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Let $h: \mathbb{S}^N\to \mathbb{S}^N$, where $\mathbb{S}^N$ is the $N$-dimensional unit hypersphere in $\mathbb{R}^{N+1}$, be a mapping with the following properties:

  1. $h$ is bijective and smooth.
  2. $\left\|h(x)\right\| = 1$ (given that we operate on a hypersphere).
  3. $h(x)^\top h(y) = h(Rx)^\top h(Ry)$ for any rotation $R$, i.e. the inner product is invariant to rotations in the domain of $h$.

Is $h$ a linear or affine transformation (i.e. is $h(x) = Qx$ where $Q$ is a rotation)?


It took me a few days, but I think I finally found a proof using that if $h(x)^\top h(y) = x^\top y$ for all $x, y$ then $h$ is linear. This can be shown by realising that if $h(x)^\top h(y) \ne x^\top y$ for any $x, y$ then $h$ cannot be bijective, see the proof below.

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  • $\begingroup$ Why work with spheres in the first place? Wouldn't it be better to ask about functions from $\mathbb{R}^N$ to $\mathbb{R}^N$ ? $\endgroup$ – Jakobian Aug 26 '20 at 16:04
  • $\begingroup$ My original problem, from which this is derived, primarily concerns mappings between hyperspheres. But I would also be interested in the $\mathbb{R}^N$ case. $\endgroup$ – w382903 Aug 26 '20 at 16:09
  • $\begingroup$ Usually the symbol $\text{“}\mapsto\text{''}$ is used in expressions like $x\mapsto x^3y^2$ (which is a different function from $y\mapsto x^3y^2$, and that shows why this symbol is used), whereas $\text{“}\to\text{''}$ is used in things like $\text{“} h: \mathbb{S}^N\to \mathbb{S}^N\text{''}.$ I edited accordingly. $\qquad$ $\endgroup$ – Michael Hardy Aug 26 '20 at 16:48
  • $\begingroup$ Thanks @MichaelHardy ! $\endgroup$ – w382903 Aug 26 '20 at 18:39
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Here is an initial step: Every such map must be a conformal diffeomorphism.

To see it consider a point $z\in S^N$ and let $R_z$ be a rotation taking $h(z)$ to $z$. We look at $\tilde h:= R_z\circ h$ to note that $$(\tilde h(Rx) , \tilde h(Ry) ) =(R_z h(Rx), R_z h(Ry)) = (h(x), h(y)) = (R_z h(x), R_z h(y))$$ and see that this enjoys the same property as $h$ with the added benefit that $z$ is fixed point of $\tilde h$. Now let $x(t)$ and $y(s)$ be two differentiable paths with $x(0)=y(0)=z$. Then for an arbitrary rotation $R$ with $R(z)=z$ one has: $$( D_z\tilde h ( R x'(0)) , D_z\tilde h(R y'(0) ) )= \frac{d}{dt}\frac d{ds}(\tilde h(R x(t)), \tilde h(Ry(t) )\lvert_{s,t=0}= \frac{d}{dt}\frac d{ds}(\tilde h( x(t)), \tilde h(y(t) )\lvert_{s,t=0}\\ =( D_z\tilde h ( x'(0)) , D_z\tilde h( y'(0) ) ).$$

Since this then holds for all tangent vectors $x'(0), y'(0)$ one has that $R^T(D_z\tilde h)^T D_z\tilde h R = (D_z\tilde h)^T D_z \tilde h$ for any rotation preserving the point $z$. This implies that $(D_z\tilde h)^T D_z\tilde h$ is a scalar multiple of the identity on $T_zS^N$, ie that $D_z\tilde h = \lambda S$ for some $\lambda\in \Bbb R$ and a rotation $S$ of $T_zS^N$.

This means that $D_z\tilde h$ is a conformal mapping $T_zS^N\to T_zS^N$. Since $R_z$ is an isometry one has that $D_zh = D_z(R_z^{-1}\tilde h)$ is then also a conformal mapping. Since $z$ was arbitrary $h$ is a conformal map.

However it is not true that every conformal diffeomorphism of the sphere satisfies the desired equation, it is easy to construct counter examples on $S^2$ via Möbius transformations. This leads me to believe that only the isometries, ie the rotations, have this property.

