5

Consider a locally connected $X\subset \mathbb R^n$. Given a point $p\in X$, consider the following condition:

  • For any $a,b\in X$, if there's a path $a\to p\to b$ in $X$, there is also a smooth path $a\to p \to b$ in $X$ with nonzero derivative at $p$ (at least once).

I think this condition may express that $X\subset\mathbb R^n$ is smooth at $p$.

Every point $p$ of Euclidean space satisfies this property since e.g the circle going through points $a,b,p$ furnishes a smooth path $a\to p\to b$. Consequently I think every point of an embedded manifold $X\subset\mathbb R^n$ satisfies this condition.

Question. Does this characterize manifolds embedded in Euclidean space? That is, if $X\subset \mathbb R^n$ is a locally connected subset whose every point satisfies the above condition, does it follow $X$ is an embedded manifold?

Added. Eric Wofsey's answer helped me realize I would like to assume $X$ is additionally locally Euclidean. His comment provides a counterexample to this case, namely the wedge in $\mathbb R^3$ formed by folding a rectangle along a line. This comment outlines the problem with the condition of my question. I have asked a follow-up question here.

Arrow
  • 13,810

1 Answers1

4

No. For instance, let $X\subset\mathbb{R}^2$ be the union of the $x$-axis and the graph of the function $f(x)=\exp(-1/x^2)$. This satisfies your conditions, but is not smooth at the origin.

In fact, such a set can be very far removed from being a manifold. For instance, the set $X=\mathbb{R}^2\setminus\mathbb{Q}^2$ satisfies your conditions. Given any three points $a,p,b\in X$, you can easily construct an uncountable family of smooth paths $a\to p \to b$ in $\mathbb{R}^2$ which have nonzero derivative at $p$ and whose images do not intersect except at $a$, $p$, and $b$. Only countably many such paths can contain a point of $\mathbb{Q}^2$, so at least one of them is in $X$.

Eric Wofsey
  • 330,363
  • Thank you for the answer. If we assume $X$ is path-connected I think the first example is eliminated. Regarding the second - I have no idea what that quotient $\mathbb R^2/\mathbb Q^2$ looks like. Is it locally connected? Could you please provide some intuition? – Arrow Jan 13 '18 at 23:55
  • Both these examples are path-connected. The first example is the union of two subsets diffeomorphic to $\mathbb{R}$ (the $x$-axis and the graph of $f(x)=\exp(-1/x^2)$ where $f(0)$ is defined as $0$) which intersect at the origin. The second example is the set difference $\mathbb{R}^2\setminus\mathbb{Q}^2$, not a quotient. It is path connected and locally path connected. – Eric Wofsey Jan 14 '18 at 00:10
  • Sorry, I read your answer very carelessly. I wonder whether there are counterexamples when $X$ is itself locally Euclidean (topologically speaking). Should I ask this separately? – Arrow Jan 14 '18 at 00:18
  • For a counterexample where $X$ is topologically locally Euclidean, consider the subspace of $\mathbb{R}^3$ you get by folding a rectangle along a line. (In other words, $X$ is something like $V\times \mathbb{R}$, where $V\subset\mathbb{R}^2$ is shaped like the letter $V$.) – Eric Wofsey Jan 14 '18 at 00:22
  • Ahh, I see. At any point $p$ on the edge of the "wedge" (i.e $X$) we can always get a path which first visits the edge at a different point $q$ and then travels along the edge though $p$ at positive velocity. Many thanks again! I will retry to formalize what I want to say later. – Arrow Jan 14 '18 at 00:28
  • Right, exactly. I was going to write that out but had trouble phrasing it succinctly and figured you could probably figure it out on your own. :) – Eric Wofsey Jan 14 '18 at 00:28