Analytic Manifolds and General Relativity I'm currently taking a course in General Relativity and We've discussed the mathematical formulation of GR. The professor told us that an analytic manifold $C^{\omega}$ is not a good choice for a manifold, but I can't think of a reason why we shouldn't choose an analytic one.
 A: Analiticity is a really strong requirement. If you are dealing with an analytic function, the function is completely defined on the entire complex plane if you manage to get information about it on an arbitrarily small neighborhood of any point of the plane. For example, if you know the function is analytic on the plane and you know $f(z)$ on the region $|z-a| < \epsilon$, then you get the function on the whole plane by writing the Taylor series
$$f(z) = \sum_{n=0}^{+\infty} \frac{f(a)^{(n)} (z-a)^n}{n!}.$$
From just a small neighborhood, you got the whole function.
A problem with using analytic manifolds in GR is similar: knowing information on a small piece of spacetime would determine it everywhere. This seems nice at first, but it is not. Suppose, for example, that you were modelling a spherically symmetric star of radius $R$. Now, the way we usually do this is of the sort
$$g_{ab} = \begin{cases}\text{vacuum solution for } r > R, \\ \text{interior solution for } r < R. \\\end{cases} \tag{1}$$
The vacuum solution is just the Schwarzschild solution, which, by Birkhoff's theorem, is the only spherically symmetric vacuum solution of the Einstein Field Equations.
Notice that the scheme of Eq. (1) applies to all spherically symmetric stars (no matter what they are made of), to black holes, to planets, and etc. If it is spherically symmetric, we can model it in that way.
However, if the spacetime was analytic, then a tiny bit of the outer vacuum solution would already be sufficient to determine the entire spacetime, in analogy with the Taylor series. This is not what we have in real life. In real life, the gravitational field of the Sun as felt on Earth won't depend on the details of the Sun's composition (if we assume it to be spherically symmetric), just on its mass, charge and angular momentum. The Earth would move in the same way if instead of the Sun we had a black hole with the same mass, charge and angular momentum. Nevertheless, inside the Sun, the gravitational field is quite different from that of a black hole.
To go back to my analogy with one-variable functions, the gravitational field of the Sun and of a black hole are like two functions $f$ and $g$ that respect $f(z) = g(z)$ for $|z| > R$, but might differ for $|z| < R$. This is only possible if they are not analytic. If they were analytic, being equal on any neighborhood would imply being equal on the whole domain, and that is certainly not interesting in gravitational physics.
Edit: an analytic manifold needs not have an analytic metric
This was pointed out in the comments and I agree with it. My answer used the metric as an example, so I should give some more detail, but the main point is still the same: analytic functions are way too restrictive. I'll first argue why one would not want to work with analytic manifolds, but on the end I'll also show some arguments of why there is no issue in doing it.
Some constructions often used in Differential Geometry are partitions of unity. A partition of unity is a collection of functions $\lbrace f_{\alpha}\rbrace$ that always add to $1$ (\sum_\alpha f_{\alpha} = 1) but for each point of the manifold there is a neighborhood in which only finitely many of them are non-vanishing. Hence, the functions belonging to a partition of unity will often have to vanish on much of the manifold and assume non-zero values on other points. See nLab an example on the real line, in case it helps to grasp the concept. The Wikipedia page also has a really nice picture.
While it can be shown that smooth partitions of unity always exist in (paracompact) smooth manifolds (see nLab). However, you'll typically won't find analytic partitions of unity in an analytic manifold, simply because analytic functions are too restrictive and refuse to give non-vanishing values at a point if they vanish on a neighborhood somewhere else.
This in principle might sound a bit too technical, but it becomes relevant because partitions of unity are used, for example, to define integration on manifolds. The trick is to write an integral over $M$ by writing it as
$$\int_M F(x) \mathrm{d}x = \sum_{\alpha} \int_{U_\alpha} F(x) f_\alpha(x) \mathrm{d}x,$$
where $\lbrace f_\alpha\rbrace$ is a partition of unity, $\lbrace U_\alpha\rbrace$ covers the manifold, and for each $\alpha$ $f_\alpha$ vanishes outside of $U_\alpha$. In the absence of analytic partitions of unity, one can't define an ``analytic integral''. According to this MathOverflow post, what one usually does is exploit the fact that analytic manifolds are smooth manifolds and define integration in a smooth, non-analytic, manner. (Let me point out that the comments to that very same question also point to other posts that apparently discuss how to define integration without the need for partitions of unity).
In short, while analyticity is nice, it is also quite restrictive. One might need to resort back to smooth, non-analytic results.
Is analyticity a problem?
I should also add a comment from Hawking & Ellis' The Large Scale Structure of Spacetime, p. 58:

If the metric is assumed to be $C^r$, the atlas of the manifold must be $C^{r+1}$. However, one can always find an analytic subatlas in any $C^s$ atlas ($s \geq 1$) (Whitney (1936), cf. Munkres (1954)). Thus it is no restriction to assume from the start that the atlas is analytic, even though one could physically determine only a $C^{r+1}$ atlas if the metric were $C^r$.

(The link for Munkres (1954) corresponds to what I could find online).
Hence, Hawking & Ellis see no issue if you want to consider an analytic manifold structure (notice they don't mention restricting the metric to only analytic metrics: that would definitely be way too restrictive). This might make you have some more of a headache working with some things such as partitions of unity (which can be done by treating the analytic manifold as a smooth manifold).
A: I will give a slightly different take on this: Not only is it generally not problematic to chose an analytic manifold as a starting point in general relativity, it is mostly desirable.
One point of confusion is that a manifold being analytic does not imply in any way that all functions on that manifold have to be analytic. In particular, it does not imply that metric or partitions of unity on the manifold be analytic (as the answer by Nicholas Alves explains that would be a unnecessarily strong restriction). It merely means that all the coordinate transformations between overlapping charts be analytic.
There is actually a somewhat opposite implication: If a manifold is not analytic, we cannot define what it means for a function on the manifold to be analytic, as functions may that are analytic at a point in one chart, may not be analytic at that same point in another chart. Not being able to define analyticity is a hugely undesirably restriction, because while analyticity is a powerful constraint, it comes with equally powerful techniques when it does apply.
For example, when we take our manifold to be analytic, the Einstein equation implies that vacuum metrics (with analytic boundary conditions) are themselves analytic. This by itself make analyticity of the base manifold desirable.
In practice analyticity of the base manifold does not lead to much of a practical constraint. In fact, I am hard pressed to think of any practical situation in general relativity where one does not use an analytical atlas of charts.
