In chapter 14 of Mirror Symmetry, one-loop renormalization of the nonlinear sigma model is performed. In Riemann normal coordinates, the interaction term in the Lagrangian is $$-\frac{1}{6}R_{IKJL}\xi^I \xi^J \partial_\mu \xi^K \partial_\mu \xi^L $$

The Feynman rule for this is easy enough to read off as $$ -\frac{1}{3} \Big[R_{IJKL} (k_1 - k_2) \cdot (k_3 - k_4) + R_{IKJL} (k_1 - k_3) \cdot(k_2 - k_4) + R_{ILJK} (k_1 - k_4) \cdot (k_2 - k_3)\Big]$$

This is understood to be for a four-point vertex with index structure $(I, k_1), (J, k_2)$, and so on. All momenta are taken to be flowing inward.

For the one-loop correction to the propagator, the book gives the result $$ \frac{1}{p^2}\frac{1}{3} \int\frac{d^2 k}{(2\pi)^2} \frac{1}{k^2} R_{IJ}$$ with $R_{IJ}$ the Ricci tensor, and $p^2$ is the external momentum. However, using the above Feynman rule, one has $$\frac{1}{p^4}\frac{1}{3} \int \frac{d^2 k}{(2\pi)^2} \frac{ k^2 + p^2}{k^2}R_{IJ} $$ Where is the extra $k^2$ coming from? I can't find any mistake in the Feynman rule or in writing down the integral. Help would be appreciated.

  • $\begingroup$ Your derivation is correct. Another way to see the Feynman rule is to integrate by part and get $\sim R \phi^3\partial^2 \phi$, and therefore for the diagram that we are concerning we should get $\sim \frac{1}{6}(2k^2+2p^2)$ from the vertex. I presume in the book the quadratically divergent term is simply ignored. See comments below page 321. $\endgroup$ – user110373 Jul 9 '18 at 21:56
  • $\begingroup$ If you split $\frac{k^2+p^2}{k^2}=1+\frac{p^2}{k^2}$, the term that is missing becomes $\propto \int R_{IJ}\propto g_{IJ}\int R$, a surface term, right? $\endgroup$ – AccidentalFourierTransform Jul 9 '18 at 23:53
  • $\begingroup$ Thanks for pointing out the footnote! I didn't even see it. @AccidentalFourierTransform: I'm not sure about that; $\int R_{IJ}$ will not be proportional to $\int R$ in general. However, by power counting, the quadratically divergent term should be a mass renormalization. Can't we just kill it by setting the renormalized mass to zero? $\endgroup$ – Spencer Tamagni Jul 10 '18 at 2:43
  • $\begingroup$ @SpencerTamagni Sorry, I didn't pay much attention and thought that $R_{IJ}$ was the two-dimensional Ricci tensor. In two dimensions, $R_{ab}=R g_{ab}$. $\endgroup$ – AccidentalFourierTransform Jul 10 '18 at 2:49
  • $\begingroup$ @AccidentalFourierTransform Yeah, I assumed that was the case. Do you know if the mass renormalization argument is valid? It seems odd to me, because the inclusion of such a term in the bare action has no natural geometric interpretation. $\endgroup$ – Spencer Tamagni Jul 10 '18 at 2:54

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