Minimal measured foliation
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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 $tilde{M}$.
I saw a claim saying "every regular leaf (a leaf in $tilde{mathcal{F}}$ that does not contain a separatrix or the union of a singularity and all separatrices ending in that singularity) is a closed subset of $tilde{M}$".
I could not think of a proof of this claim right away or find a place where this claim is proved neither. Just wondering if this is generally true, or if I missed some conditions. How about in higher dimensions? We do not have measured foliations in higher dimensions, but is the pullback of a leaf of a minimal foliation to the universal cover always a closed subset of $tilde{M}$?
Thanks in advance!
general-topology dynamical-systems geometric-topology
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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 $tilde{M}$.
I saw a claim saying "every regular leaf (a leaf in $tilde{mathcal{F}}$ that does not contain a separatrix or the union of a singularity and all separatrices ending in that singularity) is a closed subset of $tilde{M}$".
I could not think of a proof of this claim right away or find a place where this claim is proved neither. Just wondering if this is generally true, or if I missed some conditions. How about in higher dimensions? We do not have measured foliations in higher dimensions, but is the pullback of a leaf of a minimal foliation to the universal cover always a closed subset of $tilde{M}$?
Thanks in advance!
general-topology dynamical-systems geometric-topology
1
By the way, I see that you have a few questions recently that received good answers. One way this site works is that if you get one (or more) good answers to one of your questions, then you choose an answer and you accept it. This has many good benefits for the growth of math.stackexchange as a whole.
– Lee Mosher
Nov 25 at 15:25
add a comment |
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1
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favorite
up vote
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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 $tilde{M}$.
I saw a claim saying "every regular leaf (a leaf in $tilde{mathcal{F}}$ that does not contain a separatrix or the union of a singularity and all separatrices ending in that singularity) is a closed subset of $tilde{M}$".
I could not think of a proof of this claim right away or find a place where this claim is proved neither. Just wondering if this is generally true, or if I missed some conditions. How about in higher dimensions? We do not have measured foliations in higher dimensions, but is the pullback of a leaf of a minimal foliation to the universal cover always a closed subset of $tilde{M}$?
Thanks in advance!
general-topology dynamical-systems 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 $tilde{M}$.
I saw a claim saying "every regular leaf (a leaf in $tilde{mathcal{F}}$ that does not contain a separatrix or the union of a singularity and all separatrices ending in that singularity) is a closed subset of $tilde{M}$".
I could not think of a proof of this claim right away or find a place where this claim is proved neither. Just wondering if this is generally true, or if I missed some conditions. How about in higher dimensions? We do not have measured foliations in higher dimensions, but is the pullback of a leaf of a minimal foliation to the universal cover always a closed subset of $tilde{M}$?
Thanks in advance!
general-topology dynamical-systems geometric-topology
general-topology dynamical-systems geometric-topology
edited Nov 20 at 6:20
asked Nov 20 at 6:05
chikurin
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By the way, I see that you have a few questions recently that received good answers. One way this site works is that if you get one (or more) good answers to one of your questions, then you choose an answer and you accept it. This has many good benefits for the growth of math.stackexchange as a whole.
– Lee Mosher
Nov 25 at 15:25
add a comment |
1
By the way, I see that you have a few questions recently that received good answers. One way this site works is that if you get one (or more) good answers to one of your questions, then you choose an answer and you accept it. This has many good benefits for the growth of math.stackexchange as a whole.
– Lee Mosher
Nov 25 at 15:25
1
1
By the way, I see that you have a few questions recently that received good answers. One way this site works is that if you get one (or more) good answers to one of your questions, then you choose an answer and you accept it. This has many good benefits for the growth of math.stackexchange as a whole.
– Lee Mosher
Nov 25 at 15:25
By the way, I see that you have a few questions recently that received good answers. One way this site works is that if you get one (or more) good answers to one of your questions, then you choose an answer and you accept it. This has many good benefits for the growth of math.stackexchange as a whole.
– Lee Mosher
Nov 25 at 15:25
add a comment |
1 Answer
1
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oldest
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up vote
1
down vote
accepted
This is indeed generally true. The proofs I know use geometry in one way or another.
Let me explain the case of a leaf $L$ of $mathcal F$ that has no singularities.
