数理解析研究所講究録1022 巻 1997 年 71-76

71
The chordal norm of Kleinian groups
Harushi Furusawa($\mathrm{K}\mathrm{a}\mathrm{n}\mathrm{a}\mathrm{z}\mathrm{a}\mathrm{W}\mathrm{a}$Gakuin Univ.)

Let M\"ob denote the group of all orientation preserving mobius transformations of theextended complex plane $\hat{C}=C\cup\{\infty\}$ . We associate with each$f(Z)= \frac{az+b}{cz+d}\in M\ddot{o}b$, ad-bc $=1$ ,the martix$A=\in SL(\mathit{2}, c)$and set $tr(f)=tr(A)$ , where $tr(A)$ denote the trace ofto sign. The matrix norm $m(f)$ is defined by$A$. Note that $tr(f)$ is defined up$m(f)=||A-A^{-1}||=(2|a-d|^{2}+4|b|^{2}+4|c|^{2})^{\frac{1}{2}}$.

The quantity $m(f)$ is independent of the choice of representing .
For each and in M\"ob we let $[f,g]$ denote the multiplicative commutator $fgf^{-1}g-1$ .
We call the three complex numbers
$A$

$f$

$f$

$g$

$\beta(f)=tr^{2}(f)-4,\beta(g)=tr^{2}(g)-4,$ $\gamma(f,g)=tr([f,g])-2$ ,the parameters of the two generator group $<f,g>$ . These parameters are independentof the choice of representative matrices for and , and they determine
toconjugacy whenever $\gamma(f,g)\neq 0$ .
derive a lower bound for the distance in the metricof$f$$<f,g>\mathrm{u}\mathrm{p}$$g$$\mathrm{W}..\mathrm{e}$(1)$d(f,g)= \sup\{q(f(_{\mathcal{Z}}),g(Z));Z\in\hat{C}\}$

where
distance in$<f,g>\mathrm{i}\mathrm{s}$$\hat{C}$a Kleinian$\mathrm{g}\mathrm{r}\mathrm{o}\mathrm{u}_{\mathrm{P}}..$,ge,nerated.

$\mathrm{b}\mathrm{y}$.

$f,g$ in M\"ob and $q$ denotes the chordal$q^{2}(z,w)= \frac{4|z-w|^{2}}{(|z|^{2}+1)(|w|2+1)}$

A mobius transformation$h$.

is said to be a chordal isometry if$q(h(z), h(w))=q(Z,w)$

for all $z,w\in\hat{C}$ , then $m(f)$ is invariant with respect to conjugation by chordal isometries.
If
with$f\in M\ddot{o}b$$fi_{X}(f)=\{z_{1}, Z_{2}\},\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{n}$72Lemma 1. Let$f$be. loxodromic or elliptic with

(2)$f_{\dot{i}}x(f)=\{z_{1}, z_{2}\}$$2| \beta(f)|=\frac{q(z_{1},Z_{2})^{2}}{8-q(z_{1},Z_{2})^{2}}m(f)^{2}$

then.

Proof. All quantities in (2) are invariant with respect to conjugation by chordal isometries. Therefore by means of such a conjugation we may arrange that $z_{1}=-r$ and $z_{2}=r$where $0<r\leq 1$ . Then is represented by$f$$A=,$$b^{2}=a^{2}-1$.

Hence $\beta(f)=4b^{2},q(-r,.r)=\frac{4r}{r^{2}+1},\mathrm{a}\mathrm{n}\mathrm{d}m(f)^{2}=||A-A^{-1}||^{2}=(r^{2}+r^{-2})|2b|2$. Thus wehave$m(f)^{2}=2 \frac{8-q(-r,r)2}{q(-r,r)^{2}}|\beta(f)|$.

with $d(f_{\dot{i}},d)<2$ . If
Lemma 2. Suppose that is inand if
is the argument of its multiplier, then$f$$M\ddot{o}b\backslash \{id\}$$f$

has two fixed points$2\theta$(3)$m(f)^{2} \leq 8\cos\theta 2\frac{d(f,id)^{2}}{4-d(f_{\dot{i}d})^{2}}$with equality if and only if$f$,Lemma 3. If$f$

has antipodal fixed points.