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  • $\begingroup$ Interesting perspective, thanks @s.harp ! Quick question: how do you get $(R_z h(Rx), R_z h(Ry)) = (h(x), h(y))$? $\endgroup$ – w382903 Aug 27 '20 at 14:30
  • $\begingroup$ Make use of $R_z$ being an orthogonal transformation: $$(R_z h(Rx), R_z h (Ry) ) = (h(Rx), R_z^T R_z h(Ry) ) = (h(Rx), h(Ry) ) = (h(x), h(y) )$$ $\endgroup$ – s.harp Aug 27 '20 at 14:42
  • $\begingroup$ Of course... sorry, I was blind. $\endgroup$ – w382903 Aug 27 '20 at 15:02
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Here is an attempt for a formal proof using Theorem 1 in [1] which basically states that $h$ is linear if $h(x)^\top h(y) = x^\top y$. We proof by contradiction that $h$ has this property. Let $\alpha(x,y) = \cos^{-1}(x^\top y)$ be the angular distance between $x$ and $y$. The proof has four main steps:

  1. If $\exists x, y\,\,\alpha(h(x), h(y)) > \alpha(x, y)$, then there exists a $z$ on the geodesic between $x$ and $y$ such that $\alpha(h(x), h(z)) > \alpha(x, z) < \epsilon$ for a given $\epsilon > 0$.

  2. If $\exists x, y\,\,\alpha(h(x), h(y)) > \alpha(x, y)$ then there exists an $h'$ fulfilling all desired properties (1-3) such that $\alpha(h'(x), h'(y)) < \alpha(x, y)$ (and vice versa).

  3. If $\exists x, y\,\,\alpha(h(x), h(y)) < \alpha(x, y)$ then $\alpha(h(x), h(z)) < \alpha(x, z)$ for all $z$ with $\alpha(x, y) = \alpha(x, z)$.

  4. If $\alpha(h(x), h(y)) < \alpha(x, y) < \epsilon$ for any $\epsilon > 0$ then $h$ cannot be bijective, contradicting the first assumption on $h$.

This concludes the proof. I now prove each part individually:

Proof of part 1

Assume there exists $x, y$ s.t. $\alpha(h(x), h(y)) > \alpha(x, y)$. Now consider a point $z$ and a rotation $R$ such that $z = Rx$ and $y = Rz$ (i.e. $z$ is a midpoint of $x$ and $y$ on the hypersphere). Then

\begin{align} 2\alpha(x, z) &= \alpha(x, y), \\ &< \alpha(h(x), h(y)), \\ &\leq \alpha(h(x), h(z)) + \alpha(h(z), h(y)) , \\ &= \alpha(h(x), h(z)) + \alpha(h(Rx), h(Rz)), \\ &= 2\alpha(h(x), h(z)) \end{align}

where I used that the angular distance is a metric (and thus I can use the triangle inequality). Hence, $\alpha(x, z) < \alpha(h(x), h(z))$. We now just repeat this argument several times (each time halving the angle between $x$ and $z$ until $\alpha(x, z) < \epsilon$).

Proof of part 2

For any given $h$ fulfilling assumptions (1-3), its inverse $h^{-1}$ also fulfills (1-3). If there exists $x, y$ such that $\alpha(h(x), h(y)) > \alpha(x, y)$, then $\alpha(h^{-1}(x'), h^{-1}(y')) < \alpha(x', y')$ for $x' = h(x)$ and $y' = h(y)$.

Proof of part 3

For any $x^\top y = x^\top z$ we can find a rotation $R$ such that $x = Rx$ (i.e. $x$ is a fixed point of $R$) and $y = Rz$. Then if $\alpha(h(x), h(y)) < \alpha(x, y)$ we find that

$$\alpha(h(x), h(z)) = \alpha(h(Rx), h(Rz)) = \alpha(h(x), h(y)) < \alpha(x, y).$$

