MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

I present a simplified version of the problem, and if that's easy to answer then someone can try answering the more complicated version.

Simplified version

Suppose $X$ is a tangent vector field on a complex manifold. Then we have the Dolbeault operator $\bar\partial$, and a well-defined $\bar\partial X \in \Gamma(T\oplus T_{0,1}^*)$. If $\bar\partial X = 0$, then the Lie derivative $\mathcal{L}_X$ commutes with $\bar\partial$. (They are acting on some holomorphic tensor bundle.)

But what if $X$ is not (quite) holomorphic? Then what is the operator $\mathcal{L}_X \bar\partial - \bar\partial \mathcal{L}_X$? Can we express it in terms of $\bar\partial X$ in a way that makes it manifestly clear that it vanishes when $X$ is holomorphic? If we don't have a nice expression, can we at least bound its $C^k$ norms in terms of the $C^{k+s}$ norms of $\bar\partial X$?

Different version

In the version I'm really interested in, $X$ is actually a complex Poisson bivector acting by the poisson differential $d_X$, i.e., by the Schouten bracket. In this case, we wouldn't expect that $\bar\partial X = 0 \;\implies\; \bar\partial d_X = d_X \bar\partial$, but rather $$\bar\partial X = 0 \;\implies\; \bar\partial d_X + d_X \bar\partial =0$$ and I would be looking for some information about $\bar\partial d_X + d_X \bar\partial$ when $\bar\partial X \neq 0$.

share|cite|improve this question
up vote 2 down vote accepted

I'll answer the simpler question. Since $J^2=-1$, it follows that for any vector field $X$, the endomorphism $L_X(J)$ anti-commutes with $J$. In other words, $L_X(J)$ is a section of $\bar{T}^* \otimes T$. This is the same space where $\bar{\partial}X$ lives and in fact they are equal, at least up to a constant factor which I forget right now.

Now we can work out $[L_X, \bar{\partial}]$ in terms of $[L_X, J]$: $$ [L_X, \bar{\partial}] = \frac{1}{2}[L_X, d + i J d] = \frac{i}{2}[L_X,J]\circ d $$ since $[L_X,d] = 0$. So up to a factor then, $[L_X, \bar{\partial}]$ is just $\bar\partial X \circ d$. From this it seems that the $C^k$-norm of $[L_X,\bar\partial](\alpha)$ is controlled by the $C^k$ norm of $\bar\partial X$ and the $C^{k+1}$-norm of $\alpha$.

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.