Let $E$ be a vector bundle, with $X$ base space and $p:E\to X$ a surjective projection.

Let $x\in X$ be given. Let $U_1, U_2$ be two neighborhoods of $x$ in $X$ which carry local trivializations, that is, $$\varphi_1:p^{-1}(U_1)\cong U_1 \times V_1$$and $$ \varphi_2:p^{-1}(U_2)\cong U_2 \times V_2$$ are two homeomorphisms, where $V_i\in obj(Vect_{\mathbb{C}})$ for $i=1,2$.

Then clearly, $$\left.\varphi_1\right|_{U_1\cap U_2}:p^{-1}(U_1)\cap p^{-1}(U_2)\cong U_1\cap U_2 \times V_1$$ and $$\left.\varphi_2\right|_{U_1\cap U_2}:p^{-1}(U_1)\cap p^{-1}(U_2)\cong U_1\cap U_2 \times V_2$$ are two homeomorphisms and thus $$\left.\varphi_2\right|_{U_1\cap U_2}\circ \left(\left.\varphi_1\right|_{U_1\cap U_2}\right)^{-1}: U_1\cap U_2 \times V_1 \cong U_1\cap U_2 \times V_2 $$ is also a homeomorphism.

If $p_2:U_1\cap U_2\times V_2 \to V_2$ is the projection onto the second component, define the map $t_x:V_1\to V_2$ by $$ V_1\ni v_1 \mapsto p_2(\left.\varphi_2\right|_{U_1\cap U_2}\circ \left(\left.\varphi_1\right|_{U_1\cap U_2}\right)^{-1}((x,v_1))) \in V_2 $$ My question is: how do you see that $t_x$ is a linear homeomorphism and not merely a homeomorphism?


Note that $\varphi_i : p^{-1} (U_i) \to U_i \times V_i$ are not mere homeomorphisms: they are local trivialization. That is, the following diagram commute $\require{AMScd}$ \begin{CD} p^{-1}(U_i) @>\varphi_i>> U_i \times V_i\\ @V p V V\ @VV p_i V\\ U_i @>>id> U_i \end{CD}

and $\phi_i$ is linear when restricted to each fiber. This latter fact implies that the transition map is linear.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.