## Endomorphisms of vector bundles

I'm a bit stuck, and I'm hoping someone can help me out. I have a vector bundle $E$ on an algebraic curve (the ones I am interested in are holomorphic, but I'm sure that doesn't matter so much for the purposes of this question...), and I also have an endomorphism $\Psi\in\Gamma(\mbox{End}(E)).$

How in general does one compute the determinant of $\Psi$? If $E$ splits as a sum of rank-1 bundles then so too does the endomorphism bundle associated with $E$, and then it's obvious to me how to compute the determinant in that case. But, when $E$ is more general, what is an invariant way to go about computing the determinant? (I'm looking for something similar to the invariant way of computing the trace, for which I can rewrite $\Psi$ as a map $E\otimes E^*\rightarrow\mathcal{O}$ and then take the image of the identity morphism of $E$.)

I'm sure there is a quick answer to this, that I am just not seeing. Any help is greatly appreciated.

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## 3 Answers

The determinant of $\Psi$ is the top exterior power $\bigwedge^{\mathrm{rk}(E)}\Psi$ (identifying endomorphisms of the line bundle $\bigwedge^{\mathrm{rk}(E)}E$ with functions).

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Expanding a-fortiori's answer a little, this comes from the following linear algebra fact. Let $\psi$ be an endomorphism of a finite-dimensional vector space $V$ (say, $N$-dimensional). Taking exterior powers is functorial, so that $\psi$ will induce an endomorphism of any $\Lambda^p V$. In particular, consider the top exterior power $\Lambda^N V$. This is a one-dimensional vector space. Let $\omega \in \Lambda^N V$ be a nonzero vector there. Then $\psi \omega$ will be proportional to $\omega$ and that proportionality constant is the determinant.

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I can't help but repeat the earlier answers but in my own way:

This is really just a question about linear algebra. It is worth remembering that a vector bundle is just a parameterized family of vector spaces (I like to call differential geometry "parameterized linear algebra"). So anything you can do naturally to a vector space, you can do to a vector bundle.

So to extend the definition of a determinant of an endomorphism of vector spaces to one for an endomorphism of vector bundles, you just need a good natural definition of determinant:

An endomorphism of a vector space $V$ naturally induces an endomorphism of $\Lambda^nV$, where $n$ is the dimension of $V$. Since $\Lambda^nV$ is 1-dimensional, an endomorphism of $\Lambda^nV$ must consist of multiplication by a fixed scalar. That scalar is the determinant of the endomorphism.

The extension to vector bundles then becomes obvious, because you just do the same thing to each fiber.

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