MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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" and N. Markarian's paper "The Atiyah class, Hochschild cohomology and the Riemann-Roch theorem".

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" 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?

share|cite|improve this question
  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).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.