Proof of part 4

Given part 1., 2. and 3. we know that if there exists any $h$ for which there exists $x, y$ such that $h(x)^\top h(y) \ne x^\top y$, there must be a $\tilde h$ s.t. $\alpha(\tilde h(x), \tilde h(y)) < \alpha(x, y) < \epsilon$ for all $z$ with $x^\top y = x^\top z$. Choose any $x', y'$ such that $\tilde h(x') = -\tilde h(y')$ (which must exist because $\tilde h$ is bijective). Let further $z'$ be on the geodesic between $x'$ and $y'$ such that $\alpha(\tilde h(x'), \tilde h(z')) < \alpha(x', z') < \epsilon$ and let $N$ be such that $N\alpha(x', z') = \alpha(x', y')$ (this can be made precise in the limit $\epsilon\to 0$ and $N\to\infty$). Then,

\begin{align} \pi &= \alpha(\tilde h(x'), -\tilde h(x')), \\ &= \alpha(\tilde h(x'), \tilde h(y')), \\ &\leq N \alpha(\tilde h(x'), \tilde h(z')), \\ &< N\alpha(x', z'), \\ &= \alpha(x', y'), \\ &\leq \pi. \end{align}

This contradiction completes the proof.

[1] Linear mappings approximately preserving orthogonality, https://reader.elsevier.com/reader/sd/pii/S0022247X04007474?token=8EDC87750F4B5D379A98264CBF42E0CC2F4B73DFF6217BE5CB03B27A4ABD495E189DF508816E4E3DCC3E9ACEC5DA5B73

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The given conditions can be weakened. We can prove the following by mathematical induction.

CLAIM. Let $n\ge1$ and $h:S^n\to S^n\,(\subset\mathbb R^{n+1})$ be a continuous injective function such that $$ \langle h(x),h(y)\rangle=\langle h(Rx),h(Ry)\rangle\tag{1} $$ for all $x,y\in S^n$ and for all $R\in SO(n+1,\mathbb R)$. Then $h=q|_{S^n}$ for some linear isometry $q:\mathbb R^{n+1}\to\mathbb R^{n+1}$.

THE BASE CASE $n=1$. Let $e_1=(1,0)^T,\ R(\phi)$ denotes the rotation matrix for an angle $\phi$, and $\angle(x,y)$ denotes the angle between two unit vectors $x$ and $y$, i.e. $\angle(x,y)=\arccos(\langle x,y\rangle)\in[0,\pi]$. By composing $h$ with an appropriate linear isometry on $\mathbb R^2$, we may assume that $h(e_1)=e_1$. By the continuity of $h$, there exists some $\delta\in(0,\frac{\pi}{2})$ such that $\angle\left(h(e_1),\,h(R(\phi)e_1)\right)<\frac{\pi}{2}$ whenever $|\phi|<\delta$.

Let $\theta\in(0,\delta)$ be fixed and let $x_k=R(k\theta)e_1$ for every $k\ge0$. Since $|\theta|<\delta<\frac{\pi}{2}$, the points $x_0,x_1$ and $x_2$ are distinct. By condition $(1)$, $\angle(h(x_0),h(x_1))=\angle(h(x_1),h(x_2))$. The injectiveness of $h$ and our choice of $\delta$ thus imply that $h(x_1)=R(\alpha)h(x_0)$ and $h(x_2)=R(\alpha)h(x_1)$ for some $0<|\alpha|<\frac{\pi}{2}$. If we repeat the same argument for $x_k,x_{k+1},x_{k+2}$ for each $k\ge1$, we get $$ h\left(R(k\theta)e_1\right)=R(k\alpha)e_1\quad \forall k\ge0.\tag{2} $$ Likewise, if we put $x_{1/2}=h\left(R(\frac{\theta}{2})x_0\right)$, the inner product condition $(1)$ will give $\langle x_0,x_{1/2}\rangle=\langle x_{1/2},x_1\rangle$. Thus $\angle(x_0,x_{1/2})$ is equal to either $\frac12\angle(x_0,x_1)$ or $\pi+\frac12\angle(x_0,x_1)$, but the latter is impossible because $\frac{\theta}{2}<\delta$. It follows that $x_{1/2}=R(\frac{\alpha}{2})e_1$. So, if we repeat the derivation of $(2)$ but with $\theta$ replaced by $\frac{\theta}{2}$, we get $h\left(R(\frac{k\theta}{2})e_1\right)=R(\frac{k\alpha}{2})e_1$ for each $k\ge0$. Continue in this manner, we obtain $h\left(R(\frac{k\theta}{2^m})e_1\right)=R(\frac{k\alpha}{2^m})e_1$ for all integers $k,m\ge0$. It follows from the continuity of $h$ that $h(R(\phi)e_1)=R(c\phi)e_1$ for every $\phi\ge0$, with $c=\frac{\alpha}{\theta}$. As $h$ is injective, we cannot have $|c|>1$, or else $h\left(R(\frac{2\pi}{c})e_1\right)=R(2\pi)e_1=e_1=h(e_1)$. We cannot have $|c|<1$ either, otherwise $e_1=h(e_1)=h\left(R(2\pi)e_1\right)=R(2\pi c)e_1$. Thus $c=\pm1$ and $h$ is either the identity map or a reflection.

