How to prove this Weitzenbock formula?

Question

In Hutchings and Taubes lecture note on Seiberg-Witten equation HERE, above equation (4.20) the authors claim that there is a version of Weitzenbock formula reads (where $\beta \in \Omega^{0,2}(M, E)$, M is a symplectic manifold with compatible $J$, $E$ is a line bundle with $U(1)$ connection $a$)

\begin{equation} \int {{{\left| {\bar \partial _a^*\beta } \right|}^2}} = \frac{1}{2}\int {\left( {{{\left| {\nabla _a^*\beta } \right|}^2} - i\left\langle {\omega ,{F_a}} \right\rangle {{\left| \beta \right|}^2}} \right)} \end{equation}

I try to prove it, by first guessing that it might come from something like a Kahler identity \begin{equation} {{\bar \partial }_a}\bar \partial _a^*\beta = \frac{1}{2}{\nabla _a}\nabla _a^*\beta + \left( {...} \right) \end{equation} and try to figure out if the ... matches the formula in the note. But I could not reproduce the result: the ... I got is of the form $\Lambda \left( {F_a^{2,0} \wedge \beta } \right)$, where $\Lambda = (\omega\wedge)^*$, and a (2,0)-piece of $F$ rather than a (1,1) piece shows up.

So I wonder how to prove the Weitzenbock formula above? Any reference is welcome; I google for a while but may be I am not looking at the write place.

Answer 1

Perhaps this is the proof, at least for the Kahler case. I use complex coordinate system: \begin{equation} \beta = \frac{1}{{2!}}{\beta _{\bar \mu \bar \nu }}d{{\bar z}^{\bar \mu }} \wedge d{{\bar z}^{\bar \nu }} = {\beta _{\bar 1\bar 2}}d{{\bar z}^{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{1} }} \wedge d{{\bar z}^{\bar 2}} \end{equation}

Then \begin{equation} {{\bar \partial }^*}\beta = - {\nabla ^{\bar \mu }}{\beta _{\bar \mu \bar \nu }}d{{\bar z}^{\bar \nu }} \to \bar \partial {{\bar \partial }^*}\beta = - {\nabla _{\bar \lambda }}{\nabla ^{\bar \mu }}{\beta _{\bar \mu \bar \nu }}d{{\bar z}^{\bar \lambda }} \wedge d{z^{\bar \nu }} \end{equation}

Commuting the covariant derivatives, one gets the curvature and another second order derivative: \begin{equation} \bar \partial {{\bar \partial }^*}\beta = - \bar \partial \left( {{\nabla ^{\bar \mu }}{\beta _{\bar \mu \bar \nu }}} \right)d{z^{\bar \nu }} = {F_{\bar \lambda }}^{\bar \mu }{\beta _{\bar \mu \bar \nu }}d{{\bar z}^{\bar \lambda }} \wedge d{z^{\bar \nu }} - \frac{1}{2}{\nabla ^{\bar \mu }}{\nabla _{\bar \mu }}{\beta _{\bar \nu \bar \lambda }}d{{\bar z}^{\bar \lambda }} \wedge d{z^{\bar \nu }} \end{equation}

In passing, one needs to use the fact that we are sitting on a 4-manifold, and therefore $\bar \partial \beta = 0$, namely \begin{equation} {\nabla ^{\bar \mu }}{\nabla _{\bar \lambda }}{\beta _{\bar \mu \bar \nu }}d{{\bar z}^{\bar \lambda }} \wedge d{z^{\bar \nu }} = - \frac{1}{2}{\nabla ^{\bar \mu }}{\nabla _{\bar \mu }}{\beta _{\bar \nu \bar \lambda }}d{{\bar z}^{\bar \lambda }} \wedge d{z^{\bar \nu }} \end{equation}

Finally,${F_{\bar \lambda }}^{\bar \mu }{\beta _{\bar \mu \bar \nu }}d{{\bar z}^{\bar \lambda }} \wedge d{z^{\bar \nu }} = \left( {{F_{\bar 1}}^{\bar 1} + {F_{\bar 2}}^{\bar 2}} \right)\beta \sim\left\langle {\omega ,F} \right\rangle \beta $

So up to some convention of $i$, the formula is obtained.