Let's fix a hyperbolic structure on $M$, and so the universal cover $tilde M$ is isometric to the hyperbolic plane $mathbb H^2$. If you understand the connection between measured foliations and geodesic laminations then you will know that corresponding to $mathcal F$ there is a geodesic lamination $mathcal L$ on $M$, there is a self-homotopy $h : M times [0,1] to M$, and there is a leaf $L'$ of $mathcal L$, such that $h mid L' times [0,1]$ is a homotopy from the identity map to $L$. It follows that in the universal covering space there are lifted leaves $tilde L$ and $tilde L'$ are properly homotopic. Since $tilde L'$ is a bi-infinite geodesic in $mathbb H^2$, it is a closed subset, hence $tilde L$ is a closed subset.
Another proof uses geometry in a different way. Using that $mathcal F$ is minimal, one can prove that there exists another measured foliation $mathcal F'$ which is transverse to $mathcal F$. The two transverse measure of $mathcal F$ and $mathcal F'$, taken together, determine a conformal structure $mu$ on $M$ and a quadratic differential $q$ of $mu$. There is an induced singular Euclidean metric on $M$: in regular local coordinates where $q = dz^2$ the metric is given by the formula $dx^2 + dy^2$ where $z=x+iy$; in singular local coordinates where $q = z^k dz^2$ ($k ge 1$), the formula is a bit more complicated. The leaf $L$ is a bi-infinite geodesic in this singular Euclidean structure. The lift to $tilde M$ of the singular Euclidean structure is a complete $text{CAT}(0)$ geodesic metric, and $tilde L$ is a complete bi-infinite geodesic in this structure, hence $tilde L$ is a closed subset.
Thanks!! I wasn't clear about the connection between measured foliations and geodesic laminations, since most of what I know was about pseudo-Anosov from the book Automorphisms of Surfaces after Nielsen and Thurston where the foliations are from geodesic laminations after some quotient, but I just read from a book called Hyperbolic Manifolds and Discrete Groups and think I could kind of understand the first proof... However, it's still not so clear to me where we used the minimality, or is the lamination just minimal by its nature? I think I need more time to understand the second proof...
– chikurin
Nov 22 at 0:37
There is a general statement with a similar conclusion that holds for any measured foliation whatsoever, but it is complicated to state the definition of a "leaf" in that generality. The only role of minimality is to enable a simpler definition of "leaf".
– Lee Mosher
Nov 22 at 2:17
I don't think I get that... The definition I feel more familiar with is like, immersed manifolds or maximal connected integral manifolds, or I just think the leaves as those line fields on surfaces. Does what you are mentioning have anything to do with minimal sets? Actually before I posted this question, I even thought these are just properties of pseudo-Anosov diffeomorphisms...
– chikurin
Nov 22 at 4:39
Also, if you don't mind, I'd like to ask another related question. For the dual tree of $tilde{mathcal{F}}$ (if I'm correct with this notion), I've seen that "when genus$geq 2$, it's never complete", why is this true? (I'm sorry, I just don't know whom to ask around me, and I would also be happy if you know any reference that I could probably read a bit from.) Thanks again!!
– chikurin
Nov 22 at 4:41
Let me make a suggestion. I can add to this answer to address your question as stated. But if you have followups that you want to ask (say about general measured foliations, versus the ones in your question as stated; or about incompleteness of the dual tree to $tilde{mathcal F}$), it is best to formulate them as separate questions, so that the whole math.stackexchange community can see them. New questions buried in comments to answers to old questions are invisible to the community.
– Lee Mosher
Nov 22 at 15:34
|
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
This is indeed generally true. The proofs I know use geometry in one way or another.
Let me explain the case of a leaf $L$ of $mathcal F$ that has no singularities.
Let's fix a hyperbolic structure on $M$, and so the universal cover $tilde M$ is isometric to the hyperbolic plane $mathbb H^2$. If you understand the connection between measured foliations and geodesic laminations then you will know that corresponding to $mathcal F$ there is a geodesic lamination $mathcal L$ on $M$, there is a self-homotopy $h : M times [0,1] to M$, and there is a leaf $L'$ of $mathcal L$, such that $h mid L' times [0,1]$ is a homotopy from the identity map to $L$. It follows that in the universal covering space there are lifted leaves $tilde L$ and $tilde L'$ are properly homotopic. Since $tilde L'$ is a bi-infinite geodesic in $mathbb H^2$, it is a closed subset, hence $tilde L$ is a closed subset.
