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Jun 2, 2020 at 9:21 history edited solus0684 CC BY-SA 4.0
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Jun 2, 2020 at 7:33 history edited solus0684 CC BY-SA 4.0
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Jun 1, 2020 at 18:05 vote accept solus0684
Jun 1, 2020 at 9:52 answer added Ville Salo timeline score: 0
Jun 1, 2020 at 2:13 comment added solus0684 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 $R'=\pm I$.
May 31, 2020 at 17:46 comment added solus0684 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.
May 31, 2020 at 17:41 history edited solus0684 CC BY-SA 4.0
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May 31, 2020 at 2:44 comment added Ville Salo 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 $\mathbb{S}^2$?
May 31, 2020 at 1:05 review First posts
May 31, 2020 at 5:44
May 31, 2020 at 1:02 history asked solus0684 CC BY-SA 4.0