# Renormalizability of the Polyakov Action

I was told today that the Polyakov action for a $p$-brane is (superficially) re-normalizable iff $p\leq 1$. Of course, when I went to check for myself, I screwed up my power-counting, and I'm having trouble seeing why.

We work in units with $c=1=\hbar$, so that $L=T=M^{-1}$. In these units, any action must have dimension $1$, so from looking at the Nambu-Goto action, $$S_{\text{NG}}:=-T_p\int d\sigma ^{1+p}\sqrt{-g},$$ we see that $[T_p]=L^{-(1+p)}=M^{1+p}$. From the Polyakov action, $$S_{\text{P}}:=-\frac{T_p}{2}\int d\sigma ^{1+p}\sqrt{-h}h^{\alpha \beta}\partial _\alpha X^\mu \partial _\beta X^\nu G_{\mu \nu}(X),$$ we see that the the coupling constant of the interaction of of the scalar fields $X^\mu$ with $h_{\alpha \beta}$ is precisely $\frac{T_p}{2}$, which has dimensions $M^{1+p}$, and so is going to be (superficially) re-normalizable iff $1+p\geq 0$ . . . But this, of course, is not the result I was looking for . . . Where is my mistake?

• Not sure but consider following argument. Suppose G be of the form 1+aX+(aX)^2+ .. where a has units of mass. Now define new variable Y=\sqrt(T) X. Lagrangian will now have a free part, and interacting part. Coefficient of the first interaction term will be g=a/sqrt(T), of second interaction term g^2 and so on. Now note that g has mass dimension (1-p)/2. Now for renormalizability we should require that it be non negative and hence that p<=1 Sep 14, 2013 at 5:59
• @user10001 This works (in fact, with just a bit more detail, I would accept this as an answer), but there is at least one thing that worries me about this. Presumably by $1$ in $G=1+aX+(aX)^2+\cdots$, you mean $\eta _{\mu \nu}$. If this is the case, you find that the kinetic term is given by $-\frac{1}{2}\partial _\alpha Y^\mu \partial ^\alpha Y^\nu \eta _{\mu \nu}$, which means that the field $Y^0$ has the wrong sign for its kinetic term. Sep 23, 2013 at 19:33
• Yes, 1 means $\eta_{\mu\nu}$. Similarly $aX$ means $af^{(1)}_{\mu\nu\rho}X^{\rho}$ for some 'tensor' $f^{(1)}$, $(aX)^2$ means $a^2 f^{(2)}_{\mu\nu\rho\sigma}X^{\rho}X^{\sigma}$ for some 'tensor' $f^{(2)}$ and so on. Sign of kinetic term for $X^0$ would be wrong. This gives rise to negative normed (unphysical) states in the spectrum which are then eliminated by applying constraints (which come from the equation of motion of the worldsheet metric h ). Sep 24, 2013 at 4:17
• @user10001 Thanks much. Do you want to write up what you said in a comment as an answer? At this point, I could write it up myself, but I feel as if it would make more sense to give you the credit for it. Sep 27, 2013 at 19:11

The problem with my analysis before was that $T_p/2$ is not the relevant coupling constant. To read off the coupling constant, we must put the action into a slightly different form, in which the kinetic and interaction terms are separate and apparent.
To do that, we assume that everything is (real) analytic, we work locally and pick coordinates $x^\mu$ on space-time (the codomain manifold) such that $G(0)=\eta _{\mu \nu}$ in these coordinates. Then, $G(x)=\eta _{\mu \nu}+\varepsilon f^{(1)}_{\mu \nu \rho}x^\rho +\cdots$, where we are taking $f^{(1)}$ to be dimensionless, so that $[\varepsilon ]=L^{-1}=M$. Plugging this into the action, we find $$S_{\text{P}}=\int d\sigma ^{1+p}\, \left[ -\frac{T_p}{2}\partial _\alpha X^\mu \partial ^\alpha X_\mu -\varepsilon \frac{T_p}{2}f^{(1)}_{\mu \nu \rho}(\partial _\alpha X^\mu )(\partial ^\alpha X^\nu )X^\rho +\cdots \right] .$$ To get the kinetic term in the usual form so that we can read-off the appropriate coupling constant, we define $Y:=\sqrt{T_p}X$ and write the action in terms of $Y$: $$S_{\text{P}}=\int d\sigma ^{1+p}\, \left[ -\frac{1}{2}\partial _\alpha Y^\mu \partial ^\alpha Y_\mu -\frac{\varepsilon}{2\sqrt{T_p}}f^{(1)}_{\mu \nu \rho}(\partial _\alpha Y^\mu )(\partial ^\alpha Y^\nu )Y^\rho +\cdots \right] .$$ From this, we see that the coupling constant of the lowest-order interaction is $$\frac{\varepsilon}{2\sqrt{T_p}}f^{(1)}_{\mu \nu \rho},$$ which has mass dimension $$1-1/2(1+p)=1/2-1/2p.$$ The theory will thus be superficially renormalizable iff $1/2-1/2p\geq 0$, that is, iff $p\leq 1$.