Question: What is the finest topology on $\mathrm{CAlg}$ (commutative ring spectra) for which THH (Topological Hochschild Homology) satisfies descent?

Adaptations of the arguments appearing in Section 3 of BMS2 show that THH has flat descent for simplicial commutative rings. However, the methods therein cannot be used to check whether THH has flat (or any weaker form of) descent for commutative ring spectra; simplicial $R$-algebras have the special property of being the nonabelian derived category of the category of finitely generated polynomial $R$-algebras.

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    $\begingroup$ Let me guess, CAlg here means is the category of (connective/nonconnective) commutative ring spectra? $\endgroup$ – crystalline Sep 16 '19 at 9:02
  • $\begingroup$ @crystalline This is correct. Maybe I should specify I care about connective things. $\endgroup$ – Liam Keenan Sep 16 '19 at 12:02

In Theorem 1.2 of B. I. Dundas and J. Rognes: "Cubical and cosimplicial descent", Journal of the London Mathematical Society (2) 98 (2018) 439-460, DOI 10.1112/jlms.12141, we showed that for each $1$-connected map $\phi : A \to B$, of connective commutative $S$-algebras, the map from $THH(A)$ to the homotopy limit of the cosimplicial spectrum $$ [q] \mapsto THH(B \wedge_A \dots \wedge_A B) $$ (with $q+1$ copies of $B$) is an equivalence. Here $1$-connected means that $\pi_0(\phi)$ is an isomorphism and $\pi_1(\phi)$ is surjective. This should let you reduce the general descent question to the one for (simplcial) commutative rings.

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  • $\begingroup$ This is perfect, thanks. Are the connectivity requirements absolutely essential for your result? $\endgroup$ – Liam Keenan Sep 17 '19 at 13:19
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    $\begingroup$ The connectivity requirement is necessary for our proof, but not for the conclusion, as [BMS2] show in the case of (discrete) commutative rings. If $A \to B$ is a map of connective commutative $S$-algebras and $THH$ satisfies descent for $\pi_0(A) \to \pi_0(B)$, then $THH$ satisfies descent for $A \to H\pi_0(A) \to H\pi_0(B)$ and $B \to H\pi_0(B)$. I would not be surprised if you can use this to deduce that $THH$ satisfies descent for $A \to B$. $\endgroup$ – John Rognes Sep 17 '19 at 18:40
  • $\begingroup$ Indeed this works! I found a way to manage this using the "May filtration" on THH which appears Angeltveit (arxiv.org/abs/1101.1866) and Angelini-Knoll--Salch (arxiv.org/abs/1612.00541) in combination with Theorem 1.2 of "Cubical and cosimplicial descent." $\endgroup$ – Liam Keenan May 12 at 4:05

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