# Multiplicativity twisted Hochschild Kostant Rosenberg isomorphism

Let $X$ be a smooth projective variety over $\mathbb{C}$. I call (following Swan) Hochschild cohomology of $X$ the graded algebra: $$\mathrm{HH}^{\bullet}(X) := \mathrm{Ext}^{\bullet}_{X \times X}(\Delta_* \mathcal{O}_X, \Delta_* \mathcal{O}_X),$$

where $\Delta : X \rightarrow X \times X$ is the diagonal embedding. By adjunction, we have $\mathrm{HH}^{\bullet}(X) = \mathrm{Ext}^{\bullet}_{X}(\Delta^* \Delta_* \mathcal{O}_X, \mathcal{O}_X)$. Using the Koszul complex locally for $X \subset X \times X$ and $N_{X/X \times X} \simeq \Omega_X$, one proves:

$$\mathrm{HH}^{\bullet}(X) \simeq \bigoplus_{p+q = \bullet} H^p(X, \bigwedge^q T_X).$$

This isomorphism of graded vector spaces is often called the Hochschild-Kostant-Rosenberg isomorphism. Note that this is NOT a ring isomorphism in general. One of Kontsevich's (numerous) insights was that twisting it with the square root of the Todd class of $X$ makes it a ring isomorphism.

I know that there are many references on the subject (Markarian, Ramadoss, Caladararu, Yekutieli, Calaque-VandenBergh etc...), but I am not able to follow their proofs.

Could someone give some hints on how to prove that twisting with the square root of the Todd class makes this isomorphism multiplicative? If possible, I would like to have a proof which is as low-tech as one can be.

I really thank you in advance for your help!

I am far from being expert in this subject, but I will try to present my understading of there this multiplicativity comes from. I wiil refer to authors you mention but only to the parts which I hope you will find readable.

First, let me give an interpretation of isomorphism $$\mathrm{HH}^{\bullet}(X) \simeq \bigoplus_{p+q = \bullet} H^p(X, \bigwedge^q T_X).$$ due to Markarian(https://arxiv.org/abs/math/0610553). Shifted tangent bundle $T_X[-1]$ is a Lie algebra object, the bracket given bu Atiyah class. Markarian shows that its universal enveloping algebra is precisely the algebra of Hohschild cochains $$U(T_X[-1])=\mathcal{RHom}^{\bullet}(\Delta_*\mathcal{O}_X,\Delta_*\mathcal{O}_X)$$

He defines universal enveloping algebra as an algebra admitting PBW isomorphism(Def 4) and in this case PBW is precisely the quasi-isomorphism between $Sym(T_X[-1])=\bigoplus_i\Lambda^i T_X[-i]$(symmetric power becomes exterior because of sign rule applied to shift by $(-1)$) and $\mathcal{RHom}^{\bullet}(\Delta_*\mathcal{O}_X,\Delta_*\mathcal{O}_X)$ that you sketched. See Section 7 of https://arxiv.org/abs/math/0602653 for an elaboration of Markarian's arguement. There is also proven a general statement that a(we do not a priori know that it is unique) universal enveloping algebra in the sense of Markarian actually satisfies the usual universal property.

Now, let us interpret taking hypercohomology in terms of Lie algebras. Any object $\mathcal{F}\in D(X)$ carries a functorial action of $T_X[-1]$ given by Atiyah class $$T_X[-1]\otimes \mathcal{F}\to \mathcal{F}$$ In particular, сonsidering $\mathcal{O}_X$ as a module over $T_X[-1]$ we get $$R\Gamma(X,\mathcal{F})=RHom_{T_X[-1]}(\mathcal{O}_X,\mathcal{F})$$ The RHS is precisely the Lie algebra cohomology of $T_X[-1]$ with coefficents in $\mathcal{F}$ because $Hom$ space from $\mathcal{O}_X$ is equal to $T_X[-1]$-invariants.

This reduces our problem to the following purely algrebaic problem: for a Lie algebra $\mathfrak{g}$ how one should change the PBW isomorphism so that it induces an isomorphism of cohomology rings $$H^{\bullet}(\mathfrak{g},S(\mathfrak{g}))\cong H^{\bullet}(\mathfrak{g},U(\mathfrak{g}))$$ The faсt that twisting by square root of Todd class does the work is proven in M. Pevzner and C. Torossian, "Isomorphisme de Duflo et cohomologie tangentielle", see http://math.univ-lyon1.fr/~calaque/LectureNotes/LectETH.pdf for an exposition.

• Thanks, this is a nice answer. But still I was wondering: isn't it possible to get a proof which stays in the realm of Algebraic Geometry? I mean, twisting with the square root of Todd class looks so close to Grothendieck-Riemann-Roch. Isn't it possible to get a proof of multiplicativity of twisted HKR using techniques similar to that in the proof of GRR? – Libli Sep 11 '16 at 12:46
• @Libli In Markarian's approach to Riemann-Roch Todd class comes as the measure of incompatibility of PBW with action of $\mathfrak{g}$ by multiplication, so maybe you can reverse his arguement, to obtain a proof which does not touch general Lie algebras, though to do this we need some fornal argument which relates appearence of Todd class in Markarian's and Kontsevich's theories. – SashaP Sep 11 '16 at 13:50
• @Libli: to this date, no one has succeeded in proving the existence of this ring isomorphism in a way similar to GRR. Note that GRR has to do with Hochschild homology (as opposed to Hochschild cohomology). – DamienC Jan 15 at 13:06