Can a finitely generated discrete group $Gammasubset I(mathbbH^n)$ contain infinitely many elliptic elements with common fixed point?

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Let $Gammasubset I(mathbbH^n)$ be a finitely generated discrete group of isometries of the hyperbolic $n$-space.

Let $Gamma_infty$ be the stabilizer of $infty$, and assume it contains only elliptic and parabolic elements.

Let $Asubset Gamma$ be the set of elliptic elements of $Gamma$.

My question is whether or not can the generated subgroup $B=langle AcapGamma_inftyrangle$ be infinite.

---

My intuition says it cannot. In some manner that would imply that there are too many elliptic elements in $Gamma$, which would contradict the Selberg lemma. But I'm not sure if it is true, and how to formalize it. I guess the Selberg lemma shows that if $B$ is infinite it must contain a parabolic element. Question is if it is a contradiction or not.

hyperbolic-geometry

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Let $Gammasubset I(mathbbH^n)$ be a finitely generated discrete group of isometries of the hyperbolic $n$-space.

Let $Gamma_infty$ be the stabilizer of $infty$, and assume it contains only elliptic and parabolic elements.

Let $Asubset Gamma$ be the set of elliptic elements of $Gamma$.

My question is whether or not can the generated subgroup $B=langle AcapGamma_inftyrangle$ be infinite.

---

My intuition says it cannot. In some manner that would imply that there are too many elliptic elements in $Gamma$, which would contradict the Selberg lemma. But I'm not sure if it is true, and how to formalize it. I guess the Selberg lemma shows that if $B$ is infinite it must contain a parabolic element. Question is if it is a contradiction or not.

hyperbolic-geometry

share|cite|improve this question

edited Jul 21 at 12:04

asked Jul 21 at 2:17

The way of life

709216

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Let $Gammasubset I(mathbbH^n)$ be a finitely generated discrete group of isometries of the hyperbolic $n$-space.

Let $Gamma_infty$ be the stabilizer of $infty$, and assume it contains only elliptic and parabolic elements.

Let $Asubset Gamma$ be the set of elliptic elements of $Gamma$.

My question is whether or not can the generated subgroup $B=langle AcapGamma_inftyrangle$ be infinite.

---

My intuition says it cannot. In some manner that would imply that there are too many elliptic elements in $Gamma$, which would contradict the Selberg lemma. But I'm not sure if it is true, and how to formalize it. I guess the Selberg lemma shows that if $B$ is infinite it must contain a parabolic element. Question is if it is a contradiction or not.

hyperbolic-geometry

share|cite|improve this question

edited Jul 21 at 12:04

asked Jul 21 at 2:17

The way of life

709216

Let $Gammasubset I(mathbbH^n)$ be a finitely generated discrete group of isometries of the hyperbolic $n$-space.

Let $Gamma_infty$ be the stabilizer of $infty$, and assume it contains only elliptic and parabolic elements.

Let $Asubset Gamma$ be the set of elliptic elements of $Gamma$.

My question is whether or not can the generated subgroup $B=langle AcapGamma_inftyrangle$ be infinite.

---

My intuition says it cannot. In some manner that would imply that there are too many elliptic elements in $Gamma$, which would contradict the Selberg lemma. But I'm not sure if it is true, and how to formalize it. I guess the Selberg lemma shows that if $B$ is infinite it must contain a parabolic element. Question is if it is a contradiction or not.

hyperbolic-geometry

share|cite|improve this question

edited Jul 21 at 12:04

asked Jul 21 at 2:17

The way of life

709216

share|cite|improve this question

share|cite|improve this question

edited Jul 21 at 12:04

edited Jul 21 at 12:04

edited Jul 21 at 12:04

asked Jul 21 at 2:17

The way of life

709216

asked Jul 21 at 2:17

The way of life

709216

asked Jul 21 at 2:17

The way of life

709216

709216

add a comment | 

add a comment | 

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There's one restricted case in which this is true, namely if you assume that $n=2$ and that $Gamma$ consists only of orientation preserving isometries. In that case, every nontrivial finite order element of $Gamma$ is a rotation of some rational angle $2pi k/n$ around some point of $mathbb H^2$, and hence has no fixed points on the circle at infinite, so $A cap Gamma_infty$ consists solely of the identity.

