Determinant of endomorphism bundle

Question

We have that the endomorphism bundle of a smooth vector bundle $E$ is $\text{End}(E) = \text{Hom(E,E)}$. Is it necessarily true that $\det(\text{End}(E))$ is always trivial for any smooth vector bundle $E$?

I'm unsure on how to prove this. I have seen and proved that a smooth $n$-dimensional manifold is orientable if and only if $\Lambda^nM$ is trivial or if it admits a smooth nowehere-vanishing $n$-form. Given the fibres of $\det E$ are exactly $\Lambda^n E_x$ perhaps this would give some way to prove it, but I don't quite see how.

Answer 1

$\newcommand\End{\operatorname{End}}$Any vector space $V$ has an identity map $I_V: V \rightarrow V$ and $I_V \in \End(V)$. Therefore, for each $x \in M$, $I_{E_x} \in \End(E_x)$. This defines a section $I_E$ of $E$. You can check using either a smooth frame or trivialization that $I_E$ is a smooth section.

Now observe that if $E$ is a rank 1 vector bundle (i.e., line bundle), then so is $\End(E)$. Therefore, since the identity section $I_E$ is everywhere nonzero, it defines a global trivialization of $\End(E)$.

Let me include the definition of the determinant bundle. First, given any linear map $L: V \rightarrow V$, where $\dim(V) = k$, there is an naturally (i.e., independent of any choice of basis) induced map $$ \det(L): \Lambda^kV \rightarrow \Lambda^kV, $$ where for any $v_1, \dots, v_k \in V$, $$ L(v_1\wedge\cdots\wedge v_k) = L(v_1)\wedge\cdots\wedge L(v_k). $$ In particular, $\det(L) \in \End(\Lambda^kV)$.

The determinant line bundle of $\End(E)$ is $$\det(\End(E)) = \End(\Lambda^kE). $$ If $L: E \rightarrow E$ is a bundle map, i.e., a section of $\End(E)$, then $\det(L)$ is a section of $\det(\End(E))$.

My original comment can be made even simpler. There is no need to mention the identity section $I_E$ at all. Since $\Lambda^kE$ is a line bundle, the bundle $\det(\End(E)) = \End(\Lambda^kE)$ is trivial.

Answer 2

This is indeed true.

First, note that the smooth structure is irrelevant -- a smooth line bundle is smoothly trivial iff it is topologically trivial. To see this, note that topological triviality just says that there exists a nonvanishing continuous section. But we can always homotope this to a nonvanishing smooth section, yielding smooth triviality. So, we'll ignore the smooth structure here. This result turns out to be true for all vector bundles on paracompact Hausdorff spaces!

Suppose $E$ is a direct sum of line bundles $L_1 \oplus \cdots \oplus L_r$. Then $$\operatorname{End}(E) \cong (L_1 \oplus \cdots \oplus L_r) \otimes (L_1 \oplus \cdots \oplus L_r)^* \cong \bigoplus_{i,j} L_i \otimes L_j^*,$$ so $$\det \operatorname{End}(E) \cong \det \bigoplus_{i,j} L_i \otimes L_j^* = \bigotimes_{i,j} L_i \otimes L_j^*.$$ This large tensor product has each $L_i$ appearing $r$ times, and each $L_i^*$ appearing $r$ times, so overall this bundle is trivial.

Now, let $E$ be an arbitrary vector bundle of rank $r$ on a paracompact Hausdorff space $X$. By the splitting principle, there exists a map $f : Y \to X$ from a paracompact Hausdorff space such that

(i) $f^* : H^\bullet(X) \to H^\bullet(Y)$ is injective;
(ii) $f^*E \cong L_1 \oplus \cdots \oplus L_r$ for some line bundles $L_1, \dots, L_r$ on $Y$.

We know that $f^* \det \operatorname{End}(E) = \det \operatorname{End}(f^* E)$ is trivial. So, the first Chern class of $\det \operatorname{End}(f^* E)$ is trivial. Since $f^*$ is injective on cohomology, the first Chern class of $\det \operatorname{End}(E)$ is trivial. Since $\det \operatorname{End}(E)$ is a line bundle, we conclude that $\det \operatorname{End}(E)$ is trivial.

There probably exists a better proof :)