. Introductory books in algebraic topology provide a number of such tools like long exact sequences to name just one. If one proceeds working in that field and reaches research level more and more tools are just treated as "common knowledge". They are used in papers according to the current situation and often left without quotation. Over the time one gathers plenty of those tools, but I for my part still take many of them as black boxes. When I use them I always have the feeling of walking on very thin ice. Most advanced books have some of those tools scattered in their body and finding a particular one is often harder than it should be. There they are used to build up a certain theory and often don't reveal themselves as useful tools with applications beyond the topic of the respective book.
Moreover it is one thing to find the reference for a statement one knows to be more or less true, but realising which tool one has to use when one isn't even aware of the precise statement is a different story. So the first question: Are there any good books which provide a box of tools used in modern algebraic topology? Maybe something like "AT for the working mathematician". They should come with a proof but not necessarily with applications (for the above reason). Special: The above is incredibly imprecise and there are so many ways to interpret the question. Hence one example of a statement I actually want to know about, which might also give a hint at what I am looking for. Second question: What is the precise statement/where can I find a proof Given a commutative square of fibrations (cofibrations). Then the fibers (cofibers) in the horizontal direction are homotopy equivalent if and only if the fibers (cofibers) in the vertical direction are homotopy equivalent. The square is then cartesian, cocartesian, bicartesian? Edit 1: Now that I think about it it looks like the second question is just an application of the snake lemma. I have to work out the details. Still this statement may stand as an example of what I am looking for. Edit 2: A book which seems to go in the direction of what I describe might be Goerss/Jardine: simplicial homotopy theory.
|
The subject is really way too big (as are so many others of course). I worry a lot about students not in Cambridge or Chicago or Stanford or other places where there are people with folklore at their fingertips. For spectral sequences as a tool, there is a lot to be said for McCleary's guide. Kate Ponto and I just published a book this year, More Concise Algebraic Topology, that may be usable for localizations and completions (just the old-fashioned localize or complete at a set of primes) and that also gives a reasonable start on model categories. Even with that limited scope, the book is much longer than we would like: there were just too many basic details and tools not well enough documented in the literature. There are quite a few other books that go into one or another aspect of the subject (Goerss-Jardine, Neisendorfer, Strom, or, earlier, Whitehead), but it is not to be expected that a single source will cover the ground.
|
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|
102,386 |
One might say, "a random subset of $\mathbb{R}$ is not Lebesgue measurable" without really thinking about it. But if we unpack the standard definitions of all those terms (and work in ZFC), it's not so clear. Let $\Sigma \subset 2^\mathbb{R}$ be the sigma-algebra of all Lebesgue measurable sets. Give $2^\mathbb{R}$ the product measure. (It's a product of continuum many copies of the two-point set.) We want to say that $\Sigma$ is a null set in $2^\mathbb{R}$...but is $\Sigma$ even measurable? Laci Babai posed this question casually several years ago, and no one present knew how to go about it, but it might be easy for a set theorist. Also, a related question: Think of $2^\mathbb{R}$ as a vector space over the field with two elements and $\Sigma$ as a subspace. (Addition is xor, that is, symmetric set difference.) What is $\dim\left(2^\mathbb{R}/\Sigma\right)$? It's not hard to see that $\dim\left(2^\mathbb{R}/\Sigma\right)$ is at least countable, so if $\Sigma$ were measurable, it would be a null set. But that's as far as I made it.
|
The answer to your second question (assuming the axiom of choice, to dodge Asaf's comment) is that $2^{\mathbb R}/\Sigma$ has dimension $2^{\mathfrak c}$, where $\mathfrak c=2^{\aleph_0}$ is the cardinality of the continuum. The main ingredient of the proof is a partition of $[0,1]$ into $\mathfrak c$ subsets, each of which intersects every uncountable closed subset of $[0,1]$. To get such a partition, first note that there are only $\mathfrak c$ closed subsets of $[0,1]$, so you can list them in a sequence of length (the initial ordinal of cardinality) $\mathfrak c$ in such a way that each closed set is listed $\mathfrak c$ times. Second, recall that every uncountable closed subset of $[0,1]$ has cardinality $\mathfrak c$. Finally, do a transfinite inductive construction of $\mathfrak c$ sets in $\mathfrak c$ steps as follows: At any step, if the closed set at that position in your list is $C$ and if this is its $\alpha$-th occurrence in the list, then put an element of $C$ into the $\alpha$-th of the sets under construction, being careful to use an element of $C$ that hasn't already been put into another of the sets under construction. You can be this careful, because fewer than $\mathfrak c$ points have been put into any of your sets in the fewer than $\mathfrak c$ preceding stages, while $C$ has $\mathfrak c$ points to choose from. At the end, if some points in $[0,1]$ remain unassigned to any of the sets under construction, put them into some of these sets arbitrarily, to get a partition of $[0,1]$. Once you have this partition, notice that every piece has outer measure 1, because otherwise it would be disjoint from some closed set that has positive measure and is therefore uncountable. This implies that, among the $2^{\mathfrak c}$ sets that you can form as unions of your partition's pieces, only $\varnothing$ and $[0,1]$ can be measurable. In particular, no finite, nonempty, symmetric difference of these pieces is measurable. That is, they represent linearly independent elements of $2^{\mathbb R}/\Sigma$.
|
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|
102,394 |
Let $S=\mathrm{Spec}(R)$, $s=\mathrm{Spec}(k)$ and $\eta=\mathrm{Spec}(K)$, where $R$ is a d.v.r. with fraction field $K$. Let $j:\eta\rightarrow S$ Now how to compute the sheaf $R^1j_*(\mathbb{G}_{m,\eta})$ in the fppf topology? The case for etale topology is zero by considering the stalks and use the Hilbert 90.
|
The answer to your second question (assuming the axiom of choice, to dodge Asaf's comment) is that $2^{\mathbb R}/\Sigma$ has dimension $2^{\mathfrak c}$, where $\mathfrak c=2^{\aleph_0}$ is the cardinality of the continuum. The main ingredient of the proof is a partition of $[0,1]$ into $\mathfrak c$ subsets, each of which intersects every uncountable closed subset of $[0,1]$. To get such a partition, first note that there are only $\mathfrak c$ closed subsets of $[0,1]$, so you can list them in a sequence of length (the initial ordinal of cardinality) $\mathfrak c$ in such a way that each closed set is listed $\mathfrak c$ times. Second, recall that every uncountable closed subset of $[0,1]$ has cardinality $\mathfrak c$. Finally, do a transfinite inductive construction of $\mathfrak c$ sets in $\mathfrak c$ steps as follows: At any step, if the closed set at that position in your list is $C$ and if this is its $\alpha$-th occurrence in the list, then put an element of $C$ into the $\alpha$-th of the sets under construction, being careful to use an element of $C$ that hasn't already been put into another of the sets under construction. You can be this careful, because fewer than $\mathfrak c$ points have been put into any of your sets in the fewer than $\mathfrak c$ preceding stages, while $C$ has $\mathfrak c$ points to choose from. At the end, if some points in $[0,1]$ remain unassigned to any of the sets under construction, put them into some of these sets arbitrarily, to get a partition of $[0,1]$. Once you have this partition, notice that every piece has outer measure 1, because otherwise it would be disjoint from some closed set that has positive measure and is therefore uncountable. This implies that, among the $2^{\mathfrak c}$ sets that you can form as unions of your partition's pieces, only $\varnothing$ and $[0,1]$ can be measurable. In particular, no finite, nonempty, symmetric difference of these pieces is measurable. That is, they represent linearly independent elements of $2^{\mathbb R}/\Sigma$.
|
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|
102,879 |
Is there some natural bijection between irreducible representations and conjugacy classes of finite groups (as in case of $S_n$)?
|
This is a different take on Steven Landsburg's answer. The short version is that conjugacy classes and irreducible representations should be thought of as being dual to each other. Fix an algebraically closed field $k$ of characteristic not dividing the order of our finite group $G$. The group algebra $k[G]$ is a finite-dimensional Hopf algebra, so its dual is also a finite-dimensional Hopf algebra of the same dimension; it is the Hopf algebra of functions $G \to k$, which I will denote by $C(G)$. (The former is cocommutative but not commutative in general, while the latter is commutative but not cocommutative in general.) The dual pairing
$$k[G] \times C(G) \to k$$ is equivariant with respect to the action of $G$ by conjugation, and it restricts to a dual pairing
$$Z(k[G]) \times C_{\text{cl}}(G) \to k$$ on the subalgebras fixed by conjugation; $Z(k[G])$ is the center of $k[G]$ and $C_{\text{cl}}(G)$ is the space of class functions $G \to k$. Now: The maximal spectrum of $Z(k[G])$ can be canonically identified with the irreducible representations of $G$, and the maximal spectrum of $C_{\text{cl}}(G)$ can be canonically identified with the conjugacy classes of $G$. The second identification should be clear; the first comes from considering the central character of an irreducible representation. Now, the pairing above is nondegenerate, so to every point of the maximal spectrum of $Z(k[G])$ we can canonically associate an element of $C_{\text{cl}}(G)$ (the corresponding irreducible character) and to every point of the maximal spectrum of $C_{\text{cl}}(G)$ we can canonically associate an element of $Z(k[G])$ (the corresponding sum over a conjugacy class divided by its size).
|
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|
103,132 |
I'm not very experienced in this topic, but I read a short description of the Yang-Mills existence and mass gap problem, and as long as I understood it has mainly physical consequences and implications. Therefore, I was just curious what it has to do with mathematics? And what are the mathematical and general consequences of a possible solution to it?
|
There is a long, long list of mathematical subjects that were either pioneered or significantly inspired by results in quantum field theory. However, while physicists may trust the manipulations they do in QFT, and the results of those manipulations have been spectacularly successful, for almost every interesting quantum field theory, there isn't even a rigorous definition or existence proof, much less a justification behind the manipulations that led to the invention of, for example, Seiberg-Witten invariants. Solving the mass gap problem in Yang-Mills would represent the successful rigorous existence of a very non-trivial quantum field theory and the demonstration of a very nontrivial result about that field theory (that hasn't even been adequately demonstrated using physical techniques). While there probably aren't many direct mathematical consequences to the existence of a mass gap, the techniques involved would almost assuredly be applicable towards the large number of QFT inspired results in mathematics.
|
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|
103,530 |
One of the nicest results I know of is (Auslander-Buchsbaum-)Serre's theorem asserting that a (commutative!) local ring is regular iff it has finite global dimensional . I'd like to ask a somewhat vague question: what is the history and what was the context of this result? By this I mean: presumably the above characterization did not come out of thin air (or just out of Serre's mind!), and there was a buildup of ideas which lead to such an elegant and, I guess, surprising characterization of regularity. Nowadays, such a thing appears almost natural to our minds brought up in the nice set-up constructed by the founders of homological algebra and tended to by a few generations already, but I suspect it was less «evident» at the time.
|
In 1953-54 E. Artin asked me to describe homological algebra to him. In addition to Ext and Tor, I decided to show him the "homological proof" of the Hilbert Basis Theorem and indicated that the same proof showed that a regular local ring had finite global dimension. Artin then mentioned the open localization problem, and I said that if the converse of the theorem I just showed him were true, the localization result would be trivial. He asked if I could prove the converse, I said no, and we both agreed that it would be nice to prove it. The question of factoriality in regular local rings also came up in that conversation, and so I set myself the goal of proving those two theorems. I persuaded Auslander to join me in that project. When Auslander and I had almost all the results, and an outline of the string of inequalities needed, Eilenberg asked to see them, we wrote them up for him, and he went to Paris with them. It was there that Serre saw the outline of our project, and he beat us to the final proof by around a week (or the time it took an air mail letter to travel from Paris to Princeton). This may clear up the questions initially posed.
|
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|
103,837 |
Let $X$ be a set and $(T,\cdot)$ an abelian group. Is there a category of $T$-dynamical systems on $X$ which yields useful information about $X$ and $T$? More precisely, is there a category whose objects are dynamical systems, i.e., $\phi:X \times T \to X$ such that $\phi(x,t + s) = \phi(\phi(x,t),s)$? And if so, what are morphisms $\phi \to \phi'$ of $T$-dynamical systems on $X$? I assume that in general this would involve imposing structure on $X$: for example, a topology so that one could consider homotopically perturbing $\phi$. Mostly, I am interested in asking when two such dynamical systems may be considered equivalent, and what it would take to have functors from $T$-dynamical systems on $X$ to $T'$-dynamical systems on $X'$, and to have natural transformations of those functors, etc. I promise I've done (some) homework by looking at this . But note that this document
only provides candidates for equivalent dynamical systems which presumably only accounts for isomorphisms in the desired category rather than all morphisms. This may be too basic a question for the folks here; in this case I will delete it.
|
For every category $\mathfrak C$ and every monoid $\mathcal S$, you can define the category of $\mathcal S$-flows on $\mathfrak C$ as follows: -- Objects are pairs $(X, \alpha:\mathcal S\to \mathrm{End}(X))$, where $\alpha$ is a monoid homomorphism (that is, multiplication in $\mathcal S$ becomes composition of morphisms in $\mathfrak C$ and the neutral element goes to identity); -- Morphisms in the category between two objects $(X,\alpha)$ and $(X',\alpha')$ are just morphisms $\phi:X\to X'$ that commute with the actions, that is $\alpha'(s)\phi=\phi\alpha(s)$ for all $s\in \mathcal S$. Clearly an isomorphism in this category is just a morphism that is an isomorphism in $\mathfrak C$. This is standard and probably does not answer completely your question as you probably want a weaker form of equivalence than really isomorphism in the category of $\mathcal S$-flows. Anyway I think this is a good starting point. Just to understand what this category is, I can give you some examples. If you consider $\mathbb N$-flows on the category of modules $\mathrm {Mod}(R)$ over a ring $R$, you just obtain the category of modules over the polynomial ring $R[X]$. In fact, it is a classical way of looking at modules over $R[X]$ as $R$-modules with a distinguished $R$-linear endomorphism acting on them, that is, discrete-time dynamical systems. If you take $\mathbb Z$-flows you obtain the ring of Laurent polynomials $R[X^{\pm 1}]$. This is easily generalized to $\mathbb N^k$ and $\mathbb Z^k$, giving rise to polynomials in $k$ commuting variables. This point of view is generally adopted by K. Schmidt in his book "Dynamical Systems of Algebraic Origin". In fact, the general approach there is to study dynamical systems of the form $(G,\phi_1,\dots,\phi_k)$, where $G$ is a compact abelian topological group and $\phi_1,\dots,\phi_k$ are commuting topological automorphisms of $G$. This is the category of $\mathbb Z^k$-flows on the category of compact abelian groups. Via Pontryagin duality, this category can be seen to be dual to the category of $\mathbb Z^k$-flows on discrete Abelian groups, that, by what we said above, is exactly $\mathrm{Mod}(\mathbb Z[X_1^{\pm 1},\dots,X_k^{\pm 1}])$. Generalizing more, you can easily prove that $\mathcal S$-flows on $\mathrm {Mod}(R)$ are the category of modules over the monoid ring $R[\mathcal S]$.
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|
104,322 |
Recently I've been studying the problem of integer representation as sum of three squares. Most of the articles that I've found study the function $r_m(n)$ which counts the number of representations of $n$ as the sum of $m$ squares. However, this is not what I am interested in. What I'm looking for is an efficient way (for some given $n$ ) to find $x$ , $y$ and $z$ such that $n = x^2 + y^2 + z^2$ . I need to find at least one such representation. Can you recommend me some articles that study this problem? P.S. I believe that Emil Grosswald's book "Representation of integers as Sums of Squares" contains the answer. However, I could not find this book on my university's web-site.
|
This problem is discussed in my paper with Rabin, Randomized algorithms in number theory,
Commun. Pure Appl. Math. 39, 1985, S239 - S256. We give an algorithm that, assuming a couple of reasonable conjectures, will produce a representation as a sum of three squares in polynomial time.
|
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|
104,750 |
In Cushman and Bates, Global Aspects of Classical Integrable Systems, 1997, I have read In a widely circulated but unpublished letter in 1965, Palais explained the symplectic formulation of Hamiltonian mechanics. I would like to know if, in the meanwhile, this letter was made available.
|
I haven't thought about that letter for a very long time, but as far as I can recall I didn't ever make it publicly available, and I don't think any of the friends to whom I sent it did either. However, I am a bit of a pack-rat, so after Ryan Budney alerted me that this question had appeared on MO I did some searching in my piles of old papers, and I found and scanned what I am pretty sure was the mathematical content of the letter, and here is a link to the scan. http://vmm.math.uci.edu/PalaisLetterOnSymplectic.pdf (I'm not sure if I just sent out copies of this with a cover letter or rather wrote a letter in which I copied the contents of the above.) Note, that the first page of the above is a scan of page 159 of "Foundations of Mechanics" by Ralph Abraham and Jerry Marsden. It says that the letter was from around 1962, which is what I recall. Ralph was one of the recipients. Dick Palais
|
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|
104,777 |
Let $k$ be a field, and let $\mathbf{Vect}$ denote the category of vector spaces (possibly infinite-dimensional) over $k$. Taking duals gives a functor $(\ )^*\colon \mathbf{Vect}^{\mathrm{op}} \to \mathbf{Vect}$. This contravariant functor is self-adjoint on the right, since a linear map $X \to Y^*$ amounts to a bilinear map $X \times Y \to k$, which is essentially the same thing as a bilinear map $Y \times X \to k$, which amounts to a linear map $Y \to X^*$. It therefore induces a monad $(\ )^{**}$ on $\mathbf{Vect}$. What are the algebras for this monad? Remarks I assume this is known (probably since a long time ago). The first paper I came across when searching for the answer was Anders Kock, On double dualization monads , Math. Scand. 27 (1970), 151-165. I'm pretty sure it doesn't contain the answer explicitly, but it's possible that it contains some results that would help. The monad isn't idempotent (that is, the multiplication part of the monad isn't an isomorphism). Indeed, take any infinite-dimensional vector space $X$. Write our monad as $(T, \eta, \mu)$. If $\mu_X$ were an isomorphism then $\eta_{TX}$ would be an isomorphism, since $\mu_X \circ \eta_{TX} = 1$. But $\eta_{TX}$ is the canonical embedding $TX \to (TX)^{**}$, and this is not surjective since $TX$ is not finite-dimensional. There's another way in which the answer might be somewhat trivial, and that's if $(\ )^*$ is monadic. But it doesn't seem obvious to me that $(\ )^*$ even reflects isomorphisms (which it would have to if it were monadic). There's a sense in which answering this question amounts to completing the analogy: sets are to compact Hausdorff spaces as vector spaces are to ????? Indeed, the codensity monad of the inclusion functor (finite sets) $\hookrightarrow$ (sets) is the ultrafilter monad, whose algebras are the compact Hausdorff spaces. The codensity monad of the inclusion functor (finite-dimensional vector spaces) $\hookrightarrow$ (vector spaces) is the double dualization monad, whose algebras are... what? (Maybe this will help someone to guess what the answer is.)
|
Tom, I believe $(-)^\ast: \mathbf{Vect}^{op} \to \mathbf{Vect}$ is monadic, essentially because all objects in $\mathbf{Vect}$, in particular $k$ as a module over $k$ as ground field, are injective. For instance, to check that $(-)^\ast$ reflects isomorphisms, suppose $f: V \to W$ is any linear map. We have two short exact sequences $$0 \to \ker(f) \to V \to im(f) \to 0$$ $$0 \to im(f) \to W \to coker(f) \to 0$$ Because $k$ is injective, the functor $(-)^\ast = \hom(-, k)$ preserves short exact sequences: $$0 \to im(f)^\ast \to V^\ast \to \ker(f)^\ast \to 0$$ $$0 \to coker(f)^\ast \to W^\ast \to im(f)^\ast \to 0$$ and if $f^\ast$, the composite $W^\ast \to im(f)^\ast \to V^\ast$, is an isomorphism, then $W^\ast \to im(f)^\ast$ is injective, which forces $coker(f)^\ast = 0$ and therefore $coker(f) = 0$. By a similar argument, $\ker(f) = 0$. Therefore $f$ is an isomorphism. The remaining hypotheses of Beck's theorem (in the form given in Theorem 2, page 179, of Mac Lane-Moerdijk) are similarly easy to check. Obviously $\mathbf{Vect}^{op}$ has coequalizers of reflexive pairs since $\mathbf{Vect}$ has equalizers. And $(-)^\ast: \mathbf{Vect}^{op} \to \mathbf{Vect}$ (which has a left adjoint, as pointed out) preserves coequalizers; this is equivalent to saying that $\hom(-, k)$, as a contravariant functor on $\mathbf{Vect}$, takes equalizers to coequalizers, or takes kernels to cokernels, but that's the same as saying that $k$ is injective, so we're done. Oh, incidentally, double dualization is not a commutative or monoidal monad, if I recall correctly. Edit: In a comment below, Tom asks for a more concrete description of $\mathbf{Vect}^{op}$ along the lines of topological algebra. I suspect the way to go is to see $\mathbf{Vect}$ as the Ind-completion (or Ind-cocompletion) of the category of finite-dimensional vector spaces, and therefore $\mathbf{Vect}^{op}$ as the Pro-completion of the opposite category, which is again $\mathbf{Vect}_{fd}$. I think I've seen before a result that this is equivalent to the category of topological $k$-modules which arise as projective limits of (cofiltered diagrams of) finite-dimensional spaces with the discrete topology, or something along similar lines, but I'd have to look this up to be sure. There might be pertinent material in Barr's Springer Lecture Notes on $\ast$-autonomous categories, but again I'm not sure. Edit 2: Ah, found it. $\mathbf{Vect}^{op}$ is equivalent to the category of linearly compact vector spaces over $k$ and continuous linear maps. See Theorem 3.1 of this paper for example: arxiv.org/pdf/1202.3609. The result is credited to Lefschetz.
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|
105,021 |
Given a pseudo-differential operator $P$ of order zero, Seeley showed that the holomorphic family of operators $\lbrace P^{z} : z\in \mathbb{C} \rbrace$ of all complex powers is contained in the class of pseudo-differential operators. Apart from knowing that we can take powers of these operators, is there any application of this theory? I understand the utility of raising operators to fractional powers. But, irrational and complex powers are not clear.
|
There are many reasons to do this. A principal one (that was Seeley's motivation) is that the meromorphic operator family
$P^{-s}$ is trace-class when the real part of $s$ is greater than $-n/m$ where $n$ is the
dimension and $m$ is the order of $P$. This gives a meromorphic continuation of the spectral zeta function $\zeta(s) = \sum_{j=1}^\infty \lambda_j^{-s}$. This is an interesting generalization of the Riemann zeta function. Furthermore, its behaviour sheds light on the distribution of eigenvalues (the Weyl law). This zeta function is equivalent to the heat trace
$\sum_{j=1}^\infty e^{-\lambda_j t}$ via the Mellin transform. There are lots of other things that come out of this, however -- take a look at the recent book by Scott ``Traces and Determinants of Pseudodifferential Operators''.
|
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|
105,221 |
Given a semisimple Lie algebra $\mathfrak g$ with Cartan matrix $a_{ij}$, the quantum group $U_q(\mathfrak g)$ is usually defined as the $\mathbb Q(q)$-algebra with generators $K_i$, $E_i$, $F_i$ (the $K_i$ are invertible and commute with each other) and relations
$$
\begin{split}
K_iE_j &K_i^{-1}=q^{\langle\alpha_i,\alpha_j\rangle}E_j\qquad\qquad
K_iF_j K_i^{-1}=q^{-\langle\alpha_i,\alpha_j\rangle}F_j\,
\\\
&[E_i,F_j]=\delta_{ij}\frac{K_i-K_i^{-1}}
{\quad q^{\langle\alpha_i,\alpha_i\rangle/2}
-q^{-\langle\alpha_i,\alpha_i\rangle/2}\quad}\,
\end{split}
$$
along with two more complicated relations that I won't reproduce here. One then defines the comultiplication, counit, and antipode by some more formulas. Is there a way of defining $U_q(\mathfrak g)$ that doesn't involve writing down all those formulas? In other words, is there a procedure that takes $\mathfrak g$ as input, produces $U_q(\mathfrak g)$ as output, and doesn't involve the choice of a Cartan subalgebra of $\mathfrak g$?
|
$\newcommand\g{\mathfrak{g}}$The answer to your question "is there a procedure that takes $\g$ as input, produces $U_q(\g)$ as output, and doesn't involve the choice of a Cartan subalgebra of $\g$?" is No. Not if you want it "canonical" in any sense. (Of course, if I wanted to cheat my way to a "yes," I could make choices that are equivalent to choosing a Cartan but not stated as such — I would construct for you a certain canonical homogeneous space, and then ask you to pick a point in it, ....) The problem is that the automorphism group of $\g$ does not lift to $U_q(\g)$. Recall that the inner automorphism group is precisely the simplest group $G$ integrating $\g$ (take any connected group integrating $\g$ and mod out by its center). On the other hand, the inner automorphism "group" of $U_q(\g)$ is $U_q(\g)$ (or "$\operatorname{spec}(\operatorname{Fun}_q(G))$", depending on your point of view) itself. This certainly deforms the automorphisms. But there is not a procedure like you ask for, because you have to break some symmetry. Here's a way to say this correctly: In a precise sense $U_q(\g)$ degenerates to $U(\g)$ as $q\to 1$, and this is part of the structure that I take you to mean when you write "$U_q(\g)$". In this degeneration, you can also study $\frac{\partial}{\partial q}(\dots)$ at $q=1$. In particular, looking at $\frac{\partial}{\partial q}\bigr|_{q=1}$ of the comultiplication on $U_q(\g)$ recovers the Lie cobracket on $\g$. But the Lie cobracket knows the Cartan subalgebra: it is precisely the kernel of the Lie cobracket. So $\operatorname{Aut}(\g)$ cannot lift to $U_q(\g)$ compatibly with all of this structure. What does exist without choosing a Cartan subalgebra is the braided category of representations of $U_q(\g)$, although I would have to think a moment to recall how to write it down explicitly. (In the asymptotic limit $q = e^\hbar$ with formal $\hbar$, I do know how to write down $\operatorname{Rep}(U_{e^\hbar}\g)$ explicitly for any choice of Drinfel'd associator.) In particular, as a braided monoidal category, this category does have an action by $\operatorname{Aut}(\g)$. But the category is strictly less data than the Hopf algebra. Namely, "Tannakian reconstruction" is the statement that $U_q(\g)$ is the Hopf algebra of endomorphisms of a certain braided coalgebra in $\operatorname{Rep}(U_{q}\g)$ (or it is the Hopf dual of the Hopf algebra of co-endomorphisms of a certain braided algebra in the category of Ind-objects in $\operatorname{Rep}(U_{q}\g)$, if for you representations are finite-dimensional), and you cannot choose this coalgebra canonically. This coalgebra is unique up to isomorphism, but certainly not up to unique isomorphisms (or else $U_q(\g)$ would be trivial). The failure of this coalgebra to exists up to canonical isomorphism is essentially the same problem as above. In a precise way, this is failure of there to exist a canonical isomorphisms between different choices of the coalgebra is analogous of the failure of the "fundamental group" of a topological space to be an honest group. Recall that a pointed topological space has a fundamental group, which is a group determined up to canonical isomorphism. For comparison, a non-pointed but path-connected topological space has a group assigned to each point, and these are non-canonically isomorphic. Thus a non-pointed path-connected topological space has a "fundamental group up to conjugation," also known as a connected groupoid . What this should all mean, although I'm not going to try to write down the correct definition, is that there does exist canonically associated to $\g$ a "Hopf algebroid" which is noncanonically equivalent as a Hopf algebroid to any choice of $U_q(\g)$. Or, at least, I'm confident of everything in my answer in the $\hbar\to 0$ asymptotics, and have thought less about the finite-$q$ case, and so I'm generalizing intuition from that setting, but I think it's all correct.
