# Understanding the quadratic part in Seiberg Witten equation

Lets take a closed four manifold $$M:=\Sigma_1\times \Sigma_2,$$ where $$\Sigma_i$$s are compact Riemann surfaces. Now if $$V$$ and $$W$$ are Spin$$^\mathbb{C}$$ bundles on $$\Sigma_1$$ and $$\Sigma_2$$ respectively, then one can define Spin$$^\mathbb{C}$$ bundle on $$M$$ with positive Spin$$^\mathbb{C}$$ bundle defined as $$\begin{equation*} V_+\otimes W_+\oplus V_-\otimes W_- \end{equation*}$$ and negative Spin$$^\mathbb{C}$$ bundle defined as $$\begin{equation*} V_+\otimes W_-\oplus V_-\otimes W_+ \end{equation*}$$ In general the quadratic term in Seiberg Witten equation acts like this:

For positive spinors $$\phi,\psi,q(\phi)(\psi):=\langle\psi,\phi\rangle\phi-\frac{|\phi|^2}{2}\psi$$.

In the above situation for $$\phi\in V_+\otimes W_+,$$ I can understand $$q(\phi)$$ acting on spinors in $$V_+\otimes W_+$$, but as the spaces are perpendicular, I am confused about the action of $$q(\phi)$$ on spinors in $$V_-\otimes W_-$$. Is it the trivial action ($$0$$)? Can we say something about what kind of self dual two form it represents?

• @ChrisGerig, I am asking if using the induced metric on the direct sum, $q(\phi)$ acts trivially on $V_-\otimes W_-$, is that correct? If yes can we say what kind of two form that will represent? Oct 10, 2021 at 12:39
• @ChrisGerig here's what confusing me. So the quadratic term is basically $\phi^*\otimes \phi-$ its trace. If the first term is not there we shouldn't put the trace part. Ultimately the action should be trace-free. Oct 10, 2021 at 16:54
• @ChrisGerig I see now, thanks. Oct 10, 2021 at 17:28

So your $$\phi$$ in this special case really means $$(\phi,0)$$ in the direct sum, while $$\psi$$ means $$(0,\psi)$$. Then $$q(\phi)$$ is the 2x2 matrix with vanishing off-diagonal entries and nontrivial diagonal entries ($$\frac12|\phi|^2,-\frac12|\phi|^2)$$, which is traceless. Particular values are $$q(\phi)\phi=\frac12|\phi|^2\phi$$ and $$q(\phi)\psi=-\frac12|\phi|^2\psi$$. (But if we looked at the full endomorphism $$\phi\otimes\phi^*$$ then it would vanish on $$\psi$$.) Now just compute the inverse (or adjoint) of the Clifford multiplication map to get your induced 2-forms.