# Tamarkin-Tsygan Formalism

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 $G, [,]$ 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)?

• By "i is a DG-action of G onto M", I presume that you mean an action as a commutative algebra? Jul 27, 2011 at 2:05
• You're absolutely right. I edited the question Jul 27, 2011 at 9:00

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 $$(a+\epsilon b)*m=(-1)^{|a|}i_am\quad\textrm{and}\quad [a+\epsilon b,m]=L_am+i_bm$$

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...

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...
• 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. Jul 29, 2011 at 16:22