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

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.


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

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


Is this correct?

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.