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Why can't the monoidal Dold–Kan correspondence be extended to non-connected CDGAs over a field of characteristic 0?

I understand that there is a technical problem with the original proof due to Quillen given in "Rational Homotopy Theory" (Remark on p.223). However, I don't understand what is the conceptual reason for this.

Edit: In the initial post I mistakenly used the term connective to denote CDGAs that have the ground field in dim 0 and 0 in negative degrees. It was pointed out by Ben Wieland in the comments that these algebras should be called connected instead.

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    $\begingroup$ It sounds like you are proposing two generalizations at once. One generalization, to CDGAs over a field of characteristic zero, is no problem, e.g., see Corollary 5.4.1 in arxiv.org/abs/1606.01803. The other generalization, to the non-connective setting, is less clear. Classically, Dold-Kan is between simplicial abelian groups and connective chain complexes. If you move to non-connective chain complexes, what do you have on the other side? $\endgroup$ Commented Oct 30, 2022 at 16:14
  • $\begingroup$ @DavidWhite I believe that Quillen stated that the equivalence works over any field of characteristic 0 (after all we only need a retraction of the tensor algebra onto the symmetric algebra). I think that Theorem 2.5 from here math.uchicago.edu/~amathew/doldkan.pdf states the correspondence between non-negatively graded chain complexes and simplicial abelian groups. Perhaps, I don't understand what is your definition of a connective chain complex. In my question, I assume that connective CDGA are ones that have the ground field in dimension $0$ and have nothing in dimensions below $0.$ $\endgroup$ Commented Oct 31, 2022 at 13:54
  • $\begingroup$ Right but in your question you wrote that you wanted it for nonconnective cdgas. Was that a typo? Non connective usually means it's allowed to have stuff in dim below zero $\endgroup$ Commented Oct 31, 2022 at 14:57
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    $\begingroup$ Hi Grisha. Ok. I got it. I will write up an answer tonight. There is no obstruction. Everything works in characteristic zero. The paper I linked above proves it but I can explain in an answer here. $\endgroup$ Commented Oct 31, 2022 at 16:26
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    $\begingroup$ Hi Grisha. Sorry it has taken me so long to write an answer. We are hiring and it's been a ton of on-campus interview candidates, eating up pretty much all my free time. I haven't forgotten your question. $\endgroup$ Commented Nov 7, 2022 at 15:32

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The answer to the question:

Why can't the monoidal Dold–Kan correspondence be extended to non-connected CDGAs over a field of characteristic 0?

is that in fact the Dold-Kan correspondence can be extended to this setting.

The classical Dold-Kan correspondence is an equivalence of categories between the category $ch^+$ of non-negatively graded chain complexes of abelian groups, and the category $sAb$ of simplicial abelian groups. The normalization functor $N: sAb \to ch^+$ is monoidal. So is the inverse functor $\Gamma: ch^+\to sAb$, thanks to the Alexander-Whitney map. The natural isomorphism $N\Gamma \cong Id_{Ch^+}$ is monoidal, but the natural isomorphism $\Gamma N \cong Id_{sAb}$ is not, as explained in the paper Equivalences of monoidal model categories by Schwede and Shipley.

The Dold-Kan equivalence can be defined for any commutative ring $R$, analogously to the classical case $R = \mathbb{Z}$, leading to a Quillen equivalence between non-negatively graded $R$-modules and simplicial $R$-modules (where $R$ is regarded as a monoid in the category of simplicial abelian groups). This Quillen equivalence is weak monoidal rather than strong monoidal. The monoidality is enough to induce an adjoint pair on categories of $O$-algebras, for any operad $O$ (e.g., the commutative monoid operad).

Let $k$ be a field of characteristic zero. Then the Quillen equivalence $(N,\Gamma)$ is nice in the sense of Definition 3.5.5 of the paper Homotopical Adjoint Lifting Theorem by me and Donald Yau. This is the content of Example 3.5.6. Consequently, Corollary 5.4.1 proves that the induced adjunction between CDGAs over $k$ and simplicial commutative $k$-algebras is a Quillen equivalence. Note that we place no restriction at all on the values of the CDGAs in degree zero, so we're in the "non-connected" setting of the question.

For completeness, let me unpack why this setting is "nice":

  1. In $\mathcal{N} = ch^+_k$ (with the projective model structure, which here coincides with the injective model structure), for any symmetric group $\Sigma_n$, every object of $\mathcal{N}^{\Sigma_n}$ is projectively cofibrant (essentially, $\Sigma_n$ acts freely due to the characteristic being zero). So all the conditions of niceness are automatic for $\mathcal{N}$.
  2. Let $\mathcal{M}$ denote the category of simplicial $k$-modules. Whenever $X\in \mathcal{M}^{\Sigma_n}$ is cofibrant, so is the object of $\Sigma_n$-coinvariants $X_{\Sigma_n}$. Furthermore, the domains of the generating cofibrations of $\mathcal{M}$ are cofibrant.
  3. For cofibrant objects $W$ and $X$, the map $N^2: N(W\otimes X)\to NW \otimes NX$ is a weak equivalence and remains a weak equivalence after taking $\Sigma_n$-coinvariants (if $W$ and $X$ have $\Sigma_n$-actions), again, because $\mathcal{N}$ is so nicely behaved.
  4. For any $X\in \mathcal{M}^{\Sigma_n}$ cofibrant in $\mathcal{M}$, if $f$ is a (trivial) cofibration, then so is $X\otimes_{\Sigma_n} f^{\Box n}$, where $f^{\Box n}$ is the $n$-fold iterated pushout product. This condition guarantees that the homotopy theory of $\mathcal{M}$ plays nicely with symmetric powers, so that commutative monoids (and algebras over more general operads) inherit transferred model structures. It holds because we're in the characteristic zero setting.

As for the proof of Corollary 5.4.1, it comes down to Theorem 4.2.1, which works by a transfinite induction over the $Sym$ functor, where $Sym^n(X) = X^{\otimes n}/\Sigma_n$. Because each of these functors preserves weak equivalences between cofibrant objects (again using the characteristic zero assumption), the transfinite induction works and proves the relevant morphism is a weak equivalence, to know that the induced adjoint pair is a Quillen equivalence.

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    $\begingroup$ Sorry for being a full month late with this answer. It has been a hectic semester. $\endgroup$ Commented Dec 8, 2022 at 18:57

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