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Definition. Let $X\subset \mathbb R^n$ be a locally Euclidean subset. Say it has a tangent space at $p\in X$ if there exists a linear subspace $V\leq \mathbb R^n$ satisfying the following conditions.

  1. $\dim V=\dim_pX$.
  2. There exists a neighborhood $U\subset X$ of $p$ in $X$ for which $$\lim_{\substack{h\to 0\\\text{in }U-p}}\frac{\|\pi_{V^\perp}(h)\|}{\|h\|}=0.$$ Here $U-p$ is the neighborhood of the origin in $X-p$ obtained by translation of $U$ by $(-p)\in X\subset\mathbb R^n$.
  3. The orthogonal projection of $X$ onto $V$ is locally injective at $p$.

Question. Is $V\leq \mathbb R^n$ unique?

The usual argument for the uniqueness of the derivative of a map on an open subset of Euclidean spaces relies on (more or less) convexity of open balls, so is not applicable here.

I am hoping the local injectivity at $p$ of the orthogonal projection will eliminate singular cases by preventing $X$ from "wrapping around $V$" about $p$ as in the cubic cusp (a sort of ramification). Without this condition the other two do not suffice. The example I was given is the surface of revolution of a cusp, e.g $\{(x,y,\sqrt[4]{x^2+y^2})\mid x,y\in \mathbb R\}\subset\mathbb R^3$.

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  • $\begingroup$ Since my last answer/comment that was incomplete, I jotted down an actual detailed proof, but I see now that the user Dap has already provided a proof which amounts to exactly the same idea. Let me know if you want to see also my version. $\endgroup$
    – Qfwfq
    Commented Jan 31, 2018 at 17:07
  • $\begingroup$ Dear @Qfwfq, I would love to see (and upvote) your version. I will surely learn from your perspective. $\endgroup$
    – Arrow
    Commented Jan 31, 2018 at 23:48

2 Answers 2

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By invariance of domain, the condition $\dim V=\dim_pX$ implies that the local injectivity is locally a bijection to a neighborhood of $p.$ So there are open sets $V_0\subseteq V$ and $X_0\subseteq X$ with $V_0$ containing $0,$ and a function $g:V_0\to X_0$ such that $\pi_V(g(v)-p)=v$ and $g(0)=p.$ By condition 2 we have $g(v)=p+v+o(\|v\|)$ with implicit constants depending only on $X,V$ and $p.$

Assume there's another $V'$ satisfying the same conditions. For each $v\in V$ we have $\lim_{\lambda\to 0^+}\frac{\|\pi_{V'^\perp}(g(\lambda v)-p)\|}{\|g(\lambda v)-p\|}=0.$ Using $g(\lambda v)=p+\lambda v+o(\lambda\|v\|)$ we get $\pi_{V'^\perp}(v)=0,$ so $v\in V'.$ Since $\dim V=\dim V'$ we must have $V=V'.$

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  • $\begingroup$ Dear Dap, I have two questions regarding your elegant answer. 1) Do we really need invariance of domain? I think orthogonal projections are open so local injectivity gives local homeomorphy. 2) Won't the inverse $g:V_0\to X_0$ in fact be affine, as the set-theoretical inverse of an affine map? $\endgroup$
    – Arrow
    Commented Jan 28, 2018 at 18:44
  • $\begingroup$ @Arrow: 1) orthogonal projections restricted to subspace $X\subset\mathbb R^n$ aren't always open, so you need some extra property. 2) no, that doesn't make sense since $X_0$ isn't a vector space - try an example like the unit sphere. $\endgroup$
    – Dap
    Commented Jan 28, 2018 at 19:23
  • $\begingroup$ Understood. Forgive my ineptitude but I am not sure why $g(v)=p+v+o(\|v\|)$. We have $\| gv-p-v\|=\| \pi_{V^\perp} \circ (\theta_{-p}\circ g)(v) \|$ so it seems condition 2 only gives $gv-p-v\in o(\| \theta_{-p}\circ g(v) \|)$. What am I missing here? ($\theta_{-p}$ is subtraction by $p$.) $\endgroup$
    – Arrow
    Commented Jan 28, 2018 at 22:05
  • $\begingroup$ @Arrow: we have $\|v\|^2+\|g(v)-p-v\|^2=\|g(v)-p\|^2.$ If $s:=\frac{\|g(v)-p-v\|}{\|g(v)-p\|}\to 0$ (which follows from condition 2 + continuity of $g$) then $\frac{\|g(v)-p-v\|}{\|v\|}=\frac{s}{\sqrt{1-s^2}}\to 0$ since $s\mapsto \frac{s}{\sqrt{1-s^2}}$ is continuous at $s=0.$ $\endgroup$
    – Dap
    Commented Jan 29, 2018 at 8:59
  • $\begingroup$ Just a remark: I think invariance of domain is not needed: since we know $X$ is locally Euclidean, there is a compact topological disk $D\subseteq X$ (of the same dimension as $X$) around $p$, and, up to shrinking $D$ if necessary, we can assume the restriction of the orthogonal projection to $V$ is injective on $D$. So we have a continuous bijection $\pi_{V}|_D:D\to V$ between a compact space and a Hausdorff space, hence it must be a homeomorphism onto its image $\pi_{V}(D)$. $\endgroup$
    – Qfwfq
    Commented Jan 31, 2018 at 17:05
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This is a long comment, rather than a complete answer. It shows, for any subset $X$ near $0$ and any dimension of the tangent(s), that if there are two tangent spaces in the sense of the OP then they must have nonzero intersection.

