# Is a bijective, norm-preserving and rotation-invariant mapping linear?

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.

• Why work with spheres in the first place? Wouldn't it be better to ask about functions from $\mathbb{R}^N$ to $\mathbb{R}^N$ ? Aug 26, 2020 at 16:04
• 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. Aug 26, 2020 at 16:09
• 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. $\qquad$ Aug 26, 2020 at 16:48
• Thanks @MichaelHardy ! Aug 26, 2020 at 18:39

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.

• Interesting perspective, thanks @s.harp ! Quick question: how do you get $(R_z h(Rx), R_z h(Ry)) = (h(x), h(y))$? Aug 27, 2020 at 14:30
• Make use of $R_z$ being an orthogonal transformation: $$(R_z h(Rx), R_z h (Ry) ) = (h(Rx), R_z^T R_z h(Ry) ) = (h(Rx), h(Ry) ) = (h(x), h(y) )$$ Aug 27, 2020 at 14:42
• Of course... sorry, I was blind. Aug 27, 2020 at 15:02

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

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 $$x\in B_{v_2}$$.

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.