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

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!

Let $X$ be a smooth projective variety over $\mathbb{C}$. I call (following Swan) Hochschild cohomology of $X$ the garded 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 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 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!

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!

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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 garded 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 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 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!