Tangent cones and $C^1$ regularity of definable sets

TL;DR

This paper characterizes when a connected definable set in an o-minimal structure is a $C^1$ manifold by examining tangent cones, their continuity, and density conditions, establishing equivalences among these geometric properties.

Contribution

It proves the equivalence of three geometric conditions that characterize $C^1$ regularity for definable sets in o-minimal structures.

Findings

Tangent cone and paratangent cone coincide at all points for $C^1$ manifolds.

Tangent cone varies continuously and has constant dimension on $X$.

Density $ heta(X, x) < 3/2$ characterizes smoothness.

Abstract

Let $X\subset \mathbb R^n$ be a connected locally closed definable set in an o-minimal structure. We prove that the following three statements are equivalent: (i) $X$ is a $C^1$ manifold, (ii) the tangent cone and the paratangent cone of $X$ coincide at every point in $X$, (iii) for every $x \in X$, the tangent cone of $X$ at the point $x$ is a $k$-dimensional linear subspace of $\mathbb R^n$ ($k$ does not depend on $x$) varies continuously in $x$, and the density $\theta(X, x) < 3/2$.

Equations

$${\mathrm{tg}}_{x}X:=\{au\mid a\in \mathbb{R},a\ge 0,u=\underset{i\to \infty }{\lim }\frac{x_i-x}{\Vert x_i-x\Vert },\{x_i\}\subset X,\{x_i\}\to x\},$$

$${\mathrm{ptg}}_{x}X:=\{au\mid a\ge 0,u=\underset{i\to \infty }{\lim }\frac{x_i-y_i}{\Vert x_i-y_i\Vert },X\supset \{x_i\}\to x,X\supset \{y_i\}\to x\}.$$

$$\delta (P,Q)=\underset{v\in P,\Vert v\Vert =1}{\sup }\{\Vert v-{\pi }_Q(v)\Vert \},$$

$$E_x:=\{v\in {\mathbb{R}}^n:(x,v)\in E\}$$

$$E{|}_U:=\{(x,v)\in E:x\in U\}$$

$$\pi (q)\notin \pi (\partial W)$$

$${\overline{\mathbf{B}}}^k(\pi (q),s)\cap \pi (\partial W)=\varnothing .$$

$$\Vert p-\pi (q)\Vert \le \Vert p-q^{\prime }\Vert +\Vert q^{\prime }-\pi (q)\Vert \le s^{\prime }+s/2\le s.$$

$$\dim ({\mathrm{tg}}_{y}X\cap H^{\perp })\ge \dim {\mathrm{tg}}_{y}X+\dim H^{\perp }-\dim {\mathrm{tg}}_y\partial C=k+(n-k)-(n-1)=1.$$

$${\mathrm{tg}}_{(x,f(x))}{\mathrm{\Gamma }}_{f_x^i}\subset {\mathrm{tg}}_{(x,f(x))}{\mathrm{\Gamma }}_f\cap \mathrm{span}(a_i,b_1,\dots ,b_{n-k})=:l_x.$$

$$v+w=\underset{i\to \infty }{\lim }(a\frac{x_i}{\Vert x_i\Vert }-b\frac{y_i}{\Vert y_i\Vert })=\underset{i\to \infty }{\lim }\frac{a}{\Vert x_i\Vert }(x_i-y_i)\in {\mathrm{ptg}}_{0}X={\mathrm{tg}}_{0}X.$$

$$O_k:=\{z\in X:\dim {\mathrm{tg}}_{x}X\le k\}.$$

$${\mathrm{tg}}_{x}X\supset {\mathrm{tg}}_x\mathcal{O}\supset {\mathrm{tg}}_{x}X\setminus {\mathrm{tg}}_x{O}_{d-1}.$$

$${\mathrm{tg}}_{x}X\supset {\mathrm{tg}}_x\mathcal{O}\supset \overline{{\mathrm{tg}}_{x}X\setminus {\mathrm{tg}}_x{O}_{d-1}}.$$

$${\mathrm{tg}}_{x}X={\mathrm{tg}}_x\mathcal{O}\subset {\mathrm{ptg}}_x\mathcal{O}\subset {\mathrm{ptg}}_{x}X={\mathrm{tg}}_{x}X.$$

$${\pi }_x(W\cap X)={\pi }_x(W\cap \mathcal{O})=V.$$

$$X:=\overline{\{(x,\sin \frac{1}{x}),x\ne 0\}}\setminus \{(0,1),(0,-1)\}\subset {\mathbb{R}}^2,$$

$$Y:=\{-1-\frac{1}{n},n\in \mathbb{N}\}\cup [-1,1]\cup \{1+\frac{1}{n},n\in \mathbb{N}\}.$$

$$\Theta (X,x)=\underset{r\to 0}{\lim }\inf \frac{{\mathcal{H}}^k(X\cap {\overline{\mathbf{B}}}^n(x,r))}{r^k{\mu }_k}$$

$$h(z):=\frac{z^2}{|z{|}^{\frac{3}{2}}}\text{ if }z\ne 0,\text{ and }h(0)=0.$$

$$\frac{x}{\Vert x\Vert }=\frac{(z,h(z))}{\Vert (z,h(z))\Vert }=\left(\frac{z}{\sqrt{|z|}}\frac{1}{\sqrt{1+|z|}},\frac{z^2}{|z^2|}\frac{1}{\sqrt{1+|z|}}\right).$$

$${\partial }_{1}h$$

$${\partial }_{2}h$$

$${\partial }_{1}h=\frac{5}{4}\frac{z}{|z{|}^{\frac{3}{2}}}{\partial }_{1}z-\frac{3}{4}\frac{z^3}{|z{|}^{\frac{7}{2}}}{\partial }_1\bar{z}=\frac{5}{4}\frac{z}{|z{|}^{\frac{3}{2}}}-\frac{3}{4}\frac{z^3}{|z{|}^{\frac{7}{2}}}$$

$${\partial }_{2}h=\frac{5}{4}\frac{z}{|z{|}^{\frac{3}{2}}}{\partial }_{2}z-\frac{3}{4}\frac{z^3}{|z{|}^{\frac{7}{2}}}{\partial }_2\bar{z}=\frac{5}{4}\frac{z}{|z{|}^{\frac{3}{2}}}+\frac{3}{4}\frac{z^3}{|z{|}^{\frac{7}{2}}}.$$

$$\Vert u\Vert$$

$$\Vert v\Vert$$

$${\pi }_0{|}_X:X\ni (z,h(z))\mapsto h(z)\in {\mathrm{tg}}_{0}X.$$

$$\theta (X,x)=\underset{r\to 0}{\lim }\frac{{\mathcal{H}}^k(X\cap {\overline{\mathbf{B}}}^n(x,r))}{r^k{\mu }_k}.$$

$$X:=\{(x,y);y=0\}\cup \{(x,y);x>0,y=x^2\}$$

$$S_{\kappa }:=\{z\in {\mathbf{B}}^k(0,r):\mathrm{card}({\pi }^{-1}(z)\cap U)=\kappa \}.$$

