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Let $f, g$ be two homomorphisms from $\mathrm{SO}(3)$ to $PGL(2,\mathbb{C})$. Does there exist $S \in \mathrm{PGL}(2,\mathbb{C})$ such that for all $X\in \mathrm{SO}(3)$ we have $g(X)=Sf(X)S^{-1}$?

Background: I have images which are stereographic projections of subparts of a sphere. According to Wikipedia the images transform into each other using Möbius transforms. I want to parametrize these transforms in terms of the rotations given by the actuator. If the above is true the parametrization would be quite simple: i can pick any homomorphism and then just need to parametrize $S$.

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    $\begingroup$ If you don't specify continuous homomorphism, then this is false, since the automorphism group of $PGL_2(\mathbb{C})$ contains $Gal(\mathbb{C}/\mathbb{Q})$. $\endgroup$
    – Ian Agol
    Commented Jun 20, 2020 at 22:03

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There are two continuous group morphisms $SO_3\to PSL_2\mathbb{C}$ up to conjugacy: the obvious one as rotations of the Riemann sphere, and the trivial one with image the identity element. The proof: We see from the Lie algebra of $SO_3$ (cross product of vectors in $\mathbb{R}^3$) that $SO_3$ has simple Lie algebra. Hence the image $G$ of any morphism is a 0-dimensional or a 3-dimensional compact and connected group. If 0-dimensional, it is the identity element, because it is connected. So suppose that $G$ is 3-dimensional.

By $QR$ decomposition, $PSL_2\mathbb{C}$ retracts to $SO_3$, so $SO_3$ is a maximal compact subgroup, and is connected. All maximal compact subgroups are conjugate. So any maximal compact subgroup containing $G$ is a conjugate of $SO_3$. By dimension count, and connectivity, $G$ is a maximal compact subgroup, so a conjugate of $SO_3$, so unique up to conjugacy. We thus reduce to the problem of proving that $SO_3$ automorphisms are inner, which is clear from its Dynkin diagram, as it is rank one, so has a Dynkin diagram with only one node, hence no noninner automorphisms.

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    $\begingroup$ Why does the image have to be maximal compact? The image of the trivial homomorphism is not maximally compact $\endgroup$
    – Sebastian Schulz
    Commented Jun 20, 2020 at 21:43
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    $\begingroup$ @SebastianSchulz, thanks for asking this! Certainly the argument meant to include the caveat 'non-trivial' on the morphism, but even then I can't see it. A non-trivial morphism from $\operatorname{SO}_3(\mathbb R)$ will have image isomorphic either to $\operatorname{SO}_3(\mathbb R)$ or its adjoint quotient, but is it obvious that the only way to embed either of those in $\operatorname{PSL}_2(\mathbb C)$ is as a maximal compact subgroup? $\endgroup$
    – LSpice
    Commented Jun 21, 2020 at 2:07
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    $\begingroup$ @LSpice for reasons of dimension. If $SO_3$ or its adjoint quotient is embedded, the image's dimension is the dimension of a maximal compact. The image is compact, so it must be maximal. $\endgroup$
    – Joshua Mundinger
    Commented Jun 21, 2020 at 3:20
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    $\begingroup$ @JoshuaMundinger, but, still, connected components...? Some details... $\endgroup$
    – paul garrett
    Commented Jun 22, 2020 at 20:42
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    $\begingroup$ @LSpice what you mean by "its adjoint quotient"? $\mathrm{SO}_3$ is its own adjoint quotient; it's abstractly a simple group. The whole thing is clear. If the (continuous) homomorphism is nontrivial, its image is 3-dimensional, compact, and since the maximal compact subgroups in $\mathrm{PSL}_2(\mathbf{C})$ are 3-dimensional and connected [the maximal compact subgroups of every connected Lie group are connected!], the image is maximal compact. $\endgroup$
    – YCor
    Commented Jun 22, 2020 at 20:54
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Any continuous homomorphism $SO_3 \to SL_2(\mathbb C)$ is trivial because $SO_3$ is compact and has only one irreducible representation of dimension at most two: the trivial representation.

EDIT: the revised question may still be answered by representation theory. A continuous homomorphism $f: SO_3 \to PGL_2(\mathbb C)$ has a unique lift to a homomorphism $\tilde f: SU_2 \to SL_2(\mathbb C)$. By the representation theory of $SU_2$, there are two representations of dimension two: the trivial representation and the defining representation on $\mathbb C^2$. Thus if $f,g: SO_3 \to PGL_2(\mathbb C)$ are both nontrivial and continuous, their lifts $\tilde f, \tilde g$ are conjugate and thus $f$ and $g$ are conjugate.

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  • $\begingroup$ I had to change $SL$ to the Möbius group, sry. The special unitary subgroup is isomorphic to SO(3). Hence there are nontrivial homomorphisms. Furthermore by taking any $S$ and doing the transformation as above we can make new homomorphisms. The question is then if we can create all of them that way. $\endgroup$
    – user35593
    Commented Jun 20, 2020 at 20:07
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    $\begingroup$ (Note: this addressed an early wrong version of the question). Here's an immediate proof that the only homomorphism $f$ [not assumed continuous] $\mathrm{SO}(3)\to\mathrm{SL}_2(\mathbf{C})$ is trivial. Since the only elements of order dividing $2$ in $\mathrm{SL}_2(\mathbf{C})$ are $\pm I$, all elements of order $2$ in $\mathrm{SO}(3)$ map into $\pm 1$. Since $\mathrm{SO}(3)$ is generated by its elements of order $2$, $f$ maps into $\{\pm 1\}$. Since moreover these elements of order 2 are commutators, everything maps to $1$. (I'm not even using abstract simplicity of $\mathrm{SO}(3)$.) $\endgroup$
    – YCor
    Commented Jun 21, 2020 at 11:29

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