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Let $R$ be a (possibly non-commutative) unital ring and $M$ be a perfect left $R$-module. Then, we have the trace map $$ \operatorname{Tr}\colon \mathrm{Hom}_R(M,M)\to R/[R,R]\,. $$

According to the answer to my previous question, there are maps \begin{align*} \mathrm{RHom}_R(M,M) \leftarrow \mathrm{RHom}_R(M,R)\otimes^L_R M\to R\otimes^L_{R\otimes^L R^{op}}R \end{align*} with the left arrow being a quasi-isomorphism, and taking the 0-th (co)homology yields the trace map above. So far, so good.

I thought the maps on other degrees vanish for the following reason: the cohomology of $\mathrm{RHom}$ is just $\mathrm{Ext}$, so only the non-negative degree part survives. On the other hand, the homology of $R\otimes^L_{R\otimes^L R^{op}}R$ is exactly the Hochschild homology. Since homology is negatively graded cohomology, the induced map becomes trivial except for the degree-0 part.

Question: Is it really the case?

At the same time, I found it strange because, on a complex manifold $X$, we have the trace map of the form $$ \operatorname{Tr}\colon\mathrm{Ext}^n(X;F,F) \to H^n(X,\mathcal{O}_X) $$ for a coherent sheaf $F$. I expected something similar in my setting but couldn't work it out.

I would appreciate any ideas or references. Thank you.

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    $\begingroup$ When you have a map $\mathcal M\to\mathcal O_X$ in the quasicoherent derived category $D^{\ge0}\operatorname{QCoh}(X)$ on a (non-affine) scheme $X$, which is zero on the sheafified $\mathcal H^i$ for $i>0$, it could induce a nontrivial map on after taking $H^i(X;-)$ (derived global sections). This is what is happening in your situation. $\endgroup$
    – Z. M
    Commented Jun 14 at 5:20

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