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The following question came up in a discussion the other day and I have been wondering whether something is known about it. Everything below takes place over $\mathbb{C}$. I don't have the expertise to know if this is trivial or of interest. Suppose a commutative dga has a free-commutative model $(\wedge V , d)$ where V is a finite dimensional vector space.

Recall that $T^{poly}$ is the Lie-algebra of polyvector fields on $\wedge V$ (yes, everything is superized as V will be in general graded) with Schouten bracket. Part of Kontsevich's formality theorem says that the HKR map $ T^{poly} \to HC^*$(Hochschild cochains) is the first Taylor coefficient in an $L_\infty$ quasi-isomorphism between the two.

We can think of the derivation $d$ as corresponding to a vector-field $v$. It follows from a spectral sequence argument that the HKR map gives a quasi-isomorphism: $$ (T^{poly},[v,-]) \to HC( \wedge V,d)$$

Question: Can this map be upgraded to a map of $L_\infty$ algebras?

Certainly, the Taylor coefficients in the usual formality map must be doctored.

A related statement that does seem to be true and standard is that there is an $L_\infty$ quasi-isomorphism $(T^{poly}[[t]],[tv,-]) \to HC^*(\wedge V[[t]],td )$ Thus, the question is in some reasonable sense about convergence of this isomorphism. Maybe one can prove the claim by a close inspection of Kontsevich's integral formulas. Based upon these facts, however, it seems plausible to me that that the statement is in general false, but I was unable to come up with a counterexamples or an a priori reason (I didn't try too hard however). Is it true for some more restrictive group of commutative dg algebras, for example pure Sullivan algebras?

Update: Having finally looked at the Kontsevich formulas, I'm beginning to think there are some simple counting reasons that make the above formula converge, but am not sure that $f_1$ stays the same (though I believe it remains a quasi-iso). Any confirmation or help would be great. Otherwise, I'll keep thinking and update again.

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Hi Daniel - I edited this to replace the L($\infty$) by $L_\infty$, because that is how it is usually written, and also to add $-$ where you had a blank argument in some brackets. Please check these latter edits haven't messed anything up. –  David Roberts Feb 23 '11 at 5:31
    
One more thing, if you figure it out, you can add an answer to your own question (and even accept it!), which would be better than just updating the question to include the answer you found. –  David Roberts Feb 23 '11 at 5:32

1 Answer 1

up vote 1 down vote accepted

I think this does only work for so-called homological vector fields, i.e. vector fields of degree 1 which self-commute. Then you have $[v,v]=0$, which is the Maurer-Cartan equation in $T_{poly}$, and which insures that the corresponding derivative $d$ squares to zero.

Dealing with a Maurer-Cartan element you can simply use it to twist Kontsevich's formality $L_\infty$-quasi-isomorphism. When the Maurer-Cartan element is a vector field we know the explicit form of the first Taylor component of the twisted $L_\infty$-morphism: it is not HKR, but involve Bernoulli numbers (see e.g. https://www.math.ethz.ch/u/calaqued/research/LecturesDufloETH.pdf).

Anyway, the $k$-th taylor component of the $v$-twisted $L_\infty$-morphism will be given by the series $$ \phi_v^{(k)}(u_1,\dots,u_k):=\sum_{l\geq0}\frac{t^{l+1}}{(l+1)!}\phi^{(k+l)}(u_1,\dots,u_k,\underbrace{v,\dots,v}_{l~times}) $$ To conclude one just have to observe that $\phi^{(k)}$ preserves the grading given by the arity minus $2$ (arity means the number of arguments for poly-vectors and poly-differential operators). Therefore each time $v$ appears in the formula it decreases this degree by $1$.

EDIT: I include a comment into the main answer about the shape of $\phi^{(1)}_v$ in specific cases. Let me denote coordinates by $u^k$, and write $\partial_k=\frac{\partial}{\partial u^k}$ and $v=\sum_iv^i\partial_i$. We consider a matrix-valued one-form $\Xi$ given by $$ \Xi_i^j=\sum_k\partial_k\partial_iv^jdu^k, $$ and define $$ \Theta=\sum_{n>0}c_n\iota_{tr(\Xi^n)}. $$ Here $c_n$ are rational coefficients that do not matter.

Then $\phi_v^{(1)}$ is given by the precompsition of HKR with $e^{\Theta}$.

Now if I split the coordinates into even $x^i$ and odd $e^i$, and if I assume that $v=\sum_iv^i(x^1,\dots,x^m)\frac{\partial}{\partial e^i}$, then the matrix $\Xi$ has non-zero entries only in the right-up block. In particular it is upper triangular, the trace of any power of it is therefore zero, and thus $\phi^{(1)}_v=HKR$.

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Hi Damien, thanks for your great reply! This was the argument I found in my last paragraph(it's great to have a specialist confirming this) I had suspected that the maps f_1 were different than HKR but wasn't able to get started with the calculations. I have a specific case of interest, when my algebra is pure Sullivan e.g is of the form $k[x_1,\ldots,x_n]\otimes \wedge(e_1,\ldots e_m)$ with $v= \sum f_i(x_1,\ldots x_n)d/de_i $ where the variables $x_i$ are even and the $e_i$ are odd Do you know off hand if the map $f_1$ fails to agree with HKR even in this case? –  Daniel Pomerleano Apr 26 '11 at 23:39
    
I guess if I have interpreted your beautiful formula 9.8 correctly, the answer is "yes"? –  Daniel Pomerleano Apr 27 '11 at 2:37
    
Hi Daniel. If your vector fields does only derivates in the odd directions with coefficients depending only on even coordinates, then it seems ot me that in this specific case the corrections to HKR are trivial. Namely, we define the matrix valued 1-form $\Xi$ by $\Xi_i^j=\sum_k\partial_k\partial_iv_ju^k$ (where $u^k$ are coordinates and $v=\sum_j\partial_j$). Now if you split your variables into odd and even ones, and if $v$ the above specific form, then you see that this matrix has non-zero entries only in the right-up block. In particular the trace of any power of it is zero. –  DamienC Apr 27 '11 at 12:11

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