Q:

Vector bundles and tensor operations

Why does on a scheme $X$ of characteristic zero for a vector bundle $F$ (of finite rank) the operation $Sym^n(.)$ commute with taking the dual bundle $(.)^v$, i.e.
$Sym^n(F^v) \simeq Sym^n(F)^v$ canonically ?
How does the iso work explicitly?

A:

There is always a map $Sym^{n}(F^{\vee}) \rightarrow Sym^{n}(F)^{\vee}$, regardless of the characteristic, and this map is an isomorphism in characteristic $0$. The map is induced
by the natural pairing $Sym^{n}(F^{\vee}) \times Sym^{n}(F) \rightarrow \mathcal{O}_{X}$, given locally by $(\phi_{1}\cdots \phi_{n}, f_{1}\cdots f_{n}) \mapsto \sum_{\sigma \in S_{n}} \phi_{\sigma(1)}(f_{1}) \cdots \phi_{\sigma(n)}(f_{n})$. 
If you write out the induced map in local coordinates, you'll see that some of the coefficients will be divisible by various primes, sometimes causing problems in positive characteristic. In characteristic zero, however, you can easily check that this map gives an isomorphism over a trivializing cover for $F$ and hence over all of $X$.
There is a similar pairing for alternating powers, by putting in appropriate signs. This map is always an isomorphism, regardless of characteristic, because no pesky coefficients appear.

