# Is a coherent locally free sheaf isomorphic it's dual?

Hartshorne chapter II problem 5.1 a) is to prove that the double dual of a coherent locally free sheaf $\mathscr{E}$ over a ringed space $(X,O_X)$ is isomorphic to $\mathscr{E}$. This can be done by defining an analog of the evaluation map and showing it is an isomorphism on a cover of open sets on which $\mathscr{E}$ is free.

Now we can apply the non-canonical isomorphism of free modules of finite rank with their dual locally to see that there is a cover by open sets under which the restriction of $\mathscr{E}$ and the dual of $\mathscr{E}$ are isomorphic. However, I believe the general philosophy is that maps which are canonical patch together to form morphisms of sheaves while maps that are not canonical don't necessarily patch together. So in the case of coherent locally free sheaf and it's dual, do the local isomorphisms not patch together in general?

• No, a locally free sheaf is not in general isomorphic to its dual, not even in the very special case of line bundles. Here, the dual $\mathcal L^\ast$ of a line bundle $\mathcal L$ serves as an inverse to $\mathcal L$ in the Picard group. This might be a good case to study to understand what's going on. Commented Oct 11, 2014 at 21:48
• It is interesting to note that, however, on a paracompact hausdorff space, every line bundle is isomorphic to its dual. The isomorphism uses non-algebraic functions and therefore doesn't work for algebraic line bundles on schemes. Also, if you are given a vector bundle on a manifold, if I remember correctly we may choose a metric, which then makes the vector bundle isomorphic to its dual. Commented Oct 12, 2014 at 10:03

As Jesko says, already for line bundles there are problems. The easiest example is probably $$\mathscr{O}_{\mathbf{P}^n_k}(1)$$ and its dual $$\mathscr O(-1)$$. The first has lots of nonzero global sections, but the second has none; they can't even be non-canonically isomorphic.
One interesting thing to think about, which I saw pointed out in a book review by Kollár, is that in differential geometry one can choose a metric on any vector bundle and use that to identify $$E \simeq E^\vee$$. So this is a good example of how algebraic geometry differs.