Pseudocoherent analogue of compact + nuclear = dualizable? $\DeclareMathOperator\RHom{RHom}\DeclareMathOperator\Map{Map}\DeclareMathOperator\id{id}\DeclareMathOperator\colim{colim}$Let $(\mathcal A,\mathcal M)$ be a (normalized) analytic ring defined in Lectures on Analytic Geometry. We review two basic concepts there:

*

*We say that an object $N\in D(\mathcal A,\mathcal M)$ is nuclear if, for every compact object $P\in D(\mathcal A,\mathcal M)$, the canonical map $(P^\vee\otimes_{(\mathcal A,\mathcal M)}^LN)(*)\to\RHom_{D(\mathcal A)}(P,N)$ is an equivalence, where $P^\vee:=\underline{\RHom}_{D(\mathcal A)}(P,\mathcal A)$.

*Let $P\in D(\mathcal A,\mathcal M)$ be a compact object, $M\in D(\mathcal A,\mathcal M)$ an object, and $f\colon P\to M$ a map in $D(\mathcal A,\mathcal M)$. We say that the map $f$ is trace-class if there exists a map $g\colon\mathcal A\to P^\vee\otimes_{D(\mathcal A,\mathcal M)}^LM$ such that $f$ is equivalent to the composite
\[
P\xrightarrow{1\otimes g}P\otimes_{(\mathcal A,\mathcal M)}^LP^\vee\otimes_{(\mathcal A,\mathcal M)}^LM\to M
\]
where the second map contracts the first two terms. Consequently, if $M\in D(\mathcal A,\mathcal M)$ is nuclear, then every map $P\to M$ from a compact object $P$ is trace-class.

We have the following simple result mentioned in Copenhagen Masterclass: an object $M\in D(\mathcal A,\mathcal M)$ is dualizable if and only if it is compact and nuclear. A summary of the proof: if $M$ is dualizable, then the functor $\Map_{D(\mathcal A,\mathcal M)}(M,-)=\Omega^\infty(M\otimes_{(\mathcal A,\mathcal M)}^L-)$ preserves small colimits therefore $M$ is compact. Under compactness, the nuclearity is equivalent to the identity $\id_M$ being trace-class.

Question. What happens if we replace the compactness by the pseudocoherence, defined as follows?


*We say that an object $Q\in D(\mathcal A,\mathcal M)$ is pseudocoherent if there exists a (left bounded) exhaustive filtration
\[
0\to\cdots\to P_{-m}\to P_{-m+1}\to\cdots\to P_n\to P_{n+1}\to\cdots
\]
of $Q$ in $D(\mathcal A,\mathcal M)$ such that each $P_i$ is compact and the connectivity of fibers of maps $(P_i\to Q)_{i\in\mathbb Z}$ tends to $+\infty$ as $i\to+\infty$.

Assume that $Q\in D(\mathcal A,\mathcal M)$ is pseudocoherent and nuclear. We pick a filtration $(P_i)_{i\in\mathbb Z}$ as above. Fix $i\in\mathbb Z$. It follows that the map $P_i\to Q$ correspond uniquely to a map $\mathcal A\to P_i^\vee\otimes_{(\mathcal A,\mathcal M)}^LQ\simeq\colim_{j\ge i}P_i\otimes_{D(\mathcal A,\mathcal M)}^LP_j$ in $D(\mathcal A,\mathcal M)$, which factors through some $P_i^\vee\otimes_{D(\mathcal A,\mathcal M)}^LP_j$ by compactness of $\mathcal A\simeq\mathcal M[*]$. The map $\mathcal A\to P_i^\vee\otimes_{(\mathcal A,\mathcal M)}^LP_j$ gives rise to a trace-class map $P_i\to P_j$, and we get a factorization $P_i\to P_j\to Q$ of the canonical map $P_i\to Q$ where $P_j\to Q$ is the canonical map in the filtration. A priori, the trace-class map $P_i\to P_j$ does not necessarily coincide with the map canonical map $P_i\to P_j$ appears in the filtration, but by compactness of $P_i$, there exists some $k\ge j$ such that $P_j\to P_k$ (co)equalizes $P_i\rightrightarrows P_j\to P_k$.
In short, for each $i\in\mathbb Z$, there exists $k\ge i$ such that the canonical map $P_i\to P_k$ is trace-class. We deduce that
Lemma. Let $Q\in D(\mathcal A,\mathcal M)$ be an object which is pseudocoherent and nuclear. Then there exists a (left bounded) exhaustive filtration
\[
0\to\cdots\to P_{-m}\to P_{-m+1}\to\cdots\to P_n\to P_{n+1}\to\cdots
\]
of $Q$ in $D(\mathcal A,\mathcal M)$ such that each $P_i$ is compact, every transition map $P_i\to P_{i+1}$ is trace-class, and the connectivity of fibers of maps $(P_i\to Q)_{i\in\mathbb Z}$ tends to $+\infty$ as $i\to+\infty$.
Note that this condition is sufficient, since it implies that $Q$ is basic nuclear and pseudocoherent. However, I would like to relate $Q$ to dualizable objects. The following conjecture seems to be a reasonable guess:
Conjecture. (pseudocoherent + nuclear = "almost dualizable") Let $Q\in D(\mathcal A,\mathcal M)$ be an object which is pseudocoherent and nuclear. Then there exists a (left bounded) exhaustive filtration
\[
0\to\cdots\to Q_{-m}\to Q_{-m+1}\to\cdots\to Q_n\to Q_{n+1}\to\cdots
\]
of $Q$ in $D(\mathcal A,\mathcal M)$ such that each $Q_i$ is dualizable, and the connectivity of fibers of maps $(Q_i\to Q)_{i\in\mathbb Z}$ tends to $+\infty$ as $i\to+\infty$.
Applications if this conjecture is true: Suppose that $(\mathcal A,\mathcal M)$ is Fredholm mentioned in Masterclass, then this gives rise to a weaker sufficiant condition for relative discreteness.
 A: Yes, what you write is correct, at least provided that the analytic ring $A$ has the very mild property that the internal RHom from $A[S]$ to $A$ is connective for every extremally disconnected profinite set $S$.  (In other words, there should be no higher cohomology with values in $A$ on any product of two extremally disconnected profinite sets.  In fact, in all examples I know there is no higher cohomology on any profinite set.)  Since Peter doesn't add this technical condition in his statement of Theorem 14.9 in Analytic.pdf, it's likely he knows how to remove it.  Let's hope he comes along to clarify :)
In fact, for such an analytic ring $A$, the nuclear pseudocoherent $A$-modules are exactly those which admit a resolution by modules of the form $cone(1-f)$ where $f$ is a trace-class endomorphism of a compact projective connective $A$-module.  Such a cone is both compact and nuclear (one can first invert $f$ then take the cone of $1-f$ and it gives the same result), hence dualizable.
To prove this, it suffices to start with a connective nuclear pseudocoherent $A$-module $M$ and make a map from such a cone which is surjective on $H_0$.  Let $g:P\rightarrow M$ be a map from a compact projective which is surjective on $H_0$; this exists because $M$ is pseudocoherent.  Since $M$ is nuclear, $g$ is represented by a class in $P^\vee \otimes M$.  If it were always true that $P^\vee$ is connective (as it is in examples), we would see that the map $id\otimes g$ from $P^\vee \otimes P$ to $P^\vee \otimes M$ is also surjective on $H_0$ and hence there is a trace-class map $f:P\rightarrow P$ such that $g\circ f=g$.  But this exactly means that we can factor $g$ through the cone of $1-f$ as desired.