THE INDUCTIVE STEP. We will split the proof into two parts.

Part I. For any $u\in S^n$, denote the "great belt" (an analogy of great circle) normal to $u$ by $B_u:=u^\perp\cap S^n=\{x\in S^n:\langle x,u\rangle=0\}$. The purpose of this part of the proof is to show that for every unit vector $u$, we have $$ \begin{align} &h(B_u)=B_{h(u)} \text{ and}\tag{3}\\ &h|_{B_u}=q|_{B_u} \text{ for some linear isometry } q:u^\perp\to h(u)^\perp.\tag{4} \end{align} $$ Let $v=h(u)$. Then $\langle h(x),v\rangle$ is a constant on $B_u$, for, if $x,x'\in B_u$, there is some $R\in SO(n+1,\mathbb R)$ such that $Ru=u$ and $Rx=x'$. Hence the inner product condition $(1)$ implies that $\langle h(x'),v\rangle=\langle h(x),v\rangle$.

So, let $\langle h(x),v\rangle\equiv c$ on $B_u$ and let $s=\sqrt{1-c^2}$. Then $h(B_u)\subseteq cv+sB_v$. (When $n=2$, one can visualise $cv+sB_v$ as a circle with centre $cv$ and radius $s$ that is parallel to the great circle $B_v$.) Since $h$ is injective, $s\ne0$. Pick an linear isometry $q_1:\mathbb R^{n+1}\to\mathbb R^{n+1}$ such that $q_1(u)=v$ (hence $q_1(u^\perp)=v^\perp$ and $q_1(B_u)=B_v$) and define $g:B_u\to B_u$ by $$ g(x)=\frac{1}{s}q_1^{-1}\left(h(x)-cv\right).\tag{5} $$ Then $g$ is continuous and injective on $B_u$ and $\langle h(x),h(x')\rangle=c^2+s^2\langle g(x),g(x')\rangle$ for all $x,x'\in B_u$. Therefore $\langle g(x),g(x')\rangle=\langle g(Rx),g(Rx')\rangle$ for every linear isometry $R$ on $u^\perp$ with determinant $1$. Since $B_u$ is isomorphic to $S^{n-1}$, by induction assumption we have $$ g=q_2|_{B_u}\ \text{ for some linear isometry }\ q_2:u^\perp\to u^\perp.\tag{6} $$ In particular, $g$ is bijective on $B_u$ and hence $$ h(B_u)=cv+sB_v.\tag{7} $$ It follows that for any $y\in cv+sB_v$, there exists $x\in B_u$ such that $y=h(x)$. Pick any $R\in SO(n+1,\mathbb R)$ such that $Ru=-u$ and $Rx=x$. Then $ \langle h(-u),y\rangle =\langle h(Ru),h(Rx)\rangle =\langle h(u),h(x)\rangle =\langle v,y\rangle=c. $ That is, $$ \langle h(-u),y\rangle=c\quad\forall y\in cv+sB_v. $$ Since $\|h(-u)\|=1$ and $h(-u)\ne v$, this is possible only if $(c,s)=(0,1)$ (and $h(-u)=-v$). Hence $(3)$ follows from $(7)$. Also, definition $(5)$ now gives $h|_{B_u}=q_1\circ g$ and hence by $(6)$, $h|_{B_u}=q_1\circ q_2|_{B_u}=(q_1\circ q_2)|_{B_u}$. Thus $(4)$ is also satisfied.

