Proposition 64.10.2. Let $K\in D_{perf}(\Lambda )$ and $f\in \text{End}_{D(\Lambda )}(K)$. Then the trace $\text{Tr}(f)\in \Lambda ^\natural $ is well defined.
Proof. We will use Derived Categories, Lemma 13.19.8 without further mention in this proof. Let $P^\bullet $ be a bounded complex of finite projective $\Lambda $-modules and let $\alpha : P^\bullet \to K$ be an isomorphism in $D(\Lambda )$. Then $\alpha ^{-1}\circ f\circ \alpha $ corresponds to a morphism of complexes $f^\bullet : P^\bullet \to P^\bullet $ well defined up to homotopy. Set
\[ \text{Tr}(f) = \sum _ i (-1)^ i \text{Tr}(f^ i : P^ i \to P^ i) \in \Lambda ^\natural . \]
Given $P^\bullet $ and $\alpha $, this is independent of the choice of $f^\bullet $. Namely, any other choice is of the form $\tilde{f}^\bullet = f^\bullet + dh +hd$ for some $h^ i : P^ i \to P^{i-1}(i\in \mathbf{Z})$. But
\begin{eqnarray*} \text{Tr}(dh) & = & \sum _ i (-1)^ i \text{Tr}(P^ i\xrightarrow {dh} P^ i) \\ & = & \sum _ i (-1)^ i \text{Tr}(P^{i-1}\xrightarrow {hd} P^{i-1}) \\ & = & -\sum _ i (-1)^{i-1}\text{Tr}(P^{i-1}\xrightarrow {hd} P^{i-1}) \\ & = & - \text{Tr}(hd) \end{eqnarray*}
and so $\sum _ i (-1)^ i \text{Tr} ((dh+hd)|_{P^ i})=0$. Furthermore, this is independent of the choice of $(P^\bullet , \alpha )$: suppose $(Q^\bullet , \beta )$ is another choice. The compositions
\[ Q^\bullet \xrightarrow {\beta } K \xrightarrow {\alpha ^{-1}} P^\bullet \quad \text{and}\quad P^\bullet \xrightarrow {\alpha } K \xrightarrow {\beta ^{-1}} Q^\bullet \]
are representable by morphisms of complexes $\gamma _1^\bullet $ and $\gamma _2^\bullet $ respectively, such that $\gamma _1^\bullet \circ \gamma _2^\bullet $ is homotopic to the identity. Thus, the morphism of complexes $\gamma _2^\bullet \circ f^\bullet \circ \gamma _1^\bullet : Q^\bullet \to Q^\bullet $ represents the morphism $\beta ^{-1}\circ f\circ \beta $ in $D(\Lambda )$. Now
\begin{eqnarray*} \text{Tr}(\gamma _2^\bullet \circ f^\bullet \circ \gamma _1^\bullet |_{Q^\bullet }) & = & \text{Tr}(\gamma _1^\bullet \circ \gamma _2^\bullet \circ f^\bullet |_{P^\bullet })\\ & = & \text{Tr}(f^\bullet |_{P^\bullet }) \end{eqnarray*}
by the fact that $\gamma _1^\bullet \circ \gamma _2^\bullet $ is homotopic to the identity and the independence of the choice of $f^\bullet $ we saw above. $\square$
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Comments (2)
Comment #164 by Pieter Belmans on
Comment #166 by Johan on