most recent 30 from http://mathoverflow.net 2013-06-19T18:06:08Z http://mathoverflow.net/feeds/question/8014 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/8014/endomorphisms-of-vector-bundles stuck 2009-12-06T18:52:19Z 2009-12-07T00:44:08Z

<p>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 $\Psi\in\Gamma(\mbox{End}(E)).$</p> <p>How <strong>in general</strong> 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 $E$.)</p> <p>I'm sure there is a quick answer to this, that I am just not seeing. Any help is greatly appreciated.</p>

http://mathoverflow.net/questions/8014/endomorphisms-of-vector-bundles/8018#8018 a-fortiori 2009-12-06T19:18:36Z 2009-12-06T19:18:36Z

<p>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).</p>

http://mathoverflow.net/questions/8014/endomorphisms-of-vector-bundles/8021#8021 José Figueroa-O'Farrill 2009-12-06T19:28:36Z 2009-12-06T19:28:36Z

<p>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.</p>

http://mathoverflow.net/questions/8014/endomorphisms-of-vector-bundles/8053#8053 Deane Yang 2009-12-07T00:44:08Z 2009-12-07T00:44:08Z

<p>I can't help but repeat the earlier answers but in my own way:</p> <p>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.</p> <p>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:</p> <p>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.</p> <p>The extension to vector bundles then becomes obvious, because you just do the same thing to each fiber.</p>