Mixing
Why is this sequence of isometries contained in a compact set?
I'm reading through a proof of the Mostow rigidity theorem (pages 738-740) in https://www.math.ucdavis.edu/~kapovich/EPR/ggt.pdf
but I'm stuck at a certain part.
In Step 2, we consider the vertical geodesic $L subset mathbb{H}^n$ emanating from $0$ (here we are viewing $mathbb{H}^n$ in the upper half-space model), and we fix a base-point $y_0 in L$. We have that $0$ is a conical limit point of $Gamma$, which is a lattice in $text{Isom}(mathbb{H}^n)$. So there is a sequence of isometries $gamma_i in Gamma$ such that
$$lim_{itoinfty} gamma_i(y_0) = 0$$
and
$$text{dist}(gamma_i(y_0),L) le C$$
for some constant $C$. We let $y_i$ denote the nearest-point projection of $gamma_i(y_0)$ to $L$, and we define $T_i$ to be the hyperbolic translation $y mapsto t_iy$ along the axis $L$, such that $T_i(y_0) = y_i$. The author then proceeds to say that the sequence $k_i := gamma_i^{-1} circ T_i$ lies in a compact subset $C subset text{Isom}(mathbb{H}^n)$. However it is not clear to me why this is the case.
I am hoping there is an argument which doesn't directly appeal to the explicit topology of $text{Isom}(mathbb{H}^n)$ (which, by the way, is what?), but rather I am hoping that this will follow from a general statement about group actions.
Some observations I have (which may be irrelevant) are that
$left{k_i(y_0)right} subset B_C(y_0)$ (i.e. the orbit of $y_0$ under $k_i$ is bounded)
since $text{Isom}(mathbb{H}^n)$ acts transitively on $mathbb{H}^n$, we have that $mathbb{H}^n cong text{Isom}(mathbb{H}^n) / text{Stab}(y_0)$ (whether this isomorphism is more than a set bijection I'm not sure)
group-actions geometric-group-theory
asked Nov 20 '18 at 2:26
suchan
217110
add a comment |
I'm reading through a proof of the Mostow rigidity theorem (pages 738-740) in https://www.math.ucdavis.edu/~kapovich/EPR/ggt.pdf
but I'm stuck at a certain part.
In Step 2, we consider the vertical geodesic $L subset mathbb{H}^n$ emanating from $0$ (here we are viewing $mathbb{H}^n$ in the upper half-space model), and we fix a base-point $y_0 in L$. We have that $0$ is a conical limit point of $Gamma$, which is a lattice in $text{Isom}(mathbb{H}^n)$. So there is a sequence of isometries $gamma_i in Gamma$ such that
$$lim_{itoinfty} gamma_i(y_0) = 0$$
and
$$text{dist}(gamma_i(y_0),L) le C$$
for some constant $C$. We let $y_i$ denote the nearest-point projection of $gamma_i(y_0)$ to $L$, and we define $T_i$ to be the hyperbolic translation $y mapsto t_iy$ along the axis $L$, such that $T_i(y_0) = y_i$. The author then proceeds to say that the sequence $k_i := gamma_i^{-1} circ T_i$ lies in a compact subset $C subset text{Isom}(mathbb{H}^n)$. However it is not clear to me why this is the case.
I am hoping there is an argument which doesn't directly appeal to the explicit topology of $text{Isom}(mathbb{H}^n)$ (which, by the way, is what?), but rather I am hoping that this will follow from a general statement about group actions.
Some observations I have (which may be irrelevant) are that
$left{k_i(y_0)right} subset B_C(y_0)$ (i.e. the orbit of $y_0$ under $k_i$ is bounded)
since $text{Isom}(mathbb{H}^n)$ acts transitively on $mathbb{H}^n$, we have that $mathbb{H}^n cong text{Isom}(mathbb{H}^n) / text{Stab}(y_0)$ (whether this isomorphism is more than a set bijection I'm not sure)
group-actions geometric-group-theory
asked Nov 20 '18 at 2:26
suchan
217110
There are many ways to describe the topology of isometry groups(in terms of common topology in function spaces, lie groups, unit tangent bundle in this case) which you can look up. Without getting into details of a definition you basically have a proof though once you settle on a definition for the topology. Consider all isometries which move $y_0$ at most $C$, and any sequence of isometries with that property, show that under any reasonable definition you should find a subsequence which converges to an isometry which moves $y_0$ at most $C$.
– Paul Plummer
Nov 20 '18 at 6:17
In fact you can define a topology on the isometry group doing something similar
– Paul Plummer
Nov 20 '18 at 6:19
Using not the upper half space model but instead the hyperboloid model, $text{Isom}(mathbb H^n)$ is identified with the Lie group $SO(n,1)$, and the topology on it is simply the restriction of the standard topology on the space of $n times n$ matrices, which is identified with $mathbb R^{n^2}$. It is also the restriction of the compact open topology on the set of functions $mathbb H^n mapsto mathbb H^n$.
– Lee Mosher
Nov 24 '18 at 21:32
1
So, if you have a subset $A subset text{Isom}(mathbb H^n)$, and a point $p in mathbb H^n$, and if $A(p) = {alpha(p) mid alpha in A}$ is a bounded subset of $mathbb H^n$, then the closure of $A$ is a compact subset of $text{Isom}(mathbb H^n)$.
– Lee Mosher
Nov 24 '18 at 21:35
@LeeMosher Thanks for your comment. Is the intermediate argument that $A(p)$ being bounded implies $A$ is bounded? If we are viewing $text{Isom}(mathbb{H}^n)$ as a matrix group then I readily believe this to be true, but I'm failing to come up with a rigorous reason why there must be a uniform bound on the entries of each matrix in $A$.
– suchan
Nov 25 '18 at 3:46
add a comment |
I'm reading through a proof of the Mostow rigidity theorem (pages 738-740) in https://www.math.ucdavis.edu/~kapovich/EPR/ggt.pdf
but I'm stuck at a certain part.
In Step 2, we consider the vertical geodesic $L subset mathbb{H}^n$ emanating from $0$ (here we are viewing $mathbb{H}^n$ in the upper half-space model), and we fix a base-point $y_0 in L$. We have that $0$ is a conical limit point of $Gamma$, which is a lattice in $text{Isom}(mathbb{H}^n)$. So there is a sequence of isometries $gamma_i in Gamma$ such that
$$lim_{itoinfty} gamma_i(y_0) = 0$$
and
$$text{dist}(gamma_i(y_0),L) le C$$
for some constant $C$. We let $y_i$ denote the nearest-point projection of $gamma_i(y_0)$ to $L$, and we define $T_i$ to be the hyperbolic translation $y mapsto t_iy$ along the axis $L$, such that $T_i(y_0) = y_i$. The author then proceeds to say that the sequence $k_i := gamma_i^{-1} circ T_i$ lies in a compact subset $C subset text{Isom}(mathbb{H}^n)$. However it is not clear to me why this is the case.
I am hoping there is an argument which doesn't directly appeal to the explicit topology of $text{Isom}(mathbb{H}^n)$ (which, by the way, is what?), but rather I am hoping that this will follow from a general statement about group actions.
Some observations I have (which may be irrelevant) are that
$left{k_i(y_0)right} subset B_C(y_0)$ (i.e. the orbit of $y_0$ under $k_i$ is bounded)
since $text{Isom}(mathbb{H}^n)$ acts transitively on $mathbb{H}^n$, we have that $mathbb{H}^n cong text{Isom}(mathbb{H}^n) / text{Stab}(y_0)$ (whether this isomorphism is more than a set bijection I'm not sure)
group-actions geometric-group-theory
asked Nov 20 '18 at 2:26
suchan
217110
I'm reading through a proof of the Mostow rigidity theorem (pages 738-740) in https://www.math.ucdavis.edu/~kapovich/EPR/ggt.pdf
but I'm stuck at a certain part.
In Step 2, we consider the vertical geodesic $L subset mathbb{H}^n$ emanating from $0$ (here we are viewing $mathbb{H}^n$ in the upper half-space model), and we fix a base-point $y_0 in L$. We have that $0$ is a conical limit point of $Gamma$, which is a lattice in $text{Isom}(mathbb{H}^n)$. So there is a sequence of isometries $gamma_i in Gamma$ such that
$$lim_{itoinfty} gamma_i(y_0) = 0$$
and
$$text{dist}(gamma_i(y_0),L) le C$$
for some constant $C$. We let $y_i$ denote the nearest-point projection of $gamma_i(y_0)$ to $L$, and we define $T_i$ to be the hyperbolic translation $y mapsto t_iy$ along the axis $L$, such that $T_i(y_0) = y_i$. The author then proceeds to say that the sequence $k_i := gamma_i^{-1} circ T_i$ lies in a compact subset $C subset text{Isom}(mathbb{H}^n)$. However it is not clear to me why this is the case.
I am hoping there is an argument which doesn't directly appeal to the explicit topology of $text{Isom}(mathbb{H}^n)$ (which, by the way, is what?), but rather I am hoping that this will follow from a general statement about group actions.
Some observations I have (which may be irrelevant) are that
$left{k_i(y_0)right} subset B_C(y_0)$ (i.e. the orbit of $y_0$ under $k_i$ is bounded)
since $text{Isom}(mathbb{H}^n)$ acts transitively on $mathbb{H}^n$, we have that $mathbb{H}^n cong text{Isom}(mathbb{H}^n) / text{Stab}(y_0)$ (whether this isomorphism is more than a set bijection I'm not sure)
group-actions geometric-group-theory
group-actions geometric-group-theory
asked Nov 20 '18 at 2:26
suchan
217110
asked Nov 20 '18 at 2:26
suchan
217110
asked Nov 20 '18 at 2:26
suchan
217110
asked Nov 20 '18 at 2:26
suchan
217110
asked Nov 20 '18 at 2:26
suchan
217110
217110
There are many ways to describe the topology of isometry groups(in terms of common topology in function spaces, lie groups, unit tangent bundle in this case) which you can look up. Without getting into details of a definition you basically have a proof though once you settle on a definition for the topology. Consider all isometries which move $y_0$ at most $C$, and any sequence of isometries with that property, show that under any reasonable definition you should find a subsequence which converges to an isometry which moves $y_0$ at most $C$.
– Paul Plummer
Nov 20 '18 at 6:17
In fact you can define a topology on the isometry group doing something similar
– Paul Plummer
Nov 20 '18 at 6:19
Using not the upper half space model but instead the hyperboloid model, $text{Isom}(mathbb H^n)$ is identified with the Lie group $SO(n,1)$, and the topology on it is simply the restriction of the standard topology on the space of $n times n$ matrices, which is identified with $mathbb R^{n^2}$. It is also the restriction of the compact open topology on the set of functions $mathbb H^n mapsto mathbb H^n$.
– Lee Mosher
Nov 24 '18 at 21:32
1
So, if you have a subset $A subset text{Isom}(mathbb H^n)$, and a point $p in mathbb H^n$, and if $A(p) = {alpha(p) mid alpha in A}$ is a bounded subset of $mathbb H^n$, then the closure of $A$ is a compact subset of $text{Isom}(mathbb H^n)$.
– Lee Mosher
Nov 24 '18 at 21:35
@LeeMosher Thanks for your comment. Is the intermediate argument that $A(p)$ being bounded implies $A$ is bounded? If we are viewing $text{Isom}(mathbb{H}^n)$ as a matrix group then I readily believe this to be true, but I'm failing to come up with a rigorous reason why there must be a uniform bound on the entries of each matrix in $A$.
– suchan
Nov 25 '18 at 3:46
add a comment |
There are many ways to describe the topology of isometry groups(in terms of common topology in function spaces, lie groups, unit tangent bundle in this case) which you can look up. Without getting into details of a definition you basically have a proof though once you settle on a definition for the topology. Consider all isometries which move $y_0$ at most $C$, and any sequence of isometries with that property, show that under any reasonable definition you should find a subsequence which converges to an isometry which moves $y_0$ at most $C$.
– Paul Plummer
Nov 20 '18 at 6:17
In fact you can define a topology on the isometry group doing something similar
– Paul Plummer
Nov 20 '18 at 6:19
Using not the upper half space model but instead the hyperboloid model, $text{Isom}(mathbb H^n)$ is identified with the Lie group $SO(n,1)$, and the topology on it is simply the restriction of the standard topology on the space of $n times n$ matrices, which is identified with $mathbb R^{n^2}$. It is also the restriction of the compact open topology on the set of functions $mathbb H^n mapsto mathbb H^n$.
– Lee Mosher
Nov 24 '18 at 21:32
1
So, if you have a subset $A subset text{Isom}(mathbb H^n)$, and a point $p in mathbb H^n$, and if $A(p) = {alpha(p) mid alpha in A}$ is a bounded subset of $mathbb H^n$, then the closure of $A$ is a compact subset of $text{Isom}(mathbb H^n)$.
– Lee Mosher
Nov 24 '18 at 21:35
@LeeMosher Thanks for your comment. Is the intermediate argument that $A(p)$ being bounded implies $A$ is bounded? If we are viewing $text{Isom}(mathbb{H}^n)$ as a matrix group then I readily believe this to be true, but I'm failing to come up with a rigorous reason why there must be a uniform bound on the entries of each matrix in $A$.
– suchan
Nov 25 '18 at 3:46
There are many ways to describe the topology of isometry groups(in terms of common topology in function spaces, lie groups, unit tangent bundle in this case) which you can look up. Without getting into details of a definition you basically have a proof though once you settle on a definition for the topology. Consider all isometries which move $y_0$ at most $C$, and any sequence of isometries with that property, show that under any reasonable definition you should find a subsequence which converges to an isometry which moves $y_0$ at most $C$.
– Paul Plummer
Nov 20 '18 at 6:17
There are many ways to describe the topology of isometry groups(in terms of common topology in function spaces, lie groups, unit tangent bundle in this case) which you can look up. Without getting into details of a definition you basically have a proof though once you settle on a definition for the topology. Consider all isometries which move $y_0$ at most $C$, and any sequence of isometries with that property, show that under any reasonable definition you should find a subsequence which converges to an isometry which moves $y_0$ at most $C$.
– Paul Plummer
Nov 20 '18 at 6:17
In fact you can define a topology on the isometry group doing something similar
– Paul Plummer
Nov 20 '18 at 6:19
In fact you can define a topology on the isometry group doing something similar
– Paul Plummer
Nov 20 '18 at 6:19
Using not the upper half space model but instead the hyperboloid model, $text{Isom}(mathbb H^n)$ is identified with the Lie group $SO(n,1)$, and the topology on it is simply the restriction of the standard topology on the space of $n times n$ matrices, which is identified with $mathbb R^{n^2}$. It is also the restriction of the compact open topology on the set of functions $mathbb H^n mapsto mathbb H^n$.
– Lee Mosher
Nov 24 '18 at 21:32
Using not the upper half space model but instead the hyperboloid model, $text{Isom}(mathbb H^n)$ is identified with the Lie group $SO(n,1)$, and the topology on it is simply the restriction of the standard topology on the space of $n times n$ matrices, which is identified with $mathbb R^{n^2}$. It is also the restriction of the compact open topology on the set of functions $mathbb H^n mapsto mathbb H^n$.
– Lee Mosher
Nov 24 '18 at 21:32
1
1
So, if you have a subset $A subset text{Isom}(mathbb H^n)$, and a point $p in mathbb H^n$, and if $A(p) = {alpha(p) mid alpha in A}$ is a bounded subset of $mathbb H^n$, then the closure of $A$ is a compact subset of $text{Isom}(mathbb H^n)$.
– Lee Mosher
Nov 24 '18 at 21:35
So, if you have a subset $A subset text{Isom}(mathbb H^n)$, and a point $p in mathbb H^n$, and if $A(p) = {alpha(p) mid alpha in A}$ is a bounded subset of $mathbb H^n$, then the closure of $A$ is a compact subset of $text{Isom}(mathbb H^n)$.
– Lee Mosher
Nov 24 '18 at 21:35
@LeeMosher Thanks for your comment. Is the intermediate argument that $A(p)$ being bounded implies $A$ is bounded? If we are viewing $text{Isom}(mathbb{H}^n)$ as a matrix group then I readily believe this to be true, but I'm failing to come up with a rigorous reason why there must be a uniform bound on the entries of each matrix in $A$.
– suchan
Nov 25 '18 at 3:46
@LeeMosher Thanks for your comment. Is the intermediate argument that $A(p)$ being bounded implies $A$ is bounded? If we are viewing $text{Isom}(mathbb{H}^n)$ as a matrix group then I readily believe this to be true, but I'm failing to come up with a rigorous reason why there must be a uniform bound on the entries of each matrix in $A$.
– suchan
Nov 25 '18 at 3:46
add a comment |
1 Answer
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Let me write an answer under the assumption (which is implicit in your question) that you have some particular scheme for expressing $text{Isom}(mathbb H^n)$ as a matrix group, but I'll also say exactly what to do in the case $n=2$.
You have already observed that the orbit of $y_0$ under $k_i$ is bounded, in that $k_i$ moves the point $y_0$ a distance $le C$. You can therefore write each $k_i$ as a product of matrices the form
$$k_i = P_i Q_i R_i
$$
where each of $P_i$ and $R_i$ is a matrix that fixes $y_0$ and where $Q_i$ is a matrix which translates $y_0$ upward on the axis $A$ by a distance at most $C$. So you just need to convince yourself of three things:
- The set of matrices fixing $y_0$ has bounded entries;
- The set of matrices which translates $y$ upward along $A$ by distance at most $C$ has bounded entries;
- If one has a bound on the entries of three matrices $P,Q,R$ (of a fixed size) then one can derive a bound on the entries of their product $PQR$.
For an example of the last point, if the matrices are $n times n$ and the entries of $P,Q,R$ are $le b$ in absolute value then a bound on the absolute values of the entries of the matrix $PQR$ is $n^2b^3$.
For the first two points, let me describe exactly what happens in the special case $n=2$, working in the upper half plane model with orientation preserving isometry group $text{PSL}(2,mathbb R)$ (which is probably not the matrices that you are working with, but it's the matrices I know best):
- The set of matrices fixing $(0,1)$ is simply the compact set of rotation matrices
$$SO(2,mathbb R) = begin{pmatrix} cos theta & sin theta \ - sintheta & costheta end{pmatrix} quad text{where $0 le theta le 2pi$}
$$
which is obviously compact - The set of matrices which translates $(0,1)$ upward along the line $x=0$ a distance at most $C$ is simply the compact set of translation matrices
$$begin{pmatrix} e^{c/2} & 0 \ 0 & e^{-c/2} end{pmatrix} quadtext{where $0 le c le C$}
$$
answered Nov 26 '18 at 17:41
Lee Mosher
48.1k33681
add a comment |
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Let me write an answer under the assumption (which is implicit in your question) that you have some particular scheme for expressing $text{Isom}(mathbb H^n)$ as a matrix group, but I'll also say exactly what to do in the case $n=2$.
You have already observed that the orbit of $y_0$ under $k_i$ is bounded, in that $k_i$ moves the point $y_0$ a distance $le C$. You can therefore write each $k_i$ as a product of matrices the form
$$k_i = P_i Q_i R_i
$$
where each of $P_i$ and $R_i$ is a matrix that fixes $y_0$ and where $Q_i$ is a matrix which translates $y_0$ upward on the axis $A$ by a distance at most $C$. So you just need to convince yourself of three things:
- The set of matrices fixing $y_0$ has bounded entries;
- The set of matrices which translates $y$ upward along $A$ by distance at most $C$ has bounded entries;
- If one has a bound on the entries of three matrices $P,Q,R$ (of a fixed size) then one can derive a bound on the entries of their product $PQR$.
For an example of the last point, if the matrices are $n times n$ and the entries of $P,Q,R$ are $le b$ in absolute value then a bound on the absolute values of the entries of the matrix $PQR$ is $n^2b^3$.
For the first two points, let me describe exactly what happens in the special case $n=2$, working in the upper half plane model with orientation preserving isometry group $text{PSL}(2,mathbb R)$ (which is probably not the matrices that you are working with, but it's the matrices I know best):
- The set of matrices fixing $(0,1)$ is simply the compact set of rotation matrices
$$SO(2,mathbb R) = begin{pmatrix} cos theta & sin theta \ - sintheta & costheta end{pmatrix} quad text{where $0 le theta le 2pi$}
$$
which is obviously compact - The set of matrices which translates $(0,1)$ upward along the line $x=0$ a distance at most $C$ is simply the compact set of translation matrices
$$begin{pmatrix} e^{c/2} & 0 \ 0 & e^{-c/2} end{pmatrix} quadtext{where $0 le c le C$}
$$
answered Nov 26 '18 at 17:41
Lee Mosher
48.1k33681
add a comment |
Let me write an answer under the assumption (which is implicit in your question) that you have some particular scheme for expressing $text{Isom}(mathbb H^n)$ as a matrix group, but I'll also say exactly what to do in the case $n=2$.
You have already observed that the orbit of $y_0$ under $k_i$ is bounded, in that $k_i$ moves the point $y_0$ a distance $le C$. You can therefore write each $k_i$ as a product of matrices the form
$$k_i = P_i Q_i R_i
$$
where each of $P_i$ and $R_i$ is a matrix that fixes $y_0$ and where $Q_i$ is a matrix which translates $y_0$ upward on the axis $A$ by a distance at most $C$. So you just need to convince yourself of three things:
- The set of matrices fixing $y_0$ has bounded entries;
- The set of matrices which translates $y$ upward along $A$ by distance at most $C$ has bounded entries;
- If one has a bound on the entries of three matrices $P,Q,R$ (of a fixed size) then one can derive a bound on the entries of their product $PQR$.
For an example of the last point, if the matrices are $n times n$ and the entries of $P,Q,R$ are $le b$ in absolute value then a bound on the absolute values of the entries of the matrix $PQR$ is $n^2b^3$.
For the first two points, let me describe exactly what happens in the special case $n=2$, working in the upper half plane model with orientation preserving isometry group $text{PSL}(2,mathbb R)$ (which is probably not the matrices that you are working with, but it's the matrices I know best):
- The set of matrices fixing $(0,1)$ is simply the compact set of rotation matrices
$$SO(2,mathbb R) = begin{pmatrix} cos theta & sin theta \ - sintheta & costheta end{pmatrix} quad text{where $0 le theta le 2pi$}
$$
which is obviously compact - The set of matrices which translates $(0,1)$ upward along the line $x=0$ a distance at most $C$ is simply the compact set of translation matrices
$$begin{pmatrix} e^{c/2} & 0 \ 0 & e^{-c/2} end{pmatrix} quadtext{where $0 le c le C$}
$$
answered Nov 26 '18 at 17:41
Lee Mosher
48.1k33681
add a comment |
Let me write an answer under the assumption (which is implicit in your question) that you have some particular scheme for expressing $text{Isom}(mathbb H^n)$ as a matrix group, but I'll also say exactly what to do in the case $n=2$.
You have already observed that the orbit of $y_0$ under $k_i$ is bounded, in that $k_i$ moves the point $y_0$ a distance $le C$. You can therefore write each $k_i$ as a product of matrices the form
$$k_i = P_i Q_i R_i
$$
where each of $P_i$ and $R_i$ is a matrix that fixes $y_0$ and where $Q_i$ is a matrix which translates $y_0$ upward on the axis $A$ by a distance at most $C$. So you just need to convince yourself of three things:
- The set of matrices fixing $y_0$ has bounded entries;
- The set of matrices which translates $y$ upward along $A$ by distance at most $C$ has bounded entries;
- If one has a bound on the entries of three matrices $P,Q,R$ (of a fixed size) then one can derive a bound on the entries of their product $PQR$.
For an example of the last point, if the matrices are $n times n$ and the entries of $P,Q,R$ are $le b$ in absolute value then a bound on the absolute values of the entries of the matrix $PQR$ is $n^2b^3$.
For the first two points, let me describe exactly what happens in the special case $n=2$, working in the upper half plane model with orientation preserving isometry group $text{PSL}(2,mathbb R)$ (which is probably not the matrices that you are working with, but it's the matrices I know best):
- The set of matrices fixing $(0,1)$ is simply the compact set of rotation matrices
$$SO(2,mathbb R) = begin{pmatrix} cos theta & sin theta \ - sintheta & costheta end{pmatrix} quad text{where $0 le theta le 2pi$}
$$
which is obviously compact - The set of matrices which translates $(0,1)$ upward along the line $x=0$ a distance at most $C$ is simply the compact set of translation matrices
$$begin{pmatrix} e^{c/2} & 0 \ 0 & e^{-c/2} end{pmatrix} quadtext{where $0 le c le C$}
$$
answered Nov 26 '18 at 17:41
Lee Mosher
48.1k33681
Let me write an answer under the assumption (which is implicit in your question) that you have some particular scheme for expressing $text{Isom}(mathbb H^n)$ as a matrix group, but I'll also say exactly what to do in the case $n=2$.
You have already observed that the orbit of $y_0$ under $k_i$ is bounded, in that $k_i$ moves the point $y_0$ a distance $le C$. You can therefore write each $k_i$ as a product of matrices the form
$$k_i = P_i Q_i R_i
$$
where each of $P_i$ and $R_i$ is a matrix that fixes $y_0$ and where $Q_i$ is a matrix which translates $y_0$ upward on the axis $A$ by a distance at most $C$. So you just need to convince yourself of three things:
- The set of matrices fixing $y_0$ has bounded entries;
- The set of matrices which translates $y$ upward along $A$ by distance at most $C$ has bounded entries;
- If one has a bound on the entries of three matrices $P,Q,R$ (of a fixed size) then one can derive a bound on the entries of their product $PQR$.
For an example of the last point, if the matrices are $n times n$ and the entries of $P,Q,R$ are $le b$ in absolute value then a bound on the absolute values of the entries of the matrix $PQR$ is $n^2b^3$.
For the first two points, let me describe exactly what happens in the special case $n=2$, working in the upper half plane model with orientation preserving isometry group $text{PSL}(2,mathbb R)$ (which is probably not the matrices that you are working with, but it's the matrices I know best):
- The set of matrices fixing $(0,1)$ is simply the compact set of rotation matrices
$$SO(2,mathbb R) = begin{pmatrix} cos theta & sin theta \ - sintheta & costheta end{pmatrix} quad text{where $0 le theta le 2pi$}
$$
which is obviously compact - The set of matrices which translates $(0,1)$ upward along the line $x=0$ a distance at most $C$ is simply the compact set of translation matrices
$$begin{pmatrix} e^{c/2} & 0 \ 0 & e^{-c/2} end{pmatrix} quadtext{where $0 le c le C$}
$$
answered Nov 26 '18 at 17:41
Lee Mosher
48.1k33681
answered Nov 26 '18 at 17:41
Lee Mosher
48.1k33681
answered Nov 26 '18 at 17:41
Lee Mosher
48.1k33681
answered Nov 26 '18 at 17:41
Lee Mosher
48.1k33681
48.1k33681
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There are many ways to describe the topology of isometry groups(in terms of common topology in function spaces, lie groups, unit tangent bundle in this case) which you can look up. Without getting into details of a definition you basically have a proof though once you settle on a definition for the topology. Consider all isometries which move $y_0$ at most $C$, and any sequence of isometries with that property, show that under any reasonable definition you should find a subsequence which converges to an isometry which moves $y_0$ at most $C$.
– Paul Plummer
Nov 20 '18 at 6:17
In fact you can define a topology on the isometry group doing something similar
– Paul Plummer
Nov 20 '18 at 6:19
Using not the upper half space model but instead the hyperboloid model, $text{Isom}(mathbb H^n)$ is identified with the Lie group $SO(n,1)$, and the topology on it is simply the restriction of the standard topology on the space of $n times n$ matrices, which is identified with $mathbb R^{n^2}$. It is also the restriction of the compact open topology on the set of functions $mathbb H^n mapsto mathbb H^n$.
– Lee Mosher
Nov 24 '18 at 21:32
1
So, if you have a subset $A subset text{Isom}(mathbb H^n)$, and a point $p in mathbb H^n$, and if $A(p) = {alpha(p) mid alpha in A}$ is a bounded subset of $mathbb H^n$, then the closure of $A$ is a compact subset of $text{Isom}(mathbb H^n)$.
– Lee Mosher
Nov 24 '18 at 21:35
@LeeMosher Thanks for your comment. Is the intermediate argument that $A(p)$ being bounded implies $A$ is bounded? If we are viewing $text{Isom}(mathbb{H}^n)$ as a matrix group then I readily believe this to be true, but I'm failing to come up with a rigorous reason why there must be a uniform bound on the entries of each matrix in $A$.
– suchan
Nov 25 '18 at 3:46