Metric form of $AdS_5 \times S^5$

Question

I want to know the metric form of $AdS_5 \times S^5$.

I know there are two forms (maybe more?) Poincare patch and global patch.

And what is the difference between these two patches?

Can you state the two forms (?) of the metric and explain the difference?

I want to know, for example, Poincare patch is used in some areas due to something and so on.

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I noticed for global patch, identifying $AdS_5$ and $S_5$ have the same radius $R$, I have \begin{align} ds^2 = R^2 \left( - \cosh^2(\rho) dt^2 + d\rho^2 + \sinh^2(\rho) d \Omega_3^2 + \cos^2(\theta) d \phi^2 + d\theta^2 + \sin^2(\theta) d \tilde{\Omega}_3^2 \right) \end{align} where $d\Omega_3^2, d\tilde{\Omega}_3^2$ denote the two separate $3-$sphere.

Answer 1

For the ADS parts, the global patch is given by \begin{align} ds^2 = R^2 \left( -\cosh^2(\rho) d\tau^2 + d\rho^2 + \sinh^2(\rho) d \Omega_{n-2}^2 \right) \end{align} This covers the entire hyperboloid with $\tau\in [0,2\pi]$ and $\rho \in \mathbb{R}^+$.

And, the Poincare patch is given by \begin{align} ds^2 = \frac{R^2}{z^2} \left( dz^2 + dx_\mu dx^\mu \right) \end{align} where $dx_\mu dx^\mu = - dt^2 + d\vec{x}^2$.

Now the $S^5$ can be described by \begin{align} ds^2 &= d \theta^2 + \cos^2(\theta) d \phi^2 + \sin^2(\theta) d \tilde{\Omega}_3^2 \\ &=d \theta^2 + \cos^2(\theta) d \phi^2 + \sin^2(\theta) \left( d \phi_1^2 + \cos^2(\phi_1) d \phi_2^2 + \sin^2(\phi_1) d \phi_3^2 \right) \end{align}

So $AdS_5 \times S^5$ metric in global patch is given by \begin{align} ds^2 = R^2 \left( - \cosh^2(\rho) d\tau^2 + d\rho^2 + \sinh^2(\rho) d \Omega_3^2 + \cos^2(\theta) d \phi^2 + d\theta^2 + \sin^2(\theta) d \tilde{\Omega}_3^2 \right) \end{align} In Poincare coordinates, we have \begin{align} ds^2 = R^2 \left( \frac{dz^2}{z^2} - \frac{dt^2}{z^2} + \frac{d\vec{x}^2}{z^2}+ \cos^2(\theta) d \phi^2 + d\theta^2 + \sin^2(\theta) d \tilde{\Omega}_3^2 \right) \end{align} where $d\vec{x}^2 = dx_1^2 + dx_2^2 + dx_3^2$.