Understanding proof that $E$ and $E^\ast$ are isomorphic rank $1$ bundles.

Question

I would like to prove the following:

Proposition. Let $E$ be any real line bundle over $M$. Then $E$ and $E^\ast$ are isomorphic line bundles.

I have sketched what I believe works, but am having trouble filling in the "why".

Proof Sketch. Let $E$ be any real line bundle over $M$. Consider the tensor product bundle $E\otimes E$. There is an open $\{U_\alpha\}$ of $M$ such that $E\otimes E$ admits a trivialization over each $U_\alpha$ so that the corresponding transition functions $\{\{\tau_{\beta\alpha}\colon U_\alpha\cap U_\beta\to\text{GL}(1,\mathbb{R})\cong\mathbb{R}\setminus\{0\}\}\}$ are positive. It follows that $E\otimes E$ is trivial (why?). Since we have the isomorphisms \begin{align*} \text{Hom}(E^\ast,E)\cong(E^\ast)^\ast\otimes E\cong E\otimes E \end{align*} it follows that $\text{Hom}(E^\ast,E)$ is trivial. This implies $E$ and $E^\ast$ are isomorphic line bundles.

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The line bundle $E\otimes E$ is trivial if there is a global frame. How can I use the (positive) transition functions $\tau_{\beta\alpha}$ to construct a global frame? Corresponding to the open cover $\{U_\alpha\}$, I know we also have a smooth partition of unity $\{f_\alpha\colon M\to\mathbb{R}\}$.

Answer 1

The problem reduces to show the following:

Let $V$ be a rank $1$ vector bundle over $M$. Let $\{U_{\alpha}\}_{\alpha}$ be a locally finite open cover trivializing $V$. Assume that on each $U_{\alpha\beta}$, the transition function $\tau_{\alpha\beta} \colon U_{\alpha\beta} \to GL(1,\Bbb R)$ has values in $(0,+\infty)$. Then $V$ is trivializable.

Proof. Let $\varphi_{\alpha} \colon V|_{U_{\alpha}} \to U_{\alpha}\times \Bbb R$ be the associated local trivializations. Recall that if $U_{\alpha}\cap U_{\beta} \neq \varnothing$, then $\tau_{\alpha\beta}$ is given by the relation $$ \begin{array}{r|ccc} \varphi_{\alpha}\circ \varphi_{\beta}^{-1} \colon &U_{\alpha\beta}\times \Bbb R &\longrightarrow &U_{\alpha\beta}\times \Bbb R \\ & (x,t) & \longmapsto & (x, \tau_{\alpha\beta}(x)t) \end{array} $$ Let $s_{\alpha} \colon U_{\alpha} \to V$ be the local section given by $$ s_{\alpha}(x) = \varphi_{\alpha}^{-1}(x,1). $$ Let $\{\phi_{\alpha}\}$ be a partition of unity associated with the open cover $\{U_{\alpha}\}$. Let $s = \sum_{\alpha} \phi_{\alpha}s_{\alpha}$, which is a global section.

Let us show that $s$ is nowhere vanishing. Let $x \in M$, and let $\{\alpha_j\}_{j=1,\ldots,n}$ be the finite set of indexes $\alpha$ such that $\phi_{\alpha}(x) >0$. Then $$ \varphi_{\alpha_1}(s(x)) = \left(x, \sum_{j=1}^n\phi_{\alpha_j}(x)\tau_{\alpha_1\alpha_j}(x)\right). $$ Since $\phi_{\alpha_j}(x) >0$ and $\tau_{\alpha_1\alpha_j}(x) >0$, it follows that $s(x) \neq 0$. Hence, $s$ is a nowhere vanishing section of $V$, and induces a trivialization. Indeed, the map $$ \begin{array}{ccc} M\times \Bbb R & \longrightarrow & V \\ (x,t) & \longmapsto & ts(x) \end{array} $$ is a global trivialization of $V$.

Answer 2

By patching locally defined metrics on $E$ using partition of unity, one can assume that there is a metric (notion of length) on each of the fibers, varying smoothly as a function of the fiber. The metric then gives a canonical identification of $E$ with $E^*$. This works in all (finite) dimensions; there is nothing special about dimension 1. If one wants to take into account a complex structure for example, this will no longer work (because the isomorphism will not preserve it).