User cheyne - MathOverflow

Mayer-Vietoris on Fibered Products

2013-04-23T12:43:47Z

Suppose $M \xrightarrow{p} X$ is a surjective submersion of smooth manifolds. Define the fibered product in this case: $$M \times_X M = \{ (m_1, m_2) \in M \times M | p(m_1) = p(m_2) \}$$ and let $U = \coprod\limits_{\alpha}U_{\alpha}$ such that $\bigcup\limits_{\alpha}U_{\alpha} = M$ . Then we have an induced map $U \xrightarrow{\pi}X$ and the following diagram: $$U \times_X U \times_X U \xrightarrow{\pi_{12}, \pi_{23}, \pi_{13}} U \times_X U \xrightarrow{\pi_{1}, \pi_{2}} U \xrightarrow{\pi} X$$ (please excuse my notation on the multiple arrows, I don't know how to nicely tex that in here.) I am looking for a version of Mayer-Vietoris on differential forms induced by this diagram. Perhaps I can split it into two diagrams (the left three and the right three) and that would still be sufficient. The notes by Moerdijk I am reading through used Bott-Tu as a reference but I couldn't find anything useful in there. If I need to generate my own spectral sequence, that is fine, and I would relish learning opportunity but I wasn't sure if somehow Mayer-Vietoris on these fibered products was already well-known.

Edit: Specifically, given a closed 2-form $\kappa$ on $U \times_X U$, which satisfies $$\pi^*_{12}(\kappa) + \pi^*_{23}(\kappa) = \pi^*_{13}(\kappa)$$ on $U \times_X U \times_X U$, I want to show there exists (via "M-V", according to Moerdijk) a 2-form $\lambda$ on $U$ for which $\kappa = \pi^*_2(\lambda) - \pi^*_1 (\lambda)$. Then, since $d \kappa = 0$, I would have that $\pi^*_2(d \lambda) = \pi^*_1 (d \lambda)$ and so (again by "M-V", according to Moerdijk) I would have $d \lambda = \pi^*(\xi)$ for some closed 3-form $\xi$ on X.

If I was working with fibered products $U \times_M U$ I could think of this all in terms of the Cech-DeRham Complex and I would be comfortable using M-V. But as $U \times_X U$ does not seem quite like "the intersections of the open sets in the cover", I am not quite sure how to approach this.

The context in which I am working is the construction of a 3-curvature via an $S^1$-bundle gerbe, with connection.

The Abelian Group of Equivalence Classes of Gerbes

2013-03-14T03:09:53Z

Are there any good references/explanations for understanding the "contracted product" (as Brylinski calls it) of a gerbe with abelian band? I am finding it difficult, using Brylisnki's definition. to show that the map from the equivalence class of a gerbe to the degree 2 Cech cocycle is in fact a group homomorphism.

Thanks in advance.

Answer by Cheyne for Stiefel–Whitney classes in the spirit of Chern-Weil

2013-02-26T15:33:08Z

I want to say the short answer is no.

But in certain contexts, we can get things that are analogous. For example if you take a principal $B$-bundle $Q$ over $M$ and then suppose you can have a "nice" :) central extension of your lie group $B$ by $$1 \to \underline{\mathbb{C}^*}_M \to \tilde{B} \to B \to 1,$$where $\underline{\mathbb{C}^*}_M$ is the sheaf of smooth functions into $\mathbb{C}^*$, then you can define a cohomology class in $H^1(M, \underline{B})$ by seeing how well you can lift your bundle to a $\tilde{B}$-bundle. Now by the central extension, we would have $$H^1(M, \underline{B}) \overset{\sim}{=} H^2(M , \underline{\mathbb{C}^*}_M)$$ and then by the exponential sequence you would have $$H^1(M, \underline{B}) \overset{\sim}{=} H^2(M , \underline{\mathbb{C}^*}_M) \overset{\sim}{=} H^3(M, \mathbb{Z}).$$

And so what I am saying (since principal bundles have associated vector bundles) that your vector bundle in the right conditions could give you a degree three cohomology class in the integers. I'm sure you could pluck your Z/2 coefficients out of this (I don't know why you would want to be so rigid, nor am I claiming you are asking for that). Then there is an actual geometric interpretation of this integer related to a certain curvature form in this construction that I am not quite ready to add to this answer yet :)

Answer by Cheyne for Twisting an object P by an H-Torsor I

2013-02-23T22:58:16Z

I am going to put as much detail as possible in this answer without writing any diagrams (too many diagrams!).

