Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Commutative Poisson algebras $A$ can be thought of as commutative algebras equipped with a first-order deformation into a noncommutative algebra given by the Poisson bracket. A simple example is the symmetric algebra $S(\mathfrak{g})$ of a Lie algebra, which can be deformed into the universal enveloping algebra $U(\mathfrak{g})$.

Today I learned that Poisson algebras also appear in algebraic topology as follows:

  • If $X$ is a pointed space, the iterated loop space $\Omega^d(X)$ is an algebra over the little $d$-disks operad.
  • Hence the homology $H_{\ast}(\Omega^d(X))$ is an algebra over the homology of the little $d$-disks operad.
  • The homology of the little $d$-disks operad is an operad $\text{Pois}^d$ whose algebras are (graded commutative) Poisson algebras where the bracket has degree $1 - d$.

(I may have that last statement slightly wrong.)

Can these two points of view be related?

For example, is it known whether $H_{\ast}(\Omega^d(X))$ is naturally the associated graded of some filtered algebra $F(X)$ such that the Poisson bracket arises from the (super) commutator in $F(X)$?

share|improve this question
1  
I'm not entirely sure what you're looking for in the body of the question, but your "For example" admits a silly affirmative answer as follows. Going from chains to homology is a passage to an associated graded (for the "Postnikov filtration" $F_k V= \tau_{\geq k} V$). The $d$-Poisson algebra $H_*(\Omega^d X)$ is the associated graded of the $E_d$-algebra $C_*(\Omega^d X)$. (Here by $E_d$ I mean the operad of chains on the little $d$-disks operad. It's a filtered operad in the manner above, $C_*(\Omega^d X)$ is then a filtered algebra over this operad $E_d$, etc. etc.) –  Anatoly Preygel Sep 3 '11 at 1:57
    
@Anatoly: thanks! I don't think I asked the question I meant to ask; I specifically want the Poisson bracket to come from the commutator of this filtered algebra. Can you explain a little more about the Postnikov filtration? –  Qiaochu Yuan Sep 3 '11 at 5:12
add comment

3 Answers

up vote 6 down vote accepted

Let me start by rephrasing what is already in the answers of David Ben-Zvi and Theo Johnson-Freyd. The DG $\mathbb{Q}$-linear operad $\mathbb{E}_n:=C_{-\bullet}(E_n,\mathbb{Q})$ is filtered. For $n\geq2$ the filtration is the degree filtration, and thus $gr(\mathbb{E}_n)=H_{-\bullet}(E_n,\mathbb{Q})={\rm Pois}^n$.

The situation for $n=1$ is a bit different. We know that $\mathbb{E}_1\cong {\rm As}$ (this is the formality theorem for $E_1$ which, contrary to the case when $n\geq2$, is easy to prove). The operad ${\rm As}$ of associative algebras is also filtered, but in a less obvious way. To be short, one assigns the following two-step filtration onto ${\rm As}(2)=\mathbb{Q}[\Sigma_2]$ (which generates ${\rm As}$): $$ F^0{\rm As}(2)=\mathbb{Q}(1-\sigma)\subset F^1{\rm As}(2)={\rm As}(2). $$ It then an exercise to check that $gr({\rm As})={\rm Pois}^1$.

Then, in order to relate the two stories, I have the feeling that one does not need to invoque the formality of $E_n$ for $n\geq2$. Given a filtered $\mathbb{E}_n$-algebra $A$ (i.e. a filtered DG $\mathbb{Q}$-vector space equipped with an action of $\mathbb{E}_n$ that is compatible with the above filtration), then $gr(A)$ is a ${\rm Pois}^n$-algebra.

Concerning the last example in the question, one has to take $A=C_{-\bullet}(\Omega^d(X),\mathbb{Q})$ equipped with the degree filtration. Then $gr(A)=H_{-\bullet}(\Omega^d(X),\mathbb{Q})$ is going to be a ${\rm Pois}^d$-algebra.

