Anti-de Sitter $AdS_n$ may be defined by the quadric $$-(x^0)^2-(x^1)^2+\vec{x}^2=-\alpha^2\tag{1}$$ embedded in ${\mathbb{R}^{2,n-1}}$, where I write ${\vec{x}^2}$ as the squared norm ${|\vec{x}|^2}$ of ${\vec{x}=(x^2,\ldots,x^n)}$. Now, I don't quite understand how is it justified that the topology of this space is $S^1\times\mathbb{R}^{n-1}$. As I understand it informally, I could write $(1)$ as $$(x^0)^2+(x^1)^2=\alpha^2+\vec{x}^2\tag{2}$$ and then fix the ${(n-1)}$ terms ${\vec{x}^2}$, each one on $\mathbb{R}$, such that ${(2)}$ defines a circle ${S^1}$.

This is actually a reasoning I came up to later, based on the case of ${dS_n}$ (in which one just fixes the time variable) and when I saw what the topology was meant to be, but actually I first wrote ${(1)}$ as $$\vec{x}^2=(x^0)^2+(x^1)^2-\alpha^2$$ which for fixed ${x^0,x^1}$, both in $\mathbb{R}$, defines a sphere ${S^{n-2}}$, so the topology would be something like ${S^{n-2}\times\mathbb{R}^2}$, (which is indeed similar to that of ${dS_n}$) right? I even liked this one better, since I could relate it as the 2 temporal dimensions on ${\mathbb{R}^2}$ and the spatial ones on ${S^{n-2}}$.

I don't really know topology, so I would like to know what is going on even if it's pretty basic and how could I interpret topological differences physically.

Update: I originally used $\otimes$ instead of $\times$ in the question. My reference to do this is page 4 of Ingemar Bengtsson's notes on Anti-de Sitter space; so is that simply a typo in the notes?

Update 2: I'm trying to understand this thing in simpler terms. If I write Minkowski 4-dimensional space in spherical coordinates, could I say that it's topology is ${\mathbb{R}\times{S}^3}$? If so, how come?

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    $\begingroup$ The tensor product symbol $\otimes$ is sometimes used in physics when the product $\times$ should be. I don't know why. For instance, one sometimes encounters papers saying that the gauge group of the Standard Model is SU(3)$\otimes$SU(2)$\otimes$U(1). I don't know what that means. I think people use the notation because it makes them feel more "mathy" somehow. $\endgroup$
    – Matt Reece
    May 3, 2014 at 18:05
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    $\begingroup$ @MattReece Yes, with the unfortunate side effects that it's just confusing and makes them look less mathy (imho). $\endgroup$ May 3, 2014 at 21:00

1 Answer 1


Sketched proof: One may define a homotopy via the constraint

$$x_0^2+x_n^2~=~ \alpha^2 +\lambda \sum_{i=1}^{n-1}x_i^2,\quad \alpha >0,$$

where $\lambda\in[0,1]$ is the homotopy parameter. Then $\lambda=1$ corresponds to $AdS_n \subset \mathbb{R}^{n+1}$, while $\lambda=0$ corresponds to $S^1\times \mathbb{R}^{n-1} \subset \mathbb{R}^{n+1}$.

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    $\begingroup$ Does this mean that the tensor product $\otimes$ should be replaced by the Cartesian product $\times$ in the original question? $\endgroup$
    – Hunter
    May 3, 2014 at 11:30
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    $\begingroup$ $\uparrow$ @Hunter: Yes, that's a typo in the notes. Btw, a tensor product $V\otimes W$ of vector spaces wouldn't make natural sense here since $V=S^1$ is not a vector space. $\endgroup$
    – Qmechanic
    May 3, 2014 at 13:02
  • $\begingroup$ Concerning OP's 2nd update: A polar coordinate patch $\mathbb{R}\times S^3$ is diffeomorphic to $\mathbb{R}^4\backslash\{\bf 0\}$ not to $\mathbb{R}^4$. More generally, a coordinate patch $U$ may not cover the whole manifold $M$, and thus the topology of $U$ and $M$ may differ. In particular we can not deduce the topology of $M$ from knowing the topology of $U$. $\endgroup$
    – Qmechanic
    May 5, 2014 at 7:16
  • $\begingroup$ @Qmechanic $S^1$ is still an Abelian group though... ;) $\endgroup$
    – Danu
    Dec 3, 2016 at 11:09

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