2
$\begingroup$

Given a real vector bundle $E\to M$ with fiber $V$, there is an open cover $\{U_\alpha\}$ of $M$ and a collection of smooth transition maps $g_{\alpha\beta}: U_\alpha\cap U_\beta\to GL(V)$. Suppose $\rho: GL(V)\to GL(W)$ is a group homomorphism where $W$ is a vector space. How do I show that the collection of functions $\widetilde g_{\alpha\beta}=\rho\circ g_{\alpha\beta}$ is a collection of transition functions for a vector bundle $E'\to M$ with fiber $W$?

I know that if the maps $\widetilde g_{\alpha\beta}: U_\alpha\cap U_\beta\to GL(W)$ satisfy $\widetilde g_{\alpha\alpha}=id, \widetilde g_{\alpha\beta}\widetilde g_{\beta\gamma}\widetilde g_{\gamma\alpha}=id$, then from this one can construct a vector bundle with fiber $W$. But I don't see why these $\widetilde g_{\alpha\beta}$ satisfy the above conditions.

$\endgroup$
6
  • $\begingroup$ Did you try making use of the fact that $\rho$ is a group homomorphism? $\endgroup$
    – Notone
    Commented Apr 17, 2018 at 20:47
  • $\begingroup$ @Notone No -- have no idea how to use it (at least when trying to verify the cocycle conditions, I didn't see how to use it ..) $\endgroup$
    – user557
    Commented Apr 17, 2018 at 20:54
  • $\begingroup$ Well, I will write it out for you a bit: $\rho\circ(g_{a,b}\cdot g_{b,c})=(\rho\circ g_{a,b})\cdot (\rho\circ g_{b,c}) $. Does that help? $\endgroup$
    – Notone
    Commented Apr 17, 2018 at 20:57
  • $\begingroup$ @user500094: If $\bar{g}_{\alpha \beta}$ satisfy cocycle conditions, then how does it follow a way to construct a vector bundle? $\endgroup$
    – Extremal
    Commented Apr 18, 2018 at 17:02
  • 1
    $\begingroup$ @Extremal see math.stonybrook.edu/~azinger/mat566-spr15/vectorbundles.pdf p.7 $\endgroup$
    – user557
    Commented Apr 18, 2018 at 17:33

1 Answer 1

1
$\begingroup$

well, $$g_{\alpha \beta} \circ g_{\beta \gamma}=g_{\alpha \gamma}$$ by assumption, so $\rho(g_{\alpha \beta}g_{\beta \gamma}=\rho(g_{\alpha \gamma})$, and on the LHS, the homomorphism property tells that $$\tilde{g_{\alpha \beta}} \circ \tilde{g_{\beta \gamma}}=\rho(g_{\alpha \beta}) \rho(g_{\alpha \gamma})=\rho(g_{\alpha \beta}g_{\beta \gamma})=\rho(g_{\alpha \gamma}) =\tilde{g_{\alpha \gamma}}$$ as desired.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .