# Differentiable structures on R^3

This is for the sake of completeness(for references, understanding).

I ask for references for proofs that:

-There is exactly one differentiable(ie $C^\infty$) structure on $\mathbb R$, upto diffeomorphism.

-Ditto for $\mathbb R^2$.

-Ditto for $\mathbb R^3$.

-
Perhaps you could tell us where you have looked and failed to find a reference, so as to narrow our search? –  Sam Nead May 16 '10 at 22:06
Thurston's Geometry and Topology of 3-manifolds I writes something about this in chapters 3.9 and 3.10, though I haven't gone through this. –  Lennart Meier May 17 '10 at 8:49
@Sam Nead: I looked in Warner's book, Spivak(in which the $1$-dim case is an exercise, and I googled and wikipedia-ed it.). –  Akela May 17 '10 at 13:01
This question basically duplicates part of this one mathoverflow.net/questions/16035/… and in particular Michael Hoffman's answer to that question ought to be sufficient –  j.c. May 17 '10 at 16:59

UPDATE : I found some precise references that answer the OP's question and fill in some details in my original answer. See the end for them.

I don't have a precise reference for the cases of $n=2$ or $n=3$, though I suspect that they could be found in Moise's "Geometric Topology in Dimensions 2 and 3".

However, I do have a nice reference for the amazing theorem (due to Stallings) that $\mathbb{R}^n$ has a unique $C^{\infty}$ structure for $n \geq 5$. It is written up very nicely in Steve Ferry's Geometric Topology Notes. See Chapter 10 starting on page 56. What he proves is that $\mathbb{R}^n$ has a unique PL structure, but that is the key to the result. The rest of the notes are also wonderful.

This theorem is surprising for a number of reasons. For instance, it says that if $\Sigma$ is an exotic $n$-sphere for $n \geq 5$, then $\Sigma \setminus \{p\}$ is diffeomorphic to $\mathbb{R}^n$ for any point $p \in \Sigma$. In other words, the "exoticness" is "concentrated at a point". Of course, this also follows from the usual proof of the high dimensional Poincare conjecture using the $h$-cobordism theorem, which constructs a homeomorphism between a homotopy sphere and the usual sphere which is differentiable except at one point (that one point giving trouble due to the "Alexander trick").

Another remark that should be made is that Freedman and Donaldson proved that $\mathbb{R}^4$ has uncountably many $C^{\infty}$ structures.

UPDATE : OK, here are some precise references. In

MR0121804 (22 #12534) Munkres, James Obstructions to the smoothing of piecewise-differentiable homeomorphisms. Ann. of Math. (2) 72 1960 521--554.

it is proven in Corollary 6.6 that two PL-isomorphic differentiable manifolds that are homeomorphic to $\mathbb{R}^n$ are actually diffeomorphic. To make sense of this, recall that Cairns proved that differentiable manifolds have canonical PL structures; the original reference for this is his paper

MR0017531 (8,166d) Cairns, Stewart S. The triangulation problem and its role in analysis. Bull. Amer. Math. Soc. 52, (1946). 545--571.

I think JHC Whitehead might have also proven this, but I don't have a reference for that.

This reduces us to showing that $\mathbb{R}^n$ has a unique PL structure. The cases $n=2$ and $n=3$ can be found in Moise's book "Geometric Topology in Dimensions 2 and 3". As far as the original results go, the 2-dimensional case is classical, while the 3-dimensional case was originally proven by Moise in his paper

MR0048805 (14,72d) Moise, Edwin E. Affine structures in $3$-manifolds. V. The triangulation theorem and Hauptvermutung. Ann. of Math. (2) 56, (1952). 96--114.

For dimensions at least $5$, the original reference is the following

MR0149457 (26 #6945) Stallings, John The piecewise-linear structure of Euclidean space. Proc. Cambridge Philos. Soc. 58 1962 481--488.

Do you mean $\Sigma\setminus p$ is diffeomorphic to $\mathfrak R^n, or am I misundersanding? – Emerton May 17 '10 at 3:55 Whoops! That's what I meant. I'll fix it now. Thanks! – Andy Putman May 17 '10 at 4:22 Great notes! Finally I'll understand what this geometric topology is all about:) – Victor Protsak May 17 '10 at 7:11 Thank you very much, this is really good reference! – Martin Jul 4 '10 at 16:46 add comment There are infinitely many differentiable structures on$\mathbb R$: take any homeomorphism which is no diffeomorphism (such as$x\mapsto x^3$), and you get an non-usual differentiable structure on$\mathbb R$! Even better : there exist uncountably many different real analytic structures on$\mathbb R$. But this example is general : for every$k\in \mathbb N \cup \{\infty,\omega\}$, the group Homeo($\mathbb R$) acts transitively on the set of$C^k$differentiable structures on$\mathbb R$. EDIT : If you want to show that up to$\textit{diffeomorphism of differentiable manifold}$, there exists only one differentiable structure on$\mathbb R$, just remember that (more generally) all$1$-dimensional non-compact manifolds are diffeomorphic to$\mathbb R$. [this is proved in Lafontaine e.g] - Oops! Now I have added "upto diffeomorphism" in the criterion. Sorry for the lapse! – Akela May 16 '10 at 21:06 Yes, but$x\mapsto x^3$is no diffeomorphism! You surely mean "homeomorphism". – Henri May 16 '10 at 21:09$x \mapsto x^3$is a diffeomorphism from$\mathbb R$with the usual differentiable structure to$\mathbb R$with the differentiable structure corresponding to the atlas you gave above. – Akela May 16 '10 at 21:13 What Henri means, I think, is that on$\mathbb{R}$there are infinitely many non-compatible (maximal) atlases each giving the topological space$\mathbb{R}$the structure of a differentiable manifold. – Qfwfq May 16 '10 at 21:42 But the fact that two atlases$\mathfrak{A}$and$\mathfrak{B}$are not compatible just means that$id:(\mathbb{R},[\mathfrak{A}])\rightarrow (\mathbb{R},[\mathfrak{B}])$is not a diffeomerphism, but it does'n exclude that there are other diffeomerphisms between the differentiable structures: in fact, up to diffeomorphism, there's only one smooth structure on$\mathbb{R}\$. –  Qfwfq May 16 '10 at 21:42