|
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|
105,269 |
Can someone describe explicitly an abelian group $A$ such that the extension $$0 \to \mathbb{Z} \to A \to \mathbb{Q} \to 0$$ doesn't split ? Background: The Stein-Serre theorem (Hilton, Stammbach: A course in homol. algebra, Theorem 6.1) states that if $A$ is abelian of countable rank (=maximal number of linear independent elements), then $Ext(A,\mathbb{Z})=0$ implies $A$ free. When applied to $A= \mathbb{Q}$, I conclude $Ext(\mathbb{Q},\mathbb{Z})\neq 0$. Moreover, by interpreting $A \in Ext(\mathbb{Q},\mathbb{Z})$ as extension $0 \to \mathbb{Z} \to A \to \mathbb{Q} \to 0$ of abelian groups, there must be an $A$ such that the extension doesn't split. The problem is that the proof of the theorem isn't constructive and doesn't show how to construct such an $A$.
|
Nice question. For a prime $p$ let $\mathbb{Z}_{(p)} = \lbrace \frac{a}{b}\in \mathbb{Q}\mid p \nmid b\rbrace$ and $\mathbb{Z}[p^{-1}] = \lbrace \frac{a}{p^n}\in \mathbb{Q}\mid n \ge 0 \rbrace$. Then $$A := \lbrace (x,y) \in \mathbb{Q} \times \mathbb{Z}[p^{-1}] \mid x-y \in \mathbb{Z}_{(p)} \rbrace$$
has the desired non-split extension. Informal $A$ consists of all pairs $(x,y) \in \mathbb{Q} \times \mathbb{Q}$ where $y$ is the $p$-part in the partial fraction decomposition of $x$ (up to an integer summand). Proof: Let $\rho: A \to \mathbb{Q}$ be projection onto the first factor and $i: \mathbb{Z} \hookrightarrow A$ inclusion into the second factor. By partial fraction decomposition of the rationals, $\rho$ is surjective and $$\ker(\rho)=0 \times \big(\mathbb{Z}[p^{-1}] \cap \mathbb{Z}_{(p)}\big) = 0 \times \mathbb{Z} = \operatorname{im}(i).$$ Next, let $j: \mathbb{Q} \to A$ be a splitting hom. of $\rho$. Composing $j$ with the projection onto the second factor yields a hom. $f: \mathbb{Q} \to \mathbb{Z}[p^{-1}] \le \mathbb{Q}$. Since each endomorphism of $\mathbb{Q}$ is multiplication with some $q \in \mathbb{Q}$ we have $f(x)=qx \in \mathbb{Z}[p^{-1}]$ for all $x \in \mathbb{Q}$ which is only possible for $q=0$. Hence $j(x)=(x,0) \in A$ for all $x \in \mathbb{Q}$. Setting $x=1/p$ we obtain $1/p \in \mathbb{Z}_{(p)}$ which is the desired contradiction. QED
|
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105,304 |
If $f, g\in \mathbb C[a,b]$ are polynomials in two variables, are there easy criteria that allow to see if $f(x,y)-g(t,z)\in \mathbb C[x,y,t,z]$ is irreducible? Thank you very much, best
|
There has been a lot of work on establishing when polynomials of the form
$$f(x_1, \dots, x_r) - g(y_1, \dots, y_s)$$
are reducible (over the complex numbers, but also over other fields). (Negating them, or special cases thereof would thus yield criteria, in the sense of conditions, when such a polynomial is irreducible.) To a considerable extent one can reduce this problem to the case $r=s=1$. Namely, Davenport and Schinzel (Two Problems Concerning Polynomials, J. reine angew. Math., 1964) proved (this is Theorem 2, up to signchange to match the question, of the paper. The polynomial, over a field of charateristic $0$,
$$f(x_1, \dots, x_r) - g(y_1, \dots, y_s)$$
is reducible if and only if
$f(x_1, \dots, x_r) = F(R(x_1, \dots, x_r))$ and $g(y_1, \dots, y_s) = G(S(y_1, \dots, y_s))$
with polynomials $F,G,R,S$ over the same field and
$$F(u) - G(v)$$ is reducible over the same field. Note that the main case the authors care about is indeed the complex one. In particular, it follows that if one cannot write $f,g$ in such a way with nontrivial (that is degree at least 2) $F$ and $G$ then the original polynomial is irreducible [except if one of $f,g$ is constant, but this should not be the case in view of the question]. (I am not sure how to check this most efficiently, but if the polynomials are not too complex, just starting from maximal terms and inferring conditions on the rank and then working ones way down could be a viable, though likely not optimal, strategy.) If one wishes to have more complete information one is now faced with the question when a polynomial $$F(u)-G(v)$$ is (ir)reducible. Various interesting results on this problem where obtained (see below for some recent of them), but if one wishes to have an answer for specific polynomials one migt get by via using not these results, but general irreducibilty criteria for polynomials in two variables or easier to apply criteria for this polynomial. For the former the question pointed out in a comment by Camilo Sarmiento seems like a good resource (I reproduce the link for simplicity Irreducibility of polynomials in two variables ). For example the Ehrenfeucht criterion mentioned there might allow to directly exclude further cases.
Also the Eisenstein (like) criteria could be quite useful.
Also the paper by Davenport, Lewis, Schinzel mentioned in my comment below should contain some test, but as I have no access to the paper I do not know what exactly. Now for more recent results specifically on this problem: Under the assumption that the $F$ and $G$ are indepecomposable (indepcomposable meaning the polynomial is not the composition of two polynomials) there is a complete answer know (over the complex numbers). In particular, Pierrette Cassou-Noguès and Jean-Marc Coveignes ( Factorisations explicites de $g(y)-h(z)$ , Acta Arith. 87 (1999)) based on earlier work by Fried and Feit and others established an explict finite set of pairs of polynomials such that any pair $(F,G)$, with $F,G$ indecomposable and not linearly related (this means $F(x)$ is not of the form $AG(ax+b)+B$ with constants $A,a,B,b$ and $A,a$ non-zero, to avoid corner cases), with $F(u)-G(v)$ reducible is weakly linearly realted (I skip the def, but similar to lin related) to one of them. This sets is too large to give here (yet the paper is linked anyway) but what can be said briefly (and was already known before) is that the degrees of both polynomials are equal, and equal to one of $7, 11, 13, 15, 21, 31$. Note that this result uses the Classification of Finite Simple Groups. There are also other results related to this. For example Yuri Bilu (Acta Arith. 90 (1999))
studied when $F(u)-G(v)$ has a factor of degree at most two (where there is no assumption of indecomposability). Roughly, there can essentially only be a quadratic factor if both are Chebyshev polynolmials of degree a power of two.
|
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|
105,339 |
Let $M$ a right simple module and $N$ be a left simple module over a ring $R$. I'm seeking a kind of Schur's lemma, with $\mathrm{Hom}_R (M,N)$ replaced by $M \otimes_R N$. So my questions are: Can we describe $M \otimes_R N$ explicitly? In particular, for a fixed $M$, is $N$ such that $M \otimes_R N \neq 0$ unique up to isomorphism? If not, can we classify such $N$'s in a reasonable way?
|
There has been a lot of work on establishing when polynomials of the form
$$f(x_1, \dots, x_r) - g(y_1, \dots, y_s)$$
are reducible (over the complex numbers, but also over other fields). (Negating them, or special cases thereof would thus yield criteria, in the sense of conditions, when such a polynomial is irreducible.) To a considerable extent one can reduce this problem to the case $r=s=1$. Namely, Davenport and Schinzel (Two Problems Concerning Polynomials, J. reine angew. Math., 1964) proved (this is Theorem 2, up to signchange to match the question, of the paper. The polynomial, over a field of charateristic $0$,
$$f(x_1, \dots, x_r) - g(y_1, \dots, y_s)$$
is reducible if and only if
$f(x_1, \dots, x_r) = F(R(x_1, \dots, x_r))$ and $g(y_1, \dots, y_s) = G(S(y_1, \dots, y_s))$
with polynomials $F,G,R,S$ over the same field and
$$F(u) - G(v)$$ is reducible over the same field. Note that the main case the authors care about is indeed the complex one. In particular, it follows that if one cannot write $f,g$ in such a way with nontrivial (that is degree at least 2) $F$ and $G$ then the original polynomial is irreducible [except if one of $f,g$ is constant, but this should not be the case in view of the question]. (I am not sure how to check this most efficiently, but if the polynomials are not too complex, just starting from maximal terms and inferring conditions on the rank and then working ones way down could be a viable, though likely not optimal, strategy.) If one wishes to have more complete information one is now faced with the question when a polynomial $$F(u)-G(v)$$ is (ir)reducible. Various interesting results on this problem where obtained (see below for some recent of them), but if one wishes to have an answer for specific polynomials one migt get by via using not these results, but general irreducibilty criteria for polynomials in two variables or easier to apply criteria for this polynomial. For the former the question pointed out in a comment by Camilo Sarmiento seems like a good resource (I reproduce the link for simplicity Irreducibility of polynomials in two variables ). For example the Ehrenfeucht criterion mentioned there might allow to directly exclude further cases.
Also the Eisenstein (like) criteria could be quite useful.
Also the paper by Davenport, Lewis, Schinzel mentioned in my comment below should contain some test, but as I have no access to the paper I do not know what exactly. Now for more recent results specifically on this problem: Under the assumption that the $F$ and $G$ are indepecomposable (indepcomposable meaning the polynomial is not the composition of two polynomials) there is a complete answer know (over the complex numbers). In particular, Pierrette Cassou-Noguès and Jean-Marc Coveignes ( Factorisations explicites de $g(y)-h(z)$ , Acta Arith. 87 (1999)) based on earlier work by Fried and Feit and others established an explict finite set of pairs of polynomials such that any pair $(F,G)$, with $F,G$ indecomposable and not linearly related (this means $F(x)$ is not of the form $AG(ax+b)+B$ with constants $A,a,B,b$ and $A,a$ non-zero, to avoid corner cases), with $F(u)-G(v)$ reducible is weakly linearly realted (I skip the def, but similar to lin related) to one of them. This sets is too large to give here (yet the paper is linked anyway) but what can be said briefly (and was already known before) is that the degrees of both polynomials are equal, and equal to one of $7, 11, 13, 15, 21, 31$. Note that this result uses the Classification of Finite Simple Groups. There are also other results related to this. For example Yuri Bilu (Acta Arith. 90 (1999))
studied when $F(u)-G(v)$ has a factor of degree at most two (where there is no assumption of indecomposability). Roughly, there can essentially only be a quadratic factor if both are Chebyshev polynolmials of degree a power of two.
|
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|
105,438 |
Suppose $f$ is a $C^\infty$ function from the reals to the reals that is never negative. Does it have a $C^\infty$ square root? Clearly the only problem points are those at which $f$ vanishes.
|
The answer is "no". This is covered in great detail here: http://www.math.polytechnique.fr/~bony/BBCP_jfa.pdf
|
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|
105,575 |
Since the work of Serre in the early 50's on homotopy groups of spheres, it is known that the homotopy group $\pi_k(S^n)$ is finite, except when $k=n$ (in which case the group is $\mathbb{Z}$), or when $n$ is even and $k=2n-1$ (in which case the group is the direct sum of $\mathbb{Z}$ and a finite group). As a consequence, the stable homotopy groups $\pi_k^s$ are finite groups for $k>0$, and $\pi_0^s \cong \mathbb{Z}$. The work of Serre was done before anyone knew about stable homotopy theory and chromatic methods, and this makes me wonder about the following questions. Question 1: Is it possible to use methods from stable/chromatic homotopy theory to prove finiteness of stable homotopy groups of spheres directly, without having to compute any unstable homotopy groups of spheres? Question 2: Is there any philosophical or conceptual reason for why these groups should be finite?
|
I agree with Ryan that Serre's proof can be viewed as perfectly conceptual, but here is a modern version. Accept from Serre that the homotopy groups of spheres are finitely generated. Let
$k\colon S^n \longrightarrow K(\mathbf{Z},n)$ be the canonical map. We know how to rationalize
spaces and maps. The rationalization of $k$ is a map $k_{0}\colon S^n_{0}\longrightarrow K(\mathbf{Q},n)$. If $n$ is odd, $k_0$ is an isomorphism on rational cohomology and therefore
an equivalence. If $n$ is even, a very little cohomological calculation shows that the fiber of $k_{0}$ is $K(\mathbf{Q},2n-1)$. Since the homotopy groups of spheres are finitely generated,
the kernel of the map on homotopy groups induced by the rationalization $S^n\longrightarrow S^n_{0}$
is finite in each degree. The rank of the free part is immediate from what I've said about $S^n_{0}$. Serre's theorem follows. This uses no calculation of unstable homotopy groups except maybe deep down that $\pi_n(S^n) = \mathbf{Z}$.
|
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105,577 |
The analytic rank of the Mordell elliptic curve $y^2=x^3-86069^5$ indicates that it has rank 2. However, deriving a set of generators, and hence the regulator, is proving to be a little bit of an intractable problem. I'm posting this question in the hope that someone may have investigated this curve already and, as such, may be able to help me out with some of the points on the curve. Any help with this would be very much appreciated. As further background, I have been involved in a small group searching for values of #Sha > 1 up through the ranks of Mordell curves, currently from ranks 0 to 9. To date we have been able to progress to a rank 9 curve were #Sha would seem to be at least equal to 9. The details of which may be seen at the NMBRTHRY Archives . Kevin.
|
This particular curve, which I'll call $E$, may be quite challenging. The analytic rank is probably 2, but as far as I know, the only way to prove this is to show that the algebraic rank is not 0. Assuming the analytic rank is 2 and the full BSD formula, the product of the Regulator and the order of the Tate-Shafarevich group is approximately 1435241.110225344. (Using the command ConjecturalRegulator(E); in MAGMA). Since the the rank is probably 2, I would guess that the Shafarevich-Tate group is trivial, so that the regulator is quite large. In this case a 4-Descent was feasible. Below are the genus 1 curves in $\Bbb P^3$ that represent the elements of the $4$-Selmer group modulo torsion. If we can find points on these curves, they will map to points of infinite order on $E$. > E := EllipticCurve([0,-86069^5]);
> ConjecturalRegulator(E);
1435241.11022534407264592039437 2
> TD := TwoDescent(E);
> SetClassGroupBounds("GRH");
> time FD := [FourDescent(C) : C in TD];
Time: 3.890
> FD;
[
[
Curve over Rational Field defined by
15*x1^2 + 111*x1*x2 + 4*x1*x3 + 10*x1*x4 - 38*x2^2 - 95*x2*x3 + 67*x2*x4
- 54*x3^2 - 14*x3*x4 - 71*x4^2,
31*x1^2 - 71*x1*x2 - 69*x1*x3 - 23*x1*x4 + 9*x2^2 - 92*x2*x3 - 24*x2*x4
+ 35*x3^2 - 148*x3*x4 + 168*x4^2,
Curve over Rational Field defined by
35*x1^2 + 26*x1*x2 + 41*x1*x3 + 61*x1*x4 + 54*x2^2 + 11*x2*x3 + 25*x2*x4
+ 68*x3^2 + 3*x3*x4 - 78*x4^2,
36*x1^2 - 138*x1*x2 - 38*x1*x3 + 127*x1*x4 + 25*x2^2 + 42*x2*x3 +
47*x2*x4 + 81*x3^2 - 12*x3*x4 + 60*x4^2
],
[
Curve over Rational Field defined by
21*x1^2 + 13*x1*x2 + 13*x1*x3 + 44*x1*x4 - 32*x2^2 - x2*x3 - 18*x2*x4 -
45*x3^2 + 24*x3*x4 - 238*x4^2,
5*x1^2 - 14*x1*x2 + 122*x1*x3 + 268*x1*x4 + 26*x2^2 + 6*x2*x3 +
149*x2*x4 - 57*x3^2 - 23*x3*x4 - 78*x4^2,
Curve over Rational Field defined by
4*x1^2 + 49*x1*x2 + 34*x1*x3 + 26*x1*x4 + 26*x2^2 - 33*x2*x3 - 74*x2*x4
+ 53*x3^2 - 74*x3*x4 + 111*x4^2,
38*x1^2 - 84*x1*x2 + 3*x1*x3 - 88*x1*x4 - 29*x2^2 + 27*x2*x3 - 154*x2*x4
+ 5*x3^2 - 234*x3*x4 - 120*x4^2
],
[
Curve over Rational Field defined by
7*x1^2 + 78*x1*x2 + 106*x1*x3 + 62*x1*x4 - 21*x2^2 - 26*x2*x3 + 22*x2*x4
+ 34*x3^2 - 25*x3*x4 - 118*x4^2,
33*x1^2 + 2*x1*x2 - 14*x1*x3 + 106*x1*x4 + 48*x2^2 - 33*x2*x3 +
165*x2*x4 + 69*x3^2 + 31*x3*x4 - 26*x4^2,
Curve over Rational Field defined by
7*x1^2 + 46*x1*x2 + 33*x1*x3 + 23*x1*x4 + 13*x2^2 + 36*x2*x3 - 108*x2*x4
- 69*x3^2 - 88*x3*x4 + 145*x4^2,
19*x1^2 - 28*x1*x2 - 14*x1*x3 + 8*x1*x4 + 150*x2^2 - 52*x2*x3 +
190*x2*x4 - 46*x3^2 + 33*x3*x4 + 248*x4^2
]
] I searched for points on these curves up to height $10^9$ and didn't find any, which isn't that surprising given that the regulator is probably so large. Maybe an $8$- $9$- or $12$-descent would help, but I'm not sure since the regulator looks to be so large. I've come across a lot of curves like this one, and honestly I don't know what to do other than spend a great deal of time implementing higher and higher descents, which will require better and better architecture for working with algebraic number fields. Finding points on curves with rank at least 2 is much harder than the rank 1 case, where a non-torsion Heegner point can be constructed. I've often wondered if the notion of visibility of Mordell-Weil groups could be useful here to prove that the rank is 2. There might be some hope since this curve is just a sextic twist of a very simple elliptic curve, but I have no idea what other abelian variety one would try to use to "visualize" the Mordell-Weil group. EDIT: Found a rational point! Since this is a Mordell curve, it has a 3-isogeny. So ThreeDescentByIsogeny(E); works in magma. This, together with the 4-Descent above, can be patched together with Tom Fisher's 12-Descent code. Running PointSearch(C, 10^15) on all the 12-covers, I found one rational point on a 12-cover. This maps back to the HUGE point $P = (r/t^2, s/t^3) \in E(\Bbb Q)$ where $$t = 1064315237527062416197829497356636659645269584461099593669088031745089037672135800872679057089531240249644479256790731579010363138838881840905955995370777995601208434305913644468575606,$$ $$r = 280330358182289626756155234063598323321079390462341827366518987565228009421332474964681906370265513552399459534408574388270067641670339702167066942312699829200035334349465315027488431491497462250163493458119988220147933027219004129214673758569410719992144923411319943555067857245550021843686621974171705862665984304126540657354963027685841117202627482318291749321065520239017,$$ and $$s = -3894169495021322706664690332034015776766200723677334488117565017206832575610394605401029654682981124772659459298812689312208671041841435640288530899795853801784491292694434680924857836480106291120205903428248497270005655966231499766892812223437056078000735443855989892506051021805114412318184420256429821942583648074219767873044937972437645324252633026228735728785740079828040374108087622426272823877167554852034435686184476368398239806496596492675365799418315372050294056476779196364271993469777717240193414071588739725730054647942389914059081838403974737198837$$ Magma (and sage) says the canonical height of $P$ is 862.589016739449. $P$ has infinite order and is not a multiple of another point. I'll keep looking for more points on 12-covers, but there should be a smarter way to look for a second generator since we know conjecturally what the regulator is supposed to be, at least up to division by an integer square (the order of Sha). Second Edit: Second generator found by Tom Fisher! I received email from Tom Fisher who got wind of this example thanks to Kevin Acres and Mark Watkins. He used some more careful minimization techniques for 12-covers which made it much easier for him to find the second generator $Q = (u/w^2, v/w^3)$ where $$w = 182114807484521200807106885195433046250677033531559359037472093313104122418454868285361195199297225042175078422885282555317033956772998359062866824462985656413093380882725866443379879289096885109176415930832786071510257551923931848041894889001647111552913336376307990237676516910188887260245394301062683134535223719019999940520792408074037957510825778941886955164545102$$ $$u = 6409818948420000148009253515371033320765674658334995509069337486116156301741657451140832523716129416946553724548364793698626405777843956622076380997023235142487299844326278568312577933328949209325079504842164930964499614652180484249884310193722360960206856049327696413372088438499522988880226096525828673972002212550418445757426791463589511241682169507191857404060127705996714164059762587639453970388047710151991644286292989545241436751637355865317439825495449344823977758797766025325026099986880721224779341838599953004314077314458400203457162779096001177667625960285991372461677348744634118572760595736047817251613300845834732694598202906328385669708044981270667425075403152129513088853355196780998723571290752400352784381236905647783762282233$$ $$v = 9541721267275526440336656075081968921795973155189044266523956895286489666750185996900611973660561530367870689843326743006933458752307224811373451969681421106849987021781748740069623362388923316032840282833074591791294185661671902447470420594105932470690611184875968605655279379274234568020070054315280341961704988604471445169462057158108739193383808091705876229256447451432215635271166605138776098284834213663487614216411244364417456782497135614069447487775516105502956819954881548171764933506408613533330700718054166278531808088754093650937927404096948372079871885150489168059803000137727292308698902319300888120656605108547686297443217509881949442132801485412503095358513515604711600740777338910658971774981045705755311582771691405245163713860189075823123694400673272454942433819801867842344237200886745700934179591344744409814700762619336056693409131813436072830610006822421568399180878828865624130779460438617259828853398661927306856293129558704629256045989464076110208592569978237366668166654822438426674566555490387538893656476099175185704061248678944708728574308512767720849000403950411614279621166869036373101$$ This height of $Q$ is 1715.46805605884712533431816634 and the regulator is actually 1435241.11022534407264592039437 as BSD predicts if Sha(E/Q) is trivial.
|
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|
105,753 |
Hey Is there a universal property for the smash product (of pointed spaces or pointed CW-complexes or something of that ilk)? I've seen the smash product of spectra defined with a universal property in terms of the smash product of pointed-spaces, but I was wondering if there was just some simple universal property you could put on these somewhat mysterious (to me) space-level operations. EDIT: I was hoping for something more satisfying than the internal Hom adjoint. The tensor product can be defined similarly, but I find the universal product in terms of bilinear maps more intuitive (although, when unraveled, they are the same thing). I was hoping for something similar for smash. Thanks :)
|
$X \wedge Y$ represents maps from $X \times Y$ that are base-point-preserving separately in each variable, just as the tensor product represents maps that are linear separately in each variable.
|
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|
105,755 |
For a while I've been reading J.E.Humphreys's book "Representations of semisimple Lie algebras in the BGG category $\mathcal O$" under the impression that any module in $\mathcal O$ has a finite generating set composed of highest weight vectors. Now I've realised that I'm lacking a proof. So what would serve as a counterexample? Or has my assumption been correct for some reason? I apologize if the answer can be found further on in the book itself!
|
$X \wedge Y$ represents maps from $X \times Y$ that are base-point-preserving separately in each variable, just as the tensor product represents maps that are linear separately in each variable.
|
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|
105,756 |
Consider the complex 3-fold $SL(2,\mathbb C)/SL(2,\mathbb Z)$ (just for clarity: note that $SL(2,\mathbb Z)$ acts without stabilizers, so this is a complex manifold, not a complex orbifold). Is $SL(2,\mathbb C)/SL(2,\mathbb Z)$ a quasi-projective variety? The natural generalization of this question seems to be the following. Let $G$ be a semisimple linear algebraic group over $\mathbb Q$. Then $G(\mathbb Z)$ is well-defined up to taking a finite-index subgroup. Thus we can ask the same question: is the complex manifold $G(\mathbb C)/G(\mathbb Z)$ a quasi-projective variety?
|
No, the quotient is not quasi-projective. In the paper Invariant meromorphic functions on complex semisimple Lie groups by D. N. Ahiezer you can find the following result. Theorem. Let $G$ be a connected semisimple linear algebraic group
defined over the rationals and
$\Gamma$ be a subgroup of $G(\mathbb
> Q)$ which is Zariski dense in $G$.