In the Sketch of the Proof as outlined above, I was using local homeomorphisms instead of open subsets and open covers. I will switch to open (sets/covers) in this solution for simplicity and variety.

Let $P$ be an objecto of $C(U)$, where $U$ is an open subset of $X$, let $I$ be an $H$-torsor over $U$, and let $(U_{\alpha})$ be an open cover of $U$ (refined enough such that whenever there is a local property being used we need not refine any more). Then we can, by assumption of the cover, pick a section $s_{\alpha} \in I_{\alpha}$ for each $\alpha$. On $U_{\alpha \beta}$, define $\mu_{\alpha \beta} \in H(U_{\alpha \beta})$ by $\mu_{\alpha \beta}s_{\beta} = s_{\alpha}$, using the torsor there. Then since $C$ is a gerbe we have a unique way ( unique by commutativity of the band) to locally identify our torsor element with an automorphism $\mu_{\alpha \beta} : P_{\alpha \beta} \to P_{\alpha \beta} $ satisfying the cocycle condition (up to natural transformations on double and triple intersections; I will avoid these natural transformations in this answer at all costs. Having diagrams is the only way for me to feel comfortable explaining them).

By the descent property that $C$ satisfies, since we have objects $P_{\alpha}$ and isomorphisms $\mu_{\alpha \beta}: P_{\beta}|_{U_{\alpha \beta}} \to P_{\alpha}|_{U_{\alpha \beta}}$ (so here I combine the natural transformations between restricting once to the double intersection and restricting two times to the double intersection, along with my automorphism $\mu$ on the double intersection to get an isomorphism between my two $P$'s. ), then I have an object $Q$, with isomoprhisms $\psi_{\alpha}: P_{\alpha} \to Q_{\alpha}$ such that (up to natural transformations on $U_{\alpha \beta}$) $$\psi_{\alpha} = \mu_{\alpha \beta} \psi_{\beta}$$ Claim: We have an isomorphism of $H$-torsors $w: I \to \underline{\text{Isom}}(P,Q)$. Let $ \sigma \in I(V)$ and define $V_{\alpha} := V \cap U_{\alpha}$. For each $V_{\alpha}$, define $h_{\alpha} \in H(V_{\alpha})$ by $h_{\alpha}s_{\alpha} = \sigma_{\alpha}$. Then we can think of $h_{\alpha}$ as an automorphism of $P_{\alpha}$ (now thinking of $P_{\alpha}:= P_{V_{\alpha}}$) so we define our corresponding isomorphism $w(\sigma): P_V \to Q_V$ by $$w(\sigma)_{\alpha}:= h_{\alpha}\psi_{\alpha}$$.

It is straightforward to check that this map is well-defined, a morphism of torsors, and thus an isomorphism.

Note: the bit about the band being abelian helps with a different choice of $Q$ resulting in a uniquely isomorphic situation; thus allowing a bijective correspondence with the isomorphism classes of objects in $C(U)$ and $H^1(U, H)$!

Twisting an object P by an H-Torsor I

2013-02-21T16:59:50Z

I am reading Brylinski's Loop Spaces, Characteristic Classes, and Geometric Quantization.

