Mixing
Torsion element of a non-elementary Hyperbolic group
$begingroup$
Let $Gamma$ be a non elementary hyperbolic group acting on the Gromov boundary $partialGamma$. Let $a in Gamma$ be a torsion element i.e $langle arangle$ is finite. Does $a$ fix every element of $partialGamma$?
I know that that every element $a$ is either a parabolic element in which case it has one fixed point, or a hyperbolic element in which case it has two fixed points or an elliptic element in which case it has finite order. For non-elliptic elements $a$ also there is a north pole-south pole property: for each such $a in Gamma$ there are two points $x_a^+$ and $x_a^{-}$ on the boundary, not necessarily distinct such that $$a^nxxrightarrow{n to infty} x_a^+, forall x ne x_a^-$$
and $ax_a^+=x_a^+$ and $ax_a^-=x_a^-$.
This convergence property is common to all convergence groups, where I looked in as well but found no information about the fixed points of elliptic elements. Can anything be said about the fixed points of elliptic elements? Am I missing something?
Thanks for the help!!
gromov-hyperbolic-spaces hyperbolic-groups convergencegroups
edited Feb 2 at 18:48
Martin Sleziak
45k10122277
asked Feb 2 at 17:36
tattwamasi amrutamtattwamasi amrutam
8,29621643
$endgroup$
add a comment |
$begingroup$
Let $Gamma$ be a non elementary hyperbolic group acting on the Gromov boundary $partialGamma$. Let $a in Gamma$ be a torsion element i.e $langle arangle$ is finite. Does $a$ fix every element of $partialGamma$?
I know that that every element $a$ is either a parabolic element in which case it has one fixed point, or a hyperbolic element in which case it has two fixed points or an elliptic element in which case it has finite order. For non-elliptic elements $a$ also there is a north pole-south pole property: for each such $a in Gamma$ there are two points $x_a^+$ and $x_a^{-}$ on the boundary, not necessarily distinct such that $$a^nxxrightarrow{n to infty} x_a^+, forall x ne x_a^-$$
and $ax_a^+=x_a^+$ and $ax_a^-=x_a^-$.
This convergence property is common to all convergence groups, where I looked in as well but found no information about the fixed points of elliptic elements. Can anything be said about the fixed points of elliptic elements? Am I missing something?
Thanks for the help!!
gromov-hyperbolic-spaces hyperbolic-groups convergencegroups
edited Feb 2 at 18:48
Martin Sleziak
45k10122277
asked Feb 2 at 17:36
tattwamasi amrutamtattwamasi amrutam
8,29621643
$endgroup$
add a comment |
$begingroup$
Let $Gamma$ be a non elementary hyperbolic group acting on the Gromov boundary $partialGamma$. Let $a in Gamma$ be a torsion element i.e $langle arangle$ is finite. Does $a$ fix every element of $partialGamma$?
I know that that every element $a$ is either a parabolic element in which case it has one fixed point, or a hyperbolic element in which case it has two fixed points or an elliptic element in which case it has finite order. For non-elliptic elements $a$ also there is a north pole-south pole property: for each such $a in Gamma$ there are two points $x_a^+$ and $x_a^{-}$ on the boundary, not necessarily distinct such that $$a^nxxrightarrow{n to infty} x_a^+, forall x ne x_a^-$$
and $ax_a^+=x_a^+$ and $ax_a^-=x_a^-$.
This convergence property is common to all convergence groups, where I looked in as well but found no information about the fixed points of elliptic elements. Can anything be said about the fixed points of elliptic elements? Am I missing something?
Thanks for the help!!
gromov-hyperbolic-spaces hyperbolic-groups convergencegroups
edited Feb 2 at 18:48
Martin Sleziak
45k10122277
asked Feb 2 at 17:36
tattwamasi amrutamtattwamasi amrutam
8,29621643
$endgroup$
Let $Gamma$ be a non elementary hyperbolic group acting on the Gromov boundary $partialGamma$. Let $a in Gamma$ be a torsion element i.e $langle arangle$ is finite. Does $a$ fix every element of $partialGamma$?
