# Why aren't complex vector bundles isomorphic to their duals?

For real vector bundles the argument goes like this: pick a metric on the bundle (i.e. a continuously varying metric on the fibers, which exists if the base space is paracompact), and map a vector $v$ to $\langle v, \cdot \rangle$, which is an iso (if the fibers are finite-dimensional). Under the same hypotheses (paracompactness, finite dimension), choosing a hermitian metric on a complex vector $E$ bundle yields an isomorphism $E \simeq \overline{E}^*$, the vector bundle whose fibers are the complex conjugates of the fibers of the dual (in other words, each fiber consists in the space of all antilinear maps $E_x \to \mathbb{C}$): if the hermitian metric is linear in the first argument and antilinear in the second, $v \mapsto \langle v, \cdot \rangle$ is the iso (notice trying to do the same in the other coordinate, $\langle \cdot, v \rangle$ gives a bijection between $E$ and $E^*$, but this is fiberwise antilinear, not linear, so it isn't an isomorphism!).

What I don't understand is why this construction can't be carried out to prove $E \simeq E^*$, where instead of choosing a hermitian metric, one chooses a continuously parametrized nondegenerate bilinear form (which can be seen to exist as in the real or hermitian case, using partitions of unity) and define the iso as in the real case.

There are specific examples of complex vector bundles which are not isomorphic to their duals, like the tautological line bundle over the Riemann sphere. So what am I missing?

Thank you very much for any help.

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If a vector bundle $E$ is isomorphic to it's dual then $E\otimes E$ has a nonvanishing section given by the pairing. In the case of a line bundle this means $E\otimes E$ is trivial. However the first chern number of complex line bundles over an algebraic curve is a function from line bundles to $\mathbb{Z}$ that takes tensor product to addition. A line bundle over a curve is isomorphic to it's dual exactly then it's first chern number is $0$.
...But when considering bilinear (not hermitian) forms from a complex vector space, it makes no sense to talk about positive definedness, since the associated quadratic forms takes its values in $\mathbb{C}$. There is no guarantee that a linear combination of generic nondegenerate bilinear forms is still nondegenerate, so the proof breaks down. Thanks a lot for your answer. –  Emilio Ferrucci May 10 at 21:28