4
$\begingroup$

We can think hyperbolic 5-space as, $$\mathcal{H}^5=SO^+_{5,1}(\mathbb{R})/SO_5(\mathbb{R})=SL_2(\mathbb{H})/Sp^*_2(\mathbb{H}),$$$\mathbb{H}$ is real quaternion algebra. By Iwasawa Decomposition the orientation preserving isometry group of $\mathcal{H}^5$ is $$G=PSL_2(\mathbb{H})=NAK,$$where $$N=\left\lbrace n(x):=\begin{pmatrix}1&x\\&1\end{pmatrix}|x\in\mathbb{H}\right\rbrace\ \ \text{and}\ \ A=\left\lbrace a(y):=\begin{pmatrix}\sqrt y\\&\sqrt y^{-1}\end{pmatrix}|y\in\mathbb{R}^+\right\rbrace.$$

The quotient $G/K$, which could be realized as $$\left\lbrace \begin{pmatrix} y&x\\&1\end{pmatrix}|y\in\mathbb{R}^+,x\in\mathbb{H}\right\rbrace$$gives a realization of $\mathcal{H}^5$. Define $I(P)=y$ for $P\simeq n(x)a(y)$

Many literatures use that $G$ acts on $\mathcal{H}^5$ by Mobius Transformation, i.e. $$gP=\begin{pmatrix} a&b\\c&d\end{pmatrix}P=(aP+b)(cP+d)^{-1}.$$ My questions are,

1) What are the multiplication $(aP)$ or addition $(aP+b)$ rules being used here?

2) Can we think $P=x+*y$ (for some $*$) as we do in $\mathcal{H}^2 \ (*=i)$ and $\mathcal{H}^3 \ (*=j)$?

3) Is it true that $\frac{I(gP)}{I(P)}=\frac{||\det(g)||}{||cx+d||^2+||cy||^2}$ where ||.|| is quaternionic norm? (This happens in $\mathcal{H}^2$ and $\mathcal{H}^3$).

Thanks.

$\endgroup$
1
$\begingroup$

The "rule" for the generalized linear fractional transformation in these coordinates is simply the re-Iwasawa-decomposition of $g\cdot n(x)a(y)$.

I think it is just at this point that the writing of $x,y$ as $z=x+*y$ becomes much less tenable. Indeed, for hyperbolic $n$-space as $SO(n,1)/O(n)$ there seems to be no useful such expression. (Of course, this does not exclude expressing such things in terms of various associated Clifford algebras, but the happy coincidences that occur in the very small cases cease.)

If that $I(gp)$ is meant to be the analogue of "imaginary part", i.e., the split-Levi part $y$, then you can obtain the formula for it from the re-Iwasawa-decomposition. The $O(n,1)/SO(n)$ model for hyperbolic $n$-space still does admit a reasonable expression, as well.

$\endgroup$
  • $\begingroup$ Thanks. Regarding answer 1) I am little confused. What do you mean by $g.n(x)a(y)$? I mean what is the action of $g$ on $\begin{pmatrix}y&x\\&1\end{pmatrix}$ in matrix form? $\endgroup$ – Subhajit Jana May 24 '14 at 15:51
  • $\begingroup$ @Kunnysan, what I mean is to write the group product $g\cdot n(x)a(y)$ as $n(x') a(y')k$ with $k$ in the maximal compact. Then $g(x,y)=(x',y')$. Any "linear fractional transformations" that occur in such situation as just re-Iwasawa-decompositions after left group multiplication. $\endgroup$ – paul garrett May 24 '14 at 20:00

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.