Full text

1. Introduction

Let $X$ be a subset of $\mathbb{R}^{n}$ and let $x\in\mathbb{R}^{n}$. The tangent cone ${\rm tg}_{x}X$ and paratangent cone ${\rm ptg}_{x}X$ of $X$ at the point $x$ are defined as follows: if $x\not\in\overline{X}$, ${\rm tg}_{x}X={\rm ptg}_{x}X=\{0\}$, and otherwise,

[TABLE]

[TABLE]

Note that ${\rm tg}_{x}X$ and ${\rm ptg}_{x}X$ are closed sets in $\mathbb{R}^{n}$. We denote by ${\rm tg}X:=\{(x,v),x\in X,v\in{\rm tg}_{x}X\}$ and ${\rm ptg}X:=\{(x,v),x\in X,v\in{\rm ptg}_{x}X\}$.

Characterizing $C^{1}$ submanifolds of $\mathbb{R}^{n}$ in terms of their tangent cones has been studied by many authors, see for example [8], [10], [2], [6], or a survey of Bigolin and Golo [1]. In this paper we restrict ourselves to this problem in the context of o-minimal structures. We first prove that a connected locally closed definable subset of $\mathbb{R}^{n}$ is a $C^{1}$ manifold if and only if its tangent cone and paratangent cone coincide at every point (Theorem 3.7). This result is a strong version of the two-cones coincidence theorem (Theorem 3.6) which was initially proved by Tierno [10]. The result is no longer true if definability is omitted (Remark 3.9).

Next, we discuss a result recently established by Ghomi and Howard [6] that if $X\subset\mathbb{R}^{n}$ is a locally closed set such that for each $x\in X$, ${\rm tg}_{x}X$ is a hyperplane (i.e., a $(n-1)$ linear subspace of $\mathbb{R}^{n}$), and varies continuously in $x$, then $X$ is a union of $C^{1}$ hypersurfaces. Moreover, if the lower density $\Theta(X,x)$ is at most $m<\frac{3}{2}$ for every $x\in X$ then $X$ is a $C^{1}$ hypersurface. A natural question thus arises here is whether the result remains true if ${\rm tg}_{x}X$ are $k$-planes with $k<n-1$.

In section 4, we show in Example 4.3 that the first statement in the result of Ghomi-Howard is not always true if $k<(n-1)$. We also prove that the second statement is still valid, more precisely that if $X$ is a locally closed definable set such that for every $x\in X$, ${\rm tg}_{x}X$ is a $k$-dimensional linear subspace ($k$ is independent of $x$) varying continuously in $x$ and the density $\theta(X,x)<3/2$ (need not be upper bounded by an $m<\frac{3}{2})$, then $X$ is a $C^{1}$ manifold (Theorem 4.7). Notice that, in general, notions of lower density $\Theta(X,x)$ and density $\theta(X,x)$ are different. Nevertheless, with the conditions on tangent cones as above, they coincide. Therefore, our result can be considered as a generalization of the result of Ghomi-Howard.

Throughout the paper, $\mathbb{R}^{n}$ denotes the $n$-dimensional Euclidean space equipped with the standard norm $\|x\|=\sqrt{x_{1}^{2}+\ldots+x_{n}^{2}}$ where $x=(x_{1},\ldots,x_{n})$; $\overline{\mathbf{B}^{n}}(x,r)$, $\mathbf{B}^{n}(x,r)$ and $\mathbf{S}^{n-1}(x,r)$) denote respectively the closed ball, the open ball and the sphere in $\mathbb{R}^{n}$ of radius $r$ centered at $x$. Let $X$ be a subset in $\mathbb{R}^{n}$. Denote by $\overline{X}$ the closure of $X$ in $\mathbb{R}^{n}$ and by $\partial X:=\overline{X}\setminus X$ the boundary of $X$. Let $f$ be a map. We denote by ${\rm\Gamma}_{f}$ the graph of $f$.

The Grassmanian $\mathbb{G}_{n}^{k}$ of all $k$-dimensional linear subspaces of $\mathbb{R}^{n}$ is endowed with the metric $\delta$ defined as follows: for $P,Q$ in $\mathbb{G}_{n}^{k}$,

[TABLE]

where $\pi_{Q}:\mathbb{R}^{n}\to Q$ is the orthogonal projection from $\mathbb{R}^{n}$ to $Q$. Following [6], we say that $P\in\mathbb{G}_{n}^{k}$ and $Q\in\mathbb{G}_{n}^{k}$ are orthogonal when $\delta(P,Q)=1$. Remark that this terminology does not coincide with the usual notion of orthogonality for general subspaces in Euclidean geometry : $P=\{(x,y,z);\;x=0\}$ and $Q=\{(x,y,z);\;y=0\}$ are orthogonal according to our definition but not all vectors in $P$ are orthogonal to any vector in $Q$.

By a $k$-dimensional $C^{1}$ manifold in $\mathbb{R}^{n}$ (or $C^{1}$ manifold for simplicity) we mean a subset of $\mathbb{R}^{n}$, locally $C^{1}$ diffeomorphic to $\mathbb{R}^{k}$; a hypersurface in $\mathbb{R}^{n}$ is a $C^{1}$ manifold in $\mathbb{R}^{n}$ of dimension $n-1$.

Let $X\subset\mathbb{R}^{n}$. In the paper, we often denote by $\pi_{x}:\mathbb{R}^{n}\to{\rm tg}_{x}X$ the orthogonal projection. By abuse of notation, here we identify ${\rm tg}_{x}X$ with its translation $\{x+{\rm tg}_{x}X\}$.

