8
$\begingroup$

According to a result of Whitney any $C^r$-manifold, $r\geq 1$, is $C^r$-homeomorphic to a smooth embedded submanifold of some Euclidean space; see Theorem 1 in

Hassler Whitney, "Differentiable manifolds", Annals of Mathematics (2) 37 (1936), no. 3, pp. 645–680, JFM 62.1454.01, MR1503303, Zbl 0015.32001.

I expect that an analogous result should be true for $C^{r,\alpha}$-Hölder manifolds, $r\geq 1$, $\alpha\in[0,1]$, and I am wondering if there exists a reference for such a statement. Does perhaps anybody know such a reference?

Actually, I am not so much interested in the embedding, but in a refinement of the $C^{r,\alpha}$-Hölder structure to a smooth structure, i.e. in a $C^{r,\alpha}$-Hölder homeomorphism to a smooth manifold.

In particular, I am interested in the case in which the manifold to start with is itself a ($C^{r,\alpha}$-)embedded submanifold of some Euclidean space and in the case $\alpha=1$. Maybe it is easier to say something here?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
9
$\begingroup$

Every $C^1$ manifold admits a compatible $C^\infty$ structure. You can find a proof in Hirsch's "Differential topology". It is actually quite easy and based on a fact that smoothing a $C^1$ diffeomorphism by convolution leads to a smooth diffeomorphism (because derivatives converge uniformly) on a slightly smaller domain.

In particular $C^{r,\alpha}$ manifolds, $r\geq 1$, admit a compatible $C^\infty$-structure and hence such a manifold can be embedded as a smooth submanifold of an Euclidean space. However, the embedding will only be $C^{r,\alpha}$ smooth since the original manifold is only $C^{r,\alpha}$.

Improve this answer
$\endgroup$
2
  • $\begingroup$ The $C^{\infty}$ structure might be compatible with the $C^1$ structure, but why should it be compatible with the $C^{r,\alpha}$ structure? If it is not, the resulting embedding is not $C^{r,\alpha}$ smooth with respect to the $C^{r,\alpha}$ structure we started with, only with respect to a distinct $C^{r,\alpha}$ structure (that is induced by the $C^{\infty}$ structure). $\endgroup$
    – Christian Lange
    Commented Oct 2, 2022 at 17:54
  • $\begingroup$ I have the impression that one needs e.g. a Hölder analogue of Theorem 2.7 in Hirsch's book: "Let $M$ and $N$ be $C^s$ manifolds $1\leq s \leq \infty$. Then $\mathrm{Diff}^s(M,N)$ is dense in $\mathrm{Diff}^r(M,N)$ in the strong $C^r$ topology, $1\leq r <s$." The proof of this theorem using convolutions seems to go through also in the Hölder case. Thanks for the reference. $\endgroup$
    – Christian Lange
    Commented Oct 3, 2022 at 8:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .