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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?

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

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$G$ 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.

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  • $\begingroup$ Given p, the g such that p=gN is unique? $\endgroup$ – Andrea Ratto Mar 18 '19 at 18:57
  • $\begingroup$ 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 $p = hN$. $\endgroup$ – Klaus Niederkrüger Mar 18 '19 at 23:32
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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.

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