The Statement

Let $C$ be a gerbe on a space $X$ with "abelian" band $H$ , $f: Y \to X$ a local homeomorphism (this context uses local homeomorphisms in place of open subset sets and covers of open sets). Given an object $P$ of $C(Y)$ , and an $H$ torsor $I$ over $Y$ , there exists an object $Q$ of $C(Y)$ such that $I \overset{\sim}{\to} \underline{\text{Isom}}(P,Q)$ as $H$ -torsors. The object $Q$ is called "the object obtained from $P$ by twisting by the H-torsor $I$".

My Question

I can get a local ismorphism $I \overset{\sim}{\to} \underline{\text{Isom}}(P,Q)$ but how can I see it is, in fact, global?

Some Details / A sketch of the proof

So you take a surjective local homeomorphism $g: Z \to Y$ so that there is a global section of $g^{-1}I$ (i.e. take a fine-enough open cover so that you have sections of $I$ ). Then pulling back this section $s$ under the maps $p_1, p_2: Z \times_Y Z \to Z$ we get two different sections of a sheaf which is a torsor under the pullback of $H$ . So you get this section of the pullback of $H$ and since the pullback of $P$ is also a torsor under this sheaf, you have an automorphism of the pullback of $P$. This automorphism could also be thought of as an isomorphism $u: p_1^{-1}g^{-1}P \overset{\sim}{\to} p_2^{-1}g^{-1}P$ which satisfies the cocycle condition. Then by the descent property which this gerbe satisfies, we have an object $Q$ of $C(Y)$ with $\psi: g^{-1}Q \overset{\sim}{\to} g^{-1}P$ .
So we started with a section of the pullback of $I$ (i.e a local section of $I$) and ended with an isomorphism between $P$ and $Q$'s pullbacks (i.e. a local isomorphism).

What information can I use to show that this local isomorphism is in fact an isomorphism? Or is Brylinski simply abusing language?

EDIT (2/22/13): I'm fairly certain I just need to show that on double intersections these local isomorphisms differ by an inner automorphism, but since my band is abelian this would say they glue together. I just haven't wrapped my head around this idea in the context of local homeomorphisms ans pullback diagrams quite yet.

Connection Transformation Formula; Degree 3 Cech Cohomology

2012-11-20T15:44:58Z

While reading through Brylinski, as in all of my posts, I am trying to understand the following equation:

$ g_* \tilde{\theta} = \tilde{\theta} - g^{-1} dg$

Setting

I have a principal $B$ -bundle, $Q$ , over my space M, with connection $\theta$ (values in the Lie Algebra $\mathfrak{b}$ of the lie group $B$ ), and a central group extension by $\mathbb{C}^*$ ,

$$ 0 \to \mathbb{C}^* \to \tilde{B} \to B \to 0$$ where the map $p:\tilde{B} \to B$ is a principal $\mathbb{C}^*$ fibration, yielding an exact sequence of sheaves of groups:

$$ 0 \to \underline{\mathbb{C}}_M^* \to \underline{\tilde{B}}_M \to \underline{B}_M \to 0$$ where $M$ is a paracompact space.

Long story short, I restrict my attention of the space to an open set $ U \subset M$ , small enough so that I can "lift" the bundle to a $\tilde{B}$ -bundle, $\tilde{Q}$ , and the connection to a $\tilde{\mathfrak{b}}$ -valued connection $\tilde{\theta}$ on the bundle $\tilde{Q}$. The lifted bundle must satisy the condition that we have a bundle isomorphism$f: \tilde{Q}/\mathbb{C}^* \tilde{\to} Q_{U} $. The lifted connection must satisfy$$q \circ \tilde{\theta} = f^*\theta$$as 1-forms with values in $\mathfrak{b} = \tilde{\mathfrak{b}}/\mathbb{C}$; where $q$ is simply the map which quotients out by $\mathbb{C}$.

Finally, $g$ is simply a $\mathbb{C}^*$ -valued function, which can be thought of as a bundle isomorphism given our local picture is a trivialization. By $g_*$ we simply mean the pullback of $g^{-1}$ .

Warning: Matrices Beware!

