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?

  • $\begingroup$ @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? $\endgroup$
    – Partha
    Oct 10, 2021 at 12:39
  • 1
    $\begingroup$ @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. $\endgroup$
    – Partha
    Oct 10, 2021 at 16:54
  • $\begingroup$ @ChrisGerig I see now, thanks. $\endgroup$
    – Partha
    Oct 10, 2021 at 17:28

1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .