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I have read that transition maps $g_{\alpha\beta}:U_\alpha\cap U_\beta\to GL(n)$ of a vector bundle of rank $n$ are related to the Čech cohomology group $H^1\left(M,GL(n,\mathcal{C}^\infty_M)\right)$ (for the sheaf of functions with values in $GL(n)$). Could somebody provide a precise statement with an explanation or a reference?

Thank you!

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3 Answers 3

up vote 5 down vote accepted

The result you're thinking of is the following:

Proposition. Let $M$ be a smooth manifold, and let $\mathscr{C}^\infty_M$ be the sheaf of smooth functions on $M$.

  • For any positive integer $n$, there is a natural bijection between isomorphism classes of $n$-dimensional smooth vector bundles on $M$ and principal $\mathrm{GL}(n, \mathscr{C}^\infty_M)$-bundles on $M$, or equivalently, smooth principal $\mathrm{GL}(n, \mathbb{R})$-bundles on $M$.

  • For any sheaf of groups $\mathscr{G}$, there is a natural bijection between isomorphism classes of principal $\mathscr{G}$-bundles and elements of the Čech cohomology group $H^1 (M, \mathscr{G})$.

Putting the two bijections together gives us what we want.


The connection between the three ideas is more-or-less straightforward, even if the verification of the details is tedious. First of all, suppose $E$ is a $n$-dimensional smooth vector bundle on $M$. Then, there is an open cover $\mathfrak{U}$ of $M$ with respect to which $E$ is trivial, and – fixing a local trivialisation – the transition maps give us a Čech 1-cocycle for the sheaf $\mathrm{GL}(n, \mathscr{C}^\infty_M)$ with respect to the cover $\mathfrak{U}$. A Čech 1-coboundary for a sheaf is the same thing as a global section, and it is not hard to check that global sections of $\mathscr{G} = \mathrm{GL}(n, \mathscr{C}^\infty_M)$ act on the set of local trivialisations of $E$. Moreover, any two trivialisations of $E$ over $\mathfrak{U}$ are related by such a global section. Thus, we obtain a "characteristic" element of $H^1(\mathfrak{U}, \mathscr{G})$ for each trivialising cover $\mathfrak{U}$ – but we're not done yet.

Suppose we have another trivialising cover $\mathfrak{V}$; then we get an element of $H^1(\mathfrak{V}, \mathscr{G})$. But $H^1(\mathfrak{U}, \mathscr{G})$ and $H^1(\mathfrak{V}, \mathscr{G})$ are different groups, so how do we compare the characteristic elements we get? Well, first of all, we can take a common refinement $\mathfrak{W}$ of $\mathfrak{U}$ and $\mathfrak{V}$, and a trivialisation over $\mathfrak{U}$ or $\mathfrak{V}$ induces a trivialisation over $\mathfrak{W}$ by restriction. Similarly, cocycles and coboundaries can be refined, so we get homomorphisms $$H^1(\mathfrak{U}, \mathscr{G}) \rightarrow H^1(\mathfrak{W}, \mathscr{G}) \leftarrow H^1(\mathfrak{V}, \mathscr{G})$$ and it turns out that the two characteristic elements we got become equal in $H^1(\mathfrak{W}, \mathscr{G})$ under these homomorphisms. Thus we can pass all the way to $H^1(M, \mathscr{G})$ – and the characteristic element we get there only depends on the isomorphism class of $E$.

In the reverse direction, we must find a vector bundle whose characteristic element is some chosen element of $H^1(M, \mathscr{G})$. Since $H^1(M, \mathscr{G})$ is defined as the filtered colimit of $H^1(\mathfrak{U}, \mathscr{G})$ over all the open covers $\mathfrak{U}$, we can pick an open cover $\mathfrak{U}$ and a Čech 1-cocycle w.r.t. $\mathfrak{U}$ that represents the chosen cohomology class. A standard argument then shows that we can construct a vector bundle $E$ which is trivialised over $\mathfrak{U}$ with transition maps given by the chosen cocycle.

