What is $T^1(mathbb H^2/PSL_2(mathbb Z))$?
$begingroup$
Let $mathbb H^2$ be the upper-half plane.
The group $PSL_2(Z)$ acts on $mathbb H^2$ by isometries, and hence we get an action on $T^1(mathbb H^2)$.
This action is free, smooth, and proper, and thus $X=T^1(mathbb H^2)/PSL_2(Z)$ is a smooth manifold (with unique smooth structure such that the projection $T^1(mathbb H^2)to T^1(mathbb H^2)/PSL_2(Z)$ is a submersion).
In these notes, on page 3, the first line reads that: Note that $X$ is a unit tangent bundle of non-compact hyperbolic surface (with two "singular points").
I am unable to make sense of this statement. Can somebody please explain how we can see $X$ as the unit tangent bundle of a Riemannian manifold?
Also, what is meant by 'singular points' here?
Thank you.
Edit. Definition of $T^1$: Let $(M, g)$ be a Riemannian manifold. Then $T^1M$, the unit tangent bundle of $M$, is the collection of all the members of $TM$ which are of unit length. Now since $mathbb H^2$ is a Riemannian manifold (the metric being $(dx^2+dy^2)/y^2$), we can talk about $T^1(mathbb H^2)$.
riemannian-geometry hyperbolic-geometry ergodic-theory diophantine-approximation
$endgroup$
|
show 4 more comments
$begingroup$
Let $mathbb H^2$ be the upper-half plane.
The group $PSL_2(Z)$ acts on $mathbb H^2$ by isometries, and hence we get an action on $T^1(mathbb H^2)$.
This action is free, smooth, and proper, and thus $X=T^1(mathbb H^2)/PSL_2(Z)$ is a smooth manifold (with unique smooth structure such that the projection $T^1(mathbb H^2)to T^1(mathbb H^2)/PSL_2(Z)$ is a submersion).
In these notes, on page 3, the first line reads that: Note that $X$ is a unit tangent bundle of non-compact hyperbolic surface (with two "singular points").
I am unable to make sense of this statement. Can somebody please explain how we can see $X$ as the unit tangent bundle of a Riemannian manifold?
Also, what is meant by 'singular points' here?
Thank you.
Edit. Definition of $T^1$: Let $(M, g)$ be a Riemannian manifold. Then $T^1M$, the unit tangent bundle of $M$, is the collection of all the members of $TM$ which are of unit length. Now since $mathbb H^2$ is a Riemannian manifold (the metric being $(dx^2+dy^2)/y^2$), we can talk about $T^1(mathbb H^2)$.
riemannian-geometry hyperbolic-geometry ergodic-theory diophantine-approximation
$endgroup$
$begingroup$
See page $5$ here, why the unit tangent bundle of the modular surface $M=Bbb H^2/PSL_2(Bbb Z)$ is just the quotient space $X=PSL_2(Bbb R)/PSL_2(Bbb Z)$.
$endgroup$
– Dietrich Burde
Jan 26 at 11:43
$begingroup$
The document mentions that $mathbb H^2/PSL_2(mathbb Z)$ is not a manifold but an "orbifold". In that case it is not clear what is the meaning of $T^1(mathbb H^2/PSL_2(mathbb Z))$. I looked up the definition of an orbifold on wikipedia but didn't help. Can you say what is going on here?
$endgroup$
– caffeinemachine
Jan 26 at 12:10
$begingroup$
Yes, the fibered bundle has singularities at the elliptic fixed points, but this can be fixed, see S. Katok, Fuchsian groups, $§ 3.6$.
$endgroup$
– Dietrich Burde
Jan 26 at 12:31
$begingroup$
$mathbb{H}to mathbb{H}/PSL_2(mathbb{Z})$ is a submersion? What is the differential at $i$ or $1/2+isqrt{3}/2$?
$endgroup$
– Sir Wilfred Lucas-Dockery
Jan 31 at 13:07
$begingroup$
@Unpud For $mathbb Hto mathbb H/PSL_2(mathbb Z)$ to be a submersion the target has to be a manifold. But it is not. Or were you trying to say something else?
$endgroup$
– caffeinemachine
Jan 31 at 13:13
|
show 4 more comments
$begingroup$
Let $mathbb H^2$ be the upper-half plane.
The group $PSL_2(Z)$ acts on $mathbb H^2$ by isometries, and hence we get an action on $T^1(mathbb H^2)$.
This action is free, smooth, and proper, and thus $X=T^1(mathbb H^2)/PSL_2(Z)$ is a smooth manifold (with unique smooth structure such that the projection $T^1(mathbb H^2)to T^1(mathbb H^2)/PSL_2(Z)$ is a submersion).
