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There are several interesting equivalences of "Dold-Kan type" in the setting of stable $\infty$-categories. Namely, let $\mathcal C$ be a stable $\infty$-category. Then the following 3 stable $\infty$-categories are known to be equivalent:

  1. The $\infty$-category $Fun(\mathbb N, \mathcal C)$ of filtered objects in $\mathcal C$ (where $\mathbb N$ is the poset of natural numbers).

  2. The $\infty$-category $Fun(\Delta, \mathcal C)$ of cosimplicial objects in $\mathcal C$.

  3. The $\infty$-category $Ch(\mathcal C)_{\leq 0}$ of nonpositively-homologically-graded chain complexes in $\mathcal C$.

$(1) \Leftrightarrow (2)$ is due to Lurie (see HA 1.2.3) and $(1) \Leftrightarrow (3)$ is due to Ariotta. $(2) \Leftrightarrow (3)$ is meant to be reminiscent of the classical Dold-Kan theorem.

There's another place where chain complex structures can come from though, namely from actions by the circle group $S^1$. For instance, in the HKR theorem, the differential on the de Rham complex arises directly from the $S^1$-action on Hochschild cohomology. (I'm not familiar enough with the literature to have a reference for this, but I gather that the idea is to look at the map $X \oplus \Sigma X = \Sigma^\infty_+ S^1 \wedge X \to X$ coming from a circle action; the second component is a map $\Sigma X \to X$ which is exactly the data needed for a differential; that it squares to zero comes from the associativity of the circle action, since the top cell of $S^1 \times S^1$ splits off.)

In other words, we are led to consider

  1. The $\infty$-cateogry $Fun(\mathbb C\mathbb P^\infty, \mathcal C)$ of $\mathcal C$-objects with $S^1$-action.

Now, I think that (4) lives in the unbounded world -- the right things to compare to are

  1. ' $Fun(\mathbb Z, \mathcal C)$ (filtrations extending in both directions)

  2. ' (omitted -- but Kan's combinatorial spectra might be relevant)

  3. ' $Ch(\mathcal C)$ (unbounded chain complexes)

Question: Are the $\infty$-categories (1') and (3') equivalent to (4), for an arbitrary stable $\infty$-category $\mathcal C$?

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    $\begingroup$ I think for spectra it doesn't square to $0$ and is rather related to $\eta$ $\endgroup$
    – Maxime Ramzi
    Commented May 5, 2022 at 19:11
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    $\begingroup$ A chain complex is a family of objects $X_n$ and differentials between them, so your sketch doesn't really match up: from an object with $S^1$ action you can only get a specific kind of periodic chain complex. This functor exists, it is a bit easier to describe as functor to filtered objects and takes $X$ to the Tate construction $X^{tS^1}$ with a natural filtration. Of course, it's far from fully faithful. $\endgroup$
    – Achim Krause
    Commented May 6, 2022 at 3:20
  • $\begingroup$ @AchimKrause this sounds really cool, do you mind saying which filtration we’re supposed to taking on the Tate construction? Also is it easy to see that the HKR example mentioned arises as a Tate construction? $\endgroup$
    – Bbb
    Commented May 6, 2022 at 11:18
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    $\begingroup$ If you write $X^{tS^1} = \operatorname{colim}_{n\in \mathbb{Z}}(X\otimes S^{\mathbb{C}^n})^{hS^1}$, where $S^{\mathbb{C}^n}$ denotes the corresponding representation sphere, and the maps are the canonical inclusions, it's the filtration by the terms in this sequence (which make sense for negative $n$ as well, by Spanier-Whitehead duality). This is roughly filtering the Tate construction by the columns of the Tate spectral sequence. $\endgroup$
    – Achim Krause
    Commented May 6, 2022 at 12:27
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    $\begingroup$ In the HKR example, the "first order" information in the coherent chain complex associated to this filtration on $HP=HH^{tS^1}$, i.e. the actual differentials, does indeed have something to do with the de Rham differential. Their action on homotopy groups is precisely captured by the $d_2$ differential in the Tate spectral sequence, which is the de Rham differential under the HKR isomorphism. $\endgroup$
    – Achim Krause
    Commented May 6, 2022 at 12:34

1 Answer 1

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No, they are not equivalent, even for $C = Sp$.

Indeed, the category of spectra with $S^1$-action is also the category of $\mathbb S[S^1]$-modules, and is compactly generated by a single object.

On the other hand, compact objects of $Fun(\mathbb Z, Sp)$ are retracts of finite colimits of representables, and any finite set $S$ of representables cannot generate the whole thing - e.g. because if $n$ is below all the elements in $S$, then $F(n) = 0$ for any $F$ generated under colimits by $S$ . So it is not compactly generated by a single object (you have to change the proof a bit, but the same holds for $Fun(\mathbb N, Sp)$)

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  • $\begingroup$ Thanks, this is a great obstruction! I'm not fully satisfied, because it doesn't rule out that the incoherent construction I indicated above might underly a functor which might be fully faithful or something.... Also, could you explain your comment above? I agree it seems reasonable that $\eta$ should appear somehow in what it means to have a $S^1$-action, but I'm having trouble seeing how exactly. $\endgroup$
    – Tim Campion
    Commented May 5, 2022 at 19:26
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    $\begingroup$ It doesn't rule it out, indeed :) As for my comment, the point is that as you indicated, the top cell of $S^1\times S^1$ splits off. Actually it splits off after one suspension, and so the multiplication $S^1\times S^1\to S^1$, after one suspension, looks like some map $S^2\vee S^2 \vee S^3\to S^2$. The $S^2\to S^2$ maps are just identities, and the $S^3\to S^2$ map is $\eta$ $\endgroup$
    – Maxime Ramzi
    Commented May 5, 2022 at 19:29
  • $\begingroup$ In particular, if you have an $S^1$-action on $X$, $\mathbb S[S^1]\otimes \mathbb S[S^1]\otimes X\to \mathbb S [S^1] \otimes X\to X$ is going to be, "on the top cell", not $0$ but the action of $\eta$ on $X$. I should be more precise here, I specifically mean the $\eta$ coming from $S^1$, not coming from $\mathbb S$. So, precisely, $\Sigma^2X\to \Sigma X \to X$ is going to be $\mathbb S^2\otimes X \xrightarrow{\Sigma\eta \otimes X} \mathbb S^1\otimes X\to \mathbb S[S^1]\otimes X\to X$ $\endgroup$
    – Maxime Ramzi
    Commented May 5, 2022 at 19:32
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    $\begingroup$ Okay, I think I understand your statement. Is this easy to see? And isn't it dependent on the choice of splitting? $\endgroup$
    – Tim Campion
    Commented May 5, 2022 at 19:34
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    $\begingroup$ For a proof that s^2 = \eta s (where s is the degree 1 operation induced by the S^1-action) you might look at Proposition 3.3 of arxiv.org/abs/2008.09095 , where there are also references to earlier discussions. Here \eta arises as the Hopf construction for the multiplication S^1 x S^1 --> S^1. $\endgroup$
    – John Rognes
    Commented May 7, 2022 at 20:45

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