Let $X$ be a complex manifold and $TX$ its tangent bundle. The Atiyah class $\alpha(E)\in \text{Ext}^1(E\otimes TX, E)$ for a vector bundle $E$ is defined to be the obstruction of the global existence of holomorphic connections on $E$. In particular if we take $E=TX$ to be the tangent bundle of $X$ itself, we can prove that the Atiyah class $\alpha(TX)\in \text{Ext}^1(TX\otimes TX, TX)$ and this gives a Lie algebra structure $$ TX[-1]\otimes_{\mathcal{O}_X} TX[-1]\rightarrow TX[-1] $$ in the derived category $D(X)$

Now we can form the symmetric algebra of $TX[-1]$. Since $TX$ has been shifted by degree $1$, we get the symmetric algebra in odd parity, which is anti-symmetric: $S(TX):=\bigoplus_i \bigwedge^i_{\mathcal{O}_X}TX[-i]$

For the details see M, Kapranov's paper "Rozansky–Witten invariants via Atiyah classes http://arxiv.org/abs/alg-geom/9704009" and N. Markarian's paper "The Atiyah class, Hochschild cohomology and the Riemann-Roch theorem http://arxiv.org/abs/math/0610553".

We remember that for an ordinary Lie algebra $\mathfrak{g}$, the Lie bracket on $\mathfrak{g}$ extend to a natural Poisson bracket on $S(\mathfrak{g})$, by the Leibniz rule.

This Poisson bracket is useful. For example, it gives the first order deformation from $S(\mathfrak{g})$ to the universal enveloping algebra $U(\mathfrak{g})$. An easy property of the Poisson bracket is that it vanishes on the invariant sub algebra $S(\mathfrak{g})^{ \mathfrak{g}}$

On $S(TX)$ we can also define the invariant part. Here it is given by the derived Hom $$ RHom(\mathcal{O}_X,\oplus\wedge^i_{\mathcal{O}_X}TX[-i])=\bigoplus_{i,j}H^j(\wedge^i_{\mathcal{O}_X}TX) $$ The reason we define the invariant part like this is that $\mathcal{O}_X$ is the "trivial $TX[-1]$ module" in $D(X)$. See D. Calaque, A. Caldararu and J. Tu "PBW for an inclusion of Lie algebras http://arxiv.org/abs/1010.0985" for more details.

$\textbf{My question}$ has two part.

First, could we extend the Lie bracket given by the Atiyah class to $S(TX)=\bigoplus\bigwedge^i_{\mathcal{O}_X}TX[-i]$? Intuitively we can process using Leibniz rule as we did $S(\mathfrak{g})$, but since we are working in the derived category, I don't know whether there will be any problem in our construction.

Secondly, if we can extend the Lie bracket to a Poisson bracket on $\bigoplus\bigwedge^i_{\mathcal{O}_X}TX[-i]$, does it vanish on the "invariant part"?

Now I don't even know how to make sense of the statement "The Poisson bracket vanishes on the invariant part" because the invariant part is not even a subset of $\bigoplus\bigwedge^i_{\mathcal{O}_X}TX[-i]$. So we cannot simply "restrict" the Poisson bracket to the invariant part. But could we make the statement in some other sense?


1 Answer 1

  1. For the first part of your question, the answer is yes. It is simply that $D(X)$ is a symmetric monoidal $k$-linear category, hence for any Lie algebra object $V$, $Sym(V)$ is a Poisson algebra object.

  2. For the first part of the second part of your question, here is how to get a Poisson algebra structure on $R\Gamma(Sym(TX[-1]))$: it comes from the fact that $R\Gamma(-)=H^*(-)$ is a symmetric monoidal functor from $D(X)$ to $D(k-mod)$, hence it sends Poisson algebras to Poisson algebras.

  3. For the second part of the second part of your question, I would have to think more about it, but it should be a refinment of Kapranov's proof that $R\Gamma(TX[-1])$ is abelian (this is proposition 2.3.2 in Kapranov's paper).


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