I know that that every element $a$ is either a parabolic element in which case it has one fixed point, or a hyperbolic element in which case it has two fixed points or an elliptic element in which case it has finite order. For non-elliptic elements $a$ also there is a north pole-south pole property: for each such $a in Gamma$ there are two points $x_a^+$ and $x_a^{-}$ on the boundary, not necessarily distinct such that $$a^nxxrightarrow{n to infty} x_a^+, forall x ne x_a^-$$
and $ax_a^+=x_a^+$ and $ax_a^-=x_a^-$.
This convergence property is common to all convergence groups, where I looked in as well but found no information about the fixed points of elliptic elements. Can anything be said about the fixed points of elliptic elements? Am I missing something?
Thanks for the help!!
gromov-hyperbolic-spaces hyperbolic-groups convergencegroups
gromov-hyperbolic-spaces hyperbolic-groups convergencegroups
edited Feb 2 at 18:48
Martin Sleziak
45k10122277
asked Feb 2 at 17:36
tattwamasi amrutamtattwamasi amrutam
8,29621643
edited Feb 2 at 18:48
Martin Sleziak
45k10122277
asked Feb 2 at 17:36
tattwamasi amrutamtattwamasi amrutam
8,29621643
edited Feb 2 at 18:48
Martin Sleziak
45k10122277
edited Feb 2 at 18:48
Martin Sleziak
45k10122277
edited Feb 2 at 18:48
Martin Sleziak
45k10122277
45k10122277
asked Feb 2 at 17:36
tattwamasi amrutamtattwamasi amrutam
8,29621643
asked Feb 2 at 17:36
tattwamasi amrutamtattwamasi amrutam
8,29621643
asked Feb 2 at 17:36
tattwamasi amrutamtattwamasi amrutam
8,29621643
8,29621643
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A finite-order element of a hyperbolic group may have no fixed point at infinity. Consider for instance the free product $mathbb{Z}_2 ast mathbb{Z}_2$ (or $mathbb{Z}_2 ast mathbb{Z}_2 ast mathbb{Z}_2$ is you want a non-elementary example). You have an action on a simplicial tree and the finite-order elements inverse edges.
In fact, to answer your first question, it is not too difficult to show that the kernel of the action $Gamma curvearrowright partial Gamma$ is the unique maximal finite normal subgroup of $Gamma$. A possible sketch of proof goes as follows:
Let $K$ denote the kernel of the action $Gamma curvearrowright partial Gamma$. Because $Gamma$ is non-elementary, the boundary contains three pairwise distinct points $alpha,beta,gamma in partial Gamma$. Notice that, for every $N geq 1$, the intersection $$[alpha,beta]^{+N} cap [beta,gamma]^{+N} cap [alpha,gamma]^{+N}$$ is bounded, where $[cdot,cdot]$ denotes the union of all the geodesics between two points at infinity. Taking $N$ large enough so that the intersection is non-empty, it follows that $K$ stabilises a bounded subset, proving that $K$ must be finite. Moreover, as a kernel, it is automatically normal.
Next, let $F lhd Gamma$ be a finite normal subgroup of $Gamma$. Notice that, for every $x in Gamma$, the orbit $F cdot x$ has the same diameter as the the orbit $F cdot 1$. Indeed, $$F cdot x = xx^{-1}Fx cdot 1= x (F cdot 1).$$ It follows that $F$ fixes pointwise $partial Gamma$, hence $F leq K$.
answered Feb 3 at 20:40
AGenevoisAGenevois
2013
$endgroup$
$begingroup$
What do you mean by "unique maximal finite normal subgroup of $Gamma$? Why should the kernel be finite?
$endgroup$
– tattwamasi amrutam
Feb 3 at 21:38
$begingroup$
I mean that $K:= mathrm{ker}(Gamma curvearrowright partial Gamma)$ is a finite normal subgroup, and that, If $N lhd Gamma$ is another finite normal subgroup of $Gamma$, then $N leq K$.
$endgroup$
– AGenevois
Feb 4 at 6:58
$begingroup$
To see that $K$ is finite, take three distinct points $alpha, beta, gamma in partial Gamma$. If $[zeta,xi]$ denotes the union of all the geodesics between $zeta,xi in partial Gamma$, then notice that, for every sufficiently large $N geq 0$, the intersection $$[alpha,beta]^{+N} cap [beta,gamma]^{+N} cap [alpha,gamma]^{+N}$$ is a non-empty bounded set which is $K$-invariant.
$endgroup$
– AGenevois
Feb 4 at 7:02
$begingroup$
But if $Gamma$ is a $C^*$-simple group, then $K$ being normal is going to be $C^*$-simple and hence, non-amenable. But $K$ finite would imply that it is amenable, hence not finite.
$endgroup$
– tattwamasi amrutam
Feb 4 at 14:52
$begingroup$
It just means that, if $Gamma$ is $C^ast$-simple, then necessarily $K={1}$.
$endgroup$
– AGenevois
Feb 4 at 17:43
|
show 2 more comments
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$begingroup$
A finite-order element of a hyperbolic group may have no fixed point at infinity. Consider for instance the free product $mathbb{Z}_2 ast mathbb{Z}_2$ (or $mathbb{Z}_2 ast mathbb{Z}_2 ast mathbb{Z}_2$ is you want a non-elementary example). You have an action on a simplicial tree and the finite-order elements inverse edges.
In fact, to answer your first question, it is not too difficult to show that the kernel of the action $Gamma curvearrowright partial Gamma$ is the unique maximal finite normal subgroup of $Gamma$. A possible sketch of proof goes as follows:
Let $K$ denote the kernel of the action $Gamma curvearrowright partial Gamma$. Because $Gamma$ is non-elementary, the boundary contains three pairwise distinct points $alpha,beta,gamma in partial Gamma$. Notice that, for every $N geq 1$, the intersection $$[alpha,beta]^{+N} cap [beta,gamma]^{+N} cap [alpha,gamma]^{+N}$$ is bounded, where $[cdot,cdot]$ denotes the union of all the geodesics between two points at infinity. Taking $N$ large enough so that the intersection is non-empty, it follows that $K$ stabilises a bounded subset, proving that $K$ must be finite. Moreover, as a kernel, it is automatically normal.
Next, let $F lhd Gamma$ be a finite normal subgroup of $Gamma$. Notice that, for every $x in Gamma$, the orbit $F cdot x$ has the same diameter as the the orbit $F cdot 1$. Indeed, $$F cdot x = xx^{-1}Fx cdot 1= x (F cdot 1).$$ It follows that $F$ fixes pointwise $partial Gamma$, hence $F leq K$.
answered Feb 3 at 20:40
AGenevoisAGenevois
2013
$endgroup$
$begingroup$
What do you mean by "unique maximal finite normal subgroup of $Gamma$? Why should the kernel be finite?
$endgroup$
– tattwamasi amrutam
Feb 3 at 21:38
$begingroup$
I mean that $K:= mathrm{ker}(Gamma curvearrowright partial Gamma)$ is a finite normal subgroup, and that, If $N lhd Gamma$ is another finite normal subgroup of $Gamma$, then $N leq K$.
$endgroup$
– AGenevois
Feb 4 at 6:58
$begingroup$
To see that $K$ is finite, take three distinct points $alpha, beta, gamma in partial Gamma$. If $[zeta,xi]$ denotes the union of all the geodesics between $zeta,xi in partial Gamma$, then notice that, for every sufficiently large $N geq 0$, the intersection $$[alpha,beta]^{+N} cap [beta,gamma]^{+N} cap [alpha,gamma]^{+N}$$ is a non-empty bounded set which is $K$-invariant.
$endgroup$
– AGenevois
Feb 4 at 7:02
$begingroup$
But if $Gamma$ is a $C^*$-simple group, then $K$ being normal is going to be $C^*$-simple and hence, non-amenable. But $K$ finite would imply that it is amenable, hence not finite.
$endgroup$
– tattwamasi amrutam
Feb 4 at 14:52
$begingroup$
It just means that, if $Gamma$ is $C^ast$-simple, then necessarily $K={1}$.
$endgroup$
– AGenevois
Feb 4 at 17:43
|
show 2 more comments
$begingroup$
A finite-order element of a hyperbolic group may have no fixed point at infinity. Consider for instance the free product $mathbb{Z}_2 ast mathbb{Z}_2$ (or $mathbb{Z}_2 ast mathbb{Z}_2 ast mathbb{Z}_2$ is you want a non-elementary example). You have an action on a simplicial tree and the finite-order elements inverse edges.
