$SO(m+1)$-equivariant maps from $S^m$ to $S^m$

Let $$G=SO(m+1)$$ , $$m \geq 2$$, act in the standard way on $$S^m$$.

Let $$F:S^m \to S^m$$ be a $$G$$-equivariant map, i.e., $$g F(g^{-1}x) =F(x)$$ for all $$x \in S^m$$ and $$g \in G$$.

Question 1: Is F the identity map?

If the answer is negative: Is $$F$$ an isometry?

• Welcome to MathOverflow. You can (and you should) use LeTeX notation when writing questions and answers. – Piotr Hajlasz Mar 18 '19 at 17:54
• I'm pretty sure that for $m$ odd, there's an obvious counterexample... – user44191 Mar 18 '19 at 18:05

Theorem. $$F:\mathbb{S}^m\to \mathbb{S}^m$$, $$m\geq 2$$, is $$SO(m+1)$$ equivariant if and only if $$F=\operatorname{Id}$$ or $$F=-\operatorname{Id}$$.

Let me write a very detailed proof that only requires a basic knowledge of linear algebra.

Proof. It is easy to see that both $$F=\operatorname{Id}$$ and $$F=-\operatorname{Id}$$ are $$SO(m+1)$$ equivariant so it remains to prove that if $$F$$ is equivariant, then $$F=\operatorname{Id}$$ or $$F=-\operatorname{Id}$$.

Let $$e_1,e_2,\ldots, e_{m+1}$$ be the standard orthogonal basis of $$\mathbb{R}^{m+1}$$. If $$[\rho_{jk}]$$ is the matrix representation of $$\rho\in SO(m+1)$$, then the condition $$F(\rho (x))=\rho (F(x))$$ reads as $$(*)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ F_j(\rho(x))=\sum_{k=1}^{m+1}\rho_{jk}F_k(x), \quad j=1,2,\ldots,n,$$ where $$F(x)=(F_1(x),\ldots,F_n(x))$$.

Let $$F_1(e_1)=c$$. Consider all $$\rho\in SO(m+1)$$ such that $$\rho(e_1)=e_1$$. This condition means that the first column of the matrix $$[\rho_{jk}]$$ equals $$e_1$$, i.e. $$\rho_{11}=1$$, $$\rho_{j1}=0$$, for $$j>1$$. Since columns are orthogonal, for $$k>1$$ we have $$0=\sum_{j=1}^{m+1}\rho_{j1}\rho_{jk}=\rho_{1k}\, .$$ Thus $$\rho = \left[ \begin{array}{cccc} 1 & 0 & \ldots & 0 \\ 0 & \rho_{22} & \ldots & \rho_{2,m+1} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & \rho_{m+1,2} & \ldots & \rho_{m+1,m+1} \end{array} \right]\, ,$$ where $$[\rho_{jk}]_{j,k=2}^{m+1}$$ is the matrix of an arbitrary transformation in $$SO(m)$$ (rotation in the $$m$$-dimensional subspace orthogonal to $$e_1$$).

For $$x=e_1=\rho(e_1)=\rho(x)$$ and $$j\geq 2$$ identity ($$*$$) yields $$F_j(e_1)=\sum_{k=1}^{m+1} \rho_{jk} F_k(e_1) = \sum_{k=2}^{m+1} \rho_{jk}F_k(e_1)\, ,$$ and hence $$\left[ \begin{array}{c} F_2(e_1) \\ \vdots \\ F_{m+1}(e_1) \end{array} \right] = \left[ \begin{array}{ccc} \rho_{22} & \ldots & \rho_{2,m+1} \\ \vdots & \ddots & \vdots \\ \rho_{m+1,2} & \ldots & \rho_{m+1,m+1} \end{array} \right]\, \left[ \begin{array}{c} F_2(e_1) \\ \vdots \\ F_{m+1}(e_1) \end{array} \right]\, .$$ That means the vector $$[F_2(e_1),\ldots,F_{m+1}(e_1)]^T$$ is fixed under any transformation $$SO(m)$$ of $$\mathbb{R}^{m}$$, so it must be a zero vector, i.e. $$F_2(e_1)=\ldots=F_{m+1}(e_1)=0\,$$ so $$F(e_1)=(c,0,\ldots,0), \quad c=\pm 1.$$ Now formula ($$*$$) for any $$\rho\in SO(m+1)$$ and $$x=e_1$$, takes the form $$F_j(\rho(e_1))=\rho_{j1}F_1(e_1)=\pm\rho_{j1}\, .$$ Let $$x\in \mathbb{S}^m$$ and let $$\rho\in SO(m+1)$$ be such that $$\rho(e_1)=x$$. Then $$\rho_{j1}=x_j$$, $$j=1,2,\ldots,n$$ and hence $$F_j(x)=\pm\rho_{j1}=\pm x_j, \quad F(x)=\pm x.$$ Therefore $$F(x)=x$$ or $$F(x)=-x$$.

