Why is the dual tree to a measured geodesic lamination in a compact hyperbolic surface not complete?
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Let $M$ be a closed connected surface and $mathcal{F}$ a minimal (every leaf is dense) measured foliation (as, for example, in Thurston's work on surfaces) on $M$. Let $tilde{M}$ be the universal cover of $M$ and $tilde{mathcal{F}}$ the pullback of $mathcal{F}$ to $tilde{M}$.
Consider the leaf space of $tilde{mathcal{F}}$, $mathcal{T}=tilde{M}/tilde{mathcal{F}}$. Define the distance between two leaves of $tilde{mathcal{F}}$ as the minimum of the transverse measures of arcs joining the two leaves, we obtain a distance on $mathcal{T}$ which turns $mathcal{T}$ into a tree.
I have seen the following statement.
If the genus of $M$ is $geq 2$, $mathcal{T}$ is not complete for this distance.
Why is this true?
I also saw a notion "dual tree", which also shows up from time to time when I tried to search for related topics, from the book Hyperbolic manifolds and Discrete Groups (for geodesic laminations there). Is this indeed the concept that I am looking for?
Thanks in advance!
dynamical-systems hyperbolic-geometry geometric-topology
add a comment |
up vote
3
down vote
favorite
Let $M$ be a closed connected surface and $mathcal{F}$ a minimal (every leaf is dense) measured foliation (as, for example, in Thurston's work on surfaces) on $M$. Let $tilde{M}$ be the universal cover of $M$ and $tilde{mathcal{F}}$ the pullback of $mathcal{F}$ to $tilde{M}$.
Consider the leaf space of $tilde{mathcal{F}}$, $mathcal{T}=tilde{M}/tilde{mathcal{F}}$. Define the distance between two leaves of $tilde{mathcal{F}}$ as the minimum of the transverse measures of arcs joining the two leaves, we obtain a distance on $mathcal{T}$ which turns $mathcal{T}$ into a tree.
I have seen the following statement.
If the genus of $M$ is $geq 2$, $mathcal{T}$ is not complete for this distance.
Why is this true?
I also saw a notion "dual tree", which also shows up from time to time when I tried to search for related topics, from the book Hyperbolic manifolds and Discrete Groups (for geodesic laminations there). Is this indeed the concept that I am looking for?
Thanks in advance!
dynamical-systems hyperbolic-geometry geometric-topology
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $M$ be a closed connected surface and $mathcal{F}$ a minimal (every leaf is dense) measured foliation (as, for example, in Thurston's work on surfaces) on $M$. Let $tilde{M}$ be the universal cover of $M$ and $tilde{mathcal{F}}$ the pullback of $mathcal{F}$ to $tilde{M}$.
Consider the leaf space of $tilde{mathcal{F}}$, $mathcal{T}=tilde{M}/tilde{mathcal{F}}$. Define the distance between two leaves of $tilde{mathcal{F}}$ as the minimum of the transverse measures of arcs joining the two leaves, we obtain a distance on $mathcal{T}$ which turns $mathcal{T}$ into a tree.
I have seen the following statement.
If the genus of $M$ is $geq 2$, $mathcal{T}$ is not complete for this distance.
Why is this true?
I also saw a notion "dual tree", which also shows up from time to time when I tried to search for related topics, from the book Hyperbolic manifolds and Discrete Groups (for geodesic laminations there). Is this indeed the concept that I am looking for?
Thanks in advance!
dynamical-systems hyperbolic-geometry geometric-topology
Let $M$ be a closed connected surface and $mathcal{F}$ a minimal (every leaf is dense) measured foliation (as, for example, in Thurston's work on surfaces) on $M$. Let $tilde{M}$ be the universal cover of $M$ and $tilde{mathcal{F}}$ the pullback of $mathcal{F}$ to $tilde{M}$.
Consider the leaf space of $tilde{mathcal{F}}$, $mathcal{T}=tilde{M}/tilde{mathcal{F}}$. Define the distance between two leaves of $tilde{mathcal{F}}$ as the minimum of the transverse measures of arcs joining the two leaves, we obtain a distance on $mathcal{T}$ which turns $mathcal{T}$ into a tree.
