10
$\begingroup$

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.

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ Let me guess, CAlg here means is the category of (connective/nonconnective) commutative ring spectra? $\endgroup$
    – crystalline
    Commented Sep 16, 2019 at 9:02
  • $\begingroup$ @crystalline This is correct. Maybe I should specify I care about connective things. $\endgroup$
    – Liam Keenan
    Commented Sep 16, 2019 at 12:02

1 Answer 1

Reset to default
5
$\begingroup$

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.

Improve this answer
$\endgroup$
2
  • $\begingroup$ This is perfect, thanks. Are the connectivity requirements absolutely essential for your result? $\endgroup$
    – Liam Keenan
    Commented Sep 17, 2019 at 13:19
  • 1
    $\begingroup$ 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
    Commented Sep 17, 2019 at 18:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .