Groups acting on trees
Clash Royale CLAN TAG#URR8PPP
$begingroup$
Assume that $X$ is a tree such that every vertex has infinite degree, and a discrete group $G$ acts on this tree properly (with finite stabilizers) and transitively.
Is it true that $G$ contains a non-abelian free subgroup?
gr.group-theory trees
$endgroup$
add a comment |
$begingroup$
Assume that $X$ is a tree such that every vertex has infinite degree, and a discrete group $G$ acts on this tree properly (with finite stabilizers) and transitively.
Is it true that $G$ contains a non-abelian free subgroup?
gr.group-theory trees
$endgroup$
$begingroup$
Properly implies finite stabilizers
$endgroup$
– YCor
Mar 1 at 23:09
add a comment |
$begingroup$
Assume that $X$ is a tree such that every vertex has infinite degree, and a discrete group $G$ acts on this tree properly (with finite stabilizers) and transitively.
Is it true that $G$ contains a non-abelian free subgroup?
gr.group-theory trees
$endgroup$
Assume that $X$ is a tree such that every vertex has infinite degree, and a discrete group $G$ acts on this tree properly (with finite stabilizers) and transitively.
Is it true that $G$ contains a non-abelian free subgroup?
gr.group-theory trees
gr.group-theory trees
edited Mar 9 at 2:51
Shaun
12316
12316
asked Mar 1 at 22:54
Maria GerasimovaMaria Gerasimova
34118
34118
$begingroup$
Properly implies finite stabilizers
$endgroup$
– YCor
Mar 1 at 23:09
add a comment |
$begingroup$
Properly implies finite stabilizers
$endgroup$
– YCor
Mar 1 at 23:09
$begingroup$
Properly implies finite stabilizers
$endgroup$
– YCor
Mar 1 at 23:09
$begingroup$
Properly implies finite stabilizers
$endgroup$
– YCor
Mar 1 at 23:09
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes.
You're assuming more than what's necessary.
For an isometric group action on a Gromov-hyperbolic space, you have 5 possibilities (Gromov's classification):
- (a) bounded orbits
- (b) horocyclic (fixes a unique point at infinity, no loxodromic element; preserves "horospheres")
- (c) axial (preserves an axis, on which some element acts loxodromically)
- (d) focal (fixes a unique point at infinity, existence of a loxodromic element)
- (e) general type (= other): implies the existence of a non-abelian free subgroup acting metrically properly.
(a), (b), (c) are clearly ruled out in your setting (transitivity, and valency; valency $ge 3$ would be enough). (d) is ruled out because discrete groups have no metrically proper focal action at all. So (e) holds.
The case of actions on trees is a useful motivating baby case illustrating the above "classification"; all cases can actually occur.
$endgroup$
$begingroup$
Thank you! And if I assume that stabilizers are amenable, will it be the same?
$endgroup$
– Maria Gerasimova
Mar 1 at 23:40
2
$begingroup$
@MariaGerasimova you have to discard the focal case. Writing, for instance, the solvable group $mathbfZwrmathbfZ$ as an ascending HNN extension (w.r.t. an endomorphism with image of finite index), you make it act vertex-transitively on a tree with infinite valency (with abelian stabilizers). So you need to explicitly exclude the case of a fixed point at infinity.
$endgroup$
– YCor
Mar 1 at 23:56
add a comment |
$begingroup$
The geometry of simplicial trees can be extended in two directions: Gromov-hyperbolic spaces and CAT(0) cube complexes. Yves' answer is based on the first point of view. About cube complexes, we have the following statement (essentially due to Caprace and Sageev):
Theorem: Let $G$ be a group acting on a finite-dimensional CAT(0) cube complex $X$. Either $G$ contains a non-abelian free subgroup or it contains a finite-index subgroup which decomposes as a short exact sequence
$$1 to L to G to mathbbZ^n to 1$$
where $n leq dim(X)$ and where $L$ is locally $X$-elliptic (ie., any finitely generated subgroup of $L$ has a bounded orbit in $X$).
As a consequence, if a Tits' alternative is known for cube-stabilisers, then it is possible to deduce a Tits' alternative in the entire group. In the particular case of trees, we have:
Corollary: Let $G$ be a group acting on a tree $T$. Either $G$ contains a non-abelian free subgroup or it contains a finite-index* subgroup which decomposes as a short exact sequence
$$1 to L to G to Z to 1$$
where $L$ is locally $T$-elliptic and where $Z$ is trivial or infinite cyclic.
Of course, a direct argument can be given here.
Sketch of proof of the corollary. We distinguish two cases.
First, assume that $G$ has a finite orbit in $partial T$. So it contains a finite-index subgroup $H$ which fixes a point $xi in partial T$. Fix a basepoint $x in T$. Notice that, for every $g in H$, the intersection $g [x,xi) cap [x,xi)$ contains an infinite geodesic ray (as $g$ fixes $xi$). Consequently, $g$ defines a translation of length $tau(g)$ on a subray of $[x,xi)$. (Here $tau(g)$ is positive if the translation is directed to $xi$ and negative otherwise.) The key observation is that
$$tau : H to mathbbZ$$
defines a homomorphism whose kernel contains only elliptic elements (as each such element fixes pointwise a subray of $[x,xi)$). Thus, we get the desired short exact sequence.
Next, assume that $G$ has infinite orbits in $partial T$. Play ping-pong to construct a non-abelian free subgroup. $square$
*As noticed by Yves Cornulier in the comments, the subgroup can be always taken with index at most two.
$endgroup$
$begingroup$
You seem to assume (metric) properness at some point. Both the theorem and its corollary are false otherwise. Anyway I guess that the Caprace-Sageev theory is powerful enough to encompass non-proper actions.
$endgroup$
– YCor
Mar 2 at 9:58
$begingroup$
I am not assuming anything about the action. Why do you think the statements shoudn't be correct?
$endgroup$
– AGenevois
Mar 2 at 10:25
$begingroup$
Ah, sorry, it's confusing (at least to me) that "locally elliptic", which I usually know as an intrinsic property of the group, is used here as a property of the action, especially since the conclusion is a mixture of intrinsic properties of $G$ (the homomorphism onto an abelian group is not related to the action, as stated here) and extrinsic (being locally elliptic).
$endgroup$
– YCor
Mar 2 at 10:32
$begingroup$
Still there's a minor error in the corollary: it can happen that $G$ is already acting locally elliptically (horocyclic case). Side remark: in the corollary, no need to pass to a finite index subgroup, except precisely in the case $G$ preserves an axis, where one might have to pass to a subgroup of index 2.
$endgroup$
– YCor
Mar 2 at 10:34
1
$begingroup$
I replaced "elliptic" with "$X$-elliptic" to avoid confusion. I also corrected the mistake in the corollary, thank you.
$endgroup$
– AGenevois
Mar 2 at 11:03
|
show 5 more comments
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes.
You're assuming more than what's necessary.
For an isometric group action on a Gromov-hyperbolic space, you have 5 possibilities (Gromov's classification):
- (a) bounded orbits
- (b) horocyclic (fixes a unique point at infinity, no loxodromic element; preserves "horospheres")
- (c) axial (preserves an axis, on which some element acts loxodromically)
- (d) focal (fixes a unique point at infinity, existence of a loxodromic element)
- (e) general type (= other): implies the existence of a non-abelian free subgroup acting metrically properly.
(a), (b), (c) are clearly ruled out in your setting (transitivity, and valency; valency $ge 3$ would be enough). (d) is ruled out because discrete groups have no metrically proper focal action at all. So (e) holds.
The case of actions on trees is a useful motivating baby case illustrating the above "classification"; all cases can actually occur.
$endgroup$
$begingroup$
Thank you! And if I assume that stabilizers are amenable, will it be the same?
$endgroup$
– Maria Gerasimova
Mar 1 at 23:40
2
$begingroup$
@MariaGerasimova you have to discard the focal case. Writing, for instance, the solvable group $mathbfZwrmathbfZ$ as an ascending HNN extension (w.r.t. an endomorphism with image of finite index), you make it act vertex-transitively on a tree with infinite valency (with abelian stabilizers). So you need to explicitly exclude the case of a fixed point at infinity.
$endgroup$
– YCor
Mar 1 at 23:56
add a comment |
$begingroup$
Yes.
You're assuming more than what's necessary.
For an isometric group action on a Gromov-hyperbolic space, you have 5 possibilities (Gromov's classification):
- (a) bounded orbits
- (b) horocyclic (fixes a unique point at infinity, no loxodromic element; preserves "horospheres")
- (c) axial (preserves an axis, on which some element acts loxodromically)
- (d) focal (fixes a unique point at infinity, existence of a loxodromic element)
- (e) general type (= other): implies the existence of a non-abelian free subgroup acting metrically properly.
(a), (b), (c) are clearly ruled out in your setting (transitivity, and valency; valency $ge 3$ would be enough). (d) is ruled out because discrete groups have no metrically proper focal action at all. So (e) holds.
The case of actions on trees is a useful motivating baby case illustrating the above "classification"; all cases can actually occur.
$endgroup$
$begingroup$
Thank you! And if I assume that stabilizers are amenable, will it be the same?
$endgroup$
– Maria Gerasimova
Mar 1 at 23:40
2
$begingroup$
@MariaGerasimova you have to discard the focal case. Writing, for instance, the solvable group $mathbfZwrmathbfZ$ as an ascending HNN extension (w.r.t. an endomorphism with image of finite index), you make it act vertex-transitively on a tree with infinite valency (with abelian stabilizers). So you need to explicitly exclude the case of a fixed point at infinity.
$endgroup$
– YCor
Mar 1 at 23:56
add a comment |
$begingroup$
Yes.
You're assuming more than what's necessary.
For an isometric group action on a Gromov-hyperbolic space, you have 5 possibilities (Gromov's classification):
- (a) bounded orbits
- (b) horocyclic (fixes a unique point at infinity, no loxodromic element; preserves "horospheres")
- (c) axial (preserves an axis, on which some element acts loxodromically)
- (d) focal (fixes a unique point at infinity, existence of a loxodromic element)
- (e) general type (= other): implies the existence of a non-abelian free subgroup acting metrically properly.
(a), (b), (c) are clearly ruled out in your setting (transitivity, and valency; valency $ge 3$ would be enough). (d) is ruled out because discrete groups have no metrically proper focal action at all. So (e) holds.
The case of actions on trees is a useful motivating baby case illustrating the above "classification"; all cases can actually occur.
$endgroup$
Yes.
You're assuming more than what's necessary.
For an isometric group action on a Gromov-hyperbolic space, you have 5 possibilities (Gromov's classification):
- (a) bounded orbits
- (b) horocyclic (fixes a unique point at infinity, no loxodromic element; preserves "horospheres")
- (c) axial (preserves an axis, on which some element acts loxodromically)
- (d) focal (fixes a unique point at infinity, existence of a loxodromic element)
- (e) general type (= other): implies the existence of a non-abelian free subgroup acting metrically properly.
(a), (b), (c) are clearly ruled out in your setting (transitivity, and valency; valency $ge 3$ would be enough). (d) is ruled out because discrete groups have no metrically proper focal action at all. So (e) holds.
The case of actions on trees is a useful motivating baby case illustrating the above "classification"; all cases can actually occur.
edited Mar 2 at 11:27
user21820
834717
834717
answered Mar 1 at 23:15
YCorYCor
28.5k484139
28.5k484139
$begingroup$
Thank you! And if I assume that stabilizers are amenable, will it be the same?
$endgroup$
– Maria Gerasimova
Mar 1 at 23:40
2
$begingroup$
@MariaGerasimova you have to discard the focal case. Writing, for instance, the solvable group $mathbfZwrmathbfZ$ as an ascending HNN extension (w.r.t. an endomorphism with image of finite index), you make it act vertex-transitively on a tree with infinite valency (with abelian stabilizers). So you need to explicitly exclude the case of a fixed point at infinity.
$endgroup$
– YCor
Mar 1 at 23:56
add a comment |
$begingroup$
Thank you! And if I assume that stabilizers are amenable, will it be the same?
$endgroup$
– Maria Gerasimova
Mar 1 at 23:40
2
$begingroup$
@MariaGerasimova you have to discard the focal case. Writing, for instance, the solvable group $mathbfZwrmathbfZ$ as an ascending HNN extension (w.r.t. an endomorphism with image of finite index), you make it act vertex-transitively on a tree with infinite valency (with abelian stabilizers). So you need to explicitly exclude the case of a fixed point at infinity.
$endgroup$
– YCor
Mar 1 at 23:56
$begingroup$
Thank you! And if I assume that stabilizers are amenable, will it be the same?
$endgroup$
– Maria Gerasimova
Mar 1 at 23:40
$begingroup$
Thank you! And if I assume that stabilizers are amenable, will it be the same?
$endgroup$
– Maria Gerasimova
Mar 1 at 23:40
2
2
$begingroup$
@MariaGerasimova you have to discard the focal case. Writing, for instance, the solvable group $mathbfZwrmathbfZ$ as an ascending HNN extension (w.r.t. an endomorphism with image of finite index), you make it act vertex-transitively on a tree with infinite valency (with abelian stabilizers). So you need to explicitly exclude the case of a fixed point at infinity.
$endgroup$
– YCor
Mar 1 at 23:56
$begingroup$
@MariaGerasimova you have to discard the focal case. Writing, for instance, the solvable group $mathbfZwrmathbfZ$ as an ascending HNN extension (w.r.t. an endomorphism with image of finite index), you make it act vertex-transitively on a tree with infinite valency (with abelian stabilizers). So you need to explicitly exclude the case of a fixed point at infinity.
$endgroup$
– YCor
Mar 1 at 23:56
add a comment |
$begingroup$
The geometry of simplicial trees can be extended in two directions: Gromov-hyperbolic spaces and CAT(0) cube complexes. Yves' answer is based on the first point of view. About cube complexes, we have the following statement (essentially due to Caprace and Sageev):
Theorem: Let $G$ be a group acting on a finite-dimensional CAT(0) cube complex $X$. Either $G$ contains a non-abelian free subgroup or it contains a finite-index subgroup which decomposes as a short exact sequence
$$1 to L to G to mathbbZ^n to 1$$
where $n leq dim(X)$ and where $L$ is locally $X$-elliptic (ie., any finitely generated subgroup of $L$ has a bounded orbit in $X$).
As a consequence, if a Tits' alternative is known for cube-stabilisers, then it is possible to deduce a Tits' alternative in the entire group. In the particular case of trees, we have:
Corollary: Let $G$ be a group acting on a tree $T$. Either $G$ contains a non-abelian free subgroup or it contains a finite-index* subgroup which decomposes as a short exact sequence
$$1 to L to G to Z to 1$$
where $L$ is locally $T$-elliptic and where $Z$ is trivial or infinite cyclic.
Of course, a direct argument can be given here.
Sketch of proof of the corollary. We distinguish two cases.
First, assume that $G$ has a finite orbit in $partial T$. So it contains a finite-index subgroup $H$ which fixes a point $xi in partial T$. Fix a basepoint $x in T$. Notice that, for every $g in H$, the intersection $g [x,xi) cap [x,xi)$ contains an infinite geodesic ray (as $g$ fixes $xi$). Consequently, $g$ defines a translation of length $tau(g)$ on a subray of $[x,xi)$. (Here $tau(g)$ is positive if the translation is directed to $xi$ and negative otherwise.) The key observation is that
$$tau : H to mathbbZ$$
defines a homomorphism whose kernel contains only elliptic elements (as each such element fixes pointwise a subray of $[x,xi)$). Thus, we get the desired short exact sequence.
