# Why do differential geometry textbooks bother with equivalence classes of smooth structures?

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.

• 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? Jan 4, 2014 at 22:26
• @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 $\Bbb R^2$. Jan 4, 2014 at 22:36
• @Dominik, supposed that the category of one-atlas manifolds and traditional manifolds are indeed equivalent, you absolutely convinced me. Jan 4, 2014 at 23:31

• 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 $x\mapsto x^3:\Bbb R\to \Bbb R$. Jan 5, 2014 at 21:45