# When is bar-cobar duality an equivalence?

Let $A$ be an augmented differential graded algebra over a field $k$. I will write $BA$ for its bar construction (whose homology is $Tor^A(k, k)$). This is a co-augmented differential graded coalgebra over $k$; write $\Omega BA$ for its cobar construction. There is a natural dga map $\Omega BA \to A$ which uses the fact that $\Omega C$ is the tensor algebra on a shift of $C$, and the projection map $BA \to A$.

I think I'm only going to highlight my ignorance here, but my question is: what assumptions on $A$ imply that the map $\Omega BA \to A$ is an equivalence?

If I am reading any of the following references correctly:

• Husemoller-Moore-Stasheff's "Differential homological algebra and homogeneous spaces" (section II.4)
• Felix-Halperin-Thomas "Rational homotopy theory" (exercise 2, section 19)
• Keller "A-infinity algebras, modules and functor categories" (Theorem 4.3, attributed to Lefevre),

then I think that the answer is supposed to be: "no assumptions are required at all."

I seem to have two possible counterexamples to this claim. But their being counterexamples seems to rely on the positive answer to another question: when is $\Omega$ a homotopy-invariant functor? That is: if $f: C \to D$ is an equivalence of dg coalgebras, is $\Omega f$ an equivalence of dga's?

Let me explain the (counter?)examples. Let $G$ be a finite group, $k[G]$ its group ring, and $k^G$ the dual Hopf algebra (the ring of functions on $G$ with values in $k$). Then both $k[G]$ and $k^G$ are augmented algebras, where the augmentation is the counit.

1. $B(k[G])$ is isomorphic to the simplicial chain complex of the bar construction $BG$ (with coefficients in $k$). If $k$ has characteristic zero, then this is is equivalent to $k$, via a transfer argument. If $\Omega$ is a homotopy invariant, then $\Omega B(k[G]) \simeq \Omega k = k$, which is certainly not $k[G]$ unless $G$ is trivial.

2. $k^G$ is isomorphic as a $k$-algebra to a product of fields $$k^G \cong \prod_{g \in G} k,$$ and the augmentation can be identified with the map which projects onto the factor corresponding to the identity $e \in G$. The kernel of the augmentation is a summand, and hence a projective $k^G$-module. From this, you can write down a very short resolution of $k$ over $k^G$ and compute $Tor^{k^G}(k, k) = k$, concentrated in $Tor_0$. So $B(k^G) \simeq k$. Again, if $\Omega$ is homotopy-invariant, then $\Omega B(k^G) \simeq \Omega k = k$, which is not equivalent to $k^G$.

So either I have misunderstood the references above, or $\Omega$ is not a homotopy invariant functor. I'm guessing it's the latter; can anyone point me to a reference which gives criteria for this to hold?

What the references are saying is correct, and you are right. Yes, $\Omega BA \to A$ is always a quasi-isomorphism. No, $\Omega$ does not in general take quasi-isomorphisms to quasi-isomorphisms.

A sufficient condition for $\Omega$ transforming a DG-coalgebra morphism to a quasi-isomorphism of DG-algebras is a filtered quasi-isomorphism of conilpotent DG-coalgebras. This also generalizes to CDG-coalgebras (which correspond to nonaugmented DG-algebras, and for which the conventional notion of quasi-isomorphism does not even exist, but filtered quasi-isomorphisms make perfect sense).

For a reference, see my 2011 AMS Memoir "Two kinds of derived categories, Koszul duality, and comodule-contramodule correspondence", http://arxiv.org/abs/0905.2621 , Section 6.10.

• Thanks! This is very helpful. Can you clarify my confusion about my example #1 in this context, though? Using $B(k[G]) = C_*^{simp}(BG)$, I can try to filter this by the skeletal filtration on $BG$, and I can filter $k$ trivially. Then I think that the inclusion on the basepoint in $BG$ induces a filtered quasi-isomorphism $k \to C_*^{simp}(BG)$ when $k$ is rational. Is the issue that the target is not conilpotent? – Craig Westerland Jun 6 '16 at 19:09
• No, the coalgebra $B(k[G])$ is conilpotent, and the skeletal filtration on it in itself satisfies all the required conditions. The point is that the associated graded complex to the skeletal filtration is the graded coalgebra $B(k[G])$, endowed with the zero differential. The passage to the associated graded object with respect to the skeletal filtration just kills the differential for you and leaves you with the graded vector space of chains. This is not quasi-isomorphic to $k$. So it is not a filtered quasi-isomorphism. – Leonid Positselski Jun 6 '16 at 20:38
• I see; that makes a lot more sense. Thanks again! – Craig Westerland Jun 6 '16 at 20:43
• @LeonidPositselski I have a question related to Lefèvre Hasegawa's thesis, please have a look at this – Victor TC Apr 17 '19 at 18:34

See Hasegawa's thesis, available here. Lemma 1.3.2.3 is what you want, on page 35 of the pdf. To see that $\Omega B A \to A$ is a quasi-isomorphism, one filters $BA$ by its primitive elements (i.e. by the length of bar elements) which then induces on $\Omega BA$ a filtration. One gives $A$ the trivial filtration (i.e. $F^0A=0$ and all higher $F^i$ are $A$) and sees the counit is a filtered morphism. The associated graded map is the identity of $A$ on degree one, and then one shows that $\text{Gr}_i(\Omega BA)$ is contractible for $i>1$.

Another way to remember why the counit is a quasi-isomorphism is that it corresponds to the trivial twisting cochain $\beta:BA\longrightarrow A$ whose total space $BA\otimes_\beta A$ is acyclic. Then one fits the counit into the following sequence

$$1\otimes \varepsilon_A : BA\otimes_\gamma \Omega B A\to BA\otimes_\beta A$$

Both twisted complexes above are acyclic so $1\otimes \varepsilon_A$ is a quasi-isomorphism. Now $BA$ is simply connected so a spectral sequence argument works to show that, because $1$ is of course a quasi-isomorphism, so is $\varepsilon_A$.

This argument fails for $\Omega$ because it decreases the connectivity of your space. Lemma 1.3.2.2 in the thesis above gives a sufficient criterion for $\Omega$ to preserve quasi-isomorphisms. The bar construction, on the other hand, preserves quasi-isomorphisms by a direct argument using the filtration by length and the Kunneth formulas. Again, no connectivity issues arise here, which renders the desired spectral sequence convergent.