User kevin lin - MathOverflow

Answer by Kevin Lin for What are some applications of other fields to mathematics?

2010-02-09T17:09:23Z

Here's a nice paper by Sturmfels, on the question Can biology lead to new theorems?

Definition of "simplicial complex"

2009-11-20T16:40:04Z

When I think of a "simplicial complex", I think of the geometric realization of a simplicial set (a simplicial object in the category of sets). I'll refer to this as "the first definition".

However, there is another definition of "simplicial complex", e.g. the one on wikipedia: it's a collection $K$ of simplices such that any face of any simplex in $K$ is also in $K$, and the intersection of two simplices of $K$ is a face of both of the two simplices. There is also the notion of "abstract simplicial complex", which is a collection of subsets of $\{ 1, \dots, n \}$ which is closed under the operation of taking subsets. These kinds of simplicial complexes also have corresponding geometric realizations as topological spaces. I'll refer to both of these definitions as "the second definition".

The second definition looks reasonable at first sight, but then you quickly run into some horrible things, like the fact that triangulating even something simple like a torus requires some ridiculous number of simplices (more than 20?). On the other hand, you can triangulate the torus much more reasonably using the first definition (or alternatively using the definition of "Delta complex" from Hatcher's algebraic topology book, but this is not too far from the first definition anyway).

I believe you can move back and forth between the two definitions without much trouble. (I think you can go from the first to the second by doing some barycentric subdivisions, and going from the second to the first is trivial.)

Due to the fact that the second definition is the one that's listed on wikipedia, I get the impression that people still use this definition. My questions are:

Are people still using the second definition? If so, in which contexts, and why?
What are the advantages of the second definition?

How should one present curl and divergence in an undergraduate multivariable calculus class?

2010-04-19T19:58:13Z

I am a TA for a multivariable calculus class this semester. I have also TA'd this course a few times in the past. Every time I teach this course, I am never quite sure how I should present curl and divergence. This course follows Stewart's book and does not use differential forms; we only deal with vector fields (in $\mathbb{R}^3$ or $\mathbb{R}^2$). I know that div and curl and gradient are just the de Rham differential (of 2-forms, 1-forms, and 0-forms respectively) in disguise. I know that things like curl(gradient f) = 0 and div(curl F) = 0 are just rephrasings of $d^2 = 0$. However, these things are, understandably, quite mysterious to the students, especially the formula for curl, given by $\nabla \times \textbf{F}$, where $\nabla$ is the "vector field" $\langle \partial_x , \partial_y , \partial_z \rangle$. They always find the appearance of the determinant / cross product to be quite weird. And the determinant that you do is itself a bit weird, since its second row consists of differential operators. The students usually think of cross products as giving normal vectors, so they are lead to questions like: What does it mean for a vector field to be perpendicular to a "vector field" with differential operator components?! Incidentally, is the appearance of the "vector field" $\nabla = \langle \partial_x , \partial_y , \partial_z \rangle$ just some sort of coincidence, or is there some high-brow explanation for what it really is?

Is there a clear (it doesn't have to necessarily be 100% rigorous) way to "explain" the formula for curl to undergrad students, within the context of a multivariable calculus class that doesn't use differential forms?

I actually never quite worked out the curl formula myself in terms of fancier differential geometry language. I imagine it's: take a vector field (in $\mathbb{R}^3$), turn it into a 1-form using the standard Riemannian metric, take de Rham d of that to get a 2-form, take Hodge star of that using the standard orientation to get a 1-form, turn that into a vector field using the standard Riemannian metric. I imagine that the appearance of the determinant / cross product comes from the Hodge star. I imagine that one can work out divergence in the same way, and the reason why the formula for divergence is "simple" is because the Hodge star from 3-forms to 0-forms is simple. Is my thinking correct?

Stewart's book provides some comments about how to give curl and divergence a "physical" or "geometric" or "intuitive" interpretation; the former gives the axis about which the vector field is "rotating" at each point, the latter tells you how much the vector field is "flowing" in or out of each point. Is there some way to use these kinds of "physical" or "geometric" pictures to "prove" or explain curl(gradient f) = 0 and div(curl F) = 0? Is there some way to explain to undergrad students how the formulas for curl and div do in fact agree with the "physical" or "geometric" picture? Though such an explanation is perhaps less "mathematical", I would find an explanation of this sort satisfactory for my class.

Thanks in advance!

When are Hilbert schemes smooth?

2009-10-09T22:43:45Z

I know that Hilbert schemes can be very singular. But are there any interesting and nontrivial Hilbert schemes that are smooth? Are there any necessary conditions or sufficient conditions for a Hilbert scheme to be smooth?

