# 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: $$\begin{split} \bar{\star}:\Omega^i(\mathfrak{g}^{\mathbb{C}})&\rightarrow \Omega^{4-i}(\mathfrak{g}^{\mathbb{C}})\\ \bar{\star}\alpha&=\star \bar{\alpha}. \end{split}$$

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

• Every $SL(2,\mathbb{C})$-bundle admits a reduction of structure group to an $SU(2)$-bundle. Perhaps that would help you. – Ben McKay Oct 5 '16 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. – Deane Yang Oct 5 '16 at 11:33
• @BenMcKay@DeaneYang Thank you very much for your helpful comments. – Siqi He Oct 6 '16 at 7:02

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