I know that this formula is true for vector bundles, or when the group automorphisms are represented by invertible matrices; I have read that literature already. Unless you are prepared to help me understand why this situation is most certainly in the land of finite vector bundles or matrix groups, such an answer would be redundant. The formula also makes sense up to the fact that the transformed connection and original connection must differ by a complex-valued form, which is exactly what this formula prescribes.

How to Draw Complex Line Bundles

2012-06-24T16:40:28Z

I am giving a presentation soon on the Classification of Complex Line Bundles and I would like to have some very "basic" visualizations to use as examples.

Background and Context

I am considering the Cech-cohomology of a principal $ \mathbb{C}^{*} $ bundle, where my sheaf $\underline{\mathbb{C}}_M^{*}$ is the sheaf of smooth $\mathbb{C}^{*}$ valued functions on the manifold $M$. Using the exponential sequence of sheaves $$ 0 \to \mathbb{Z}(1) \to \underline{\mathbb{C}}_M \to \underline{\mathbb{C}}_M^{*} \to 0$$ we get an isomorphism (via properties of cohomology and the connecting homomorphism) $$H^1(M, \underline{\mathbb{C}}_M^{*}) \cong H^2(M, \mathbb{Z}(1)) $$

It turns out that $H^1(M, \underline{\mathbb{C}}_M^{*}) $ is also isomorphic to the group of isomorphism classes of principal- $\mathbb{C}^{*}$ bundles over $M$. Since the principal- $\mathbb{C}^{*}$ bundles are in one-to-one correspondence with the complex line bundles, it should be evident how this all relates to my title.

My Questions

(1) Given the above information, and some knowledge of cohomology, there should be only a trivial principal- $\mathbb{C}^{*}$ bundle on the circle $S^1$. How can we see this visually?

*See my example/analogue below.

(2) Similarly, how can we visualize a non-trivial principal- $\mathbb{C}^{*}$ bundle on the standard 2-dimensional torus?

*Example/Analogue:

So consider a circle bundle on $S^1$, then we can consider a section of the bundle like so:

Now, given two sections on adjacent trivializations,

We can imagine deforming one section into another, to get our transition functions. Now, I can also believe that any such family of sections can be deformed into a global section, so again I want to know why this necessarily doesn't work on the Hopf bundle via pictures.

Weil Kostant Integrality Result as Stated by Brylisnki

2012-04-20T16:46:00Z

I'm reading through Brylinski's "Loop Spaces, Characteristic Classes and Geometric Quantization" and I am stuck on a piece of Theorem 2.2.15, which asserts that If $K$ is a closed complex-valued 2-form on a paracompact manifold $M$ whose cohomology class is in the image of $H^2(M, \mathbb{Z}(1)) \rightarrow H^2(M, \mathbb{C})$ then there exists a line bundle $L$ with connection on $M$, whose curvature is equal to $K$.

The proof goes like so:

Let $K^{\bullet} = \underline{\mathbb{C}}^* \ \underrightarrow{d\ log} \ \underline{A}^1_{M, \mathbb{C}}$ and so we have an exact sequence of complexes of sheaves: $$0 \rightarrow {\mathbb{C}}^* \rightarrow K^{\bullet} \rightarrow \underline{A^2}{M, cl}[-1] \rightarrow 0 $$

Where that nastily noted $\underline{A^2}{M, cl}[-1]$ means the two term complex with 0 in the first slot and closed 2 forms on $M$ in the second slot. The fact that this sequence is exact is beautifully explained in my other question: here.

Ok well given the above exact sequence, we would thus be able to get the exact sequence: $$0 \rightarrow H^1(M,{\mathbb{C}}^*) \rightarrow H^1(M,K^{\bullet}) \rightarrow \underline{A^2}{M, cl} \rightarrow H^2(M, \mathbb{C}^ * ) $$

The above sequence makes sense to me. HOWEVER The claim is now that since our closed 2-form K is in the image of

$H^2(M, \mathbb{Z}(1)) \rightarrow H^2(M, \mathbb{C})$

Then the cohomology class $[K] \in H^2(M, \mathbb{C})$ has zero image in $H^2(M, \mathbb{C}^*)$ and so by the above sequence it must come from $H^1(M, K^{\bullet})$, i.e. it comes from being the curvature of a line bundle with connection.

