vector bundles and cocycles

Question

I need a detailed solution to a self-study book's exercise (Bott & Tu's Differential forms in algebraic topology, exercise 6.2).

Show that two vector bundles on $M$ are isomorphic iff their cocycles relative to some open cover are equivalent.

I can show it in one direction: isomorphism implies cocycles equivalent, but not the other. Please give detail answer; I am totally burnt out by and do not need "hints" :)

Answer 1

The proof below is a rewriting of the proof of Theorem 2.7 in Fibre Bundles by Dale Husemoller.

Let $M$ be a manifold, $\{U_\alpha\}$ an open cover of $M$, and $\pi:E\to M$ and $\pi':E'\to M$ two vector bundles of rank $n$ over $M$, with trivializations $\{\phi_\alpha\}$ and $\{\phi_\alpha'\}$, whose transition functions are denoted by $g_{\alpha\beta}$ and $g_{\alpha\beta}'$, respectively. By assumption, $g_{\alpha\beta} (x) = \lambda_\alpha g_{\alpha\beta}' \lambda_\beta^{-1} (x)$ for all $x\in U_\alpha \cap U_\beta$; here $\lambda_\alpha:U_\alpha\to GL(n,\mathbb{R})$ for all $\alpha$. We wish to construct a bundle isomorphism $f:E\to E'$.

For each $\alpha$, we define $f_\alpha:U_\alpha \times \mathbb{R}^n \to U_\alpha \times \mathbb{R}^n$ by $f_\alpha(x,y) = (x,\lambda_\alpha(x)^{-1}y)$, and we define $f:E\to E'$ by requiring that $f=\phi_\alpha' f_\alpha \phi_\alpha^{-1}$, or $\phi_\alpha' f_\alpha = f \phi_\alpha$ on $U_\alpha \times \mathbb{R}^n$. By construction, $f$ is linear on corresponding fibers.

So far $f$ is only defined locally; to prove that $f$ is a globally defined bundle map, let $(x,y)\in (U_\alpha \cap U_\beta) \times \mathbb{R}^n$, and we will check two definitions of $f$ (using the indices $\alpha$ and $\beta$, respectively) agree. That is, we want to show $\phi_\beta' f_\beta(x,y) = f \phi_\beta(x,y)$ implies (hence equivalent to) $\phi_\alpha' f_\alpha(x,y) = f \phi_\alpha(x,y)$. For this we make the following calculation: $$ \phi_\beta' f_\beta(x,y) = \phi_\beta'(x,\lambda_\beta(x)^{-1} y) =\phi_\alpha'(x,g_{\alpha\beta}'(x) \lambda_\beta(x)^{-1} y) =\phi_\alpha'(x,\lambda_\alpha(x)^{-1} g_{\alpha\beta}(x) y) =\phi_\alpha' f_\alpha(x, g_{\alpha\beta}(x) y) . $$ Using $f\phi_\beta(x,y) = f\phi_\alpha(x,g_{\alpha\beta}(x) y)$, we have $f\phi_\alpha(x,g_{\alpha\beta}(x) y) = \phi_\alpha' f_\alpha(x,g_{\alpha\beta}(x) y)$, or $f \phi_\alpha = \phi_\alpha' f_\alpha$. Therefore, $f$ is a well-defined global bundle map.

It is clear that $f$ is locally invertible (all $\phi_\alpha$, $\phi_\alpha'$ and $\lambda_\alpha$ are), and its local inverses paste together to give a bundle map by the same consideration. Hence $f$ is a bundle isomorphism.

Answer 2

I'm quite late, but here is a slightly different point of view: Note my convention will be that $f_{ij}$ maps from the $i$ chart to the $j$ chart, i.e., $f_{jk}f_{ij} = f_{ik}$ and then also $f_{ij} = \lambda_j^{-1} f_{ij}'\lambda_i$. Given the transition functions $f_{ij}$, we know $E \cong (\bigsqcup U_i \times \mathbb{R}^n)/\sim$ where $(u \in U_i,v) \sim (u \in U_j,f_{ij}(u)v)$. Similarly, $E' \cong (\bigsqcup U_i \times \mathbb{R}^n)/\sim'$. I like quotient spaces, so in situations like this, it may be easier to just forget $\pi$ and $E$ and instead use the viewpoint that a bundle is just a disjoint union of trivializations with some gluing (of course, one should prove this as well).

In the forwards direction when $E \cong E'$, one finds $\lambda(u)$ is just the function $(u,v) \mapsto (u,\lambda(u)v)$. Therefore, it is very natural to define $$f\colon \bigsqcup U_i \times \mathbb{R}^n \to \bigsqcup U_i \times \mathbb{R}^n$$ $$(u_i,v) \mapsto (u_i,\lambda_i(u_i)v).$$ For free, we get $\bigsqcup U_i \times \mathbb{R}^n \to (\bigsqcup U_i \times \mathbb{R}^n)/\sim'$ given by projection. Thus, we have $f \colon \bigsqcup U_i \times \mathbb{R}^n \to (\bigsqcup U_i \times \mathbb{R}^n)/\sim'$ which we want to show respects $\sim$. For such a pair $(u \in U_i,v) \sim (u \in U_j,f_{ij}(u)v)$, by definition $f$ of the LHS is $(u,\lambda_i(u) v) \sim' (u,f_{ij}'\lambda_i(u) v) $ and $f$ of the RHS is $(u,\lambda_j f_{ij}(u)v)$. By the assumption on the $\lambda's$, these images are indeed equal. Thus, $f$ is well defined (and clearly surjective). By doing everything in the other direction, clearly we can construct an inverse of $f$, showing the isomorphism.