Hi, 
as the title says i'm wondering if there's a "simple" and known commutation relation between the following two differential operators. Let $E$ be a holomorphic vector bundle over a compact kahler manifold $X$, fix a Hermitian metric $h$ on $E$ and a kahler metric $g$ on $X$. Denote with $D=D^{'}+D^{''}$ the associated Chern connection and its (1,0) and (0,1) parts. Denote with $\delta=\delta^{'}+\delta^{''}$ the adjoint respect to $h$ of $D$.
So my question is what is $[D^{''},\delta^{'}]$? Is there a relation like kahler relations?


Thank you in advance.



Is there a "simple commutation" relation between  $D^{''}$ and $\delta^{'}$,  with $D^{''}$ the (0,1) part of the chern connection of a vector bundle and $\delta^{'}$ the adjoint of the (1,0) part? 
Yes, the relation is that they anti-commute. 


Let's see this very briefly (you can find it in almost all books on Kähler geometry and Hodge theory).


We want to compute $[D''\_E,\delta'\_E]$, where $[\bullet,\bullet]$ is the graded commutator and let me call $\omega$ (instead of $g$) the Kähler form. Then one has that
$$
[\Lambda\_\omega,D''\_E]=-i\delta'\_E,
$$
so that 
$$
[D''\_E,\delta'\_E]=i[D''\_E,[\Lambda\_\omega,D''\_E]],
$$
where $\Lambda\_\omega=\*^{-1}L\_\omega\*$ is the formal adjoint of the operator of degree $(1,1)$ given by $L\_\omega\bullet=\omega\wedge\bullet$.


Now, the (graded) Jacobi identity gives
$$
-[D''\_E,[\Lambda\_\omega,D''\_E]]+[D''\_E,\underbrace{[D''\_E,\Lambda\_\omega]}\_{=-[\Lambda\_\omega,D''\_E]}]+[\Lambda\_\omega,\underbrace{[D''\_E,D''\_E]}\_{=0}]=0,
$$
thus $-2[D''\_E,[\Lambda\_\omega,D''\_E]]=0$ and 
$$
D''\_E\delta'\_E+\delta'\_ED''\_E=[D''\_E,\delta'\_E]=0.
$$


