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I am working on a project that I suspect requires calculations involving Möbius transformations on 3-dimensional space $\mathbb{R}^3$, identified with the quaternions $\mathbb{H}$ with $k$-component zero (in other words, the action of 4-dimensional hyperbolic isometries on the 3-dimensional boundary of $H^4$). My understanding is that such transformations can be identified with $2 \times 2$ quaternionic matrices $\left(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix}\right)$ satisfying $ab^*,cd^*,c^*a,d^*b \in \mathbb{R}^3$ and $ad^*-bc^*=1$.

I would like a proof that this is the correct group. More importantly, I would like to check that the usual nice properties carry over from the real and complex cases (any Möbius transformation is fully determined by its action on 3 points, it can be decomposed as a composition of simple transforms, etc.), and what things need to be adjusted. If anyone knows a good reference for these sort of questions, it would probably save me a lot of time, and I would be most grateful.

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    $\begingroup$ Ahlfors has several survey papers on this topic I believe, e.g.: tandfonline.com/doi/abs/10.1080/17476938608814142 $\endgroup$
    – Ian Agol
    Dec 24, 2014 at 22:01
  • $\begingroup$ Alan F. Beardon, The Geometry of Discrete Groups -- may be your reference (possibly you already know it). $\endgroup$ Dec 24, 2014 at 23:43
  • $\begingroup$ I haven't considered the quaternionic description yet, but simply from the geometry (i.e. inversions in spheres) I'd be surprised if this “fully determined by its action on 3 points” could work here. Any three points define a circle, any four points a sphere. So if you fix three points, there should still be a one-parameter family of spheres through that circle, i.e. a one-parameter family of corresponding transformations, and only one of them would be the identity. You'd need the action on 4 points to determine the transformation, I'd say. $\endgroup$
    – MvG
    Dec 27, 2014 at 22:12
  • $\begingroup$ @MvG, I would assume that there might be more freedom than just "you can map any 3 points to any 3 points," although it is certainly not true that you can map any 4 points to any other 4 points (which is really what you would hope to be true), so it would be good to figure out precisely what degrees of freedom exist. Also, there is something odd here, because the action of $PSL_2(\mathbb{R})$ is determined by three points and so is $PSL_2(\mathbb{C})$. Algebraically, it feels like $PSL_2(\mathbb{H})$ should behave similarly---granted, we probably lose a lot jumping to the non-commutative case. $\endgroup$ Dec 28, 2014 at 2:15
  • $\begingroup$ @Arseniy: You are right, mapping any 4 points to any other won't work. Perhaps the best way to see this is in terms of the $H^4$ isometries. Choose the image of any point – 4 real degrees of freedom. Choose the image of a point at unit distance from the first – a point on a three-sphere, 3 RDOF. Image of a point at unit distance from both of these – $S^2$, 2 RDOF. Fourth point at unit distance of all three – 1 RDOF. Total of 4+3+2+1=10 RDOF, which in terms of $\mathbb R^3$ makes 3⅓ points. So you have three points and then one more real parameter, as my 1-parameter family already suggested. $\endgroup$
    – MvG
    Dec 28, 2014 at 10:19

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