MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

So it is relatively easy to show that there exists only one smooth structure on the real line $\mathbb{R}$. So here are 2 natural questions:

Q1: Up to equivalence, is there only one real analytic structure on $\mathbb{R}$? If so, then do we have a simple proof of that?

Q2: Where can I find the simplest proofs that there exists only one smooth structure on $\mathbb{R}^2$ and $\mathbb{R}^3$?

So I've heard that on $\mathbb{R}^4$ there are infinitly (in fact uncountably) many non-equivalent smooth structures, so what about real analytic strucutres? Is there some kind of moduli space of smooth structures on $\mathbb{R}^4$. if so, in how many ways is it possible to deform a smooth structure into a real analytic one?

share|cite|improve this question
Thanks a lot Steven for the link. So do you know of a simple proof that there exists only one real analytic structure on $\mathbb{R}$ which is compatible with its smooth structure? – Hugo Chapdelaine Jan 20 '11 at 18:59
Grauert-Remmert is probably irrelevant for this simple case. – Hugo Chapdelaine Jan 20 '11 at 18:59
It was my (offhand) impression that every topological group with underlying space an $\mathbb{R}$-manifold had a unique structure as a real-analytic Lie group (and that this is part of the Gleason-Montgomery-Zippin theory). This is at least one attractive uniqueness result. – Pete L. Clark Jan 20 '11 at 22:35
up vote 2 down vote accepted

Regarding Q1, put an analytic Riemann metric on your 1-manifold. Integrating a unit speed vector field gives an analytic diffeomorphism to $\mathbb R$. Another way to prove analytic structures are unique is to notice the same argument that one uses to prove that the group of $C^k$-diffeomorphisms of $\mathbb R$ has the homotopy type of $\mathbb Z_2$ works for analytic diffeomorphisms -- simply take the straight-line homotopy between your original diffeomorphism and either the identity or the negative identity, appropriately.

Regarding Q2, I don't know much in the way of really simple proofs. But when $n=2$ you've got the Uniformization Theorem from complex analysis. That's relatively simple.

share|cite|improve this answer
Hi Ryan, may be I miss something here, but how do you put an analytic Riemann metric on a smooth manifold? Usually you piece your local inner products using a partition of unity which forces you to work in the smooth setting. – Hugo Chapdelaine Jan 21 '11 at 2:29
You put the analytic Riemann metric on the analytic manifold. The flow from the ODE gives you an analytic diffeomorphism to $\mathbb R$ with its standard (analytic) structure. – Ryan Budney Jan 21 '11 at 3:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.