Is there an isomorphism from a smooth vector bundle $E$ and its double dual bundle $E^{**}$ which doesn't depend on a bundle metric on $E$?

Question

I want to show that the map $F:(E,\pi)\rightarrow (E^{**},\pi^{**})$ define fiberwise to be $F\vert_{E_p}(v)=ev_v$ (where $ev_v:E_p^*\rightarrow\mathbb{R}$ sends $w\mapsto w(v)$ ) is a smooth bundle homomorphism. I know there already exists an isomorphism by virtue of the fact that every vector bundle admits a metric, but I want to see if this $F$ works. I'm stuck on the smooth part, it's clear, at least to me, that it's fiberwise linear, and that $\pi^{**}F=\pi$

Answer 1

Of course. This is just the vector space double duality elevated to the bundle level; it is a completely metric-independent notion.

The canonical map $\iota_E:E\to E^{**}$ is clearly bijective (because on each fiber we have a linear isomorphism). Smoothness is a local condition, so we can work one local trivialization at a time; we may thus assume $M$ is open in some $\Bbb{R}^k$, $E=M\times V$ is a trivial vector bundle and then $E^{**}=M\times V^{**}$, and the map $\iota_E$ becomes (i.e once you compose with the vector bundle chart on the domain and target) $M\times V\to M\times V^{**}$, $(x,v)\mapsto (x,\iota_V(v))$. This is clearly a smooth map (it’s even linear in the second component).

Hence, $\iota_E:E\to E^{**}$ is a smooth vector bundle map over $M$ which is a fiberwise isomorphism. By abstract reasoning, you can deduce the inverse is also smooth; though in this case, you can see easily that the inverse is locally given by $M\times V^{**}\to M\times V$, $(x,\zeta)\mapsto (x,\iota_V^{-1}(\zeta))$ which is smooth (again it’s even linear in the second argument). Hence, $\iota_E$ gives an isomorphism of vector bundles.