In fact, to answer your first question, it is not too difficult to show that the kernel of the action $Gamma curvearrowright partial Gamma$ is the unique maximal finite normal subgroup of $Gamma$. A possible sketch of proof goes as follows:
Let $K$ denote the kernel of the action $Gamma curvearrowright partial Gamma$. Because $Gamma$ is non-elementary, the boundary contains three pairwise distinct points $alpha,beta,gamma in partial Gamma$. Notice that, for every $N geq 1$, the intersection $$[alpha,beta]^{+N} cap [beta,gamma]^{+N} cap [alpha,gamma]^{+N}$$ is bounded, where $[cdot,cdot]$ denotes the union of all the geodesics between two points at infinity. Taking $N$ large enough so that the intersection is non-empty, it follows that $K$ stabilises a bounded subset, proving that $K$ must be finite. Moreover, as a kernel, it is automatically normal.
Next, let $F lhd Gamma$ be a finite normal subgroup of $Gamma$. Notice that, for every $x in Gamma$, the orbit $F cdot x$ has the same diameter as the the orbit $F cdot 1$. Indeed, $$F cdot x = xx^{-1}Fx cdot 1= x (F cdot 1).$$ It follows that $F$ fixes pointwise $partial Gamma$, hence $F leq K$.
answered Feb 3 at 20:40
AGenevoisAGenevois
2013
$endgroup$
$begingroup$
What do you mean by "unique maximal finite normal subgroup of $Gamma$? Why should the kernel be finite?
$endgroup$
– tattwamasi amrutam
Feb 3 at 21:38
$begingroup$
I mean that $K:= mathrm{ker}(Gamma curvearrowright partial Gamma)$ is a finite normal subgroup, and that, If $N lhd Gamma$ is another finite normal subgroup of $Gamma$, then $N leq K$.
$endgroup$
– AGenevois
Feb 4 at 6:58
$begingroup$
To see that $K$ is finite, take three distinct points $alpha, beta, gamma in partial Gamma$. If $[zeta,xi]$ denotes the union of all the geodesics between $zeta,xi in partial Gamma$, then notice that, for every sufficiently large $N geq 0$, the intersection $$[alpha,beta]^{+N} cap [beta,gamma]^{+N} cap [alpha,gamma]^{+N}$$ is a non-empty bounded set which is $K$-invariant.
$endgroup$
– AGenevois
Feb 4 at 7:02
$begingroup$
But if $Gamma$ is a $C^*$-simple group, then $K$ being normal is going to be $C^*$-simple and hence, non-amenable. But $K$ finite would imply that it is amenable, hence not finite.
$endgroup$
– tattwamasi amrutam
Feb 4 at 14:52
$begingroup$
It just means that, if $Gamma$ is $C^ast$-simple, then necessarily $K={1}$.
$endgroup$
– AGenevois
Feb 4 at 17:43
|
show 2 more comments
$begingroup$
A finite-order element of a hyperbolic group may have no fixed point at infinity. Consider for instance the free product $mathbb{Z}_2 ast mathbb{Z}_2$ (or $mathbb{Z}_2 ast mathbb{Z}_2 ast mathbb{Z}_2$ is you want a non-elementary example). You have an action on a simplicial tree and the finite-order elements inverse edges.
In fact, to answer your first question, it is not too difficult to show that the kernel of the action $Gamma curvearrowright partial Gamma$ is the unique maximal finite normal subgroup of $Gamma$. A possible sketch of proof goes as follows:
Let $K$ denote the kernel of the action $Gamma curvearrowright partial Gamma$. Because $Gamma$ is non-elementary, the boundary contains three pairwise distinct points $alpha,beta,gamma in partial Gamma$. Notice that, for every $N geq 1$, the intersection $$[alpha,beta]^{+N} cap [beta,gamma]^{+N} cap [alpha,gamma]^{+N}$$ is bounded, where $[cdot,cdot]$ denotes the union of all the geodesics between two points at infinity. Taking $N$ large enough so that the intersection is non-empty, it follows that $K$ stabilises a bounded subset, proving that $K$ must be finite. Moreover, as a kernel, it is automatically normal.
