Let $M$ be a non-empty subset of $\mathbb R^n$, $n \geq 2$.

Recall that a vector $v$ is tangent to $M$ at the point $m \in M$ if it exists a differentiable curve $\gamma : I \to M$ such that $\gamma(0) = m$ and $\gamma'(0) = v$, where $I \subset \mathbb R$ is an interval that contains a neighborhood of $t=0$.

Suppose that it exists an integer $k \geq 1$ such that, for every $m \in M$, the set of vectors that are tangent to $M$ at $m$ is a linear space of dimension $k$ (the same $k$ for every $m$).

Is it true that $M$ is a differential manifold of dimension $k$?

  • $\begingroup$ en.wikipedia.org/wiki/Tangent_bundle $\endgroup$
    – sharpend
    May 14 at 1:12
  • 1
    $\begingroup$ @sharpend excuse-me but I failed to see how this link answers my question. In this article, $M$ is always assumed to be a manifold... $\endgroup$
    – Héhéhé
    May 14 at 1:27
  • $\begingroup$ Okay, now I see that you really do mean $M$ to be a general subset of $\mathbb{R}^n$, not necessarily open. I read too quickly and gave an unhelpful link. :) $\endgroup$
    – sharpend
    May 14 at 1:31
  • $\begingroup$ Hmm, can one get a chart from considering the flow of vectors in the tangent space close to the origin? $\endgroup$ May 14 at 1:31
  • 3
    $\begingroup$ The answer is no, for all $k$. This is sort of a homework-y problem. I've voted to close. $\endgroup$ May 14 at 1:38

1 Answer 1


I found that the answer is no.

For $n=2$: take $M$ as the union of the two circles of radius 1 centered at $(\pm 1,0)$, at any point one has a tangent space of dimension $k=1$ but it is not a manifold (double point at $(0,0)$).

This idea should be generalizable to arbitrary $n$ and $k$.

  • $\begingroup$ Indeed, two tangent spheres of dimension $k$ give a counterexample for all $0<k<n$. For $0=k<n$, take a Cantor set; for $k=n>1$, the complement of a sphere of dimension $k-1$, together with one point on this sphere (i.e. $\{(1,0,\ldots,0)\}\cup(\mathbb R^n\setminus\mathbb S^{n-1})$). The cases $k=n=0$ and $k=n=1$ remain, but in this case it clearly works. (I am working with $0\leq k\leq n$ and $M$ non-empty). $\endgroup$
    – Pierre PC
    May 14 at 9:04
  • $\begingroup$ For something more wild take $\mathbb{R}^k \times \mathbb{Q}^{n-k}$. This is not an manifold near any point, but clearly has tangent space of dimension $k$ everywhere. $\endgroup$ May 14 at 13:15

Not the answer you're looking for? Browse other questions tagged or ask your own question.