I have seen the following statement.
If the genus of $M$ is $geq 2$, $mathcal{T}$ is not complete for this distance.
Why is this true?
I also saw a notion "dual tree", which also shows up from time to time when I tried to search for related topics, from the book Hyperbolic manifolds and Discrete Groups (for geodesic laminations there). Is this indeed the concept that I am looking for?
Thanks in advance!
dynamical-systems hyperbolic-geometry geometric-topology
dynamical-systems hyperbolic-geometry geometric-topology
edited Nov 23 at 8:04
asked Nov 23 at 7:46
chikurin
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Here's a rough description of why $mathcal T$ is incomplete.
Start at some point $x_0$ of $tilde M$. Let $ell_0$ be a lonnnng leaf segment of $tilde{mathcal F}$ with one endpoint on $x_0$, and let $y_0$ be the opposite endpoint. Let $tau_0$ be a transverse arc of transverse measure $2^{-0}=1$ with one endpoint at $y_0$. Let $x_1$ be the opposite endpoint. Continue by induction: assuming $x_n$ has been defined, let $ell_n$ be a lonnnng leaf segment with one endpoint on $x_n$, let $y_n$ be the opposite endpoint, let $tau_n$ be a transverse arc of transverse measure $2^{-n}$, and let $x_{n+1}$ be the opposite endpoint. You get an infinite path of the form
$$ell_0 tau_0 ell_1 tau_1 ell_2 tau_2 cdots
$$
The total transverse measure of the initial segment $ell_0 tau_0 cdots ell_n tau_n$ that connects $x_0$ to $x_n$ is equal to $2-2^{-n}$, and this converges to $2$. By careful choice of the $tau$'s in the induction step you can guarantee that this initial segment is "quasitransverse", implying that $2-2^{-n}$ is the minimum of the transverse measure of any arc from $x_0$ to $x_n$. So, in the dual tree $mathcal T$, you get an isometric embedding $[0,2) mapsto mathcal T$. But, this embedding has no point to converge to as the parameter in $[0,2)$ approaches $2$; in fact, one can check that this infinite path escapes all compact subsets of $tilde M$. So $mathcal T$ is incomplete.
By the way, another place I know, where the dual tree is explained very clearly, is in Otal's book "Le théorème d'hyperbolisation pour les variétés fibrées de dimension 3", in the chapter on Skora's theorem.
I think I can understand what you are saying, but what I am confused about is, after taking quotient, isn't a lonnng leaf got identified to a single point in the tree?
– chikurin
Nov 24 at 16:14
1
That's exactly correct. And therefore the infinite concatenation $ell_0tau_0ell_1tau_1ell_2tau_2cdots$, after passing to the quotient $mathcal T$, could (by abusing notation) be written as $tau_0 tau_1 tau_2cdots$, which is the image of the isometric embedding $[0,2) mapsto mathcal T$.
– Lee Mosher
Nov 24 at 16:21
Ohhh, I now see... Thanks a lot!!
– chikurin
Nov 25 at 6:18
Another thing, I tried to read the book "The Hyperbolization Theorem for Fibered 3-Manifolds" with my lame French for an hour and then found an English translation that day, in case there is anyone who does not read French and is not aware of the existence of an English translation. :)
– chikurin
2 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Here's a rough description of why $mathcal T$ is incomplete.
