Is there any 'elementary' proof of the uniqueness of smooth structures on $\mathbb{R}$? By elementary, I mean that the proof does not use any sophisticated topological machinery. In particular, I'm looking for a proof that only assumes undergraduate real analysis of one variable.
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4$\begingroup$ The very concept of smooth structure and diffeomorphism is way beyond undergraduate real analysis. $\endgroup$– Alex DegtyarevCommented Jan 13, 2015 at 12:27
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3$\begingroup$ The classification of 1-manifolds up to diffeomorphisms has a proof that uses little more than existence and uniqueness to solutions of C^1 ODEs. That gives a perhaps more pleasant proof than thinking about charts. $\endgroup$– Ryan BudneyCommented Jan 13, 2015 at 22:06
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$\begingroup$ Use bump functions to make a complete metric, a 2-1 cover to make a unit vector field, and then map a time variable to the manifold by flowing a point along the vector field: your manifold has the real line as universal covering space, and the covering map is an isometry. Work out the groups of isometries of the real line that act freely without fixed points. $\endgroup$– Ben McKayCommented Jan 22, 2015 at 13:45
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$\begingroup$ Uniqueness means "uniqueness modulo homeomorphism" i.e. any two smooth structures are conjugate by some homeomorphism, or equivalently that any two smooth structures define diffeomorphic smooth manifolds. $\endgroup$– YCorCommented Jun 28 at 5:55
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2 Answers
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You can assume that you have an atlas where you have charts on countably many open intervals. Then you need to check that you can replace two adjacent intervals with one interval. Iterating this, you can construct a diffeomorphism between the whole thing and an open subset of $\mathbb R$. Using some standard diffeomorphisms, you get one with all of $\mathbb R$.
So the key step is done by gluing two intervals together. This can be done with bump functions. If you glue together the intervals $(0,2)$ and $(3,5)$ by some smooth map $(1,2) \to (3,4)$ you can change the smooth structure on $(0,2)$ by using a new smooth map $(0,2) \to (0,2)$ that is equal to the identity on small values and equal to the gluing map on large values. Do the same thing to $(3,5)$, and the gluing map becomes the identity.
answered Jan 13, 2015 at 13:56
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You can find an elementary proof in Corollary 4.4 of our joint paper:
- Mykola Lysynskyi, Sergiy Maksymenko, Classification of differentiable structures on the non-Hausdorff line with two origins, arxiv:2406.09576
It follows the line similar to the one mentioned in the answer by Will Sawin.
We needed that proof for having an equivalent statement which at first sight might look more unusual:
If $M$ is a one-dimensional (not necesarily Hausdorff) $C^{k}$-manifold and $U \subset M$ is an open subset homeomorphic with $\mathbb{R}$, then there exists a homeomorphism $h: U \to \mathbb{R}$, such that the chart $(h,U)$ is compatible with the $C^k$-structure on $M$.
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1$\begingroup$ I guess the conclusion of your statement in italics could instead be "there exists a $C^k$-diffeomorphism $h\colon U\to \mathbb{R}$"? $\endgroup$– David Roberts ♦Commented Jun 28 at 7:09
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1$\begingroup$ Yes, exactly. The reason of that formulation is that in the paper we look at the classification of C^k structures on M as a description of orbits of the action of homeomorphisms of M on the set of C^k structures. $\endgroup$ Commented Jun 28 at 7:32
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$\begingroup$ Oh, that is a nice way to think about it, thanks for clarifying. $\endgroup$– David Roberts ♦Commented Jun 28 at 12:04