is elliptic of order(4)$n,\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{n}$$d(f_{\dot{i}},d) \geq 2\sin(\frac{\pi}{n})$with equality if and only if $n=2$ or is primitive with antipodal fixed points.
Proof. If has order and $fiX(f)=\{-r, r\}$ , then $m(f)^{2}=4(r^{2}+r^{-2})\sin^{2}\theta$ where$\theta=k\pi/n$ and $1\leq k<n$ . Thus if $d(f, id)<2$ , we have
$f$$f$$n$$d(f,id)^{2} \geq\frac{16(r^{2}+r-2)\sin^{2}\theta}{8\cos^{2}\theta+4(r^{2}+r-2)\sin^{2}\theta}\geq 4\sin^{2}\theta$from Lemma 2, and the right hand side of above inequality is an incresing in $(r^{2}+r^{-2})$ . We
derive $d(f,\dot{i}d)^{2}\geq 4\sin^{2}\theta\geq 4\sin^{2}\pi/n$ with equality when $n>2$ if and only if $r=k=1$ .
Lemma 4. If$f$and(5)$g$are in M\"ob, then$\max\{d(f_{\dot{i}d},), d(g,\dot{i}d)\}\geq\{\frac{4|\gamma(f,g)|^{\frac{1}{2}}}{2+.|\gamma(f,g)|^{\frac{1}{2}}}\}^{\frac{1}{2}}$Proof. If and are in $SL(2, C)$ , thenSuppose that and are represented by
$A$

$B$

$f$

$g$

$||$

$A$

AB–BA
and$B$$||^{2} \leq\frac{1}{8}||A-A^{arrow\iota}||^{2}||B-B^{-1}||^{2}$.

in $SL(2, c)$ , then we have $|\gamma(f,g)|=$73$|ir([A, B])-2|=|det((AB)(BA)-1-I)|=|det(AB-BA)| \leq\frac{1}{2}||$ AB–BA$A-A^{-1}||^{2}||B-B^{-1}||^{2}= \frac{1}{16}m(f)2m(g)^{2}.\mathrm{I}\mathrm{t}$is$\mathrm{e}\mathrm{a}s$$||^{2} \leq\frac{1}{16}||$ily seen that$m(f)^{2} \leq\frac{8d(f,\dot{i}d)^{2}}{4-d(f_{\dot{i}d})^{2}},$’from Lemma 2 where equality holds if is either the identity or hyperbolic with antipodal
fixed points. Without loss of generality, set $m(g)\leq m(f)$ , then we have$f$

$4| \gamma(f,g)|^{\frac{1}{2}}\leq m(f)m(g)\leq m(f)^{2}\leq\frac{8d^{2}}{4-d^{2}}$.

Therefore we have the result.
Lemma 5 Suppose that and are in M\"ob and that $f(H)=g(H)=H$ . If andare hyperbolic with $\gamma(f,g)<0$ , then there exists in M\"ob such that $h(H)=H$ and$2\beta(f_{1})=m(f_{1})^{2},2\beta(g1)=m(g1)^{2}$ , where $f_{1}=hfh^{-}1,g_{1}=hgh-1$ and each ofhave
antipodal fixed points.
Theorem 6 Suppose thata Kleinian subgroup of M\"ob and that and haveno common fixed point and are not both of order 2. Then
$f$

$f$

$g$

$g$

$h$

$f_{1},g_{1}$$f$$<|f,g>\mathrm{i}\mathrm{s}$

(6)

$g$

$\max\{d(f,g),d(f^{-}1,g-1)\}\geq k_{1}$’whereis an absolute constant, $0.853\leq k_{1}\leq 0.911..$ .
Proof. Let $\gamma=\gamma(f,g)=\gamma(fg^{-1},g-1)$ and $\beta=\beta(fg^{-1})=\beta(g^{-1}f)$ . If $fg^{-1}$ isof order and where $n=2,3,4$ , or 6, then $d(f,g)=d(fg^{-}, i1d)\geq 2\sin(\pi/n)$ fromLemma 3. If $\gamma(fg^{-1},g-1)=\beta(fg^{-1})$ , then [M2]’ lmplies that
elliptic of order 2,or is elliptic of order 2. Therefore we assume thatconsider the subgroup $<fg^{-1-1},gf>=<fg^{-1},g^{-}(1fg-1)g>$ of $<fg^{-1},g^{-1}>$ with$\gamma(fg^{-1},g-1f)=\gamma(fg^{-1},g^{-1})\{\gamma(fg^{-1},g-1)-\beta(fg^{-1})\}=\gamma(\gamma-\beta)\neq 0.\mathrm{T}\mathrm{h}\mathrm{u}\mathrm{s}$we have$k_{1}$$\mathrm{n}$$fg^{-1}\dot{\mathrm{i}}\mathrm{s}$$3,4,\mathrm{o}\mathrm{r}6$$\gamma\neq\beta,\mathrm{a}\mathrm{n}\mathrm{d}$$\mathrm{g}$(7)$|\gamma(fg^{-1-},g1f)|\geq 2-2\cos(\pi/7)$

Therefore we have$\max\{d(f,g), d(f-1,g^{-1}\}$$=$$\geq$$\max\{d(fg^{-}, i1d), d(g-1f, id)\}$.

$( \frac{4(2\cos(\frac{\pi}{7})-2\cos(\frac{\pi}{7})+1)}{2\cos(\frac{\pi}{7})-2\cos(\frac{\pi}{7})+3})^{\frac{1}{2}}\geq 0.853$

From now on, we show an upper bound for , let $<\phi,$the (2,3,7) triangle groupacting on the upper half plane $H$ with $\phi^{2}=\psi^{3}=(\phi\psi)^{7}=id$ and set $f=\phi\psi$ and $g=\psi\phi$Then $\gamma=tr([\phi,\psi])-2=2\cos(\frac{2\pi}{7})-1,$ $\beta=tr^{2}([\phi, \psi])-4=2(\cos(\frac{2\pi}{7})+\cos(\frac{\pi}{7})-1)>0$
whereand $g^{-1}f$ are hyperbolic with $\gamma(fg^{-1},g^{-1}f)=\gamma(\gamma-\beta)<$. Then Lemma 5 show that there exists a mobius transformation of $H$ such that$\mathrm{d}$$\psi>\mathrm{d}\mathrm{e}\mathrm{n}\mathrm{o}\mathrm{t}\mathrm{e}$$fg^{-1}=[\phi, \psi].\mathrm{H}\mathrm{e}\mathrm{n}\mathrm{C}\mathrm{e}fg^{-1}$

$0$

$h$

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fix $(hfg-1h-1)=\{z_{1}, z2\}$ and fix $(hg^{arrow 1}fh^{-1})=\{w_{1},w_{2}\}$ where $q(z_{1}, z_{2})=q(w_{1,2}w)=2$ .
Then $2\beta(hfg^{-}h^{-1}1)=m(hfg-1h^{-1})^{2},2\beta(hg^{-1}fh^{-1})=m(hg^{-1}fh^{-}1)^{2}$ and $hfg^{-1}h^{-1}$and $hg^{-1}fh-1$ are both hyperbolic with antipodal fixed points. Therefore we have theresult$d(f,g)$$=$$d(f^{-1},g^{-1})$$=$$d(fg^{-1},id)=2( \frac{\cos(\frac{2\pi}{7})+\cos(\frac{\pi}{7})-1}{\cos(\frac{2\pi}{7})+\cos(\frac{\pi}{7})+1})^{\frac{1}{2}}$$d(g^{-1}f,id)=2( \frac{\cos(\frac{2\pi}{7})+\cos(\frac{\pi}{7})-1}{\cos(\frac{2\pi}{7})+\cos(\frac{\pi}{7})+1})^{\frac{1}{2}}$from Lemma 2 and hence that $k_{1}\leq 0.911$ .
Theorem 7 Suppose thatare not both of order 2,$f$

and are elements of a Kleinian subgroup. Then and commute or
$g$

$f$

$3,4,\mathrm{o}\mathrm{r}6$(8)$G$of M\"ob which$g$$\max\{d(f,g), d(f^{-}1,g-1)\}\geq k_{1}$Proof. It suffices to consider the case where $\gamma(f,g)=0$ and$fg\neq gf,\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{n}$$fiX(f)\cap fi_{X}(g)\neq\phi,$ $fiX(f)\neq f_{\dot{i}}x(g)$Then $<f,g>$ is elementary and $h=[f,g]$ is parabolic. We complete the proof by.
showin.g that $fg^{-1}$ is elliptic of order ( $=2,3,4$or 6) and $d(f,g)=d(fg^{-1},\dot{i}d)$.
Suppose that fix $(h)=\{\infty\}$ . If or , say , is parabolic, then is elliptic of order
and2,3,4 or 6 and an elementary .lculation shows that the same is true ofare both elliptic, then$\geq 1$

$n$

$f$

$g$

$f$

$g$

$fg^{-1}.\mathrm{I}\mathrm{f}f$$\mathrm{c}\mathrm{a}$$g$$f(z)=\nu z+a,g(Z)=\mu z+a,$ $fg^{-1}(Z)= \frac{\nu}{\mu}z+c$where$\nu^{p}=\mu^{q}=(\frac{\nu}{\mu})^{r}=1,$ $p,$$\nu\neq\mu \mathrm{a}\mathrm{n}\mathrm{d}fg^{-}1$$q\in\{2,3,4,6\}$

and $r\in\{1,2,3,4,6\}$ . By hypothesis $p\neq q$ ,is of order 2,3,4,6.

Example. If $f(z)=\lambda z$ and $g(z)=\lambda z-c$ where $\lambda^{p}=1,0<|c|\leq 2$ and $p=2,3,4$ or 6,
discrete whilethen$<f,g>\mathrm{i}\mathrm{s}$$d(f,g)=d(f-1,g-1)= \frac{8|c|}{4+|_{C|^{2}}}arrow 0$. It is necessary to make the hypothesis thatasor 6 in Theorem 7.
$carrow \mathrm{O}$$f$and$g$

are not both of oder 2,3,4Remark. For each $1<b<a<\infty$ let $f=f_{0}g0$ and $g=g_{0}$ where$f_{0}(z)=a^{2}Z,g_{0}(z)= \frac{(b^{2}+1)z+2b}{2bz+(b^{2}+1)}$

75
Then

$<f,g>\dot{\mathrm{i}}\mathrm{s}$nonelementary Kleinian while(9)$d(f,g)–d(f0, id)=2( \frac{a^{2}-1}{a^{2}+1})arrow 0$

as $aarrow 1$ . Hence there exists no universal lower bound for $d(f,g)$ and $d(f^{-1},g^{-1})$ .
These theorems give a geometric estimate of how different two mobius transformations
must be in order to generate a nonelementary Kleinian group.
Theorem 8 Suppose that $<f,$ $g>$ is a Kleinian group and and have no common
fixed point and are not both of order 2. If $fg$ is also not of order 2, then
$f$

(10)

$\max\{d(fg,gf), d((fg)-1, (gf)^{-1})\}\geq k_{1}$$g$.

Proof. Suppose that is not of order 2 and let $\gamma=\gamma(f,g)$ and $\beta=\beta(fg)$ . If,then $\beta([f,g])=\gamma(\gamma+4)=-3,$ $-4$ and thus that $[f,g]$ is elliptic of order 2 or 3. Hence$\gamma=\beta$$g$$d(fg,gf)=d([f,g], id)\geq\sqrt{3}$.

