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Andrey Feldman
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It depends on the dimensionality of the space one works with. In $2D$2d, the kinetic term that you presented appears after the integration over 4 Grassmann coordinates in the expression $\int \mathrm{d} \ \sigma^2 \mathrm{d} \ \theta^4 K(\Phi, \overline{\Phi})$$\int \mathrm{d}^2 \sigma \ \mathrm{d}^4 \theta K(\Phi, \overline{\Phi})$ which means $\cal{N}=2$. Perhaps it is the reason of your confusion?

It depends on the dimensionality of the space one works with. In $2D$, the kinetic term that you presented appears after the integration over 4 Grassmann coordinates in the expression $\int \mathrm{d} \ \sigma^2 \mathrm{d} \ \theta^4 K(\Phi, \overline{\Phi})$ which means $\cal{N}=2$. Perhaps it is the reason of your confusion?

It depends on the dimensionality of the space one works with. In 2d, the kinetic term that you presented appears after integration over 4 Grassmann coordinates in the expression $\int \mathrm{d}^2 \sigma \ \mathrm{d}^4 \theta K(\Phi, \overline{\Phi})$ which means $\cal{N}=2$. Perhaps it is the reason of your confusion?

Source Link
Andrey Feldman
  • 2k
  • 1
  • 11
  • 24

It depends on the dimensionality of the space one works with. In $2D$, the kinetic term that you presented appears after the integration over 4 Grassmann coordinates in the expression $\int \mathrm{d} \ \sigma^2 \mathrm{d} \ \theta^4 K(\Phi, \overline{\Phi})$ which means $\cal{N}=2$. Perhaps it is the reason of your confusion?