In these notes, on page 3, the first line reads that: Note that $X$ is a unit tangent bundle of non-compact hyperbolic surface (with two "singular points").
I am unable to make sense of this statement. Can somebody please explain how we can see $X$ as the unit tangent bundle of a Riemannian manifold?
Also, what is meant by 'singular points' here?
Thank you.
Edit. Definition of $T^1$: Let $(M, g)$ be a Riemannian manifold. Then $T^1M$, the unit tangent bundle of $M$, is the collection of all the members of $TM$ which are of unit length. Now since $mathbb H^2$ is a Riemannian manifold (the metric being $(dx^2+dy^2)/y^2$), we can talk about $T^1(mathbb H^2)$.
riemannian-geometry hyperbolic-geometry ergodic-theory diophantine-approximation
$endgroup$
Let $mathbb H^2$ be the upper-half plane.
The group $PSL_2(Z)$ acts on $mathbb H^2$ by isometries, and hence we get an action on $T^1(mathbb H^2)$.
This action is free, smooth, and proper, and thus $X=T^1(mathbb H^2)/PSL_2(Z)$ is a smooth manifold (with unique smooth structure such that the projection $T^1(mathbb H^2)to T^1(mathbb H^2)/PSL_2(Z)$ is a submersion).
In these notes, on page 3, the first line reads that: Note that $X$ is a unit tangent bundle of non-compact hyperbolic surface (with two "singular points").
I am unable to make sense of this statement. Can somebody please explain how we can see $X$ as the unit tangent bundle of a Riemannian manifold?
Also, what is meant by 'singular points' here?
Thank you.
Edit. Definition of $T^1$: Let $(M, g)$ be a Riemannian manifold. Then $T^1M$, the unit tangent bundle of $M$, is the collection of all the members of $TM$ which are of unit length. Now since $mathbb H^2$ is a Riemannian manifold (the metric being $(dx^2+dy^2)/y^2$), we can talk about $T^1(mathbb H^2)$.
riemannian-geometry hyperbolic-geometry ergodic-theory diophantine-approximation
riemannian-geometry hyperbolic-geometry ergodic-theory diophantine-approximation
edited Jan 31 at 16:03
caffeinemachine
asked Jan 25 at 20:09
caffeinemachinecaffeinemachine
6,63221355
6,63221355
$begingroup$
See page $5$ here, why the unit tangent bundle of the modular surface $M=Bbb H^2/PSL_2(Bbb Z)$ is just the quotient space $X=PSL_2(Bbb R)/PSL_2(Bbb Z)$.
$endgroup$
– Dietrich Burde
Jan 26 at 11:43
$begingroup$
The document mentions that $mathbb H^2/PSL_2(mathbb Z)$ is not a manifold but an "orbifold". In that case it is not clear what is the meaning of $T^1(mathbb H^2/PSL_2(mathbb Z))$. I looked up the definition of an orbifold on wikipedia but didn't help. Can you say what is going on here?
$endgroup$
– caffeinemachine
Jan 26 at 12:10
$begingroup$
Yes, the fibered bundle has singularities at the elliptic fixed points, but this can be fixed, see S. Katok, Fuchsian groups, $§ 3.6$.
$endgroup$
– Dietrich Burde
Jan 26 at 12:31
$begingroup$
$mathbb{H}to mathbb{H}/PSL_2(mathbb{Z})$ is a submersion? What is the differential at $i$ or $1/2+isqrt{3}/2$?
$endgroup$
– Sir Wilfred Lucas-Dockery
Jan 31 at 13:07
$begingroup$
@Unpud For $mathbb Hto mathbb H/PSL_2(mathbb Z)$ to be a submersion the target has to be a manifold. But it is not. Or were you trying to say something else?
$endgroup$
– caffeinemachine
Jan 31 at 13:13
|
show 4 more comments
$begingroup$
See page $5$ here, why the unit tangent bundle of the modular surface $M=Bbb H^2/PSL_2(Bbb Z)$ is just the quotient space $X=PSL_2(Bbb R)/PSL_2(Bbb Z)$.
$endgroup$
– Dietrich Burde
Jan 26 at 11:43
$begingroup$
The document mentions that $mathbb H^2/PSL_2(mathbb Z)$ is not a manifold but an "orbifold". In that case it is not clear what is the meaning of $T^1(mathbb H^2/PSL_2(mathbb Z))$. I looked up the definition of an orbifold on wikipedia but didn't help. Can you say what is going on here?
$endgroup$
– caffeinemachine
Jan 26 at 12:10
$begingroup$
Yes, the fibered bundle has singularities at the elliptic fixed points, but this can be fixed, see S. Katok, Fuchsian groups, $§ 3.6$.
