determinant of a perfect complex - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T15:00:04Z http://mathoverflow.net/feeds/question/7124 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/7124/determinant-of-a-perfect-complex determinant of a perfect complex Jonathan Wise 2009-11-29T09:55:14Z 2011-04-29T14:38:25Z <p>Say $K_\bullet$ is a bounded complex of vector bundles. I seem to want the determinant of $K_\bullet$ to be the alternating tensor product of the terms of the complex:</p> <p>$\det(K) = \bigotimes_n \det(K_n)^{(-1)^n}$.</p> <p>Is there a reason why this is the right definition (or the wrong definition)? Is there a better definition?</p> http://mathoverflow.net/questions/7124/determinant-of-a-perfect-complex/7125#7125 Answer by Gonçalo Marques for determinant of a perfect complex Gonçalo Marques 2009-11-29T10:53:37Z 2009-11-29T10:53:37Z <p>You should take a look at the appendix A of "Discriminants, Resultants, and Multidimensional Determinants" by Gelfand, Kapranov and Zelevinsky</p> http://mathoverflow.net/questions/7124/determinant-of-a-perfect-complex/7136#7136 Answer by YBL for determinant of a perfect complex YBL 2009-11-29T13:46:33Z 2009-11-29T13:46:33Z <p>It is a theorem of Deligne that this is essentialy the only possible formula if you ask for the determinant functor to satisfy some natural properties (mainly det has to be compatible with exact sequences). See theses <a href="http://personal.us.es/fmuro/splitbeamer.pdf" rel="nofollow">Slides</a> for example.</p> http://mathoverflow.net/questions/7124/determinant-of-a-perfect-complex/7143#7143 Answer by David Ben-Zvi for determinant of a perfect complex David Ben-Zvi 2009-11-29T16:47:52Z 2009-11-29T17:35:59Z <p>I now realize this is covered beautifully in the notes by Muro linked to by YBL, but anyway here's a brief summary: Given a perfect complex of vector spaces (let's work first over a point) we get a homotopy point of the K-theory spectrum, and we can start asking "which point is it" in more and more refined fashion. First we ask for which component it's in (i.e. look at pi_0) - these are labeled by the integers, ie by the Euler characteristic of your complex. Next you can ask to describe it as an object of the fundamental groupoid of K-theory (i.e. give also pi_1 information). This fundamental groupoid is canonically identified with the (Picard) groupoid of graded (super)lines. The grading is given by the Euler characteristic (ie project on pi_0), and the superline is the determinant line of your complex. If you give concrete realizations of the higher fundamental groupoids of the K-theory spectrum you get concrete K-theoretic invariants of your complex of a higher and higher categorical nature (the ultimate one being of course just giving your complex itself as a homotopy point (or contractible subset) of K-theory). You can do the same in families, i.e. over a base, giving a locally constant function on the base from pi_0 (the Euler characteristic of your complex), and a Z-graded super line bundle on the base (determinant line), and so on..</p> <p>One place this is used beautifully (and where I learned it) is <a href="http://arxiv.org/abs/math/0610055" rel="nofollow">Beilinson's work on epsilon factors</a>. </p> http://mathoverflow.net/questions/7124/determinant-of-a-perfect-complex/20501#20501 Answer by Dennis for determinant of a perfect complex Dennis 2010-04-06T14:13:37Z 2010-04-06T14:13:37Z <p>Is it really a theorem of Deligne that this is "canonical"? I thought it was Knudsen-Mumford.</p> http://mathoverflow.net/questions/7124/determinant-of-a-perfect-complex/63421#63421 Answer by Yujia Qiu for determinant of a perfect complex Yujia Qiu 2011-04-29T14:38:25Z 2011-04-29T14:38:25Z <p>As I understand the construction of the determinant of a perfect complex, this definition is quite straightforward, following from the fact that in a short exact sequence, say $$0\rightarrow S\rightarrow E\rightarrow Q\rightarrow 0$$ defining the determinant of the sequence to be the alternating tensor is the canonical way to make it isomorphic to $\mathbb{1}$.</p> <p>Also, I think good references to this may be the original paper by Knudsen-Mumford, a book by Kato, and also a paper by kings which are listed below:</p> <p>Knudsen, Finn Faye; Mumford, David: The projectivity of the moduli space of stable curves. I. Preliminaries on ''det'' and ''Div''. The part about determinants appears in Chapter I, but note that there is a typo defining the determinant, namely in the map of the transposition of tensor product, there should be $\alpha\cdot\beta$ instead the sum of these two as a power of $-1$;</p> <p>Guido Kings: An introduction to the equivariant Tamagawa number conjecture: the relation to the Birch-Swinnerton-Dyer conjecture</p> <p><a href="http://www.uni-regensburg.de/Fakultaeten/nat_Fak_I/FGAlgZyk/index-en.html" rel="nofollow">http://www.uni-regensburg.de/Fakultaeten/nat_Fak_I/FGAlgZyk/index-en.html</a></p> <p>There is a part about determinants in lecture 1 section 5, where there are not a lot of details but it provides a good view towards the construction of determinant.</p> <p>Kato: Lectures on the approach to Iwasawa theory for Hasse-Weil L-functions via B_dR, part I, which mentions determinant in 2.1.</p>