The Stacks project

Proposition 63.10.2. 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. $\square$


Comments (2)

Comment #164 by on

There are two missing references here, one indicated by "(insert reference here)" and the other by "???". I am not familiar with the matter, so I have no clue as to where they should point.


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 03TI. Beware of the difference between the letter 'O' and the digit '0'.