Then there are no invariant analytic
hypersurfaces in $G$ invariant by the
action of $\Gamma$. In particular, if
the quotient $G/ \Gamma$ exists as a
complex variety then every
meromorphic function on it is constant. This implies that the quotient is not quasi-projective and even more: it cannot be holomorphically embedded in any algebraic variety.
|
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|
105,878 |
I am studying orbifold fundamental group (or more generally orbifold homotopy groups). In a nutshell, my questions is: what are they intuitively? In what follows I give definitions and more precise questions. My definition of orbifold fundamental group is via classifying space of groupoid, which is explained in the next paragraph (so you may want to skip it if you know the definition). Let $\mathcal{G}$ be a topological groupoid consisting a topological spaces $G_{0}$ of $objects $ and $G_{1}$ of $arrows$ together with usual continuous structure maps. Let $|\mathcal{G}|$ denote the associated topological space $G_{0}/G_{1}$. Let $G_{n}$ be the iterated fibered product $G_{n}=G_{1}\times_{s,t} G_{n-1}$. These $G_{n}$ have the structure of a simplicial topological space, called the $nerve$ of $\mathcal{G}$. Face operads $d_{i}:G_{n}\rightarrow G_{n-1}$ for $i=0,\dots,n$ are given by
$$
d_{i}(g_{1},\dots,g_{n})=(g_{1},\dots,g_{i}g_{i+1},\dots,g_{n})
$$
for $i=1,\dots,n-1$ and
$$
d_{0}(g_{2},\dots,g_{n})=(g_{2},\dots,g_{n}), \ \
d_{n}(g_{1},\dots,g_{n})=(g_{2},\dots,g_{n-1}).
$$
The classifying space $B\mathcal{G}$ of $\mathcal{G}$ is then defined as
$$
B\mathcal{G}=\bigsqcup_{n}(G_{n}\times \Delta^{n})((d_{i}(g),x)\sim(g,\delta_{i}(x))),
$$
where $\Delta^{n}$ is the topological $n$-simplex and $\delta_{i}:\Delta^{n-1}\rightarrow \Delta^{n}$ is the standard facemap.\ The $n$-th orbifold homotopy group of $\mathcal{G}$ based at $x\in |\mathcal{G}|$ is defined to be
$$
\pi_{n}^{orb}(\mathcal{G},x)=\pi_{n}(B\mathcal{G},y),
$$
where $y\in G_{0}$ maps to $x$ under the quotient map $G_{0}\rightarrow |\mathcal{G}|$. The following are my questions: Why is this a reasonable definition? Any manifold $M$ can be thought of topological groupoid via its chart i.e. $G_{0}=\bigsqcup_{i}U_{i}$ and $G_{1}=\bigsqcup_{i,j}U_{i}\times_{M} U_{j}$. It is not clear to me that the definition above reproduce $\pi_{n}(M)$. I am aware of an explicit description the orbifold fundamental groups of the orbifold Riemann surface $\Sigma_{g,n,k}$ of genus $g$ and $n$ orbifold points $p_{i}$ of order $k_{i}$:
$$
\pi_{n}^{orb}(\Sigma_{g,n,k})
=\langle \alpha_{i},\beta_{i},\sigma_{j} \ (1\le i \le g,1\le j \le n)\ | \
\sigma_{1}\dots\sigma_{n}\prod_{i=1}^{g}[\alpha_{i},\beta_{i}]=1,,\sigma_{i}^{k_{i}}=1\rangle
$$
Is it easy to see this explicit presentation by the definition above? It seems there are several ways to define the fundamental group of an orbifold, such as covering space etc. How should one understand orbifold fundamental groups? Thank you for your assistance.
|
The definition of an orbifold in terms of a groupoid is flexible and technically useful and gives very clean definitions, but it's not so close to geometric intuition. Perhaps it is once you've mastered the art of thinking simplicially, which I probably haven't. I tend to think of orbifolds like this: the simplest orbifolds are the global quotients $[X/G]$. In this case the formalism has been cooked up so that geometry of $[X/G]$ is exactly the same as $G$-equivariant geometry on $X$ (whatever this means in a given context). Another catchphrase for this is that all group actions behave like free group actions. All other orbifolds can be obtained by gluing together ones of the above form. This leads to the definition in terms of coordinate patches of the form $[U/G]$. The easiest example of an orbifold fundamental group is when your orbifold has the form $[X/G]$ with $X$ simply connected. Since every group action behaves like a free one, the map $X \to [X/G]$ is a covering map, which exhibits $X$ as the universal cover of the orbifold and $G$ as the group of deck transformations. So $\pi_1([X/G]) = G$. The second simplest example is, I think, an (effective) orbifold Riemann surface. Fortunately you asked precisely about this one. Here you can really think concretely about loops on your surface. Fix an orbifold point $x$ of order $n$. The idea is that an orbifold point of order $n$ is $(1/n)$th of a point, so that an orbifold point is something inbetween an ordinary point and a puncture. The higher order stabilizer the point has, the closer it is to being an actual puncture. More concretely, what this means is that a loop on your surface that winds exactly $n$ times around $x$ can be shrunk across $x$ . It's like the $n$ turns together add up to one whole point, which your loop can then slide across. To see this slightly more formally, think in a chart centered on $x$, where your orbifold looks like $[D/\mu_n]$, with $D$ the unit disk and $\mu_n$ the group of $n$th roots of unity. Any loop in $D$ around the origin can be shrunk to a point, which should imply that the image of this loop in our orbifold is also homotopically trivial. But the image is just a loop that goes $n$ times around $x$. To see that no fewer than $n$ turns suffice, you need to think a bit about the definition of a loop on an orbifold. In any case, once we accept this fact we can obtain the presentation of the fundamental group that you gave in your question, in the same way as for the ordinary fundamental group of a punctured Riemann surface. An example is $[S^2 / \mu_n]$ with $\mu_n$ acting by rotations along the equator. You have orbifold points at the north and south pole. Either of the two descriptions above immediately imply that the fundamental group is cyclic of order $n$: in terms of the second description, a generator is a simple loop around one of the poles, which becomes trivial when it is wound around itself $n$ times. You might also find my recent question, and the answer by Jeffrey Giansiracusa, useful: Homotopy theory of topological stacks/orbifolds
|
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|
105,971 |
(And a related question: Where should an analytic number theorist learn about Bessel functions?) Bessel functions occur quite frequently in analytic number theory. One example, Corollary 4.7 of Iwaniec and Kowalski, says the following. Let $r(m)$ be the number of representations of $m$ as two squares, and suppose that $g$ is smooth and compactly supported in $(0, \infty)$. Then, $$\sum_{m = 1}^{\infty} r(m) g(m) = \pi \int_0^{\infty} g(x) dx + \sum_{m = 1}^{\infty} r(m) h(m),$$ where
$$h(y) = \pi \int_0^{\infty} g(x) J_0(2 \pi \sqrt{xy}) dx.$$ $J_0(x)$ is a Bessel function, and I+K follow with four equivalent integral expressions -- the equivalence of which is not at all obvious by looking at them. Looking at Lemma 4.17, the relevance appears to be that Bessel functions arise when you take Fourier transforms of radially symmetric functions. Another example comes from (3.8) of this paper of Miller and Schmid, and the relevance comes from the identity $$\int_0^{\infty} J_0(\sqrt{x}) x^{s - 1} dx = 4^s \frac{\Gamma(s)}{\Gamma(1 - s)},$$ where the gamma factors come from functional equations of $L$-functions. Okay, if this is true, then I understand why we care, but it seemed a bit deus ex machina to me. There are many other examples too, for example the Petersson formula in the theory of modular forms, etc. There are $I$-Bessel functions, $K$-Bessel functions, $Y$-Bessel functions, etc., all of which seem to satisfy a dizzying number of highly nontrivial identities, and reading Iwaniec and Kowalski one gets the sense that an expert should have the ability to recognize and manipulate them on sight. They also provide references to, e.g., (23.451.1) of a book by Gradhsteyn and Rizhik, and although I confess I have not looked at it, I can infer from the formula number that it is not the sort of thing I might read on an airport layover. Meanwhile, Wikipedia tells me that they naturally arise as solutions of certain partial differential equations. Looks extremely interesting, although I'm afraid I am not an expert in PDE. As an analytic number theorist, how might I make friends with these objects? How should I look at them, and what conceptual frameworks do they fit in? Thank you! (ed. Thanks to everyone for informative answers! I could only accept one answer but +1 all around)
|
Radial Fourier transforms provide a good, consistent perspective on most of the theory. The Fourier transform $\widehat{f}(t)$ of a function $f \colon \mathbb{R}^n \to \mathbb{R}$ is given by the integral of $f(x) e^{2\pi i \langle x,t \rangle} \, dx$ over $x \in \mathbb{R}^n$. If $f$ is a radial function (i.e., $f(x)$ depends only on $|x|$), then we can radial symmetrize everything and the exponential function averages out to a radial function. Specifically, we get
$$
\widehat{f}(t) = 2\pi |t|^{-(n/2-1)} \int_0^\infty f(r) J_{n/2-1} (2 \pi r |t|) r^{n/2} \, dr.
$$
The precise factors are a little annoying, but basically this just means $J_{n/2-1}$ is what you get when you radially symmetrize an exponential function in $n$ dimensions. It's easy to see that if you symmetrize $e^{2\pi i \langle x,t \rangle}$ by averaging over all $x$ on a sphere, then you get a radial function of $t$, and furthermore as you vary the radius of the sphere you just rescale the function. So the one function $J_{n/2-1}$ captures all of this, modulo scaling. One consequence is that Bessel functions inherit the orthogonality of the exponential functions (i.e., the different scalings are orthogonal), so they also inherit all the consequences of orthogonality. For example, this is really where the differential equation comes from. There's a strong analogy between Bessel functions and orthogonal polynomials, where rescaling the Bessel function corresponds to varying the degree of the polynomial. You also get certain qualitative results for free: for example, the product of two Bessel functions should be an integral of Bessel functions with positive coefficients, since this corresponds to saying the product of two radial, positive-definite functions remains positive definite. You can write down the coefficients explicitly, but sometimes all you need is nonnegativity, and in any case this point of view makes it easy to believe that there should be an explicit formula. This is basically a low-brow version of the representation theory approach. Basically, ordinary Fourier analysis studies $L^2(\mathbb{R}^n)$ under the action of the translation group $\mathbb{R}^n$. If you look at the full group of isometries of $\mathbb{R}^n$ (including the orthogonal group), then it's just a little more elaborate, and the Bessel functions arise as zonal spherical functions. It's worthwhile working through this perspective, but in practice just thinking about radial Fourier analysis gives you most of the benefits with less machinery.
|
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|
106,191 |
This is mostly a reference request, as this must be well-known! Let $A$ and $B$ be two real symmetric matrices, one of which is positive definite. Then it is easy to see that the product $AB$ (or $BA$, which has the same eigenvalues) is similar to a symmetric matrix, so has real eigenvalues. Take the vectors of eigenvalues of $A$ and of $B$, sorted in decreasing order, and let their componentwise product be $ab$. What is known about the relationship (e.g., inequalities) between $ab$ and the vector of eigenvalues of the product $AB$ (also taken in decreasing order)? Some experimentation gives the conjecture that there is a majorization order between them, for instance. This must be well-known!
|
Here are the results that you are probably looking for. The first one is for positive definite matrices only (the theorem cited below fixes a typo in the original, in that the correct version uses $\prec_w$ instead of $\prec$). Theorem (Prob.III.6.14; Matrix Analysis, Bhatia 1997). Let $A$ and $B$ be Hermitian positive definite. Let $\lambda^\downarrow(X)$ denote the vector of eigenvalues of $X$ in decreasing order; define $\lambda^\uparrow(X)$ likewise. Then,
\begin{equation*}
\lambda^\downarrow(A) \cdot \lambda^\uparrow(B) \prec_w \lambda(AB) \prec_w \lambda^\downarrow(A) \cdot \lambda^\downarrow(B),
\end{equation*} where $x \cdot y := (x_1y_1,\ldots ,x_ny_n)$ for $x,y \in \mathbb{R}^n$ and $\prec_w$ is the weak majorization preorder. However, when dealing with matrix products, it is more natural to consider singular values rather than eigenvalues. Therefore, the relation that you might be looking for is the log-majorization \begin{equation*}
\log \sigma^\downarrow(A) + \log\sigma^\uparrow(B) \prec \log\sigma(AB) \prec \log\sigma^\downarrow(A) + \log\sigma^\downarrow(B),
\end{equation*}
where $A$ and $B$ are arbitrary matrices, and $\sigma(\cdot)$ denotes the singular value map. Reference R. Bhatia. Matrix Analysis . Springer, GTM 169. 1997.
|
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|
106,497 |
The proof that the set of classes of vector bundles is homotopy invariant relies on the paracompactness and the Hausdorff property of the base space. Are there any known examples of: Non trivial vector bundles over a paracompact non-Hausdorff contractible space Non trivial vector bundles over a Hausdorff non-paracompact contractible space Non trivial vector bundles over a non-Hausdorff non-paracompact contractible space
|
Let $U$ and $V$ be two copies of the real line and make a space $X$ by gluing them by the identity along the strictly positive half-line: $x\in U$ equals $x\in V$ for $x>0$. Now make a rank one vector bundle over this space by taking a trivial bundle over each of the lines: glue $U\times\mathbb R$ to $V\times \mathbb R$ by identifying $(x,y)\in U\times\mathbb R$ with $(x,f(x)y)\in V\times\mathbb R$ for $x>0$. Let's choose the clutching function to be $f(x)=x$, or any other nowhere zero continuous function of positive real $x$ that cannot be extended to a nowhere zero function of all real $x$. Surely $X$ is contractible, but the bundle can't be trivial since the function $f:U\cap V\to GL_1(\mathbb R)$ cannot be expressed as the quotient of a function extendible to $U$ and a function extendible to $V$, since that would make it itself extendible to all of $\mathbb R$.
|
{
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"https://mathoverflow.net/questions/106497",
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|
106,560 |
Mochizuki has recently announced a proof of the ABC conjecture. It is far too early to judge its correctness, but it builds on many years of work by him. Can someone briefly explain the philosophy behind his work and comment on why it might be expected to shed light on questions like the ABC conjecture?
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I'll take a stab at answering this controversial question in a way that might satisfy the OP and benefit the mathematical community. I also want to give some opinions that contrast with or at least complement grp. Like others, I must give the caveats: I do not understand Mochizuki's claimed proof, his other work, and I make no claims about the veracity of his recent work. First, some background which might satisfy the OP. For years, Mochizuki has been working on things related to Grothendieck's anabelian program. Here is why one might hope this is useful in attacking problems like ABC: Begin with the Neukirch-Uchida theorem. See "Über die absoluten Galoisgruppen algebraischer Zahlkörper," by J. Neukirch, Journées Arithmétiques de Caen (Univ. Caen, Caen, 1976), pp. 67–79. Asterisque, No. 41-42, Soc. Math. France, Paris, 1977. Also "Isomorphisms of Galois groups," by K. Uchida, J. Math. Soc. Japan 28 (1976), no. 4, 617–620. The main result of these papers is that a number field is determined by its absolute Galois group in the following sense: fix an algebraic closure $\bar Q / Q$, and two number fields $K$ and $L$ in $\bar Q$. Then if $\sigma: Gal(\bar Q / K) \rightarrow Gal(\bar Q / L)$ is a topological isomorphism of groups, then $\sigma$ extends to an inner automorphism $Int(\tau): g \mapsto \tau g \tau^{-1}$ of $Gal(\bar Q / Q)$. Thus $\tau$ conjugates the number field $K$ to the number field $L$, and they are isomorphic. So while class field theory guarantees that the absolute Galois group $Gal(\bar Q / K)$ determines (the profinite completion of) the multiplicative group $K^\times$, the Neukirch-Uchida theorem guarantees that the entire field structure is determined by the profinite group structure of the Galois group. Figuring out how to recover aspects of the field structure of $K$ from the profinite group structure of $Gal(\bar Q / K)$ is a difficult corner of number theory. Next, consider a (smooth) curve $X$ over $Q$; suppose that the fundamental group $\pi_1(X({\mathbb C}))$ is nonabelian. Let $\pi_1^{geo}(X)$ be the profinite completion of this nonabelian group. Basic properties of the etale fundamental group give a short exact sequence:
$$1 \rightarrow \pi_1^{geo}(X) \rightarrow \pi_1^{et}(X) \rightarrow Gal(\bar Q / Q) \rightarrow 1.$$ Now, just as one can ask about recovering a number field from its absolute Galois group ($Gal(\bar Q / K)$ is isomorphic to $\pi_1^{et}(K)$), one can ask how much one can recover about the curve $X$ from its etale fundamental group. Any $Q$-point $x$ of $X$, i.e. map of schemes from $Spec(Q)$ to $Spec(X)$ gives a section $s_x: Gal(\bar Q / Q) \rightarrow \pi_1^{et}(X)$. One case of the famous "section conjecture" of Grothendieck states that this gives a bijection from $X(Q)$ to the set of homomorphisms $Gal(\bar Q / Q) \rightarrow \pi_1^{et}(X)$ splitting the above exact sequence. One hopes, more generally, to recover the structure of $X$ as a curve over $Q$ from the induced outer action of $Gal(\bar Q / Q)$ on $\pi_1^{geo}(X)$. (take an element $\gamma \in Gal(\bar Q / Q)$, lift it to $\tilde \gamma \in \pi_1^{et}(X)$, and look at conjugation of the normal subgroup $\pi_1^{geo}(X)$ by $\tilde \gamma$, well-defined up to inner automorphism independently of the lift.) As in the case of the Neukirch-Uchida theorem, there is an active and difficult corner of number theory devoted to recovering properties of rational points of (hyperbolic) curves from etale fundamental groups. Here are two dramatically difficult problems in the same spirit: How can you describe the regulator of a number field $K$ from the structure of the profinite group $Gal(\bar Q / K)$? Given a section $s: Gal(\bar Q / Q) \rightarrow \pi_1^{et}(X)$, how can one describe the height of the corresponding point in $X(Q)$? I would place Mochizuki's work in this anabelian corner of number theory; I have always kept a safe and respectful distance from this corner. Now, to say something not quite as ancient that I gleaned from flipping through Mochizuki's recent work: Many people here on MO and elsewhere have been following research on the field with one element. It is a tempting object to seek, because analogies between number fields and function fields break down quickly when you realize there is no "base scheme" beneath $Spec(Z)$. But I see Mochizuki's work as an anabelian approach to this problem, and I'll try to describe my understanding of this below. Consider a smooth curve $X$ over a function field $F_p(T)$. The anabelian approach suggests looking at the short exact sequence
$$1 \rightarrow \pi_1^{et}(X_{\overline{F_p(T)}}) \rightarrow \pi_1^{et}(X) \rightarrow Gal(\overline{F_p(T)} / F_p(T)) \rightarrow 1.$$
But much more profitable is to look instead at $X$ as a surface over $F_p$ which corresponds in the anabelian perspective to studying
$$1 \rightarrow \pi_1^{et}(X_{\bar F_p}) \rightarrow \pi_1^{et}(X) \rightarrow Gal(\bar F_p / F_p) \rightarrow 1.$$
But this is pretty close to looking at $\pi_1^{et}(X)$ by itself; there's just a little profinite $\hat Z$ quotient floating around, but this can be characterized (I think) group theoretically within the study of $\pi_1^{et}(X)$ itself. I would understand (after reading Mochizuki) that looking at curves $X$ over function fields $F_p(T)$ as surfaces over $F_p$ is like looking at only the etale fundamental group $\pi_1^{et}(X)$ without worrying about the map to $Gal(\overline{F_p(T)} / F_p(T))$. So, the natural number field analogue would be the following. Consider a smooth curve $X$ over $Q$. In fact, let's make $X = E - \{ 0 \}$ be a once-punctured elliptic curve over $Q$. Then the absolute anabelian geometry suggests that to study $X$, it should be profitable to study the etale fundamental group $\pi_1^{et}(X)$ all by itself as a profinite group. This is the anabelian analogue of what others might call "studying (a $Z$-model of) $X$ as a surface over the field with one element". Without understanding any of the proofs in Mochizuki, I think that his work arises from this absolute anabelian perspective of understanding the arithmetic of once-punctured elliptic curves over $Q$ from their etale fundamental groups. The ABC conjecture is equivalent to Szpiro's conjecture which is a conjecture about the arithmetic of elliptic curves over $Q$. Now here is a suggestion for number theorists who, like myself, have unfortunately ignored this anabelian corner. Let's try to read the papers of Neukirch and/or Uchida to get a start, and let's try to understand Minhyong Kim's work on Siegel's Theorem ("The motivic fundamental group of $P^1 \backslash ( 0, 1, \infty )$ and the theorem of Siegel," Invent. Math. 161 (2005), no. 3, 629–656.) It would be wonderful if, while we're waiting for the experts to weight in on Mochizuki's work, we took some time to revisit some great results in the anabelian program. If anyone wants to start a reading group / discussion blog on these papers, I would enjoy attending and discussing.
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106,705 |
It sometimes happens that 1D problems are easier to solve by somehow adding a dimension. For example, we convert linear differential equations for a real unknown to a complex unknown (to use complex exponentials), or we compute a power series' radius of convergence by thinking in the complex plane (or use complex analytic properties in path integrals), or we evaluate $\int^\infty_{-\infty} e^{-x^2}\ dx$ by squaring it... So, are any 2D problems easier to solve in even higher dimensions? I can't think of any.
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Of course, there is one such problem! This is the Cauchy problem for the wave equation
$$\frac{\partial^2u}{\partial t^2}=\Delta u,\qquad u(x,0)=f(x),\quad \frac{\partial u}{\partial t}(x,0)=g(x),$$
where $x\in{\mathbb R}^d$. To solve it, it is enough to know the case where $f\equiv0$. If $d=3$, this problem is solved by using spherical means. We obtain
$$u(x,t)=tM_{t,x}[g],$$
where $M_{t,x}$ denotes the mean over the sphere of radius $t$ and center $x$. The two-dimensional case is way more complicated. The formula can only be found by considering that a $2$D-solution is a special case of a $3$D-solution. Then the solution involves a complicated integral over the disk $D(x;t)$ instead of the circle. This is why the Huyghens principle holds true in $3$ space dimensions but not in $2$ space dimensions.
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106,838 |
Suppose $n > 1$ is a natural number. Suppose $K$ and $L$ are fields such that the general linear groups of degree $n$ over them are isomorphic, i.e., $GL(n,K) \cong GL(n,L)$ as groups. Is it necessarily true that $K \cong L$? I'm also interested in the corresponding question for the special linear group in place of the general linear group. NOTE 1: The statement is false for $n = 1$, because $GL(1,K) \cong K^\ast$ and non-isomorphic fields can have isomorphic multiplicative groups. For instance, all countable subfields of $\mathbb{R}$ that are closed under the operation of taking rational powers of positive elements have isomorphic multiplicative groups. NOTE 2: It's possible to use the examples of NOTE 1 to construct non-isomorphic fields whose additive groups are isomorphic and whose multiplicative groups are isomorphic.
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The answer is "yes", see below. Dieudonné in his book "La géométrie des groupes classiques"
considers the abstract group $SL_n(K)$ for a field $K$, not necessarily commutative,
and writes $PSL_n(K)$ for $SL_n(K)$ modulo the center.
In Ch. IV, Section 9, he considers the question whether $PSL_n(K)$ can be isomorphic to $PSL_m(K')$ for $n\ge 2,\ m\ge 2$.
He writes that they can be isomorphic only for $n=m$, except for $PSL_2(\mathbb{F}_7)$ and $PSL_3(\mathbb{F}_2)$.
If $n=m>2$, then the isomorphism is possible only if $K$ and $K'$ are isomorphic or anti-isomorphic.
The same is true for $m=n=2$ if both $K$ and $K'$ are commutative, except for the case $K=\mathbb{F}_4$,
$K'=\mathbb{F}_5$. Dieudonné gives ideas of proof and references to Schreier and van der Waerden (1928), to his paper "On the automorphisms of classical groups"
in Mem. AMS No. 2 (1951) and to the paper of Hua L.-K. and Wan in J. Chinese Math. Soc. 2 (1953), 1-32. This answers affirmatively the question for $SL_n$, because if $SL_n(K)\cong SL_n(K')$, then $PSL_n(K)\cong PSL_n(K')$. In the case $n=2$, $K=\mathbb{F}_4$,
$K'=\mathbb{F}_5$, the orders $|SL_2(\mathbb{F}_4)|=60$ and $|SL_2(\mathbb{F}_5)|=120$ are different, and therefore these groups are not isomorphic. This also answers affirmatively the question for $GL_n$, because $SL_n(K)$ is the commutator subgroup of $GL_n(K)$, except for $GL_2(\mathbb{F_2})$, see Dieudonné, Ch. II, Section 1. In the case $n=2$, $K=\mathbb{F}_2$, we have $|GL_2(\mathbb{F}_2)|=6$ , which is less than $|GL_2(\mathbb{F}_q)|=q(q-1)(q^2-1)$ for any $q=p^r>2$, hence
$GL_2(\mathbb{F}_2)\not\cong GL_2(\mathbb{F}_q)$ for $q>2$.
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106,943 |
What is the history behind the colorful name of this result? Cartan-Eilenberg states it without any particular fanfare.
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I suspect the name just arose naturally (for obvious reasons) but that it would be tough to trace back to any single person. After Cartan-Eilenberg proved it in 1956 (Homological Algebra, p.40) the first mention I see in English is by Tate in 1966/67 (p-divisible groups, p.178) followed by Hartshorne in 1968 (Cohomological Dimension of Algebraic Varieties, p.446), neither of which bother with a citation, reference, or quotation marks ( 1 ). However, it was used a bit earlier - also without citation or quotation marks - by Begueri-Poitou in 1965 ( 2 ) as 'lemme du serpent'; mentioned early on in their abstract. [NB: the first page of the linked pdf incorrectly lists the second author's surname as Poiton.] Edit: I used numdam to search for 'serpent'. Cartan has an unrelated quotation about a snake nearly biting its own tail (1965, pdf p.16/17), but actually relevant is a paper by Grothendieck dating to 1964 mentioning a snake diagram ("le diagramme du serpent", pdf p.195/258) that he attributes to (Bourbaki, Alg. comm , chap. I, $\S$1, no 4, prop. 2). You can see the term snake diagram in the much later English translation , and (confirmed by the comments below) it is found in the 1961 original, as well. As far as I know, this is the first published instance that uses the snake terminology. My guess for the time being: The term snake diagram originated (in French) around 1961 and was first used by one of the Bourbaki members (possibly Cartan, Eilenberg, or Grothendieck). Snake lemma almost certainly has a similar origin.
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107,043 |
The extension of the 2-adic valuation to the reals used in the usual proof clearly uses AC. But is this really necessary? After all, given an equidissection in $n$ triangles, it is finite, so it should be possible to construct a valuation for only the algebraic numbers, and the coordinates of the summits (with a finite number of "choices"), and then follow the proof to show that $n$ must be even. Or am I badly mistaken?
