When the Wess-Zumino-Witten model

$$S_{WZW}=\frac{k}{4\pi}\int d^2 z \, \, \mathrm{Tr}[\partial u \bar{\partial}u^{-1} ]+ \frac{k}{12\pi}\int d^3 \sigma \epsilon^{ijk}\, \mathrm{Tr}[(u^{-1}\partial_i u)(u^{-1}\partial_j u)(u^{-1}\partial_k u)]$$

is expanded around a solution of the equations of motions $u=u_0 e^{iT^a\pi^a}$ one gets

$$S_{WZW}=\frac{k}{4\pi}\int d^2 z \lbrace \, \mathrm{Tr}[\partial u_0 \bar{\partial}u_0^{-1} ]+ \frac{1}{2}\partial_\mu \pi^a\partial ^\mu \pi^a \\ +\frac{1}{2}(\eta^{\mu\nu}-\epsilon^{\mu\nu}) \, \mathrm{Tr}\lbrace (u_0^{-1}\partial_\mu u_0)[T^a\pi^a,T^b\partial_\nu \pi^b]\rbrace + \mathcal{O}(\pi^3)\rbrace$$

The one loop renormalization diagram is like

enter image description here

In "Non-Perturbative Field Theory" by Y.Frishman and J.Sonnenschein (chapter 4.2 page 65) I read that the non vanishing contributions come only from the diagrams with both vertices proportional to $\eta^{\mu\nu}$ or to $\epsilon^{\mu\nu}$. Could someone explain me how one comes to that conclusion?


1 Answer 1


One comes to this conclusion due to the fact that the contraction of a symmetric tensor with an antisymmetric one vanishes. Writing down the loop diagrams involves a contraction of both vertices. If you get expressions proportional to $\epsilon_{\mu\nu}\eta^{\mu\nu}$, this will be zero due to the fact that the metric is symmetric and the epsilon tensor is antisymmetric.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.