3
$\begingroup$

$\require{AMScd}$

Background: This question is about the bar and cobar constructions, and their relationship with the indecomposables of a dg-algebra. A brief summary of the bar and cobar constructions on ncatlab can be found at [1]. I will also mention the more general case of $A_\infty$ algebras; this case is discussed in this survey [2] or in Proute's thesis [3]. I posted a related question a few months ago in the mathoverflow post [4]. There I called the dg-module of indecomposables the ``linearized chain complex.'' It seems that the former is more standard terminology so I've switched for the time being.

Setup: Let $(A,\partial,\epsilon)$ be an augmented dg-algebra over a field $k$, i.e. a dg-algebra $(A,\partial)$ along with dg-algebra map $\epsilon:A \to k$. Here $k$ is a dga with trivial grading and differential.

The module of indecomposables $(IA,\partial)$ of $(A,\partial,\epsilon)$ is a dg-module whose underlying graded $k$ vector-space is $IA = \ker(\epsilon)/\ker(\epsilon)^2$. The differential $\partial$ on $A$ descends to a differential $\partial$ on $IA$. Furthermore, the correspondence $(A,\partial,\epsilon) \mapsto (IA,\partial)$ is functorial. That is, given a map: $$f:(A,\partial,\delta) \to (B,\partial,\epsilon)$$ of augmented dg-algebras, i.e. a dg-algebra map with $\epsilon \circ f = \delta$, we get a map of dg-modules: $$If:(IA,\partial) \to (IB,\partial)$$

The bar complex $BA$ of $(A,\partial,\epsilon)$ to a dg-coalgebra whose underlying graded vector-space is: $$ BA := k \oplus BA^+ \qquad BA^+ := \bigoplus_{i=1}^\infty \ker(\epsilon)[1]^{\otimes i} $$ The differential combines multiplication and differentiation from $(A,\partial)$ (see [1] for an actual definition) and the coaugmentation $\iota:k \to BA$ is just the inclusion. Similarly, the cobar complex $\Omega C$ of an augmented dg-coalgebra $C$ is an augmented dg-algebra admitting an analogous description to that of $BA$, as a certain tensor algebra. Both constructions are functorial, and there are generalizations of to $A_\infty$ algebras and coalgebras (see [2] or [3]).

The functors $B$ and $\Omega$ form an adjoint pair ($B$ is right adjoint to $\Omega$) and we have natural adjoint maps $\Omega B A \to A$ and $C \to B \Omega C$, which are quasi-isomorphisms of augmented dg-algebras and coaugmented dg-coalgebras, respectively. There is a reference for this in the post [5]. In particular, we get a natural map of dg-modules between the indecomposables: $$ I\Omega B A \to IA $$ I believe that it is true that $I\Omega C \simeq \text{coker}(\kappa)$ for any coaugmented dg-coalgebra $(C,\partial,\kappa)$ (an actual isomorphism of chain groups, not just a quasi-isomorphism) so that the above map yields a canonical map of dg-modules and a corresponding map on homology: $$ BA^+ \to IA \qquad H(BA^+) \to H(IA) $$

Question: Under what circumstances is the natural map $I\Omega B A \to IA$ an isomorphism on homology? When is it surjective on homology? I'm particularly interested in the case where the underlying graded algebra of $(A,\partial,\epsilon)$ is the free graded commutative algebra over some generators. Any partial answers are much appreciated.

Examples: In some cases this map is simple to understand. For instance, when $A$ is an algebra with the trivial differential and grading, then $H(IA) = IA$ and I believe that the map on homology $H(BA^+) \to H(IA)$ is just projection onto $H^1(BA^+) \simeq IA$.

Another context where something similar happens is when $A = \Omega C$ where $(C,\partial,\kappa)$ is a coaugmented dg-coalgebra (or even an $A_\infty$ coalgebra). Then the map: $$\text{coker}(\iota) \simeq BA^+ \to IA \simeq \text{coker}(\kappa)$$ is just the $A_\infty$ quasi-isomorphism on the cokernels of the coaugmentations induced by the quasi-isomorphism of $A_\infty$ coalgebras of $B\Omega C \to C$, itself induced by the adjunction map $\Omega B \Omega C \to \Omega C$. Since $A_\infty$ quasi-isomorphisms descend to isomorphisms on homology, we know in particular $H(BA^+) \simeq H(IA)$ under the map of interest.

$\endgroup$
4
$\begingroup$

The functor of indecomposables is the left adjoint of a Quillen adjunction between dg-algebras and dg-modules. (For a general reference, see Section 12.1.3 of the book Algebraic Operads by Loday and Vallette, though this was certainly known before the book – I just happen to have it on my desk.) As such it preserves quasi-isomorphisms between cofibrant algebras. The bar-cobar construction on $A$ is always cofibrant, so a sufficient condition for your claim is that $A$ should be cofibrant.

Since you are interested in the case of free symmetric algebra, if $A$ is free symmetric on some space of generators $V$ equipped with a filtration $$0 = F_0 V \subset F_1 V \subset \dots \subset V = \bigcup_{i \ge 0} F_i V$$ such that $\partial(F_i V) \subset S(F_{i-1}(V))$ (i.e. the differential of an element in $F_i V$ is a product of elements in lower filtration), then $A$ is cofibrant. In fact you can assume that the filtration is indexed by an ordinal, not just integers, if you need.

The homology of $I \Omega B A$ is the Hochschild homology of $A$ with constant coefficients $HH_*(A;\Bbbk)$. (Again, see the book of Loday and Vallette.) I wouldn't expect your map to be a quasi-isomorphism if $A$ is not cofibrant – it certainly can happen, but it's basically a coincidence. As for surjective on cohomology... I have no idea.

$\endgroup$
  • $\begingroup$ Dear Najib: unless I am missing something, the indecomposables of $\Omega BA$ are the generators $BA$. It's homology is $\operatorname{Tor}^A(\mathbb k,\mathbb k)$. To get Hochschild cohomology of $A$, you want to take the complex of derivations of $\Omega BA$ plus $\Omega BA$, i.e. the cone of the map $\Omega BA\to\operatorname{Der}(\Omega BA)$ that includes it as inner derivations. Or is $\Omega BA$ being augmented over something other than $\mathbb k$? $\endgroup$ – Pedro Tamaroff Jul 16 '18 at 14:13
  • $\begingroup$ @Pedro I should have been more precise. I meant the Hochschild homology of $A$ with constant coefficients, i.e. $HH_*(A; \Bbbk)$. $\endgroup$ – Najib Idrissi Jul 16 '18 at 14:26
  • $\begingroup$ Ah, sure. I even misread homology for cohomology, hence my comment. $\endgroup$ – Pedro Tamaroff Jul 16 '18 at 14:32
  • $\begingroup$ @NajibIdrissi Thank you, this is exactly the kind of answer that I was looking for! I have some follow up questions for you: $\endgroup$ – Julian Chaidez Jul 16 '18 at 18:33
  • $\begingroup$ (1) Most of the references that I've found online make a point of constructing a model category structure for non-negatively graded dg-algebras. I need the general Z-graded case, which I have found a few references for. I'm assuming that the section of Loday-Vallette that you mention is general enough for this? $\endgroup$ – Julian Chaidez Jul 16 '18 at 18:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.