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No choice is needed. If, in a choiceless universe, there were a counterexample, then that counterexample amounts to finitely many real numbers (the coordinates of the relevant points). It would still be a counterexample in the sub-universe of sets constructible (in Gödel's sense) from those finitely many reals. But that sub-universe satisfies the axiom of choice, so your favorite ZFC proof of the theorem applies there.
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107,168 |
I asked this question at Maths Stack Exchange , but I haven't received any replies yet (I'm not sure how long I should wait before it is acceptable to ask here, assuming there is such a period of time). The Wikipedia page for Weitzenböck identities is explicitly example based. I am looking for a reference which takes a more rigorous approach (as well as a discussion of the Bochner technique). In particular, I am interested in references which focus on these identities in complex geometry. I have already consulted Griffiths & Harris which is mentioned in the article, but it only contains one example. Berger's A Panoramic View of Riemannian Geometry doesn't have much more. My interest in Weitzenböck identities has been motivated by a question arising from the following theorem: Let $X$ be a Kähler manifold and $E$ a hermitian holomorphic vector bundle with Chern connection $\nabla$. Then for the Laplacians $\Delta_{\bar{\partial}} = \bar{\partial}\bar{\partial}^* + \bar{\partial}^*\bar{\partial}$, $\Delta_{\partial} = \partial\partial^* + \partial^*\partial$, we have $\Delta_{\bar{\partial}} = \Delta_{\partial} + [iF_{\nabla}, \Lambda]$. Is $\Delta_{\bar{\partial}} = \Delta_{\partial} + [iF_{\nabla}, \Lambda]$ an example of a Weitzenböck identity?
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Here is roughly the philosophy of the Weitzenbock technique. (Most of what follows is taken from Berline-Getzler-Vergne book.) Suppose that $E_0,E_1\to M$ are vector bundles on an oriented Riemann manifolds $M$ equipped with hermitian metrics. Denote by $C^\infty(E_i)$ the space of smooth sections of $E_i$. A symmetric 2nd order differential operator $L: C^\infty(E_0)\to C^\infty(E_0)$ is called a generalized Laplacian on $E_0$ if its principal symbol $\sigma_L$ coincides with the principal symbol of a Laplacian. Concretely this means the following. For a smooth function $f\in C^\infty(M)$ denote by $M_f$ the linear operator $C^\infty(E_0)\to C^\infty( E_0)$ defined by the multiplication with $f$. Then $L$ is a generalized Laplacian if for any $f_0,f_1\in C^\infty(M)$ and any $u\in C^\infty(E_0)$ we have $$ [\; [\; L,M_{f_0}\; ], M_{f_1}\; ]u = -2g( df_0,df_1)\cdot u $$ where $[-,-]$ denotes the commutator of two operators. Equivalently, this means $$[[L,M_{f_0}],M_{f_1}]=-2M_{g(df_0,df_1)}. $$ One can show that if $L$ is a generalized Laplacian on $E_0$, then there exists a connection $\nabla$ on $E_0$, compatible with the metric on $E_0$, and a symmetric endomorphism $W$ of $E_0$ such that $$ L =\nabla^*\nabla +W. $$ The classical Weitzenbock formulas give explicit descriptions to the Weitzenbock remainder $W$ and the connection $\nabla$. Usually the generalized Laplacians are obtained through Dirac type operators which are first order differential operators $D: C^\infty(E_0)\to C^\infty(E_1)$ such that both operators $D^\ast D$ and $D D^\ast$ are generalized Laplacians on $E_0$ and respectively $E_1$. We can rewrite this in a compact form by using the operator $$\mathscr{D}: C^\infty(E_0)\oplus C^\infty(E_1)\to C^\infty(E_0)\oplus C^\infty(E_1), $$ $$\mathscr{D}(u_0\oplus u_1)= (D^*u_1)\oplus (D u_0). $$ Then $D$ is Dirac type iff $\mathscr{D}^2$ is a generalized Laplacian. The Weitzenbock remainders of $D^\ast D$ and $D D^\ast$ involve curvature terms. If the Weitzenbock remainder of $D^*D$ happens to be a positive endomorphism of $E_0$, then one can conclude that $$\ker D=\ker D^\ast D=0. $$ The Hodge-Dolbeault operator $$\frac{1}{\sqrt{2}}(\bar{\partial}+\bar{\partial}^*): \Omega^{0,even}(M)\to \Omega^{0,odd}(M) $$ on a Kahler manifold $M$ is a Dirac type operator. For more details and examples you can check Sec. 10.1 and Chap 11 of my lecture notes .
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107,327 |
Given the Dedekind eta function $\eta(\tau)$. Define $m = (p-1)/2$ and a $24$th root of unity $\zeta = e^{2\pi i/24}$. Let p be a prime of form $p = 12v+5$. Then for $n = 2,4,8,14$:
$$\sum_{k=0}^{p-1} \Big(\zeta^{m k}\, \eta\big(\tfrac{\tau+m k}{p}\big)\Big)^n = -\big(\sqrt{p}\;\eta(p\tau)\big)^n\tag1$$ Let p be a prime of form $p = 12v+11$. Then for $n = 2,6,10,14, and\; 26$:
$$\sum_{k=0}^{p-1} \Big(\zeta^{m k}\, \eta\big(\tfrac{\tau+m k}{p}\big)\Big)^n = \big(\sqrt{p}\;\eta(p\tau)\big)^n\tag2$$ Note : One can also use the identity $\zeta^N \eta(\tau) = \eta(\tau+N)$ for a further simplification. It is easily checked with Mathematica that $(1)$ and $(2)$ hold for hundreds of decimal digits, but are they really true? (Kindly also see this related post . I have already emailed W. Hart, and he replied he hasn't seen such identities yet.) ----EDIT---- The $n = 26$ for proposed identity $(2)$ was added courtesy of W.Hart's answer below. (I'm face-palming myself for not checking $n = 26$.)
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This question asks in effect to show that $\eta^n$ is a $\pm p^{n/2}$
eigenfunction for the Hecke operator $T_p$. The claim holds because
each of these $\eta^n$ happens to be a CM form of weight $n/2$,
and $p$ is inert in the CM field ${\bf Q}(i)$ or ${\bf Q}(\sqrt{-3})$.
In plainer language, the sum over $k$ takes the $q$-expandion
$$
\eta(\tau)^n = \sum_{m \equiv n/24 \phantom.\bmod 1} a_m q^m
$$
and picks out the terms with $p|m$, multiplying each of them by $p$;
and the result is predictable because the only $m$ that occur are
of the form $(a^2+b^2)/d$ or $(a^2+ab+b^2)/d$ where $d = 24 / \gcd(n,24)$,
and the congruence on $p$ implies that $p|m$ if and only if $p|a$ and $p|b$. For $n=2$ this is immediate from the pentagonal
number identity ,
which states in effect that $\eta(\tau)$ is the sum of $\pm q^{c^2/24}$
over integers $c \equiv 1 \bmod 6$, with the sign depending on $c \bmod 12$
(and $q = e^{2\pi i \tau}$ as usual). Thus
$$
\eta^2 = \sum_{c_1^{\phantom0},c_2^{\phantom0}}
\pm q^{(c_1^2+c_2^2)/24} = \sum_{a,b} \pm q^{(a^2+b^2)/12}
$$
where $c,c' = a \pm b$. Once $n>2$ there's a new wrinkle: the coefficient of each term
$q^{(a^2+b^2)/d}$ or $q^{(a^2+ab+b^2)/d}$ is not just $\pm 1$
but a certain homogeneous polynomial of degree $(n-2)/2$ in $a$ and $b$
(a harmonic polynomial with respect to the quadratic form $a^2+b^2$ or
$a^2+ab+b^2$). Explicitly, we may obtain the CM forms $\eta^n$ as follows: @ For $n=4$, sum $\frac12 (a+2b) q^{(a^2+ab+b^2)/6}$ over all $a,b$ such that
$a$ is odd and $a-b \equiv1 \bmod 3$. [This is closely related with the
fact that $\eta(6\tau)^4$ is the modular form of level $36$ associated to
the CM elliptic curve $Y^2 = X^3 + 1$, which happens to be isomorphic with
the modular curve $X_0(36)$.] @ For $n=6$, sum $(a^2-b^2) q^{(a^2+b^2)/4}$ over all $a \equiv 1 \bmod 4$
and $b \equiv 0 \bmod 2$. @ For $n=8$, sum $\frac12 (a-b)(2a+b)(a+2b) q^{(a^2+ab+b^2)/3}$
over all $(a,b)$ congruent to $(1,0) \bmod 3$. @ For $n=10$, sum $ab(a^2-b^2) q^{(a^2+b^2)/12}$
over all $(a,b)$ congruent to $(2,1) \bmod 6$. @ Finally, for $n=14$, sum
$\frac1{120} ab(a+b)(a-b)(a+2b)(2a+b)q^{(a^2+ab+b^2)/12}$
over all $a,b$ such that $a \equiv 1 \bmod 4$ and $a-b \equiv 4 \bmod 12$. I can't give a reference for these identities, but once such a formula
has been surmised it can be verified by showing that the sum over $a,b$
gives a modular form of weight $n/2$ and checking that it agrees with
$\eta^n$ to enough terms that it must be the same.
|
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107,825 |
The Thom class and Thom isomorphism theorem for oriented vector bundles are proven ( at least to my knowledge) by induction on the open covers and some manipulation with Mayer-Vietoris sequences. What is the "actual reason" behind the existence of Thom class? It seems strange that such an interesting class would exist just because some Mayer-Vietoris sequences routinely produce it.
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It is easy to understand the existence of a Thom class by considering cellular cohomology. Let the given vector bundle be $E\to B$ with fibers of dimension $n$. One can assume without significant loss of generality that $B$ is a CW complex with a single 0-cell. The Thom space $T(E)$ is the quotient $D(E)/S(E)$ of the unit disk bundle of $E$ by the unit sphere bundle. One can give $T(E)$ a CW structure with $S(E)/S(E)$ as the only 0-cell and with an $(n+k)$-cell for each $k$-cell of $B$. These cells in $T(E)$ arise from pulling back the bundle $D(E)\to B$ via characteristic maps $D^k\to B$ for the $k$-cells of $B$. These pullback are products since $D^k$ is contractible. In particular, $T(E)$ has a single $n$-cell and an $(n+1)$-cell for each 1-cell of $B$. There are no cells in $T(E)$ between dimension $0$ and $n$. The cellular boundary of an $(n+1)$-cell is 0 if $E$ is orientable over the corresponding 1-cell of $B$, and it is twice the $n$-cell in the opposite case. Thus $H^n(T(E);{\mathbb Z})$ is $\mathbb Z$ if $E$ is orientable and $0$ if $E$ is non-orientable. In the orientable case a generator of $H^n(T(E);{\mathbb Z})$ restricts to a generator of $H^n(S^n;{\mathbb Z})$ in the "fiber" $S^n$ of $T(E)$ over the 0-cell of $B$, hence the same is true for all the "fibers" $S^n$ and so one has a Thom class.
|
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|
107,945 |
I am quite new at nonstandard analysis, and recently I became aware of its use in probability theory mainly through the following two books: Nelson (1987). Radically Elementary Probability Theory Geyer (2007). Radically Elementary Probability and Statistics Although Nelson's book is several decades old, as far as I can see, its approach has not yet caught on. Also, I couldn't find a lot of papers published in the leading probability journals on that topic. I am quite intrigued by that phenomenon. My questions are the following Why hasn't nonstandard analysis been widely adopted by probabilists? Were there some success stories in some particular sub-fields of probability theory or statistics? Does there exist some known fundamental objections in probability theory to the approach in there?
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Non-standard analysis has been quite successful in settling existence questions in probability theory. Hyperfinite Loeb spaces allow for several constructions that cannot be done on standard probability spaces. In particular, NSA was quite useful for the construction of certain adapted processes. There is a paper by Hoover and Keisler, Adapted Probability Distributions , from 1984, in which the authors show that many of the properties that make hyperfinite Loeb spaces so useful where due to a property they called saturation: A probability space $(\Omega,\Sigma,\mu)$ is saturated if whenever $\nu$ is a Borel probability measure on $[0,1]^2$ and $f:\omega\to[0,1]$ a random variable with distribution equal to the marginal of $\nu$ on the first coordinate, then there exists a random variable $g:\Omega\to[0,1]$ such that the distribution of $(f,g)$ is $\nu$. An example of a saturated probability space that is not a hyperfinite Loeb space is the coin-flipping measure on $\{0,1\}^\kappa$ when $\kappa$ is uncountable. A relatively readable exposition of this approach can be found in the small book Model Theory of Stochastic Processes by Fajardo and Keisler . There are also several related papers and surveys on Keisler's homepage . In a sense, we nowadays understand fairly well how certain powerful techniques of non-standard analysis work below the surface, so we can use a lot of the constructions freed of NSA. There isn't really anything where it is necessary to use NSA. Still, NSA is a rather powerful and useful tool. A good overview over what it can do for probability theory, mainly the theory of sochastic processes, is in the article by Osswald and Sun in Nonstandard Analysis for the Working Mathematician by Loeb.
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107,955 |
Given a graph $G$, let $H$ be a $k$-degenerate (not necessarily induced) subgraph of maximal size. Are there any known lower bounds on $|E(H)|$ for particular classes of $G$ and values of $k$? I've seen several papers on the subject, but only for induced subgraphs $H$. I'm particularly interested in the case where $k=2$ and $G$ is both regular and bipartite, but any additional information would be helpful. (I'm aware of the naive lower bound found by arbitrarily removing edges until no vertex has degree $(k+1)$ or more...I'm looking for something better).
|
Non-standard analysis has been quite successful in settling existence questions in probability theory. Hyperfinite Loeb spaces allow for several constructions that cannot be done on standard probability spaces. In particular, NSA was quite useful for the construction of certain adapted processes. There is a paper by Hoover and Keisler, Adapted Probability Distributions , from 1984, in which the authors show that many of the properties that make hyperfinite Loeb spaces so useful where due to a property they called saturation: A probability space $(\Omega,\Sigma,\mu)$ is saturated if whenever $\nu$ is a Borel probability measure on $[0,1]^2$ and $f:\omega\to[0,1]$ a random variable with distribution equal to the marginal of $\nu$ on the first coordinate, then there exists a random variable $g:\Omega\to[0,1]$ such that the distribution of $(f,g)$ is $\nu$. An example of a saturated probability space that is not a hyperfinite Loeb space is the coin-flipping measure on $\{0,1\}^\kappa$ when $\kappa$ is uncountable. A relatively readable exposition of this approach can be found in the small book Model Theory of Stochastic Processes by Fajardo and Keisler . There are also several related papers and surveys on Keisler's homepage . In a sense, we nowadays understand fairly well how certain powerful techniques of non-standard analysis work below the surface, so we can use a lot of the constructions freed of NSA. There isn't really anything where it is necessary to use NSA. Still, NSA is a rather powerful and useful tool. A good overview over what it can do for probability theory, mainly the theory of sochastic processes, is in the article by Osswald and Sun in Nonstandard Analysis for the Working Mathematician by Loeb.
|
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|
108,505 |
I am trying to prepare a "mathematics talk" for five year olds from my daughter's elementary school. I have given many mathematics talks in my life but this one feels
very tough to prepare. Could the members of the community share their experience
with these kind of lectures. I was thinking to talk about some theorems of Euclidean geometry which will include some old fashion compass, straight edge construction with some kind "magical outcome" and then try to give kids some logical reasons for the "magic". Any ideas? Edit: I would like to thank one more time to the members of the mathoverflow community for their generous input and support as well to report the outcome of my talk. I just got out from my daughter's elementary school where I ended up teaching four class periods today instead of the one I originally prepared for. I taught two sections of 5 year olds (26 kids a section) as well as two large group of fifth graders (close to 100 kids in total). I was "over prepared" to talk to 5 year olds which came handy with fifth graders. Inspired by the answers from this forum I chose to talk about Platonic solids and have kids mostly engaged in practical activities as oppose of "teaching" them. My assistant chair at Augusta State University Georgia has generously shared her large collection of POLYDRON blocks. I had three bags full of equilateral triangles, squares, and pentagons. I have also pre-build one set of all five Platonic solids (Tetrahedron, Cube,
Octahedron, Dodecahedron, and Icosahedron). I have also printed out cut and fold maps for all solids from this website and gave them to kids together with building blocks. We identified first the properties of polygons (number of sides, vertices, and angles) of each of building blocks were to use as well as the fact they were regular (sides and angles of equal length). I was rather surprised that five year old children have no problem identifying pentagon as it is the shape of rather important building in Washington DC. Then we introduced the rules of our "game": Only the same "shapes" were to be used for building solids. Two faces could meet only in one edge. Each vertex of the solid had to meet the same number of faces. Five year old kids had no problem assembling Tetrahedron, Cube,
Octahedron however not a single group (they were allowed to work alone of in groups of 2-3) was able to assemble Dodecahedron, and Icosahedron. This was not the case with fifth graders (older kids) where several groups (4-5 out of 100 kids) successfully assembled Dodecahedron, and Icosahedron. Even 5 year olds were able to identify number of faces, edges, and vertices by counting from the cut and fold charts. They had harder time identifying Schläfli symbols for each Platonic surface due to the fact that they had to count them on my pre-build models but they have never the less accomplished the task. We were able to come up with Euler characteristic (magic number as I referred) but the real focus was on subtracting numbers which we did using our fingers. Obviously the kids got lost after the cube due to the size of numbers involved. I was not able to convey any information about further combinatorial properties of Platonic solids related to Schläfli symbols to five year olds. On another hand fifth graders had no problems identifying $$pF=2E=qV$$ but had hard time solving equations as $pF=2E$ for $F$ and $2E=qV$ for $V$ and substituting
into $$V-E+F=2$$ not a single fifth grader could follow my computation for the estimate $$\frac{1}{p}+\frac{1}{q}>\frac{1}{2}$$ where were effectively ended our little lecture. In both sections kids asked me to preform some more "magic tricks". I glued a long strip of paper for them creating a cylinder and Mobius bend. Many kids thought of cylinder as a circle and Mobius bend as a figure eight (few fifth graders mentioned infinity symbols) even that they could not give any logical explanation why they think that way. We cut cylinder and Mobius bend and children start cheering my name when Mobius bend "broke" into just another bigger Mobius bend. Five year olds wanted to hug me after the lecture and sit at my table in cafeteria. The fifth graders were either indifferent or came to me after the talk to shake my hand and ask if I can teach another class. It is also worth noting that while playing with blocks many fifth graders made prisms, pyramids while some try to pass non-platonic solids for Platonic solids bending rules of our game. Teachers were trusty for this kind of experience. They are in a bad need for professional development after years of budget cuts and fear for their jobs. The school is going to buy blocks. I hope to make visits semi-regular and help them as much as I can (obviously out of selfish interest to improve my daughter's education). I have already planed to introduce some other games like tangram, pentominoes, and Hanoi tower. I will also install GeoGebra on their computers. I might edit this post in next few days and add few details. Most Kind Regards, Predrag
|
I'm going to quote Bill Thurston from his interview for More Mathematical People : Thurston: ... One thing that is very important is the education of children... In the elementary schools in Princeton that my kids have attended, there is an annual event called Science Day. They bring in scientists from the community, and we spend a day going around from class to class talking about things. I have enjoyed doing that quite a bit. MMP: What have you talked about? Thurston: I have done different things every year for ten years or so; for example, topology, symmetry, binary counting on fingers... I find that kids are really ready to pick up mathematics in the way that I myself think about it. Of course, it's toned down. MMP: Can you be a little bit more specific about the way you think about mathematics? Thurston: That's a tough question. It might be nice to give an example. At one time I went into a class of kids and made lots of equilateral triangles. We made a tetrahedron by putting three triangles at each vertex. Then I asked what happens if you put four triangles, and they constructed an octahedron. Then with five triangles at each vertex they constructed an icosahedron. But with six triangles they found that the construction just lays flat. And then I asked about seven triangles at each vertex. They pieced it together and they got these hyperbolic tesselations in four-space. They loved that. The kids did. But the teacher really felt ill at ease. She didn't know what was happening.
|
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|
108,598 |
I'm teaching Calculus and my students asked me to calculate the integral of $\ x\! \cdot\!\tan(x)$ . I spent quite a lot of effort to do this, but I'm now even not sure if the integral could be presented in elementary functions. Does anybody know how to calculate it, or otherwise prove it is not integrable in elementary functions?
|
Finding an anti-derivative of $x\tan x$ amounts to finding an anti-derivative of $f=\frac{x}{e^x+1}$. Consider the field $K=\mathbb C(x,e^x)$. Note that $K$ is closed under taking derivatives. If $f$ is elementary integrable, then Liouville's Theorem gives elements $u_i\in K$, $\gamma_i\in\mathbb C$, $v\in K$ with
\begin{equation}
\frac{x}{e^x+1}=\sum\gamma_i\frac{u_i'}{u_i}+v'.
\end{equation}
Consider the $u_i$ and $v$ as rational functions in $e^x$ with coeffcients in $\mathbb C(x)$. By the property of the logarithmic derivative we may assume that the $u_i$ are actually distinct irreducible monic polynomials with respect to $e^x$, or elements from $\mathbb C(x)$. Looking at poles (with respect to the `variable' $e^x$) shows that at most one of the $u_i$ is $e^x+1$, and the other $u_i$'s are in $\mathbb C(x)$. Similarly, we see that $v\in\mathbb C(x)$. So there indeed must be one index $i$ with $u_i=e^x+1$. However, $\frac{x}{e^x+1}-\gamma_i\frac{u_i'}{u_i}=\frac{x}{e^x+1}-\gamma_i\frac{e^x}{e^x+1}$ isn't in $\mathbb C(x)$, a contradiction. Remark: The argument given here is somewhat sketchy, some routine details need to be filled in, like that $u_i'$ and $u_i$, as polynomials in $e^x$, are relatively prime. A beautiful paper about Liouville's Theorem is Rosenlicht's article Integration in finite terms . My argument somewhat follows Rosenlicht's example of finding an anti-derivative of $f(x)e^{g(x)}$, where $f$ and $g$ are rational functions.
|
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|
108,912 |
Let $A_n=\{a\cdot b : a,b \in \mathbb{N}, a,b\leq n\}$. Are there any estimates for $|A_n|$? Will it be $o(n^2)$?
|
This question is known as the multiplication table problem, and was originally posed by Erdős in 1955. Erdős proved that $|A_n|=o(n^2)$ , and this was sharpened by Tenenbaum in 1984. In 2008, Ford gave the exact magnitude and proved that $$\left|\lbrace a\cdot b:\ a,b\leq N\rbrace\right|\asymp \frac{N^2}{(\log N)^c(\log\log N)^{3/2}},$$ where $$c=1-\frac{(1+\log \log 2)}{\log 2}.$$ In 2010 Koukoulopoulos gave multidimensional generalizations of Ford's result, proving that $$\left|\lbrace a_1\cdots a_{k+1}\ :\ a_i\leq N \text{ for all } \ i\rbrace\right|\asymp \frac{N^{k+1}}{(\log N)^{c_k}(\log\log N)^{3/2}},$$ where $$c_{k}=\int_{1}^{\frac{k}{\log(k+1)}}\log x\text{d}x=\frac{\log(k+1)+k\log\left(k\right)-k\log\log(k+1)-k}{\log(k+1)}.$$ Some references: Ford 2008: The distribution of integers with a divisor in a given interval. arXiv link Koukoulopoulos 2010: Localized Factorization of Integers. arXiv link Koukoulopoulos 2012: On the number of integers in a generalized multiplication table. arXiv link Remark: The dates used above refer to the publication dates (not necessarily the date posted to the arXiv).
|
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|
109,149 |
Let me begin with what looks like a joke. According to a Bourbaki member, the following conversation occurred during a meeting dedicated to polishing the but-last version of an Algebra Bourbaki volume: (a Bourbaki member) Why not state explicitly that the coefficients of cyclotomic polynomials are $0,\pm 1$ ? (another member) Because it's false. Here is what I am aware: if $n$ has at most two distinct odd prime factors, then the coefficients of $\Phi_n(X)$ are $0,\pm1$ (Migotti, 1883). In other words, this holds true for $n=2^mp^kq^\ell$, where $p,q$ are primes. On the other hand, it is false for $n=105=3\cdot5\cdot7$, because the coefficient of $X^7$ (or of $X^{41}$ as well) is $-2$. My question is whether there is a complete characterization of those $n$ for which the coefficients of $\Phi_n(X)$ are $0,\pm1$ ? If not, are there other infinite lists of cyclotomic polynomials with this property?
|
There are other families, but there is by no means a complete characterization known.
Even for products of three primes there is no complete answer known. A relevant keyword is flat cyclotomic polynomial. Some results to give a flavor of the problem. The follwing is due to N. Kaplan (from some years ago, Journal of Number Theory): Let $p,q,r$ be primes (strictly increasing in size), if $r$ is $1$ or $-1$ modulo $pq$ then the coefficients of the $pqr$ cyclotomic pollynomial are only $0,1, -1$. He also has some periodicity result that allows constructions: let $n$ be an integer and $s,t$ primes strictly greater $n$, and congruent modulo $n$ then the set of coefficients of the $nt$ and $ns$ cyclotomic polynomial coincide. In 2010 S. Elder obtained further results in this direction. Let $p,q,r$ be primes (strictly increasing in size), if $r$ is $2$ or $-2$ modulo $pq$ then the coefficients of the $pqr$ cyclotomic polynomial are only $0,1, -1$ if and only if $q$ is $1$ modulo $p$. Elder also has some additional results for products of three primes, and also for products of four and five primes; see this presentation of Elder where also the above mentioned results can be found.
|
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|
109,160 |
The function $F(x) = \sum_{0}^{\infty} x^n/n^n$ may be familiar to many readers as an example sometimes used when teaching tests for absolute convergence of entire functions defined by power series. I know of no name for it, nor any use for it aside from pedagogical, so this is a pure curiosity question which I hope is acceptable. The function seems to have one real zero around $x = -1.40376$; a single extremum, a minimum around $x = -5.71837$; and then to approach the $x$-axis from below asymptotically as $x$ goes to negative infinity. Is this true?
|
[ Edited to outline the end of the argument that $f(-M) \rightarrow 0$
(and to correct a few typos etc. while I'm at it) ] Yes, $F(x) \rightarrow 0$ from below as $x \rightarrow -\infty$.