I hope it is understandable why this is not obvious to me. There seems to be many different identifications going on with all of the possible places you can get a map $H^2(M, \mathbb{Z}(1)) \rightarrow H^2(M, \mathbb{C}) $, *not all of which are clearly linked with a map * $H^2(M, \mathbb{C}) \rightarrow H^2(M, \mathbb{C}^*)$

I think that this mysterious map $H^2(M, \mathbb{Z}(1)) \rightarrow H^2(M, \mathbb{C}) $ comes from first thinking of the $\mathbb{Z}(1)$-valued Cech cocyle living in $\mathbb{R}(2)$ Deligne Cohomology, obtaining a 2-form and then using the De-Rham Cech isomorphism! phew! That still doesn't explain the mapping properties and identifications made above. I will update this last bit and clarify as soon as possible.

Answer by Cheyne for Blackbox Theorems

2012-06-26T17:15:42Z

Cech and Sheaf (derived-functor) cohomologies are isomorphic on a paracompact space $X$ with the sheaf being, for example, $\underline{\mathbb{C}}^*_M$ , the sheaf of $\mathbb{C}^*$ -valued functions on $X$.

The proof uses partititions of unity along with hypercohomology and results from spectral sequences. If this ISN'T a black box theorem, I'd love a concise explanation :)

Descent of Morphisms of Sheaves

2012-05-09T21:12:12Z

While reading Brylinski I am trying to understand the descent of morphisms of sheaves.

In trying to form a new definition of a presheaf $A$ over a space $X$, we associate to each surjective local homeomorphism $f:Y \to X$ a set, denoted $A(Y\xrightarrow{f}X)$. The "restriction" condition of a presheaf amounts to: given a surjective local homeomorphism $g:Z \to Y$ we have a pullback map $g^{-1}:A(Y\xrightarrow{f}X) \to A(Z \xrightarrow{fg}X)$. The transitivity property for these "restriction" (pullback) maps is that given any diagram $$W \xrightarrow{h} Z \xrightarrow{g} Y \xrightarrow{f} X$$ having $(gh)^{-1} = h^{-1} \circ g^{-1}$ as pullbacks $A(Y\xrightarrow{f}X) \to A(W \xrightarrow{fgh} Z)$. $\ \$

If $A$ is already a presheaf, in the good 'ol fashioned sense, then we can define our assignment $A(Y\xrightarrow{f}X)$ to be the global sections of $Y$ given by the inverse image of $A$ on $X$, i.e. $\Gamma(Y, f^{-1}A)$

I have 2 questions:

Is it true that if $A$ is already a sheaf in the good 'ol fashioned sense, then the above property (transitivity of the "restriction") is satisfied? My proof feels trivial, hence my worry. Also, I am uneasy since Brylisnki doesn't state this fact but instead says it "should" be true.
He later comments that as functors from the category of sheaves on $Y$ to the category of sheaves on $W$ , $h^{-1}\circ g^{-1}$ and $(gh)^{-1}$ are NOT equal; but there is a natural transformation. Why are these two functors not equal? It seems like they send the same sheaves to the same places, unless of course I am making identifications of categories that I don't realize?

Is this Sequences of Complexes of Sheaves Exact?

2012-04-27T12:21:50Z

So in another question of mine there is a sequence of complexes of sheaves which the author asserts is exact.

Where that nastily noted $\underline{A^2}{M, cl}[-1]$ means the two term complex with 0 in the first slot and closed 2 forms on $M$ in the second slot.