By a definable set we mean a set which is definable (with parameters) in an o-minimal expansion $(\mathbb{R},<,+,\cdot,\dots)$ of the ordered field of real numbers. Definable sets form a large class of subsets of $\mathbb{R}^{n}$: for instance, any semi-algebraic set, any sub-analytic set is definable. We refer the reader to [11], [5] for the basic properties of o-minimal structures. In the paper, we will use Curve selection Lemma ([5], Theorem 3.2), Uniform finiteness on fibers ([5] Theorem 2.9) and Hardt’s definable triviality Theorem ([5], Theorem 5.22) without repeating the references.

2. Bundle of vector spaces

Let $X$ be a subset of $\mathbb{R}^{n}$. Let $E\subset X\times\mathbb{R}^{n}$. For $x\in X$ we denote by

[TABLE]

the fiber of $E$ at the point $x$. For $U\subset X$, we set

[TABLE]

and call it the restriction of $E$ to $U$. If every fiber of $E$ is a linear subspace of $\mathbb{R}^{n}$ we call $E$ a bundle of vector spaces over $X$, or a bunlde over $X$, or just a bundle if the base $X$ is clear from the context. We call $E$ a* trivial bundle* if all its fibers have the same dimension, and a closed bundle if it is a closed set in $\mathbb{R}^{2n}$.

Suppose $E$ is a trivial bundle over $X$. If the map $X\to\mathbb{G}_{n}^{k}$ defined by $x\mapsto E_{x}$ is continuous, we say that $E$ is a continuous trivial bundle.

Lemma 2.1** ([7], Propositions I, II, pages 39, 40).**

Let $E\subset X\times\mathbb{R}^{n}$.

1. (i)

If $E$ is a closed trivial bundle then $E$ is continuous. 2. (ii)

If $E$ is a closed bundle then the function $x\mapsto\dim E_{x}$ is upper-semicontinuous, i.e., for $x\in X$ there is an open neighborhood $U_{x}$ of $x$ in $X$ such that $\dim E_{x}\geq\dim E_{y}$ for every $y\in U_{x}$.

Proof.

$(i)$ - Suppose that $k$ is the dimension of fibers of $E$. If $E$ is not continuous, there exists a sequence $\{x_{i}\}\subset X$ tending to $x$, $\lim_{i\to\infty}E_{x_{i}}\to\tau\in\mathbb{G}_{n}^{k}$ and $\tau\not\subset E_{x}$. By the closedness of $E$, $\tau\subset E_{x}$, which is a contradiction.

$(ii)$ - Suppose the assertion is not true, i.e., there exist a point $x\in X$ and a sequence $\{x_{m}\}$ in $X$ tending to $x$ such that $\dim E_{x_{m}}>\dim E_{x}$. We may assume that $\lim_{m\to\infty}E_{x_{m}}=P\in\mathbb{G}_{n}^{k}$, since $\mathbb{G}_{n}^{k}$ is compact. Note that $k>\dim E_{x}$. Since $E$ is closed, $P\subset E_{x}$. This implies $k\leq\dim E_{x}$, which is a contradiction. ∎

Lemma 2.2**.**

Let $X$ be a locally closed subset of $\mathbb{R}^{n}$. If ${\rm ptg}X$ is a trivial bundle then ${\rm ptg}X$ is continuous.

Proof.

It follows directly from the definition of the paratangent cone that ${\rm ptg}(\overline{X})$ is a closed set in $\mathbb{R}^{2n}$ and ${\rm ptg}_{x}X={\rm ptg}_{x}\overline{X}$ for every $x\in X$. If $V\subset X$ is a closed set in $\mathbb{R}^{n}$ then ${\rm ptg}X|_{V}={\rm ptg}\overline{X}|_{V}$ is a closed set in $\mathbb{R}^{2n}$.

Let $x\in X$. Since $X$ is locally closed, there is $W_{x}$, a neighborhood of $x$ in $X$, which is closed in $\mathbb{R}^{n}$. The restriction ${\rm ptg}X|_{W_{x}}$ is then a closed set in $\mathbb{R}^{2n}$. Since ${\rm ptg}X$ is a trivial bundle, so is ${\rm ptg}X|_{W_{x}}$. By $(i)$ in Lemma 2.1 ${\rm ptg}X|_{W_{x}}$ is continuous, ${\rm ptg}X$ is, therefore, a continuous trivial bundle. ∎

3. Two-cones coincidence Theorem

The aim of this section is to prove Theorem 3.7, a strong version of two-cones coincidence theorem of Tierno for definable sets.

We need the following two lemmas which generalize Lemma 3.3 and Lemma 3.1 in [6].

Lemma 3.1**.**

Let $X$ be a locally closed subset of $\mathbb{R}^{n}$ such that ${\rm tg}X$ is a continuous trivial bundle of $k$-dimensional vector spaces. Let $x\in X$ and $H$ be a $k$-plane in $\mathbb{R}^{n}$ which is not orthogonal to ${\rm tg}_{x}X$. Let $\pi:\mathbb{R}^{n}\to H$ the orthogonal projection. Then, there exists an open set $U$ of $x$ in $X$ such that $\pi|_{U}:U\to H$ is an open map.

Proof.

The proof follows closely the proof of Lemma 3.3, [6].

By the continuity of ${\rm tg}X$, we can choose an open neighborhood $U$ of $x$ in $X$ such that for all $q\in U$, ${\rm tg}_{q}X$ is not orthogonal to $H$, or equivalently that ${\rm tg}_{q}X$ is transverse to $H^{\perp}$, the orthogonal complement of $H$ in $\mathbb{R}^{n}$. We will prove that $\pi|_{U}$ is an open map. Fix $q\in U$. By the local closedness of $X$, there is an $r>0$ small enough such that $W:=X\cap\overline{\mathbf{B}}^{n}(q,r)\subset U$ is a compact set. Moreover, the boundary $\partial W$ is in $\partial\overline{\mathbf{B}}^{n}(q,r)$, meaning $q\not\in\partial W$. With $r$ sufficiently small, we may assume that

[TABLE]

because, otherwise, there exists a sequence of positive numbers $\{r_{i}\}$ tending to [math] such that for each $i$, there is a point $q_{i}\in X\cap\partial\overline{\mathbf{B}}^{n}(q,r_{i})$ with $\pi(q_{i})=\pi(q)$. This implies that ${\rm span}(q-q_{i})\subset H^{\perp}$. As $i\to\infty$ we have $q_{i}\to q$ and the sequence ${\rm span}(q-q_{i})$ (extracting a subsequence if necessary) tends to a line $l\in{\rm tg}_{x}X$. Thus, $l\in H^{\perp}\cap{\rm tg}_{q}X$; but $\dim(H^{\perp}\cap{\rm tg}_{q}X)=\dim H^{\perp}+\dim{\rm tg}_{q}X-n=0$ since $H^{\perp}$ is transverse to ${\rm tg}_{q}X$, which is a contradiction.

