Conformal properties of AdS space

Question

My current research is connected with anti-de Sitter ṣpace, which is why I am interested in the following question. It is well known that the metric in Poincare patch is conformally equivalent to that in half of Minkowski space. The question is if there is a similar statement about full AdS (specifically, AdS4) : is there a somewhat simple manifold to which full AdS is conformally equivalent? Topological reasons tell that it might be something like $R^3 \times S_1$; is this indeed the case? It should be simple enough, but it eludes me and I haven't found such result in literature.

Answer 1

Indeed, there is such a manifold conformally equivalent to the AdS. As explained for example here in chapter 2, we find $\text{AdS}_2$ to be equivalent to a strip, $\text{AdS}_3$ to a cylinder and in general $\text{AdS}_{n+1}$ equivalent to $\mathbb{R}\times D_n$. Here, $D_n$ denotes the $n$-dimensional open ball. However, the metric of this space is not the one obtained by the standard embedding in Euklidean space.

In this convention time is infinitly extended, thus the $\mathbb{R}$ part. Sometimes AdS is considered compact in time, for example when defined as quotient space $\text{AdS}_{n+1}=O(2,n)/O(1,n)$. In this case, the real line $\mathbb{R}$ has to be replaced by the circle $S_1$. I guess this is the case in your question.

So to conclude, you are right with AdS being homeomorphic to $S_1\times\mathbb{R}^n$, as $\mathbb{R}^n$ is topologically equivalent to $D_n$.

I hope this could answer your question!