# Scalar fields in AdS$_3$

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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 $\frac{1}{\cos^2\rho}$?

This post imported from StackExchange Physics at 2016-03-29 19:59 (UTC), posted by SE-user Demosthene

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The determinant of the metric is

$$\sqrt{|g|} = \frac{\sin \rho }{\cos^3 \rho} = \frac{\sin \rho}{\cos \rho} \frac{1}{\cos^2 \rho}$$

which gives you the desired factor of $1/\cos^2 \rho$.

answered Apr 3, 2016 by Bootstrapper
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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 $d=1$ 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 $$P_\phi=\frac{sin\rho}{cos\rho}\dot{\phi}.$$

answered Apr 3, 2016 by (35 points)
reshown Apr 19, 2016

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