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$$\text{AdS}_n = \{\vec x\in\mathbb R^{n-1,2}\mid\vec x\cdot\vec x=-1\}$$

Using an orthonormal basis $\{e_i\}$ for $\mathbb R^{n-1,2}$, with ${e_1}^2={e_2}^2=-1$ and $(e_{\geq3})^2=1$, this space can be parametrized as

$$x = (e_1\cos t+e_2\sin t)\cosh r+\sigma(\theta)\sinh r$$

where $\sigma$ is a point on a unit $(n-2)$-sphere (in the span of $\{e_3,\cdots,e_{n+1}\}$), and $\theta=(\theta_1,\cdots,\theta_{n-2})$ are the ordinary spherical coordinates.

https://en.wikipedia.org/wiki/Anti-de_Sitter_space#Global_coordinates

The metric is obtained from the partial derivatives:

$$dx = (-e_1\sin t+e_2\cos t)dt\cosh r+(e_1\cos t+e_2\sin t)\sinh r\,dr+d\sigma\sinh r+\sigma\cosh r\,dr$$

$$ds^2=dx\cdot dx = -dt^2\cosh^2r-\sinh^2r\,dr^2+d\sigma^2\sinh^2r+\cosh^2r\,dr^2$$

$$= -dt^2\cosh^2r+dr^2+d\sigma^2\sinh^2r$$

The coordinates $t$ and $t+2\pi$ represent the same point. In the universal cover, all values of $t$ should represent different points.


I think I found a way to embed the universal cover in $\mathbb R^{n,2}$. Again, $\sigma$ is on a unit $(n-2)$-sphere, but now it's in the span of $\{e_4,\cdots,e_{n+2}\}$.

$$x = \big(e_1t+e_2(\tfrac12t^2-1)+e_3(\tfrac12t^2)\big)\cosh r+\sigma(\theta)\sinh r$$

$$x\cdot x = \big(-t^2-(\tfrac14t^4-t^2+1)+(\tfrac14t^4)\big)\cosh^2r+\sinh^2r = -1$$

$$dx = \big(e_1+e_2t+e_3t\big)dt\cosh r+\big(e_1t+e_2(\tfrac12t^2-1)+e_3(\tfrac12t^2)\big)\sinh r\,dr+d\sigma\sinh r+\sigma\cosh r\,dr$$

$$ds^2 = \big(-1-t^2+t^2\big)dt^2\cosh^2r+2\big(-t-t(\tfrac12t^2-1)+t(\tfrac12t^2)\big)dt\,dr\cosh r\sinh r+\big(-t^2-(\tfrac14t^4-t^2+1)+(\tfrac14t^4)\big)\sinh^2r\,dr^2+d\sigma^2\sinh^2 r+\cosh^2r\,dr^2$$

$$= -dt^2\cosh^2r+dr^2+d\sigma^2\sinh^2r$$

The metric is the same as before.

And this new manifold is contained in $\text{AdS}_{n+1}$. It's implicitly defined by the quadratic equations

$$x\cdot x=-{x_1}^2-{x_2}^2+{x_3}^2+\cdots+{x_{n+2}}^2=-1$$

$$x_3-x_2=\sqrt{{x_1}^2+{x_2}^2-{x_3}^2}$$


Is this correct?

If so, what is its significance? (I hope that's not too "opinion based".)

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