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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?

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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
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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.

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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
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