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This is a crosspost of this MSE question.

Given a locally Euclidean (locally homeomorphic to some Euclidean space) topological subspace $X\subset\mathbb R^n$ and $p\in X$, let $\mathrm{T}_pX$ denote the tangent set of $X$ at $p$, namely the set of derivatives of curves in $X$ based at $p$ and differentiable at zero.

Questions.

  1. Are the following conditions equivalent? What are the implications between them?
  2. Does the second condition imply $\pi_V$ is a local homeomorphism about $p$?
  3. Does the first condition naturally furnish a local homeomorphism at $p$ between a neighborhood of $p$ in $\mathrm T_pX$ and a neighborhood of $p$ in $X$?

(For 3, the only natural candidate I see here is orthogonal projection. I don't know how to prove it's locally injective. The conditions of the everywhere differentiable inverse function theorem may be false as exemplified e.g by $f(x)=(x^5\cos(\frac 1x))^{\frac13}$, so that won't work.)


  1. $\mathrm{T}_pX$ is a linear subspace of $\mathbb R^n$ of dimension $\dim_pX$.
  2. There's a linear subspace $V\subset\mathbb R^n$ of dimension $\dim_pX$ for which we have the following equality, where the limit is taken in a translated neighborhood in $X$ of $p$. $$\lim_{h\to 0} \frac{\pi_{V^\perp}(h)}{\|h\|}=0$$ (Here $\pi_{V^\perp}$ is the orthogonal projection onto the orthogonal complement of $V\subset\mathbb R^n$.)
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  • $\begingroup$ Why the downvote? $\endgroup$
    – Arrow
    Jan 25, 2018 at 13:58
  • $\begingroup$ boy it strikes me as problematic terminology that a tangent space is not a space of tangents. I thought every diagram commutes... (!) $\endgroup$
    – Samantha Y
    Jan 25, 2018 at 21:18

1 Answer 1

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I don't think the two conditions are equivalent. Take for example

$$X:=\{ (x,y)\in \mathbb{R}^2\mid x\geq0,y=x^2 \}\cup\{ (x,y)\in \mathbb{R}^2\mid x\geq0,y=2x^2 \}$$

(or the cusp $X:=\{(x,y)\in\mathbb{R}^2\mid y^2=x^3\}\;$).

In this case, $T_p X=0$ at $p=(0,0)\in\mathbb{R}^2$. And $V=\{(x,y)\mid y=0\}$.

Also, in this case the second condition is true but $\pi$ is not a local diffeo around the origin.


Added: Maybe I have a counterexample that shows that 1. does not imply 2.

Basically: take a continuous function $f:\mathbb{R}^2\to\mathbb{R}$ that has directional derivatives (existing and) equal to zero along every direction through the origin $(0,0)$ but its restriction along the $y=x^2$ parabola in $\mathbb{R}^2$ goes like $|x|$, i.e. $f(x,x^2)=|x|$. Set $X:=$ the graph of $f$ in $\mathbb{R}^2$. Then, if I'm not mistaken, $T_p X=\{z=0\}$. Also, the only possible $V$ could only be $V=\{z=0\}$ too, but $\pi(h)/||h||$ would not tend to $0$ as $h$ approaches the origin along the "approaching" subset $\Sigma=\{(x,x^2,|x|)\}\subset X$.

Now I'll try to formalize this. Set

$$\Sigma:=\{(x,x^2,|x|)\in\mathbb{R}^3\mid x\in\mathbb{R}\}\subset \mathbb{R}^3.$$

For every $x\in\mathbb{R}$,

  • let $\ell'_x$ be the straight segment in $\mathbb{R}^3$ joining $(x,\frac{1}{2}x^2,0)$ to $(x,x^2,|x|)$
  • let $\ell''_x$ be the straight segment in $\mathbb{R}^3$ joining $(x,2x^2,0)$ to $(x,x^2,|x|)$.

Set

$$S:=\{(x,y)\in\mathbb{R}^2\mid \frac{1}{2}x^2\leq y \leq 2x^2\}\subset\mathbb{R}^2$$

and $D:=\mathbb{R}^2\smallsetminus S$, so $D\times\{0\}\subset\mathbb{R}^3$.

Finally

$$X:=(D\times\{0\})\cup \bigcup_{x\in\mathbb{R}}(\ell'_x\cup\ell''_x)\subset\mathbb{R}^3.$$

If I got this right, $X$ should be the graph of a function $f:\mathbb{R}^2\to\mathbb{R}$ with the following properties:

  • $f$ is continuous.
  • For every direction $v\in\mathbb{R}^2$, the directional derivative $\partial_v f(0)$ at the origin exists and is zero.
  • $T_p X=\{z=0\}$ for $p=(0,0,0)$.
  • $\Sigma\subset X$.

If a linear subspace $V\subseteq\mathbb{R}^3$ as in condition 2. of the OP exists, then, looking at the straight lines through the origin, I think it would be forced to be $V=\{z=0\}=T_p X$. But the condition $\pi(h)/||h||\to 0$ for $h\to 0$ is not met for $h\in\Sigma$ since

$$\pi(x,x^2,|x|)=|x|,\quad h=(x,x^2,|x|)\in\Sigma$$

hence it's not met for $h\in X$.

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  • $\begingroup$ Thank you for the beautifully simple counterexample! What is your opinion on the first condition implying the second? $\endgroup$
    – Arrow
    Jan 24, 2018 at 23:36
  • $\begingroup$ Dear Qfwfq, is this a more-or-less correct picture of your second $X$? If so, $\mathrm T _pX$ is the $xy$-plane because $X$ contains the germ of every straight line through the origin in the $xy$-plane, and contains only horizontal tangents to a curve? What if we define the tangent set with one-sided derivatives of curves? $\endgroup$
    – Arrow
    Jan 25, 2018 at 15:09
  • $\begingroup$ 1Yes. 2Yes. 3 I think you would get the same result with "one sided derivatives" because, still, there would be no smooth curves passing through the origin and not lying in the xy plane. But, instead, if by "one sided derivative" you actually mean the derivative of "one sided curves" $[0,\varepsilon)\to X$, then probably $T_p X$ would be no longer a linear subspace. $\endgroup$
    – Qfwfq
    Jan 25, 2018 at 23:11

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