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The functor that embeds the category of commutative algebras to associative algebras has the left adjoint - the quotient by the commutant ideal. For any dg-algebra $A$ let $A_{Ab}$ denote the derived functor of the quotient by the commutant ideal, i. e. take a free resolution of the algebra and then take the quotient by two-sided ideal generated by commutators. There is the canonical map $A \to A_{Ab}$.

Suppose that $A$ is the algebra of rational cohomology of a homotopy type $X$.

My question is: does $A_{Ab}$ has some geometric interpretation? More precisely: does there exist a functor $M$ from homotopy types to homotopy types and a natural transformation $e: M \to id$ such that $H^*(M(X))=A_{Ab}$ and $e$ gives the canonical map?

There is a similar question on the geometric meaning of $A_{Ab}$ for $A=H_*(\Omega X)$, the homology of a loop space with the Pontryagin product. The answer to this question is simple: $A_{Ab}=H_*(\Omega^{\infty+1}\Sigma^{\infty} X)$ and the canonical map is induced by the canonical $\Omega X\to \Omega^{\infty+1}\Sigma^{\infty} X$.

Note that the Lie algebra of rational homotopy groups $\pi_{*-1}M(X)$ must be $L(U(\pi_{*-1} X))$, where $U$ is the universal enveloping algebra and $L$ makes an associative algebra into Lie algebra with the commutator as the Lie bracket.

It seems that my question is related with this one.

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1 Answer 1

[I have replaced an earlier and less complete answer]

The question depends on the precise meaning of $A_{ab}$. In the derived world, if we kill commutators then we create new potential commutators and so do not immediately end up with something commutative. Killing commutators once is the same as taking Hochschild homology. I think it is now known from work of Smith and McClure that when you apply HH to something with an action of the little $k$-cubes operad $C(k)$, you always get something with an action of $C(k+1)$ (ie it is "one step more commutative"). An associative algebra has an action of $C(1)$, so you can apply HH repeatedly and pass to a colimit to get something with an action of $C(\infty)$, which is an $E_\infty$ operad. If we are working over $\mathbb{Q}$ then $E_\infty$ algebras are essentially the same as commutative algebras. I think that this procedure converts free associative algebras to free commutative algebras.

UPDATE: David Ben-Zvi is right that I have misremembered results that actually apply to Hochschild cohomology (rather than homology) here. Nonetheless, what I said about repeatedly killing commutators makes some kind of intuitive sense, so I still wonder whether something along those lines could be true.

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Dear Neil, thank you for your answer. It seems that my question was not crystal clear, I already edited it. The functor I need is not derived quotient by the space generated by commutators, which is the Hochschild homology, but quotient by the two–sided ideal generated by commutators. For example, it sends a free associative algebra to free commutative algebra generated by the same set of generators. –  nikitamarkarian Apr 21 '12 at 18:15
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Dear Neil, the edited version is very interesting. But could you clarify some points, please. Firstly, as far as I know the Hochschild homology functor makes an algebra less commutative, from a $e_n$ algebra one get a $e_{n-1}$-algebra. Then, it is not clear for me how does it related with my question. The algebra in hand is cohomology of something, it is already commutative. Then I forget about it, take its free resolution and then take quotient by the ideal generated by commutators. The question is: does this procedure have some geometric sense? –  nikitamarkarian Apr 25 '12 at 13:10
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The results being discussed making algebras more commutative (generalized Deligne conjecture) are for Hochschild cohomology rather than homology (ie center, rather than cocenter or trace). The question about abelianization in this stronger sense is very interesting. Essentially it seems you're asking about the derived algebraic geometry version of Kapranov's picture of formal NC geometry via commutator expansions. –  David Ben-Zvi Apr 25 '12 at 18:10
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Dear David, thank you for your comment. I thought about a connection between abelianization and Kapranov's picture. The abelianization functor has an adjoint one, it's simply inclusion of commutative algebras into associative ones. Composition of this adjoint functors gives us a comonad in the category of commutative algebras, and any associative algbra gives a comodule over this comonad. This is another way to look at associative algebras "from point of view" of commutative ones. I don't know how this observation could be useful. –  nikitamarkarian Apr 27 '12 at 16:07

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