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Assume g,f:ARMS2 are two bijective functions defined on the set A. Now assume a constraint C: x,yA,RSO(3):Rf(x)=f(y)Rg(x)=g(y). Does C imply f=±g?

[Update]: The answer is "yes" when C is modified to x,yA,RSO(3):Rf(x)=f(y)Rg(x)=g(y).

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  • 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:AS2 such that f±g. I also don't get the second paragraph, what is the product on S2?
    – Ville Salo
    Commented May 31, 2020 at 2:44
  • My original notation might be wrong. What I meant was that if for any RSO(3) for which Rf(x)=f(y) holds Rg(x)=g(y) also holds and vice versa. I think I have to change R to R. Right? I removed the second paragraph for the sake of clarity.
    – solus0684
    Commented May 31, 2020 at 17:46
  • This is what I came up with if C is modified to x,yA,RSO(3):Rf(x)=f(y)Rg(x)=g(y). Let θ(R) denote the rotation angle corresponding to rotation matrix R. We can show that f(x)f(y)=max. 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

1 Answer 1

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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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