Suppose $M$ is a compact four manifold and $P$ is an $SU(2)$ bundle, let $\mathfrak{g}$ be the adjoint bundle of $P$, given a connection $A$ on this bundle. Given $\phi\in \Omega^1(\mathbb{g})$, we have the following Weitzenbock formula:


$$d\_A d^{\star}\_A\phi+d^{\star}\_Ad\_A\phi=\nabla^{\star}\_A\nabla\_A\phi+\star[\star F\_A,\phi]+Ric(\phi).$$


My question is: $\textbf{does this formula works for an $SL(2,\mathbb{C})$ connection}$? 


To be explicitly: suppose we have another $SL(2,\mathbb{C})$ bundle $P'$ and an $SL(2,\mathbb{C})$ connection $\mathbb{A}$, take $\mathfrak{g}^{\mathbb{C}}$ be the adjoint bundle of $P'$.


For complex forms, we have the complex Hodge star operator $\bar{\star}$ as follows: 
\begin{equation}
\begin{split}
\bar{\star}:\Omega^i(\mathfrak{g}^{\mathbb{C}})&\rightarrow \Omega^{4-i}(\mathfrak{g}^{\mathbb{C}})\\
\bar{\star}\alpha&=\star \bar{\alpha}.
\end{split}
\end{equation}


In addition, for $\alpha,\beta\in\Omega^i(\mathfrak{g}^{\mathbb{C}}),$ we have the inner product: $$<\alpha,\beta>:=-\int Tr(\alpha\wedge\bar{\star} \beta).$$


For the derivative $d\_{\mathbb{A}}$, with respect to this inner product, we have the adjoint operator in four dimensional $d\_{\mathbb{A}}^{\bar{\star}}:=-\bar{\star}d\_{\mathbb{A}}\bar{\star}$.


Given $\Phi\in\Omega^1(\mathfrak{g}^{\mathbb{C}})$, what can we say about the Weitzenbock formula for $$(d\_{\mathbb{A}}d\_{\mathbb{A}}^{\bar{\star}}+d\_{\mathbb{A}}^{\bar{\star}}d\_{\mathbb{A}})\Phi?$$


Does the following thing still holds:
$$d\_{\mathbb{A}}d\_{\mathbb{A}}^{\bar{\star}}\Phi+d\_{\mathbb{A}}^{\bar{\star}}d\_{\mathbb{A}}\Phi=\nabla^{\bar{\star}}\_{\mathbb{A}}\nabla\_{\mathbb{A}}\Phi+\bar{\star}[\bar{\star} F\_\mathbb{A},\Phi]+Ric(\Phi)?$$


Here are two things I am worried:


(1) I go through the prove of Weinzenbock formula for SU(2) in [1], I am worried about this is only true for the operator $d\_{\mathbb{A}}^{\star}:=-\star d\_{\mathbb{A}}\star$ not for $d\_{\mathbb{A}}^{\bar{\star}}$, these two are really different operators.


(2) For the order zero term action, will it be $\bar{\star}[\bar{\star} F\_\mathbb{A},\Phi]$ or $\star[\star F\_{\mathbb{A}},\Phi]$?


Thank you very much.


[1] J.Bourguignon and H.Lawson, Stability and isolation phenomena for Yang-Mills fields



About the Weitzenböck Formula for $SL(2,\mathbb{C})$ connection
I think a get a complete answer for this question. 


As $SL(2,\mathbb{C})$ bundle P' can reduce to an $SU(2)$ bundle denote as P and $sl(2,\mathbb{C})$ is $su(2)\oplus isu(2)$, under this decomposition, our connection $\mathbb{A}$ can be write as $\mathbb{A}=A+iB$, here $A$ is an $SU(2)$ connection and $B\in\Omega^1(\mathfrak{g})$. 


Under this identification, $F\_{\mathbb{A}}=F\_A-B\wedge B+id\_A B$


If we denote $d\_{\mathbb{A}}^{\star}\Phi=-\star d\_{\mathbb{A}}\star\Phi$, we have the Weitzenbock formula:


$$d\_{\mathbb{A}}d\_{\mathbb{A}}^{\star}\Phi+d\_{\mathbb{A}}^{\star}d\_{\mathbb{A}}\Phi=\nabla^{\star}\_{\mathbb{A}}\nabla\_{\mathbb{A}}\Phi+\star[\star F\_{\mathbb{A}},\Phi]+Ric(\Phi).$$ The proof for this are exactly the same as shown in [1].


However, for the $d\_{\mathbb{A}}^{\bar{\star}}$ operator, we are taking a conjugate for the connection $\mathbb{A}$, and the order zero terms are not some pleasant and here is the result:


$$d\_{\mathbb{A}}d\_{\mathbb{A}}^{\bar{\star}}\Phi+d\_{\mathbb{A}}^{\bar{\star}}d\_{\mathbb{A}}\Phi=\nabla^{\bar{\star}}\_{\mathbb{A}}\nabla\_{\mathbb{A}}\Phi+Ric(\Phi)+\star[\star F\_A+\star B\wedge B,\Phi]+G\_{\mathbb{A}}(\Phi).$$


Choose a locally orthogonal bases, the $G\_{\mathbb{A}}(\Phi)\_j=\sum^4\_{i=1} [\nabla\_i B\_j+\nabla\_jB\_i, \Phi\_i].$


Two things to remark about the order zero terms:


1 the plus signature before $B\wedge B$ in $F\_A+B\wedge B$ appears because we take a conjugate when we define $d\_\mathbb{A}^{\bar{\star}}$.


2 $G\_\mathbb{A}(\Phi)$ is completely different from $\star[d\_A B,\Phi]$.


