convention of a default atlas Recently, I have been studying the basics of differential geometry and te necessary preliminaries. I arrived at the construction of differential structure on topological manifolds, where the non-uniqueness of these structures is mentioned. After this, the precence of a (pick of) structure seems to be assumed in most of what follows.
From what I can tell, it often looks like there is an obvious choice, while the others are labled 'exotic'. This seems to be most obvious when looking at $\mathbb{R^4}$. There is an entire wikipedia page dedicated to exotic versions of $\mathbb{R^4}$, that are homeomorphic, but not diffeomorphic to $\mathbb{R^4}$ as a smooth manifold. My question then becomes: when no differential structure is given, is there a convention of a structure to pick? How do I distinguish between the exotic structures and the 'usual' ones. In the case of $\mathbb{R^4}$, I would bet the default structure would/could consist of an atlas, build from all of the open sets, equipped with the identity from $\mathbb{R^4}$ into $\mathbb{R^4}$.
As my background is in physics, I am in particular interested in the differential structure that one would usually establish on a manifold in the context of four dimentional General Relativity. But perhaps this should be a sepparate question on the physics SE. I apologize in advance for improper use of jargon or other, possibly trivial mistakes. Feel free to correct me.
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To further clairify the source of this question as well as my confusion:
From my experience with GR it seems physicists usually take a (pseudo?) Riemanian metric as a starting point. To me it seems this fixes only a piece of the manifold, described by the single coordinate map. The physicist would then declare that this describes a manifold, but i am uncertain how to pick the additional contributions to the atlas. On top of this, the physicist might sometimes perform a coordinate transformation, which seems tricky to me when the rest of the chart was not well comunicated. How would I know if the new coordinates would be in the same chart? In addition, I am uncertain weither this should be discussed here, or on the physiscs SE.
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I will ay some point pose a related question about GR on the physics SE. The question then reduces to one of mathematical convention: weither or not $\mathbb{R}^n$ is usually assumed to be equipped with the identity on open sets, as an atlas, and weither or not other 'assumptions' exist when the atlas definition is left out.
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new related thread in physics SE https://physics.stackexchange.com/questions/566643/how-to-select-which-differentiable-manifold-to-use-to-model-spacetime/661820
 A: Yes you're right; if $M$ is presented to you as being embedded in $\mathbb R^n$, the default atlas would be some covering of $M$ with open sets in $\mathbb R^m$ (standard topology) and the identity map on $\mathbb R^m$, where $m=\dim(M)$.
A: First of all, to comment on Hoot's comment, spacetime is (unless you assume there is no gravity, but then you are really doing special relativity and the whole manifold aspect becomes a bit obsolete as you have global coordinates available) definitively not $\mathbb{R}^4$. It is a 4-dimensional connected (and usually taken to be) non-compact smooth manifold which is equipped with a Lorentzian metric, i.e. a Minkowski metric at each tangent space.
To see what physicists do and don't do when discussing GR you have to realise what assumptions we make. We assume, or rather demand (following Einstein), that spacetime is locally a Minkowski space, and that's probably what you have interpreted as [insert source] taking a Lorentzian metric as a starting point locally and then working from there. We also assume that spacetime is an actual manifold, so there is an atlas, we just don't really worry about because we consider stuff locally (although the global structure is eventually certainly relevant).