However, even when $n=2$ there are counterexamples if you allow elements of $Gamma$ to reverse orientation. For instance, the infinite dihedral group $D_infty = langle a,b mid a^2 = b^2 = abrangle$ acts on $mathbb H$ by the formula $a cdot (x+iy) = -x+iy$, $b cdot (x+iy) = 1-x+iy$. Under this action, the point $infty$ is fixed by every element, and all of the infinitely many conjugates of $a$ and $b$ are reflection isometries each of which has a whole line of fixed points. For example, $a$ fixes every point on the imaginary axis $x=0$, and more generally $b^k a b^-k$ fixes every point on the line $x=k$.

In higher dimensions there are orientation preserving counterexamples as well. The general idea is that if you use the "upper half space" model $mathbb H^n = mathbb R^n-1 times mathbb R_+$ then the action of the group of Euclidean isometries of $mathbb R^n-1$ extends to an isometric action on $mathbb H^n$ that fixes $infty$. Its easy to produce infinite groups of orientation preserving Euclidean isometries having infinitely many torsion elements.

For a specific example acting on $mathbb H^3$, start with the infinite dihedral group of orientation preserving Euclidean isometries of $mathbb R^2$ generated by the rotations $a(x,y) = (-x,-y)$ and $b(x,y) = (1-x,-y)$.

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answered Jul 21 at 13:42

Lee Mosher

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There's one restricted case in which this is true, namely if you assume that $n=2$ and that $Gamma$ consists only of orientation preserving isometries. In that case, every nontrivial finite order element of $Gamma$ is a rotation of some rational angle $2pi k/n$ around some point of $mathbb H^2$, and hence has no fixed points on the circle at infinite, so $A cap Gamma_infty$ consists solely of the identity.

However, even when $n=2$ there are counterexamples if you allow elements of $Gamma$ to reverse orientation. For instance, the infinite dihedral group $D_infty = langle a,b mid a^2 = b^2 = abrangle$ acts on $mathbb H$ by the formula $a cdot (x+iy) = -x+iy$, $b cdot (x+iy) = 1-x+iy$. Under this action, the point $infty$ is fixed by every element, and all of the infinitely many conjugates of $a$ and $b$ are reflection isometries each of which has a whole line of fixed points. For example, $a$ fixes every point on the imaginary axis $x=0$, and more generally $b^k a b^-k$ fixes every point on the line $x=k$.

In higher dimensions there are orientation preserving counterexamples as well. The general idea is that if you use the "upper half space" model $mathbb H^n = mathbb R^n-1 times mathbb R_+$ then the action of the group of Euclidean isometries of $mathbb R^n-1$ extends to an isometric action on $mathbb H^n$ that fixes $infty$. Its easy to produce infinite groups of orientation preserving Euclidean isometries having infinitely many torsion elements.

For a specific example acting on $mathbb H^3$, start with the infinite dihedral group of orientation preserving Euclidean isometries of $mathbb R^2$ generated by the rotations $a(x,y) = (-x,-y)$ and $b(x,y) = (1-x,-y)$.

share|cite|improve this answer

answered Jul 21 at 13:42

Lee Mosher

45.6k33478

add a comment | 

up vote
2
down vote



accepted

There's one restricted case in which this is true, namely if you assume that $n=2$ and that $Gamma$ consists only of orientation preserving isometries. In that case, every nontrivial finite order element of $Gamma$ is a rotation of some rational angle $2pi k/n$ around some point of $mathbb H^2$, and hence has no fixed points on the circle at infinite, so $A cap Gamma_infty$ consists solely of the identity.

However, even when $n=2$ there are counterexamples if you allow elements of $Gamma$ to reverse orientation. For instance, the infinite dihedral group $D_infty = langle a,b mid a^2 = b^2 = abrangle$ acts on $mathbb H$ by the formula $a cdot (x+iy) = -x+iy$, $b cdot (x+iy) = 1-x+iy$. Under this action, the point $infty$ is fixed by every element, and all of the infinitely many conjugates of $a$ and $b$ are reflection isometries each of which has a whole line of fixed points. For example, $a$ fixes every point on the imaginary axis $x=0$, and more generally $b^k a b^-k$ fixes every point on the line $x=k$.

In higher dimensions there are orientation preserving counterexamples as well. The general idea is that if you use the "upper half space" model $mathbb H^n = mathbb R^n-1 times mathbb R_+$ then the action of the group of Euclidean isometries of $mathbb R^n-1$ extends to an isometric action on $mathbb H^n$ that fixes $infty$. Its easy to produce infinite groups of orientation preserving Euclidean isometries having infinitely many torsion elements.