Lemma. Let $\{0\}\neq X\subseteq\mathbb{R}^n$ be any subset that has the origin as an accumulation point. Let $V$ and $V'$ be two nonzero linear subspaces (of arbitrary dimension) of $\mathbb{R}^n$ satisfying the limit conditions $$\lim_{\stackrel{h\in X}{h\to 0}}\frac{|\pi_{V^\perp}(h)|}{|h|}=0\quad\quad(\star_V)$$ $$\lim_{\stackrel{h\in X}{h\to 0}}\frac{|\pi_{V'^\perp}(h)|}{|h|}=0\quad\quad(\star_{V'}).$$

Then $V\cap V'\neq\{0\}.$

Proof. Assume by contradiction that $V\cap V'=0$. Write $h\in\mathbb{R}^n$ as $h=(x,y)\in V\oplus V^\perp$. Since the $\ell^2$ (usual Euclidean) norm $|h|$ and the norm given by $|(x,y)|_{V,V^\perp}:=\sqrt{|x|^2+|y|^2}$ are equivalent norms on $\mathbb{R}^n$, the condition $$\frac{|y|}{\sqrt{|x|^2+|y|^2}}\to0\;,$$ for $h=(x,y)\to0$ in a given subset having $0$ as accumulation point, is equivalent to the condition $|y|/|x|\to 0$ for $h=(x,y)\to 0$ in the same subset.

For $\varepsilon>0$ let $$\mathcal{C}_{V,\varepsilon}:=\{ (x,y)\in\mathbb{R}^n\mid |y|\leq \varepsilon |x|\}.$$ This is a closed cone (invariant under dilations) containing $V$. Likewise, let $$\mathcal{C}_{V',\varepsilon'}:=\{h\in\mathbb{R}^n\mid |\pi_{V'^{\perp}}(h)|\leq\varepsilon' |\pi_{V'}(h)|\}$$ be the analogous cone containing $V'$.

By the remark on equivalence of norms, condition $(\star_V)$ is easily seen to imply the following: for any given $\varepsilon>0$, there is a $\delta<<1$ such that $$X\cap B_\delta\subseteq \mathcal{C}_{V,\varepsilon}.$$ Here $B_\delta$ is the ball or radius $\delta$ in $\mathbb{R}^n$ centered at the origin. In the same way, condition $(\star_{V'})$ implies that, for any given $\varepsilon'>0$, there is a $\delta'<<1$ ensuring $$X\cap B_{\delta'}\subseteq \mathcal{C}_{V',\varepsilon'}.$$ In particular, given two $\varepsilon,\varepsilon'$, for $\delta<<1$ we have $X\cap B_\delta\subseteq \mathcal{C}_{V,\varepsilon}\cap\mathcal{C}_{V',\varepsilon'}$.

Now, assume by contradiction that neither $V\subseteq V'$ nor $V'\subseteq V$. In what follows we are going to choose $\varepsilon$ and $\varepsilon'$ such that $\mathcal{C}_{V,\varepsilon}\cap\mathcal{C}_{V',\varepsilon'}=\{0\}$. By the previous remarks, for any $\delta<<1$ we would then have $X\cap B_\delta\subseteq\{0\}$ which is a contradiction because the origin is an accumulation point for $X$.

Let $S_{V'}$ be the (compact) Euclidean unit sphere in $V'$, and let $v_0'$ minimize the quantity $|\pi_{V^\perp}(v')|$ for $v'\in S_{V'}$. Since $V'$ is not contained in $V$, $|\pi_{V^\perp}(v_0')|>0$. If we choose $\varepsilon<|\pi_{V^\perp}(v_0')|/|v_0'|$, we have $v_0'\notin\mathcal{C}_{V,\varepsilon}$ and hence $$V'\cap\mathcal{C}_{V,\varepsilon}=\{0\}.$$

Let $S_V$ be the Euclidean unit sphere in $V$ centered at the origin, and let $v_0$ minimize the quantity $|\pi_{V'^\perp}(v)|$ on the compact set $S_V\cap\mathcal{C}_{V,\varepsilon}$. Again, $|\pi_{V'^\perp}(v_0)|>0$, and we choose an $\varepsilon'<|\pi_{V'^\perp}(v_0)|/|v_0|$. For this choice of $\varepsilon'$, then, we indeed have $$\mathcal{C}_{V,\varepsilon}\cap\mathcal{C}_{V',\varepsilon'}=\{0\},$$ and this gives the desired contradiction. $\square$


(Edit: I deleted a corollary because it was not correct)

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  • $\begingroup$ Dear Qfwfq, thank you very much for this detailed answer. At first glance I do not notice the local injectivity assumption anywhere. Does this mean the subset $\{(x,y,\sqrt[4]{x^2+y^2})\mid x,y\in \mathbb R\}\subset\mathbb R^3$ is not a counterexample as I claimed in my post? As stated in the link, any vector space containing the $z$-axis would satisfy the limit condition, particularly two distinct planes intersecting at the $z$-axis, in seeming contradiction to the last proposition of your answer. $\endgroup$
    – Arrow
    Commented Jan 28, 2018 at 1:11
  • $\begingroup$ Clearly my lemma didn't make much sense, as stated. I have now edited it; the linked MSE answer is now (I think) not a counterexample to my lemma. But it is still a counterexample to its corollary (i.e. my "Proposition"). $\endgroup$
    – Qfwfq
    Commented Jan 28, 2018 at 1:43

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