The convergence is slow, and precise asymptotic analysis seems to be
somewhat annoying because it involves the lower branch of the Lambert W function . The massive cancellations in $\sum_{n=0}^\infty x^n/n^n$ for
$x \rightarrow -\infty$ can be tamed by the familiar device of writing
$$
\frac1{n^n} = \frac1{(n-1)!} \int_0^\infty t^{n-1} e^{-nt} dt
$$
for $n=1,2,3,\ldots$. Multiplying by $x^n$, summing over $n>0$,
and restoring the $n=0$ term
$x^0/0^0=1$ yields
$$
f(x) = 1 + x \int_0^\infty e^{txe^{-t}} e^{-t} dt.
$$
Hence if $x=-M$ then
$$
f(x) = f(-M) = 1 - M \int_0^\infty e^{-Mte^{-t}} e^{-t} dt,
$$
and as $M \rightarrow +\infty$
the integral naturally splits into the parts $t \leq 1$ where
$t e^{-t}$ is increasing and $t \geq 1$ where $t e^{-t}$ is decreasing.
We let $u = t e^{-t}$, so the integrand becomes
$e^{-Mu} du/(1-t)$. For $t<1$ we use Abel's power series $t = \sum_{m=1}^\infty m^{m-1} u^m/m!$ to expand the integral in
an asymptotic series:
$$
\int_0^1 e^{-Mte^{-t}} e^{-t} dt \sim
\frac1M + \frac1{M^2} + \frac{2^2}{M^3} + \frac{3^3}{M^4} + \frac{4^4}{M^5} + \cdots
$$
which is already enough to get $f(-M) < 0$ for large $M$.
[Curiously the asymptotics of $\sum_{n=0}^\infty (-M)^n/n^n$
have led us to the divergent series $\sum_{n=1}^\infty n^n/M^n$.] But the resulting bound $f(-M) < -1/M$ underestimates $|f(-M)|$: numerically
$f(-100) \simeq -.1826$, $\phantom.$ $f(-1000) \simeq -.1180$, and
$\phantom.$ $f(-10000) \simeq -.0899$, suggesting that $f(-M)$ decays
only as $-1/\log M$ or so. The reason must be the $t>1$ part of the integral.
On this part, $t = \log(1/u) + \log\log(1/u) + o(1)$ as $t \rightarrow \infty$,
so the integral behaves to first order like
$\int_0^{1/e} e^{-Mu} du / \log(1/u)$. Now $\log(1/u) \rightarrow 0$
as $u \rightarrow 0+$, but the convergence is slower than any positive
power of $u$. Therefore, the integral is $o(1/M)$, which completes
the argument that $f(x) \rightarrow 0$ as $x \rightarrow -\infty$;
but the integral is not $O(1/M^\theta)$ for any $\theta > 1$, so $f(-M)$
decays slower than any positive power of $M$. A more thorough asymptotic analysis of the $t>1$ integral as
$M \rightarrow \infty$ looks routine but unpleasant, so I'll stop
at this point; perhaps somebody else here will be interested in
pursuing it further.
|
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|
109,395 |
I first got concious of the notion of normal varieties around 3 years ago and despite the fact that by now I can manipulate with it a bit, this notion still puzzles me a lot.
One thing that strikes me is that the definition of normality is so entirely algebraic. From my common sense understanding the notion of normal varieties restricts the class of spaces that we consider to more-less reasonable ones. It looks to me that this definition is analogous to the definition of pseudo-manifold. At least the obvious similarity is that in both cases the set of non-singular points is connected. Normality pops up everywhere and its definition is very short. But it is hard for me to imagine that a differential topologist or differential geometer could come up with such a definition.
Why is the notion of normatilty is so omnipresent?
What is "geometric" meaning of normality? Maybe a more concrete question would be like this. Suppose $X$ is an irreducible algebraic subvariety in $\mathbb C^n$ with singularities in co-dimension $2$. Can one somehow looking on singularities, their stratification and the way $X$ lies in $\mathbb C^n$ say if it is normal or not? Added. Who was the person who invented this notion? I would like to thank everybody for useful comments and links.
|
This is basically the same as roy smith's excellent comment, but I'd like to put a slightly different spin on it. A normal variety is a variety that has no undue gluing of subvarieties or tangent spaces. Let me explain what I mean by gluing.
Given a variety $X$, a closed sub- scheme $Y \subseteq X$ and a finite (even surjective) map $Y \to Z$, you can glue $X$ and $Z$ along $Y$ (identifying points and tangent information). This is the pushout of the diagram $X \leftarrow Y \rightarrow Z$. You might not always get a scheme (although you do in the affine case) but you always get an algebraic space. In the affine case, this just corresponds to the pullback in the category of rings. Example 1: $X = \mathbb{A}^1$ glued to $Z = \bullet$ (one point) along $Y = \bullet, \bullet$ (two points) is a nodal curve. Example 2: $X = \mathbb{A}^1$ glued to $Z = \bullet$ (one point) along $Y = \star = \text{Spec } k[x]/x^2$ a fuzzy point gives you a cuspidal curve. Example 3: $X = \mathbb{A}^2$ glued to $Z = \mathbb{A}^1$ along one of the axes $Y = \mathbb{A}^1$ via the map $Y \to Z$ corresponding to $k[t^2] \subseteq k[t]$ gives you the pinch point / Whitney's umbrella = $\text{Spec } k[x^2, xy, y]$. If I recall correctly, all non-normal varieties $W$ come about this way for some appropriate choice of normal $X$ (the normalization of $W$) and $Y$ and $Z$ (NOT UNIQUE). Roughly speaking, if you are given $W$ and want to construct $X, Y, Z$, do the following: Let $X$ be the normalization, let $Z$ be some sufficiently deep thickening of the non-normal locus of $X$ and let $Y$ be some appropriate pre-image scheme of $Z$ in $X$. Edit: There is a proof available now HERE Assuming this is true, you can see that all non-normal things are non-normal because they either have some points identified (as in 1 or 3) or some tangent space information killed / collapsed (as in example 2), or some combination of the two.
|
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|
109,779 |
Theorem 1B.9 in Hatcher's Algebraic Topology says that for a (pointed) connected CW complex $X$ and group $G$, there is a bijection $\text{Hom}(\pi_1(X), G) \cong [X,K(G,1)]$, where $\pi_1(X)$ is the first fundamental group of $X$, and $K(G,1)$ is the first Eilenberg-MacLane space of $G$. I guess he is describing an adjunction of functors here, between the category of homotopy classes of maps between connected pointed CW complexes and groups. This surprised me. If $\pi_1$ is a left-adjoint functor, then we should conclude that it is cocontinuous, i.e. takes pushouts to pushouts. But I had understood the van Kampen theorem to say something like "$\pi_1$ takes certain pushouts in $\text{hTop}_*$ to pushouts in groups". For example, van Kampen requires the morphisms to be inclusions, among other things. Presumably then not all pushouts are preserved under $\pi_1$, for example if the maps are not inclusions. I tried to come up with a pushout of non-injective pointed topological spaces which would give a counterexample to van Kampen, but I could not. Is there one? Can you give one? And if there is one, why doesn't that contradict the status of $\pi_1$ as a left-adjoint? And if there isn't one, then why can't the hypotheses of the van Kampen theorem be weakened?
|
The problem is that there are not a lot of actual colimits in the homotopy category of (connected) CW complexes, so knowing that $\pi_1$ preserves them (which is true) is pretty much useless. The pushouts appearing in the van Kampen theorem are pushouts in $Top$ but not in the homotopy category, so the van Kampen theorem does not follow from this adjunction. On the other hand, the functor $\pi_1$ preserves all homotopy colimits, and the hypotheses in the van Kampen theorem guarantee that the pushout in Top is a homotopy pushout.
|
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|
109,817 |
A standard Carleman-type estimate is of the form $$
\sum_{|\alpha|<m}{\tau^{2(m-|\alpha|-1)}\int{|D^{\alpha}u|^{2}e^{2\tau\phi}}dx}\leq K\int{|Pu|^{2}e^{2\tau\phi}dx},\quad u\in C_{0}^{\infty},
$$ where $\phi$ is some weight function. This formula turns to be very useful in the study of uniqueness of Cauchy problems, and many mathematicians have considered this (such as Calderon, Hormander, Kenig, Sogge, and Tataru...) For a first look at this inequality, I'm wondering whether the weight fuction has an essential role, and furthermore, what's the original idea of it? Are there some very simple but illuminating examples that show the reasonability of the Carleman estimates? One example in my mind is the first order operator $P=D+ix$ , where $D = \frac{1}{i} \frac{d}{dx}$ .
It's easy to see that $P^*=D-ix$ , and $$
P^*P-I=PP^*+I=-\frac{d^2}{dx^2}+x^2,
$$ which is the so-called harmonic oscillator. Here we have $$
2\|u\|_{L^2}\leq \|Pu\|_{L^2},\quad u\in C_{0}^{\infty}.
$$ But in this simple example, there is no need to use a weight function.
From the proof I guess the decomposition $P=\frac{P+P^*}{2}+\frac{P-P^*}{2}$ may be one of the general ideas.
|
The weight function $\phi$ indeed plays an (I would even say the ) essential role in Carleman estimates, and constructing one such that the estimate holds is one of the major challenges in proving and applying Carleman estimates. I will give a very informal overview, and refer to the references below for the technicalities. The original motivation for introducing Carleman estimates was to prove unique continuation theorems, which can informally be stated as follows: Given a partial differential operator $P$, an oriented hypersurface $\Sigma$, and a function $u$ satisfying $Pu=0$, then if $u=0$ (locally) on one side $\Sigma^+$ of $\Sigma$, it vanishes on the other side $\Sigma^-$ as well. This can be interpreted in the sense that the complete information about $u$ in $\Sigma^-$ can be retrieved from information in $\Sigma^+$, i.e., the complete information flows across $\Sigma$. For linear hyperbolic equations, such information flow is described by the characteristics, and the idea behind Carleman estimates is to extend this concept to general (pseudo-)differential operators via microlocal analysis. This leads to considering so-called bicharacteristic rays , which need to satisfy the geometric property that all such rays passing near $\Sigma$ in $\Sigma^+$ must cross into $\Sigma^-$. Furthermore, solutions to $Pu=0$ should not decay exponentially along such rays towards $\Sigma$ (otherwise the information gets lost before it crosses $\Sigma$). A surface satisfying these conditions is called strongly pseudo-convex (with respect to $P$). To make these conditions precise, one considers $\Sigma$ as the level set $\{\phi(x)=0\}$ (and $\Sigma^+$ as $\{\phi(x)>0\}$). The pseudo-convexity conditions above then become positivity conditions for the derivative of $\phi$ along the bicharacteristic flow (expressed using Poisson brackets). The connection is now that if $\Sigma=\{\phi(x)=0\}$ is a strongly pseudo-convex surface, then $e^{\lambda\phi}$ is strongly pseudo-convex function (in the sense of these positivity conditions) for sufficiently large $\lambda$. The parameter $\lambda$ is tied to the decay rate along the bicharacteristic rays, and hence the Carleman weight can be seen as accounting for this decay: The weight vanishes as the information flow across $\Sigma$ decays. The weight thus localizes the estimate to that part of $\Sigma^-$ where you still have usable information from $\Sigma^+$. For example, for elliptic operators of second order, every smooth surface is strongly pseudo-convex (the decay condition is always satisfied); similarly, for hyperbolic operators with constant (real) coefficients, every non-characteristic surface is strongly pseudo-convex. In particular, for the wave operator $Pu=u_{{{tt}}} - c^2\Delta u$, any convex surface is strongly pseudo-convex, as well as the zero level sets for $\phi = x^2 - \beta t^2$ for $\beta < c^2$ (this bound is sharp). [1] A good overview of unique continuation using Carleman estimates (along the lines I've given) is Daniel Tataru, Unique Continuation Problems for Partial Differential Equations, The IMA Volumes in Mathematics and its Applications Volume 137, 2004, pp 239-255 . [2] Carleman estimates are also useful for inverse problems for partial differential equations; this side is discussed in Victor Isakov, Inverse Problems for Partial Differential Equations, 2nd ed., 2006, Springer . (The example for the wave equation is Theorem 3.4.1.) [3] David Dos Santos Ferreira has written a habilitation thesis on Carleman estimates.
|
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|
110,057 |
This is a atypical question for the forum. I'd like to get some advice on whether I should keep pursuing Math in the traditional route, i.e. get a PhD, do research & teach, etc. Due to financial constraints, I worked almost full time and did not really explore much during my undergrad years. I got my BS in Finance in 2.5 years from one of the top schools in the country. Then I worked for a couple years in the investment industry, and decided it is not what I wanted - very flat learning curve and to be a leader in the industry, it's really not how much you know, but how well you can sell. So after I saved enough money to live frugally for a few years, I quit this industry and see if I can find another career path. I then went to take various classes at community college to explore, while working part-time on and off. In about a year or so, I decided to do math. I liked the subject very much, and seemed capable in it. Later, I did almost two more years of Math at a state university, all the way to advanced real analysis and advanced linear algebra. I also participated in some research with professors. I struggled in some classes, but overall I got As and Bs. Everyone expects me to pursue a PhD. This is my struggle: although I like the subject and mostly enjoy the thought-process, I hate the lifestyle. I was either studying or in the lab, because in order to do well, I have to do it all the time, including almost all weekends. I feel very very lonely. A few years ago, I had a very busy undergraduate schedule, but I could predict how long a homework/project would take and schedule time to make and keep friends accordingly. Now, I don't know if I can be done with a proof in 3 hours or 3 days! I feel bad if I abandon my unfinished work and just go out. I do try to talk to my current peers and professors, but they are mostly about the coursework or the projects we or they are working on. I look at my professors, they are almost always there to do research, day and night, weekday and weekends, even the ones with family. The more leisurely ones are the ones who either got their professorship or published some great papers a while ago. Otherwise, everyone else is very busy. I start to doubt that if everyone else is okay with it, it's me who don't love Math enough to infiltrate it into most parts of my life. After all, they are satisfied with seeing their friends once in a while, have a little chit-chat here and there, but I am very unhappy. I also realize that while I have to spend hours on a proof, some people get it very quickly. Maybe I am just incapable. People also work on a very irregular schedule - sometimes they come in at 2am and leave at noon, sometimes vice-versa. I don't mind when they work, but somehow the social activity, if any, is held very late at night, out of a spontaneous mode... "oh, my brain can't function, let's watch a movie/play frisbee/poker!" at midnight! And sadly, I don't want to be out at midnight. Again, other people seem to be fine with it as well. Should I continue this path? If I do it, I don't want to do it in a rush way; so it'll take me another 2 years to do a master; and another 4 years to do a PhD, plus many more years in this research environment. (I also would like to learn how you cope with the financial challenges to support a family if you are pursuing the research route. I am afraid that my research is not good enough to get good funding; and positions at universities are very very competitive.) Should I somehow engage it part-time only? I think my professors will welcome me to continue doing research for them, even just on the weekend. However, without formerly enrolling in school, my work will be mostly unpaid since their funding comes from NASA and other institutions, and I don't have a PhD to qualify me as a visiting scholar to let me access school's supercomputers legally. Also, I feel like I won't grow much in this part-time manner. I had considered doing this on my own, at my own pace, but I am not at a level where I can do/learn on myself yet. I need someone to discuss and check if I understand something correctly. I much prefer doing it in person than digging through online forums. This is long, and I appreciate your time in reading and I really look forward to some guidance or experience-sharing. Thank you. [Technical comments:] Question asked by Flora . Originally tagged mp.mathematical-physics , ca.analysis-and-odes , ra.rings-and-algebras , to provide more focused mathematical context. [end technical comments, by quid].
|
One famous mathematician answered to all such questions from young people: No, you should not do math.
You will do it successfully only in the case that you cannot do anything else.
In which case you do not ask these questions. Your observations are correct: mathematicians work overtime, at irregular hours, think about their problems continously, even
at the time which should belong to their families, and so on. And sometimes you cannot solve
a problem for months and years.
And the way to a permanent academic position
is long and hard. And the reward in terms of money is not really so great. I understand that there can be various opinions, but I am of the same opinion as
this mathematician. I know many people who quit mathematics on various stages of their careers,
because of various obstacles. Usually they make more money than they could make if they pursued an
academic math career. And have more time for all kinds of activities not related to their jobs. On the other hand, if you really like mathematics, you can always do it as a hobby,
without making it your profession. This hobby, unlike many other hobbies, costs you almost nothing,
and you can dedicate as much of your time to it as you wish. EDIT. Let me expand a little, and I stress that this is only my personal point of view. Doing research in mathematics is not a profession:-) It is a hobby. My profession is teaching
(I am a professor). Of course, to qualify for this position, I had to show some ability to do research. But in the beginning, when I had to choose my career, I did not know whether I am able
to do research or not. And I believe nobody knows, with possibly very rare exceptions.
So when I made my choice (to pursue a degree in "pure" mathematics), I knew that in any case I can
be a high school teacher, possibly a college teacher, and I did research as a hobby.
The results eventually permitted me to become a professor at a university. In my youth, I even had
for some time a "research position", with no teaching duties. But this caused discomfort:
I never wanted my living (salary) to depend only on research results. Because when you do real
research, you are never sure what (and when) will come out of it:-)
It is nice of course when they give you some extra money (I mean grants) for your hobby, but I would
not like to depend on this for my living.
|
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|
110,062 |
And what else can be said, if so? ( Original math.SE post ) In more detail: Say $(G,\mathscr{T})$ is a topological group. It has a left uniformity $\mathscr{L}$ and a right uniformity $\mathscr{R}$. (It also has a two-sided uniformity $\mathscr{U}$, which is the join of the two.) Now, uniformities on a given set form a complete lattice, so we can also consider the meet of the two, $\mathscr{V}$. However, the meet of two uniformities that yield the same topology does not necessarily again yield the same topology, so it's possible that $\mathscr{T}'$, the topology coming from $\mathscr{V}$, is coarser than our original topology $\mathscr{T}$. (Obviously, this does not happen if the group is balanced, i.e. $\mathscr{L}=\mathscr{R}$; it also does not happen if $\mathscr{T}$ is locally compact, since the meet of two uniformities yielding the same locally compact topology does again yield the same topology. Actually, I don't know an actual case where this does happen, so I guess a first question I can ask is, are there any actual examples of this?) So my question is, is $(G,\mathscr{T}')$ again a topological group? Obviously inversion is continuous, since $\mathscr{V}$ makes inversion uniformly continuous, but it's not clear what would happen with multiplication. If it is a topological group, then we can ask things like, how does $\mathscr{V}$ compare to $\mathscr{L}'$, $\mathscr{R}'$, $\mathscr{U}'$, and $\mathscr{V'}$? (Well, obviously it's coarser than the last of these.) And considering $\mathscr{T} \mapsto \mathscr{T}'$ as an operation on group topologies on $G$, what happens when we iterate it? When we iterate it transfinitely?
|
One famous mathematician answered to all such questions from young people: No, you should not do math.
You will do it successfully only in the case that you cannot do anything else.
In which case you do not ask these questions. Your observations are correct: mathematicians work overtime, at irregular hours, think about their problems continously, even
at the time which should belong to their families, and so on. And sometimes you cannot solve
a problem for months and years.
And the way to a permanent academic position
is long and hard. And the reward in terms of money is not really so great. I understand that there can be various opinions, but I am of the same opinion as
this mathematician. I know many people who quit mathematics on various stages of their careers,
because of various obstacles. Usually they make more money than they could make if they pursued an
academic math career. And have more time for all kinds of activities not related to their jobs. On the other hand, if you really like mathematics, you can always do it as a hobby,
without making it your profession. This hobby, unlike many other hobbies, costs you almost nothing,
and you can dedicate as much of your time to it as you wish. EDIT. Let me expand a little, and I stress that this is only my personal point of view. Doing research in mathematics is not a profession:-) It is a hobby. My profession is teaching
(I am a professor). Of course, to qualify for this position, I had to show some ability to do research. But in the beginning, when I had to choose my career, I did not know whether I am able
to do research or not. And I believe nobody knows, with possibly very rare exceptions.
So when I made my choice (to pursue a degree in "pure" mathematics), I knew that in any case I can
be a high school teacher, possibly a college teacher, and I did research as a hobby.
The results eventually permitted me to become a professor at a university. In my youth, I even had
for some time a "research position", with no teaching duties. But this caused discomfort:
I never wanted my living (salary) to depend only on research results. Because when you do real
research, you are never sure what (and when) will come out of it:-)
It is nice of course when they give you some extra money (I mean grants) for your hobby, but I would
not like to depend on this for my living.
|
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|
110,208 |
I have a really hard time "feeling" what it means for a group to fail to be linear. Vaguely, I'd like to know how one should think about such groups. More precisely: What are some interesting examples of groups that aren't linear? Are there general constructions that one can use to cook up a group or a family of groups that isn't linear? Are there general techniques that one can use to show that a given group isn't linear? More loosely, what is it about these groups that makes them interesting?
|
Consider the class of finitely generated linear groups. Such groups $G$ satisfy certain well-known restrictions, for instance: Every such $G$ is residually finite (Malcev, 1940). Thus, most Baumslag-Solitar groups, e.g.
$$
\langle a, b| a b^2 a^{-1} =b^3\rangle
$$
are not linear. This is the simplest example of a nonlinear f.g. group I know. $G$ is virtually torsion-free (Selberg, 1960). In particular, if $G$ is torsion then it is finite (which was known to Burnside). Note that there are infinite torsion residually finite groups (first examples are due to Golod and Shafarevich); such groups have to be nonlinear. $G$ satisfies Tits' alternative (Tits, 1972): Either $G$ contains a free nonabelian subgroup or contains a solvable subgroup of finite index. (Thus, for instance, Thompson group is not linear.) Tarski mosters will violate all of the above restrictions. There are more subtle restrictions, for instance, $Aut(F_n), n\ge 3$ is not linear (Formanek and Procesi, 1992). Consider reading Wehrfritz' book "Infinite linear groups" or this survey to get a better idea of what linearity means for f.g. groups, specially, Lubotzky's criterion of linearity. Concerning your question of why nonlinear groups are interesting: Many of them occur naturally (like $Aut(F_n)$), the rest push the boundaries of our understanding of the class of f.g. groups. For many "natural" groups, linearity is unknown, e.g., the mapping class group $Mod_g$, $g\ge 3$.
|
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|
110,345 |
Consider a complex power series $\sum a_n z^n \in \mathbb C[[z]]$ with radius of convergence $0\lt r\lt\infty$ and suppose that for every $w$ with $\mid w\mid =r$ the series $\sum a_n w^n $ converges . We thus obtain a complex-valued function $f$ defined on the closed disk $\mid z\mid \leq r$ by the formula $f(z)=\sum a_n z^n$. My question: is $f$ continuous ? This is a naïve question which looks like it should be answered in any book on complex analysis. But I checked quite a few books, among which the great treatises : Behnke-Sommer, Berenstein-Gay, Knopp, Krantz, Lang, Remmert, Rudin, Stein-Shakarchi, Titchmarsh, ... . I couldn't find the answer, and yet I feel confident that it was known in the beginning of the twentieth century. Edit Many thanks to Julien who has answered my question: Sierpinski proved (in 1916) that there exists such a power series $\sum a_n z^n $ with radius of convergence $r=1$ and associated function $f(z)=\sum a_n z^n $ not bounded on the closed unit disk and thus certainly not continuous. It is strange that not a single book on complex functions seems to ever have mentioned this example. On the negative side, I must confess that I don't understand Sierpinski's article at all! He airdrops a very complicated, weird-looking power series and proves that it has the required properties in a sequence of elementary but completely obscure computations. I would be very grateful to anybody who would write a new answer with a few lines of explanation as to what Sierpinski is actually doing.
|
I searched all over for an answer to this question back in my student days. I found the answer in a paper by Sierpinski, "Sur une série potentielle qui, étant convergente en tout point de son cercle de convergence,représente sur ce cercle une fonction discontinue ", which is featured in his collected works, see here , p282) and apparently was published in 1916. It does confirm your expectation that this was known in the beginning of the twentieth century (I don't know whether it's the first proof or not, but from the paper it's clear that Sierpinski thought the result to be new). EDIT: I just realized that not everybody speaks French ;-) so, to be clear: Sierpinski produces an example where the function converges everywhere on the unit circle but is discontinuous on the circle.
|
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|
110,797 |
This is a vague question, and I will no doubt be (properly!) chastised for posing it.
I would like to generate a set $S$ of points in $\mathbb{R}^3$—$|S|$ finite or infinite—which
has the property that, viewing $S$ under orthogonal projection along a random direction $\vec{u}$ results
in a more-or-less generic, undistinguished cloud of points. But , there is some specific projection
direction $\vec{u^*}$, where suddenly (if one were 3D-rotating the points under mouse control) the cloud resolves itself, through unlikely point alignments, to paint a recognizable image, e.g., Is this an impossible :-) hope? Update .
Following Michael Murray's recipe, with $10,000$ points within a cube in $\mathbb{R}^3$, three different viewpoints: (Somehow my analytical smiley has a Halloween evil glint!) PS ( 31 Oct 2012 ). Happy Halloween! Another addition ( 23Jun2018 ): (Image from John Urschel / MoMath video .)
|
Surely you can draw the 2-D image in the XY plane so it consists of points of the form (x, y, 0) and then give each point in it a random non-zero Z co-ordinate. So it should look like a mess except viewed looking in along the Z-axis.
|
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|
111,059 |
as far as I know, there are two main ways to have a relative version of De Rham Cohomology for a pair (M,N), where M and N are smooth manifolds and N is a closed (as a topological subspace) submanifold of M: 1) Godbillon, Elements de topologie algébrique: $\Omega^p(M,N)$ is the space of all forms on $M$ whose restriction to $N$ is zero. This is a subalgebra of $\Omega^p(M)$ so it defines a cohomological space $H^p(M,N)$. 2) Bott-Tu, Differential forms in algebraic topology: this times, $\Omega^p(M,N)=\Omega^p(M)\oplus \Omega^{p-1}(N)$ with differential $d(\omega,\theta)=(d\omega,i^*(\omega)-d\theta)$, where $i:N\to M$ is the inclusion. Does these two cohomologies give the same results? Otherwise, are they related and how are they related? Bott and Tu's paragraph on the relative De Rham cohomology is very short. Does someone know a good reference on this subject? Thank you in advance.
|
A chain map $\Theta$ from the Godbillon theory to the Bott-Tu version is given by $\omega \mapsto (\omega,0)$ (note that is a chain map only on $\Omega^{p} (M;N)_{G}$).
