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Why is $S^1\times\mathbb{R}^{n-1}$ the topology of $AdS_n$?

+ 5 like - 0 dislike
227 views

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][1]; 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?

  [1]: http://www.fysik.su.se/~ingemar/Kurs.pdf


This post imported from StackExchange Physics at 2014-05-04 11:13 (UCT), posted by SE-user Pedro Figueroa

asked May 3, 2014 in Theoretical Physics by Pedro Figueroa (85 points) [ revision history ]
edited May 7, 2014 by Pedro Figueroa
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.

This post imported from StackExchange Physics at 2014-05-04 11:14 (UCT), posted by SE-user Matt Reece
@MattReece Yes, with the unfortunate side effects that it's just confusing and makes them look less mathy (imho).

This post imported from StackExchange Physics at 2014-05-04 11:14 (UCT), posted by SE-user joshphysics

2 Answers

+ 4 like - 0 dislike

The Cartesian product of two topological spaces \(M\)and \(N\) can be thought of as the set of all ordered pairs selected from \(M\) and \(N\).  (There's more to the definition than this, but this is what's pertinent for our purposes.)  For example, the Cartesian product of two open/closed intervals is an open/closed rectangle.  This means that if you're going to tell me that AdSn has the same topology as \(\mathbb{R}^n \times S^m\), I should always be able to pick a point from the Euclidean space and a point from the m-sphere, and you can show me the point in AdSn that corresponds to that ordered pair.

With that in mind:  if you write your constraint as 

\((x^0)^2 + (x^1)^2 = \alpha^2 + \vec{x}^2\),

then it's not too hard to see that this defines a circle in the \(x^0\text{-}x^1\) plane for any value of \(\vec{x}\), with radius \(\sqrt{\alpha^2 + \vec{x}^2}\).  This means that if I pick a point on the circle \(S^1\), and pick a point in the Euclidean space \(\mathbb{R}^{n-1}\), then there will always be a unique point in AdSn corresponding to that choice.

On the other hand, if you write 

\(\vec{x}^2 = (x^0)^2 + (x^1)^2 - \alpha^2\),

then this does not define an (n-2)-sphere for all possible choices of \(x^0\) and \(x^1\), since the right-hand side can become negative.  Thus, your proposed map from ordered pairs in \(\mathbb{R}^2\) and \(S^{n-2}\) to AdSn doesn't work, since I can choose points from the base spaces that have no corresponding point in AdSn (namely, any point from \(\mathbb{R}^2\) that has \((x^0)^2 + (x^1)^2 < \alpha^2\), and any point at all from \(S^{n-2}\).)  

Of course, this flaw in your second argument doesn't prove on its own that AdSn isn't homeomorphic to \(\mathbb{R}^2 \times S^{n-2}\)—merely that this map isn't a homeomorphism.  But since we already found another argument that AdSn was homeomorphic to \(\mathbb{R}^{n-1} \times S^1\), and since these two product spaces aren't homeomorphic to each other (they're topologically distinct, so they can't be), we can conclude that AdSn isn't homeomorphic to \(\mathbb{R}^2 \times S^{n-2}\).  (Unless, of course, \(n = 3\).)

answered May 5, 2014 by Johnny Assay (70 points) [ revision history ]
I like these nice rather intuitive explanations, +1. Just recently I was reading up about AdS/CFT a bit, so this post came in quite handy :-)

Great, it is indeed pretty simple, thanks.

+ 3 like - 0 dislike

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

This post imported from StackExchange Physics at 2014-05-04 11:14 (UCT), posted by SE-user Qmechanic
answered May 3, 2014 by Qmechanic (2,790 points) [ no revision ]
Does this mean that the tensor product $\otimes$ should be replaced by the Cartesian product $\times$ in the original question?

This post imported from StackExchange Physics at 2014-05-04 11:14 (UCT), posted by SE-user Hunter
$\uparrow$ @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.

This post imported from StackExchange Physics at 2014-05-04 11:14 (UCT), posted by SE-user Qmechanic

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