Another proof uses geometry in a different way. Using that $mathcal F$ is minimal, one can prove that there exists another measured foliation $mathcal F'$ which is transverse to $mathcal F$. The two transverse measure of $mathcal F$ and $mathcal F'$, taken together, determine a conformal structure $mu$ on $M$ and a quadratic differential $q$ of $mu$. There is an induced singular Euclidean metric on $M$: in regular local coordinates where $q = dz^2$ the metric is given by the formula $dx^2 + dy^2$ where $z=x+iy$; in singular local coordinates where $q = z^k dz^2$ ($k ge 1$), the formula is a bit more complicated. The leaf $L$ is a bi-infinite geodesic in this singular Euclidean structure. The lift to $tilde M$ of the singular Euclidean structure is a complete $text{CAT}(0)$ geodesic metric, and $tilde L$ is a complete bi-infinite geodesic in this structure, hence $tilde L$ is a closed subset.
Thanks!! I wasn't clear about the connection between measured foliations and geodesic laminations, since most of what I know was about pseudo-Anosov from the book Automorphisms of Surfaces after Nielsen and Thurston where the foliations are from geodesic laminations after some quotient, but I just read from a book called Hyperbolic Manifolds and Discrete Groups and think I could kind of understand the first proof... However, it's still not so clear to me where we used the minimality, or is the lamination just minimal by its nature? I think I need more time to understand the second proof...
– chikurin
Nov 22 at 0:37
There is a general statement with a similar conclusion that holds for any measured foliation whatsoever, but it is complicated to state the definition of a "leaf" in that generality. The only role of minimality is to enable a simpler definition of "leaf".
– Lee Mosher
Nov 22 at 2:17
I don't think I get that... The definition I feel more familiar with is like, immersed manifolds or maximal connected integral manifolds, or I just think the leaves as those line fields on surfaces. Does what you are mentioning have anything to do with minimal sets? Actually before I posted this question, I even thought these are just properties of pseudo-Anosov diffeomorphisms...
– chikurin
Nov 22 at 4:39
Also, if you don't mind, I'd like to ask another related question. For the dual tree of $tilde{mathcal{F}}$ (if I'm correct with this notion), I've seen that "when genus$geq 2$, it's never complete", why is this true? (I'm sorry, I just don't know whom to ask around me, and I would also be happy if you know any reference that I could probably read a bit from.) Thanks again!!
– chikurin
Nov 22 at 4:41
Let me make a suggestion. I can add to this answer to address your question as stated. But if you have followups that you want to ask (say about general measured foliations, versus the ones in your question as stated; or about incompleteness of the dual tree to $tilde{mathcal F}$), it is best to formulate them as separate questions, so that the whole math.stackexchange community can see them. New questions buried in comments to answers to old questions are invisible to the community.
– Lee Mosher
Nov 22 at 15:34
|
show 1 more comment
up vote
1
down vote
accepted
This is indeed generally true. The proofs I know use geometry in one way or another.
Let me explain the case of a leaf $L$ of $mathcal F$ that has no singularities.
Let's fix a hyperbolic structure on $M$, and so the universal cover $tilde M$ is isometric to the hyperbolic plane $mathbb H^2$. If you understand the connection between measured foliations and geodesic laminations then you will know that corresponding to $mathcal F$ there is a geodesic lamination $mathcal L$ on $M$, there is a self-homotopy $h : M times [0,1] to M$, and there is a leaf $L'$ of $mathcal L$, such that $h mid L' times [0,1]$ is a homotopy from the identity map to $L$. It follows that in the universal covering space there are lifted leaves $tilde L$ and $tilde L'$ are properly homotopic. Since $tilde L'$ is a bi-infinite geodesic in $mathbb H^2$, it is a closed subset, hence $tilde L$ is a closed subset.
Another proof uses geometry in a different way. Using that $mathcal F$ is minimal, one can prove that there exists another measured foliation $mathcal F'$ which is transverse to $mathcal F$. The two transverse measure of $mathcal F$ and $mathcal F'$, taken together, determine a conformal structure $mu$ on $M$ and a quadratic differential $q$ of $mu$. There is an induced singular Euclidean metric on $M$: in regular local coordinates where $q = dz^2$ the metric is given by the formula $dx^2 + dy^2$ where $z=x+iy$; in singular local coordinates where $q = z^k dz^2$ ($k ge 1$), the formula is a bit more complicated. The leaf $L$ is a bi-infinite geodesic in this singular Euclidean structure. The lift to $tilde M$ of the singular Euclidean structure is a complete $text{CAT}(0)$ geodesic metric, and $tilde L$ is a complete bi-infinite geodesic in this structure, hence $tilde L$ is a closed subset.