Side remark: Observe that the story for $E_0$ is even more degerated. Nevertheless,deformation theory of $E_0$-algebras is still very interesting (for a discussion about this issue and its relation to the BV formalisms, see Costello-Gwilliam work-in-progress http://math.northwestern.edu/~costello/factorization_public.html - especially 5b and 5c).

share|improve this answer
    
I'm not familiar enough with any of this material that my opinion is trustworthy, but I think Ben Wieland's comment on David Ben-Zvi's answer still applies: it sounds like this abstract machinery produces a Poisson bracket that doesn't come from the (super?) commutator on an $\mathbb{E}_n$-algebra, but that comes from some kind of special commutator already present in the $\mathbb{E}_n$-structure. Is this accurate? $A = C(\Omega^d(X), \mathbb{Q})$ comes with some kind of (super?) commutator which passes to $\text{gr}(A)$. Is this the $\text{Pois}^d$ bracket? –  Qiaochu Yuan Sep 12 '11 at 16:01
2  
@Qiaochu: I said it badly in my answer, but the Pois^d bracket absolutely comes from a "commutator" in the E_n algebra. You just have to modify what you mean by "commutator". The correct definition of "the commutator" in an n-algebra is the 2-ary operation corresponding to the fundamental class of the (n-1)-sphere (this sphere is the space of 2-ary operations). When n=1, the 0-sphere consists of two points, and the fundamental class is the one that signs the two points differently. –  Theo Johnson-Freyd Sep 12 '11 at 17:25
    
@Theo: thanks! Alright, so it looks like the answer is something like "this is true in a more meaningful sense than the sense in which you wanted it to be true," which I guess I can get behind. –  Qiaochu Yuan Sep 12 '11 at 22:05
add comment

I should begin by apologizing for rambling on a bit. You've asked questions that are closely related to things I've been thinking about, but I don't know all the literature well, and I clearly haven't figured out how to say things concisely.

I will write $E_d$ for the operad (in spaces) of little $d$-disks, and $G_d$ ("$G$" for "Gerstenhaber") for what you're calling $\operatorname{Pois}^d$. (Because the word "$d$-Poisson algebra" appears at least in some papers by Cattaneo and collaborators, but at least the early ones get the sign wrong, and call by "$0$-Poisson" what should be called "$1$-Poisson".) Namely, $G_d$ is an operad in graded vector spaces generated by a binary "multiplication" in (homological) grading $0$ and a binary "bracket" in grading $d-1$ (and a zero-ary "unit" in grading $0$), such that the multiplication (and unit) is unital commutative, and such that the bracket is Lie up to shifting by $d-1$, and such that the Leibniz rule is satisfied. (In the correct sign conventions, saying that the bracket is "Lie after shifting" mean that it satisfies a Jacobi identity and is antisymmetric for odd $d$ and symmetric for even $d$.)

Begin with the operad $E_d$. The space of binary operations is a $d$-sphere. When $d\geq 1$, the $d$-sphere is nonempty, and its (integer!) homology is free on two generators, one in degree $0$ and the other in homological degree $d-1$. When $d=1$, please work in a setting where $2$ is invertible, and switch from the basis of points to the basis consisting of the average of the two points in the $0$-sphere and the difference. Let me call by "the bracket" any binary operation representing the degree-$(d-1)$ generator in homology (the fundamental class) and by "the multiplication" any binary operation representing the degree-$0$ class in homology. Let me also linearize by taking chains; so I will work with the operad of chain complexes $\mathrm C_\bullet(E_d)$, but write "$E_d$-algebra" for "$\mathrm C_\bullet E_d$-algebra".

Then the first observation is that the bracket in a fairly precise way measures the failure of the multiplication is be commutative. The second observation is that the bracket distributes over itself up to a contractible space, and is (anti)$^d$commutative, where "(anti)$^d$" means "anti" when $d$ is odd, and is empty when $d$ is even. So in fact if you throw away the multiplication and use only the bracket, there is a map to $\mathrm C_\bullet(E_d)$ from $\operatorname{Lie}[d-1]$, where by "$\operatorname{Lie}[d-1]$" I mean the Lie operad with the bracket shifted into homological degree $d-1$. And I mean that this map exists in some infinity sense. The point is that any $E_d$-algebra, if you remember only the action of the $d$-sphere, is a Lie algebra up to shifting by homological degree $(d-1)$, and up to replacing points with contractible spaces and all that. "Strongly homotopy" is a word that should now be bandied about.

