How to select which differentiable manifold to use to model spacetime? I'm studying differential geometry basics for general relativity. I know that spacetime is modeled as a 4-dimensional smooth manifold. Smooth manifold means that we consider a restriction of the maximal atlas such that all charts in it are compatible. A smooth manifold is specified once we choose an equivalence class of compatible atlases. Among all coordinate charts among all of those atlases (in the equivalence class), the differentiability notion is well-defined - a curve that's differentiable in one coordinate system of one atlas will be differentiable in any other coordinate system of any other atlas in the equivalence class.
This much is clear.
I'm watching a lecture on the same topic and the lecturer discusses about an issue here (the link starts at the relevant timestamp and it's just a 1.5 min watch till 24:00). For the 4D case, there exist uncountably many smooth atlases up to diffeomorphism. See also page 44 of this paper (thanks to Chiral Anomaly for this).
At 23:25 timestamp, he mentions that the differentiability of curves depends on the atlas we're using. So a curve may be differentiable w.r.t. one atlas and not to another (not in the same equivalence class but another smooth atlas nonetheless).
So this amounts to saying that depending on the choice of smooth atlas equivalence class (which amounts to choosing what differentiable manifold to use), a curve may be differentiable in one choice and not in another.
This makes it seem that the differentiability notion is ill-defined. The lecturer talks about the same issue briefly here as well, but doesn't get around to discussing the resolution. How is this issue resolved?
 A: I put this here since it doesnt fit in a comment. Its not an answer exactly, but I hope it may clear things up. Its pretty speculative too. Perhaps someone more knowledgeable can re-affirm or deny this. And point to the right citations.
I asked a very similar question a few years ago, inspired by the same lecturer. I think the answers tend to be confusing, because different people read the question differently. I certainly didn't end up any wiser. The question posed by the lecturer is one of differentiable structure, which then effectively becomes the question from this title:
What manifold can I use?
This is not to be confused with choices of metrics/connections, we are talking about the atlas here. When discussing what differentiable manifold to use (and only later equip with a metric) the most obvious answer is "whatever manifold you like"; you can pick one exactly as exotic as you please. This is not a helpful answer to me, because in most physics textbooks I am familiar with, a lot of calculations use a predetermined chart and deal with business locally. Making you wonder how to check if such a chart was compatible in the first place. With these sort of exercises with charts given, of course you are to presume compatibility. But what if I want to come up with my own chart, without relying on a given one?
But what if we don't want an exotic manifold?
I am not familiar enough with the subject in order to advise on curved spaces in general and expanding universes and so on. But there is an obvious choice for a 'standard' differentiable structure on $\mathbb{R}^4$ which doesn't strike me as a poor choice for flat space. For any $\mathbb{R}^d$ there is a trivial chart given by the identity on $\mathbb{R}^d$ on any open set (even globally). This satisfies any conditions you have for a $C^0$ chart: its a continuous map from the manifold into $\mathbb{R}^d$ for the right value of $d$. Thus there must be a unique maximal $C^\infty$-atlas that contains this chart. I have not seen this mentioned as 'the standard structure' anywhere, but I think it is. This is not a surprising structure for $\mathbb{R}^d$ because its equivalent to saying "I define being compatible with my altlas on $\mathbb{R}^d$ as being compatible with the notion of differentiation we already have on $\mathbb{R}^d$ (by virtue of it being a field and having a well defined subtraction and so on)."
Additionally, as far as I can tell, taking $\mathbb{R}^4$ with its standard topology and this differentiable structure still allows for curvature which is then necessarily encoded by/in the metric. This discussion is likely confused by the fact that coordinates used by physicists seem to depend on the (curvature) situation at hand. Possibly to get nice expressions of geodesics or something like that. Notice that the Lagrangian should probably depend on the curvature, but not on the coordinates used. In the end picking the 'natural' differentiable structure out of several shouldn't be shocking. We do the same thing with the topology where there is also a uncountable number of choices. The main difference seems to be that, for the case of the topology, this is well discussed.
Also notice that physicists tend to work locally. Which means its good enough when the atlas locally looks like the standard $\mathbb{R}^d$ one. In practice this strikes me as a distinction without a difference. Your predictions are the same.
Again, I want to be careful here: I am not familiar enough with the subject to be able to tell you what structure is normally picked in what case. But picking the standard one in the above sense and going with it unless you have a reason to pick something else, seems like the pragmatic move.
Finally
From what I can piece together from the lecturer it sounds like mathematicians expected there to be a unique structure on $\mathbb{R}^4$ (up to isomorphism), before this was disproven. Perhaps this is why I have yet to find a mention of 'standard structure' in this context: people thought any structure would be equivalent, so point in explicitly picking one. Even though most people would probably use the above one when an explicit choice was needed.
I saw a similar confusing discussion on my own SE question and your related physicsformums thread. What I think is going on is the following: when this question pops up, people who are intimate with this subject might be more familiar with GR on different manifolds as well. As such they might disagree with a statement like "GR is done on standard $\mathbb{R}^4$" because it may not hold in general. They might then start arguing with the proponents of such a view who interpreted it in a more pragmatic way (what structure does one normally pick in the more mainstream cases?) This later group of people  is usually less sure of their answer and might be dissuaded. I am basically suggesting their answer might actually be the one you might be looking for.
A: This is not really an answer, but aims to clear up some of the confusions that can happen with this problem.
There are three levels of structure that are at play here :

*

*an underlying topological space (usually $\mathbb R^4$, but in the case of a Schwarzschild BH, we need to remove the worldline of the singularity for everything to work out properly later)

*endowed with a smooth structure, ie an equivalence class of compatible smooth atlases.

*a metric, which induces a linear connection, curvature tensor, etc.

GR usually takes as an axiom "space-time is a $4$-dimensional smooth manifold", meaning that the choice of the first two structures is free.
The issue of exotic 4-manifolds : In $4$ dimension, some topological spaces can be equipped with several distinct smooth structure.  First, it is important to note that in the definition of a $4$-manifold, it is always the standard euclidean $\mathbb R^4$ which is used. Then, remains the question on whether we could do GR on an exotic $4$-manifold. In principle there is nothing preventing us from doing so, and only the physical consequences of such exotic space-times could tell if this is useful or not. This is discussed in this article and this one for example.
A: From a physics point of view, the answer is "whichever combination of manifold, atlas, and connection is simplest but adequate to make sense of experimental data". By "simplest" here one means "simplest to human minds and discussions among a community of people".
When we do General Relativity we do not start by setting out the most general definition of what a manifold could possibly be. Rather, we give some cursory attention to that and then just jump in with some reasonably general idea and hope that it will do the job. The job being to make sense of experimental data and to make an intellectually satisfying whole.
