# About the Weitzenböck Formula for $SL(2,\mathbb{C})$ connection

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 , 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.

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

• Every $SL(2,\mathbb{C})$-bundle admits a reduction of structure group to an $SU(2)$-bundle. Perhaps that would help you. Oct 5, 2016 at 6:13
• Have you already tried to do the calculation? It seems to me that it might work, but the operator probably will not be elliptic. It's probably hyperbolic. Oct 5, 2016 at 11:33

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 .

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 $\star[d_A B,\Phi]$.