Since $\pi(\partial W)$ is a compact set, there is $s>0$ such that

[TABLE]

It suffices to show that $\pi(W)$ contains an open neighborhood of $\pi(q)$ in $H$.

Suppose on the contrary that $\overline{\mathbf{B}}^{k}(\pi(q),\varepsilon)\not\subset\pi(W)$, $\forall\varepsilon>0$. Choose a point $q^{\prime}\in\overline{\mathbf{B}}^{k}(\pi(q),s/2)\setminus\pi(W)$ and let $s^{\prime}$ be the distance from $q^{\prime}$ to $\pi(W)$. Note that $s^{\prime}\leq s/2$. Since $\pi(W)$ is compact, $\overline{\mathbf{B}}^{k}(q^{\prime},s^{\prime})\cap\pi(W)\neq\emptyset$ and $\mathbf{B}^{k}(q^{\prime},s^{\prime})\cap\pi(W)=\emptyset$. For every $p\in\overline{\mathbf{B}}^{k}(q^{\prime},s^{\prime})$,

[TABLE]

This means that $\overline{\mathbf{B}}^{k}(q^{\prime},s^{\prime})\subset\overline{\mathbf{B}}^{k}(\pi(q),s)$. By (3.1), $\overline{\mathbf{B}}^{k}(q^{\prime},s^{\prime})\cap\pi(\partial W)=\emptyset$. Take $y^{\prime}\in\overline{\mathbf{B}}^{k}(q^{\prime},s^{\prime})\cap\pi(W)$ and $y\in\pi^{-1}(y^{\prime})\cap W$. Note that $y\not\in\partial W$, so $y$ is an interior point of $W$, and hence ${\rm tg}_{y}W={\rm tg}_{y}X$ which is a linear subspace.

Since $\mathbf{B}^{k}(q^{\prime},s^{\prime})\cap\pi(W)=\emptyset$, no point of $W$ is contained in the cylinder $C:=\pi^{-1}(\mathbf{B}^{k}(q^{\prime},s^{\prime}))$. This implies that ${\rm tg}_{y}X\subset{\rm tg}_{y}\partial C$. Both ${\rm tg}_{y}X$ and $H^{\perp}$ are included in ${\rm tg}_{y}\partial C$, so

[TABLE]

This shows that ${\rm tg}_{y}X$ is orthogonal to $H$, which is a contradiction. ∎

Lemma 3.2**.**

Let $U\subset\mathbb{R}^{k}$ be an open set and $f:U\to\mathbb{R}^{n-k}$ be a map. Suppose that ${\rm\Gamma}_{f}$ is locally closed. If ${\rm tg}{\rm\Gamma}_{f}$ is a continuous trivial bundle of $k$-dimensional vector spaces and ${\rm tg}_{(x,f(x))}{\rm\Gamma}_{f}$ is not orthogonal to $\mathbb{R}^{k}$ for every $x\in U$, then $f$ is $C^{1}$.

Proof.

We first prove that $f$ is continuous. Suppose on the contrary that $f$ is not continuous, meaning that there are $x\in U$ and a sequence $\{x_{i}\}$ in $U$ tending to $x$ such that $\lim_{i\to\infty}f(x_{i})=y\neq f(x)$.

Since ${\rm\Gamma}_{f}$ and the orthogonal projection $\pi:\mathbb{R}^{n}\to\mathbb{R}^{k}$ satisfy the hypothesis of Lemma 3.1, there is an open set $V$ of $(x,f(x))$ in ${\rm\Gamma}_{f}$ such that $\pi|_{V}$ is an open map. Set $W:=\pi(V)$, which is an open neighborhood of $x$ in $\mathbb{R}^{k}$. Since $\{x_{i}\}$ tends to $x$, there is $N\in\mathbb{N}$ such that $x_{i}\in W$ for all $i>N$. This implies that $(x_{i},f(x_{i}))\in V$ for all $i>N$.

If $y\neq\infty$, shrinking $V$ so that $(x,y)\not\in\overline{V}$, there is a neighborhood $K$ of $(x,y)$ in $\mathbb{R}^{n}$ such that $K\cap V=\emptyset$. Since $f(x_{i})$ tends to $y$, we have $(x_{i},f(x_{i}))\in K$ for all $i>N$ when $N$ is large enough. This shows that $K\cap V\neq\emptyset$, which is a contradiction.

If $y=\infty$, $(x_{i},f(x_{i}))\not\in V$ for all $i>N$ when $N$ is large enough. This again gives a contradiction.

We now show that $f$ is a $C^{1}$ map.

Let $\{a_{1},\ldots,a_{k},b_{1},\ldots,b_{n-k}\}$ be the canonical basis of $\mathbb{R}^{n}$. For $x\in U$, consider the function $f^{i}_{x}(t):=f(x+ta_{i})$. The graph ${\rm\Gamma}_{f^{i}_{x}}$ of $f^{i}_{x}$ is the intersection ${\rm\Gamma}_{f^{i}_{x}}={\rm\Gamma}_{f}\cap(x,0)+{\rm span}(a_{i},b_{1},\dots,b_{k})$. This implies that

[TABLE]

But $l_{x}$ is a line, because ${\rm tg}_{(x,f(x))}{\rm\Gamma}_{f}$ is not orthogonal to $\mathbb{R}^{k}\times\{0\}^{n-k}$. On the other hand, since $f$ is continuous, ${\rm\Gamma}_{f^{i}_{x}}$ is a continuous curve, so ${\rm tg}_{(x,f(x))}{\rm\Gamma}_{f^{i}_{x}}$ has dimension at least $1$. Then ${\rm tg}_{(x,f(x))}{\rm\Gamma}_{f^{i}_{x}}=l_{x}$, so $f^{i}_{x}$ is differentiable at $t=0$. Thus, $f$ has partial derivatives at any point.

The bundle ${\rm tg}_{(x,f(x))}{\rm\Gamma}_{f}$ is continuous, hence $l_{x}$, its transverse intersection with ${\rm span}(a_{i},b_{1},\dots,b_{n-k})$ is continuous. Therefore $f$ has continuous partial derivatives on $U$, so $f$ is $C^{1}$.

