# The image of every strictly non-expanding map between spheres is contained in an open hemisphere?

$\renewcommand{\S}{\mathbb{S}^n}$ $\renewcommand{\R}{\mathbb{R}}$ $\newcommand{\conv}{\operatorname{Conv}}$ $\newcommand{\Int}{\operatorname{Int}}$

This is a follow-up of this question.

Let $f:A \subseteq \mathbb{S}^n \to \mathbb{S}^n$ be a map which strictly decreases distances, i.e $$d(f(x),f(y)) <d(x,y) \, \text{ for every } x,y \in A.$$

($d$ can be either the intrinsic distance or the Euclidean one, it does not matter)

Is it true that $f(A)$ is contained in the interior of some hemisphere? Is it contained in a disk with (intrinsic) radius $\frac{\pi}{2}-\epsilon$ for some $\epsilon$?

Edit:

It turns out that $f(A)$ is not necessarily contained in a disk with radius $\frac{\pi}{2}-\epsilon$.

George Lowther gave the following nice example:

Fix a meridian line joining north and south poles on the sphere $\mathbb{S}^2$. Let $f:\mathbb{S}^2 \to \mathbb{S}^2$ map each point to the unique point on the meridian with the same latitude. Then $f$ strictly reduces the distance between any two points which do not have the same longitude*. Let $A$ be any set containing points arbitrarily close (but not equal) to both the north and south poles, all of which have distinct longitudes. Restricting $f$ to $A$ gives the counterexample:

Indeed, $f(A)$ contains points on a great semi-circle (the fixed meridian) which are arbitrarily close to the north and south pole. If $f(A)$ was contained in a disk of radius $\frac{\pi}{2}-\epsilon$, then by triangle inequality the distance of every two points in it would be less than or equal to $\pi-2\epsilon$, a contradiction.

Note however that $f(A)$ does lie in the interior of a hemisphere -look at the hemisphere centered in the middle point of the meridian, and recall the north and south poles were left out from $A$, and hence also from $f(A)$.

Finally, if we try to include one of the poles in $A$, in an attempt to produce a counter-example for the weaker conjecture, we will destroy the strict distance decreasing property of $f$. (Distance of a point to the pole depends only on the latitude).

*The fact that $f$ is strictly decreasing distance between points which do not have the same longitude follows immediately from the formula for the spherical distance:

$\Delta\sigma=\arccos\bigl(\sin\phi_1\cdot\sin\phi_2+\cos\phi_1\cdot\cos\phi_2\cdot\cos(\Delta\lambda)\bigr)$

Partial Results:

An argument here (pg 7, lemma 2.8), shows that the convex hull of $f(A)$ does not contain the origin. Hence, $\, f(A)$ is contained in a closed hemisphere. (This follows from the hyperplane separation theorem).

Moreover, if $A \subseteq \mathbb{S}^n$ is compact then the answer is positive:

Since $f(A)$ is compact, $\operatorname{conv}(f(A))$ is compact.

Any closed convex set in $\mathbb{R}^n$ which does not contain the origin $\bar O$, is contained in a closed half-space which does not contain $\bar O$ (by the strong hyperplane separation theorem).

In particular, $\operatorname{conv}(f(A))$ is contained in the interior of a half-space whose boundary does contain $\bar O$, so $f(A)$ is contained in the interior of a hemisphere.

Thus, if $A$ is not closed, a naive approach would be to try to extend it to the closure of $A$. However, we might lose the strict distance-decreasing nature of $f$ in the extension, which is necessary for the argument showing $\operatorname{conv}(f(A))$ does not contain the origin.

• I'm not sure about the main statement, but I can give a counterexample to the second, stronger statement: Fix a meridian line joining north and south poles on the sphere. Let $f\colon{\mathbb S}^2\to{\mathbb S}^2$ map each point to the unique point on the meridian with the same latitude. Then $f$ strictly reduces the distance between any two points which do not have the same longitude. Let $A$ be any set containing points arbitrarily close (but not equal) to both the north and south poles, all of which have distinct longitudes. Restricting $f$ to $A$ gives the counterexample. Apr 1, 2017 at 13:13
• Great example! Very beautiful, really. Apr 1, 2017 at 14:51

Here's a proof of the main statement. I will use induction on the dimension $n$ so, suppose that the result holds in $\mathbb S^m$ all $0\le m < n$.

