Constructing a vector bundle using Vector bundle construction lemma

Question

Given are:

• an open cover of $\{U_\alpha\}_{\alpha\in A}$ of a smooth manifold $M$.

• smooth maps $\tau_{\alpha\beta}\colon U_{\alpha}\cap U_{\beta}\rightarrow \text{GL}(k,\mathbb{R})$ with $\tau_{\alpha\beta}(p)\tau_{\gamma\alpha}(p)=\tau_{\gamma\beta}(p)$ for all $p\in U_{\alpha}\cap U_{\beta}\cap U_{\gamma}$

I have to construct from this a vector bundle $E$ over $M$. The hint they gave in the book is to consider $X=\cup_{\alpha\in A}U_{\alpha}\times\mathbb{R^k}$ and define here on an appropriate equivalence relation. I found a page here where someone else asked the same but I don't understand it yet:

" Let $X=\coprod_{\alpha\in A}(U_\alpha\times\mathbf R^k)$. You want to find a relation ${\sim}$ on $X$ such that $X/{\sim}$ and $E$ are in bijection. The obvious candidate for a bijection is the one defined on each $U_\alpha\times\mathbf R^k$ by $\phi:(p,v)\in U_\alpha\times\mathbf R^k \mapsto (p,v)\in E_p$. This $\phi$ is surjective, but not injective: if $p$ is in $U_\alpha\cap U_\beta$, then both $(p,v)\in U_\alpha\times\mathbf R^k$ and $(p,v)\in U_\beta\times\mathbf R^k$ are sent to $(p,v)\in E_p$. The obvious solution is to define the relation ${\sim}$ by: for $(p,v)\in U_\alpha\times\mathbf R^k$ and $(q,w)\in U_\beta\times\mathbf R^k$, say that $(p,v)\sim(q,w)$ when $p=q$ and $w=\tau_{\alpha\beta}(p)v$. Then $\phi$ gives a bijection $X/{\sim}\to E$, and you can use it to build the bijections $\Phi_\alpha$ in the construction lemma."

The confusing thing is that if we define the equivalence relation like that, we can not say whether $(p,v)\in U_{\alpha}\times\mathbb{R}^k$ and $(p,v)\in U_{\beta}\times\mathbb{R}^k$ are equivalent to each other, because then we would have $v=\tau_{\alpha\beta}(p)v$, but how can we know if this is true? Also can you give me hints of how to construct the $\Phi_{\alpha}$ in the lemma from this, because I don't see it.

Answer 1

The confusion is with the meaning of disjoint union. Write elements of $X$ as triples $(p,v,\alpha)$ for $p\in U_\alpha$. Denote elements of $X/\sim$ as $[p,v,\alpha]$. Then you have $[p,v,\alpha]=[p,\tau_{\alpha\beta} v,\beta]$.

Answer 2

Let $X:=\bigcup_{\alpha\in A}\{\alpha\}\times U_\alpha\times\mathbb{R}^k$. For $(\alpha, p, v), (\beta, q, w)\in X$, Let \begin{eqnarray} (\alpha, p, v)\sim(\beta, q, w)\stackrel{\text{def}}\Longleftrightarrow p=q, w=\tau_{\alpha\beta}(p)v. \end{eqnarray} This gives a equivalence relation on $X$. Let $E:=X/\sim$, $\pi([\alpha,p,v]):=p$. Then $\pi:E\to M$ is the vector bundle that we want to construct. $\Phi_\alpha([\alpha,p,v]):=(p,v)$ gives a local trivialization $\Phi_\alpha:\pi^{-1}(U_\alpha)\to U_\alpha\times\mathbb{R}^k$.

Answer 3

The idea in the previous reply was very helpful to me. Yet some more rigour about the well-definedness of the $\Phi_{\alpha}$ in the last reply could be necessary(and I needed):

Suppose $[\beta,q,w]=[\alpha,p,v]$. Then obviously $(p,v)\neq(p,w)$ so that the map does not seem well-defined. I will try to handle this by showing that each fiber $\pi^{-1}$ can be identified with $ \mathbf{R}^n$ or $\mathbf{C}^n$.

$\pi^{-1}(p)=\{[\alpha,p,v]:p\in U_{\alpha}, v\in \mathbf{R}^n\}$

Let f:$\pi^{-1}(p)\mapsto \mathbf{R}^n$ be defined as $f([\alpha,p,v])=v$

However, this way of definition is not well-defined as I said at the beginning. It is better to think $\mathbf{R}^n$ as the quotient of the set $\sqcup\mathbf{R}^n_{\alpha}$ (where $\alpha$ are chosen such that $p\in U_{\alpha}$)by the following relation:

$v\in\mathbf{R}^n_{\alpha} \backsim w\in\mathbf{R}^n_{\beta} \Leftrightarrow \tau_{\alpha\beta}(p)v=w$. Properties of transition maps imply that this quotient can be identified with one of the vector spaces $\mathbf{R}^n_{\alpha}$.

Now take f:$\pi^{-1}(p)\mapsto \sqcup\mathbf{R}^n_{\alpha}/\backsim$ and $f([\alpha,p,v])=[v]$ (here $[v]$ denotes the equivalence class of $v\in\mathbf{R}^n_{\alpha}$)so that each fiber can be given a vector space structure.