# Does $\mathcal{M} = AdS_2 \otimes S_2$ makes any sense as a manifold?

I'm not a topologist or a group theorist and I need a clarification about some notations.

Consider the Bertotti-Robinson metric in General Relativity (relativity students should study this metric, by the way, it's a really nice one!): $$$$\tag{1} ds^2 = dt^2 - a^2 \, (d\vartheta^2 + \sin^2 \vartheta \, d\varphi^2) - \sin^2{\!\omega t} \: dz^2,$$$$ where $$a$$ and $$\omega$$ are some constants. This metric is often described as the direct product of an ordinary sphere ($$S_2$$) and a 2 dimensional Anti-deSitter spacetime ($$AdS_2$$). It's usually described as $$AdS_2 \otimes S_2$$. I have three simple questions:

1. In this example, is the direct product $$\otimes$$ the same as a cartesian product $$\times$$ ? Does it make sense to write $$AdS_2 \times S_2$$ instead? While I know what is the direct product of matrices and cartesian product of vector spaces, I'm a bit confused here!

2. If the whole 4D spacetime manifold is $$\mathcal{M}^4$$, does it make sense to write $$\mathcal{M}^4 = AdS_2 \otimes S_2$$ ? What about $$\mathcal{M}^4 = AdS_2 \times S_2$$ ?

3. Is $$S_2 \otimes AdS_2$$ the same thing as $$AdS_2 \otimes S_2$$? I know that the direct product of matrices isn't commutative ($$A \otimes B \ne B \otimes A$$), but I wonder if this is pertinent to the description of manifolds, not matrices.

It is actually the direct product, not the tensor product (physicists frequently get too sloppy and end up using one or another without acknowledging the difference between the two). It is trivial to show that a direct product of manifolds is a manifold.

Direct product is commutative, so $$S_2 \times AdS_2 = AdS_2 \times S_2$$ as manifolds in the sense that there exists a 1:1 diffeomorphism.

Also, tensor product of vector spaces is commutative as well (in the sense that the two resulting vector spaces are isomorphic as vector spaces, but as pointed out by Accidental@ in the comments, this doesn't hold for the elements of vector spaces which are also usually tensored together by abuse of notation). For every two vector spaces $$U$$ and $$V$$, there exists a canonical isomorphism between $$U \otimes V$$ and $$V \otimes U$$: $$u \otimes v \mapsto v \otimes u.$$

• Ahaa! Thanks, this is already clarifying what I believed, and was confused in reading everywhere the direct product of spaces instead of cartesian product. So $AdS_2 \otimes S_2$ doesn't make any sense, like what I thought, right? – Cham Aug 9 '19 at 22:28
• @Cham it is really hard to answer in the affirmative because there is always a possibility that any mathematical construction might have been defined in some weird way by some author which I'm not aware of. Let's say it is much more probable that the authors who use $\otimes$ with two manifolds as operands are just being sloppy and they actually meant $\times$, than it is that they actually assign some sort of a meaning to that expression. – Prof. Legolasov Aug 9 '19 at 22:31
• I believe that using $\otimes$ in $AdS_2 \otimes S_2$ is really a source of confusion. And they are all writing "direct product", instead of "cartesian product". This is a big source of confusion, I believe. Maybe there's something important that I'm missing here. – Cham Aug 9 '19 at 22:35
• @Cham afaik direct product and Cartesian product are essentially the same thing (Wikipedia says the former is a generalization of the latter), whereas the tensor product is qualitatively different. They are using $\otimes$ though which stands for tensor product. The fact that they call it a direct product further reinforces my (already strong) belief that they're using the wrong symbol and it should be $\times$, not $\otimes$. – Prof. Legolasov Aug 9 '19 at 22:38
• @Prof.Legolasov tensor products are not commutative. There is indeed an isomorphism, but that does not mean it is commutative. (cf. e.g. Symmetric monoidal category for the categories where there is such an isomorphism; not all categories have one though). – AccidentalFourierTransform Aug 10 '19 at 0:37

Take a look at:

Extremal limits and black hole entropy Sean M. Carroll, Matthew C. Johnson, Lisa Randall

In this paper they show the extremal limit of a Kerr black hole where $$r_\pm=m\pm\sqrt{m^2-a^2cos^2\theta}$$ and $$r_+\rightarrow r_-$$ discontinuously pushes the spacelike region between the horizons into $$AdS_2\times\mathbb S^2$$.

I illustrate something about $$AdS/black~hole$$ correspondence by showing how the near horizon condition for an accelerated observer is approximately $$AdS_2\times\mathbb S^2$$.

• Do you agree that using the symbol $\otimes$ instead of the cartesian product $\times$ is a confusing mistake, in $AdS_2 \otimes S_2$? – Cham Aug 10 '19 at 1:12
• Yes I agree, I just did not make comment on that as it had already been done. A Cartesian product of A and B takes elements of those and forms ordered pairs, A direct product is such that with $AdS_2\times\mathbb S^2$ there is an $AdS_2$ at every point of $\mathbb S^2$, and the converse holds as well. – Lawrence B. Crowell Aug 10 '19 at 10:27
• Then I don't understand the difference between direct product and cartesian product. They shouldn't be exactly the same. If we take two vector spaces: $\mathscr{E}_1$ of dimension $p$ and $\mathscr{E}_2$ of dimension $q$, then the direct product $\mathscr{E}_1 \otimes \mathscr{E}_2$ should give a new vector space of dimension $p q$, while the cartesian product $\mathscr{E}_1 \times \mathscr{E}_2$ gives a space of dimension $p + q$, isn't? – Cham Aug 10 '19 at 15:08
• Also, if we take two metrics of dimension $p$ and $q$, and add them together (like the metric (1) in my question), we then get a new metric of dimension $p + q$ (on some augmented spacetime). Thus this is a cartesian product, not a direct product. – Cham Aug 10 '19 at 15:11