For a specific example acting on $mathbb H^3$, start with the infinite dihedral group of orientation preserving Euclidean isometries of $mathbb R^2$ generated by the rotations $a(x,y) = (-x,-y)$ and $b(x,y) = (1-x,-y)$.

share|cite|improve this answer

answered Jul 21 at 13:42

Lee Mosher

45.6k33478

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

There's one restricted case in which this is true, namely if you assume that $n=2$ and that $Gamma$ consists only of orientation preserving isometries. In that case, every nontrivial finite order element of $Gamma$ is a rotation of some rational angle $2pi k/n$ around some point of $mathbb H^2$, and hence has no fixed points on the circle at infinite, so $A cap Gamma_infty$ consists solely of the identity.

However, even when $n=2$ there are counterexamples if you allow elements of $Gamma$ to reverse orientation. For instance, the infinite dihedral group $D_infty = langle a,b mid a^2 = b^2 = abrangle$ acts on $mathbb H$ by the formula $a cdot (x+iy) = -x+iy$, $b cdot (x+iy) = 1-x+iy$. Under this action, the point $infty$ is fixed by every element, and all of the infinitely many conjugates of $a$ and $b$ are reflection isometries each of which has a whole line of fixed points. For example, $a$ fixes every point on the imaginary axis $x=0$, and more generally $b^k a b^-k$ fixes every point on the line $x=k$.

In higher dimensions there are orientation preserving counterexamples as well. The general idea is that if you use the "upper half space" model $mathbb H^n = mathbb R^n-1 times mathbb R_+$ then the action of the group of Euclidean isometries of $mathbb R^n-1$ extends to an isometric action on $mathbb H^n$ that fixes $infty$. Its easy to produce infinite groups of orientation preserving Euclidean isometries having infinitely many torsion elements.

For a specific example acting on $mathbb H^3$, start with the infinite dihedral group of orientation preserving Euclidean isometries of $mathbb R^2$ generated by the rotations $a(x,y) = (-x,-y)$ and $b(x,y) = (1-x,-y)$.

share|cite|improve this answer

answered Jul 21 at 13:42

Lee Mosher

45.6k33478

There's one restricted case in which this is true, namely if you assume that $n=2$ and that $Gamma$ consists only of orientation preserving isometries. In that case, every nontrivial finite order element of $Gamma$ is a rotation of some rational angle $2pi k/n$ around some point of $mathbb H^2$, and hence has no fixed points on the circle at infinite, so $A cap Gamma_infty$ consists solely of the identity.

However, even when $n=2$ there are counterexamples if you allow elements of $Gamma$ to reverse orientation. For instance, the infinite dihedral group $D_infty = langle a,b mid a^2 = b^2 = abrangle$ acts on $mathbb H$ by the formula $a cdot (x+iy) = -x+iy$, $b cdot (x+iy) = 1-x+iy$. Under this action, the point $infty$ is fixed by every element, and all of the infinitely many conjugates of $a$ and $b$ are reflection isometries each of which has a whole line of fixed points. For example, $a$ fixes every point on the imaginary axis $x=0$, and more generally $b^k a b^-k$ fixes every point on the line $x=k$.

In higher dimensions there are orientation preserving counterexamples as well. The general idea is that if you use the "upper half space" model $mathbb H^n = mathbb R^n-1 times mathbb R_+$ then the action of the group of Euclidean isometries of $mathbb R^n-1$ extends to an isometric action on $mathbb H^n$ that fixes $infty$. Its easy to produce infinite groups of orientation preserving Euclidean isometries having infinitely many torsion elements.

For a specific example acting on $mathbb H^3$, start with the infinite dihedral group of orientation preserving Euclidean isometries of $mathbb R^2$ generated by the rotations $a(x,y) = (-x,-y)$ and $b(x,y) = (1-x,-y)$.

share|cite|improve this answer

answered Jul 21 at 13:42

Lee Mosher

45.6k33478

share|cite|improve this answer

share|cite|improve this answer

answered Jul 21 at 13:42

Lee Mosher

45.6k33478

answered Jul 21 at 13:42

Lee Mosher

45.6k33478

answered Jul 21 at 13:42

Lee Mosher

45.6k33478

45.6k33478

add a comment | 

add a comment | 

 

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