Set $B=f(A)$ and, by uniform continuity, extend $f$ continuously to the closure $\bar A$ of $A$. Then, $f$ is 1-Lipschitz and $\bar B=f(\bar A)$. Using ${\rm c}(\bar B)$ to denote the convex cone spanned by $\bar B$ (i.e., the set of positive linear combinations of $\bar B$), set $S=\bar B\cap(-c(\bar B))$. We can suppose that $c(\bar B)$ contains the origin otherwise, by the hyperplane separation theorem for compact sets, there is a $u\in\mathbb R^{n+1}$ such that $\bar B$ is contained in the hemisphere $\{x\in\mathbb S^n\colon\langle u,x\rangle > 0\}$. So, we suppose that $\sum_{i=1}^m\lambda_ix_i=0$ for some $\lambda_i > 0$ and $x_i\in\bar B$. Then, $x_1=-\sum_{i=2}^m\lambda_1^{-1}\lambda_ix_i\in S$ and $S$ is nonempty.

Supposing that $x\in S$, then $x=-\sum_{i=1}^m\lambda_i x_i$ for some $\lambda_i > 0$ and $x_i\in\bar B$. Setting $\lambda_0=1$ and $x_0=x$ then, $\sum_{i=0}^m\lambda_ix_i=0$. Choosing $y_i\in \bar A$ with $f(y_i)=x_i$, \begin{align} &\sum_{i=0}^m\lambda_i^2+2\sum_{0\le i < j\le m}\lambda_i\lambda_j\langle x_i,x_j\rangle=\lVert\sum_{i=0}^m\lambda_ix_i\rVert^2=0,\\ &\sum_{i=0}^m\lambda_i^2+2\sum_{0\le i < j\le m}\lambda_i\lambda_j\langle y_i,y_j\rangle=\lVert\sum_{i=0}^m\lambda_iy_i\rVert^2\ge0. \end{align} The 1-Lipschitz property says that $\langle x_i,x_j\rangle\ge \langle y_i,y_j\rangle$ so, by the inequality above, we have equality. So, $\sum_{i=0}^m\lambda_iy_i=0$. Next, for any $y^\prime\in\bar A$ set $x^\prime=f(y^\prime)$. \begin{align} &\sum_{i=0}^m\lambda_i\langle x^\prime,x_i\rangle=\langle x^\prime,\sum_{i=0}^m\lambda_ix_i\rangle=0,\\ &\sum_{i=0}^m\lambda_i\langle y^\prime,y_i\rangle=\langle y^\prime,\sum_{i=0}^m\lambda_iy_i\rangle=0. \end{align} Using the 1-Lipschitz property again, $\langle x^\prime,x_i\rangle\ge\langle y^\prime,y_i\rangle$ so, by the equalities above, this is an equality.

Hence, $f$ preserves the distance between any element of $f^{-1}(S)$ and any element of $\bar A$. We see that $f$ restricted to $f^{-1}(S)$ preserves distances. As any isometry on a subset of $\mathbb S^n$ extends to an isometry on $\mathbb S^n$, we can compose $f$ with an isometry to ensure that $f^{-1}(S)=S$ and $f$ fixes all points of $S$.