$$G$$ acts transitively on $$S^m$$, so if you know the image of the north pole $$N = (0,\dotsc,0,1)$$, there is for every other point $$p\in S^m$$ an element $$g\in S^m$$ such that $$p = g\cdot (0,\dotsc,0,1)$$. Thus you will have $$F(p) = F\bigl(g\cdot (0,\dotsc,0,1)\bigr) = g\cdot F(0,\dotsc,0,1)$$, i.e., knowing the image under $$F$$ of one single point already determines the map $$F$$.

Now for the question whether $$F$$ is the identity. $$N$$ is fixed by the subgroup $$H = SO(m)$$, thus it follows that $$F(N)$$ also needs to be fixed by the same subgroup, but the only elements in $$S^{m+1}$$ fixed by $$H$$ are the north and the south pole.

Now if you take into account that $$g\cdot (-p) = - g\cdot p$$, it follows should follow that $$F$$ is either the identity or the antipodal map.

• Given p, the g such that p=gN is unique? – Andrea Ratto Mar 18 '19 at 18:57
• No, certainly not: If $H$ is the stabilizer of $N$, and if $g$ is such that $p = gN$, then any element in $h\in gH$ will also satisfy $p = hN$. – Klaus Niederkrüger Mar 18 '19 at 23:32

Proposition. Let $$G$$ be a group (acting on itself on the left), $$X$$ a $$G$$-set, and $$f:G\to X$$ a $$G$$-equivariant map. Then there exists a unique $$x\in X$$ such that $$f(g)=gx$$.

Proof: immediate.

Corollary: Let $$G$$ be a group, $$H$$ a subgroup, and $$f$$ a $$G$$-equivariant map $$G/H\to G/H$$. Then $$f$$ is a permutation of $$G/H$$, and has the form $$f(gH)=gqH$$, where $$q\in N_G(H)$$, and $$q$$ is well-determined as element of $$N_G(H)/H$$.

Proof: First compose as a map $$G\to G/H$$: then we deduce from the proposition that $$f(gH)=gqH$$ for some $$q\in G$$ and all $$g\in G$$; we see that $$q$$ is determined modulo right multiplication by $$H$$. Since $$f(gH)=f(ghH)$$ for all $$h\in H$$, we deduce $$gqH=ghqH$$ for all $$h\in H$$ and $$g\in G$$, that is, $$q^{-1}hq\in H$$ for all $$h\in H$$. This precisely means $$q\in N_G(H)$$.

Corollary: in the setting of the question, we only have $$\pm$$ identity.

Proof: write $$S^m=\mathrm{SO}(m+1)/\mathrm{SO}(m)$$. By the corollary, we need to determine the normalizer of $$\mathrm{SO}(m)$$. Since $$m\ge 2$$, the latter preserves a unique line, hence this line is preserved by this normalizer, which equals $$\mathrm{S}(\mathrm{O}(m)\times\mathrm{O}(1))$$ and contains $$\mathrm{SO}(m)$$. This corresponds to the two desired equivariant maps.

Such a reasoning extends to various other homogeneous spaces.