Start at some point $x_0$ of $tilde M$. Let $ell_0$ be a lonnnng leaf segment of $tilde{mathcal F}$ with one endpoint on $x_0$, and let $y_0$ be the opposite endpoint. Let $tau_0$ be a transverse arc of transverse measure $2^{-0}=1$ with one endpoint at $y_0$. Let $x_1$ be the opposite endpoint. Continue by induction: assuming $x_n$ has been defined, let $ell_n$ be a lonnnng leaf segment with one endpoint on $x_n$, let $y_n$ be the opposite endpoint, let $tau_n$ be a transverse arc of transverse measure $2^{-n}$, and let $x_{n+1}$ be the opposite endpoint. You get an infinite path of the form
$$ell_0 tau_0 ell_1 tau_1 ell_2 tau_2 cdots
$$
The total transverse measure of the initial segment $ell_0 tau_0 cdots ell_n tau_n$ that connects $x_0$ to $x_n$ is equal to $2-2^{-n}$, and this converges to $2$. By careful choice of the $tau$'s in the induction step you can guarantee that this initial segment is "quasitransverse", implying that $2-2^{-n}$ is the minimum of the transverse measure of any arc from $x_0$ to $x_n$. So, in the dual tree $mathcal T$, you get an isometric embedding $[0,2) mapsto mathcal T$. But, this embedding has no point to converge to as the parameter in $[0,2)$ approaches $2$; in fact, one can check that this infinite path escapes all compact subsets of $tilde M$. So $mathcal T$ is incomplete.
By the way, another place I know, where the dual tree is explained very clearly, is in Otal's book "Le théorème d'hyperbolisation pour les variétés fibrées de dimension 3", in the chapter on Skora's theorem.
I think I can understand what you are saying, but what I am confused about is, after taking quotient, isn't a lonnng leaf got identified to a single point in the tree?
– chikurin
Nov 24 at 16:14
1
That's exactly correct. And therefore the infinite concatenation $ell_0tau_0ell_1tau_1ell_2tau_2cdots$, after passing to the quotient $mathcal T$, could (by abusing notation) be written as $tau_0 tau_1 tau_2cdots$, which is the image of the isometric embedding $[0,2) mapsto mathcal T$.
– Lee Mosher
Nov 24 at 16:21
Ohhh, I now see... Thanks a lot!!
– chikurin
Nov 25 at 6:18
Another thing, I tried to read the book "The Hyperbolization Theorem for Fibered 3-Manifolds" with my lame French for an hour and then found an English translation that day, in case there is anyone who does not read French and is not aware of the existence of an English translation. :)
– chikurin
2 hours ago
add a comment |
up vote
2
down vote
accepted
Here's a rough description of why $mathcal T$ is incomplete.
Start at some point $x_0$ of $tilde M$. Let $ell_0$ be a lonnnng leaf segment of $tilde{mathcal F}$ with one endpoint on $x_0$, and let $y_0$ be the opposite endpoint. Let $tau_0$ be a transverse arc of transverse measure $2^{-0}=1$ with one endpoint at $y_0$. Let $x_1$ be the opposite endpoint. Continue by induction: assuming $x_n$ has been defined, let $ell_n$ be a lonnnng leaf segment with one endpoint on $x_n$, let $y_n$ be the opposite endpoint, let $tau_n$ be a transverse arc of transverse measure $2^{-n}$, and let $x_{n+1}$ be the opposite endpoint. You get an infinite path of the form
$$ell_0 tau_0 ell_1 tau_1 ell_2 tau_2 cdots
$$
The total transverse measure of the initial segment $ell_0 tau_0 cdots ell_n tau_n$ that connects $x_0$ to $x_n$ is equal to $2-2^{-n}$, and this converges to $2$. By careful choice of the $tau$'s in the induction step you can guarantee that this initial segment is "quasitransverse", implying that $2-2^{-n}$ is the minimum of the transverse measure of any arc from $x_0$ to $x_n$. So, in the dual tree $mathcal T$, you get an isometric embedding $[0,2) mapsto mathcal T$. But, this embedding has no point to converge to as the parameter in $[0,2)$ approaches $2$; in fact, one can check that this infinite path escapes all compact subsets of $tilde M$. So $mathcal T$ is incomplete.
By the way, another place I know, where the dual tree is explained very clearly, is in Otal's book "Le théorème d'hyperbolisation pour les variétés fibrées de dimension 3", in the chapter on Skora's theorem.
I think I can understand what you are saying, but what I am confused about is, after taking quotient, isn't a lonnng leaf got identified to a single point in the tree?
– chikurin
Nov 24 at 16:14
1
That's exactly correct. And therefore the infinite concatenation $ell_0tau_0ell_1tau_1ell_2tau_2cdots$, after passing to the quotient $mathcal T$, could (by abusing notation) be written as $tau_0 tau_1 tau_2cdots$, which is the image of the isometric embedding $[0,2) mapsto mathcal T$.
– Lee Mosher
Nov 24 at 16:21
Ohhh, I now see... Thanks a lot!!
– chikurin
Nov 25 at 6:18
Another thing, I tried to read the book "The Hyperbolization Theorem for Fibered 3-Manifolds" with my lame French for an hour and then found an English translation that day, in case there is anyone who does not read French and is not aware of the existence of an English translation. :)
– chikurin
2 hours ago
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Here's a rough description of why $mathcal T$ is incomplete.
Start at some point $x_0$ of $tilde M$. Let $ell_0$ be a lonnnng leaf segment of $tilde{mathcal F}$ with one endpoint on $x_0$, and let $y_0$ be the opposite endpoint. Let $tau_0$ be a transverse arc of transverse measure $2^{-0}=1$ with one endpoint at $y_0$. Let $x_1$ be the opposite endpoint. Continue by induction: assuming $x_n$ has been defined, let $ell_n$ be a lonnnng leaf segment with one endpoint on $x_n$, let $y_n$ be the opposite endpoint, let $tau_n$ be a transverse arc of transverse measure $2^{-n}$, and let $x_{n+1}$ be the opposite endpoint. You get an infinite path of the form
$$ell_0 tau_0 ell_1 tau_1 ell_2 tau_2 cdots
$$
The total transverse measure of the initial segment $ell_0 tau_0 cdots ell_n tau_n$ that connects $x_0$ to $x_n$ is equal to $2-2^{-n}$, and this converges to $2$. By careful choice of the $tau$'s in the induction step you can guarantee that this initial segment is "quasitransverse", implying that $2-2^{-n}$ is the minimum of the transverse measure of any arc from $x_0$ to $x_n$. So, in the dual tree $mathcal T$, you get an isometric embedding $[0,2) mapsto mathcal T$. But, this embedding has no point to converge to as the parameter in $[0,2)$ approaches $2$; in fact, one can check that this infinite path escapes all compact subsets of $tilde M$. So $mathcal T$ is incomplete.
By the way, another place I know, where the dual tree is explained very clearly, is in Otal's book "Le théorème d'hyperbolisation pour les variétés fibrées de dimension 3", in the chapter on Skora's theorem.
Here's a rough description of why $mathcal T$ is incomplete.
Start at some point $x_0$ of $tilde M$. Let $ell_0$ be a lonnnng leaf segment of $tilde{mathcal F}$ with one endpoint on $x_0$, and let $y_0$ be the opposite endpoint. Let $tau_0$ be a transverse arc of transverse measure $2^{-0}=1$ with one endpoint at $y_0$. Let $x_1$ be the opposite endpoint. Continue by induction: assuming $x_n$ has been defined, let $ell_n$ be a lonnnng leaf segment with one endpoint on $x_n$, let $y_n$ be the opposite endpoint, let $tau_n$ be a transverse arc of transverse measure $2^{-n}$, and let $x_{n+1}$ be the opposite endpoint. You get an infinite path of the form
$$ell_0 tau_0 ell_1 tau_1 ell_2 tau_2 cdots
$$
The total transverse measure of the initial segment $ell_0 tau_0 cdots ell_n tau_n$ that connects $x_0$ to $x_n$ is equal to $2-2^{-n}$, and this converges to $2$. By careful choice of the $tau$'s in the induction step you can guarantee that this initial segment is "quasitransverse", implying that $2-2^{-n}$ is the minimum of the transverse measure of any arc from $x_0$ to $x_n$. So, in the dual tree $mathcal T$, you get an isometric embedding $[0,2) mapsto mathcal T$. But, this embedding has no point to converge to as the parameter in $[0,2)$ approaches $2$; in fact, one can check that this infinite path escapes all compact subsets of $tilde M$. So $mathcal T$ is incomplete.
By the way, another place I know, where the dual tree is explained very clearly, is in Otal's book "Le théorème d'hyperbolisation pour les variétés fibrées de dimension 3", in the chapter on Skora's theorem.
edited 5 hours ago
answered Nov 23 at 16:37
Lee Mosher
47.4k33681
47.4k33681
I think I can understand what you are saying, but what I am confused about is, after taking quotient, isn't a lonnng leaf got identified to a single point in the tree?
– chikurin
Nov 24 at 16:14
1
That's exactly correct. And therefore the infinite concatenation $ell_0tau_0ell_1tau_1ell_2tau_2cdots$, after passing to the quotient $mathcal T$, could (by abusing notation) be written as $tau_0 tau_1 tau_2cdots$, which is the image of the isometric embedding $[0,2) mapsto mathcal T$.
– Lee Mosher
Nov 24 at 16:21
Ohhh, I now see... Thanks a lot!!
– chikurin
Nov 25 at 6:18
Another thing, I tried to read the book "The Hyperbolization Theorem for Fibered 3-Manifolds" with my lame French for an hour and then found an English translation that day, in case there is anyone who does not read French and is not aware of the existence of an English translation. :)
– chikurin
2 hours ago
add a comment |
I think I can understand what you are saying, but what I am confused about is, after taking quotient, isn't a lonnng leaf got identified to a single point in the tree?
– chikurin
Nov 24 at 16:14
1
That's exactly correct. And therefore the infinite concatenation $ell_0tau_0ell_1tau_1ell_2tau_2cdots$, after passing to the quotient $mathcal T$, could (by abusing notation) be written as $tau_0 tau_1 tau_2cdots$, which is the image of the isometric embedding $[0,2) mapsto mathcal T$.
– Lee Mosher
Nov 24 at 16:21
Ohhh, I now see... Thanks a lot!!
– chikurin
Nov 25 at 6:18
Another thing, I tried to read the book "The Hyperbolization Theorem for Fibered 3-Manifolds" with my lame French for an hour and then found an English translation that day, in case there is anyone who does not read French and is not aware of the existence of an English translation. :)
– chikurin
2 hours ago
I think I can understand what you are saying, but what I am confused about is, after taking quotient, isn't a lonnng leaf got identified to a single point in the tree?
– chikurin
Nov 24 at 16:14
I think I can understand what you are saying, but what I am confused about is, after taking quotient, isn't a lonnng leaf got identified to a single point in the tree?
– chikurin
Nov 24 at 16:14
1
1
That's exactly correct. And therefore the infinite concatenation $ell_0tau_0ell_1tau_1ell_2tau_2cdots$, after passing to the quotient $mathcal T$, could (by abusing notation) be written as $tau_0 tau_1 tau_2cdots$, which is the image of the isometric embedding $[0,2) mapsto mathcal T$.
– Lee Mosher
Nov 24 at 16:21
That's exactly correct. And therefore the infinite concatenation $ell_0tau_0ell_1tau_1ell_2tau_2cdots$, after passing to the quotient $mathcal T$, could (by abusing notation) be written as $tau_0 tau_1 tau_2cdots$, which is the image of the isometric embedding $[0,2) mapsto mathcal T$.
– Lee Mosher
Nov 24 at 16:21
Ohhh, I now see... Thanks a lot!!
– chikurin
Nov 25 at 6:18
Ohhh, I now see... Thanks a lot!!
– chikurin
Nov 25 at 6:18
Another thing, I tried to read the book "The Hyperbolization Theorem for Fibered 3-Manifolds" with my lame French for an hour and then found an English translation that day, in case there is anyone who does not read French and is not aware of the existence of an English translation. :)
– chikurin
2 hours ago
Another thing, I tried to read the book "The Hyperbolization Theorem for Fibered 3-Manifolds" with my lame French for an hour and then found an English translation that day, in case there is anyone who does not read French and is not aware of the existence of an English translation. :)
– chikurin
2 hours ago
add a comment |
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