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  A generalization of the Seiberg-Witten equations

+ 3 like - 0 dislike

Following the book of Friedrich "Dirac operators and riemannian geometry" (AMS, vol 25), I define the generalized Seiberg-Witten equations for $(A,A',\psi , \phi)$, with $A,A'$ two connections and $\psi, \phi$, two spinors:


$D_A ( \psi)=0$


$D_{A'} ( \phi)=0$


$F_+ (A)=-(1/4) \omega (\psi)$


$F_+ (A')=-(1/4) \omega (\phi)$


$ A-  A' = Im( \frac{d<\psi|\phi>}{<\psi|\phi>})$

$Im$ is the imaginary part of the complex number.

The gauge group $(h,h') \in Map(M,S^1)$ acts over the solutions of the generalized Seiberg-Witten equations:

$(h,h').(A,A',\psi,\phi)=((1/h)^* A, (1/{h'})^* A', h \psi, h' \phi )$

We have compact moduli spaces because it is a closed set in the product of two compact sets (the SW moduli spaces).

Moreover, the situation can perhaps be generalized to $n$ solutions of the Seiberg-Witten equations $(A_i ,\psi_i )$:


$D_{A_i}( \psi_i)=0$


$F_+(A_i)= -(1/4) \omega (\psi_i)$


$ A_i- A_j=Im( \frac{d<\psi_i|\psi_j>}{<\psi_i|\psi_j>})$

asked Jul 20, 2018 in Theoretical Physics by Antoine Balan (-80 points) [ revision history ]
edited Oct 2, 2018 by Antoine Balan

1 Answer

+ 0 like - 0 dislike

The situation seems to make sense only in the analytical case (all the functions are demanded to be analytical) because of a quotient which is difficult to define. Moreover, in the differential case, we can take simply the product of two spinors as a condition of coupling ($<\psi,\phi>=0$).

answered Jul 20, 2019 by Antoine Balan (-80 points) [ no revision ]

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