Extending vector bundles on a given open subscheme, reprise

2010-08-16T19:38:58Z

In this question, Ariyan asks about the question of uniqueness of extensions of vector bundles when they exist.

Sasha's answer suggests that extensions of vector bundles don't always exist.

More precisely, if $F$ is a vector bundle on an open subscheme $U$, there does not always exist a vector bundle $F'$ on the ambient space $X$ such that $F'|_U \cong F$.

Can anyone give me a simple example of such an $F$?

I am mainly interested in the case when $X$ is a variety (over $\mathbb{C}$), and $U$ is an open subvariety. Probably I want $X$ to be smooth.

Mumford conjecture: Heuristic reasons? Generalizations? ... Algebraic geometry approaches?

2010-01-10T08:44:02Z

The Mumford conjecture states that for each integer $n$, we have: the map $\mathbb{Q}[x_1,x_2,\dots] \to H^\ast(M_g ; \mathbb{Q})$ sending $x_i$ to the kappa class $\kappa_i$, is an isomorphism in degrees less than $n$, for sufficiently large $g$. Here $M_g$ denotes the moduli of genus $g$ curves, and the degree of $x_i$ is the degree of the kappa class $\kappa_i$. This conjecture was proved by Madsen-Weiss a few years ago.

What are the heuristic or moral reasons for the conjecture? (EDIT: I am particularly interested in algebraic geometric reasons, if there are any. Though algebraic topologial reasons are very welcome too.) What lead Mumford to formulating the conjecture in the first place?
I know very little about the Madsen-Weiss proof, but I know that it mainly uses algebraic topology methods. Are there any approaches to the conjecture which are more algebraic-geometric?
Is there any analogous theorem or conjecture regarding the (topological) $K$-theory of $M_g$? Or the Chow ring of $M_g$? etc.

Relations among Hodge classes?

2012-02-10T09:39:10Z

Let $\pi : C_g \to M_g$ be the universal curve over the moduli stack of genus $g$ curves. Let $\omega_\pi$ be the relative canonical bundle. Then $\mathbb{H} := \pi_\ast \omega_\pi$ is a rank $g$ vector bundle, and it is called the Hodge bundle. Its $i$-th Chern class $c_i(\mathbb{H})$ is denoted by $\lambda_i$. These classes are called Hodge classes.

I'd like to know: What are the (known) polynomial relations among the $\lambda_i$'s? Is there an exhaustive list?

A simple basic relation is $\lambda_1^2 = 2\lambda_2$. I can derive this as follows:

By Grothendieck-Riemann-Roch, we have $ch(\pi_! \omega_\pi) = ch(\mathbb{H})-1 = \pi_\ast ( ch(\omega_\pi) \cdot td(\omega_\pi^\vee))$. By a simple calculation, the degree 3 (if you use Chow groups) (degree 6 if you use cohomology) term of $ch(\omega_\pi) \cdot td(\omega_\pi^\vee)$ is zero. Thus the degree 2 term $ch_2(\mathbb{H})$ of the left hand side is zero. We also have $c_2(\mathbb{H}) = \frac{ch_1(\mathbb{H})^2}{2} - ch_2(\mathbb{H})$, hence we conclude $2c_2(\mathbb{H}) = ch_1(\mathbb{H})^2 = c_1(\mathbb{H})^2$.

Answer by Kevin Lin for Riemann's theorem on theta

2012-02-03T16:30:17Z

The locus of holomorphic line bundles in $Pic_{g-1}(\Sigma)$ with a nontrivial holomorphic section is equivalently characterized as the image of $u_{g-1} : Sym^{g-1} \Sigma \to Pic_{g-1}(\Sigma)$ under the Abel-Jacobi map. In section 4 of chapter 1 of the book of Arbarello-Cornalba-Griffiths-Harris, the theta divisor is defined via the Riemann theta function. Then in the next section, they give a pretty low-tech proof of the fact that the cohomology classes of the theta divisor and of the image of $u_{g-1}$ agree (a special case of the Poincare formula), by reducing to the case where $\Sigma$ is a product of elliptic curves. Finally, they prove Riemann's theorem precisely in the way that you suggest.

A few questions about Kontsevich formality

2010-07-22T03:35:15Z

[K] refers to Kontsevich's paper "Deformation quantization of Poisson manifolds, I".

Background

Let $X$ be a smooth affine variety (over $\mathbb{C}$ or maybe a field of characteristic zero) or resp. a smooth (compact?) real manifold. Let $A = \Gamma(X; \mathcal{O}_X)$ or resp. $C^\infty(X)$.

Denote the dg Lie algebra of polyvector fields on $X$ (with Schouten-Nijenhuis bracket and zero differential) by $T$. Denote the dg Lie algebra of the shifted Hochschild cochain complex of $A$ (with Gerstenhaber bracket and Hochschild differential) by $D$.

Then the Hochschild-Konstant-Rosenberg theorem states that there is a quasi-isomorphism of dg vector spaces from $T$ to $D$. However, the HKR map is not a map of dg Lie algebras. It is not a map of dg algebras, either (where the multiplication on $T$ is given by the wedge product and the multiplication on $D$ is given by the cup product of Hochschild cochains).

I believe "Kontsevich formality" refers to the statement that, while the HKR map is not a quasi-isomorphism --- or even a morphism --- of dg Lie algebras, there is an $L_\infty$ quasi-isomorphism $U$ from $T$ to $D$, and therefore $D$ is in fact formal as a dg Lie algebra.

The first "Taylor coefficient" of the $L_\infty$ morphism $U$ is precisely the HKR map (see section 4.6.2 of [K]).

Moreover, this quasi-isomorphism $U$ is compatible with the dg algebra structures on $T$ and $D$ (see section 8.2 of [K]), and it yields a "corrected HKR map" which is a dg algebra quasi-isomorphism. The "correction" comes from the square root of the $\hat{A}$ class of $X$. See this previous MO question.

Questions

(0) Are all of my statements above correct?

(1) In what way is the $L_\infty$ morphism $U$ compatible with the dg algebra structures? I don't understand what this means.

(2) When $X$ is a smooth (compact?) real manifold, I think that all of the statements above are proved in [K]. When $X$ is a smooth affine variety, I think that the statements should all still be true. Where can I find proofs?

(3) Moreover, the last section of [K] suggests that the statements are all still true when $X$ is a smooth possibly non-affine variety. For a general smooth variety, though, instead of taking the Hochschild cochain complex of $A = \Gamma(X;\mathcal{O}_X)$, presumably we should take the Hochschild cochain complex of the (dg?) derived category of $X$. Is this correct? If so, where can I find proofs?

In the second-to-last sentence of [K], Kontsevich seems to claim that the statements for varieties are corollaries of the statements for real manifolds, but I don't see how this can possibly be true. In the last sentence of the paper, he says that he will prove these statements "in the next paper", but I'm not sure which paper "the next paper" is, nor am I even sure that it exists, since "Deformation quantization of Poisson manifolds, II" doesn't exist.

P.S. I am not sure how to tag this question. Feel free to tag it as you wish.

Index vs. equivariant index (and then taking invariant part)?

2011-11-29T22:24:21Z

Let $C$ be a smooth projective curve and let $C^{(n)}$ be its $n$th symmetric power.

Let $E$ be a $S_n$-equivariant vector bundle over the Cartesian power $C^n$. Suppose that $E$ descends to a vector bundle $\tilde{E}$ over the symmetric power $C^{(n)}$.

Then there are two things I can do:

I can compute the $S_n$-equivariant index of $E$ to get an element $V \in K_{S_n}(\text{pt})$. Then I can take the dimension of the $S_n$-invariant part $V^{S_n}$ to get a number.
I can compute the ordinary index of $\tilde{E}$ over $C^{(n)}$, which is a number.

Do these two numbers agree?

Of course I can ask this same question for a more general setting, but this is the setting that I care about right now; I'll be happy with answers for both this setting or a more general one. I am also interested in any references that talk about this kind of "take-equivariant-index-and-then-take-invariant-part" procedure; I haven't really been able to find anything about this in the literature.

Compact Kaehler manifolds that are isomorphic as symplectic manifolds but not as complex manifolds (and vice-versa)

2009-10-09T22:23:49Z

What are some examples of compact Kaehler manifolds (or smooth complex projective varieties) that are not isomorphic as complex manifolds (or as varieties), but are isomorphic as symplectic manifolds (with the symplectic structure induced from the Kaehler structure)? Elliptic curves should be an example, but I can't think of any others. I'm sure there should be lots...
In the other direction, if I have two compact Kaehler manifolds (or smooth complex projective varieties) that are isomorphic as complex manifolds (or as varieties), then are they necessarily isomorphic as symplectic manifolds?
And one last question that just came to mind: If two smooth complex (projective, if need be) varieties are isomorphic as complex manifolds, then they are isomorphic as varieties?

References for Donaldson-Thomas theory and Pandharipande-Thomas theory?

2009-12-22T19:22:57Z

I'm looking for good introductory references for Donaldson-Thomas theory and Pandharipande-Thomas theory. I know a bit about Gromov-Witten theory, but almost nothing about Donaldson-Thomas and Pandharipande-Thomas. Are there some canonical (or good non-canonical) references for Donaldson-Thomas theory and Pandharipande-Thomas theory? References that assume knowledge of Gromov-Witten theory are fine.

