# 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?

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?

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Can I ask that you try to find a way to ask (a shortened version of) your question in the title? It will help other users know what question you want to ask, making it more likely they follow the link, and it will also help Google index your question and raise it to the top of the search page, making it more likely to help future mathematicias. Remember that you get 240 characters for the title --- more than a tweet and a half. –  Theo Johnson-Freyd Aug 19 '10 at 21:34
Sorry for the difficulties, from now on i'll do it for sure! –  Italo Aug 19 '10 at 22:20
Thanks! Titles like this are great. –  Theo Johnson-Freyd Aug 20 '10 at 5:09

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 $$D''_E\delta'_E+\delta'_ED''_E=[D''_E,\delta'_E]=0.$$