If a manifold admits a real analytic structure then the manifold is analytic?

Question

I have been reading the notion of real analytic space in nLab and found a statement that puzzles me.

That Whitney embedding theorem shows that every paracompact smooth manifold admits a real analytic structure.

Whitney embedding theorem shows that a differentiable manifold $M$ can be embedded in $\mathbb{R}^n$ for a $n$ sufficient large. We know that $\mathbb{R}^n$ is an analytic manifold but it is not clear for me from this that $M$ admits an analytic structure. Does it mean that $M$ is an analytic manifold?

If the answer to this question is true then there is a contradiction since there are smooth functions $f:\mathbb{R} \to \mathbb{R}$ whose graphs are embedded in $\mathbb{R}^2$ that are not analytic e.g. the bump function.

Moishe Kohan · Accepted Answer · 2021-04-29T12:59:02.443

Whoever wrote this nLab article seriously goofed. Their proof break down in two different ways:

It might not be possible to embed the given smooth manifold $M^m$ in $R^N$ in such a way that the image $i(M)$ is the zero-set of a regular map $f: R^N\to R^{N-m}$, where "regular" means that $Df$ is a submersion along $i(M)$. The obstruction here is that the tangent bundle of $M$ might not be "stably trivial" (i.e. $M$ itself might not be stably parallelizable). The simplest example is any non-orientable manifold. Note that regularity is the key to their sketch since if $f$ fails to be regular, then for approximating polynomial vector-functions, the zero set will not even be (in general) of the same dimension as $M$, let alone diffeomorphic to it.
Suppose, however, that $M$ is stably parallelizable; then it indeed can be embedded in some $R^N$ as the zero level set of a smooth mapping $f: R^N\to R^{N-m}$. Then everything works as they say if $M$ is compact. However, if $M$ is noncompact, you cannot use the Weierstrass approximation theorem to achieve their goal since the approximation is only uniform on compacts. Thus, if they take $p_j$ their approximating polynomial mappings, then $p_j^{-1}(0)$ will be a polynomial subvariety in $R^N$. But, in general, they cannot even guarantee this to be a submanifold (only that these are submanifolds in the given ball, when $j$ is large enough). What's worse, not every smooth manifold is diffeomorphic to a submanifold of $R^N$ given by a system of polynomial equations. The simplest example is a zero-dimensional manifold with infinitely many components. See my answer here for more details.

The actual proof of existence is quite technical. I know only of one place where it is written in all the details:

Whitney, H., Differentiable manifolds., Ann. Math., Princeton, (2) 37, 645-680 (1936). ZBL62.1454.01.

To conclude: In his paper Whitney indeed proves that every smooth $n$-dimensional manifold embeds in some $R^{2n+1}$ so that the image is a (smooth) real-analytic subset. Thus, $M$ admits a real-analytic structure.

However, the proof goes well beyond the Weierstrass approximation theorem (which is actually never used in the proof!).

Hah, I gave that nLab page much more credit that it deserved. I'm just going to delete my answer. — Lee Mosher, Apr 27 '21 at 18:15
Dear @Moishe Kohan, thanks for your answer. To summarize your answer. If $M$ is smooth, there is an embedding $i$ into $\mathbb{R}^N$ such that $i(M)$ is real-analytic. Is it correct? — jaogye, Apr 29 '21 at 12:56
Since the embedding $i$ is a diffeomorphism, we can conclude that if $M$ admits a $C^1$ structure then $M$ admits a $C^\infty$ structure. Is this right? — jaogye, May 08 '21 at 13:39

If a manifold admits a real analytic structure then the manifold is analytic?

1 Answers1