0
$\begingroup$

Related: Can we characterize the Möbius transformations that maps the unit circle into itself?

The Mobius transformation is of the form $$f(z)=\frac{az+b}{cz+d}$$ In the 3D case, all the coefficients are quaternions, right?

I found a site https://www.quora.com/How-are-quaternions-used which show which transformation map the plane z=0 onto itself. However,

  • It does not mention about the Poincare disk, and convert the coordinate between Poincare disk and Poincare upper half plane is not easy.
  • It use $(1,i,j)$ as basis vectors instead of the usual $i,j,k$. Is that have any significant difference?

Follow the way in the 2D case (I use $i,j,k$ as basis vectors) I use formula $$f(z)=q\times(z-a)\times\frac1{1-conjg(a)\times z}\times q^{-1}$$ where q is a rotation quaternion (https://en.wikipedia.org/wiki/Quaternions_and_spatial_rotation)

Is that correct? Also, I can't find any obvious reason why $\left|\frac{z-a}{1-\bar az}\right|=1$.

$\endgroup$
  • $\begingroup$ You are using a wrong definition of Moebuus transformations in 3d. $\endgroup$ – Moishe Kohan Apr 28 '16 at 11:44
  • $\begingroup$ @studiosus What point is it wrong? Is that the non-commutativity of quaternion make it impossible to define division in H, or what? $\endgroup$ – user202729 Apr 28 '16 at 12:00
  • $\begingroup$ This is a minor problem. The right definition is that a Moebuus transformations in arbitrary dimension is a composition of inversions. There are several equivalent characterizations. $\endgroup$ – Moishe Kohan Apr 28 '16 at 12:06
  • $\begingroup$ @studiosus What is considered equivalent? Isn't $\frac{az+b}{cz+d}$ satisfy? Why not? $\endgroup$ – user202729 Apr 28 '16 at 12:11
  • $\begingroup$ You should read the linked text more closely, it discusses transformations of 3d half space. General Moebuus transformations will not preserve the half space. Lastly, read the Ahlfors article mentioned in the link. It should clarify your problems. $\endgroup$ – Moishe Kohan Apr 28 '16 at 12:24
1
$\begingroup$

First of all, there is a book by Lars Ahlfors available here which treats Moebius transformations in higher dimensional spaces in great detail.

One can define a general Moebius transformation $g: S^n\to S^n$ as follows. I will think of $S^n$ as the 1-point compactification of the Euclidean n-space $E^n$. First, you define the standard inversion in $S^n$ $$ I(x)= \frac{x}{|x|^2}. $$ It sends the center of inversion (the origin) to infinity and fixes the unit sphere in $E^n$ pointwise. Composing this inversion with dilations and translations you get more inversions. In addition, reflections in hyperplanes in $E^n$ are also regarded as inversions.

Definition. The group $M(S^n)$ of Moebius transformations of $S^n$ consists of finite compositions of inversions defined as above.

For instance, the stabilizer of $\infty$ in this group is the group of Euclidean similarities, i.e. compositions of dilations, translations and orthogonal transformations of $R^n$.

There are various equivalent descriptions of this group. For instance, according to the Liouville's theorem, it equals the group of conformal transformations of $S^n$, i.e. the group of diffeomorphisms of the unit sphere in $E^{n+1}$ which preserve angles between tangent vectors.

Now, as for your question, as written it does not really make sense. To begin with, you have to specify what are $a, b, c, d,$ and $z$. Are they arbitrary quaternions? Then $z$ belongs to the $4$-dimensional real vector space, not $E^3$. Are they purely imaginary quaterions? But then $f(z)$ need not be purely imaginary, even if you define the division properly. Ok, you can ask if this describes the group $M(S^4)$. You can identify the algebra of quaternions with $E^4$ by using the standard quaternionic norm to define distances. I think, you get an index 2 subgroup of $M(4)$ this way using the formula $$ f(z)=(az+b)(cz+d)^{-1}. $$

$\endgroup$
  • $\begingroup$ Your answer is good, but the resulting function I have found is too complex. Can you provide a simple closed form like $f(z)=\frac{e^{i \theta}(z-a)}{1- \bar {a}z}$? $\endgroup$ – user202729 May 1 '16 at 12:09
  • $\begingroup$ I can do it in 4d if you like, the 3d case is unclear to me. Ahlfors article n his notes has his own way which works in all dimensions. $\endgroup$ – Moishe Kohan May 1 '16 at 12:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.