E.g. if I know that my topology is that of a hyperboloid, how much freedom do I have left for my choice of the metric? And the other way around: if my metric is some conformal factor times the unit hyperboloid how much freedom do I have left in my choice for the topology? Am I forced to conclude that the topology of this (Lorentzian) spacetime needs to be that of a hyperboloid?

  • $\begingroup$ Certainly the latter statement is at least partially wrong--your spacetime could be a manifold-with-boundary, or have disconnected subcomponents, for instance. $\endgroup$ – Jerry Schirmer Jul 30 '13 at 21:26

Let a smooth manifold $M$ be given, namely a suitably non-pathological topological space along with a smooth structure. A metric $g$ on $M$ is any symmetric, nondegenerate $(0,2)$ tensor field on $M$ having constant index. The pair consisting of $(M,g)$ is then called a semi-riemannian manifold.

So suppose you hand someone the a topological space like a hyperboloid $H$, and suppose that someone turns this hyperboloid into a smooth manifold by equipping it with a smooth structure, then you can still choose any metric on the hyperboloid you wish and make it a perfectly acceptable semi-riemannian manifold, so needless to say, there's a whole lot of freedom.

Note, however, that (especially in physics literature on, say, AdS/CFT) a hyperboloid $H$, as a smooth manifold, is often defined as an embedded submanifold of some higher dimensional smooth manifold $N$. Moreover, as it often happens, this higher dimensional smooth manifold already has a specified metric $g_N$. In this case, there is a particularly natural choice for the metric on $H$; it's called the pullback metric or the induced metric on $H$ depending on who you're talking to. The pullback metric is defined as follows. If $f:M\to N$ is the embedding through which $H$ was defined (in other words $H$ is just the image of $M$ under $f$), then the pullback metric is usually denoted $f^*g^N$ and is defined as follows: $$ (f^*g^N)_p(v,w) = g^N_{f(p)}(df_p(v), df_p(w)) $$ for all $p\in M$ and $v,w\in T_pM$. Intuitively, the pullback metric is precisely the notion of distance inherited by $H$ from $N$, a smooth manifold that already has a notion of distance defined on it. I can't remember the last time I encountered a situation in the (high energy theoretical) physics literature in which someone considered an embedded submanifold with a metric other than the induced metric.

Now, for the second question which is basically the reverse of the first, we need to be careful. When we use the word "metric" as defined above, implicit in the use of this terminology is the existence of a smooth manifold on which the metric is defined, so in this sense, the reverse direction doesn't make much sense.

However, I think (and correct me if I'm wrong) you're second question can be reformulated as follows:

For each $\mu,\nu = 1, \dots, n$, I have a matrix $g=(g_{\mu\nu})$ of functions $g_{\mu\nu} = g_{\mu\nu}(x_1, \dots, x_n)$ defined on some open subset of $\mathbb R^n$ that is nondegenerate and has constant index, then does there exist a realization of $g$ as the local coordinate representation of the induced metric of some embedded submanifold $M$ of some ambient manifold $N$? If so, is this realization unique?

If this is not a correct reformulation of your question, then let me know in the comments below.

I don't know the definitive answer to these questions, but I strongly suspect that the answers are yes and no respectively. If such a realization exists, then, as Jerry Schirmer basically points out in his comment, you can just, for example, chop $M$ into two, disconnected pieces to obtain another realization.

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  • $\begingroup$ Your correctly rephrased my question. To avoid the point @JerrySchirmer raised, I would like to put the additional assumptions that the manifold M is without boundary and connected. Would any one know the answer to this more restrictive question? $\endgroup$ – user25477 Aug 1 '13 at 14:05

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