I claim that this induces an isomorphism on cohomology. A couple of special cases is obvious: if $N=\emptyset$, then both theories agree with absolute de Rham theory.
If $N \to M$ is a homotopy equivalence, both theories are trivial by long exact sequences and homotopy invariance of the absolute theory. For the general case, pick a tubular neighborhood $U$ of $N$. You get short exact sequences of chain complexes (in both cases) $$
0\to \Omega (M;N) \to \Omega(U;N) \oplus \Omega (M-N) \to \Omega (U-N) \to 0
$$ (exactness is checked by means of a partition of unity), and $\Theta$ compares the both short exact sequences. The associated (Mayer-Vietoris) exact sequence and the $5$-lemma concludes the proof.
|
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|
111,404 |
Question 1 Given a representation of a finite group, what algorithm can be used to check is it irreducible or not ? (Main case - complex numbers, comments on other cases are also welcome. "Given" means finite set of matrices is given). Question 2 Given a representation of a finite group, what algorithms can be used to decompose it to the direct sum of irreducibles) ? For the question 1 I would do the following: rep is irrep if its commutant consists of scalar matrices. So I can try to find matrices commuting with all elements of the group and look whether I get only scalar matrices. Are there more effective ways to do it ? Related question: How to compute all irreducible representations of a finite group ? (how GAP is doing this?)
|
I think this question is not quite so trivial as Qiaochu suggests. Devising practical algorithms for problems of this type is certainly an active area of research within the computational group theory community. The first problem is that the group may be too large for it to be feasible to compute its conjugacy classes and character table. It might be one of the large sporadic simple groups for example. Over finite fields, the so-called Meataxe methods, which are actually just based on linear algebra, have been used to find the composition factors of some modules of dimensions in the hundreds of thousands. There are ongoing efforts to extend these methods to characteristic zero. If your representation is in characteristic zero, and you can compute the classes and character table, then the straightforward character theoretic methods will tell you what the absolutely irreducible constituents of the module are, but that does not immediately enable you to decompose the module explicitly. Another problem that arises in practice is that your representation may be over the rationals, for example, but its absolutely irreducible constituents might not be realizable over the rationals. In that case, the reducibility of the module will depend on the Schur Index of representation',s constituents, which can be calcualated from the character table. But again, if you find out, for example, that a rational representation is the direct sum of two isomorphic irreducibles, then it can be very difficult to find the basis change that exhibits the direct sum.
|
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|
111,519 |
According to Steven Krantz's Mathematical Apocrypha ( pg. 186 ): As was custom, Weil often attended tea
at [Princeton] University . Graduate
student Steven Weintrab one day went
about the room asking various famous
mathematicians who was the greatest
mathematician of the twentieth
century. When he asked Weil, the
answer (without hesitation) was "Carl
Ludwig Siegel (1896-1981)." As the title of Krantz's book suggests, the anecdote may be apocryphal. However, there are other better grounded accounts of great mathematicians expressing the highest admiration for Siegel: (A) In The Map of My Life Shimura wrote: I always thought that few people
really understood my work. I knew that
Chevalley, Eichler, Siegel, and Weil
understood my work, and that was
enough for me [...] Of course [Siegel]
established himself as one of the
giants in the history of mathematics
long ago [...] Among his
contemporaries, [Weil] thought highly
of Siegel [...] (B) In an published interview (pg. 30) Selberg said [Siegel] was in some ways, perhaps,
the most impressive mathematician I
have met. I would say, in a way,
devestatingly so. The things that
Siegel tended to do were usually
things that seemed impossible. Also
after they were done, they seemed
still almost impossible. Why might Weil, Shimura and Selberg have been so impressed by Siegel? I should emphasize that I'm not trying to precipitate a debate about the relative standing of historical mathematicians - rather - I'm hoping to learn about aspects of Siegel's work that I might otherwise overlook. I'm also not looking for, e.g. quotations from the Wikipedia article on him, but rather, less familiar material.
|
No one with any familiarity with his work can doubt that Siegel was one of the greatest mathematicians of the 20th century. Weil was a decisive, opinionated man -- just the type of person who would have an answer to this question ready at hand. And "Carl Ludwig Siegel" is a totally unsurprising answer from anyone. (Also "Andre Weil" would be a totally unsurprising answer from anyone: it might be my answer!) But it is especially unsurprising coming from Weil. The list of contemporary mathematicians of the Siegel-Weil caliber is short enough, and among mathematicians on that list -- e.g. Wiener, von Neumann, Kolmogorov, Godel --
the research interests of Siegel and Weil were especially close: for instance, there is a Siegel-Weil formula . Both brought their prodigious knowledge and technique to bear on number theory, but with distinct, and distinctive, styles. To be very brief and crude, Weil had a fundamentally algebraic approach, whereas Siegel had a fundamentally analytic approach. My own approach to mathematics is rather close to Weil's (although in magnitude, microscopic compared to his): I very much appreciate that finding the right bit of "structure" can make the solution of your problems self-evident. A lot -- by no means all -- of Weil's work is like that: the finished product is so tidy and efficacious that you too easily forget to ask how he thought of any of it in the first place. To someone with this "algebraic" style, Siegel's work looks like a sequence of miracles. So it is unsurprising to me that someone like Weil would select someone like Siegel to give his top regards. I think you can also gain some insight into why Weil named Siegel by considering their ages: Siegel (born in 1896) was ten years older than Weil (born in 1906). Ten years is long enough for Siegel always to have been ahead of Weil in his career and stature, but short enough for them to be true contemporaries and competitors. Most other great mathematicians that spring to mind when I think of Weil are actually quite a bit younger, e.g. Serre (born 1926), Tate (born 1925), Shimura (born 1930); it makes sense that Weil is not going to name any of these as the greatest mathematician of the 20th century. Indeed all three are alive well into the 21st century. [Added: I just remembered that Chevalley (born 1909) was a contemporary of Weil of a similar stature. But Chevalley was very close to Weil, both personally and in mathematical styles and tastes. It is psychologically natural to esteem (and fear) most that which is most different from ourselves, not that which is most similar. Anyway, for Weil to name Chevalley would have sounded arrogant, as if not being able to name himself he picked the person standing right next to him.] By the way, I think that Shimura and Siegel are quite similar in style as well as stature. I read Shimura's autobiography, and I think he is right to be profoundly disappointed that Siegel did not take more of an interest in his work. Shimura's work is closer to being a continuation of Siegel's (including a continuation of the brilliance, creativity and orginality!) than any other mathematician I can think of, so it is natural that Shimura holds Siegel in high regard. There is also something "organic" in the work of both Siegel and Shimura which naturally bristles a bit at the "Bourbakistic" influence of the French school: it seems clear enough, for instance, that the modern theory of "Shimura varieties" is both an addition and a subtraction from what Shimura himself intended. I know several of Shimura's students, and though they work in what the rest of the mathematical world thinks of as parts of algebraic number theory and arithmetic geometry, in the way they actually think about mathematics they take a more analytic approach...like Siegel. I have even fewer credentials to speak for Selberg than I do for any of these others, but I imagine that he may have felt a similar kinship to Siegel, i.e., the use of an "analytic" approach to studying problems that others regard as being more algebraic.
|
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|
111,557 |
We let $B$ be a simple algebra over $\mathbb Q$, with the usual notations for PEL type Shimura varieties. In his paper "Ordinariness in good reductions of Shimura varieties of PEL-type" (available here http://archive.numdam.org/article/ASENS_1999_4_32_5_575_0.pdf ), Wedhorn proved, under the assumption that $p$ is unramified in $B$, that the ordinary locus in (the reduction) a PEL type Shimura variety is non-empty if and only if it is dense if and only if $p$ splits completely in the reflex field. So my question is the following: Are there similar results, even partial, without the unramifiedness assumption?
|
No one with any familiarity with his work can doubt that Siegel was one of the greatest mathematicians of the 20th century. Weil was a decisive, opinionated man -- just the type of person who would have an answer to this question ready at hand. And "Carl Ludwig Siegel" is a totally unsurprising answer from anyone. (Also "Andre Weil" would be a totally unsurprising answer from anyone: it might be my answer!) But it is especially unsurprising coming from Weil. The list of contemporary mathematicians of the Siegel-Weil caliber is short enough, and among mathematicians on that list -- e.g. Wiener, von Neumann, Kolmogorov, Godel --
the research interests of Siegel and Weil were especially close: for instance, there is a Siegel-Weil formula . Both brought their prodigious knowledge and technique to bear on number theory, but with distinct, and distinctive, styles. To be very brief and crude, Weil had a fundamentally algebraic approach, whereas Siegel had a fundamentally analytic approach. My own approach to mathematics is rather close to Weil's (although in magnitude, microscopic compared to his): I very much appreciate that finding the right bit of "structure" can make the solution of your problems self-evident. A lot -- by no means all -- of Weil's work is like that: the finished product is so tidy and efficacious that you too easily forget to ask how he thought of any of it in the first place. To someone with this "algebraic" style, Siegel's work looks like a sequence of miracles. So it is unsurprising to me that someone like Weil would select someone like Siegel to give his top regards. I think you can also gain some insight into why Weil named Siegel by considering their ages: Siegel (born in 1896) was ten years older than Weil (born in 1906). Ten years is long enough for Siegel always to have been ahead of Weil in his career and stature, but short enough for them to be true contemporaries and competitors. Most other great mathematicians that spring to mind when I think of Weil are actually quite a bit younger, e.g. Serre (born 1926), Tate (born 1925), Shimura (born 1930); it makes sense that Weil is not going to name any of these as the greatest mathematician of the 20th century. Indeed all three are alive well into the 21st century. [Added: I just remembered that Chevalley (born 1909) was a contemporary of Weil of a similar stature. But Chevalley was very close to Weil, both personally and in mathematical styles and tastes. It is psychologically natural to esteem (and fear) most that which is most different from ourselves, not that which is most similar. Anyway, for Weil to name Chevalley would have sounded arrogant, as if not being able to name himself he picked the person standing right next to him.] By the way, I think that Shimura and Siegel are quite similar in style as well as stature. I read Shimura's autobiography, and I think he is right to be profoundly disappointed that Siegel did not take more of an interest in his work. Shimura's work is closer to being a continuation of Siegel's (including a continuation of the brilliance, creativity and orginality!) than any other mathematician I can think of, so it is natural that Shimura holds Siegel in high regard. There is also something "organic" in the work of both Siegel and Shimura which naturally bristles a bit at the "Bourbakistic" influence of the French school: it seems clear enough, for instance, that the modern theory of "Shimura varieties" is both an addition and a subtraction from what Shimura himself intended. I know several of Shimura's students, and though they work in what the rest of the mathematical world thinks of as parts of algebraic number theory and arithmetic geometry, in the way they actually think about mathematics they take a more analytic approach...like Siegel. I have even fewer credentials to speak for Selberg than I do for any of these others, but I imagine that he may have felt a similar kinship to Siegel, i.e., the use of an "analytic" approach to studying problems that others regard as being more algebraic.
|
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|
111,724 |
Sometime ago I read somewhere (and I don't remember where it was) that Stefan Banach--a highly creative and great mathematician--did not always write down his ideas. Allegedly, he did not write his own thesis (but of course, all the mathematics in it came from him). Is that true? And is it known who wrote it then?
|
Here is a quote from the article by Krzysztof Ciesielski: On Stefan Banach and some of his results . Banach J. Math. Anal. 1 (2007), no. 1, 1–10. There is a curious story how Banach got his Ph.D. He was being forced
to write a Ph.D. paper and take the examinations, as he very quickly
obtained many important results, but he kept saying that he was not
ready and perhaps he would invent something more interesting. At last
the university authorities became nervous. Somebody wrote down
Banach’s remarks on some problems, and this was accepted as an
excellent Ph.D. dissertation. But an exam was also required. One day
Banach was accosted in the corridor and asked to go to a Dean’s room,
as “some people have come and they want to know some mathematical
details, and you will certainly be able to answer their questions”.
Banach willingly answered the questions, not realising that he was
just being examined by a special commission that had come to Lvov for
this purpose. It is true that Banach was mainly self-taught as a mathematician, although he attended some lectures by Stanislaw Zaremba at Jagiellonian University. By the way, engineering programs in the former Austro-Hungarian monarchy (including Lvov Polytechnics) required quite an intensive training in mathematics, although of course the latest developments (Lebesgue integral etc.) were not part of the curriculum. Addendum 0: The above story is also related by Roman Kaluza in his biography of Banach. He heard it from Turowicz, who credits Nikodym as his source (he himself joined the department later, when Banach was already a professor). Well, on one hand, Nikodym was a friend of Banach and his early partner in mathematical discussions, but on the other hand, at the time of Banach's PhD, he was teaching high school in Krakow. (This point was made by Krzysztof Ciesielski in an email exchange with me.) Addendum 1: Banach's thesis, written in French (which he knew well and used before in publications) can be found here: http://kielich.amu.edu.pl/Stefan_Banach/pdf/oeuvres2/305.pdf It was published in Fundamenta Mathematicae 3 (1922), pp.133-181, and bears only Banach's name. The footnote says that it is a "Thesis presented in June 1920 at the Lvov University for obtaining the degree of the Doctor of Philosophy." On the first page there is a statement that maybe gives some evidence of Banach's tendency to wait until getting the best version of his results: ``Mr. Wilkosz and I have some results (which we propose to publish later) on operations whose domains are sets of Duhamelian functions(...)". There is no joint work with Wilkosz in the collected works of Banach... Addendum 2: Some details brought up by other users need correction. First, Steinhaus met Banach and Nikodym in Krakow, where Banach grew up, not in Lvov. This is explicitly recorded in his "Memoirs and Notes", and somewhat less explicitly in the address he gave much later at a session devoted to Banach: http://kielich.amu.edu.pl/Stefan_Banach/steinhaus63.html ("Planty" is a major green belt in the old city of Krakow; in Lvov there were "Waly").
Second, Banach's PhD supervisor (only in the formal sense) was Steinhaus. Antoni Lomnicki held a chair of mathematics at the Lvov Polytechnics (not to be confused with the Lvov University), where Banach got his first position as an assistant (pre-PhD). Addendum 3: Here is what Steinhaus wrote about working habits of Banach (see the link above): His lectures were excellent; he never lost himself in particulars, he
never covered the blackboard with numerous and complicated symbols. He
did not care for verbal perfection; all manner of personal polish was
alien for him and, throughout his life he retained, in his speech and
manners, some characteristics of a Cracow street urchin. He found it
very difficult to formulate his thoughts in writing. He used to write
his manuscripts on loose sheets torn out of a notebook; when it was
necessary to alter any parts of the text, he would simply cut out the
superfluous parts and stick underneath a piece of clean paper, on
which he would write the new version. Had it not been for the aid of
his friends and assistants, Banach's first studies would have never
got to any printing office." And also: "Banach could work at all times
and everywhere. He was not used to comfort and he did not want any. A
professor's earnings ought to have supplied all his needs amply. But
his love of spending his life in cafes and a complete lack of
bourgeois thrift and regularity in everyday affairs made him incur
debts, and, finally, he found himself in a very difficult situation.
In order to get out of it he started writing textbooks. Addendum 4: This is based on information I received from Danuta Ciesielska, a Polish mathematician and a historian of mathematics (and my classmate from Krakow). The documents from the Lvov University are now split between the Lvov District Archive and Lvov City Archive, http://www.archives.gov.ua/Eng/Archives/ra13.php - Wayback Machine link (the documents of Polytechnics were transported to Wroclaw, Poland after 1945). The catalogs underwent major reorganization, which makes it quite difficult to find particular documents there. Besides the employees' folders, the documentation of PhD and habilitation proceedings is often found in the minutes of faculty meetings. Regarding Banach's PhD, Ciesielska saw a letter from Steinhaus to dean Stanecki (dated September 28, 1920) asking him to set the date for Banach's doctoral exam, to which Stanecki replied that the date cannot be set before Messrs. Steinhaus and Zylinski (the committee members) evaluate the thesis. (Aside: Math Genealogy Project lists Kazimierz Twardowski as one of Banach's advisors. On the surface of it, this makes little sense, as Twardowski was a philosopher and a logician; his expertise was far removed from what Banach worked on. However, as a professor of Lvov University, he was on the committee and signed the papers.) She also points out that in some institutions (e.g., Jagiellonian University in Krakow), if a PhD thesis was published after the exam, the printed copy/journal offprint replaced the submitted manuscript/typescript. It is not clear if this was the case in Lvov.
|
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|
111,765 |
Motivation: Incompleteness (and various independence statements) is about unprovable statements. One natural way to make an unprovable statement provable is to assume it as a new axiom. But this feels like cheating, so people often look for "natural" axioms to add that will imply their favorite unprovable statement. The question is whether there are any statements that are unprovable in a theory where essentially the only way to get a theory which proves the statement is to assume it. Question: Let $T$ be a theory (e.g. ZFC). Say a statement $S$ is "minimally unprovable" in/over $T$ if $S$ is not provable from $T$ (and neither is its negation), and such that if $R$ is any statement that is provable from $T \cup \{S\}$ but not provable from $T$ alone, then there is a proof in $T$ that $R$ is equivalent to $S$. Does every (sufficiently powerful?)
theory have minimally unprovable
statements? Are there examples of a(n interesting) theory $T$
together with a minimally unprovable
statement $S$?
|
The following is essentially Joel's answer and also essentially the last part of Francois's answer, but its "look and feel" seems different enough to make it worth pointing out. The main point is that, if $S$ is minimally unprovable over $T$ then $T\cup\{\neg S\}$ is consistent and complete. Consistency is just your requirement that $S$ is not provable from $T$. To establish completeness, suppose $Y$ is a sentence that is not provable from $T\cup\{\neg S\}$; I'll show that $\neg Y$ must be provable from $T\cup\{\neg S\}$. The assumption means that $(\neg S)\to Y$ isn't provable from $T$, and therefore neither is its contrapositive $(\neg Y)\to S$. Therefore neither is the (propositionally equivalent) formula $(S\lor\neg Y)\to S$. Now apply the main clause in the definition of minimally unprovable, with $S\lor\neg Y$ in the role of $R$. We've just seen that the conclusion of that clause fails, so one of the hypotheses must fail. The first hypothesis says that $R$ is provable from $T\cup\{S\}$, which is clearly true because of the disjunct $S$ in $R$. So the second hypothesis must fail; that is, $R$ must be provable from $T$. But then, by propositional logic, $\neg Y$ is provable from $T\cup\{\neg S\}$, as claimed. Now if $T$ is recursively axiomatizable and contains enough arithmetic, then $T\cup\{\neg S\}$ has the same properties and, by Rosser's improvement of Goedel's second incompleteness theorem, it cannot be both consistent and complete. Therefore, $S$ cannot be minimally unprovable over $T$.
|
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|
111,851 |
Jerry Shurman has a lovely set of notes explaining the classical definition of Hecke characters, the idelic definition of Hecke characters, their relationship, and the classification of algebraic Hecke characters for $\mathbb{Q}$ as Dirichlet characters. He also gives a single family of examples of algebraic Hecke characters with infinite order, namely $\displaystyle \chi: \mathbb{Z}[i] \to \mathbb{C}^{\times}$ given by $\displaystyle \chi(z) = \left(\frac{z}{|z|}\right)^{4n}$ for integers $n$ . It's clear that one has essentially the same family for imaginary quadratic number fields with class number $1$ . But what about imaginary quadratic fields with higher class number? I imagine that one has one family analogous to the one above for each ideal class, but I don't know what they should look like... What do infinite image algebraic Hecke characters for real quadratic fields look like? Because the unit group is infinite, one can't kill the unit group as above, by putting a $4$ in the exponent...
|
The "most obvious" algebraic Hecke characters of a field $K$ are the characters of the ideal class group of $K$, which have trivial infinity-type and trivial conductor. There might be no non-trivial examples (as in the case of Q). But you can get more examples by: beefing up the conductor (which gets you Dirichlet characters, or more generally characters of ray class groups, but never anything of infinite order) changing the infinity-type (the restriction of $\chi$ to the connected component of the identity in $(K \otimes \mathbb{R})^\times \subseteq \mathbb{A}_K^\times$). Over $K=\mathbb{Q}$, the infinity-types are pretty restricted: the only algebraic characters of the group of positive reals are the maps $x \mapsto x^k$, so you just get powers of the "norm" character (the character of the ideles whose restriction to $\mathbb{R}_+$ is the identity, and which sends a uniformizer at a prime $p$ to $1/p$). Over a number field the game is more subtle. Let's first suppose $K$ is totally real of degree $n$. The infinity-type of a character looks like $z \mapsto z_1^{k_1} \dots z_n^{k_n}$ for integers $k_1, ..., k_n$, where $z_1, ..., z_n$ are the embeddings into $\mathbb{R}$, but there is a constraint that the infinity-type needs to vanish on a finite-index subgroup of the global units, and this forces the vector $k_1, ..., k_n$ to be orthogonal to the lattice in $\mathbb{R}^n$ generated by the vectors $$ \{ (\log |u_1|, ..., \log |u_n|) : u \in \mathcal{O}_K^\times\}. $$ By Dirichlet's unit theorem this lattice has rank $r_1 + r_2 - 1 = n-1$, so its orthogonal complement has rank at most 1 -- it's spanned by $(1, ..., 1)$. This tells us that every algebraic Hecke character is just a finite-order character times a power of the norm character, which is a bit boring. For non-totally-real fields the game gets more interesting because there are not so many units. If $K$ is a CM field of even degree $2d$, then the unit group has rank $d-1$, and you can show that the weights of algebraic Hecke characters span a lattice of rank $d + 1$ (spanned by the norm character and characters of the form $x \mapsto \sigma_i(x) / \overline{\sigma_i(x)}$ for each embedding $\sigma_i: K \hookrightarrow \mathbb{C}$). If $K$ is not either totally real or CM, things are more interesting still: the lattice spanned by the logs of the units has rank $r_1 + r_2 - 1$, so its orthogonal complement has dimension $1 + r_2$ over $\mathbb{R}$, but you can't find enough integer vectors in the orthogonal complement. For instance, if $K = \mathbb{Q}(\sqrt[3]{2})$ then the only possibility is the norm character, again (it is a fun exercise to check this by hand in this case). This is an instance of a theorem of [edit: Emil Artin and] Andre Weil: for any number field $K$, and any algebraic Groessencharacter of $K$, the infinity-type of the character must factor through the norm map to the maximal CM subfield of $K$ [edit: or to $\mathbb{Q}$ if there is no such subfield].
|
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|
112,029 |
Is it possible to mathematically prove that the speed of gravitational waves in general relativity equals the speed of light, without linearizing the Einstein field equations? The approach via the linearization of the EFE's, which is used in many books on relativity, does not seem to provide an exact proof that disturbances in spacetime propagate at the speed $ c $ .
|
What is the speed of a wave in a non-linear theory? Answering before considering your question is important, because that answer will tell you where to look for your answer. A useful notion is that of domain of dependence (see for example a decent book on GR for a detailed discussion, e.g., Wald or Hawking & Ellis). If $S\subseteq \Sigma$ is a subset of a Cauchy surface, then the domain of dependence $D(S)$ is the region of the spacetime where the solution is completely determined by the initial data on $S$, irrespective of what initial data is specified on the complement $\Sigma \setminus S$. Thus, the "slope" of the boundaries of $D(S)$ when represented in a spacetime diagram can be interpreted as the rate at which the influence of the initial data from $\Sigma \setminus S$ is encroaching on the spacetime region where the solution is determined by the initial data on $S$ alone. In other words, the "slope" of the boundary of $D(S)$ determines the speed of the propagation of disturbances in the solutions of the PDE you are solving. This speed also agrees with the speed of traveling wave solutions in the case of a linear PDE with constant coefficients. Note that this "speed" of propagation could be independent of the underlying solution itself (the case in electromagnetism) or may actually depend on the solution (the case in GR). It happens to be a fact of life that for both electromagnetism and GR that the boundary of $D(S)$ is always traced out by the null rays of the spacetime metric (which is dynamical in the latter case, but need not be in the former). This fact then implies that the speed of electromagnetic and gravitational waves is the same, independent of any linearization. The above result on the boundary of the domain of dependence is standard in the hyperbolic PDE literature and can be found, for instance, in the books of John ( Partial differential equations ), Lax ( Hyperbolic partial differential equations ) and Hoermander ( Lectures on nonlinear hyperbolic differential equations ). Briefly, the boundary is always "characteristic", where for PDEs of Lorentzian wave equation type the "characteristics" coincide with the null directions of the metric.
|
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|
112,127 |
In the Wikipedia article about descriptive set theory I read that $\mathbb{R}$ (with its usual topology) is a Polish space, and that every Polish space 1) can be obtained as a continuous image of the Baire space $\mathcal{N}$ 2) can be obtained as the image of a continuous bijection defined on a closed subset of the Baire space. Then I learn from the Wikipedia article on Baire space that it is actually homeomorphic to the irrational numbers with their usual subspace topology inherited from the real line. So my questions: Can we describe explicitly a surjective continuous map from the irrationals to $\mathbb{R}$ ? Same for a continuous bijection from a closed subset of the irrationals to $\mathbb{R}$ .
|
For any irrational number $x$, let $f(x)$ be the real number arising from the integer part of $x$, together with every other digit of the rest of the expansion of $x$. This is surjective, since one may interleave the digits of any real $y$ with any nonrepeating pattern, and thereby find an irrational $x$ with $f(x)=y$. This is continuous, since if $x_n\to x$, then $f(x_n)\to f(x)$.
|
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|
112,231 |
I tried to find some paper published in Notices of the American Mathematical Society 1975 By E. Calabi, On manifolds with non-negative Ricci curvature II Notices of the American Mathematical Society 22 1975 A205 But when I searched at mathscinet and click Calabi, there is no such a paper listed under his name. Also AMS's website does not have any issues from before 1995. So is it a book or something?
|
As Andreas Blass and others surmised, this is indeed an abstract. It was for a 10-minute talk at the Annual Meeting in January, 1975. Here is the entire "paper": On Manifolds with nonnegative Ricci curvature II Let $M$ be an $n$-dimensional
Riemannian manifold with nonnegative
Ricci curvature. Then the exponential
mapping $\exp_p$ for any $p\in M$ ,
restricted to the domain bounded by
the cut locus, is everywhere volume
decreasing.From this fact one deduces
the following THEOREM. Let $M$ be a
Riemannian, $n$-dimensional, complete
manifold with nonnegative Ricci
curvature. Then, if $r$ denotes the
injectivity radius and $D$ the
geodesic diameter of $M$ , the volume
$V$ of $M$ satisfies $V \ge c_nr^{n-1}D$, where $c_n$ is a positive
constant depending on $n$ . In
particular, if $M$ is not compact (i.
e. if $D=\infty$), the volume of $M$ ,
under the same assumptions, is
infinite. (Received November 6, 1974.) A couple of notes. I reproduced the capitalization and (non)hyphenation of the title as it appeared in the Notices . I also tried to preserve some oddities in punctuation in the text, but otherwise "TeX-ified" it; the original is literally typed, with a handwritten $\in$ symbol. Added 11/13/12: Out of idle curiosity, I went back to the library today, to see if Calabi ever gave a talk titled "On Manifolds with nonnegative Ricci curvature I." If he did, it wasn't at an AMS meeting (or else I didn't dig back far enough). In the process, however, I noticed that, beginning in October, 1972, the Notices ran a "Queries" column, inviting "questions from members regarding mathematical matters such as details of, or references to, vaguely remembered theorems, sources of exposition of folk theorems, or the state of current knowledge concerning published conjectures" -- i.e., a sort of snail-mail version of MathOverflow. Here's the inaugural query (the answer to which arrived in a then-speedy three months): 1 . R.P. Boas (2440 Simpson Street, Evanston, Illinois 60201). Given a
finite collection of vectors, of total
length 1, in a plane, we can always
arrange them in a polygon, starting
from 0, that at some stage gets at
least $1/\pi$ away from 0. Mitrinovic
[ Analytic inequalities , 1970, pp.