Next, assume that $G$ has infinite orbits in $partial T$. Play ping-pong to construct a non-abelian free subgroup. $square$
*As noticed by Yves Cornulier in the comments, the subgroup can be always taken with index at most two.
$endgroup$
$begingroup$
You seem to assume (metric) properness at some point. Both the theorem and its corollary are false otherwise. Anyway I guess that the Caprace-Sageev theory is powerful enough to encompass non-proper actions.
$endgroup$
– YCor
Mar 2 at 9:58
$begingroup$
I am not assuming anything about the action. Why do you think the statements shoudn't be correct?
$endgroup$
– AGenevois
Mar 2 at 10:25
$begingroup$
Ah, sorry, it's confusing (at least to me) that "locally elliptic", which I usually know as an intrinsic property of the group, is used here as a property of the action, especially since the conclusion is a mixture of intrinsic properties of $G$ (the homomorphism onto an abelian group is not related to the action, as stated here) and extrinsic (being locally elliptic).
$endgroup$
– YCor
Mar 2 at 10:32
$begingroup$
Still there's a minor error in the corollary: it can happen that $G$ is already acting locally elliptically (horocyclic case). Side remark: in the corollary, no need to pass to a finite index subgroup, except precisely in the case $G$ preserves an axis, where one might have to pass to a subgroup of index 2.
$endgroup$
– YCor
Mar 2 at 10:34
1
$begingroup$
I replaced "elliptic" with "$X$-elliptic" to avoid confusion. I also corrected the mistake in the corollary, thank you.
$endgroup$
– AGenevois
Mar 2 at 11:03
|
show 5 more comments
$begingroup$
The geometry of simplicial trees can be extended in two directions: Gromov-hyperbolic spaces and CAT(0) cube complexes. Yves' answer is based on the first point of view. About cube complexes, we have the following statement (essentially due to Caprace and Sageev):
Theorem: Let $G$ be a group acting on a finite-dimensional CAT(0) cube complex $X$. Either $G$ contains a non-abelian free subgroup or it contains a finite-index subgroup which decomposes as a short exact sequence
$$1 to L to G to mathbbZ^n to 1$$
where $n leq dim(X)$ and where $L$ is locally $X$-elliptic (ie., any finitely generated subgroup of $L$ has a bounded orbit in $X$).
As a consequence, if a Tits' alternative is known for cube-stabilisers, then it is possible to deduce a Tits' alternative in the entire group. In the particular case of trees, we have:
Corollary: Let $G$ be a group acting on a tree $T$. Either $G$ contains a non-abelian free subgroup or it contains a finite-index* subgroup which decomposes as a short exact sequence
$$1 to L to G to Z to 1$$
where $L$ is locally $T$-elliptic and where $Z$ is trivial or infinite cyclic.
Of course, a direct argument can be given here.
Sketch of proof of the corollary. We distinguish two cases.
First, assume that $G$ has a finite orbit in $partial T$. So it contains a finite-index subgroup $H$ which fixes a point $xi in partial T$. Fix a basepoint $x in T$. Notice that, for every $g in H$, the intersection $g [x,xi) cap [x,xi)$ contains an infinite geodesic ray (as $g$ fixes $xi$). Consequently, $g$ defines a translation of length $tau(g)$ on a subray of $[x,xi)$. (Here $tau(g)$ is positive if the translation is directed to $xi$ and negative otherwise.) The key observation is that
$$tau : H to mathbbZ$$
defines a homomorphism whose kernel contains only elliptic elements (as each such element fixes pointwise a subray of $[x,xi)$). Thus, we get the desired short exact sequence.
Next, assume that $G$ has infinite orbits in $partial T$. Play ping-pong to construct a non-abelian free subgroup. $square$
*As noticed by Yves Cornulier in the comments, the subgroup can be always taken with index at most two.
$endgroup$
$begingroup$
You seem to assume (metric) properness at some point. Both the theorem and its corollary are false otherwise. Anyway I guess that the Caprace-Sageev theory is powerful enough to encompass non-proper actions.
$endgroup$
– YCor
Mar 2 at 9:58
$begingroup$
I am not assuming anything about the action. Why do you think the statements shoudn't be correct?
$endgroup$
– AGenevois
Mar 2 at 10:25
$begingroup$
Ah, sorry, it's confusing (at least to me) that "locally elliptic", which I usually know as an intrinsic property of the group, is used here as a property of the action, especially since the conclusion is a mixture of intrinsic properties of $G$ (the homomorphism onto an abelian group is not related to the action, as stated here) and extrinsic (being locally elliptic).
$endgroup$
– YCor
Mar 2 at 10:32
$begingroup$
Still there's a minor error in the corollary: it can happen that $G$ is already acting locally elliptically (horocyclic case). Side remark: in the corollary, no need to pass to a finite index subgroup, except precisely in the case $G$ preserves an axis, where one might have to pass to a subgroup of index 2.
$endgroup$
– YCor
Mar 2 at 10:34
1
$begingroup$
I replaced "elliptic" with "$X$-elliptic" to avoid confusion. I also corrected the mistake in the corollary, thank you.
$endgroup$
– AGenevois
Mar 2 at 11:03
|
show 5 more comments
$begingroup$
The geometry of simplicial trees can be extended in two directions: Gromov-hyperbolic spaces and CAT(0) cube complexes. Yves' answer is based on the first point of view. About cube complexes, we have the following statement (essentially due to Caprace and Sageev):
Theorem: Let $G$ be a group acting on a finite-dimensional CAT(0) cube complex $X$. Either $G$ contains a non-abelian free subgroup or it contains a finite-index subgroup which decomposes as a short exact sequence
$$1 to L to G to mathbbZ^n to 1$$
where $n leq dim(X)$ and where $L$ is locally $X$-elliptic (ie., any finitely generated subgroup of $L$ has a bounded orbit in $X$).
As a consequence, if a Tits' alternative is known for cube-stabilisers, then it is possible to deduce a Tits' alternative in the entire group. In the particular case of trees, we have:
Corollary: Let $G$ be a group acting on a tree $T$. Either $G$ contains a non-abelian free subgroup or it contains a finite-index* subgroup which decomposes as a short exact sequence
$$1 to L to G to Z to 1$$
where $L$ is locally $T$-elliptic and where $Z$ is trivial or infinite cyclic.
Of course, a direct argument can be given here.
Sketch of proof of the corollary. We distinguish two cases.
First, assume that $G$ has a finite orbit in $partial T$. So it contains a finite-index subgroup $H$ which fixes a point $xi in partial T$. Fix a basepoint $x in T$. Notice that, for every $g in H$, the intersection $g [x,xi) cap [x,xi)$ contains an infinite geodesic ray (as $g$ fixes $xi$). Consequently, $g$ defines a translation of length $tau(g)$ on a subray of $[x,xi)$. (Here $tau(g)$ is positive if the translation is directed to $xi$ and negative otherwise.) The key observation is that
$$tau : H to mathbbZ$$
defines a homomorphism whose kernel contains only elliptic elements (as each such element fixes pointwise a subray of $[x,xi)$). Thus, we get the desired short exact sequence.
Next, assume that $G$ has infinite orbits in $partial T$. Play ping-pong to construct a non-abelian free subgroup. $square$
*As noticed by Yves Cornulier in the comments, the subgroup can be always taken with index at most two.
$endgroup$
The geometry of simplicial trees can be extended in two directions: Gromov-hyperbolic spaces and CAT(0) cube complexes. Yves' answer is based on the first point of view. About cube complexes, we have the following statement (essentially due to Caprace and Sageev):
Theorem: Let $G$ be a group acting on a finite-dimensional CAT(0) cube complex $X$. Either $G$ contains a non-abelian free subgroup or it contains a finite-index subgroup which decomposes as a short exact sequence
$$1 to L to G to mathbbZ^n to 1$$
where $n leq dim(X)$ and where $L$ is locally $X$-elliptic (ie., any finitely generated subgroup of $L$ has a bounded orbit in $X$).
As a consequence, if a Tits' alternative is known for cube-stabilisers, then it is possible to deduce a Tits' alternative in the entire group. In the particular case of trees, we have:
Corollary: Let $G$ be a group acting on a tree $T$. Either $G$ contains a non-abelian free subgroup or it contains a finite-index* subgroup which decomposes as a short exact sequence
$$1 to L to G to Z to 1$$
where $L$ is locally $T$-elliptic and where $Z$ is trivial or infinite cyclic.
Of course, a direct argument can be given here.
Sketch of proof of the corollary. We distinguish two cases.
First, assume that $G$ has a finite orbit in $partial T$. So it contains a finite-index subgroup $H$ which fixes a point $xi in partial T$. Fix a basepoint $x in T$. Notice that, for every $g in H$, the intersection $g [x,xi) cap [x,xi)$ contains an infinite geodesic ray (as $g$ fixes $xi$). Consequently, $g$ defines a translation of length $tau(g)$ on a subray of $[x,xi)$. (Here $tau(g)$ is positive if the translation is directed to $xi$ and negative otherwise.) The key observation is that
$$tau : H to mathbbZ$$
defines a homomorphism whose kernel contains only elliptic elements (as each such element fixes pointwise a subray of $[x,xi)$). Thus, we get the desired short exact sequence.
Next, assume that $G$ has infinite orbits in $partial T$. Play ping-pong to construct a non-abelian free subgroup. $square$
*As noticed by Yves Cornulier in the comments, the subgroup can be always taken with index at most two.
edited Mar 2 at 10:59
answered Mar 2 at 9:41
AGenevoisAGenevois
1,517815
1,517815
$begingroup$
You seem to assume (metric) properness at some point. Both the theorem and its corollary are false otherwise. Anyway I guess that the Caprace-Sageev theory is powerful enough to encompass non-proper actions.
$endgroup$
– YCor
Mar 2 at 9:58
$begingroup$
I am not assuming anything about the action. Why do you think the statements shoudn't be correct?
$endgroup$
– AGenevois
Mar 2 at 10:25
$begingroup$
Ah, sorry, it's confusing (at least to me) that "locally elliptic", which I usually know as an intrinsic property of the group, is used here as a property of the action, especially since the conclusion is a mixture of intrinsic properties of $G$ (the homomorphism onto an abelian group is not related to the action, as stated here) and extrinsic (being locally elliptic).
$endgroup$
– YCor
Mar 2 at 10:32
$begingroup$
Still there's a minor error in the corollary: it can happen that $G$ is already acting locally elliptically (horocyclic case). Side remark: in the corollary, no need to pass to a finite index subgroup, except precisely in the case $G$ preserves an axis, where one might have to pass to a subgroup of index 2.
$endgroup$
– YCor
Mar 2 at 10:34
1
$begingroup$
I replaced "elliptic" with "$X$-elliptic" to avoid confusion. I also corrected the mistake in the corollary, thank you.
$endgroup$
– AGenevois
Mar 2 at 11:03
|
show 5 more comments
$begingroup$
You seem to assume (metric) properness at some point. Both the theorem and its corollary are false otherwise. Anyway I guess that the Caprace-Sageev theory is powerful enough to encompass non-proper actions.
$endgroup$
– YCor
Mar 2 at 9:58
$begingroup$
I am not assuming anything about the action. Why do you think the statements shoudn't be correct?
$endgroup$
– AGenevois
Mar 2 at 10:25
$begingroup$
Ah, sorry, it's confusing (at least to me) that "locally elliptic", which I usually know as an intrinsic property of the group, is used here as a property of the action, especially since the conclusion is a mixture of intrinsic properties of $G$ (the homomorphism onto an abelian group is not related to the action, as stated here) and extrinsic (being locally elliptic).
$endgroup$
– YCor
Mar 2 at 10:32
$begingroup$
Still there's a minor error in the corollary: it can happen that $G$ is already acting locally elliptically (horocyclic case). Side remark: in the corollary, no need to pass to a finite index subgroup, except precisely in the case $G$ preserves an axis, where one might have to pass to a subgroup of index 2.
$endgroup$
– YCor
Mar 2 at 10:34
1
$begingroup$
I replaced "elliptic" with "$X$-elliptic" to avoid confusion. I also corrected the mistake in the corollary, thank you.
$endgroup$
– AGenevois
Mar 2 at 11:03
$begingroup$
You seem to assume (metric) properness at some point. Both the theorem and its corollary are false otherwise. Anyway I guess that the Caprace-Sageev theory is powerful enough to encompass non-proper actions.
$endgroup$
– YCor
Mar 2 at 9:58
$begingroup$
You seem to assume (metric) properness at some point. Both the theorem and its corollary are false otherwise. Anyway I guess that the Caprace-Sageev theory is powerful enough to encompass non-proper actions.
$endgroup$
– YCor
Mar 2 at 9:58
$begingroup$
I am not assuming anything about the action. Why do you think the statements shoudn't be correct?
$endgroup$
– AGenevois
Mar 2 at 10:25
$begingroup$
I am not assuming anything about the action. Why do you think the statements shoudn't be correct?
$endgroup$
– AGenevois
Mar 2 at 10:25
$begingroup$
Ah, sorry, it's confusing (at least to me) that "locally elliptic", which I usually know as an intrinsic property of the group, is used here as a property of the action, especially since the conclusion is a mixture of intrinsic properties of $G$ (the homomorphism onto an abelian group is not related to the action, as stated here) and extrinsic (being locally elliptic).
$endgroup$
– YCor
Mar 2 at 10:32
$begingroup$
Ah, sorry, it's confusing (at least to me) that "locally elliptic", which I usually know as an intrinsic property of the group, is used here as a property of the action, especially since the conclusion is a mixture of intrinsic properties of $G$ (the homomorphism onto an abelian group is not related to the action, as stated here) and extrinsic (being locally elliptic).
$endgroup$
– YCor
Mar 2 at 10:32
$begingroup$
Still there's a minor error in the corollary: it can happen that $G$ is already acting locally elliptically (horocyclic case). Side remark: in the corollary, no need to pass to a finite index subgroup, except precisely in the case $G$ preserves an axis, where one might have to pass to a subgroup of index 2.
$endgroup$
– YCor
Mar 2 at 10:34
$begingroup$
Still there's a minor error in the corollary: it can happen that $G$ is already acting locally elliptically (horocyclic case). Side remark: in the corollary, no need to pass to a finite index subgroup, except precisely in the case $G$ preserves an axis, where one might have to pass to a subgroup of index 2.
$endgroup$
– YCor
Mar 2 at 10:34
1
1
$begingroup$
I replaced "elliptic" with "$X$-elliptic" to avoid confusion. I also corrected the mistake in the corollary, thank you.
$endgroup$
– AGenevois
Mar 2 at 11:03
$begingroup$
I replaced "elliptic" with "$X$-elliptic" to avoid confusion. I also corrected the mistake in the corollary, thank you.
$endgroup$
– AGenevois
Mar 2 at 11:03
|
show 5 more comments
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$begingroup$
Properly implies finite stabilizers
$endgroup$
– YCor
Mar 1 at 23:09