K-theory and K-theory pushforward in topology vs. in algebraic geometry

2011-10-11T21:59:47Z

Let $f : X \to Y$ be a [fill in the blank] morphism of [fill in the blank] complex varieties. Then we have the pushforward $f_! : K(X) \to K(Y)$ which is defined by $f_!(E) = \sum_i (-1)^i [R^i f_\ast E]$, the alternating sum of the higher direct images. Here we take $K(X)$ to mean the $K$-group of coherent sheaves.

On the other hand we can also define $K(X)$ as the $K$-group of $C^\infty$ complex vector bundles on $X$ considered as a real manifold. Then we can define a Gysin map $f_!$ using the Thom isomorphism theorem for $K$-theory.

Which adjectives do I need to fill in the blanks with to make the two notions of $K(X)$ agree?

Which adjectives do I need to fill in the blanks with to make the two notions of $f_!$ agree?

If $X$ is smooth and projective, then any coherent sheaf has a finite resolution by locally free sheaves, so we have a map $K^{alg}(X) \to K^{top}(X)$. On the other hand, I don't think it's true that any $C^\infty$ complex vector bundle has a holomorphic structure, so I don't think there is a map $K^{top}(X) \to K^{alg}(X)$ ...

Answer by Kevin Lin for K-theory and K-theory pushforward in topology vs. in algebraic geometry

2011-10-12T01:12:13Z

I found a paper which I think answers #2:

Riemann-Roch and topological K-theory for singular varieties, by Baum, Fulton, MacPherson

http://www.springerlink.com/content/k284857584wp9032/

They prove that the algebraic $f_!$ and the topological $f_!$ agree in the case of proper morphisms of (not necessarily smooth) quasi-projective varieties.

In other words we have a commutative square

K^{alg}(X) ---> K^{alg}(Y)
   |               |
   V               V
K^{top}(X) ---> K^{top}(Y)

What is Chern-Simons theory?

2009-10-31T01:12:04Z

What is Chern-Simons theory? I have read the wikipedia entry, but it's pretty physics-y and I wasn't really able to get any sense for what Chern-Simons theory really is in terms of mathematics.

Chern-Simons theory is supposed to be some kind of TQFT. But what kind of TQFT exactly? When mathematicians say that it is a TQFT, does this mean that it's a certain kind of functor from a certain bordism category to a certain target category? If so, what kind of functor is it? What kind of bordism category is it? What kind of target category is it? How exactly is the functor defined?

Also, from attending talks of Michael Freeman, I know that Chern-Simons theory is supposed to describe some aspects of the fractional quantum Hall effect. How does this work? How do I take some sort of Chern-Simons computation on a 3-(or 4-?)manifold and extract from that some kind of physical prediction about some 2d electron gas? I've also heard that Witten has interpretted various knot invariants like the Jones polynomial in terms of Chern-Simons theory. So does this mean that the Jones polynomial of a knot has a physical interpretation? If so, what is it?

Derived categories and homotopy categories

2009-10-12T19:15:48Z

There are two constructions that look quite similar to me: the derived category of an abelian category, and the homotopy category of a model category. Is there any explicit relationship between these two constructions? (This question is related to, and indeed the inspiration for, one of my previous questions.)

Answer by Kevin Lin for Cohomology class of the diagonal

2011-08-31T20:21:28Z

Let $\Delta : M \to M \times M$ be the diagonal map. Since $M$ is a complex manifold, say of complex dimension $n$, it has a canonical orientation class $[M] \in H_{2n}(M, \mathbb{Z})$. Then you can take the pushforward in homology to get $\Delta_\ast [M] \in H_{2n}(M \times M, \mathbb{Z})$. If $M$ is compact then $M \times M$ is also compact and you can use Poincare duality to get an element in $H^{2n}(M \times M, \mathbb{Z})$. This is the cohomology class of the diagonal.

More generally, the words to look up are Thom isomorphism theorem or Gysin sequence or Gysin map. The inclusion $\Delta$ induces a Gysin map $\Delta_\ast: H^i(M) \to H^{i-(-2n)}(M \times M)$. The cohomology class of the diagonal is the image of $1 \in H^0(M)$ under this map. You can do the same kind of thing in $K$-theory, Chow groups, etc.

Cohomology of Theta divisor on Jacobian?

2011-08-25T16:40:05Z

Let $C$ be a curve of genus $g \geq 1$ and let $J^d$ be its degree $d$ Jacobian.

Inside of $J^{g-1}$ there is the Theta divisor $\Theta$, which can be defined in various ways; the quickest definition is probably: it's the image of the Abel-Jacobi map $C^{(g-1)} \to J^{g-1}$ sending an effective degree $g-1$ divisor to the corresponding line bundle. Picking an isomorphism $J^{g-1} \cong J^d$, we also write $\Theta$ for the corresponding divisor in $J^d$.