Otherwise $<fg,gf>=<fg,g(fg)g^{-1}>\mathrm{i}\mathrm{s}$ Kleinian with$\gamma(fg,gf)=\gamma(fg,g)\{\gamma(fg,g)-\beta(fg)\}=\gamma(\gamma-\beta)\neq 0$,then $k_{1}\geq 0.853$ from Theorem 6. Next let
the group which (6) holds withequality in Theorem 6. Then $f=\phi\psi$ and $g=\psi\phi$ where $<\phi,$the triangle groupwithand we obtain$<f,g>\mathrm{b}\mathrm{e}$$\psi>\mathrm{i}\mathrm{s}$$\phi^{2}=\psi^{3}=(\phi\psi)^{7}=\dot{i}d$$d(f,g)=d(f-1,g-1)=0.911$ .
from Theorem 5. Hence the group $<\phi,$$\psi>\mathrm{s}\mathrm{h}\mathrm{o}\mathrm{w}\mathrm{s}$that $k_{1}\leq 0.911$ .

Gehring and Martin showed that if $<f,g>$ is a nonelementary Kleinian subgroup ofM\"ob , then$m(f)m(g)\geq 4(\sqrt{2}-1)=1.656..$ ,(11)follows fromrgensen’s inequality and the proof of Lemma 4. In the proof of Lemma$16|\gamma(f,g)|<m(f)^{2}m(g)^{2}$
4, we haveand ifa nonelementary Kleinian group$m(f)m(g)\geq
4\sqrt{|\gamma(f,g)|}-\geq
1.780$
then,.
The following result shows that the average of the chordal norms of the generatorsand is always bounded below by a constantand $d(g, id)arrow 2$ as $d(f, id)arrow \mathrm{O}$
uniformly in the collection of all nonelementary Kleinian groups $<f,g>$ .
$\mathrm{J}\emptyset$$<f,g>\mathrm{i}\mathrm{s}$$f$$k_{1}$$g$

Theorem 9 Suppose that(12)$<f,g>\mathrm{i}\mathrm{s}$nonelementary Kleinian group of M\"ob. Then$d(f, id)+d(g,id)\geq 2k_{1},2d(f, id)+d(g, id)\geq 2$

76
Proof. We may assume that $2a=d(f, id)+d(g, id)\leq 2$ in the proof for both parts of(12) and that $d(g,id)\leq d(f_{\dot{i}},d)$ in the proof for the first inequality in (12). Next above
assumption together with the first inequality in (12) imply$2d(f_{\dot{i}},d)+d(g, id)$ whenever $d(f_{\dot{i}},d)\geq d(g, id)$ , hence we may also assume that $d(f, id)\leq$in the proof for the second inequality in (12). Then $d(f, id)=a-x$ and $d(g, id)=$$a+x$ where $0\leq x<a$ and we obtain$2< \frac{3}{2}\{d(f_{\dot{i}},d)+d(g,\dot{i}d)\}\leq$$d(g,\dot{i}d)$$16| \gamma(f,g)|\leq m(f)^{2}m(g)^{2}\leq\frac{8d(f,\dot{i}d)2}{4-d(f,id)2}\frac{8d(g,\dot{i}d)^{2}}{4-d(g,id)^{2}}$from Lemma 4. Let $\phi(x)=\{4(a-x)-2-1\}\{4(a+X)-2-1\}$ and $\phi(x)\leq 4\{2-2\cos(\pi/7)\}-1$
is increasing with respect to $[0, a)$ where $0<a\leq 1$ and we haveby Cao. Since$\phi(x)$$a \geq 2(\frac{\sqrt{|\gamma(f,g)|}}{2+\sqrt{|\gamma(f,g)|}})^{\frac{1}{2}}$This establishes the first part of (12) with $k_{1}\geq 0.853\ldots$If $2-d(g,\dot{i}d)>2d(f_{\dot{i}},d)$ , then $a+x<2\{1-(a-x)\}$ and$\psi(y)=(4y^{-2}-1)\{(1-y)-2-1\}<\phi(X)$where $y=a-x$ . By elementary calculus,$\psi’(y)$$\psi(y)$$=$$\geq$$-8y^{-2}(1-y)-3(5y-2)(y-2)$$\psi(2/5)>40>4\{2-2\cos(\pi/7)\}^{-1}$

for $0<y<1$ and we have a contradiction. This establishes the second part of (12).

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