What is the determinant of a complex of vector spaces?

I first met the notion of determinants of complexes of vector spaces in the book “Discriminants, Resultants, and Multidimensional Determinants“, but I just cannot understand the definition in that book. Could anyone explain it clearly or give some good references?

Answer

The appendix A of the book that you mention is probably the best reference for determinants!

  • If V is a k-vector space of dimension d, then its determinant is the one-dimensional vector space
    det(V)def=Λd(V).
    If V=0, we set
    det(0)def=k.

  • If V is a complex of finite dimensional vector spaces such that Vi=0 for almost all i (all but finitely many i), then one defines
    det(V)def=iZdet(Vi)(1)i.
    Here tensor products are over k, and the sign 1 denotes taking the dual vector space: V1def=V.

  • In the above situation, one can show that
    iZdet(Vi)(1)iiZdet(Hi(V))(1)i,
    where Hi(V) denote the cohomology spaces.

  • In general, if V is a complex that has finite dimensional cohomology spaces, and Hi(V)=0 for almost all i, then we may use the right hand side (*) as the definition:
    det(V)def=iZdet(Hi(V))(1)i.

It is worth noting that the whole point of det is that it is functorial. Namely, VW induces an isomorphism of one-dimensional vector spaces det(V)det(W).


Here’s one helpful analogy. Note that the formula (*) reminds the formula for the Euler characteristic

iZ(1)idimVi=iZ(1)idimHi(V),
and the proof of (*) actually goes along the same lines: it is based on the fact that if we have a short exact sequence of finite dimensional vector spaces
0VVV
then there is a canonical isomorphism
\det V \cong \det V’ \otimes_k \det V”
(this is analogous to the formula \dim V = \dim V’ + \dim V”.)

Attribution
Source : Link , Question Author : Jie Wang , Answer Author : Community

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