∎

Remark 3.3*.*

The statement of Lemma 3.1 [6] is similar to the statement of Lemma 3.2 except the local closedness of ${\rm\Gamma}_{f}$ is missing. This is a gap because $f$ might not be continuous if ${\rm\Gamma}_{f}$ is not locally closed. For example, consider the function $f(x)=0$ if $x$ is a rational number, and $f(x)=1$ otherwise. The tangent cone to ${\rm\Gamma}_{f}$ is the $x$-axis at any point, hence ${\rm\Gamma}_{f}$ is a continuous trivial bundle, but $f$ is not continuous.

Theorem 3.4**.**

A locally closed set $X\subset\mathbb{R}^{n}$ is a $C^{1}$ manifold if and only if ${\rm tg}X$ is a continuous trivial bundle and the restriction of the map $\pi_{x}:X\to{\rm tg}_{x}X$ to some neighborhood of $x$ in $X$ is injective.

Proof.

The necessity is a trivial fact. We now prove the sufficiency. For $x\in X$, by the hypothesis, there exists an open neighborhood $U$ of $x$ such that $\varphi:=\pi_{x}|_{U}:U\to{\rm tg}_{x}X$ is injective. Moreover, $\pi_{x}$ is open by Lemma 3.1. This implies $\varphi:U\to\varphi(U)$ is a homeomorphism. Consider the map $\psi:=\varphi^{-1}:\varphi(U)\to U\subset\mathbb{R}^{n}$. We have ${\rm tg}{\rm\Gamma}_{\psi}={\rm tg}X|_{{\rm\Gamma}_{\psi}}$, which is a continuous trivial bundle. Shrinking $U$ if necessary we may assume that ${\rm tg}_{y}X$ is not orthogonal to ${\rm tg}_{x}X$ for every $y\in U$. The map $\psi$ then satisfies the conditions of Lemma 3.2, so it is of class $C^{1}$, meaning that $U$ is a $C^{1}$ manifold. ∎

Remark 3.5*.*

Theorem 3.4 is slightly stronger than a similar result proved by Gluck (Theorem 10.1, [8]). In the result of Gluck, $X$ is assumed to be a topological manifold instead of a locally closed set as in our statement.

Theorem 3.6** (Two-cones coincidence, Tierno [10]).**

A locally closed subset $X$ of $\mathbb{R}^{n}$ is a $C^{1}$ manifold if and only if $TX$ and ${\rm ptg}X$ coincide, and both are trivial bundles of vector spaces over $X$.

Proof.

Since ${\rm ptg}X$ is a trivial bundle, it is continuous by Lemma 2.2. On the other hand, ${\rm tg}X={\rm ptg}X$, hence ${\rm tg}X$ is a continuous trivial bundle.

Let $x\in X$. We denote by $\pi_{x}:\mathbb{R}^{n}\to{\rm tg}_{x}X$ the orthogonal projection. By Theorem 3.4, it suffices to prove that the map $\pi_{x}$ is injective on some neighborhood of $x$ in $X$.

Suppose on contrary that there are sequences of points $\{z_{i}\}_{i}$ and $\{z^{\prime}_{i}\}_{i}$ in $X$ converging to $x$ such that $\pi_{x}(z_{i})=\pi_{x}(z^{\prime}_{i})$ for all $i$. This implies that ${\rm span}(z_{i}-z^{\prime}_{i})$ accumulates to a line $l\subset{\rm tg}_{x}X^{\perp}$. By the definition, $l\subset{\rm ptg}_{x}X$. Since ${\rm ptg}_{x}X={\rm tg}_{x}X$, $l\subset{\rm tg}_{x}X\cap{\rm tg}_{x}X^{\perp}=0$, a contradiction. ∎

Theorem 3.7** (Definable two-cones coincidence).**

A connected, locally closed definable subset of $\mathbb{R}^{n}$ is a $C^{1}$ manifold if and only if ${\rm tg}X$ and ${\rm ptg}X$ coincide.

Proof.

We just need to show the sufficiency. First we prove that for every $x\in X$, ${\rm tg}_{x}X$ is a linear subspace of $\mathbb{R}^{n}$, or equivalently that ${\rm tg}X$ is a bundle. Fix $x\in X$, we may identify $x$ with the origin [math]. By the hypothesis, ${\rm tg}_{0}X={\rm ptg}_{0}X$ which is symmetric, i.e., if $v\in{\rm tg}_{0}X$ so is $-v$. It is enough to verify that if $v,w\in{\rm tg}_{0}X$ then $v+w\in{\rm tg}_{0}X$. Since $v,-w\in{\rm tg}_{0}X$ and $X$ is a definable set there exist two curves $\gamma,\beta$ in $X$ starting at [math] such that $v\in{\rm tg}_{0}\gamma$ and $-w\in{\rm tg}_{0}\beta$ (see Curve selection Lemma). Choose sequences of points $\{x_{i}\}_{i}\subset\gamma$ and $\{y_{i}\}_{i}\subset\beta$ converging to [math] such that $\|x_{i}\|=\frac{a}{b}\|y_{i}\|$, where $a:=\|v\|$ and $b:=\|w\|$. Thus,

[TABLE]

From now on, we set

[TABLE]

Fix $k$ and let $z\in O_{k}$. Take $V$ to be a closed neighborhood of $z$ in $X$. Since $X$ is locally closed, we may assume $V$ to be a closed set in $\mathbb{R}^{n}$ so ${\rm ptg}X|_{V}$ is a closed bundle. By Lemma 2.1 (ii), for $y\in V$, the map $y\mapsto\dim{\rm ptg}_{y}X$ is upper-semicontinuous, so is the map $y\mapsto\dim{\rm tg}_{y}X$, meaning that there exists $U\subset V$, an open neighborhood of $z$, such that $k=\dim{\rm ptg}_{z}X\geq\dim{\rm ptg}_{y}X$ for all $y$ in $U$. This implies that $U\subset O_{k}$, hence $O_{k}$ is an open set in $X$.

Since $X$ is definable, $\dim{\rm tg}_{x}X\leq\dim X=d$ for every $x\in X$ (see [9], Lemma 1.2), hence $O_{d}=X$. Set $\mathcal{O}:=X\setminus O_{d-1}$, which is a closed set of $X$.

We denote by $X_{\rm Sing}$ the set of singular points of $X$, i.e., points at which $X$ fails to be a $C^{1}$ manifold of dimension $d$. Remark that $X_{\rm Sing}$ is a definable set of dimension less than $d$ (see for instance [11], [5]).