Ok, so we can go a bit further. If you have a family of $E_d$-algebras over the (formal) disk $\operatorname{Spec}(k[\![\hbar]\!])$ that degenerates at the origin $\hbar=0$ to an $E_\infty$-algebra (recall that there are inclusions of operads $E_d \hookrightarrow E_{d+1}$, and the direct limit $E_\infty$ has a contracible space of $k$-ary operations for any $k\geq 0$, and so is "strongly homotopically" the commutative operad), then among other things you have a family of $\operatorname{Lie}[d-1]$-algebras that degenerates to the $0$-algebra at the origin. Then rescale the Lie bracket by $\hbar^{-1}$; now it need not vanish at $\hbar = 0$. The (homotopy-)associativity in the $E_d$-algebra implies a (homotopy-)Leibniz rule. So at the origin actually you have a (strongly homotopy) $G_d$-algebra.

This is all a long story to explain something short, which David Ben-Zvi has already alluded to: $G_d$ "is" the operad of "first order" $E_d$ algebras. Of course, I haven't at all shown that there aren't maybe further relations, or (which is the same thing!) further operations that survive (maybe after rescaling by $\hbar^{-1}$) at the limit. Essentially, the proof (that I know of) that there aren't any only applies in $d\geq 2$, and only in characteristic $0$: the celebrated fact that the homology $\mathrm H_\bullet(E_d)$ is on the nose $G_d$, as operad of chain complexes. I'm pretty sure, but don't know a reference, that this fails in non-zero characteristic. Point, is, when $d \geq 2$ it is clear that $\mathrm H_0(E_d)$ is (chains on) the commutative operad, because the other operations are all far away from $0$ in homological degree, and so there is a map $\mathrm H_\bullet(E_d) \leftarrow G_d$ in any characteristic when $d\geq 2$. It is somewhat remarkable that over $\mathbb Q$ this map is an iso; evidence that it is remarkable is that this fact comes after many wrong proofs.

Note that having a map $G_d \to \mathrm H_\bullet(E_d)$ is a far cry from having a map $G_d \to \mathrm C_\bullet(E_d)$. Sure, you can chose generators of homology and try to lift from $\mathrm H_\bullet$ to $\mathrm C_\bullet$, but there are spaces of ways to do this that are not contractible, and so you won't in general be able to make the lift be one of operads-up-to-strongly-homotopy. What I've tried to explain is a map from $G_d$ to the operad of "$E_d$-algebras over the formal disk that are $E_\infty$ at the point". (Recall: maps of operads go opposite to "forgetful" maps.)

Anyway, I'd like to end with a few comments on the too-celebrated "formality theorem". Formality is stronger than but implies the statement that every $G_d$-algebra is the first-order part of an $E_d$-algebra. Here's an example. Let me work with categories enriched over $\mathbb Q$. A Casimir category is a symmetric monoidal category equipped with a natural isomorphism between $\otimes$ and $\otimes\circ \operatorname{flip}$ satisfying something like Jacobi and Leibniz. Then formality with $d=2$ implies that every Casimir category is the first-order part of a braided monoidal category.

When $d=2$, Tamarkin pointed out that formality is the same as existence of something called "Drinfeld associators". $d=2$ is the number I understand the best, because Drinfeld associators are a bit older than the general formality theorem (to which I don't believe there is a proof over $\mathbb Q$; at least, Lambrechts and Volic, in their excellent paper on Kontsevich's proof of formality, say that for the (correct) unital versions of the operads, the trick that gets from Kontsevich's proof, which is over the periods, to something over $\mathbb Q$, doesn't work). Proofs of formality, i.e. Drinfeld associators, are the points of a scheme (or something like one — maybe I want to work more homotopically, and maybe I want projective limits, and ...). By "proofs of formality" I mean more or less "morphisms of operads". Anyway, this scheme (at least for $d=2$) is non-empty over $\mathbb Q$ (and for all $d$ it is nonempty over the periods), but it is never "contractible", whatever that means for this type of scheme-like object. So you really are making a choice when you invoke formality, and some choices are better than others. For example, some are more amenable to explicit computations, some satisfy extra symmetries, etc.. And there is no getting around this, because the groups of automorphisms of $G_d$ or $E_d$ are not contractible, and the space of isomorphisms is, of course, a bitorsor for these two groups.