Almost exactly the same argument shows that there is a bijection between elements of $H^1(M, \mathscr{G})$ and isomorphism classes of principal $\mathscr{G}$-bundles.

I'm afraid I don't actually know of a reference for this situation exactly. Milne talks about the second bijection in §11 of his Lectures on étale cohomology, and Johnstone more-or-less defines $H^1(M, \mathscr{G})$ using the bijection in §8.3 of Topos theory.

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Thank you very much! –  Benjamin Sep 18 '12 at 6:52

Well, consider a Čech Complex on M, given by the open covering $\{U_\alpha\}$, such that on $U_\alpha$, M is trivial (i.e there exists a chart there making the vector bundle trivial). Let us take on each $U_\alpha$ the local trivializations there, and note that these are elements of $GL(n,C U^\infty_I)$ ( I by this mean all continuous $U \rightarrow GL(n) $). We have a Čech chain complex: $$\amalg_\alpha C(U_\alpha) \rightarrow \amalg_{\alpha, \beta} C(U_\alpha \cap U_\beta) \rightarrow \cdots,$$ where $C(U_I)$ denotes $GL(n,C U^\infty_I)$. You know that a Čech 1-cochain is an element f of $\amalg_{\alpha,\beta} C(U_\alpha \cap U_{\beta})$ such that for all $\alpha, \beta, \gamma$, $f_{\alpha,\beta}-f_{\alpha,\gamma}+f_{\beta,\gamma} = 0 $, which should amount to the relation that the transition functions satisfies the ''cocycle condition''.

I realize that this explanation is a bit sketchy, but I hope that I could get the main idea across.

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Thank you for your answer! –  Benjamin Sep 18 '12 at 6:53

As has already been said, there is a bijection betweeen $H^1(M, GL(n, \mathcal{C}^{\infty}_M))$ and the set of all isomorphism classes of smooth rank $n$ vector bundles on $M$. Lee's Introduction to Smooth Manifolds is a good book in general, but for your purposes it has some very helpful exercises. If you really want to understand the bijection, I suggest you go through the following exercises:

  • Exercise 5.3: This shows that the transition functions for a smooth rank $n$ vector bundle on $M$ define a Čech $1$-cocycle with values in $GL(n, \mathcal{C}^{\infty}_M)$.
  • Exercise 5.4: This shows that any such cocycle defines a smooth rank $n$ vector bundle on $M$.
  • Exercise 5.5: This shows the relationship between two Čech $1$-cocycles which define isomorphic smooth rank $n$ vector bundles on $M$.

One would hope that the relationship obtained in Exercise 5.5 is 'cohomologous'. However, we have to be careful because for $n > 1$, $GL(n, \mathcal{C}^{\infty}_M)$ is a sheaf of non-abelian groups; in particular, what does it mean to be cohomologous in this setting? Fortunately, Brylinski's Loop Spaces, Characteristic Classes and Geometric Quantization defines what it means (see Chapter 4, Section 4.1, Equation 4-2); note, this is just for the case of 1-cocycles, but that is all you need. Thankfully, the result you get from Exercise 5.5 tells you that two cocycles defining isomorphic smooth rank $n$ vector bundles are cohomologous. In summary:

For a given cohomology class in $H^1(M, GL(n, \mathcal{C}^{\infty}_M))$, choose a cocycle representative $(\tau_{ij})$ and construct (via Exercise 5.4) the smooth rank $n$ vector bundle $E$ on $M$. Then the map $[(\tau_{ij})] \mapsto [E]$, where square brackets indicate the corresponding cohomology and isomorphism classes respectively, is well-defined and is in fact a bijection between $H^1(M, GL(n, \mathcal{C}^{\infty}_M))$ and the set of all isomorphism classes of smooth rank $n$ vector bundles on $M$.


You can still appreciate what is happening if you don't do the exercises, but I recommend you get a copy of Lee's book and at least look at them.

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Thank you very much for these references. –  Benjamin Sep 18 '12 at 6:54

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