Part II. Let $\{e_1,\ldots,e_{n+1}\}$ be the standard basis of $\mathbb R^{n+1}$. For brevity, abbreviate $B_{e_i}$ as $B_i$. By composing $h$ with an appropriate linear isometry, we may assume that $h(e_1)=e_1$. By $(3)$ and $(4)$, $h(B_1)=B_1$ and $h|_{B_1}$ agrees with an linear isometry. So, by composing $h$ with another linear isometry that leaves $e_1$ invariant, we may assume that $$ h(e_i)=e_i\ \text{ for each } i.\tag{8} $$ For each $i$, if we apply $(3)$ and $(4)$ to $u=e_i$, we see that $h(B_i)=B_i$ and that $h|_{B_i}=q_i|_{B_i}$ for some linear isometry $q_i$ defined on $e_i^\perp$. It follows from $(8)$ that $q_i$ is the identity map and hence so is $h|_{B_i}$.

Our proof is complete if we can show that $h$ is the identity map on the whole hypersphere $S^n$. For any $x=(x^1,x^2,\ldots,x^{n+1})^T\in S^n$, normalise $(0,x^2,\ldots,x^{n+1})^T$ to a unit vector $v_{n+1}$ if $|x^1|<1$, or define $v_{n+1}=e_{n+1}$ if $|x^1|=1$. Let $v_1=e_1$ and extend $\{v_1,v_{n+1}\}$ to an orthonormal basis $\{v_1,v_2,\ldots,v_{n+1}\}$ of $\mathbb R^{n+1}$. Then $x\in B_{v_2}$.

Since $v_1=e_1$, we have $h(v_1)=v_1$ by $(8)$. We also have $h(v_i)=v_i$ for each $i\ge2$ because $v_2,\ldots,v_{n+1}\in B_1$ and $h|_{B_1}$ is the identity map. Therefore, by applying $(3)$ to $u=v_2$, we have $h(B_{v_2})=B_{h(v_2)}=B_{v_2}$ and by $(4)$, $h$ agrees with an linear isometry $q:v_2^\perp\to v_2^\perp$ on $B_{v_2}$. Hence $q(v_i)=h(v_i)=v_i$ on a basis $\{v_1,v_3,v_4,\ldots,v_{n+1}\}$ of $v_2^\perp$, i.e. $q$ is the identity map. Therefore $h|_{B_{v_2}}$ is also the identity map. Hence $h(x)=x$, because $x\in B_{v_2}$.

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Given any orthogonal linear transformation $R$, let $R'=hRh^{-1}$. One then has for every $x$ and $y$ in $\mathbb{S}^N$ that $$ ⟨R'(x),R'(y)⟩ = ⟨hRh^{-1}(x),hRh^{-1}(y)⟩ = ⟨hh^{-1}(x),hh^{-1}(y)⟩ = ⟨x, y⟩, $$ so $R'$ preserves inner-producs and hence $R'$ must be an orthogonal transformation (Theorem 1 in reference (1) cited in @w382903's answer).

The map $$ \phi: R∈O(n) \mapsto hRh^{-1}\in O(n) $$ is therefore an automorphism of the orthogonal group $O(n)$. Since $\phi$ is continuous it must preserve the connected component of the identity and hence $\phi$ restricts to an automorphism of $SO(n)$.

Since all automorphisms of $SO(n)$ are given by conjugation by an orthogonal transformation (see this post), there exists $U$ in $O(n)$ such that $$ hRh^{-1} = URU^{-1}, $$ for all $R$ in $SO(n)$. This translates to $U^{-1} hR = RU^{-1} h$, which means that $$ k:= U^{-1} h $$ commutes with every $R$ in $SO(n)$.

(We can't yet conclude that $k$ lies in the center of any matrix group in sight because we don't yet know it is linear, but we are pretty close!)

Let us next prove that for every $x$ in $\mathbb{S}^N$, one has that $k(x)=\pm x$. Assuming by contradiction that this is not so, one may find a rotation $R$ such that $R(x)=x$, and $R(k(x))\neq k(x)$ (this requires that $N\geq 3$, which we assume from now on). Then $$ R(k(x)) = k(R(x)) = k(x), $$ a contradiction.

This proves the claim and since $k$ is continuous, the choice of sign "$\pm$" must be constant for every $x$, meaning that $k$ is either the identity $I$ or $-I$, and consequently $$ h = Uk = \pm U, $$ as desired.

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