# Can every manifold be given an analytic structure?

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

(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.

edit: Some time ago Riccardo Benedetti sent me some comments to append to my answer. They appear below, with some minor MO-formatting on my part.

In the famous paper "Real algebraic manifolds" (Annals of Math 56, 3, 1952), John Nash just proved that:

"Every compact closed smooth manifold M embedded in some $R^N$, with $N$ big enough (as usually $N=2Dim(M)+1$ suffices), can be smoothly approximated by a union of components, say $M_a$, of the non-singular locus of a real algebraic subset $X$ of $R^N$."

In the same paper he stated also some conjectures/questions, in particular whether one can get $M_a = X$ (so that $X$ is a non-singular real algebraic model of $M$), or whether one can even get such an algebraic model which is rational.

A.H. Wallace (1957) solved positively the first question under the assumption that $M$ is a boundary. Finally a complete positive answer was given by A. Tognoli (1973) by using, among other things, a so called "Wallace trick" and the fact (due to Milnor) that the smooth/un-oriented bordism group is generated by real algebraic projective non singular varieties.

Starting from this Nash-Tognoli theorem, mostly in the 80's-90's of the last century, a huge activity has been developed about the existence of real algebraic models for several instances of smooth or polyhedral structures, with major contributions by S. Akbulut and H. King (in particular they proved that if M is embedded in $R^n$, an algebraic model can be realized in $R^{n+1}$; to my knowledge it is open if we can stay in the given $R^n$).

If I am not wrong, the realization of real algebraic varieties via planar linkages (with related credit to Thurston) does not provide an alternative proof of Nash-Tognoli theorem.

The "Nash rationality conjecture" is more intriguing and has been basically "solved" in dimension less or equal to 3. This is mentioned for instance in some answers to the questions:

What's the difference between a real manifold and a smooth variety?

"You might also be interested in some of the articles by Kolla'r on the Nash conjecture contrasting real varieties and real manifolds. such as "What are the simplest varieties?", Bulletin, vol 38. I like the pair of theorems 54, 51, subtitled respectively: "the Nash conjecture is true in dim 3", and "The Nash conjecture is false in dim 3".

What is known about the MMP over non-algebraically closed fields

"Another issue is the rational connectivity and its relation to Mori fiber spaces ...... To illustrate the difficulties, here is a conjecture of Nash (yes, that Nash):

Let $Z$ be a smooth real algebraic variety. Then $Z$ can be realized as the real points of a rational complex algebraic variety. This, actually, turns out to be false. Kollár calls it the shortest lived conjecture as it was stated in 1954 and disproved by Comessatti around 1914 (I don't remember the exact year). However, even if the statement is false, just the fact that it was made and no one realized for 50 years that it was false should show that these questions are by no means easy. (Comessatti's paper was in Italian and I have no idea how Kollár found it.)

Kollár showed more systematically the possible topology types of manifolds that can satisfy this statement. In particular, Kollár shows that any closed connected 3-manifold occurs as a possibly non-projective real variety birationally equivalent to $P^3$. In other words the way Nash's conjecture fails is on the verge of the difference between projective and proper again showing that these questions are not easy." (Sandor Kovacs)

As I have been even more concerned with, let me add a few comments. Comessatti's result was certainly "well known" at least to some Italian people and also to the real algebraic Russian guys of Arnol'd's and mostly Rokhlin's school. Moreover (before Kollar's work) it had been rediscovered (via modern tools) for instance by R. Silhol. This allows the following 2D solution of the Nash conjecture:

" (1) Every non orientable compact closed surface admits a rational projective non singular model which can be explicitly given by (algebraically) blowing-up $\mathbb RP^2$ at some points;

(2) $S^2$ and $T^2$ are the only orientable surfaces that admit non singular projective rational models (Comessatti);

(3) Every surface $S$ admits a projective rational model possibly having one singular point (to get it, first smoothly blow-up $S$ at one point $p$ getting a non orientable $S'$ containing a smooth exceptional curve $C$ over $p$. Take a projective non singular rational model $S'_r$ of $S$. Finally one can prove that $C$ can be approximated in $S'_r$ by a non singular algebraic curve $C_a$ and that we can perform a "singular" blow-down of $C_a$ producing the required singular rational model of $S$.)"

Inspired by this 2D discussion, in the paper (with A. Marin) Dechirures de varietes de dimension trois et la conjecture de Nash de rationalite' en dimension trois Comment. Math. Helv. 67 (4) (1992), 514-545 (a PDF file is available in http://www.dm.unipi.it/~benedett/research.html)

we got a formally similar 3D solution. More precisely we provide at first a complete classification of compact closed smooth 3-manifolds $M$ up to "flip/flop" along smooth centres (see below - these are the "Dechirures" - perhaps we were wrong to write the paper in French ... ).

Summarizing (and roughly) the results are:

• There is a complete invariant $I(M)$ for this equivalence relation. Depending only on $I(M)$, we explicitly produce a real projective non singular rational 3-fold $Z$ such that $I(M)=I(Z)$.

• There is a smooth link $L$ in $M$ and a non singular real algebraic link $L_a$ in $Z$ such that by smoothly (algebraically) blowing up $M (Z)$ along $L (L_a)$ we get the same manifold $Z'$ and furthermore the disjoint real algebraic exceptional tori over $L_a$ coincide with the exceptional tori over $L$ (thinking all within the smooth category, basically by definition this means that $Z$ and $M$ are equivalent up to flip/flop along smooth centres).

• Clearly $Z'$ is projective rational non singular and algebraically dominates $Z$. It also smoothly dominates M. Finally we can convert the smooth blow-down onto $M$ to a singular algebraic blow-down producing a projective rational model $M_r$ of $M$, possibly singular at a non singular real algebraic copy of the link $L$ in $M_r$.

• The invariant $I(M)$ is easy when $M$ is orientable; this is just the dimension of $H_2(M;\mathbb Z/2)$. In this case $Z$ is obtained by algebraically blowing up $S^3$ at some points. $I(M)$ is much more complicated if M is non orientable and involves, among other things, certain quadratic forms on $Ker(i_*:H_1(S;\mathbb Z/2) \to H_1(M;\mathbb Z/2))$, $S$ being any characteristic surface of $M$.

A combination of our work with Kollar's one roughly gives:

• (3D "a la" Comessatti) In the projective framework, in general our singular rational models cannot be improved (singularity cannot be avoided, and in a sense we provided models with very mild singularities); in other words those blowing-down $Z'\to M_r$ were intrinsically singular.

• (Non projective non singular rational models) Starting from our real projective rational non singular 3-fold $Z'$, as above, Kollar proves that one can realize a non singular blow-down $Z'\to M'_r$, provided that we leave the projective framework.

Finally it is intriguing to note another important occurrence of the opposition projective singular vs non singular (related to the existence of intrinsically singular blow-down). Going back to the original Nash-Tognoli kind of problems, for a while it was conjectured and very "desidered" (for several reasons also related to the general question of characterizing say the compact polyhedron which admit possibly singular real algebraic models) that every M as above admits a "totally algebraic model" M_a, i.e. such that $H_*(M_a;\mathbb Z/2)$ is generated by the $\mathbb Z/2$-fundamental class of algebraic sub-varieties of $M_a$. On the contrary we constructed counterexamples in:

(with M. Dedo') Counterexamples to representing homology classes by real algebraic subvarieties up to homeomorphism, Compositio Math. 53 (2) (1984), 143-151 (idem)

This contrasts with a result by Akbulut-King that M admits singular totally algebraic models.

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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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