# How much differs the category of real-analytic manifolds from $C^\infty$ ones?

I was thinking about the difference between the concept of real-analytic function (for any point the Taylor-series of $$f$$ converge to the function in a neighborhood of the point) and complex analytic (equivalent to holomorphicity). There are many important differences indeed analyticity of complex functions is a more restrictive property, as it has more restrictive necessary conditions and complex analytic functions have more structure than their real-line counterparts.

For example in the complex case we have Liouville's theorem (any bounded complex analytic function defined on the whole complex plane is constant). But in the real analytic case this is false, just take $$f(x)=\frac{1}{1+x^2}$$ Actually, this example gives to us much more differences then this simple one.

In the real case it's a standard fact that any real-analytic function is $$C^\infty$$ but the converse is not true because of the following example $$\begin{equation*} {\displaystyle f(x)={\begin{cases}e^{-{\frac {1}{x}}}&{\text{if }}x>0,\\0&{\text{if }}x\leq 0,\end{cases}}} \end{equation*}$$It is $$C^\infty$$ but it is not real-analytic at the origin. In light of this example I would like to ask this question

How does the category of real-analytic manifolds differs from that of $$C^\infty$$ ones? Or rather, are there many examples of manifolds that admit smooth but not real-analytic atlases?

## EDIT

Here I'll try to resume all the comments below. Thanks everybody for the answers!

(1) Take a look at this question Can every manifold be given an analytic structure? with much more discussion

(2) The main resut is that: Any $$C^1$$ manifold admits a compatible real analytic structure (proved by Whitney)

• Any smooth (or even $C^1$) manifold admits a compatible real analytic structure. This is a result of Whitney (see, for instance, this question mathoverflow.net/questions/8789/… with much discussion). A real analytic manifold is distinct from merely a smooth one in that there is a well defined notion of real analytic function on such a manifold, but a smooth structure does not suffice for this. Commented Feb 7, 2021 at 16:59
• It is a result of Grauert, building on prior work of Whitney. Whitney proved that every $C^1$ admits a compatible $C^{\infty}$ structure, unique up to $C^1$ diffeomorphism. Grauert improved the result, using very different methods (Stein manifolds) to replace $C^{\infty}$ by real analytic. Commented Feb 7, 2021 at 17:08
• The forgetful functor from analytic to smooth manifolds is essentially surjective and faithful, but not full, as your example illustrates. In more detail, the fact that an analytic map between connected manifolds is completely determined by its restriction to an arbitrarily small open set means that analytic manifolds are "rigid" whereas smooth manifolds are "floppy". An actual analytic geometer could say more. (e.g. I don't know but I would assume in the presence of nontrivial topology that not every smooth map is homotopic to an analytic one.) Commented Feb 7, 2021 at 17:09
• I don't have an immediate duplicate in mind, but I'm sure I've seen some closely related variant of this question before …. It's such a good and natural question that it's hard to imagine we went over a decade without asking it here! Commented Feb 7, 2021 at 17:15
• @TimCampion You are incorrect, there are analytic approximation theorems. Everything is the same after passing to the homotopy category. A short chapter discussing analytic approximation theorems is in Hirsch's differential topology book, though I don't remember if he talks about relative analytic approximation.
– mme
Commented Feb 7, 2021 at 18:40