I'm looking at lecture notes on AdS/CFT by Jared Kaplan, and in section 4.2 he claims that the action for a free scalar field in AdS$_3$ is $$S=\int dt d\rho d\theta \dfrac{\sin\rho}{\cos\rho}\dfrac{1}{2}\left[\dot{\phi}^2-\left(\partial_\rho\phi\right)^2-\dfrac{1}{\sin^2\rho}\left(\partial_\theta\phi\right)^2-\dfrac{m^2}{\cos^2\rho}\phi^2\right]$$ and that the canonical momentum conjugate to $\phi$ is $$P_\phi=\dfrac{\delta L}{\delta\dot{\phi}}=\dfrac{\sin\rho}{\cos^2\rho}\dot{\phi}$$ Now, my question is: where do the $\cos^2\rho$ terms in the action and the conjugate momentum come from?

Maybe I'm missing something obvious, but when computing the canonical momentum, shouldn't I only pick up the prefactor of $\frac{\sin\rho}{\cos\rho}$?

As for the mass term in the action, I know that the free scalar field action in AdS$_{d+1}$ is $$S=\int_{AdS}d^{d+1}x\sqrt{-g}\left[\dfrac{1}{2}\left(\nabla_A\phi\right)^2-\dfrac{1}{2}m^2\phi^2\right]$$ with the metric $$ds^2=\dfrac{1}{\cos^2{\rho}}\left(dt^2-d\rho^2-\sin^2{\rho}\ d\Omega_{d-1}^2\right)$$ so how is the mass term picking up an extra $\frac{1}{\cos^2\rho}$?


The metric for $AdS_3$ is $$ds^2=\frac{1}{cos^2\rho}(dt^2-d\rho^2-sin^2\rho d\theta^2)$$, because $d=2$ is $AdS_3$. So $$g=\frac{1}{cos^2\rho}\times\frac{-1}{cos^2\rho}\times\frac{-sin^2\rho}{cos^2\rho}=\frac{sin^2\rho}{cos^6\rho}.$$ That's why in the mass term there is an extra $\frac{1}{cos^2\rho}$.

And I think there is a typo in the expression of the momentum. It should be $$P_\phi=\frac{sin\rho}{cos\rho}\dot{\phi}.$$

  • $\begingroup$ Thanks! But isn't $d=2$ AdS$_3$ rather? $\endgroup$ – Demosthene Apr 3 '16 at 12:10
  • $\begingroup$ @Demosthene yes, you're right. $\endgroup$ – Nahc Apr 3 '16 at 12:19

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.