How to compute $H^\ast(J;\Theta)$, or $h^\ast(J;\Theta)$? Or alternatively, what is known about these groups?

I suspect this is something embarrassingly standard and/or obvious and/or well-known and/or classical, but I haven't been able to figure anything out. The only thing along these lines that I was able to figure out was how to compute the Euler characteristic $\chi(J;\Theta^k)$ where $k$ is an integer: By Hirzebruch-Riemann-Roch and the Poincare formula it's $$\int_J \operatorname{ch}(\Theta^k) = \int_J e^{k\theta} = \int_J k^g \theta^g / g! = k^g.$$

Answer by Kevin Lin for construction of the Jacobian of a curve

2011-08-24T01:15:05Z

Let $N = L \otimes (q^\ast q_\ast (L\otimes L_\gamma^{-1}))^{-1}$. It suffices to show that the zero locus $D \subset C \times T$ of $s \in \Gamma(N)$ is flat over $T$. If $T$ is nice (Noetherian, blah, blah), it then suffices to show that the fiberwise degree of $D$ is constant. Note that the restriction of $N$ to $C \times \{ t \}$ is isomorphic to the restriction of $L$ to $C \times \{ t \}$. Since $L$ is fiberwise degree $r$ by assumption, it follows that $D$ is fiberwise degree $r$.

Oh and you need to check that the section $s$ is fiberwise nonzero.

Well, here's how you do that --- first look at the map $\phi : q^\ast q_\ast (L \otimes p^\ast L_\gamma^{-1}) \to L \otimes p^\ast L_\gamma^{-1}$. What does this map look like on fibers? Well, note that $H^0 (C \times \{ t \} , (L \otimes p^\ast L_\gamma^{-1})|_{C \times \{ t \}})$ is by assumption 1 dimensional. Hence the restriction of $q^\ast q_\ast (L \otimes p^\ast L_\gamma^{-1})$ to $C \times \{ t \}$ is a trivial line bundle. So, the map $\phi$ on the fiber $C \times \{ t \}$ looks like the map $\mathcal{O}_{C \times \{ t \}} \to (L \otimes p^\ast L_\gamma^{-1})|_{C \times \{ t \}}$ corresponding to the one nonzero global section in $H^0 (C \times \{ t \} , (L \otimes p^\ast L_\gamma^{-1})|_{C \times \{ t \}})$. Clearly this map is not zero.

I'll let you do the rest...

Why should algebraic objects have naturally associated topological spaces? (Formerly: What is a topological space?)

2010-02-08T11:42:00Z

In this question, Harry Gindi states:

The fact that a commutative ring has a natural topological space associated with it is a really interesting coincidence.

Moreover, in the answers, Pete L. Clark gives a list of other "really interesting coincidences" of algebraic objects having naturally associated topological spaces.

Is there a deeper explanation of the occurrence of these "really interesting coincidences"? It seems to suggest that the standard definition of "topological space" (collection of subsets, unions, intersections, blah blah), which somehow always seemed kind of a weird and artificial definition to me, has some kind of deeper significance or explanation, since it pops up everywhere...

The (former) title of this question is meant to be provocative ;-)

Symmetric powers of a curve = projective bundle over Jacobian, and the relative version thereof

2011-06-24T02:13:14Z

I am interested in this claim:

The $n$th symmetric power $C^{(n)}$ of a genus $g$ curve $C$ is isomorphic to the projectivization $\mathbb{P}(E_n)$ of the sheaf $E_n := \pi_\ast(P_n)$ over the Jacobian $J(C)$, where $P_n$ is a degree $n$ Poincare bundle over $C \times J(C)$ and $\pi$ is the projection $C \times J(C) \to J(C)$.

Moreover, under this isomorphism, the standard line bundle $\mathcal{O}(1)$ over $\mathbb{P}(E_n)$ corresponds to the line bundle $\mathcal{O}(D)$, where $D$ is the divisor corresponding to the image of the map $C^{(n-1)} \hookrightarrow C^{(n)}$ given by $p_1 + \cdots + p_{n-1} \mapsto p_1 + \cdots + p_{n-1} + p$, where $p$ is some fixed point.

(Also, the isomorphism $\phi : C^{(n)} \to \mathbb{P}(E_n)$ is compatible with the Abel-Jacobi map $u: C^{(n)} \to J(C)$, that is, $u = p \circ \phi$, where $p : \mathbb{P}(E_n) \to J(C)$.)

