Hattori-Stallings trace

Question

Let $R$ be a (possibly non-commutative) unital ring and $M$ be a left $R$-module. If $M$ is finitely generated and projective, the natural map $$\iota:\mathrm{Hom}_R(M,R)\otimes_R M\to \mathrm{Hom}_R(M,M):f\otimes m \mapsto (m'\mapsto (f(m')m))$$ is an isomorphism and we can define the Hattori-Stallings trace by the composition $$\mathrm{Tr}:\mathrm{Hom}_R(M,M)\xrightarrow{\iota^{-1}}\mathrm{Hom}_R(M,R)\otimes_R M \xrightarrow{\mathrm{ev}} R/[R,R].$$ My first question is:

(1) Does $\mathrm{Tr}([f,g])=0$ hold for any $R$-module map $f,g:M\to M$?

If $M$ is further assumed to be free, this is true since we can take a free basis and directly compute it.

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Next, assume $M$ to be a perfect module (i.e., there exists a finite-length projective resolution $P_\bullet \to M$ with each $P_n$ being finitely generated). In this case, we can presumably define the trace of $f:M\to M$ by taking a lift $f_\bullet:P_\bullet\to P_\bullet$ and setting $$\operatorname{Tr}(f) = \sum_i(-1)^i\operatorname{Tr}(f_i).$$

My second question is:

(2) How can we prove that this is well-defined? Or, are there any references?

This MO post claims that it is well-defined without any proof.

Thank you.

Answer 1

The answer to (1) is yes. In fact this is true more generally: for any $f: M\to N, g:N\to M$, you have $tr(fg) = tr(gf)$.

The point is that $\hom_R(M,N)\otimes \hom_R(N,M)\cong \hom_R(M,R)\otimes_R N \otimes \hom_R(N,R)\otimes_R M$ and that under this identification, composition in one direction corresponds to evaluation $N\otimes\hom_R(N,R)\to R$, and in the other direction it corresponds to $\hom_R(M,R)\otimes M \to R$.

But since you're composing and then evaluating anyways, it amounts to evaluating both, and so they are equal.

For (2), the most conceptual way is to give a definition that does not depend on the chosen resolution. It works the same as what you explained, but using derived hom's and tensors. Namely, for $M$ perfect, the canonical map $\mathrm{RHom}_R(M,R)\otimes^L_R M\to \mathrm{RHom}_R(M,M)$ is a quasi-isomorphism, and the former has a map $\mathrm{RHom}_R(M,R)\otimes^L_R M\to R\otimes^L_{R\otimes^L R^{op}}R$, the $H_0$ of the latter simply being $R/[R,R]$.

Combining all of these you get that $tr$ is defined on $H_0(\mathrm{RHom}_R(M,M))$, which is $\hom_R(M,M)$; but in fact it is more properly a map $\mathrm{RHom}_R(M,M)\to HH(R)$, where $HH$ is Hochschild (or maybe Shukla) homology, i.e. $R\otimes^L_{R\otimes^L R^{op}}R$.

You can then check that this "fancy" definition of the trace agrees with the concrete one given by a projective resolution. Alternatively, you can prove it as a theorem, namely, it follows from the additivity of Hattori-Stallings traces, which itself can be proved in several different ways, e.g. by proving that $HH$ is a "localizing invariant"