Continuity of determinant of vector bundle morphism

Question

Suppose $E$ is a real topological vector space and $B$ is any topological space. We can generate a trivial vector bundle $\xi=(B\times E,B, \pi,\cdot, +)$ where the projection $\pi:B\times E\rightarrow B$ is the usual $(p,v)\mapsto p$. Now suppose $f:B\times E\rightarrow B\times E$ is a vector bundle automorphism, that is, $f$ is an homeomorphism which maps each fibre $\pi^{-1}(p)$ isomorphically onto $\pi^{-1}(q)$ provided that $q=(\pi \circ f \circ \pi^{-1} )(p)$. Now, clearly for any such $p$ and $q$ we can associate to $f$ an isomorphism $I_p$ such that $$f(p,v)=(q,I_pv) $$ We can then define $I:B\rightarrow \mathbb{R}$ as $p\mapsto \text{det}(I_p)$. My two questions are :

1. Is $I$ necessarily continuous?

2. Is $I$ non-zero?

I am no good at handling determinants but it seems to me that this should be fairly ease. Thanks in advance.

Edit (Analysis on question 2): Question 2 is clearly true, since each $I_p$ should be an isomorphism defined on $\pi^{-1}(p)$ and hence it must be non-singular.

Answer 1

Yes it necessarily is continuous! For such an $f$ $(p,v)\mapsto(q,w)$, define $g:B \rightarrow B$ as $p\mapsto q$ and $J:B\rightarrow L(E,E)$ as $p\mapsto I_p$, then we can write $$f=(g,J\circ \pi) $$

$f$ is an homeomorphism and hence continuous, which is equivalent to both $g$ and $J\circ \pi$ being continuous. I claim that $J$ is continuous, which follows from the fact that $\pi$ is an open map.

Since $I=\text{det}\circ J$, then it suffices to show that $\text{det}$ is continuous! For this we will need some machinery, namely the topology of $L(E,E)$ the idea of the proof is the following. We obviously consider $E$ to be finite dimensional, otherwise det does not make sense; we use the definition of det as $$Te^1\wedge \ldots \wedge Te^n=\text{det}(T)\;e^1\wedge \ldots \wedge e^n $$ for any given basis $e^1,\ldots, e^n$ of $E$, thus $\text{det}(J(p))$ is continuous if the map $$ e^1\wedge \ldots \wedge e^n\mapsto J(p)e^1\wedge \ldots \wedge J(p)e^n$$ is continuous. Now, the map $(v_1,...,v_n)\mapsto v_1\wedge \ldots \wedge v_n$ is multilinear on a finite dimensional vector space and thus continuous (Proof?). Now we can write the desired map as a composition of continuous maps. This completes the proof.

Answer 2

I think the standard way to verify this is to write your aggregate map as a composition of other maps that you know are continuous. Notice that your aggregate map can be written as

$$ I(p) \;\; =\;\; \det \left (\pi_E \circ f \circ \iota_p\right ) $$

where we take $\iota_p:E \to B\times E$ to be the canonical inclusion $\iota_p(v) = (p,v)$ and $\pi_E:B\times E \to E$ the canonical projection $\pi_E(q,v) = v$. You can verify that since $\iota_p$ is continuous for all $p$ and each of $f, \pi_E, \det$ are continuous, then the map $I$ will be continuous as well.