How to prove that topological Hochschild homology of a smooth proper stable k-linear infinity category is dualizable?

Question

Let $k$ be a perfect field of characteristic $p$. I heard that the Topological Hochschild homology of a smooth proper stable infinity category (or dg-category) is dualizable as a THH(k)-module spectrum in the infinity category of cyclotomic spectra. Does this follow from the fact that a smooth proper stable infinity category is dualizable in the category $\text{Cat}^{\text{perf}}_{\infty,k}$ ?

Answer 1

$\newcommand{\THH}{\mathrm{THH}} \newcommand{\Cat}{\mathrm{Cat}} \newcommand{\perf}{\mathrm{perf}} \newcommand{\Sp}{\mathrm{Sp}} \newcommand{\Mod}{\mathrm{Mod}}$ If you ask about dualizability in $\THH(k)$-modules in $\Sp$, it indeed follows from this together with the fact that $\THH: \Cat^{\perf}_{\infty,k}\to \Mod_{\THH(k)}(\Sp)$ is symmetric monoidal, hence preserves dualizability.

This symmetric monoidality follows from the following two properties of $\THH: \Cat^{\perf}_\infty\to \Sp$:

1- It is symmetric monoidal

2- It commutes with sifted colimits.

The latter ensures that if you do a relative tensor product (which is defined via a bar construction, which is a colimit of a simplicial object), you can do it before or after applying $\THH$.

The exact same argument works for modules over $\THH(k)$ in $\Sp^{BS^1}$.

If you ask, as you did, about modules in cyclotomic spectra, then this is still so, and this is still because $\THH$ has these two properties with values in cyclotomic spectra, but it is somehow not as easy to prove : for starters, I don't know of a convenient reference for $\Cat^\perf_\infty\to \mathrm{CycSp}$, although it is folklore that such a thing exists and is symmetric monoidal.

Once you have this, the rest follows in exactly the same way because colimits in $\mathrm{CycSp}$ are computed underlying, say, by corollary II.1.7. in Nikolaus-Scholze.

EDIT : let me correct a slight mistake I made here. It is not true that $\THH$ commutes with all sifted colimits; it commutes with sifted colimits of ring spectra (in particular, along maps of ring spectra, not maps in $\Cat^\perf_{\infty,k}$). It does commute with all filtered colimits, and so to prove that it is strong symmetric monoidal, this weaker property turns out to be sufficient.

Also for the purpose of recording it, let me mention here the following easy counterexample to commutation with sifted colimits: the functor $X\mapsto (\Sp^X)^\omega$ from spaces to $\Cat^\perf_\infty$ commutes with all colimits, and its composition with $\THH$ is $\mathbb{S}[LX]$, where $LX := \mathrm{map}(S^1,X)$, which clearly does not commute with sifted colimits (e.g. take a simplicial set representing $X$, and view it as an expression of the form $X\simeq \mathrm{colim}_{\Delta^{\mathrm{op}}}X_n$ and you'll find that the assembly map for this diagram is $\mathbb{S}[X]\to \mathbb{S}[LX]$ which is quite rarely an equivalence)