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 $k$?



If tangent vectors are a vector space of same dimension at every point, does one has a manifold?
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 $k$.


