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 $\bar\partial X \neq 0$.

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 $\bar\partial X \neq 0$.

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) holomrphic? 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 $\bar\partial X \neq 0$.

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 $\bar\partial X \neq 0$.