$\DeclareMathOperator\THH{THH}\DeclareMathOperator\Perf{Perf}\DeclareMathOperator\map{map}\DeclareMathOperator\tr{tr}\DeclareMathOperator\HH{HH}\DeclareMathOperator\fib{fib}\DeclareMathOperator\id{id}\DeclareMathOperator\KEnd{KEnd}$Let $R$ be a ring spectrum, and $P$ a perfect $R$-module. The definition of $\THH(R)$ with a cyclic bar construction over the whole of $\Perf(R)$ allows to define a canonical morphism $\map_R(P,P)\to \THH(R)$, where $\map_R(P,P)$ is the mapping spectrum of $P$. Specializing to $\pi_0$ yields a map $[P,P]\to \pi_0\THH(R)$.
Let me call $\tr(f)$ the image of $f: P\to P$ under this map, this is the Hattori-Stallings trace.
I have two questions about this construction:
Is it true that $\tr(f)$ is equal to $\tr(\lambda)$ for some $\lambda \in [R,R]\cong \pi_0R$ ? Does the answer change if we assume $R$ is commutative ?
If $R$ is a commutative ring spectrum ($E_2$ is enough, but if $E_\infty$ is needed for a positive answer, I would rather hear about that), then $\THH(R)$ has a canonical ring structure. Suppose $f$ is nilpotent as an endomorphism of $P$. Is $\tr(f) $ nilpotent ?
Remarks : Here are some remarks and thoughts about these questions. I will first address 1., and then 2..
Observe that if $R$ is commutative, then there is a map $\THH(R)\to R$ such that the composite $\map_R(P,P)\to R$ is (on $\pi_0$) the ordinary symmetric monoidal trace. In particular, because $R\simeq \map_R(R,R)\to \THH(R) \to R$ is the identity, if there is such a $\lambda$, then it must be the symmetric monoidal trace of $f$.
If $R$ is connective, then $\pi_0\THH(R) = \HH_0(\pi_0R)$, ordinary Hochschild homology, and so in particular the map $\pi_0R\to \pi_0\THH(R)$ is surjective, which shows that the question is only interesting for nonconnective ring spectra.
In fact, question 1. makes sense when we replace $\Perf(R)$ with an arbitrary stable $\infty$-category with a distinguished object to play the role of $R$, e.g. one could think of a monoidal stable $\infty$-category and the unit therein. However, in this generality the answer is no (there are counterexamples for example for $G$-spectra).
Note that 1. is not asking whether $\pi_0 R\to \pi_0\THH(R)$ is surjective, because this is in general wrong for non-connective rings. It's asking whether it's surjective on the image of $\pi_0\KEnd(R)$.
For $P= \bigoplus_i P_i$, the endomorphism $f$ is represented by a matrix $(f_{ij})$, more precisely it is $\sum_{ij} \iota_i f_{ij} p_j$ where $\iota_i : P_i\to P$ is the inclusion and $p_j : P\to P_j$ the projection, so that $\tr(f) = \sum_{ij} \tr(\iota_i f_{ij} p_j) = \sum_{ij} \tr(p_j\circ \iota_i \circ f_{ij})$ by cyclic invariance, and then $p_j \circ \iota_i = \delta_{ij}\id_{P_j}$ so that the usual formula for traces as $\sum_i \tr(f_{ii})$ applies. Therefore if $P$ is a direct sum of (possibly shifted) copies of $R$, any endomorphism $f$ has $\tr(f)$ in the image of $\pi_0R$.
For any $P,Q$ and endomorphism $f$ of $P$, one can find an endomorphism of $Q$ with the same trace, by point 7., $f\oplus 0$ works. It follows that we can work up to retracts if needed - e.g. we can focus on "finite" $R$-modules rather than perfect ones. As modules, these have filtrations by direct sums as in 7), but not necessarily (as far as I can tell) as modules with endomorphisms. If one could find such a filtration, we would be done.
A final remark about question 1. is that if you have counterexamples for $\HH_k$ for some other base ring $k$ than the sphere spectrum, I would also like to hear about it.
As only $\pi_0$ is concerned, for ordinary commutative rings $R$ question 2. is basically asking about symmetric monoidal traces of nilpotent endomorphisms, and we can then reduce to the case of fields, where the claim is true. A similar strategy for commutative ring spectra would require thinking about "residue fields of ring spectra" which, I think, is a complicated problem so should probably be avoided as a proof strategy.
Suppose $R$ is connective. Then by question 1., we have $\tr(f) = \tr(\lambda)$ where, by observation 3, $\lambda$ is the monoidal trace of $f$. But furthermore, $R$ being connective and commutative, we have $\pi_0\THH(R) = \pi_0R$, so $\tr(\lambda)$ is nilpotent if and only if $\lambda$ is — in other words, for connective commutative ring spectra, we are reduced to the same question for symmetric monoidal traces, which maybe is easier ?
To prove the usual statement over fields, one would use the $\ker(f^k)$ filtration of a vector space $V$. However, the $\fib(f^k)$ filtration doesn't do the trick in general, for two reasons : a- $\fib(0) = P\oplus \Omega P$ and not just $P$, b- it seems that if this filtration could be made to work, it would most likely show that the trace is $0$, and not nilpotent. Of course, if $R$ is not reduced, any nilpotent element of $R$ provides a counterexample.
Question 2. also makes sense for a general symmetric monoidal stable $\infty$-category, and I am also interested in an answer in this more general case.