My questions:

This is claimed on page 309 of the book "Geometry of Algebraic Curves" by Arbarello-Cornalba-Grifiths-Harris, for $n \geq 2g-1$ (so that $E_n$ is a vector bundle, by Riemann-Roch; for smaller $n$ it isn't necessarily a vector bundle and they don't address this case). Their proof is pretty sketchy. It basically just says that, since the fibers of $\mathbb{P}(E_n) \to J(C)$ correspond to effective degree $n$ divisors, it follows that $C^{(n)} \cong \mathbb{P}(E_n)$. But this seems to me to only prove a set theoretic bijection between the two. So, how do I prove that I actually have an isomorphism of varieties? Or, is there a(nother) reference?
I believe the claim should still be true for $n < 2g-1$. Again, how do I prove this? Is there a reference? The sheaf $E_n$ will no longer be locally free, so $\mathbb{P}(E_n)$ will no longer be a bundle of projective spaces, but one should still be able to take the projectivization of a sheaf...
On page 7 of the paper http://arxiv.org/abs/0805.3621 by Moonen and Polishchuk, they talk about the families version of this statement. To be precise, they consider a family $\pi: C \to S$ of curves, and everything is done relative to the base $S$ (take relative symmetric product, take relative Jacobian, and so on). In this case, we must have a section $s: S \to C$ of $\pi$, corresponding to picking a point in each fiber, in order for the map $C^{(n-1)} \hookrightarrow C^{(n)}$ and the divisor $D$ to make sense. Anyway, Moonen and Polishchuk claim that in this families situation, $\mathbb{P}(E_n)$ is still isomorphic to the symmetric product $C^{(n)}$, and that under this isomorphism the line bundle $\mathcal{O}(1)$ corresponds to the line bundle $\mathcal{O}(D + n\psi)$, where $\psi$ is given by $\psi = \pi^\ast s^\ast K$, where $K$ is the relative canonical class of $\pi$. But how do I prove these statements?

Answer by Kevin Lin for What is the role of contact geometry in the hamiltonian mechanics?

2011-08-09T19:08:21Z

I think the basic example is when you have a symplectic manifold $M$ with a Hamiltonian $H : M \to \mathbb{R}$. Then take a regular value $a$ of $H$, and look at the hypersurface $N := H^{-1}(a)$, which will be a smooth submanifold of $M$ of odd dimension. Then (probably with some more hypotheses that I forget now), $N$ will have a contact structure, and the corresponding Reeb vector field will agree with the Hamiltonian vector field $X_H$ corresponding to $H$. Recall that the value of the Hamiltonian function $H$ is constant along the flows of $X_H$. In terms of physics this is interpreted as conservation of energy or something like that. So in this basic example, contact geometry can be thought of as the study of Hamiltonian mechanics for a fixed value of energy.

Reference for moduli stack of principal G-bundles?

2011-08-04T19:08:07Z

Hi,

I'm looking for a reference for the fact that the moduli stack $M_{GL_r,X}$ of $GL_r$-bundles over $X$ is an algebraic (Artin) stack. I'm only interested in the case where $X$ is a curve (for now).

I think this is supposed to be in Laumon-Moret--Bailly's "Champs Algebriques", but my French is not so great and I have been unable to find it in there. If it is actually in there, can you help a non-Francophone out?

Thanks!

Answer by Kevin Lin for Picard sheaves for elliptic curves

2011-08-01T19:44:17Z

Let $J^d$ be the degree $d$ Jacobian variety, parameterizing degree $d$ line bundles. Let $P^0$ be a degree $0$ Poincare bundle on $E \times J^0$.

Let $x$ be a point of $E$. The bundle $p^\ast (\mathcal{O}_E(nx)) \otimes P^0$ on $E \times J^0$ is the pullback of a degree $n$ Poincare bundle $P^n$ on $E \times J^n$ along the map $\operatorname{id}_E \times (- \otimes \mathcal{O}_E(nx)) : E \times J^0 \to E \times J^n$. This map is clearly an isomorphism.

By Proposition 2.1 on page 309 of Arbarello-Cornalba-Griffiths-Harris [for a more comprehensive treatment of this result, see Schwarzenberger's "Jacobians and Symmetric Products"], for $n \geq 1$, (the total space of) the projectivization of the sheaf $q_\ast(P^n)$ is isomorphic to the $n$th symmetric power of $E$, and under this isomorphism, the map from this projective bundle to $J^0 \cong E$ is the Abel-Jacobi map. Hence the same is true for the projectivization of $q_\ast(p^\ast (\mathcal{O}_E(nx)) \otimes P^0)$.

