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Clarify what can be done with det_\kappa

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aisejohan committed Sep 7, 2019
1 parent 3f8bf28 commit 00b152e644895b79067baa98fb041cf337626d0a
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  1. +10 −8 chow.tex
are given a finite filtration (see are given a finite filtration (see
Homology, Definition \ref{homology-definition-filtered}) Homology, Definition \ref{homology-definition-filtered})
$$ $$
M = F^n \supset F^{n + 1} \supset \ldots \supset F^{m - 1} \supset F^m = 0. 0 = F^m \subset F^{m - 1} \subset \ldots \subset F^{n + 1} \subset F^n = M
$$ $$
Then there is a canonical isomorphism then there is a well defined and canonical isomorphism
$$ $$
\gamma_{(M, F)} : \gamma_{(M, F)} :
\bigotimes\nolimits_i \det\nolimits_\kappa(F^i/F^{i + 1}) \det\nolimits_\kappa(F^{m - 1}/F^m) \otimes_\kappa \ldots \otimes_k
\det\nolimits_\kappa(F^n/F^{n + 1})
\longrightarrow \longrightarrow
\det\nolimits_\kappa(M) \det\nolimits_\kappa(M)
$$ $$
well defined up to sign(!). One can make the sign explicit either by To construct it we use isomorphisms of Lemma \ref{lemma-det-exact-sequences}
giving a well defined order of the terms in the tensor product (starting with coming from the short exact sequences
higher indices unfortunately), and by thinking of the target category for $0 \to F^{i - 1}/F^i \to M/F^i \to M/F^{i - 1} \to 0$.
the functor $\det_\kappa$ as the category of Part (2) of Lemma \ref{lemma-det-exact-sequences} with $G = 0$ shows
$1$-dimensional super vector spaces. See \cite[Section 1]{determinant}. we obtain the same isomorphism if we use the short exact sequences
$0 \to F^i \to F^{i - 1} \to F^{i - 1}/F^i \to 0$.


\medskip\noindent \medskip\noindent
Here is another typical result for determinant functors. Here is another typical result for determinant functors.

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