Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
5  
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. –  Matt Reece May 3 at 18:05
3  
@MattReece Yes, with the unfortunate side effects that it's just confusing and makes them look less mathy (imho). –  joshphysics May 3 at 21:00

1 Answer 1

up vote 2 down vote accepted

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 $S^1\times \mathbb{R}^{n-1} \subset \mathbb{R}^{n+1}$.

share|improve this answer
1  
Does this mean that the tensor product $\otimes$ should be replaced by the Cartesian product $\times$ in the original question? –  Hunter May 3 at 11:30
2  
$\uparrow$ @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. –  Qmechanic May 3 at 13:02
    
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 $U$. –  Qmechanic May 5 at 7:16

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.