If $X$ fails to be holomorphic, what is $\mathcal{L}_X \bar\partial - \bar\partial \mathcal{L}_X$?

Question

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 $\bar\partial X \neq 0$.

Answer 1

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 $\alpha$.