Answer by Kevin Lin for Pulling back a line bundle on the Jacobian to a spin bundle on the curve

2011-07-08T08:41:07Z

For a general curve $C$ of genus $g$, it is a fact that the Neron-Severi group of the Jacobian $J$ of $C$ is generated by the class $\theta$ corresponding to the divisor $\Theta$. (I am not very strong in algebraic geometry, so I guess that I would rather prefer to work with the probably equivalent statement: The subgroup of $H^2(J;\mathbb{Z})$ generated by first Chern classes of algebraic line bundles on $J$ is generated by $\theta$.) I don't know the proof of this, but the reference seems to be Arbarello-Cornalba-Griffiths-Harris, volume II...

So, by the formula you cite, it follows that for any algebraic line bundle $L$ on $J$, the degree of $\alpha_c^\ast L$ must be an integer multiple of $g-1+1=g$. Hence there can be no $L$ such that $\alpha_c^\ast L = \kappa$, since $\kappa$ is of degree $g-1$.

Err, hmm, well, actually, if $g=1$ then it's possible, since then $\kappa$ is of degree zero: for example, put $\kappa = \mathcal{O}_C$ and $L = \mathcal{O}_J$, and then we do have $\alpha_c^\ast L = \kappa$. But that's kind of trivial, anyways.

At least this all seems to work for a general curve $C$... I have no idea about a curve for which the above statement about the Neron-Severi group doesn't hold.

(As for $\kappa \oplus \kappa^{-1}$, this argument doesn't rule out the possibility of an $E$ such that $\alpha_c^\ast E = \kappa \oplus \kappa^{-1}$. But $\kappa \oplus \kappa^{-1}$ is degree zero, so such an $E$ would have to be degree zero...)

A very basic question about Abel-Jacobi map

2011-06-29T20:20:34Z

Let $C$ be a compact Riemann surface, let $C^2$ be the cartesian square of $C$, let $J(C)$ be the degree zero Jacobian of $C$, and let $\delta : C^2 \to J(C)$ be the map $(x,y) \mapsto [\mathcal{O}(x-y)]$.

In this paper http://arxiv.org/abs/math/9810054 of Hain and Reed, page 9, they say that it is an elementary exercise in algebraic topology to show that $\delta^\ast(\phi) = \Delta + (\psi_1 + \psi_2)/2$.

Explanation of notation:

$\phi$ denotes the symplectic form $\sum dx_i \wedge dy_i$ on $J(C)$, where $x_i, y_i$ are coordinates on the torus $J(C) = H^1(C;\mathcal{O}) / H^1(C;\mathbb{Z})$ corresponding to a symplectic basis of $H_1(C;\mathbb{Z}) \cong H^1(C;\mathbb{Z})$. This is also the class of the theta divisor $\Theta$.
$\psi_i$ is the first Chern class of the relative cotangent bundle of the $i$th projection $C^{2} \to C$.
$\Delta$ is the class of the diagonal $C \to C^{2}$, $x \mapsto (x,x)$.

My question is: How to do this "elementary exercise"? It ought to be easy but I'm just not seeing it...

Answer by Kevin Lin for Is there an algebraic construction of the Quillen (determinant) Line Bundle?

2011-06-19T19:17:06Z

I hope I'm not getting any details wrong, but I think the Narasimhan-Seshadri theorem asserts that the moduli space of $G=U(n)$ representations of $\pi_1$ is the same as the moduli space $M(n)$ of semistable rank $n$ vector bundles (i.e. principal $GL(n,\mathbb{C})$-bundles). (Probably this generalizes to any compact Lie group $G$?)

I believe $L$ is defined algebraically over $M(n)$ as $\operatorname{det} \mathbb{R}\pi_\ast U$, where $U$ is the universal bundle over $\Sigma \times M(n)$ and $\pi$ is the second projection.

"Approximating" $BGL(1)$ by projective spaces

2011-05-22T20:05:04Z

Given a representation $V$ of a group $G$, we can think of $V$ as a vector bundle over the classifying stack $BG$, and we can define its index $\chi(BG; V)$ to be the dimension of the $G$-invariant part $V^G$ of the representation.

Now let $G = GL(1)$ (or $\mathbb{G}_m$ or $\mathbb{C}^\ast$ if you like). Then let $\phi_n : \mathbb{P}^n \to BG$ be the map corresponding to the bundle $\mathcal{O}(1)$ (or maybe we should take map corresponding to the bundle $\mathcal{O}(-1)$, I always mix it up).

Let $V$ be a representation of $G$. Then we can compute the index (i.e. sheaf cohomology Euler characteristic) of $\phi_n^\ast V$ over $\mathbb{P}^n$. We can show, just by doing the calculation, that for sufficiently large $n$ (how large depends on the representation $V$), this index is a polynomial in $n$. Then, plugging in $n=-1$ into this polynomial, one can show that the result is equal to $\chi(BG; V)$, defined as above.

So, in this fashion, we can recover the index over the stack $BGL(1)$ if we know the index over each $\mathbb{P}^n$ (for sufficiently large $n$).

