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

Hello to all.

There is a well-known formalism in deformation-quantization which puts the algebraic structure of polyvector fields in a noncommutative setting. Tamarkin-Tsygan define a (pre)calculus to be the data $(G, M, L, i)$ where $(G,.,[,])$ is a Gerstenhaber algebra, $M$ is a complex, $i$ is an action of the graded algebra $(G,.)$ onto the graded module $M$ and $L$ an action of the graded Lie algebra $G[1], [,]$ onto the graded module $M$.

These two actions have to satisfy the following two compatibilities: $i_{[a,b]}= [L_a, i_b]$ and $L_{ab}= L_ai_b+(-1)^{\vert a\vert}i_aL_b$. Is there a way to interpret this second rule in a more natural way or does one just impose this rule because it is satisfied by the protopical examples (ie. polyvector fields and hochschild (co)homology)?

share|improve this question
    
By "i is a DG-action of G onto M", I presume that you mean an action as a commutative algebra? –  Theo Johnson-Freyd Jul 27 '11 at 2:05
    
You're absolutely right. I edited the question –  Louis Jul 27 '11 at 9:00

2 Answers 2

Here is a sketch of topological description of a Tamarkin-Tsygan precalculus.

Consider the compactified configuration spaces $C_n$ and $D_{1,n}$ of $n$ points on $\mathbb{R}^2$ and $\mathbb{R}^2-\{(0,0)\}$, respectively. The collection $(C_n,D_{1,n})_n$ form a topological (colored) operad.

Claim: the collection $(H_{-\bullet}(C_n,\mathbb{Q}),H_{-\bullet}(D_{1,n},\mathbb{Q}))_n$ is the operad of Tamarkin-Tsygan precalculi.

It is well-known that $(H_{-\bullet}(C_n,\mathbb{Q}))_n$ is the Gerstenhaber operad. Then the operations $L$ and $i$ are the two classes (of respective degrees $1$ and $0$) in $H_{\bullet}(D_{1,1},\mathbb{Q})=H_{\bullet}(S^1,\mathbb{Q})$.

The two compatibility conditions can be understood as identities in $H_1(D_{1,2},\mathbb{Q})$.

If you want to get the operator $d$ (i.e. get a calulus rather than a precalculus) you'll have to replace $D_{1,n}$ by its semi-direct product with $S^1$.

See also Section 11 (especially $\S 11.3$) of this paper by Kontsevich and Soibelman for a similar point-of-view and its relation to a generalization of Deligne's conjecture.


EDIT: there is also an algebraic motivation for the compatibility conditions, which is explained in the original paper of Tamarkin and Tsygan. Namely, if $A$ is a Gerstenhaber algebra then $A[\epsilon]$, with $deg(\epsilon)=1$, is a Gerstenhaber algebra with modified product $a*b=a\cdot b+(-1)^{|a|}\epsilon[a,b]$. And if $(A,M)$ is a precalculus then $M$ becomes a Gerstenhaber module over $A[\epsilon]$ with $$ (a+\epsilon b)*m=(-1)^{|a|}i_am\quad\textrm{and}\quad [a+\epsilon b,m]=L_am+i_bm $$

share|improve this answer

Well, this is probably not the deep insight you are looking for, but if you consider the polyvector fields on a manifold, you can first define the Lie derivative $L_X$ of a polyvector field $X$ on differential forms by extending Cartan's formula, i.e. $L_X = [i_X, d]$. Note that there is a certain ambiguitiy here concerning signs, this version seems to work fine :)

Having this, the second formula $L_{X \wedge Y} = i_X L_Y + (-1)^Y L_X i_Y$ is a matter of computations. Interesting in this approach is that the Schouten (=Gerstenhaber) bracket of polyvector fields can be defined by the relation $i_{[X, Y]} = [L_X, i_Y]$. In fact, this gives perhaps the best definition of the Schouten bracket in differential geometry...

Added later:

a rough idea why this works should be as follows: $L_X$ is a differential operator of order one (with the usual super-signs) and $i_X$ is of order $0$. So their commutator is again an order $0$ differential operator on the module of differential forms. Now the non-trivial thing to show is that all order zero operators are such insertion operators. Hence the commutator has to be the insertion with some polyvector field, denoted by $[X, Y]$. Using the algebraic relations already known it is now a matter of computation that $[X, Y]$ satisfies the Gersterhaber bracket relations. Thus it defines a Gerstenhaber algebra structure.

Now in general this at least suggest that $[L_X, i_Y]$ defines a Gerstenhaber bracket on your algebra. Hence you only require that it coincides with the given one. Of course, here you need to know that $X \mapsto i_X$ is injective etc. but I guess this captures the idea...

share|improve this answer
    
Stefan, I've been mulling over your (very interesting) comments. First, you say that in the case of a calculus the second rule follows from the first one. As to your second observation, it indeed gives a very elegant interpretation of the Schouten bracket, I'm not sure however if this is applicable in the general setting of a precalculus. It is interesting though that the second rule is required to prove the symmetry of the bracket [X,Y] in your construction –  Louis Jul 29 '11 at 11:18
    
Hi Louis. Well, I'm not too familiar with the precise setting of the Tamarkin-Tsygan calculus. It is only in the case of a manifold, that the Schouten bracket can actually be defined this way. I guess you will need several (technical?) assumptions to yield that in the general case as well. I have not checked any sort of details. But I guess that this points already into the direction that the required axioms are not completely independent under some suitable assumptions on the nature of the module (non-degenerate enough...?) –  Stefan Waldmann Jul 29 '11 at 14:47
    
I was also thinking that if $G$ is nondegenerate and $M$ faithful, the first rule should imply the second.I'll keep scribbling, hopefully something comes out of it. Thanks for the discussion. –  Louis Jul 29 '11 at 16:22

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.