Since $O_{d-1}\subset X_{\rm Sing}$, $\dim O_{d-1}<d$. For $x\in\mathcal{O}$,

[TABLE]

Taking the closures of all sets above,

[TABLE]

Because ${\rm tg}_{x}X$ is a linear space of dimension $d$ and ${\rm tg}_{x}O_{d-1}$ is a linear subspace of ${\rm tg}_{x}X$ of dimension less than $d$, $\overline{{\rm tg}_{x}X\setminus{\rm tg}_{x}O_{d-1}}={\rm tg}_{x}X$. So,

[TABLE]

Since $\mathcal{O}$ is closed in the locally closed set $X$, it is also locally closed. As been shown above, ${\rm tg}\mathcal{O}={\rm ptg}\mathcal{O}={\rm ptg}X|_{\mathcal{O}}$ which is a trivial bundle. By Theorem 3.6, $\mathcal{O}$ is a $C^{1}$ manifold of dimension $d$. Next we will prove that $X=\mathcal{O}$, therefore, it is a $C^{1}$ manifold.

Let $x\in\mathcal{O}$ and $\pi_{x}:\mathbb{R}^{n}\to{\rm tg}_{x}\mathcal{O}$ be the orthogonal projection. It follows from Theorem 3.4 and Lemma 3.1 that there is an open neighborhood $U$ of $x$ in $\mathbb{R}^{n}$ such that the restriction of $\pi_{x}$ to $U\cap\mathcal{O}$ is injective and $V:=\pi_{x}(U\cap\mathcal{O})$ is an open set in ${\rm tg}_{x}\mathcal{O}$. Set $W:=\pi_{x}^{-1}(V)\cap U$, so that $W$ is an open neighborhood of $x$ in $\mathbb{R}^{n}$ with $\pi_{x}(W)=V$. This implies that

[TABLE]

Since ${\rm tg}_{x}\mathcal{O}={\rm ptg}_{x}X$, shrinking $U$ if necessary, the restriction of $\pi_{x}$ to $W\cap X$ is injective. The sets $W\cap X$ and $W\cap\mathcal{O}$ then are graphs of mappings over the same domain $V$. On the other hand, $W\cap\mathcal{O}\subset W\cap X$, then $W\cap\mathcal{O}=W\cap X$. This means that $W\cap\mathcal{O}$, an open neighborhood of $x$ in $\mathcal{O}$, is an open neighborhood of $x$ in $X$. Thus, $\mathcal{O}$ is an open set in $X$. Since $\mathcal{O}$ is both closed and open in $X$ and $X$ is connected, $\mathcal{O}$ is equal to $X$.

∎

A direct consequence of Theorem 3.6 is:

Corollary 3.8**.**

Let $X\subset\mathbb{R}^{n}$ be a locally closed definable set. Suppose that ${\rm tg}_{x}X={\rm ptg}_{x}X$ for every $x\in X$. Then, each connected component of $X$ is a $C^{1}$ manifold.

Remark 3.9*.*

The definability in the hypothesis of Theorem 3.7 is necessary. Consider the following locally closed sets,

[TABLE]

and

[TABLE]

The sets $X$ and $Y$ are not definable in any o-minimal structure : $X\cap\mathbb{R}\times\{0\}$ and $Y$ have infinitely many componant. The set $X$ is connected, ${\rm tg}X={\rm ptg}X$, but $X$ is not a $C^{1}$ manifold. The set $Y$ has ${\rm tg}Y={\rm ptg}Y$, but $[-1,1]$, a connected component of $Y$, is not a $C^{1}$ manifold.

4. Definable sets with continuous trivial tangent cones

Let us recall the result proved by Ghomi and Howard [6].

Definition 4.1**.**

Let $X\subset\mathbb{R}^{n}$, $x\in\mathbb{R}^{n}$ . Suppose that * the Hausdorff dimension* of ${\rm tg}_{x}X$, denoted by $\dim_{\mathcal{H}}{\rm tg}_{x}X$, is an integer $k$. The lower density $\Theta(X,x)$ of $X$ at the point $x$ is defined as follows: if $x\not\in\overline{X}$ then $\Theta(X,x)=0$, and otherwise,

[TABLE]

where $\mathcal{H}^{k}$ is the $k$-dimensional Hausdorff measure, $\mu_{k}$ is the volume of the unit ball of dimension $k$.

Theorem 4.2** (Theorem 1.1, [6]).**

Let $X$ be a locally closed subset of $\mathbb{R}^{n}$. Suppose that ${\rm tg}X$ is a $(n-1)$-dimensional continuous trivial bundle. Then,

1. (i)

$X$* is a union of $C^{1}$ hypersurfaces;* 2. (ii)

if $\Theta(X,x)$ is at most $m<3/2$ then $X$ is a $C^{1}$ hypersurface.

The following example shows that in general the statement $(i)$ of Theorem 4.2 is no longer true when the hyperplanes are replaced by $k$-planes with $k<n-1$, meaning that a locally closed subset in $\mathbb{R}^{n}$ with continuous trivial tangent cone might not be a union of $C^{1}$ manifolds.

Example 4.3**.**

We identify $\mathbb{C}$ with $\mathbb{R}^{2}$. Consider the map $h:\mathbb{C}\to\mathbb{C}$ defined as follows:

[TABLE]

Denote by $X\subset\mathbb{R}^{4}$ the graph of $h$. We have

(1) $X$ is locally closed;

(2) ${\rm tg}X$ is a $2$-dimensional continuous trivial bundle;

(3) $X$ is not a union of $C^{1}$ submanifolds of dimension $2$ of $\mathbb{R}^{4}$.

Proof.

Remark that $h$ is continuous, hence $X$ is a topological manifold and the condition (1) is automatically satisfied. Moreover, $h$ is smooth except at the origin, where statement (2) is obvious. We now calculate ${\rm tg}_{0}X$. Let $x\in X\setminus\{0\}$, we may write $x=(z,h(z))$ for some $z\in\mathbb{C}^{*}$. We have

[TABLE]

Notice that $\frac{z}{\sqrt{|z|}}\frac{1}{\sqrt{1+|z|}}\to 0$ when $z\to 0$. Hence ${\rm tg}_{0}X\subset\{z=0\}$. On the other hand, if $z=re^{i\theta}$ and $r\to 0$, $\frac{z^{2}}{|z^{2}|}\frac{1}{\sqrt{1+|z|}}\to e^{2i\theta}$. Thus, $(0,e^{2i\theta})\in{\rm tg}_{0}X$ for all $\theta$. This implies that ${\rm tg}_{0}X=\{z=0\}$.