This result seems to make sense, because at least in topology we have $B\mathbb{C}^\ast \simeq \mathbb{CP}^\infty$, and at least intuitively we can think of $\mathbb{CP}^n$ as "approximating" $\mathbb{CP}^\infty$.

But so far I don't really have a satisfactory explanation for this, other than "it follows by doing the computation" and "it seems to make intuitive sense" as I've explained above. I wonder if there is a better way to see this; does it follow by some deeper facts?; does it follow by some more general theorems? I'm sorry that my question is not very precise.

I would also be interested in any other theorems or results which relate finite dimensional projective spaces with $BGL(1)$. (There are already a few such results in the link in S. Carnahan's comment below.)

Applications of Grothendieck-Riemann-Roch?

2010-10-27T07:25:14Z

I am currently trying to learn a bit about Grothendieck-Riemann-Roch...

To try to get a better feeling for it, I am looking for examples of nice applications of GRR applied to a proper morphism $X \to Y$ where $Y$ is not a point. I already I know of a fair number of nice applications of HRR, i.e. GRR when $Y$ is a point. I've read through some of the relevant sections of Fulton's Intersection Theory book, but I've only found applications of HRR there, though it's very possible that I overlooked something.

I am also interested in seeing worked-out, explicit, concrete examples, with explicit Chow/cohomology classes.

Thanks much!

S. Agnihotri, "Quantum cohomology and the Verlinde algebra"

2011-04-26T19:54:39Z

I am looking for the Oxford PhD thesis of S. Agnihotri, "Quantum cohomology and the Verlinde algebra". I can't seem to find it online. Does anyone know how / where I can find this? Thank you!

Comment by Kevin Lin

2013-05-02T08:51:43Z

@Barbara: The word you're probably looking for is "velocity", not "speed". From wikipedia: "velocity is the rate of change of the position of an object, equivalent to a specification of its speed and direction of motion." In English at least, I think "velocity" is meant to refer to the vector and "speed" is meant to refer to the scalar (the magnitude of the vector).

Comment by Kevin Lin

2012-02-10T17:21:24Z

@Gunnar Magnusson: Thank you very much. Page 36 of Zvonkine's notes should be pretty useful for me.

Comment by Kevin Lin

2012-01-09T18:43:30Z

mathoverflow.net/faq#othersedit

Comment by Kevin Lin

2012-01-06T19:31:43Z

@Sandor Kovacs: I have edited the question to clarify. That is not what I mean by index.

Comment by Kevin Lin

2012-01-06T19:31:06Z

@Laurent Moret-Bailly: I have edited the question, hopefully it is not confusing anymore.

Comment by Kevin Lin

2011-12-15T01:52:16Z

Papers are written without any (explicit) regard for readers other than the author himself/herself? :(

Comment by Kevin Lin

2011-10-20T03:27:45Z

Hi there; sorry that we are closing your question -- it is only because this site is not meant for questions like yours. MathOverflow is intended for research level math questions. You will find math.stackexchange.com to be a great place to ask your question, though.

Comment by Kevin Lin

2011-10-12T07:48:27Z

btw, how can I do commutative diagrams here on MO?

Comment by Kevin Lin

2011-10-12T04:00:55Z

"Shing Tung Yau" is the Cantonese pronunciation; "Cheng Tong Qiu" is the Mandarin pronunciation. "Zhang" is the Mandarin pronunciation; "Chang" is the Cantonese pronunciation.

Comment by Kevin Lin

2011-10-12T01:07:30Z

@Anatoly, thanks for the elaboration! That's interesting, and helpful.

Comment by Kevin Lin

2011-10-12T00:02:36Z

Well, I'm not looking at graded anything. That is, I'm only looking at $K^0$.

Comment by Kevin Lin

2011-10-08T01:15:43Z

How are the line bundles presented/given to you?

Comment by Kevin Lin

2011-10-05T19:12:46Z

@Chris: Thanks for the correction. I was thinking of moduli stacks of holomorphic G-bundles on a Riemann surface, rather than classifying stacks. But you are right.

Comment by Kevin Lin

2011-10-04T07:46:03Z

@Theo: In the context of algebraic geometry, "algebraic stack" means a stack satisfying certain conditions, probably the most relevant one being that there is a scheme (or an "algebraic space", something close to a scheme) that surjects onto the stack. This scheme is to be thought of as an "atlas" for the stack. While the example you give is not "algebraic" in the sense that it isn't defined in terms of algebraic geometry, it is still "algebraic" in the sense that it is covered by a "scheme" -- in this case a manifold, or at least some kind of infinite dimensional manifold.

Comment by Kevin Lin

2011-09-19T00:16:55Z

ulrich, thanks!