331-332] cites Bourbaki [1955], but
the theorem was known at least in the
early 1940's, when I remember seeing a
paper on it; can anybody supply the
reference? My own favorite is from the next issue: 4 . Cleve B. Moler (Department of Mathematics, University of New Mexico,
Albuquerque, New Mexico 87106). Can
somebody recommend a good source where
I can learn about the connection of
mathematics and various biological
processes such as photosynthesis? Cleve Moler is perhaps best known as the inventor of MATLAB. In conclusion, the answer to the OP's question, "is the Notices available before 1995?" the answer seems to be yes, but only at libraries that hold onto old journals. I wonder if the AMS could be persuaded to make the early volumes of the Notices available through JSTOR.
|
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|
112,329 |
Let $X$ be a projective normal surface over $\mathbb{C}$. In this related question it is stated as soon as $X$ is smooth any vector bundle defined on the compliment of a codimension 2 subset extends to all of $X$. Does this fail when $X$ has codimension 2 singularities? If vector bundles do not always extend, is there a nice example of a surface $X$ and a vector bundles $E$ on $X - \{p_1, \dotsc, p_n\}$ that does not extend to a vector bundles on $X$? Does anyone know of references that discuss the classification of vector bundles on non smooth but normal surfaces? Variant: What about when you replaces vector bundle with principal $G$-bundle for a reductive group $G$?
|
As Piotr remarks, these kind of questions quickly lead to studying reflexive sheaves. I would add that one also better get acquainted with Serre's condition $S_2$. For more on this see this and this MO answers. Perhaps the first remark is that besides the singularities of the surface $X$ you also have to take into account the singularities of the sheaf you are considering. You are asking about normal surfaces. Normal implies $S_2$ and pretty much everything that follows is OK for $S_2$ in arbitrary dimensions. A coherent sheaf on an open set can always be extended as a coherent sheaf on the ambient space. Furthermore if $X$ is $S_2$ and $j:U\hookrightarrow X$ is an open set such that $\mathrm{codim}_X(X\setminus U)\geq 2$, then a coherent sheaf $\mathscr F$ on $X$ such that $\mathscr F|_U$ is locally free is reflexive if and only if $$\mathscr F \simeq j_*(\mathscr F|_U).$$
Notice that this means that if $X$ is $S_2$, then a locally free sheaf $\mathscr E$ on $U$ can be extended as a locally free sheaf to $X$ if and only if $j_*\mathscr E$ is locally free. (The point is, that since $X$ is $S_2$, $j_* \mathscr E$ is reflexive and if there is a locally free sheaf extending $\mathscr E$, then that is also a reflexive sheaf which agrees with $j_*\mathscr E$ on an open set with at least codimension $2$ complement, so they have to agree). This also tells you how to produce locally free sheaves that cannot be extended as a locally free sheaf: Take any reflexive sheaf that is not locally free, then take the open set where it is locally free. By the above, this sheaf on that open set is a locally free sheaf that does not have an extension as a locally free sheaf on the ambient space. Piotr's example is probably the simplest such sheaf. So now the question is: When is a reflexive sheaf locally free? In some contexts one defines the singularity set of a coherent sheaf $\mathscr F$ as the locus where it is not locally free. Then there are various results that say that the singularity set of certain sheaves have to be at least such and such codimension. Here is a short list of those: Sample statement : If $\mathscr F$ is bluh , then the singularity set of $\mathscr F$ has codimension at least boo . Actual statement If $X$ is smooth, we can substitute the following into the above sentence: If bluh = coherent , then boo = $1$. If bluh = torsion-free , then boo = $2$. If bluh = reflexive , then boo = $3$. You may recognize theorems that you know as simple consequences of these: Thm 1 For any coherent sheaf there exists a non-empty open set on which it is (locally) free. Remark You may also note that this does not require $X$ smooth, but that's essentially because the conclusion is invariant under first restricting to the open set where $X$ is smooth. Thm 2 A torsion-free sheaf on a smooth curve is locally free. Remark We know that this fails if the variety is either not smooth or has dimension at least $2$. Thm 3 A reflexive sheaf on a smooth surface is locally free. Remark This is the reason why the statement you mention at the start is true. If $X$ is a smooth surface, then a sheaf that's locally free on the complement of finitely many points pushes forward to a reflexive sheaf which by this theorem has to be locally free.
Piotr's example or any Weil divisor that's not Cartier shows that this fails if the surface is not factorial. (Meaning that every local ring is a UFD. This is a weaker condition than being smooth.) It is also true that a reflexive sheaf of rank $1$ on a smooth variety is always locally free, so you need to go to higher ranks to get a counter-example. To complete the picture here is an example of a reflexive sheaf on a smooth $3$-fold that is not locally free. Example Consider the Euler sequence of $\mathbb P^3$:
$$
0\to \mathscr O_{\mathbb P^3}\to \bigoplus_{i=0}^3 \mathscr O_{\mathbb P^3} (1) \to
T_{\mathbb P^3} \to 0
$$
and consider (one of) the induced maps
$$
\alpha: \mathscr O_{\mathbb P^3} (1) \to T_{\mathbb P^3}.
$$
This is clearly injective and let the cokernel of $\alpha$ be $\mathscr F$.
The original (Euler) sequence shows that $\alpha$ cannot be a vector bundle embedding and hence $\mathscr F$ cannot be locally free. I leave it for you to prove that it is reflexive. (This is not absolutely trivial, but it is a good exercise). So, to answer your question, the statement you cite is not true for singular varieties, at least not in the way it is stated. You can cook up some versions that work by adding additional assumptions. In any case, these are likely the notions that will help you do that. Finally, here are some references: On reflexive sheaves I would recommend Hartshorne's series of papers Stable reflexive sheaves I-II-III . You can read about the singularity set and other interesting stuff in Chapter 2 of Okonek-Schneider-Spindler's Vector bundles on complex projective spaces
|
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112,574 |
This is a duplicate of the following question to which I did not receive any answer: https://math.stackexchange.com/questions/238247/complete-but-not-cocomplete-category Let $\mathfrak C$ be an abelian, cocomplete category. If $\mathfrak C$ has a generator and colimits are exact (i.e., $\mathfrak C$ is Grothendieck) then $\mathfrak C$ is the torsion-theoretic localization of a full category of modules (by the Gabriel-Popescu Theorem) and so it is also complete. Anyway I'm not aware of any counter-example showing that a cocomplete abelian category may not be complete. So my question is: could you provide such example or a reference to a proof of the bicompleteness of cocomplete abelian categories? My first idea was to look for counterexamples in non-Grothendieck subcategories of a Grothendieck category. After some attempt I realized the following Lemma. Let $\mathfrak C$ be a Grothendieck category and $\mathcal T$ a full hereditary torsion subcategory (i.e. $\mathcal T$ is closed under taking sub-objects, quotient objects, extensions and coproducts). Then $\mathcal T$ is bicomplete. Proof. Let $T:\mathfrak C\to \mathcal T$ be the hereditary torsion functor associated to $\mathcal T$. Now, given a family {$C_i:i\in I$} of objects in $\mathcal T$ we can take the product $(P,\pi_i:P\to C_i)$ of this family in $\mathfrak C$. We claim that $(T(P), T(\pi_i))$ is a product in $\mathcal T$. Indeed, let $X\in \mathcal T$ and choose maps $\phi_i:X\to C_i$. By the universal property of products in $\mathfrak C$, there exists a unique morphism $\phi:X\to P$ such that $\pi_i\phi=\phi_i$ for all $i\in I$. Now, since $X\in\mathcal T$, there is an induced map $T(\phi):X\to T(P)$ which is clearly the unique possible map satisfying $T(\pi_i)T(\phi)=T(\phi_i)=\phi_i$. \\\ Thus there are lots of non-Grothendieck bicomplete abelian categories. EDIT: notice that in the lemma we never use the hypothesis that the subcategory $\mathcal T$ is closed under taking extensions or subobjects. In fact, if $\mathcal T$ is just closed under taking coproducts and quotients, one defines the functor $T:\mathfrak C\to \mathcal T$ such that, for all object $X\in\mathfrak C$, $T(X)\in \mathcal T$ is the direct union of all the subobjects belonging to $\mathcal T$ (image (which is a quotient) of the coproduct of all the subobject of $X$ belonging to $\mathcal T$ under the universal map induced by the inclusions of the subobjects in $X$). Clearly $T(X)$ is fully invariant as a subobject of $X$ (by the closure of $\mathcal T$ under taking quotients and the construction of $T$) and so $T$ can be defined on morphisms by restriction. It is also clear that $T(X)=X$ if $X\in\mathcal T$ so the proof of the lemma can be easily adapted to this case. REMARK: the new relaxed hypotheses of the lemma allow us to exclude other "exotic" examples... in particular, if you want to take the abelian subcategory of all the semisimple objects in a given Grothendieck category, this is closed under coproducts and quotients.
|
I think I have an example. Fix a chain of fields $k_\alpha$ indexed by ordinals $\alpha$, where $k_\alpha\subset k_\beta$ is an infinite field extension for all pairs $\alpha<\beta$ of ordinals. First I'll define an "abelian category" which has large Hom-sets. An object $V$ will consist of a $k_\alpha$-vector space $V(\alpha)$ for each ordinal $\alpha$, together with a $k_\alpha$-linear map $v_{\alpha,\beta}:V(\alpha)\to V(\beta)$ for each pair $\alpha<\beta$ of ordinals, such that $v_{\beta,\gamma}\circ v_{\alpha,\beta}=v_{\alpha,\gamma}$ whenever $\alpha<\beta<\gamma$. A morphism $\theta:V\to W$ will consist of a $k_\alpha$-linear map $\theta_\alpha:V(\alpha)\to W(\alpha)$ for each $\alpha$, such that $w_{\alpha,\beta}\circ\theta_\alpha=\theta_\beta\circ v_{\alpha,\beta}$ for all $\alpha<\beta$. Now let's say that an object $V$ is "$\alpha$-good" if, for every $\beta>\alpha$, $V(\beta)$ is generated as a $k_\beta$-vector space by the image of $v_{\alpha,\beta}$, and that $V$ is "good" if it is $\alpha$-good for some $\alpha$. If $V$ is $\alpha$-good, then any morphism $\theta:V\to W$ is determined by $\theta_\gamma$ for $\gamma\leq\alpha$, so the full subcategory $\mathfrak{C}$ of good objects has small Hom-sets. It's straightforward to check that $\mathfrak{C}$ is an abelian category, and it has small coproducts in the obvious way, where $\left(\coprod_{i\in I} V_i\right)(\alpha)=\coprod_{i\in I} V_i(\alpha)$. I claim that $\mathfrak{C}$ does not have all small products. For any $\alpha$, let $P_{\alpha}$ be the ($\alpha$-good) object with
$$P_\alpha(\beta)=\begin{cases}0&\mbox{if }\beta<\alpha\\k_\beta&\mbox{if }\alpha\leq\beta\end{cases}$$
and obvious inclusion maps. Then for any object $W$, $\operatorname{Hom}_{\mathfrak{C}}(P_\alpha,W)$ is naturally isomorphic to $W(\alpha)$ (i.e., $P_\alpha$ represents the functor $W\mapsto W(\alpha)$ from $\mathfrak{C}$ to $k_\alpha$-vector spaces), and if $\alpha<\beta$ then the map
$$W(\alpha)=\operatorname{Hom}_{\mathfrak{C}}(P_\alpha,W)\to\operatorname{Hom}_{\mathfrak{C}}(P_\beta,W)=W(\beta),$$
induced by the obvious inclusion $P_\beta\to P_\alpha$, is just $w_{\alpha,\beta}$. Suppose $W$ were the product in $\mathfrak{C}$ of of a countable number of copies of $P_0$. Since it's an object of $\mathfrak{C}$, $W$ must be $\alpha$-good for some $\alpha$. Then
$$W(\alpha)=\operatorname{Hom}_{\mathfrak{C}}(P_\alpha,W)=\prod_{i\in\mathbb{N}}k_\alpha$$
and for $\beta>\alpha$
$$W(\beta)=\operatorname{Hom}_{\mathfrak{C}}(P_\beta,W)=\prod_{i\in\mathbb{N}}k_\beta.$$ But then $W(\beta)$ is not generated as a $k_\beta$-vector space by the image of the natural map $w_{\alpha,\beta}:W(\alpha)\to W(\beta)$, since $k_\beta$ is an infinite extension of $k_\alpha$, contradicting the $\alpha$-goodness of $W$.
|
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|
112,575 |
If we set $\exp(x)=\sum x^k/k!$, then $\exp(x+y)=\exp(x)\cdot \exp(y)$. In terms of coefficients it means that $(x+y)^n=\sum \frac{n!}{k!(n-k)!} x^ky^{n-k}$, i.e. just binomial expansion. Now consider logarithm. Set $L(x):=\sum_{k>0} x^k/k$, then $L(x)=-\log(1-x)$ in a sense, and hence $$L(u+v-uv)=L(u)+L(v),$$ i.e. $\sum (u+v-uv)^n/n=\sum (u^n+v^n)/n$, or, if we pass to coefficients of $u^av^b$ ($a,b\geq 1$), we get
$$
\sum_k (-1)^k\frac{(a+b-k-1)!}{(a-k)!(b-k)!k!}=0
$$ The question is what is combinatorial meaning of this identity. Maybe, it is some exclusion-inclusion formula, as it is usual for alternating sums?
|
As David noted, since the summands aren't in general integers, it's difficult to give a combinatorial interpretation to the formula. However, if we multiply by $a$ or $b$ we get integers and we can give a combinatorial interpretation to the identity that we obtain (though doing this destroys the symmetry between $a$ and $b$). If we multiply by $b$, the sum may be written
$$\sum_{k=0}^a (-1)^k \binom{a+b-k-1}{a-k}\binom{b}{k}.$$
This is a special case ($m=a+b-1$) of the more general identity
$$\sum_{k=0}^a (-1)^k\binom{m-k}{a-k}\binom{b}{k} = \binom{m-b}{a},$$
which we can prove combinatorially. (Incidentally, this identity is a form of Vandermonde's theorem.) To prove this formula, we start with a set $M$ of size $m$, with a subset $B$ of size $b$. To interpret $\binom{m-k}{a-k}\binom{b}{k}$, we choose a $k$-subset $K$ of $B$ and then choose an $(a-k)$-subset $C$ of $M-K$. The right side $\binom{m-b}{a}$ counts pairs $(K,C)$ in which $K$ is empty and $C$ is an $a$-subset of $M-B$. To prove the identity, we find an involution on the set of pairs $(K,C)$ not of this form that changes the parity of $|K|$. The pairs $(K,C)$ to be canceled are those in which $K\cup C$ contains at least one element of $B$. Then the involution moves the smallest element of $(K\cup C) \cap B$ from $K$ to $C$ or from $C$ to $K$.
|
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|
112,679 |
The motivation is simple, as it is trivially right when $p=2$. When considering the duality between $L^p$ ($l^p$) and $L^q$ ($l^q$) when $p$ and $q$ are conjugate in the sense that $1/p+1/q=1$, I wonder if $L^p$ and $l^p$ are the same in the sense of isometry.
I tried to use the situation when $p=2$, however I find it difficult to give a linear homeomorphism.
|
Variants of this question show up often enough here on MO and over at math.SE that it seems worthwhile to collect some facts and links. I say isomorphic for linearly homeomorphic and isometric for isometrically isomorphic . One main upshot is: The family of Banach spaces $L^p, \ell^q$ for $1 \leq p, q \leq \infty$ consists of pairwise non-isomorphic spaces, except for two cases: There is the obvious isometry between $L^2$ and $\ell^2$ and There is a non-obvious isomorphism between $L^\infty$ and $\ell^\infty$, due to Pełczyński. However, $L^\infty$ and $\ell^\infty$ are not isometric. Plenty of references can be found in the following threads on MO and math.SE: Bill Johnson explains in $L^p$ vs. $L^q$ how one can tell these spaces apart for $1 \lt p \neq q \lt \infty$ using type and cotype considerations, and in the comments there is also some discussion of existence of embeddings. A detailed explanation of the non-existence of an isomorphism between the spaces $L^p$ and $\ell^q$ for $1 \leq p,q \lt \infty$ (modulo Bill Johnson's answer in 1.) is in the answer to If $1\leq p < \infty$ then show that $L^p([0,1])$ and $\ell_p$ are not topologically isomorphic . The main ingredient in that answer is Pitt's theorem stating that every operator $\ell^q \to \ell^p$ is compact for $1 \leq p \lt q \lt \infty$, the existence of an embedding $L^2 \to L^p$ plus a little bit of duality theory. See also How do you prove that $\ell^p$ is not isomorphic to $\ell^q$? for a discussion how Pitt's theorem implies non-isomorphism of $\ell^p$ and $\ell^q$ and Inclusion of $L^p$-spaces, reloaded for a discussion of embeddings of $L^2$ into $L^p$ via Rademacher functions or Gaussians. Pełczyński's isomorphism between $L^\infty$ and $\ell^\infty$ is discussed in Is there an explicit isomorphism between $L^\infty[0,1]$ and $\ell^\infty$? and non-existence of an isometry is discussed in Isometry between $L^\infty$ and $\ell^\infty$ . Let me finish by recommending the very nice book by Albiac and Kalton, Topics in Banach Space Theory , as an alternative to Lindenstrauss-Tzafriri. It contains a gentle introduction to the above ideas and much more. Edit: Further links to related topics: One crucial point used in establishing Pełczyński's isomorphism is the injectivity of $\ell^\infty$ and $L^\infty$. For $\ell^\infty$ this is a standard exercise in applying Hahn-Banach (coordinatewise) and for $L^\infty$ Bill Johnson gives a quick proof in Direct proof of injectivity of $L^\infty$ . See also Direct proof of "K is projective iff C(K) has the Hahn-Banach property"? for a generalization and related results. Complemented subspaces of $\ell_p(I)$ for uncountable $I$ Isometric embeddings of $\ell_q^m$ into $\ell_p$ and $L_p$ for $p,q\in[1,+\infty]$
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|
112,764 |
Though it's relatively clear that the characteristic classes do not characterise a vector bundle (and after looking through some books) I could not find an example of a vector bundle which is not stably trivial but whose characteristic classes (those which may be defined*) are all trivial. Could someone be so kind as to point out a reference for this? *an example with trivial Stiefel-Whitney class would already be nice.
|
You will find an answer to your question in Hatcher's book project "Vector bundles an K-theory" (p. 75-76) (available on his homepage).
Using the fact that $\pi_8(O(10))=\mathbb Z_2$, you can build a non- stably trivial vector bundle over the sphere $S^9$ (using the clutching function associated to the non-trivial homotopy class $S^8\rightarrow O(10)$ in $\pi_8(O(10))$). This vector bundle has all his Stiefel-Whitney and Pontryagin classes equal to zero. The vanishing of $w_9$ follows from Wu's formula $w_9=w_1w_8+Sq^1(w_8)$.
|
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|
113,840 |
For consecutive primes $a\lt b\lt c$, prove that $a+b\ge c$. I cannot find a counter-example to this. Do we know if this inequality is true? Alternatively, is this some documented problem (solved or unsolved)?
|
Yes, this is true. In 1952, Nagura proved that for $n \geq 25$ , there is always a prime between $n$ and $(6/5)n$ . Thus, let $p_k$ be a prime at least $25$ . Then $p_k+p_{k+1} > 2p_k$ . But by Nagura's result we have that $p_{k+2} \leq 36/25 p_k < 2p_k$ . Finally, one can easily check by hand that the result holds for small primes.
|
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|
114,034 |
Assume you are an algebraic geometry advanced student who has mastered Hartshorne 's book supplemented on the arithmetic side by the introduction of Lorenzini - " An Invitation to Arithmetic Geometry " and by Liu - " Algebraic Geometry and Arithmetic Curves ". What would be a good learning path towards the proof of the Weil Conjectures for algebraic varieties (not just curves)? What modern references are available and in which order should be studied? Besides the original article I and article II by Deligne and the results on rationality by Dwork, there is the book Freitag/Kiehl - " Étale Cohomology and the Weil Conjecture " and the online pdf by Milne - " Lectures on Étale Cohomology ". The first title is out of stock and hard to get and the second seems to me too brief and succinct. Is it better to master étale cohomology by itself elsewhere and then refer to the original articles? Is any further algebraic/arithmetic background necessary? Thank you in advance for any hints on how to approach such a study program, and for any related advice towards a self-learning path in arithmetic geometry. ( This question is cross posted to math.stackexchange so all kind of students and professionals can provide their advice regardless of their membership to these forums. )
|
I'm not an expert, but here is how I would plan my trip: There are obviously two parts: rationality + functional equation + comparison with Betti numbers (which follow from the construction of etale cohomology) and the Riemann hypothesis (which is much deeper). I would therefore follow the following plan: Learn classical algebraic topology, esp. singular cohomology, Poincare duality, Lefschetz fixed point formula, Leray-Serre spectral sequence. Learn an overview of $\ell$-adic cohomology, without technical details. First, understand how a good cohomology theory like in 1. above will prove the first part of the conjectures. Then understand etale topology and definition of $\ell$-adic cohomology groups and how the Frobenius morphism happens to act on them. Technical machinery underlying $\ell$-adic cohomology. I haven't studied this myself very well, but Milne's book seems to be a standard reference. Read Deligne's Weil I article. It's beautifully written and you don't need much more than 1. and 2. above. The main technical tool is the use of Lefschetz pencils, which is there just to make induction on dimension possible. You can just assume Lefschetz pencils exist, or look to SGA if interested. Note that in Deligne's approach it is crucial to work with constructible sheaves, not just the constant sheaf. Read Deligne's Weil II article. It reproves Weil I and adds much more, but is much longer and more difficult. Note: 3. and 4. above might be mostly independent. A very good reference is Katz's article "L-functions and monodromy: four lectures on Weil II".
|
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|
114,245 |
A bit unsure if the following vague question has enough mathematical content to be suitable upon here. In the case, please feel free to close it. In several circumstances of competition, a particular situation of partial information occurs, usually described as " I know that you know that I know... something ". We may distinguish a whole hierarchy of more and more complicate situations closer and closer to a complete information. E.g. : $I_0$: I know $X$, but you don't know that I know. $I_1$: I know $X$, you know that I know, but I don't know that you know that I know. $I_1$: I know $X$, you know that I know, I know that you know that I know, but you don't know that I know that you know that I know. .... &c. For small values of $k$, I can imagine simple situations where passing from $I_k$ to $I_ {k+1}$ really makes a difference (for instance: you are Grandma Duck, and $X$ is : "you left a cherry pie to cool on the window ledge". Clearly, $I_0$ is quite agreeable position; $I_1$ may lead to an unpleasant end (for me); $I_2$ leaves me some hope, if I behave well, and so on). But, I can't imagine how passing from $I_6$ to $I_7$ may affect my strategy, or Grandma's. Are there situations, real or
factitious, concrete or abstract, where $I_k$
implies a different strategy than
$I_{k+1}$ for the competitors? What
about $I_{\omega}$ and, more generally,
$I_\alpha$ for an ordinal $\alpha$
(suitably defined by induction)? How these situations are modeled mathematically?
|
My wife and I have a standing agreement where I pick up our son Horatio from school and she picks up our daughter Hypatia. One day, because I knew I would be near Hypatia's school, it was convenient to swap duties. I emailed her a message, "I'll pick up Hypatia today, and you get Horatio. Please confirm; otherwise it is as usual." She texted me back, "Let's do it. Let me know if you get this message, so that I know we're really on." I left her a voicemail, "OK, we're definitely on for the swap! ....as long as I know you get this message." She emailed me back, "Got the message. We're on! But let me know that you get this message so I can count on you." You see, without confirmation she couldn't be sure that I knew she had gotten my earlier confirmation of her acknowledgement of my first message, and she may have worried that the plan to swap was consequently off. And so on ad infinitum ...... How truly frustrating it was for us that at no stage of our conversation could we seem to know for certain that the other person had all the necessary information to ensure that the plan would be implemented! The result, of course, since we had time to exchange only at most a finite number of messages, was that the only rational course of action was for us each to abandon the plan to swap: we both independently decided just to pick up the usual child. To see that this was rational, observe that clearly the first message needed to have been confirmed in order for the plan to be implemented properly. Furthermore, if the $n$-th message need not have been confirmed, then it wasn't important to know that it had been received and the algorithm should have worked whether or not it was received, meaning that it didn't actually need to have been sent. So by induction, no number of confirmations suffices to implement the common knowledge that we both needed, namely, that we had each agreed to make the swap. See also the two generals problem .
