# A How to calculate the extrinsic curvature of boundary of AdS_2

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1. Jul 15, 2017

### craigthone

I have a simple but technical problem:

How to calculate the extrinsic curvature of boundary of AdS_2?
I am not very familiar with this kind of calculation.

$$ds^2=\frac{dt^2+dz^2}{z^2}$$
is given by (t(u),z(u)).
The induced metric on the boundary is
$$ds^2_{bdy}=g_{\alpha\beta}dx^{\alpha}dx^{\beta}=g_{\alpha\beta}\frac{\partial x^{\alpha}}{\partial y^a}\frac{\partial x^{\beta}}{\partial y^b} dy^ady^b==g_{\alpha\beta}e^{\alpha}_ae^{\beta}_bdy^ady^b=h_{ab} dy^ady^b$$
For $Ads_2$ case, $ds^2_{bdy}= h_{uu}dudu$ where $$h_{uu}= \frac{z'^2+t'^2}{z^2}$$

My calculation is the following:

1) compute normal vecotr (nt,nz)

From the orthogonal relation $$e^\alpha_a n_\alpha=o$$ and unit norm condition $$g_{\alpha\beta}n^{\alpha}n^{\beta}=1$$ we have $$n^t=\frac{zz'}{\sqrt{t'^2+z'^2}} , n^z=-\frac{zt'}{\sqrt{t'^2+z'^2}}$$

2) compute the extrinsic curvature $$K=\nabla_\alpha n^{\alpha}$$

$$K=\nabla_\alpha n^{\alpha}=\frac{1}{\sqrt g}[\partial_t(\sqrt g n^t)+\partial_z(\sqrt g n^z)]=\frac{1}{\sqrt g}[\frac{1}{t'}\partial_u(\sqrt g n^t)+\frac{1}{z'}\partial_u(\sqrt g n^z)]$$

I tried some times but I can not reprodue the result in the paper.
My question is whether there are some mistakes in the formulas I used above. Thanks in advance.

2. Jul 20, 2017

### PF_Help_Bot

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