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

Question

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

Comments

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

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@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

Answer 1

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 AdS_n 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 AdS_n 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 AdS_n 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 AdS_n doesn't work, since I can choose points from the base spaces that have no corresponding point in AdS_n (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 AdS_n 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 AdS_n 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 AdS_n isn't homeomorphic to $\mathbb{R}^2 \times S^{n-2}$.  (Unless, of course, $n = 3$.)

Comments

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 :-)

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Great, it is indeed pretty simple, thanks.

Answer 2

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

Comments

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

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$\uparrow$ @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