Next, let $F lhd Gamma$ be a finite normal subgroup of $Gamma$. Notice that, for every $x in Gamma$, the orbit $F cdot x$ has the same diameter as the the orbit $F cdot 1$. Indeed, $$F cdot x = xx^{-1}Fx cdot 1= x (F cdot 1).$$ It follows that $F$ fixes pointwise $partial Gamma$, hence $F leq K$.
answered Feb 3 at 20:40
AGenevoisAGenevois
2013
$endgroup$
A finite-order element of a hyperbolic group may have no fixed point at infinity. Consider for instance the free product $mathbb{Z}_2 ast mathbb{Z}_2$ (or $mathbb{Z}_2 ast mathbb{Z}_2 ast mathbb{Z}_2$ is you want a non-elementary example). You have an action on a simplicial tree and the finite-order elements inverse edges.
In fact, to answer your first question, it is not too difficult to show that the kernel of the action $Gamma curvearrowright partial Gamma$ is the unique maximal finite normal subgroup of $Gamma$. A possible sketch of proof goes as follows:
Let $K$ denote the kernel of the action $Gamma curvearrowright partial Gamma$. Because $Gamma$ is non-elementary, the boundary contains three pairwise distinct points $alpha,beta,gamma in partial Gamma$. Notice that, for every $N geq 1$, the intersection $$[alpha,beta]^{+N} cap [beta,gamma]^{+N} cap [alpha,gamma]^{+N}$$ is bounded, where $[cdot,cdot]$ denotes the union of all the geodesics between two points at infinity. Taking $N$ large enough so that the intersection is non-empty, it follows that $K$ stabilises a bounded subset, proving that $K$ must be finite. Moreover, as a kernel, it is automatically normal.
Next, let $F lhd Gamma$ be a finite normal subgroup of $Gamma$. Notice that, for every $x in Gamma$, the orbit $F cdot x$ has the same diameter as the the orbit $F cdot 1$. Indeed, $$F cdot x = xx^{-1}Fx cdot 1= x (F cdot 1).$$ It follows that $F$ fixes pointwise $partial Gamma$, hence $F leq K$.
answered Feb 3 at 20:40
AGenevoisAGenevois
2013
edited Feb 22 at 9:05
answered Feb 3 at 20:40
AGenevoisAGenevois
2013
answered Feb 3 at 20:40
AGenevoisAGenevois
2013
answered Feb 3 at 20:40
AGenevoisAGenevois
2013
2013
$begingroup$
What do you mean by "unique maximal finite normal subgroup of $Gamma$? Why should the kernel be finite?
$endgroup$
– tattwamasi amrutam
Feb 3 at 21:38
$begingroup$
I mean that $K:= mathrm{ker}(Gamma curvearrowright partial Gamma)$ is a finite normal subgroup, and that, If $N lhd Gamma$ is another finite normal subgroup of $Gamma$, then $N leq K$.
$endgroup$
– AGenevois
Feb 4 at 6:58
$begingroup$
To see that $K$ is finite, take three distinct points $alpha, beta, gamma in partial Gamma$. If $[zeta,xi]$ denotes the union of all the geodesics between $zeta,xi in partial Gamma$, then notice that, for every sufficiently large $N geq 0$, the intersection $$[alpha,beta]^{+N} cap [beta,gamma]^{+N} cap [alpha,gamma]^{+N}$$ is a non-empty bounded set which is $K$-invariant.
$endgroup$
– AGenevois
Feb 4 at 7:02
$begingroup$
But if $Gamma$ is a $C^*$-simple group, then $K$ being normal is going to be $C^*$-simple and hence, non-amenable. But $K$ finite would imply that it is amenable, hence not finite.
$endgroup$
– tattwamasi amrutam
Feb 4 at 14:52
$begingroup$
It just means that, if $Gamma$ is $C^ast$-simple, then necessarily $K={1}$.
$endgroup$
– AGenevois
Feb 4 at 17:43
|
show 2 more comments
$begingroup$
What do you mean by "unique maximal finite normal subgroup of $Gamma$? Why should the kernel be finite?