Write $z=z_{1}+iz_{2}$ and $h=h_{1}+ih_{2}$. For $x=(z,h(z))$, $z\neq 0$, ${\rm tg}_{x}X$ is generated by two vectors $u=(1,0,\frac{\partial h_{1}}{\partial z_{1}},\frac{\partial h_{2}}{\partial z_{1}})$ and $v=(0,1,\frac{\partial h_{2}}{\partial z_{1}},\frac{\partial h_{2}}{\partial z_{2}})$. Denote by $\partial_{1}$ and $\partial_{2}$ the directional derivatives in the variable $z$ along $z_{1}$-axis and $z_{2}$-axis respectively. We know that

[TABLE]

Note that $h=\frac{z^{2}}{(z\bar{z})^{\frac{3}{4}}}$. Computation gives,

[TABLE]

[TABLE]

If $z$ tends to [math], then $|\partial_{1}h|$ and $|\partial_{2}h|$ tend to $\infty$. Therefore,

[TABLE]

Hence $\lim_{x\to 0}{\rm tg}_{x}X={\rm tg}_{0}X$. This implies (2).

Now we show (3). Denote by $\pi_{0}:\mathbb{R}^{4}\to{\rm tg}_{0}X$ the orthogonal projection from $\mathbb{R}^{4}$ onto ${\rm tg}_{0}X$ :

[TABLE]

This map is not injective in any neighborhood of [math] since $\forall z,\;h(z)=h(-z)$. By Theorem 3.4, $X$ is not a $C^{1}$ manifold. Since $X$ is a connected topological manifold, it cannot be the union of two or more $C^{1}$ manifolds of dimension $2$. ∎

Definition 4.4** ([3],[4],[9]).**

Let $X\subset\mathbb{R}^{n}$ be a definable set and let $x\in\mathbb{R}^{n}$. Suppose that $\dim X=k$. It is known that the following limit always exists

[TABLE]

We call it the* density *of $X$ at the point $x$.

Remark 4.5*.*

The notions of lower density and density are not the same even for definable sets. For example, consider $X:=\{(x,y,z)\in\mathbb{R}^{3}:z^{4}=x^{2}+y^{2}\}$. It is easy to see that $\Theta(X,0)=\infty$ while $\theta(X,0)=0$. However, if $X$ is a definable set and ${\rm tg}X$ is a trivial bundle then $\dim_{\mathcal{H}}{\rm tg}_{x}X=\dim{\rm tg}_{x}X=\dim X$, and therefore $\Theta(X,x)=\theta(X,x)$ for every $x\in X$.

Lemma 4.6**.**

1. (i)

Let $X,Y$ be definable sets of the same dimension. If $\dim(X\cap Y)<\dim X$, then $\theta(X\cup Y,x)=\theta(X,x)+\theta(Y,x)$. If $X\subset Y$, then $\theta(X,x)\leq\theta(Y,x)$. 2. (ii)

If $X$ is a definable set then $\theta(X,x)\geq\theta({\rm tg}_{x}X,0)$.

Proof.

$(i)$ is a direct consequence of the definition of density, and $(ii)$ of Theorem 3.8 of [9]. ∎

Theorem 4.7**.**

Let $X$ be a locally closed definable subset of $\mathbb{R}^{n}$. If ${\rm tg}X$ is a continuous trivial bundle and $\theta(X,x)<3/2$ for every $x\in X$, then $X$ is a $C^{1}$ manifold.

Remark 4.8*.*

The condition on the density above is sharp, since

[TABLE]

satisfies all other hypothesis of Theorem 4.7 and $\theta(X,0)=3/2$.

Proof.

Denote by $\pi_{x}:\mathbb{R}^{n}\to{\rm tg}_{x}X$ the orthogonal projection. Suppose on the contrary that $X$ is not a $C^{1}$ manifold. By Theorem 3.4, there exists $x\in X$ such that there is no neighborhood of $x$ in $X$ to which the restriction of $\pi_{x}$ is injective. We may assume that $x$ coincides with the origin [math] and ${\rm tg}_{x}X=\mathbb{R}^{k}$ where $k=\dim X$. The map $\pi_{x}$ now becomes $\pi:\mathbb{R}^{n}\to\mathbb{R}^{k}$, the orthogonal projection to the first $k$ coordinates.

By Lemma 3.1, there is an open neighborhood $U$ of [math] in $X$ such that $\pi|_{U}$ is an open map. Hence there exists $r>0$ such that $\mathbf{B}^{k}(0,r)\subset\pi(U)$. Shrinking $U$ if necessary we assume that $\pi(U)=\mathbf{B}^{k}(0,r)$. We also assume that for every $x\in U$, ${\rm tg}_{x}X$ is not orthogonal to $\mathbb{R}^{k}$.

By the uniform finiteness on fibres of definable sets, there exists $N\in\mathbb{N}$ such that for every $z\in\mathbf{B}^{k}(0,r)$, $|\pi^{-1}(z)\cap U|$, the number of connected components of $\pi^{-1}(z)\cap U$, does not exceed $N$. In fact, in this case, $|\pi^{-1}(z)\cap U|={\rm card}(\pi^{-1}(z)\cap U)$ where card denotes the cardinality. If otherwise, there is a connected component of $\pi^{-1}(z)\cap U$, write $F$, such that $\dim F\geq 1$. Since $F$ is definable, there is a point $\tilde{z}\in K$ such that $\dim{\rm tg}_{\tilde{z}}F=\dim F\geq 1$. But $F\subset(\{z\}\times\mathbb{R}^{n-k})$, so ${\rm tg}_{\tilde{z}}F\subset(\{0\}\times\mathbb{R}^{n-k})$. This implies that ${\rm tg}_{\tilde{z}}F\subset{\rm tg}_{\tilde{z}}X\cap\{0\}\times\mathbb{R}^{n-k}$. Since $\tilde{z}\in U$, ${\rm tg}_{\tilde{z}}X$ is not orthogonal to $\mathbb{R}^{k}$, then ${\rm tg}_{\tilde{z}}X\cap(\{0\}\times\mathbb{R}^{n-k})=\{0\}$, which gives a contradiction.

