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Let $M$ be a (real) manifold. Recall that an analytic structure on $M$ is an atlas such that all transition maps are real-analytic (and maximal with respect to this property). (There's also a sheafy definition.) So in particular being analytic is a structure, not a property.

Q1: Is it true that any topological manifold can be equipped with an analytic structure?

Q2: Can any $C^\infty$ manifold (replace "analytic" by "$C^\infty$" in the first paragraph) be equipped with an analytic structure (consistent with the smooth structure)?

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In perusing this question, I was amused to notice that a complete answer is spread out across the three answers below: $C^0$ manifolds cannot in general be smoothed to $C^1$, and if they can, then the smoothing may not be unique in dimensions 4 and up. In contrast, for $k \geq 1$, a $C^k$ manifold admits a unique smoothing to a $C^\infty$ structure, and a $C^\infty$ manifold admits a unique smoothing to a $C^\omega$ structure. –  Tim Campion Apr 28 '13 at 17:22
Caveat: the notion of uniqueness has to be kind of subtle. Obviously a "smoothing" of a "rough" manifold $X$ is a "rough" diffeomorphism $X \to Y$, where $Y$ is a "smooth" manifold. What does it mean for the smoothing to be unique? Naively one might ask that for any other smoothing map $X \to Y'$, the induced map $Y \to Y'$ is "smooth". This would imply that any "rough" diffeo between "smooth" manifolds is automatically "smooth". This is false for the notions of "rough" and "smooth" that we're considering. –  Tim Campion Apr 28 '13 at 20:13
E.g. When "rough" = $C^\infty$ and "smooth" = $C^\omega$: Let $\phi$ be a $C^\infty$, nondecreasing "step" function on $\mathbb{R}$ (constant outside an interval, but not globally constant). Then $f(x) = x + \phi(x)$ is a $C^\infty$ diffeomorphism of $\mathbb{R}$ which is not $C^\omega$. Similarly, when "rough" = $C^1$ and "smooth" = $C^2$ or better: $f(x) = x + \int_0^x |t|dt$ is a $C^1$-diffeo of $\mathbb{R}$ which is not $C^2$. –  Tim Campion Apr 28 '13 at 20:23
At a minimum, uniqueness of smoothings ought to mean that the "smooth" diffeo type determines the "rough" diffeo type. That is, if $Y,Y'$ are smoothings of $X$, then there exists SOME "smooth" diffeo $Y \to Y'$. You'd have to think that this "smooth" diffeo is somehow close/homotopic to the canonical "rough" diffeo $Y \to Y'$ coming from the smoothing maps. Does the regularity of the homotopy just fall out of the more general theory of smoothing of maps? In any case, the embedding of the "smooth" diffeo group into the "rough" one should be dense, and a (weak?) homotopy equivalence ... right? –  Tim Campion Apr 28 '13 at 21:09
Oh-- and nobody has said anything about uniqueness of smoothings from $C^k$ to $C^{k+1}$ (although existence follows from the existence of a $C^\infty$ smoothing). Does it hold, in whatever relevant sense? I'd have to guess from the rest of the discussion that it does, but nobody has actually said it... –  Tim Campion Apr 28 '13 at 21:16

3 Answers 3

up vote 33 down vote accepted

(similar to Mariano's post)

Q1: no. There are topological manifolds that don't admit triangulations, let alone smooth structures. All smooth manifolds admit triangulations, this is a theorem of Whitehead's. The lowest-dimensional examples of topological manifolds that don't admit triangulations are in dimension 4, the obstruction is called the Kirby-Siebenmann smoothing obstruction.

Q2: $C^1$ manifolds all admit compatible $C^\infty$ and analytic ($C^\omega$) structures. This is a theorem of Hassler Whitney's, in his early papers on manifold theory, where he proves they embed in euclidean space. The basic idea is that your manifold is locally cut out of euclidean space by $C^1$-functions so you apply a smoothing operator to the function and then argue that the level-set does not change (up to $C^1$-diffeomorphism), provided your smoothing approximation is small enough in the $C^1$-sense. I'm not sure who gets the original credit but you can go much further -- compact boundaryless smooth manifolds are all realizable as components of real affine algebraic varieties, planar linkages in particular. There's a Millson and Kapovich paper on the topic available if you do a Google search. It seems people give a lot of credit to Bill Thurston.

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There is an amazing theorem of Morrey and Grauert that says that not only does every (paracompact) smooth manifold have a real analytic structure, the real analytic structure is unique. Using Whitney's ideas, you can show that two real analytic manifolds $M$ and $M'$ that are diffeomorphic are also real-analytic equivalent, if they both embed analytically in Euclidean space. And of course Whitney lets you assume that one of them does. The question boils down to whether real analytic real-valued functions on a real analytic manifold separate points. Morrey and Grauert proved that they do. Although I don't understand either proof, I remember that a key step in Grauert's proof is to make the tangent bundle of $M$ into a complex manifold, and show that it is a Stein manifold.

Anyway, the result is much harder than what Whitney did, which is a very good but expected application of the Weierstrass approximation theorem. To understand the issues, you can consider instead real algebraic manifolds in the geometric topology sense (rather than in the algebraic geometry sense). These are called Nash manifolds. A circle has an embeddable Nash structure, the one where trigonometric functions separate points. But $\mathbb{R}/\mathbb{Z}$ is another Nash circle, one of zillions that do not embed as Nash manifolds into Euclidean space.

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I'm a little confused about the "The question boils down to" comment. If the manifold is sitting in Euclidean space as an analytic submanifold, all the coordinate functions $x_1, x_2, \cdots, x_n$ on Euclidean space are analytic on Euclidean space so they're analytic on the manifold and they separate points. –  Ryan Budney Dec 13 '09 at 21:56
Oh, this is for the $M'$ that may not have an analytic embedding in Euclidean space -- ah, understood. Once you have $M'$ embedding analytically in euclidean space you include into a large enough Euclidean space and construct an analytic isotopy between the embeddings using Whitney's ideas? –  Ryan Budney Dec 13 '09 at 21:58
Yes, for instance you can make an analytic isotopy. Or, as in other steps in Whitney's stuff, you can perturb a diffeomorphism of $M'$ on $M$ into a nearby analytic map by Weierstrass approximation, and then project back onto $M$ using an embedded normal bundle (the normal exponential map). –  Greg Kuperberg Dec 13 '09 at 22:18
references (found in a 1998 posting to sci.math.research by ... Greg Kuperberg!)---- H. Grauert, On Levi's problem and the imbedding of real-analytic manifolds, Ann. of Math. 68 (1958), 460-472.---------------C. B. Morrey, The analytic embedding of abstract real-analytic manifolds, Ann. of Math. 68 (1958), 159-201. –  Tim Campion Apr 3 '13 at 9:25

Kervaire proved in [M. Kervaire , A manifold which does not admit any differentiable structure. Comment. Math. Helv. 34 (1960), pp. 257–270.] that there are topological manifolds which cannot be smoothed. On the other hand, every $C^k$ manifold with $k>0$ can be uniquely smoothed to a $C^\infty$ manifold, by a theorem of Whitney.

Finally, if I recall correctly I thought not every $C^k$ manifold had a compatible analytic structure...

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