# Proof that the superstring action is Weyl invariant

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The superstring action is:

$$S = k\int \mathrm d\sigma \sqrt{-h} \left [ h^{\alpha \beta} \partial_\alpha X^\mu \partial_\beta X_\mu + 2i {\bar{\psi}} ^\mu \rho^\alpha \partial_\alpha \psi_\mu - i {\bar \chi}_\alpha \rho^\beta \rho^\alpha \psi^\mu \left (\partial_\beta X_\mu -\frac{i}{4}{\bar \chi}_\beta \psi_\mu \right)\right ]$$

This is Weyl invariant, but the symmetry isn't manifest. All the books and lecture notes I've seen just state this outright, without proving it. Is there a reference where I can see the proof done in detail? Even when they give the action in the superconformal gauge ($h_{\alpha \beta} = \eta_{\alpha \beta}$ and $\chi_\alpha = 0$), they don't bother to prove it.

This post imported from StackExchange Physics at 2014-06-29 09:35 (UCT), posted by SE-user user46242

asked Jun 27, 2014
edited Jul 12, 2014
If we take into account quantum corrections, the theory is truly Weyl invariant only for Ricci-flat metrics.

This post imported from StackExchange Physics at 2014-06-29 09:35 (UCT), posted by SE-user JamalS

Related simpler version: physics.stackexchange.com/q/70508

This post imported from StackExchange Physics at 2014-06-29 09:35 (UCT), posted by SE-user joshphysics

@joshphysics I don't think that that is very relevant here. The OP clearly states that he is talking about the superstring action not the bosonic string action, for which, as you state, proving Weyl symmetry is simpler, and much simpler to the degree that mentioning it here is completely useless.

## 2 Answers

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Short answer: 1+1=1/2+1/2+1=1/2+1/2+1=1/2+1/2+1/2+1/2=2.

Long answer: the Weyl invariance is (almost) manifest. The proof is essentially a simple dimension counting. As the Weyl symmetry is about rescaling of the metric $h_{\alpha \beta}$, we have to make explicit the metric dependence in the various terms of the action. As the theory contains spinors, we have to use a 2d "tetrad" formalism with a "zweibein" $e^a_\alpha$ satisfying $h_{\alpha \beta} = e^a_\alpha e^b_\beta \eta_{ab}$, $e^\alpha_a e^b_\beta = \eta^b_a$ where $\alpha$, $\beta$... are curved 2d space-time coordinates indices and $a$, $b$... are 2d Minkowski coordinates indices. This formalism is implicit in the action written in the question but it is better to replace the curved metric dependant gamma matrices $\rho^\alpha$ by Minkowski, metric independant, gamma matrices $\rho^a$. So we rewrite the action as:

$S = k \int d\sigma \sqrt{-h} ( h^{\alpha \beta} \partial_\alpha X^\mu \partial_\beta X_\mu + 2i {\bar{\psi}}^\mu \rho^a e^\alpha_a \partial_\alpha \psi_\mu$

$- i {\bar \chi}_a \rho^b \rho^a \psi^\mu e^\beta_b \partial_\beta X_\mu - \frac{1}{4}{\bar \chi}_a \rho^b \rho^a \psi^\mu {\bar \chi}_b \psi_\mu )$

This action is invariant under the Weyl symmetry

$e^a_\alpha \mapsto \lambda e^a_\alpha$, $X \mapsto X$, $\psi \mapsto \lambda^{-1/2} \psi$, $\chi \mapsto \lambda^{-1/2} \chi$

where $\lambda$ is an arbitrary function on the 2d spacetime.

Indeed:

As $e^a_\alpha \mapsto \lambda e^a_\alpha$, we have $h_{\alpha \beta} \mapsto \lambda^2 h_{\alpha \beta}$ and so $det(h) \mapsto \lambda^4 det(h)$ because we are in 2d and so $\sqrt{det(h)} \mapsto \lambda^2 \sqrt{det(h)}$. So we have to show that each term between the brackets in the expression giving $S$ is multiplied by $\lambda^{-2}$ under the Weyl symmetry.

First term: term in $h^{\alpha \beta} \partial X \partial X$. As $h_{\alpha \beta} \mapsto \lambda^2 h_{\alpha \beta}$, we have $h^{\alpha \beta} \mapsto \lambda^{-2} h^{\alpha \beta}$ hence the result because $X \mapsto X$.

Third term: term in $\chi \psi e^\beta_b$. We have $\chi \mapsto \lambda^{-1/2} \chi$, $\psi \mapsto \lambda^{-1/2} \psi$, $e^\beta_b \mapsto \lambda^{-1} e^\beta_b$ hence the result because $-1/2-1/2-1=-2$.

Fourth term: term in $\chi \psi \chi \psi$. We have $\chi \mapsto \lambda^{-1/2} \chi$, $\psi \mapsto \lambda^{-1/2} \psi$, hence the result because $-1/2-1/2-1/2-1/2=-2$.

Second term: term in $\overline{\psi} \rho^a e^\alpha_a \partial \psi$. It is the unique slightly subtle point because this term contains a derivative of a field charged under the Weyl symmetry. In fact, it is a general but not completely trivial fact that in any spacetime dimension, a fermion minimally coupled to gravity is (classically) Weyl invariant: the Dirac operator is conformally invariant. Let's see what happens explicitely in our 2d case. Under , $\psi \mapsto \lambda^{-1/2} \psi$, $e^\alpha_a \mapsto \lambda^{-1} e^\alpha_a$, the term  $\overline{\psi} \rho^a e^\alpha_a \partial \psi$ gives

$\lambda^{-2}\overline{\psi} \rho^a e^\alpha_a \partial \psi + \lambda^{-3/2} \overline{\psi} \rho^a e^\alpha_a \psi \partial (\lambda^{-1/2})$.

We have to show that the undesired extra term vanishes:

$\overline{\psi} \rho^a \psi= 0$,

which is a special property of Majorana fermions in 2d (and which was already implicitely used in the expression of the action $S$ to put an apparently non covariant derivative instead of the full Dirac operator). There are many ways to prove this relation. One possibility: $\overline{\psi} \rho^a \psi = \psi^{T} \rho^0 \rho^a \psi$ because $\psi$ is real; if $a=0$, we obtain something proportional to $\psi^{T} \psi = \psi_{+}^2+\psi_{-}^2=0$ because the components $\psi_{\pm}$ of $\psi$ are anticommuting variables; if $a=1$, $\rho^0 \rho^1$ is the 2d "$\gamma_5$ matrix" and so we obtain something proportional to $\psi_{+}^2-\psi_{-}^2=0$ because the components $\psi_{\pm}$ of $\psi$ are anticommuting variables.

answered Jul 2, 2014 by (5,120 points)
edited Jul 2, 2014 by 40227

Wow @40227, thanks a lot for this patiently detailled great answer :-)!

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As you say, the Weyl symmetry in the superstring action in the RNS formalism is not manifest. This means you can’t really “prove” Weyl symmetry for the superstring action. Instead, you postulate that the superstring action obeys Weyl symmetry (is conformally invariant) and derive the constraint equations necessary for your postulate to be satisfied.

You can an overview of this process in Section 4.3 (pg 118 – 120) of Becker, Becker, and Schwarz, String theory and M-theory. The constraint amounts to just

$J_-=J+=T_{--}=T_{++}=0$

answered Jul 1, 2014 by (1,975 points)
edited Jul 1, 2014

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