The fact that this sequence is exact in itself seems to rely on the fact that the sheafification of the image of the contant $\mathbb{C}^*$ sheaf is isomorphic to the sheaf of smooth functions $\underline{\mathbb{C}^ * }$ right? Well that part is bothersome to me

Comment by Cheyne

2013-04-26T15:21:49Z

I figured it out, I just need a few days to typing it up. I don't need a spectral sequence :)

Comment by Cheyne

2013-04-23T15:38:49Z

...which I don't believe is in Bott-Tu :)

Comment by Cheyne

2013-04-23T13:35:38Z

I've added some details to address your comment, thanks!

Comment by Cheyne

2013-03-14T15:04:26Z

I believe I can show the map is a bijection; maybe I am missing this fact that the bijection is good enough? It's that I want to have this theorem about the cohomology classifying these equivalence classes of gerbes (I believe due to Giraud) in terms of intersections of open sets. Brylinski's definition of the product uses local homeomorphisms, which I've dealt with before with no problems but this time translating the definition into intersections and open covers is stumping me. I'll check out that paper thanks for the suggestions as always!

Comment by Cheyne

2013-02-26T15:42:34Z

By the way, I'm fairly certain that Weil developed some of these methods.

Comment by Cheyne

2012-12-07T18:03:59Z

By $f$, you mean $b$, right?

Comment by Cheyne

2012-12-07T01:03:46Z

Also, I don't understand your "definition" of $f^*\omega$ = f^{-1} \omega (df)$, when $f$ is $G$ -valued and $\omega$ is $\mathfrak{g}$ - valued; unless you implicitly mean an adjoint map induced by $f$ on the Lie-Algebras?

Comment by Cheyne

2012-12-07T00:46:44Z

Thank you very much for this, Johannes. I am working through the details at the moment. Firstly, I think the main thing I was missing was that any connection can be decomposed where one piece is the Maurer Cartan form; this is obvious to me now (thanks!). Secondly, I am working through your equations after "Next you have to invoke the fact" and I am just making sure that when you talk about "linear groups" you are allowing for infinite dimensions?

Comment by Cheyne

2012-12-06T22:55:40Z

Thank you, David. I apologize if I wasn't clearer, the first two lines of my post (namely the grey box) were "my question": namely, I can't verify the equation in the grey box in the beginning (Page 174, eq(4-22) in Brylisnki). I understand a bit about the Maurer-Cartan form, but not that I get $g^{-1}dg$ when my lie group is not a matrix goup. Any help or answers are greatly appreciated.

Comment by Cheyne

2012-11-27T15:05:28Z

Could someone please tell me what I can do to make this post more answerable?

Comment by Cheyne

2012-11-20T15:03:32Z

I'm assuming we mean Real line bundles? There are non-trivial complex line bundles on S^2.

Comment by Cheyne

2012-07-11T20:44:01Z

David: I finally had time to follow what you said step by step. I found that the transition function between my two natural sections was "1/z" and I am just working through the details of formally showing that this is not equivalent to the identity in this context. Thanks a lot!!

Comment by Cheyne

2012-06-27T22:40:40Z

I'm currently reading Brylisnki's Loop Spaces, Characteristic Classes, and Geometric Quantization which might be helpful.

Comment by Cheyne

2012-06-27T16:58:56Z

Thank you, David. I am trying to visualize your last paragraph, but I think I get what you're saying. I'm going to try and better understand your setup, then really believe the last paragraph before I come back to considering this problem answered. Thanks again!

Comment by Cheyne

2012-06-27T01:28:29Z

Hello Ryan, I appreciate your input and your breadth of knowledge of bundles. However, I don't have any issues with proving any of the statements in my question. I literally want pictures to go along with the statements. Now that I provided pictures for the circle bundle over $S^1$, I think (1) makes a bit more sense. I still can't draw appropriate pictures for (2) and use the pictures to show that I can't "line up the endpoints" (please don't take that literally) in the case of the sections of the sphere into the Hopf Bundle.