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In contemporary textbooks on differential geometry, the definition of smooth manifolds is given in a (IMHO) awkwardly obfuscated way, by saying that a smooth manifold is a topological space endowed with an equivalence class of compatible atlasses. Why does it not suffice to define a differentiable manifold as a topological space together with a single, not necessarily maximal, atlas?

If I'm not mistaken, when you proceed by defining smooth maps between manifolds in the usual way, you end up with a category which is at least equivalent to the usual category.

One may argue that the definition of a smooth manifold in terms of a smooth structure has the advantage that two manifolds on the same underlying set which are diffeomorphic by means of a diffeomorphism which is simply the identity on the underlying set, are in fact the very same object, but I don't see why this should be a technical advantage.

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  • $\begingroup$ I wouldn't underestimate the importance of the point you acknowledge in your final paragraph. When thinking about smooth manifolds simpliciter, do you really want to bother distinguishing $S^2$ with the atlas given by stereographic projection from $(0,0,\pm 1)$, from $S^2$ with the atlas given by stereographic projection from $(\pm 1, 0, 0)$, from $S^2$ with the atlas given by your favourite homeomorphism of $S^2 \cap \{z > -\epsilon\}$ and $S^2 \cap \{z < \epsilon\}$ with open discs? $\endgroup$ Jan 4, 2014 at 22:26
  • $\begingroup$ @BranimirĆaćić: I wouldn't bother because both are diffeomorphic. I guess that you would neither care whether $S^2$ was the unit sphere in $\Bbb R^3$ or the one-point compactification of $\Bbb R^2$. $\endgroup$
    – Dominik
    Jan 4, 2014 at 22:36
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    $\begingroup$ @Dominik, supposed that the category of one-atlas manifolds and traditional manifolds are indeed equivalent, you absolutely convinced me. $\endgroup$
    – Berci
    Jan 4, 2014 at 23:31

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If one makes the definition as in your first paragraph one obtains "too many" objects. On any manifold that carries at least one nontrivial differentiable atlas there'd be infinitely many different such atlasses (just add or remove a few maps), resulting in inifinitely many differentiable manifolds on the same underlying topological manifold [EDIT: For a concrete example see Branimir Çaçiç's comment]. Thus results that on certain topological manifolds there are only so-and-so many differential structures could not be formulated with the language of differentiable manifold. If taking a single atlas would make sense at all then it would be a maximal one. But that would make certain arguments more involved as well (and one would have to show the existence of a maximal atlas above an atlas one can easily describe).

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    $\begingroup$ Unfortunately, even with the standard definition, one cannot simply say for example that there are "28 differentiable structures on $S^7$". This is because even $\Bbb R$ has many different (but isomorphic) smooth structures such as the one given by the chart $x\mapsto x^3:\Bbb R\to \Bbb R$. $\endgroup$
    – Dominik
    Jan 5, 2014 at 21:45
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There is a reason to keep the maximal atlas in the back of your mind and that is the engineering aspect of mathematics. Certain choices of a sub-atlas, like the one you may have started with, for representing such things as functions on the manifold may not be very convenient. If you are illustrating the manifold and functions on it, or fitting functions to data, then certain charts may be much more efficient. It depends on the things you might want to represent and how the coordinate spacing is on the manifold. If the thing you are trying to represent varies rapidly just where the coordinate spacing is greatest then you may need another chart. So the wider atlas is a convenient toolkit, if you need it and can get access to the best charts.

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