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Let $A$ be a dg algebra, say over a field. The Homotopy Transfer Theorem says that $H(A)$ can noncanonically be given the structure of $A_\infty$-algebra, extending the induced multiplication on $H(A)$, in such a way that $A$ and $H(A)$ are quasi-isomorphic as $A_\infty$-algebras.

If $C$ is instead a dg coalgebra then we can also transfer the coalgebra structure to a quasi-isomorphic $A_\infty$-coalgebra structure on $H(A)$. Unfortunately the relation of quasi-isomorphism is less well behaved for coalgebras than for algebras and often one wants to consider instead the notion of weak equivalence: a coalgebra morphism $C \to C'$ is called a weak equivalence if the induced map on cobar constructions $\Omega C\to\Omega C'$ is a quasi-isomorphism. Is any dg coalgebra weakly equivalent to its cohomology as an $A_\infty$ coalgebra?

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  • $\begingroup$ "If $C$ is instead a dg coalgebra then we can also transfer the coalgebra structure to a quasi-isomorphic $A_\infty$-coalgebra structure on $H(A)$." -- is this true? $\endgroup$ – Leonid Positselski Nov 26 '18 at 0:45
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    $\begingroup$ "Is any dg coalgebra weakly equivalent to its cohomology as an $A_\infty$ coalgebra?" -- is seems that this can't be true. Say, suppose that we are working in the categories of noncounital (or equivalently, coaugmented strictly counital) dg and $A_\infty$-coalgebras. Then there exists a dg-coalgebra wihich is not weakly equivalent to zero, but whose cohomology coalgebra is zero. But there is only one (namely, zero) $A_\infty$ coalgebra structure on a zero vector space. $\endgroup$ – Leonid Positselski Nov 26 '18 at 0:54
  • $\begingroup$ Thanks, Leonid. That settles it pretty decisively. Regarding your question, I've seen this stated without proof in at least two places, and I imagined it would be easy: the way the HTT is proven in Loday-Vallette is that a "homotopy retract" from a chain complex V to W induces a map of co-operads $B\mathrm{End}(V)\to B\mathrm{End}(W)$ given by a sum over trees, from which the result follows. But "turning the picture upside down" should give instead a map $B\mathrm{coEnd}(V)\to B\mathrm{coEnd}(W)$ between bar constructions on coendomorphism operads. But I haven't checked anything carefully. $\endgroup$ – Dan Petersen Nov 26 '18 at 8:09
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We worked out some answers to this question in our paper: arXiv:1904.03585 (Edit: the following answer refers to v1 of the paper on the arXiv!)

Here's the short version. There are two possible natural definitions of an $A_\infty$-coalgebra:

(A) It is a graded vector space $V$ with maps $V \to V^{\otimes n}$, $n \geq 1$, satisfying identities exactly dual to those defining an $A_\infty$-algebra.

(B) It is a graded vector space $V$ and a square zero derivation of the tensor algebra on $V$.

They are not equivalent; (B) is stronger than (A). In the paper we call them naive and genuine $A_\infty$-coalgebras. There is a homotopy transfer theorem for naive $A_\infty$-coalgebras: for any naive $A_\infty$-coalgebra $C$ there is a transfered naive $A_\infty$-structure on $H(C)$ which is quasi-isomorphic to the one on $C$. This is false for genuine $A_\infty$-algebra structures in general.

We say that an $A_\infty$-coalgebra $C$ is conilpotent if it admits an exhaustive filtration of the form $$ 0 = F_0 C \subseteq F_1C \subseteq F_2C \subseteq \ldots $$ which is compatible with the coalgebra structure. We call such a filtration a positive filtration. In the conilpotent case, the notions of genuine and naive $A_\infty$-coalgebra coincide. Any filtered quasi-isomorphism of positively filtered $A_\infty$-coalgebras is a weak equivalence. If $C$ is a conilpotent $A_\infty$-coalgebra equipped with a positive filtration, and $C \to V$ is a filtered quasi-isomorphism of chain complexes, then it's possible to transfer the $A_\infty$-coalgebra structure on $C$ to an $A_\infty$-structure on $V$, so that the $A_\infty$-morphism $C \rightsquigarrow V$ is a filtered quasi-isomorphism and hence a weak equivalence. It follows that although $C$ is in general not weakly equivalent to $H(C)$ for any transferred structure, there is always a weak equivalence from $C$ to $H(\operatorname{Gr} C)$ with a transferred $A_\infty$-coalgebra structure on $H(\operatorname{Gr} C)$. If we fix a positive filtration on $C$ then the $A_\infty$-coalgebra $H(\operatorname{Gr} C)$ with its transferred structure is uniquely determined up to a noncanonical filtered $A_\infty$-isomorphism, and it deserves to be called the filtered minimal model of $C$.

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  • $\begingroup$ Usually (A) also comes with the condition that the operators are locally of finite support, meaning that for each $v\in V$, only finitely many co-operations are non-zero. This is indeed the definition (B). I agree that (A) is naive, but do people really use that definition more than (B)? $\endgroup$ – Pedro Tamaroff Jan 27 at 17:09
  • $\begingroup$ All three notions - naive, genuine, and conilpotent infinity-coalgebras - can be found in the literature being referred to as infinity-coalgebras without any modifiers. This was a bit surprising to me. In the updated version we changed the term "genuine" to "locally finite", and removed the modifier "naive" completely. It seemed to imply a moral judgement which wasn't really necessary, and it turned out that in the updated version we get the most traction out of the "naive" definition... $\endgroup$ – Dan Petersen Jan 28 at 20:10
  • $\begingroup$ Agreed. The "naive" one is obtained by taking the completed tensor algebra. Perhaps that helps with a name. I also agree removing the modifier is a good idea. At any rate, kudos for a really interesting paper. :) $\endgroup$ – Pedro Tamaroff Jan 28 at 20:37

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