# If tangent vectors are a vector space of same dimension at every point, does one has a manifold? [closed]

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 $$k$$?

• en.wikipedia.org/wiki/Tangent_bundle May 14 at 1:12
• @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... May 14 at 1:27
• 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. :) May 14 at 1:31
• Hmm, can one get a chart from considering the flow of vectors in the tangent space close to the origin? May 14 at 1:31
• The answer is no, for all $k$. This is sort of a homework-y problem. I've voted to close. May 14 at 1:38

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 $$k$$.
• 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). May 14 at 9:04
• 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. May 14 at 13:15