Is the non-simply connected version of AdS space a maximally symmetric spacetime?

Question

A common construction of anti-de Sitter space is the following:

1. Start with the flat five-dimensional manifold with metric $ds_5^2 = -du^2 - dv^2 + dx^2 + dy^2 + dz^2$.
2. Consider the hyperboloid submanifold given by $-u^2 - v^2 + x^2 + y^2 + z^2 = -\alpha^2$.
3. Define the hyperbolic coordinates $(t, \rho, \theta, \phi)$ on the hyperboloid. The coordinate $t$ is periodic with period $2 \pi$. The submanifold geometry inherited from the ambient space yields the metric $ds^2 = \alpha^2 \left( -\cosh^2(\rho)\, dt^2 + d\rho^2 + \sinh^2(\rho)\, d\Omega_2^2 \right)$.
4. Consider the universal cover of this hyperboloid, which has the same metric as above but the new coordinate $t$ ranges over the whole real line. This is AdS space.

My questions are about the original hyperboloid constructed in steps 1-3, before taking the universal cover.

Q1. What is the global topology of this spacetime? Is it $S^1 \times \mathbb{R}^3$?

Q2. Is this spacetime maximally symmetric? It seems like it should be, since (I think) it has the same Killing vector fields as AdS. But I've sometimes seen the claim that "AdS is the unique maximally symmetric spacetime with constant negative scalar curvature." Is that true, or is this hyperboloid a counterexample?

Answer 1

Related answers: this one by me and this one by Slereah, which has a classification of maximally symmetric spaces with one timelike dimension (citing Spaces of Constant Curvature: Sixth Edition by Joseph A. Wolf).

Your hyperboloid (which Wolf calls $\mathbb H^3_1$) is topologically $S^1\times\mathbb R^3$. The maximally symmetric spacetimes of that signature and curvature are the covers of $\mathbb H^3_1 / \mathbb Z_2$, so $\mathbb H^3_1$ and $\text{AdS}_4$ are maximally symmetric. $\text{AdS}_4$ is unique if you also require simple connectedness.