Algebraic Topology – Is There a Dold-Kan Theorem for Circle Actions?

at.algebraic-topologyct.category-theoryhigher-category-theoryhomotopy-theorystable-homotopy

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 $\mathcal C$?

Best Answer

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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