In chapter 7.1.1. in Tong's notes about String Theory could someone sketch how can I show the statements that he makes around eq. 7.5

  • That the addition of the counterterm can be absorbed by renormalization the wavefunction and the metric

  • How does he conclude from the renormalization $$G_{\mu \nu} \rightarrow G_{\mu \nu} + \dfrac{\alpha '}{\epsilon}\mathcal{R}_{\mu\nu} $$ that the beta function equals $$\beta_{\mu\nu}(G) = \alpha ' \mathcal{R}_{\mu \nu} \quad ? $$

  • 1
    $\begingroup$ Look up the computation of one-loop beta functions in dimensional regularisation. $\endgroup$
    – suresh
    Commented Sep 25, 2014 at 7:23
  • $\begingroup$ Linked. $\endgroup$ Commented May 3 at 21:23

1 Answer 1


On account of symmetry, 2d σ-models, even though beset with an infinity of counterterms, have those regiment themselves into few geometrical tensors constrained by symmetry; and thus limit themselves to tensors, such as the Ricci, and then underlie the β-function in the conventional manner: it is not zillions of couplings that evolve, it is only the geometry, and, on highly symmetric "pion" manifolds, by just a scale or two! (For the hypersphere, below, just its radius of curvature--it puffs up to noninteracting flatness, conformal invariance, with energy. cf. Tong 7.1.2)

Both are illustrated in our paper BCZ 1985, and especially in (2.41-2.49) and in Appendix A, meant to answer just these questions. But it is a long story, to which no justice can be done in this short format.

Nevertheless, your second question has a straightforward answer, implicit in Tong's notes, and, of course, section 2 of the paper cited here.

  • In ε = d-2 dimensions, taking the "pion" fields φ to be dimensionless, but the bare metric to have dimension ε, to one loop rewrite your expression as $$ (G_{\mu\nu}/\alpha')^{(0)}=M^\epsilon ( G_{\mu\nu}/\alpha' -\frac{1}{\epsilon}R^{(1)}_{\mu\nu} ). $$ But the bare α'-full metric must be independent of the RG scale M; so operating on this equation by $M \frac{d}{dM}$ at the pole $\epsilon \to 0$ nets $$ 0= M \frac{d}{dM} \frac{G_{\mu\nu}}{\alpha'} - R^{(1)}_{\mu\nu}, $$ where the superscript (1) indicates the residue at the pole, and $$ M \frac{d}{dM} \frac{G_{\mu\nu}}{\alpha'} = R^{(1)}_{\mu\nu}. $$

Thus, in our scaled conventions, in a hypersphere ($R_{\mu\nu}=2 G_{\mu\nu}$), whose inverse radius-squared, α', decreases with scale M , asymptotic freedom manifests itself: $d\alpha'/d\ln M =-2\alpha' ^2$. Asymptotically, the sphere flattens to a conformally invariant plane.

  • $\begingroup$ continuing on the expression I see using the quotient rule for derivatives and your expression for the beta function of $\alpha'$ (not sure if I'm using the right terminology) I find: $$M\frac{d}{dM}\frac{G_{\mu\nu}}{\alpha'} = \frac{1}{\alpha'}M\frac{dG_{\mu\nu}}{dM}+2G_{\mu\nu} = \frac{1}{\alpha'}\beta(G)+2G_{\mu\nu}$$ Which then gives $$\beta(G) = \alpha' R_{\mu\nu}^{(1)} + \alpha' 2 G_{\mu\nu} =? \alpha' R_{\mu\nu}$$ It looks like we got something different from what we are supposed to get, or am I missing something? $\endgroup$ Commented May 3 at 20:38
  • $\begingroup$ Could you, in particular explain where the extra term, and the superscript (1) go? $\endgroup$ Commented May 3 at 20:56
  • $\begingroup$ There is no extra term. They are explained in the linked paper, (2.44) et seq, and in any creditable QFT text on dimensional regularization, such as Schwartz's , etc. You are asking question on the iterative framework of the RG... Maybe a separate question? $\endgroup$ Commented May 3 at 21:11
  • $\begingroup$ Required reading: e.g. 26.6.1 of Schwartz. The effective coupling is G/α' ... $\endgroup$ Commented May 3 at 21:43
  • $\begingroup$ with extra term I mean $\alpha' 2 G_{\mu\nu}$, I got it based on the calculation in my comment above, but it does not appear in the anwer that Tong is giving. $\endgroup$ Commented May 4 at 15:24

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