Thanks!! I wasn't clear about the connection between measured foliations and geodesic laminations, since most of what I know was about pseudo-Anosov from the book Automorphisms of Surfaces after Nielsen and Thurston where the foliations are from geodesic laminations after some quotient, but I just read from a book called Hyperbolic Manifolds and Discrete Groups and think I could kind of understand the first proof... However, it's still not so clear to me where we used the minimality, or is the lamination just minimal by its nature? I think I need more time to understand the second proof...
– chikurin
Nov 22 at 0:37
There is a general statement with a similar conclusion that holds for any measured foliation whatsoever, but it is complicated to state the definition of a "leaf" in that generality. The only role of minimality is to enable a simpler definition of "leaf".
– Lee Mosher
Nov 22 at 2:17
I don't think I get that... The definition I feel more familiar with is like, immersed manifolds or maximal connected integral manifolds, or I just think the leaves as those line fields on surfaces. Does what you are mentioning have anything to do with minimal sets? Actually before I posted this question, I even thought these are just properties of pseudo-Anosov diffeomorphisms...
– chikurin
Nov 22 at 4:39
Also, if you don't mind, I'd like to ask another related question. For the dual tree of $tilde{mathcal{F}}$ (if I'm correct with this notion), I've seen that "when genus$geq 2$, it's never complete", why is this true? (I'm sorry, I just don't know whom to ask around me, and I would also be happy if you know any reference that I could probably read a bit from.) Thanks again!!
– chikurin
Nov 22 at 4:41
Let me make a suggestion. I can add to this answer to address your question as stated. But if you have followups that you want to ask (say about general measured foliations, versus the ones in your question as stated; or about incompleteness of the dual tree to $tilde{mathcal F}$), it is best to formulate them as separate questions, so that the whole math.stackexchange community can see them. New questions buried in comments to answers to old questions are invisible to the community.
– Lee Mosher
Nov 22 at 15:34
|
show 1 more comment
up vote
1
down vote
accepted
up vote
1
down vote
accepted
This is indeed generally true. The proofs I know use geometry in one way or another.
Let me explain the case of a leaf $L$ of $mathcal F$ that has no singularities.
Let's fix a hyperbolic structure on $M$, and so the universal cover $tilde M$ is isometric to the hyperbolic plane $mathbb H^2$. If you understand the connection between measured foliations and geodesic laminations then you will know that corresponding to $mathcal F$ there is a geodesic lamination $mathcal L$ on $M$, there is a self-homotopy $h : M times [0,1] to M$, and there is a leaf $L'$ of $mathcal L$, such that $h mid L' times [0,1]$ is a homotopy from the identity map to $L$. It follows that in the universal covering space there are lifted leaves $tilde L$ and $tilde L'$ are properly homotopic. Since $tilde L'$ is a bi-infinite geodesic in $mathbb H^2$, it is a closed subset, hence $tilde L$ is a closed subset.
Another proof uses geometry in a different way. Using that $mathcal F$ is minimal, one can prove that there exists another measured foliation $mathcal F'$ which is transverse to $mathcal F$. The two transverse measure of $mathcal F$ and $mathcal F'$, taken together, determine a conformal structure $mu$ on $M$ and a quadratic differential $q$ of $mu$. There is an induced singular Euclidean metric on $M$: in regular local coordinates where $q = dz^2$ the metric is given by the formula $dx^2 + dy^2$ where $z=x+iy$; in singular local coordinates where $q = z^k dz^2$ ($k ge 1$), the formula is a bit more complicated. The leaf $L$ is a bi-infinite geodesic in this singular Euclidean structure. The lift to $tilde M$ of the singular Euclidean structure is a complete $text{CAT}(0)$ geodesic metric, and $tilde L$ is a complete bi-infinite geodesic in this structure, hence $tilde L$ is a closed subset.
This is indeed generally true. The proofs I know use geometry in one way or another.
Let me explain the case of a leaf $L$ of $mathcal F$ that has no singularities.
Let's fix a hyperbolic structure on $M$, and so the universal cover $tilde M$ is isometric to the hyperbolic plane $mathbb H^2$. If you understand the connection between measured foliations and geodesic laminations then you will know that corresponding to $mathcal F$ there is a geodesic lamination $mathcal L$ on $M$, there is a self-homotopy $h : M times [0,1] to M$, and there is a leaf $L'$ of $mathcal L$, such that $h mid L' times [0,1]$ is a homotopy from the identity map to $L$. It follows that in the universal covering space there are lifted leaves $tilde L$ and $tilde L'$ are properly homotopic. Since $tilde L'$ is a bi-infinite geodesic in $mathbb H^2$, it is a closed subset, hence $tilde L$ is a closed subset.
