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I am reading Kontsevich' famous paper on deformation quantization of Poisson manifolds. In section 1.4.2 on page 4 he gives the general formula for the star product associated to a Poisson structure on $\mathbb{R}^n$ up to and including order 2. Now I have noticed that in the second order term, terms of the form (for example) $\partial_k(\alpha^{ij})\partial_i(\alpha^{kl})\partial_j(f)\partial_l(g)$, and I do not understand why. This term would correspond to the following graph:

graph

I.e., there is an edge going from 1 to 2 and an edge going from 2 to 1; all such terms would be of the form $\partial_j(\alpha^{i\cdot})\partial_i(\alpha^{j\cdot})\times\text{derivatives of $f$ and $g$}$, and none of them occur in his formula in section 1.4.2. As far as I can see, such graphs would be admissible under the rules on page 5 in the paper. I have yet to find a way to evaluate the weight associated to this kind of graphs, but at least the integrand of the weight for the graph above is nonzero. So my question is:

Why do terms of this form not contribute to the second order term of the star product?

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Using a remark in this recent paper by Thomas Willwacher as a hint, I found out. The problem is solved by the following: there exists a gauge transformation $D$ that transform the extra term away. To see this, we first need the second order term of a gauge transformed star product: If $\star = I + \hbar^nB_n$ and $D = I + \hbar^nD_n$, and $f\star'g = D^{-1}(Df\star Dg)$ then $$ B_2'(f,g) = B_2(f,g)+B_1(f,D_1g)+B_1(D_1f,g)-D_1(B_1(f,g))+D_1(f)D_1(g)- D_1(g D_1f)-D_1(f D_1g)+D_1(D_1(f g))+g D_2f+fD_2g-D_2(f g), $$ where, of course, $B_2'$ is the second order term of $\star'$. Setting $D_1 = 0$ reduces this to $$ B_2'(f,g) = B_2(f,g)+g D_2f+f D_2g-D_2(f g) = B_2(f,g) + d^H(D_2)(f\otimes g) $$ where $d^H$ is the Hochschild differential. (Also, $B_1' = B_1$.)

Now comes the crucial part: the term in the question $\partial_k(\alpha^{ij})\partial_i(\alpha^{kl})\partial_j\otimes\partial_l =: A$ is $d^H$-exact: indeed, $A = d^H\left(-\frac12\partial_k(\alpha^{ij})\partial_i(\alpha^{kl})\partial_j\partial_l\right)$. So if we set $D_2$ to minus the expression in the brackets, then, $A$ no longer occurs in $B_2'$ while the other terms of $B_2$ occur unmodified in $B_2'$.

Finally, expressions of the form $\partial_k(\alpha^{ij})\partial_i(\alpha^{kl})f\partial_j\partial_l(g)$ (corresponding to the diagram in the question but with both downwards arrows now pointing to $g$) also cannot occur in $B_2$ because $B_n$ is zero for any $n$ when one of its arguments is constant, and such a term would violate this rule.

Thus, the star product in section 1.4.2 in Kontsevich's paper does not equal his general formula up to order 2; instead, they are equivalent.

(By the way, the weight of the term corresponding to the diagram is, including multiplicity from similar graphs, equal to $-\frac16$ according to this paper by Giuseppe Dito. In particular, it is nonzero.)

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