I keep seeing the sentence " you can find a reagularly parametrized smooth curve through each point in each direction"

"There are three smooth paths starting there and lying on the boundary of the cube - the edges ( almost any parametrization should work)" //edit : the last sentence came from how to prove boundary of the cube isnt smooth manifold, that was an assumption

and indeed its seems correct, and by taking parametrization as graph, its seems to be correct - but this is just on one direction. what about the others ?

what is a formal explanation for this ?


The boundary of a cube is not a smooth manifold, and the question in your title is false for the boundary of a cube, as your question seems to suggest.

Assuming, as your tag suggests, you are asking about a smooth manifold $M$, the statement in your question is true because it is true in every coordinate chart. In any open subset $U \subset \mathbb{R}^n$, for any $p \in U$, and for any vector $v$, there exists $\epsilon > 0$ such that the smooth curve $t \mapsto p + t \vec v$ defined for $-\epsilon<t<+\epsilon$ is contained in $U$. Since this is true in each coordinate chart for $M$, it's true in $M$.

  • $\begingroup$ fixed the question, thanks $\endgroup$ – Idan Perez Jul 14 '17 at 14:06

Every point $x$ of a $n$-dimensional manifold has a neighborhood diffeomorphic to a ball of $\mathbb{R}^n$, and you have a smooth curve in any dimension in $\mathbb{R}^n$.


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