|
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|
114,528 |
Let $U_1, U_2$ be open subsets of $\mathbb{R}^n$. Both are naturally differentiable submanifold, getting the differentiable structure from $\mathbb{R}^n$. Further, both are natural topological manifolds, as submanifolds of $\mathbb{R}^n$. Question: If $U_1$ and $U_2$ are homeomorphic , are they also diffeomorphic ? Of course two general topological manifolds which are homoemorphic do not need to be diffeomorphic. But here the differentiable structure is a very special one. The answer might depend on the dimension $n$. For $n=1,2,3$ it is yes, as there each topological manifold has a unique differentiable structure. For $n \geq 5$ and $U_1$ an open ball the answer is yes by the uniqueness of differentiable structures on $\mathbb{R}^n$ for $n \geq 5$. Some special cases are: What happens if $U_1$ (and hence $U_2$) is contractible? What happens if $U_1$ is a ball and $n=4$? Is there an exotic $\mathbb{R}^4$ which can be realized as an open subset of the standard $\mathbb{R}^4$? (The question came up because I encountered a sloppy definition of a manifold. One can view the above manifolds as being defined by only one chart. [That of course depends on your definition of chart, if you require it to start from a ball or not.] So the questions basically asks: How do manifolds with only one chart look like?)
|
In fact, there exist uncountably many small exotic smooth $\mathbb{R}^4$'s, i.e. smooth manifolds $X$ which are homeomorphic to $\mathbb{R}^4$ but not not diffeomorphic to it and which can be smoothly embedded as open subsets of $\mathbb{R}^4$. There are discussions of this in many places; I recommend first reading the appropriate part of Scorpan's book "The Wild World of 4-Manifolds" for a brief survey (it includes a nice bibliography of more detailed sources).
|
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|
114,555 |
Observing the behaviour of a few physicists "in nature", I had the impression that among the mathematical tools they use a lot (along with possibly much more sofisticated maths, of course), there is certainly Taylor expansion. They have a quantity (function) that they need to approximate: they expand it in Taylor series, keep the order of approximation that is useful for their purposes, and discard the irrelevant terms. Appearently, there is little preoccupation for mathematically justifying this procedure, even if the to-be-approximated quantity is not given by an explicit form which is clearly known to be analytic. As Physics clearly gets no problems from the above mathematical subtleties, this may just mean that the distinction between analytic and smooth functions is somehow irrelevant to the basic equations of physics, or rather to the approximations of their solutions that are empirically testable. If non-analytic smooth functions are irrelevant to Physics, why is it so? Are there equations of physical importance in which non-analytic smooth solutions actually are important and cannot be safely considered "as if they were analytic" for the approximation purposes? Remark: analogous questions may arise about Fourier series expansions. One possible way the practice goes might be: Consider a (differential or otherwise) equation $P(f)=0$ usually with analytic coefficients. Expand the coefficients in Taylor series around a point in the scale of physical interest. Discard higher order terms obtaining an approximated equation with polynomial coefficients $\tilde{P}(f)=0$ . Make the ansatz that the solutions $f$ of interest must be analytic. Find the coefficients of $f$ by hand or by other means. This leaves open the question why the ansatz is mathematically justified, if the equation of interest was $P$ not $\tilde{P}$ . Do analytic solutions of $\tilde{P}$ aptly approximate solutions of $P$ ? Edit: I understand now that these last two lines are not very well formulated. Perhaps, ignoring the $\tilde{P}$ thing, I should have just asked something like: Given any $\epsilon>0$ , does knowing the analytic solutions (i.e. knowing their coefficients, possibly up to an arbitrarily large but finite number of digits) of $P$ give all the information about all solutions of $P$ up to $\epsilon$ -approximation? Are there physically well known classes of equations $P$ in which this may not happen (perhaps even up to taking very regular approximations of the coefficients/parameters of $P$ itself)?
|
As a physicist "in nature" perhaps I can give a few examples that illustrate how non-analytic functions
can appear in physics and counter the idea that physicists do not worry about the justification
of these procedures. Example 1 involves one of the most precise comparisons between experiment and theory known to
physics, namely the g factor of the electron. The quantity g is a proportionality factor between
the spin of the electron and its magnetic moment. Perturbation theory in QED gives a formula
$$g-2= c_1 \alpha + c_2 \alpha^2 + c_3 \alpha^3 + \cdots $$
where the coefficients $c_i$ can be computed from i-loop Feynman diagrams and
$\alpha=e^2/\hbar c \simeq 1/137$ is the fine structure constant. Including up to four
loop diagrams gives an expression for $g$ which agrees to one part in $10^{8}$ with experiment.
Yet it is known that that this perturbative series has zero radius of convergence. This is true quite generally in quantum field theory. Physicists do not ignore this, rather they regard it as evidence that QFT's are not defined by their perturbation series but must also include non-perturbative effects,
generally of the form $e^{-c/g^2}$ with $g$ a dimensionless coupling constant. Much effort has gone
into understanding these non-perturbative effects in a variety of quantum field theories. Instanton
effects in non-Abelian gauge theory are an important example of non-perturbative phenomena. Example 2 involves the Hydrogen atom in an electric field of magnitude $E$, aka the Stark effect.
One can compute
the shift in the energy eigenvalues of the Hydrogen atom Hamiltonian due to the applied electric
field as a power series in $E$ using perturbation theory and again one finds excellent agreement
with experiment. One can also prove that this series has zero radius of convergence. In fact,
the Hamiltonian is not bounded from below and does not have any normalizable energy eigenstates. The physics of this situation explains what is going on. The electron can tunnel
through the potential barrier and escape from being bound to the nucleus of the Hydrogen atom,
but for reasonable size electric fields the lifetime of these states exceeds the age of the universe.
The perturbation theory does not converge because there are no energy eigenstates to converge to,
but it still provides an excellent approximation to the energy eigenstates measured experimentally because the experiments
are done on a time scale which is very short compared to the lifetime of the metastable state. So I would say that at least in these examples there is a very nice interplay between the physics and
the mathematics. The lack of analyticity has a clear physical interpretation and this is something that
is understood by physicists. Of course I'm sure there are other example where such approximations
are made without a clear physical justification, but this just means that one should understand the
physics better.
|
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|
114,745 |
There are certainly non-monic polynomials of degree 4 with all roots on the unit circle, but no roots are roots of unity; $5 - 6 x^2 + 5 x^4$ for example. Now, for a monic polynomial of degree $n$, this is impossible (I think). So, my question is, given a monic polynomial with integer coefficients of degree $n$,
what is the maximal number of roots that can lie on the unit circle, and not be roots of unity? For example, $1 + 3 x + 3 x^2 + 3 x^3 + x^4$ has two roots on the unit circle, and two real roots.
|
There exist irreducible monic polynomials such that all their roots apart from two lie on the unique circle (and are not roots of unity). Such polynomials can be chosen among Salem polynomials and they exit in arbitrary high degree. By definition a Salem polynomial $S(x)\in \mathbb Z[x]$ is a monic irreducible reciprocal polynomial with exactly two roots off the unit circle, both real and positive. Of course non of the roots of these polynomials are roots of unity, since these polynomials are irreducible. See for example theorem 1.6 in the article Automorphisms of even unimodular lattices and unramified Salem numbers of Gross and Mcmullen: http://www.math.harvard.edu/~ctm/home/text/papers/unim/unim.pdf Theorem. For any odd integer $n\ge 3$ there exist infinitely many unramified Salem polynomials of degree $2n$.
|
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|
114,943 |
According the the Wikipedia page , the second generation proof is up to at least nine volumes: six by Gorenstein, Lyons and Solomon dated 1994-2005, two covering the quasithin business by Aschbacher and Smith in 2004, and one by Aschbacher, Lyons, Smith and Solomon in 2011. However, this latter book is really just the second part of an outline of the proof, the first part of which was written by Gorenstein in the 80s (the reason for the delay is, of course, that the quasithin case hadn't actually been settled at the time of the announcement of completion). Hence the last update on the second-generation proof is really 2005. With the recent formal proof in Coq of the Odd-order Theorem, it would be good to know where the traditional proof is up to. EDIT 6 August 2013: Any news as to the completion of that seventh volume as mentioned in the comments? EDIT 29 September 2016 Just a bump to this question in case people know more about where the progress is at. Books 7 and 8 should probably have made some progress since I asked this originally.
|
There is an interesting review by Ron Solomon of a paper in this area, which has been featured on the Beyond Reviews blog. In particular, he outlines the broad tactics that people are using in CFSG II, and some of the content that will be going into volume 7. Also, Inna Capdeboscq apparently gave an outline of volume 8, or at least a chunk of it, at the Asymptotic Group Theory conference in Budapest. This was mentioned by Peter Cameron on his blog , sadly with no detail! If anyone can get a whiff of what she said, I would be grateful. EDIT 15 October 2016 I emailed the group-pub mailing list and was told second-hand that Ron Solomon 'has hopes' volume 7 will be submitted next year. EDIT 27 March 2018 Thanks to Timothy Chow in a comment on another answer, here is the link to the published version of Volume 7 . So now the countdown to Volume 8 starts... EDIT 22 June 2018 Even better news: Volume 8 ...is near completion and promised to the AMS by August 2018. The completion of Volume 8 will be a significant mathematical milestone in our work. ( source ) Also (from the same article): We anticipate that there will be twelve volumes in the complete series [GLS], which we hope to complete by 2023. Considerable work has been done on this problem [the bicharacteristic case], originally by Gorenstein and Lyons, and more recently by Inna Capdeboscq, Lyons, and me. We anticipate that this will be the principal content of Volume 9 [GLS], coauthored with Capdeboscq. When p is odd, there is a major 600-page manuscript by Gernot Stroth treating groups with a strongly p-embedded subgroup, which will appear in the [GLS] series, probably in Volume 11. There are also substantial drafts by Richard Foote, Gorenstein, and Lyons for a companion volume (Volume 10?), which together with Stroth’s volume will complete the p-Uniqueness Case. It would be wonderful to complete our series by 2023, the sixtieth anniversary of the publication of the Odd Order Theorem. Given the state of Volumes 8, 9, 10, and 11, the achievement of this goal depends most heavily on the completion of the e(G) = 3 problem. It is a worthy goal. EDIT Mar 2019 Volume 8 has been published. The page listing the available volumes, along with links to more details is here . The summary of this volume is as follows: This book completes a trilogy (Numbers 5, 7, and 8) of the series The Classification of the Finite Simple Groups treating the generic case of the classification of the finite simple groups. In conjunction with Numbers 4 and 6, it allows us to reach a major milestone in our series—the completion of the proof of the following theorem: Theorem O: Let G be a finite simple group of odd type, all of whose proper simple sections are known simple groups. Then either G is an alternating group or G is a finite group of Lie type defined over a field of odd order or G is one of six sporadic simple groups. Put another way, Theorem O asserts that any minimal counterexample to the classification of the finite simple groups must be of even type. The work of Aschbacher and Smith shows that a minimal counterexample is not of quasithin even type, while this volume shows that a minimal counterexample cannot be of generic even type, modulo the treatment of certain intermediate configurations of even type which will be ruled out in the next volume of our series. EDIT February 2021 Volume 9 has now been published. From the preface: This book contains a complete proof of Theorem $\mathcal{C}_5$ , which covers the “bicharacteristic” subcase of the $e(G) \ge 4$ problem. The outcome of this theorem is that $G$ is isomorphic to one of the six sporadic groups for which $e(G)\ge 4$ , or one of six groups of Lie type which exhibit both characteristic 2-like and characteristic 3-like properties. Finally, in Chapter 7, we begin the proof of Theorem $\mathcal{C}_6$ and its generalization Theorem $\mathcal{C}^∗_6$ , which cover the “ $p$ -intermediate” case. $\ldots$ In the preceding book in this series, we had promised complete proofs of Theorems $\mathcal{C}_6$ and $\mathcal{C}^∗_6$ in this book, but because of space considerations, we postpone the completion of those theorems to the next volume. EDIT September 2021 In response to a question from Hugo de Garis , Ron Solomon sent the following email in January 2021: Vol. 9 is already submitted, accepted and scheduled for publication. It should be published early this year.
As for the rest, my best guess now is that there will in fact be 4 further volumes , not 3. A roughly 800 pages manuscript on the Uniqueness Theorem has been completed by Gernot Stroth. With some additional material, it will fill 2 further volumes. This could probably be readied for publication by a year from now. However, our team (Inna Capdeboscq, Richard Lyons, Chris Parker and myself) are currently focussing on the remaining work to be done for the other two volumes. It is difficult to estimate how long this will take. With luck we might have a first draft completed this calendar year, but it might take longer.
It is safe to say that the remaining volumes will not all be published before 2023. I hope it is also safe to say that they will all be published no later than 2025. (Emphasis added) EDIT 29 Dec 2021 Richard Lyons maintains an erratum for the whole published second generation CFSG on this page: https://sites.math.rutgers.edu/~lyons/cfsg/ EDIT 05 Apr 2022 In response to a further question from de Garis (see the page linked above), Solomon wrote (in March 2022): We have been working on the theorems for both Volumes 10 and 11. Just in the past few weeks, we have decided to concentrate on the completion of Volume 10. This is proceeding very well and we should be able to submit Volume 10 for publication this year, I believe. I fear that I may have been a bit overoptimistic in predicting the completion of all the volumes by the end of 2024.
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|
115,061 |
Does every profinite group arise as the étale fundamental group of a connected scheme? Equivalently, does every Galois category arise as the category of finite étale covers of a connected scheme? Not every profinite group is an absolute galois group of a field (the only finite ones have order $1$ or $2$ by Artin-Schreier). Therefore we cannot restrict to spectra of fields. Perhaps one first has to check if every finite group arises as a fundamental group of a scheme. I don't even know enough examples to answer this question for cyclic groups. At least order $3$ is possible (see here , Remark 2). If the answer turns out to be no , then I would like to know which profinite groups arise as fundamental groups.
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[Edit:] The answer should be positive, that is, every profinite group appears as the fundamental group of a scheme. Here is a sketch of proof. First of all, I claim that for any finite group $G$ there exists a complex affine simply connected variety $X$ with a free action of $G$. Start from a faithful finite-dimensional representation $G \to \mathrm{GL}(V)$ of $G$, such that there is an open subset $U \subseteq V$ where the action of $G$ is free, and such that $V \smallsetminus U$ has codimension at least 2 in $V$. Then $X$ is obtained with an easy equivariant extension of Jouanolou's trick. This yields a $G$-covering $X \to X/G$; then $X/G$ is an affine variety, and its fundamental group is $G$. Now take a profinite group $G = \projlim_{i\in I}G_i$; identify $G$ with the corresponding affine group scheme over $\mathbb C$, in the usual fashion. For each finite subset $J\subseteq I$ denote by $G_J$ the image of $G$ in $\prod_{j\in J}G_j$; clearly we have $G = \projlim_{J \subseteq I}G_J$. For each $i$ take a complex affine simply connected variety $X_i$ with a free action of $G_i$. Consider the affine scheme $X := \prod_{i \in I}X_i$, with the resulting action of $G$, and the quotient $X/G$, which is the spectrum of the ring of invariants $\mathbb C[X]^G$. For each finite subset $J\subseteq I$ set $X^J := \prod_{j \in J}X_j$. The action of $G$ on $X^J$ factors through a free action of $G_J$. Furthermore, we have $\mathbb C[X]^G = \injlim_{J \subseteq I}\mathbb C[X^J]^{G_J}$, hence $X/G = \projlim X^J/G_J$. Now, $X^J$ is simply connected, hence the fundamental group of $X^J/G_J$ is $G_J$. On the other hand, it is easy to see that the Galois category of $X$ is the inductive limit of the Galois categories of the $X^J$, so that its Galois group is precisely $\projlim G_J = G$. [Edit2:] Let me clarify what I mean by the "equivariant Jouanolou trick". The theorem of Jouanolou says that if $U$ is a quasi-projective variety, there exists an locally trivial fibration in affine spaces $X\to U$, where $X$ is affine. What we need here is the statement that if $G$ is a finite group acting on $U$, then we can construct such a map $X \to U$ that is also $G$-equivariant. This is easy: start from a fibration in affine space $Y \to U$ with $Y$ affine, and take $X$ to be the fiber product over $U$ of all the $g^*Y$ for $g \in G$, with the obvious action of $G$.
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|
115,112 |
I hope this question is appropriate for MO. It comes from a genuine desire to understand the big picture and ground my own studies "morally". I'm a graduate student with interest in number theory. I feel like I'm in danger of losing the big picture as I venture a bit deeper and reflect on where I am at. My fundamental is this: I care about the natural numbers - and thus naturally care about the Riemann Zeta function. Number theorists have embarked on various adventures in studying generalized integers (rings of integers of Q-extensions), and their associated zeta functions, and beyond (e.g. Langlands program). Some mathematicians seem to be interested in these generalized integers and zeta functions for their own sake. I am not. Given my passion for $\mathbb{N}$ and zeta, why should I study these other objects? I understand that philosophically to understand an object it's good to understand its context, and its similarities and differences to its brothers and cousins. This principle makes a lot of sense. But specifically, what new understandings of $\mathbb{N}$ and zeta have we gained thus far by studying these more general systems? Are there clearly articulated reasons why we can hope to bring back more "treasure" from these more general searches that may shed light on $\mathbb{N}$ in particular ? I worry sometimes that number theory is becoming divorced from its original "ground", though I believe (and hope) this feeling derives mainly from ignorance. EDIT: My question was probably not written very well. I am aware of some of the benefits of studying solutions of polynomials in ring extensions (e.g. solving cases of FLT). My concern is with the broad scope of number theory research today, particularly in the land of generalized zeta-functions and Langlands program. I am uncomfortable (in my ignorance, I admit!) with the apparent lack of a clear connection to the "natural" concerns of number theorists prior to the mid 20th century. I hope that my question is taken in the spirit of a naive apprentice asking masters for motivation, and a layout of the land of modern research as it connects to concerns that used to be universal.
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Where to begin? As a tiny example, suppose that you are interested in the solutions in rational numbers to an equation $Y^2=X^3+AX+B$, i.e., the rational points on an elliptic curve $E$. One naturally looks at the algebraic solutions $E(\overline{\mathbb{Q}})$, since one can do geometry over the algebraically closed field $\overline{\mathbb{Q}}$, and then picks out the rational points $E(\mathbb{Q})$ by studying the action of the Galois group $G(\overline{\mathbb{Q}}/\mathbb{Q})$ and picking out the points that are invariant for the group action. Hopefully you're also interested in rational numbers, since they're just ratios of integers, but if you insist on problems with integers, then one can look at the integer solutions $E(\mathbb{Z})$, which forms a finite set (Siegel's theorem, made effective by Baker). But even there, one way to study $E(\mathbb{Z})$ is by analyzing it as a subset of $E(\mathbb{Q})$, so you're back to rational solutions. So that's a bit long-winded, but it illustrates a general principle. If a problem involving a set is hard, for example a problem involving integers, it may be easier to solve a problem involving a larger set and then pick out the subset that you're really interested in. Having said all of this, it is also true that there are mathematicians who find studying "new" objects to be intrinsically interesting, regardless of the original applications that they were devised for. As an example, there are lots of people who study automorphic forms for their own sake. If there are applications to classical sorts of problems involving integers, that's great, but it's not their motivation or primary interest. Luckily, there's room for everyone in the big tent that makes up the mathematical research community.
|
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|
115,416 |
If $\frac{d}{dx}$ is a differential operator, what are its inputs? If the answer is "(differentiable) functions" (i.e., variable-agnostic sets of ordered pairs), we have difficulty distinguishing between $\frac{d}{dx}$ and $\frac{d}{dt}$, which in practice have different meanings. If the answer is "(differentiable) functions of $x$", what does that mean? It sounds like a peculiar hybrid of mathematical object (function) with mathematical notation (variable $x$). Does $\frac{d}{dx}$ have an interpretation as an operator, distinct from $\frac{d}{dt}$, and consistent with its use in first-year Calculus?
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(From the post on my blog :) To my way of thinking, this is a serious question, and I am not really satisfied by the other answers and comments, which seem to answer a different question than the one that I find interesting here. The problem is this. We want to regard $\frac{d}{dx}$ as an operator in the abstract senses mentioned by several of the other comments and answers. In the most elementary situation, it operates on a functions of a single real variable, returning another such function, the derivative. And the same for $\frac{d}{dt}$. The problem is that, described this way, the operators $\frac{d}{dx}$ and $\frac{d}{dt}$ seem to be the same operator, namely, the operator that takes a function to its derivative, but nevertheless we cannot seem freely to substitute these symbols for
one another in formal expressions. For example, if an instructor were to write $\frac{d}{dt}x^3=3x^2$, a student might object, "don't you mean $\frac{d}{dx}$?" and the instructor would likely reply, "Oh, yes, excuse me, I meant $\frac{d}{dx}x^3=3x^2$. The other expression would have a different meaning." But if they are the same operator, why don't the two expressions have the same meaning? Why can't we freely substitute different names for this operator and get the same result? What is going on with the logic of reference here? The situation is that the operator $\frac{d}{dx}$ seems to make sense only when applied to functions whose independent variable is described by the symbol "x". But this collides with the idea that what the function is at bottom has nothing to do with the way we represent it, with the particular symbols that we might use to express which function is meant. That is, the function is the abstract object (whether interpreted in set theory or category theory or whatever foundational theory), and is not connected in any intimate way with the symbol "$x$". Surely the functions $x\mapsto x^3$ and $t\mapsto t^3$, with the same domain and codomain, are simply different ways of describing exactly the same function. So why can't we seem to substitute them for one another in the formal expressions? The answer is that the syntactic use of $\frac{d}{dx}$ in a formal expression involves a kind of binding of the variable $x$. Consider the issue of collision of bound variables in first order logic: if $\varphi(x)$ is the assertion that $x$ is not maximal with respect to $\lt$, expressed by $\exists y\ x\lt y$, then $\varphi(y)$, the assertion that $y$ is not maximal, is not correctly described as the assertion $\exists y\ y\lt y$, which is what would be obtained by simply replacing the occurrence of $x$ in $\varphi(x)$ with the symbol $y$. For the intended meaning, we cannot simply syntactically replace the occurrence of $x$ with the symbol $y$, if that occurrence of $x$ falls under the scope of a quantifier. Similarly, although the functions $x\mapsto x^3$ and $t\mapsto t^3$ are equal as functions of a real variable, we cannot simply syntactically substitute the expression $x^3$ for $t^3$ in $\frac{d}{dt}t^3$ to get $\frac{d}{dt}x^3$. One might even take the latter as a kind of ill-formed expression, without further explanation of how $x^3$ is to be taken as a function of $t$. So the expression $\frac{d}{dx}$ causes a binding of the variable $x$, much like a quantifier might, and this prevents free substitution in just the way that collision does. But the case here is not quite the same as the way $x$ is a bound variable in $\int_0^1 x^3\ dx$, since $x$ remains free in $\frac{d}{dx}x^3$, but we would say that $\int_0^1 x^3\ dx$ has the same meaning as $\int_0^1 y^3\ dy$. Of course, the issue evaporates if one uses a notation, such as the $\lambda$-calculus, which insists that one be completely explicit about which syntactic variables are to be regarded as the independent variables of a functional term, as in $\lambda x.x^3$, which means the function of the variable $x$ with value $x^3$. And this is how I take several of the other answers to the question, namely, that the use of the operator $\frac{d}{dx}$ indicates that one has previously indicated which of the arguments of the given function is to be regarded as $x$, and it is with respect to this argument that one is differentiating. In practice, this is almost always clear without much remark. For example, our use of $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ seems to manage very well in complex situations, sometimes with dozens of variables running around, without adopting the onerous formalism of the $\lambda$-calculus, even if that formalism is what these solutions are essentially really about. Meanwhile, it is easy to make examples where one must be very specific about which variables are the independent variable and which are not, as Todd mentions in his comment to David's answer. For example, cases like $$\frac{d}{dx}\int_0^x(t^2+x^3)dt\qquad
\frac{d}{dt}\int_t^x(t^2+x^3)dt$$ are surely clarified for students by a discussion of the usage of variables in formal expressions and more specifically the issue of bound and free variables.
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|
115,549 |
I read on the nLab that in "Pursuing stacks" Grothendieck made several interesting conjectures, some of which have been proved since then. For example, as David Roberts wrote in answer to this question , Grothendieck conjectured, and Cisinski proved, that the class of weak equivalences in the Thomason model structure was the smallest basic localizer. I am interesting in knowing what other conjectures made in PS have turned out to be true, or other "ideas" that have been successfully realized/formalized. Ideally it would be nice to include references to the relevant papers.
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To dash off a quick answer, Pursuing Stacks is composed of (if memory serves correctly) three themes. The first was homotopy types as higher (non-strict) groupoids. This part was first considered in Grothendieck's letters to Larry Breen from 1975, and is mostly contained in the letter to Quillen which makes up the first part of PS (about 12 pages or so). Maltsiniotis has extracted Grothendieck's proposed definition for a weak $\infty$ -groupoid, and there is work by Ara towards showing that this definition satisfies the homotopy hypothesis. The other parts (not entirely inseparable) are the first thoughts on derivators , which were later taken up in great detail in Grothendieck's 1990-91 notes (see there for extensive literature relating to derivators, the first 15 of 19 chapters of Les Dérivateurs are themselves available), and the 'schematisation of homotopy types', which is covered by work of Toën , Vezzosi and others on homotopical algebraic geometry (e.g. HAG I , HAG II ) using simplicial sheaves on schemes. This has taken off with work of Lurie , Rezk and others dealing with derived algebraic geometry , which is going far ahead of what I believe Grothendieck envisaged. During correspondence with Grothendieck in the 80s , Joyal constructed what we now call the Joyal model structure on the category of simplicial sets simplicial sheaves to give a basis to some of the ideas being tossed around at the time. ( Edited 2022 ) Edit: I forgot something that is in PS, and that is the theory of localisers and modelisers, Grothendieck's conception of homotopy theory which you mention, which is covered in the work of Cisinski. Edit 2019: Toën has a new preprint out Bertrand Toën, Le problème de la schématisation de Grothendieck revisité , arXiv: 1911.05509 with abstract starting "The objective of this work is to reconsider the schematization problem of [ Pursuing Stacks ], with a particular focus on the global case over Z. For this, we prove the conjecture [Conj. 2.3.6 of Toën's Champs affines ]..."
|
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