Set

[TABLE]

Then $\{S_{\kappa}\}_{\kappa=1}^{N}$ becomes a definable partition of $\mathbf{B}^{k}(0,r)$. We may assume that $0\in\overline{S_{\kappa}},\forall\kappa$ and $S_{N}\neq\emptyset$. Note that $N\geq 2$ since the restriction of $\pi$ is not injective on any neighborhood of [math]. We claim that

(a) $S_{N}$ is an open set,

(b) $|(\pi^{-1}(S_{N})\cap U|=N$,

(c) Each connected component of $\pi^{-1}(S_{N})\cap U$ is a $C^{1}$ manifold.

We now give a proof of the claim.

Let $q\in S_{N}$. Since ${\rm card}(\pi^{-1}(q)\cap U)=N$, we may write $\pi^{-1}(q)\cap U=\{q_{1},\ldots,q_{N}\}$. There is $\epsilon>0$ sufficiently small such that $K_{i}\cap K_{j}=\emptyset$ for $i\neq j$, where $K_{i}:=\mathbf{B}^{n}(q_{i},\epsilon)\cap U$, $i\in\{1,\ldots,N\}$. Since the map $\pi|_{U}$ is open, there exists an open neighborhood $V_{q}$ of $q$ in $\mathbb{R}^{k}$ such that $V_{q}\subset\pi(K_{i}),\forall i\in\{1,\ldots,N\}$. For $q^{\prime}\in V_{q}$, $\forall i\in\{1,\ldots,N\}$, $\pi^{-1}(q^{\prime})\cap K_{i}\neq\emptyset$, hence ${\rm card}(\pi^{-1}(q^{\prime})\cap U)\geq N$. By the definition of $N$, ${\rm card}(\pi^{-1}(q^{\prime})\cap U)$ cannot exceed $N$, therefore ${\rm card}(\pi^{-1}(q^{\prime})\cap U)=N$, meaning $q^{\prime}\in S_{N}$. This implies $V_{q}\subset S_{N}$, or equivalently $S_{N}$ is open, (a) is proved.

Denote by $A_{1},\ldots,A_{m}$ the connected components of $\pi^{-1}(S_{N})\cap U$. Note that for every $i$, $A_{i}$ is an open set in $U$, so $\pi(A_{i})$ is an open set in $\mathbb{R}^{k}$ since $\pi|_{U}$ is an open map. Assume that (b) does not hold, i.e., $m>N$. Then, there exists an $i\leq m$ such that $\pi(A_{i})$ does not cover the whole of $S_{N}$, for simplicity we assume $i=1$. Since $\pi(A_{1})\subsetneq S_{N}$ is open, there exists $p\in S_{N}\cap\partial\pi(A_{1})$. Writing $\pi^{-1}(p)\cap U=\{p_{1},\ldots,p_{N}\}$, there is an $\alpha\in\{1,\ldots,N\}$ such that $p_{\alpha}$ belongs to $\overline{A_{1}}$. But $p_{\alpha}\not\in A_{1}$ since $\pi(p_{\alpha})\in\partial\pi(A_{1})$, then $p_{\alpha}\in A_{\beta},\beta\neq 1$. This impiles that $\overline{A_{1}}\cap A_{\beta}\neq\emptyset$, so $A_{1}$ and $A_{\beta}$ are the same connected component, which is a contradiction.

It follows from (b) that for each $i\in\{1,\ldots,N\}$ the restriction $\pi|_{A_{i}}:A_{i}\to S_{N}$ is a bijection. In other words, $A_{i}={\rm\Gamma}_{\xi_{i}}$ with $\xi_{i}:S_{N}\to\mathbb{R}^{n}$, $\xi_{i}(y)=\pi^{-1}(y)\cap A_{i}$. Note that ${\rm tg}A_{i}={\rm tg}X|_{A_{i}}$ which is a continuous trivial bundle, its fibers are, moreover, not orthogonal to $\mathbb{R}^{k}$ by the construction. This shows that the function $\xi_{i}$ satisfies conditions of Lemma 3.2, hence ${\rm\Gamma}_{\xi_{i}}$ is a $C^{1}$ manifold. This ends the proof of (c).

Let $z\in S_{N}$ such that $\|z\|<r/4$. Let $z^{\prime}$ be a point realizing the distance from $z$ to the boundary $\partial S_{N}$ of $S_{N}$. Since $0\in\partial S_{N}$, $s:=\|z^{\prime}-z\|\leq r/4$, and then $\|z^{\prime}\|\leq\|z^{\prime}-z\|+\|z\|\leq r/4+r/4=r/2$, hence $z^{\prime}\in\pi(U)=\mathbf{B}^{k}(0,r)$. Since $\mathbf{B}^{k}(z,s)\subset S_{N}$, $\pi^{-1}(\mathbf{B}(z,s))\cap U$ has exactly $N$ connected components, denoted by $\{B_{1},\ldots,B_{N}\}$. Remark that $z^{\prime}\not\in S_{N}$ (i.e, ${\rm card}(\pi^{-1}(z^{\prime})\cap U)<N$) but $z^{\prime}\in\pi(\overline{B_{i}}),\forall i\in\{1,\ldots,N\}$, so there are $i,j$, $i\neq j$ such that $\overline{B_{i}}\cap\overline{B_{j}}\cap\pi^{-1}(z^{\prime})\neq\emptyset$. Take $w\in\overline{B_{i}}\cap\overline{B_{j}}\cap\pi^{-1}(z^{\prime})$. Take $C\subset\mathbf{B}^{k}(0,r)$ a small closed ball outside $\overline{\mathbf{B}}^{k}(z,s)$ and tangent to $\overline{\mathbf{B}}^{k}(z,s)$ at $z^{\prime}$. Denote by $D$ the connected component of $\pi^{-1}(C)\cap U$ which contains $w$.

Since ${\rm tg}_{w}U$ is a $k$-dimensional linear subspace of $\mathbb{R}^{n}$ and $\pi|_{{\rm tg}_{w}U}:{\rm tg}_{w}U\to\mathbb{R}^{k}$ is a linear bijective map, $D,B_{i},B_{j}$ are disjoint definable sets of dimension $k$ in $U$ and ${\rm tg}_{w}D$, ${\rm tg}_{w}B_{i}$, ${\rm tg}_{w}B_{j}$ are half $k$-planes. By Lemma 4.6,

[TABLE]

This contradicts the hypothesis that $\theta(X,x)<\frac{3}{2}$ for every $x\in X$. ∎

Acknowledgements

This research has been supported by ANR project STAAVF. The third author has been also received the funding from NCN grant 2014/13/B/ST1/00543.