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Assume $g,f\colon A\subset\mathbb{R}^M\rightarrow\mathbb{S}^2$ are two bijective functions defined on the set $A$. Now assume a constraint $C$: $\forall x,y\in A, \exists R\in SO(3)\colon Rf(x)=f(y)\iff Rg(x)=g(y)$. Does $C$ imply $f=\pm g$?

[Update]: The answer is "yes" when $C$ is modified to $\forall x,y\in A, \forall R\in SO(3)\colon Rf(x)=f(y)\iff Rg(x)=g(y)$.

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  • $\begingroup$ I recommend you clarify your formulas a bit. Ignoring the second paragraph, the answer is "no". Constraint $C$ holds for any functions $f, g$: for all $x, y$ we can pick $R$ so that neither equation holds, thus the equivalence holds. And when $A$ is nonempty, it is of course possible to find $f, g : A \to \mathbb{S}^2$ such that $f \neq \pm g$. I also don't get the second paragraph, what is the product on $\mathbb{S}^2$? $\endgroup$
    – Ville Salo
    Commented May 31, 2020 at 2:44
  • $\begingroup$ My original notation might be wrong. What I meant was that if for any $R\in SO(3)$ for which $Rf(x)=f(y)$ holds $Rg(x)=g(y)$ also holds and vice versa. I think I have to change $\exists R$ to $\forall R$. Right? I removed the second paragraph for the sake of clarity. $\endgroup$
    – solus0684
    Commented May 31, 2020 at 17:46
  • $\begingroup$ This is what I came up with if $C$ is modified to $\forall x,y\in A, \forall R\in SO(3): Rf(x)=f(y)\iff Rg(x)=g(y)$. Let $\theta(R)$ denote the rotation angle corresponding to rotation matrix $R$. We can show that $f(x)\cdot f(y)=\max\limits_{R\in SO(3): Rf(x)=f(y)} \cos{\theta(R)}$. This implies that if $C$ holds $\forall x,y\in A f(x)f(y)=g(x)g(y)$ and consequentially we can show that $\exists R'\in RO(3): g=R'f$. By plugging this into C this implies that $\forall x,y\in A, R\in SO(3): Rf(x)=f(y) \iff RR'f(x)=R'f(y)$ which implies $\forall R\in SO(3): RR'=R'R$ equivalent to $R'=\pm I$. $\endgroup$
    – solus0684
    Commented Jun 1, 2020 at 2:13

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You are correct, if you change $C$ so $R$ is $\forall$-quantified, then $C$ implies $f = \pm g$.

Pairs $\{v, -v\}$ have distinct stabilizers under $SO(3)$ so setting $x = y$ we see $f = \pm g$ pointwise. But then if $f(x) = g(x)$ and $f(y) = -g(y)$, picking any $R$ such that $Rf(x) = f(y)$, if we were to have $Rg(x) = g(y)$ then we would have $$ Rf(x) = Rg(x) = g(y) = -f(y) = -Rf(x), $$ a contradiction. Of course neither the choice of domain nor the fact the dimension of the sphere is $2$ matters.

(Your argument is probably correct as well, but I had already written this when I realized $\cdot$ means dot product.)

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  • $\begingroup$ I suppose the only thing used about the action of $SO(3)$ on $\mathbb{S}^2$ is that the group acts transitively, and whenever $x$ and $y$ have the same stabilizer, we have $gx = y$ for some $g \in Z(G)$. $\endgroup$
    – Ville Salo
    Commented Jun 1, 2020 at 9:55

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