$endgroup$
– tattwamasi amrutam
Feb 3 at 21:38
$begingroup$
I mean that $K:= mathrm{ker}(Gamma curvearrowright partial Gamma)$ is a finite normal subgroup, and that, If $N lhd Gamma$ is another finite normal subgroup of $Gamma$, then $N leq K$.
$endgroup$
– AGenevois
Feb 4 at 6:58
$begingroup$
To see that $K$ is finite, take three distinct points $alpha, beta, gamma in partial Gamma$. If $[zeta,xi]$ denotes the union of all the geodesics between $zeta,xi in partial Gamma$, then notice that, for every sufficiently large $N geq 0$, the intersection $$[alpha,beta]^{+N} cap [beta,gamma]^{+N} cap [alpha,gamma]^{+N}$$ is a non-empty bounded set which is $K$-invariant.
$endgroup$
– AGenevois
Feb 4 at 7:02
$begingroup$
But if $Gamma$ is a $C^*$-simple group, then $K$ being normal is going to be $C^*$-simple and hence, non-amenable. But $K$ finite would imply that it is amenable, hence not finite.
$endgroup$
– tattwamasi amrutam
Feb 4 at 14:52
$begingroup$
It just means that, if $Gamma$ is $C^ast$-simple, then necessarily $K={1}$.
$endgroup$
– AGenevois
Feb 4 at 17:43
$begingroup$
What do you mean by "unique maximal finite normal subgroup of $Gamma$? Why should the kernel be finite?
$endgroup$
– tattwamasi amrutam
Feb 3 at 21:38
$begingroup$
What do you mean by "unique maximal finite normal subgroup of $Gamma$? Why should the kernel be finite?
$endgroup$
– tattwamasi amrutam
Feb 3 at 21:38
$begingroup$
I mean that $K:= mathrm{ker}(Gamma curvearrowright partial Gamma)$ is a finite normal subgroup, and that, If $N lhd Gamma$ is another finite normal subgroup of $Gamma$, then $N leq K$.
$endgroup$
– AGenevois
Feb 4 at 6:58
$begingroup$
I mean that $K:= mathrm{ker}(Gamma curvearrowright partial Gamma)$ is a finite normal subgroup, and that, If $N lhd Gamma$ is another finite normal subgroup of $Gamma$, then $N leq K$.
$endgroup$
– AGenevois
Feb 4 at 6:58
$begingroup$
To see that $K$ is finite, take three distinct points $alpha, beta, gamma in partial Gamma$. If $[zeta,xi]$ denotes the union of all the geodesics between $zeta,xi in partial Gamma$, then notice that, for every sufficiently large $N geq 0$, the intersection $$[alpha,beta]^{+N} cap [beta,gamma]^{+N} cap [alpha,gamma]^{+N}$$ is a non-empty bounded set which is $K$-invariant.
$endgroup$
– AGenevois
Feb 4 at 7:02
$begingroup$
To see that $K$ is finite, take three distinct points $alpha, beta, gamma in partial Gamma$. If $[zeta,xi]$ denotes the union of all the geodesics between $zeta,xi in partial Gamma$, then notice that, for every sufficiently large $N geq 0$, the intersection $$[alpha,beta]^{+N} cap [beta,gamma]^{+N} cap [alpha,gamma]^{+N}$$ is a non-empty bounded set which is $K$-invariant.
$endgroup$
– AGenevois
Feb 4 at 7:02
$begingroup$
But if $Gamma$ is a $C^*$-simple group, then $K$ being normal is going to be $C^*$-simple and hence, non-amenable. But $K$ finite would imply that it is amenable, hence not finite.
$endgroup$
– tattwamasi amrutam
Feb 4 at 14:52
$begingroup$
But if $Gamma$ is a $C^*$-simple group, then $K$ being normal is going to be $C^*$-simple and hence, non-amenable. But $K$ finite would imply that it is amenable, hence not finite.
$endgroup$
– tattwamasi amrutam
Feb 4 at 14:52
$begingroup$
It just means that, if $Gamma$ is $C^ast$-simple, then necessarily $K={1}$.
$endgroup$
– AGenevois
Feb 4 at 17:43
$begingroup$
It just means that, if $Gamma$ is $C^ast$-simple, then necessarily $K={1}$.
$endgroup$
– AGenevois
Feb 4 at 17:43
|
show 2 more comments
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