Ok, a conclusion. I'm pretty sure the answer to your last question is "yes". But I'm also pretty sure that it hasn't been written down, really. (I would love to be told otherwise!) At least, I'm not aware of a paper that explains well the situation when $d\geq 3$. When $d=2$, there is plenty, including all the work by Etingof and collaborators on associators and Lie bialgebra (bi)quantization; these papers tend to be long, but understandable on about the fourth read. Some of it is well-reviewed by Bar-Natan, who is an excellent writer. Tamarkin's papers are short, and I find hard to read, but have a lot in them. I do highly recommend the papers by Severa on this stuff, and in particular his write-up of Tamarkin's work. A last part of the $d=2$ story is related to words like "Alekseev Torossian" and "Kashiwara Verne", and I don't understand any of that part.

share|improve this answer
    
Actually, noncontractibility is also nuanced. When $d=2$, there is a "properad" that presents the notion of "finite-dimensional Lie bialgebra", and you can pose the quantization problem for these, and there is a scheme of quantizations, and a map from the scheme of Drinfeld associators to this scheme. Etingof and Enriquez show that this map has image just one point. But they do not, so far as I know, show that the scheme of "finite dimensional quantizations" is contractible; just that all "possibly-infinite dimensional quantizations" act the same on finite-dimensional Lie bialgebras. –  Theo Johnson-Freyd Sep 12 '11 at 4:25
    
The paper of Severa is arxiv.org/abs/0902.3576. He wrote another relevant paper with Willwacher, about formality when $d=2$: arxiv.org/abs/0905.1789 By the way, did Tamarkin really prove that formality for $d=2$ is the same as existence of Drinfeld associators ? –  DamienC Sep 12 '11 at 9:46
    
@DamienC: Well, erm, I don't remember the literature well enough. Tamarkin certainly shows that having an associator gives a proof of formality when d=2. Conversely, he shows that any proof of formality gives a Lie bialgebra quantization theorem. But that's really not very far from an associator, maybe without some side conditions; in particular, a sufficiently "universal" quantization certainly includes an associator, by an old result recently written up by Bar-Natan and Dancso. –  Theo Johnson-Freyd Sep 12 '11 at 17:18
    
Or I think you can run Tamarkin's associator->formality proof in reverse, and at least get a some slightly homotopy-ized version of Drinfeld associator. –  Theo Johnson-Freyd Sep 12 '11 at 17:19
    
@Theo: you are right! I actually think that you get a Drinfeld associator stricto sensu. Tamrakin$^{-1}$ can be guessed from the paper of Severa and Willwacher (arxiv.org/abs/0905.1789), where they compare the comstruction of Tamarkin and the one of Kontsevich. –  DamienC Sep 13 '11 at 11:47
add comment

Hopefully an expert will clarify/correct this naive answer, but it seems to me that it follows from the formality theorem [NB: in characteristic zero] that a first order deformation of a commutative algebra into an $E_n$ algebra (algebra over little n-discs, in chain complexes, for $n>1$) is equivalent to a graded Poisson structure extending the commutative multiplication (the deformation being simply given by rescaling the bracket to zero --- this is just an imprecise version of Anatoly's comment). In that sense the $E_n$ and $E_1$ (or associative) cases look very different: the "linearization" of $E_n$ at commutative algebras is $E_n$, while for $E_1$ this is manifestly not the case - associative and Poisson algebras look quite different! In other words, the formality theorem trivializes the problem of $E_n$-quantization.

Note that this is something special about chain complexes or graded vector spaces (in particular a stable phenomenon) - I don't think there's any sense in which this picture extends to $E_n$ algebras in more general ($\infty$-)categories -- for example a braided tensor (aka $E_2$) category isn't just a symmetric tensor category + a braiding operator. In fact the Lie operad itself is just a stable phenomenon AFAIK - it's not an operad in topological spaces, so one can't even formulate the question of writing "$E_n= E_\infty + Lie[n-1]$" more generally.

share|improve this answer
    
I think Qiaochu asks a completely different question. You talk about a fancy $E_n$-commutator that isn't just $ab-ba$, but involves a sphere of "orders." And you talk about ordinary defms, algebras over $Q[[e]]$, where $e$ has degree 0. I think he asks about algebras where $e$ has degree $1-d$. A $d$-Poisson structure is close to an assoc $Q[e]/e^2$-algebra with com fiber. The usual theory says that it extends to a $Q[e]$-algebra, at least if we're graded not chains. Does this defm have a nice interp? What about parity of 1-d? $Q[e]$ isn't always diff from square-zero. –  Ben Wieland Sep 3 '11 at 19:30
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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