Next, let $V$ be the subspace of $\mathbb R^{n+1}$ spanned by $S$. Choosing any $x\in \bar A$ any $y\in V$ we have $y=\sum_{i=1}^m\lambda_iy_i$ for $y_i\in S$. Then, $$\langle f(x),y\rangle=\sum_{i=1}^m\lambda_i\langle f(x),y_i\rangle=\sum_{i=1}^m\lambda_i\langle x,y_i\rangle=\langle x,y\rangle.$$ So, $f$ preserves the distance of any point of $A$ from any point of $V$. In particular, it preserves the distance between any point in $A\cap V$ and any point in $A$, which contradicts the strict non-expanding property unless $A\cap V=\emptyset$. Also, letting $y_1$ and $y_2$ be the orthogonal projections of $x$ and $f(x)$ into $V$, $$\lVert y_1-x\rVert=\lVert y_1-f(x)\rVert\ge\lVert y_2-f(x)\rVert=\lVert y_2-x\rVert.$$ As $y_1$ is the unique element of $V$ minimising $\lVert y_1-x\rVert$, it is equal to $y_2$, and $f$ preserves the orthogonal projections onto $V$.

Using $V^\perp$ to denote the orthogonal complement of $V$, define $\tilde A\subseteq \mathbb S^{n}\cap V^\perp$ to be the set of $a\in\mathbb S^{n}\cap V^\perp$ such that $\varphi(x,a)\equiv x+\sqrt{1-\lVert x\rVert^2}a\in A$ for some $x\in V$ (with $\Vert x\rVert < 1$). Supposing that $a,b\in\tilde A$ there exist $a^\prime,b^\prime\in \mathbb S^{n}\cap V^\perp$ such that \begin{align} &f\left(\varphi(x,a)\right)=\varphi(x,a^\prime),\\ &f\left(\varphi(y,b)\right)=\varphi(y,b^\prime). \end{align} for some $x,y\in V$. Looking at the inner products, $$\langle\varphi(x,a),\varphi(y,b)\rangle=\langle x,y\rangle+\langle a,b\rangle\sqrt{(1-\lVert x\rVert^2)(1-\lVert y\rVert^2)}$$ The fact that $f$ is strictly non expanding on $A$ gives $\langle a^\prime,b^\prime\rangle\ge \langle a,b\rangle$ with a strict inequality if $a\not=b$. In particular, if $a=b$ then $\langle a^\prime,b^\prime\rangle=1$ so $a^\prime=b^\prime$. Therefore, we can define $g\colon\tilde A\to\mathbb S^{n}\cap V^\perp$ such that $a^\prime=g(a)$. That is, $$f(\varphi(x,a))=\varphi(x,g(a)).$$ Now, $S$ is nonempty so $\mathbb S^{n}\cap V^\perp$ is isometric to $\mathbb S^m$ for some $m < n$. As $g$ is strictly non-expanding, the induction hypothesis states that there is a $u\in V^\perp$ such that $\langle u,a\rangle > 0$ for all $a\in g(\tilde A)$. Then $f(A)$ is contained in the open hemisphere $\{x\in\mathbb S^n\colon\langle u,x\rangle > 0\}$.

• I completely rewrote the answer, avoiding the incorrect theorem I previously stated. Barring inevitable typos, it should be good now! Apr 2, 2017 at 20:35
• @George Lowther : Thank you for your counterexample. And I have additional three questions : (1) If $A$ is finite set in $S^1$, then the statement holds ? (2) Is there a counter example on $S^3$ ? (3) On $S^2$, if $f(A)$ is not contained in a geodesic of lenth $\pi$, then the statement holds ? Thank you for your attention. Apr 3, 2017 at 2:13
• @GeorgeLowther Wow. Can you please explain why any isometry on a subset of $\mathbb S^n$ extends to an isometry on $\mathbb S^n$? (I only found a reference for the case where $\operatorname{Image}(f)$ does not lie in any closed hemisphere, which is not our case). Even assuming this is true, I do not see how do you deduce that by composing $f$ with an isometry we can ensure that $f^{-1}(S)=S$ and $f$ fixes all points of $S$. Can you elaborate on that step? Thanks. Apr 3, 2017 at 8:30
• You can extend to an isometry on all of $\mathbb R^{n+1}$ which preserves the origin, then restrict to the unit sphere. It must be a standard result, but I just found it here math.stackexchange.com/questions/1634271/…. If $g$ is an isometry agreeing with $f$ on $f^{-1}(S)$, then consider replacing $f$ by $f\circ g^{-1}$ and $A$ by $g(A)$. Apr 3, 2017 at 10:12