Another proof uses geometry in a different way. Using that $mathcal F$ is minimal, one can prove that there exists another measured foliation $mathcal F'$ which is transverse to $mathcal F$. The two transverse measure of $mathcal F$ and $mathcal F'$, taken together, determine a conformal structure $mu$ on $M$ and a quadratic differential $q$ of $mu$. There is an induced singular Euclidean metric on $M$: in regular local coordinates where $q = dz^2$ the metric is given by the formula $dx^2 + dy^2$ where $z=x+iy$; in singular local coordinates where $q = z^k dz^2$ ($k ge 1$), the formula is a bit more complicated. The leaf $L$ is a bi-infinite geodesic in this singular Euclidean structure. The lift to $tilde M$ of the singular Euclidean structure is a complete $text{CAT}(0)$ geodesic metric, and $tilde L$ is a complete bi-infinite geodesic in this structure, hence $tilde L$ is a closed subset.
answered Nov 21 at 19:15
Lee Mosher
47.6k33681
47.6k33681
Thanks!! I wasn't clear about the connection between measured foliations and geodesic laminations, since most of what I know was about pseudo-Anosov from the book Automorphisms of Surfaces after Nielsen and Thurston where the foliations are from geodesic laminations after some quotient, but I just read from a book called Hyperbolic Manifolds and Discrete Groups and think I could kind of understand the first proof... However, it's still not so clear to me where we used the minimality, or is the lamination just minimal by its nature? I think I need more time to understand the second proof...
– chikurin
Nov 22 at 0:37
There is a general statement with a similar conclusion that holds for any measured foliation whatsoever, but it is complicated to state the definition of a "leaf" in that generality. The only role of minimality is to enable a simpler definition of "leaf".
– Lee Mosher
Nov 22 at 2:17
I don't think I get that... The definition I feel more familiar with is like, immersed manifolds or maximal connected integral manifolds, or I just think the leaves as those line fields on surfaces. Does what you are mentioning have anything to do with minimal sets? Actually before I posted this question, I even thought these are just properties of pseudo-Anosov diffeomorphisms...
– chikurin
Nov 22 at 4:39
Also, if you don't mind, I'd like to ask another related question. For the dual tree of $tilde{mathcal{F}}$ (if I'm correct with this notion), I've seen that "when genus$geq 2$, it's never complete", why is this true? (I'm sorry, I just don't know whom to ask around me, and I would also be happy if you know any reference that I could probably read a bit from.) Thanks again!!
– chikurin
Nov 22 at 4:41
Let me make a suggestion. I can add to this answer to address your question as stated. But if you have followups that you want to ask (say about general measured foliations, versus the ones in your question as stated; or about incompleteness of the dual tree to $tilde{mathcal F}$), it is best to formulate them as separate questions, so that the whole math.stackexchange community can see them. New questions buried in comments to answers to old questions are invisible to the community.
– Lee Mosher
Nov 22 at 15:34
|
show 1 more comment
Thanks!! I wasn't clear about the connection between measured foliations and geodesic laminations, since most of what I know was about pseudo-Anosov from the book Automorphisms of Surfaces after Nielsen and Thurston where the foliations are from geodesic laminations after some quotient, but I just read from a book called Hyperbolic Manifolds and Discrete Groups and think I could kind of understand the first proof... However, it's still not so clear to me where we used the minimality, or is the lamination just minimal by its nature? I think I need more time to understand the second proof...
– chikurin
Nov 22 at 0:37
There is a general statement with a similar conclusion that holds for any measured foliation whatsoever, but it is complicated to state the definition of a "leaf" in that generality. The only role of minimality is to enable a simpler definition of "leaf".
– Lee Mosher
Nov 22 at 2:17
I don't think I get that... The definition I feel more familiar with is like, immersed manifolds or maximal connected integral manifolds, or I just think the leaves as those line fields on surfaces. Does what you are mentioning have anything to do with minimal sets? Actually before I posted this question, I even thought these are just properties of pseudo-Anosov diffeomorphisms...
– chikurin
Nov 22 at 4:39
Also, if you don't mind, I'd like to ask another related question. For the dual tree of $tilde{mathcal{F}}$ (if I'm correct with this notion), I've seen that "when genus$geq 2$, it's never complete", why is this true? (I'm sorry, I just don't know whom to ask around me, and I would also be happy if you know any reference that I could probably read a bit from.) Thanks again!!
