Changing the fiber in vector bundles

126 Views Asked by At

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.

1

There are 1 best solutions below

0
On

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.