How to prove this Weitzenbock formula? 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.
 A: 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.
