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Let $X$ a smooth projective scheme over a field $k$. And let $THH(X)$ denotes the topological Hochschild homology of $X$. Recall that the spectra $THH(X)$ admits an action of the of circle $S^{1}$. Let $C_{p^n}$ the cycle subgroup of $S^{1}$ with $p^{n}$ elements.

Question 1 when the map from homotopy fixed points to fixed points
$$ THH^{C_{p}}(X)\rightarrow THH^{hC_{p}}(X)$$ is an equivalence (after $p$-compeltion)?

Question 2 what is the interpretation in algebraic geometry of the the groups $$ \pi_{i}THH(X)^{hS^{1}}$$ up to $p$-completion?

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    $\begingroup$ For 1), the answer depends on the explicit model of $THH(X)$ you take. $\endgroup$
    – Fernando Muro
    Commented Dec 23, 2014 at 21:55
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    $\begingroup$ could you say more ? $\endgroup$
    – Max
    Commented Dec 23, 2014 at 21:57
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    $\begingroup$ what I mean is that the homotopy fixed points functor is the total left derived functor of the plain fixed points functor, hence if you apply both functors to a cofibrant object you obtain weakly equivalent results, whereas you don't if you apply them to a non-cofibrant object. This is not much, really, and won't be very helpful, so I include it as a comment. $\endgroup$
    – Fernando Muro
    Commented Dec 24, 2014 at 13:08
  • $\begingroup$ @Fernando: I think that people want THH to be a genuinely $S^1$-equivariant spectrum, which in particular means that it is more than a spectrum equipped with an action of $S^1$ up to homotopy, and in particular it is equipped with genuine (not homotopy) fixed point data. At least Tyler Lawson's answer below seems to only make sense under that assumption. $\endgroup$
    – Qiaochu Yuan
    Commented Dec 25, 2014 at 11:34
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    $\begingroup$ Qiaochu of course, did you maybe understand anything different from my comments? $\endgroup$
    – Fernando Muro
    Commented Dec 25, 2014 at 14:27

1 Answer 1

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The map (which actually goes from $THH^{C_p}$ to $THH^{hC_p}$) usually not an equivalence in the $p$-complete setting, at least if your input is genuinely a ring. You can detect the difference using a mapping cone; the mapping cone of this map is equivalent to the mapping cone of a map from $THH$ to the so-called Tate construction $THH^{tC_p}$ on it. For a ring $R$, the former is usually concentrated in nonnegative degrees while the latter rarely is (using homological grading).

However, there are some useful cases where it is an equivalence in sufficiently high degrees. These include, for example, the cases of $\Bbb Z/p$ and $\Bbb Z_p$, and there is a theorem of Tsalidis that (under some relatively mild extra assumptions) if this holds, then it also holds for the maps $THH^{C_{p^k}} \to THH^{hC_{p^k}}$. (This result played a critical role in Hesselholt-Madsen's calculation of the K-theory of local fields.)

If you were working rationally instead of in a $p$-complete setting, the homotopy groups of $THH^{hS^1}(R)$ correspond to "periodic" cyclic homology. In the $p$-complete setting instead, the homotopy groups of $THH^{hS^1}(R)$ are groups that are called $TF(R)$ in the literature. These are, at least in part, connected to an inverse limit of truncated portions of the de Rham-Witt complex under Frobenius maps. (My understanding of this is that, on the algebro-geometric side, there is some connection to Breuil-Kisin modules, but I understand nothing about these and only know this connection exists from watching a talk of Scholze's.)

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