– chikurin
Nov 22 at 4:41
Let me make a suggestion. I can add to this answer to address your question as stated. But if you have followups that you want to ask (say about general measured foliations, versus the ones in your question as stated; or about incompleteness of the dual tree to $tilde{mathcal F}$), it is best to formulate them as separate questions, so that the whole math.stackexchange community can see them. New questions buried in comments to answers to old questions are invisible to the community.
– Lee Mosher
Nov 22 at 15:34
Thanks!! I wasn't clear about the connection between measured foliations and geodesic laminations, since most of what I know was about pseudo-Anosov from the book Automorphisms of Surfaces after Nielsen and Thurston where the foliations are from geodesic laminations after some quotient, but I just read from a book called Hyperbolic Manifolds and Discrete Groups and think I could kind of understand the first proof... However, it's still not so clear to me where we used the minimality, or is the lamination just minimal by its nature? I think I need more time to understand the second proof...
– chikurin
Nov 22 at 0:37
Thanks!! I wasn't clear about the connection between measured foliations and geodesic laminations, since most of what I know was about pseudo-Anosov from the book Automorphisms of Surfaces after Nielsen and Thurston where the foliations are from geodesic laminations after some quotient, but I just read from a book called Hyperbolic Manifolds and Discrete Groups and think I could kind of understand the first proof... However, it's still not so clear to me where we used the minimality, or is the lamination just minimal by its nature? I think I need more time to understand the second proof...
– chikurin
Nov 22 at 0:37
There is a general statement with a similar conclusion that holds for any measured foliation whatsoever, but it is complicated to state the definition of a "leaf" in that generality. The only role of minimality is to enable a simpler definition of "leaf".
– Lee Mosher
Nov 22 at 2:17
There is a general statement with a similar conclusion that holds for any measured foliation whatsoever, but it is complicated to state the definition of a "leaf" in that generality. The only role of minimality is to enable a simpler definition of "leaf".
– Lee Mosher
Nov 22 at 2:17
I don't think I get that... The definition I feel more familiar with is like, immersed manifolds or maximal connected integral manifolds, or I just think the leaves as those line fields on surfaces. Does what you are mentioning have anything to do with minimal sets? Actually before I posted this question, I even thought these are just properties of pseudo-Anosov diffeomorphisms...
– chikurin
Nov 22 at 4:39
I don't think I get that... The definition I feel more familiar with is like, immersed manifolds or maximal connected integral manifolds, or I just think the leaves as those line fields on surfaces. Does what you are mentioning have anything to do with minimal sets? Actually before I posted this question, I even thought these are just properties of pseudo-Anosov diffeomorphisms...
– chikurin
Nov 22 at 4:39
Also, if you don't mind, I'd like to ask another related question. For the dual tree of $tilde{mathcal{F}}$ (if I'm correct with this notion), I've seen that "when genus$geq 2$, it's never complete", why is this true? (I'm sorry, I just don't know whom to ask around me, and I would also be happy if you know any reference that I could probably read a bit from.) Thanks again!!
– chikurin
Nov 22 at 4:41
Also, if you don't mind, I'd like to ask another related question. For the dual tree of $tilde{mathcal{F}}$ (if I'm correct with this notion), I've seen that "when genus$geq 2$, it's never complete", why is this true? (I'm sorry, I just don't know whom to ask around me, and I would also be happy if you know any reference that I could probably read a bit from.) Thanks again!!
– chikurin
Nov 22 at 4:41
Let me make a suggestion. I can add to this answer to address your question as stated. But if you have followups that you want to ask (say about general measured foliations, versus the ones in your question as stated; or about incompleteness of the dual tree to $tilde{mathcal F}$), it is best to formulate them as separate questions, so that the whole math.stackexchange community can see them. New questions buried in comments to answers to old questions are invisible to the community.
– Lee Mosher
Nov 22 at 15:34
Let me make a suggestion. I can add to this answer to address your question as stated. But if you have followups that you want to ask (say about general measured foliations, versus the ones in your question as stated; or about incompleteness of the dual tree to $tilde{mathcal F}$), it is best to formulate them as separate questions, so that the whole math.stackexchange community can see them. New questions buried in comments to answers to old questions are invisible to the community.
– Lee Mosher
Nov 22 at 15:34
|
show 1 more comment
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By the way, I see that you have a few questions recently that received good answers. One way this site works is that if you get one (or more) good answers to one of your questions, then you choose an answer and you accept it. This has many good benefits for the growth of math.stackexchange as a whole.
– Lee Mosher
Nov 25 at 15:25