Why does the dual of a vector bundle use the inverse transpose?

Question

I would expect that the dual of a vector bundle would be defined by the inverse conjugate transpose, as that would be the inverse of the adjoint. When $\alpha_{ij}:X\to Y$ is a transition matrix in $E$, we have $\alpha_{ij}^*:Y^*\to X^*$, so $(\alpha_{ij}^*)^{-1}:X^*\to Y^*$ would be (in my opinion) the most canonical choice for a transition matrix in $E^*$. So why do we use $(\alpha_{ij}^t)^{-1}$ instead of $(\alpha_{ij}^*)^{-1}=(\overline{\alpha_{ij}}^t)^{-1}$?

Answer 1

In general you like various objects to be holomorphic, so why introduce complex conjugations ad hoc? Furthermore, if you change basis in a complex vector space with transition matrix $\alpha$, then the dual basis in the dual space is change by the inverse transpose of $\alpha$. Remember that the dual pairing is bilinear, not sesqulinear.

I think the reason you think conjugation may be useful is that you have experience with hermitian forms on complex vector spaces (or bundles). But the main reason these beasts are so nice is positive-definitness, which is not possible for complex bilinear scalar products. For a pairing between two different spaces, positive definitness is undefined, so why bother with sesqulinearity?