$endgroup$
– Dietrich Burde
Jan 26 at 12:31
$begingroup$
$mathbb{H}to mathbb{H}/PSL_2(mathbb{Z})$ is a submersion? What is the differential at $i$ or $1/2+isqrt{3}/2$?
$endgroup$
– Sir Wilfred Lucas-Dockery
Jan 31 at 13:07
$begingroup$
@Unpud For $mathbb Hto mathbb H/PSL_2(mathbb Z)$ to be a submersion the target has to be a manifold. But it is not. Or were you trying to say something else?
$endgroup$
– caffeinemachine
Jan 31 at 13:13
$begingroup$
See page $5$ here, why the unit tangent bundle of the modular surface $M=Bbb H^2/PSL_2(Bbb Z)$ is just the quotient space $X=PSL_2(Bbb R)/PSL_2(Bbb Z)$.
$endgroup$
– Dietrich Burde
Jan 26 at 11:43
$begingroup$
See page $5$ here, why the unit tangent bundle of the modular surface $M=Bbb H^2/PSL_2(Bbb Z)$ is just the quotient space $X=PSL_2(Bbb R)/PSL_2(Bbb Z)$.
$endgroup$
– Dietrich Burde
Jan 26 at 11:43
$begingroup$
The document mentions that $mathbb H^2/PSL_2(mathbb Z)$ is not a manifold but an "orbifold". In that case it is not clear what is the meaning of $T^1(mathbb H^2/PSL_2(mathbb Z))$. I looked up the definition of an orbifold on wikipedia but didn't help. Can you say what is going on here?
$endgroup$
– caffeinemachine
Jan 26 at 12:10
$begingroup$
The document mentions that $mathbb H^2/PSL_2(mathbb Z)$ is not a manifold but an "orbifold". In that case it is not clear what is the meaning of $T^1(mathbb H^2/PSL_2(mathbb Z))$. I looked up the definition of an orbifold on wikipedia but didn't help. Can you say what is going on here?
$endgroup$
– caffeinemachine
Jan 26 at 12:10
$begingroup$
Yes, the fibered bundle has singularities at the elliptic fixed points, but this can be fixed, see S. Katok, Fuchsian groups, $§ 3.6$.
$endgroup$
– Dietrich Burde
Jan 26 at 12:31
$begingroup$
Yes, the fibered bundle has singularities at the elliptic fixed points, but this can be fixed, see S. Katok, Fuchsian groups, $§ 3.6$.
$endgroup$
– Dietrich Burde
Jan 26 at 12:31
$begingroup$
$mathbb{H}to mathbb{H}/PSL_2(mathbb{Z})$ is a submersion? What is the differential at $i$ or $1/2+isqrt{3}/2$?
$endgroup$
– Sir Wilfred Lucas-Dockery
Jan 31 at 13:07
$begingroup$
$mathbb{H}to mathbb{H}/PSL_2(mathbb{Z})$ is a submersion? What is the differential at $i$ or $1/2+isqrt{3}/2$?
$endgroup$
– Sir Wilfred Lucas-Dockery
Jan 31 at 13:07
$begingroup$
@Unpud For $mathbb Hto mathbb H/PSL_2(mathbb Z)$ to be a submersion the target has to be a manifold. But it is not. Or were you trying to say something else?
$endgroup$
– caffeinemachine
Jan 31 at 13:13
$begingroup$
@Unpud For $mathbb Hto mathbb H/PSL_2(mathbb Z)$ to be a submersion the target has to be a manifold. But it is not. Or were you trying to say something else?
$endgroup$
– caffeinemachine
Jan 31 at 13:13
|
show 4 more comments
2 Answers
2
active
oldest
votes
$begingroup$
The key word here is Seifert fiber spaces (don't be fooled by the Wikipedia link, Seifert fiber spaces do not have to be closed manifolds).
Oriented 3-manifolds $M$ that fiber over a connected 2-manifold $S$ with circle fiber $C$ have a pretty simple classification theory. That theory has a few intricacies in the case that $M$ (and $S$) is closed, centered on the concept of the Euler number of the fibration. However, if $M$ (and $S$) are not closed then the classification is very simple: each is fiber isomorphic to the product $C times S$.
More generally, an oriented Seifert fibered 3-manifold $M$ has the structure of a fibration in the category of orbifolds, where the base 2-orbifold $O$ is oriented and hence is a surface with a collection of singularities each determined by its order (a natural number $ge 2$), and where the generic fiber is a circle. The Siefert fiber structure has a local invariant over each singularity of $O$, which describes how the generic circles over points near $O$ wind around the singular circle wind over $O$; this invariant is a rational number modulo $1$, whose denominator is equal to the order of the singularity. Again, the full classification theory is complicated by Euler number if $M$ is closed, but when $M$ is not closed (equivalently $S$ is not closed), the Siefert fiber structure is completely determined by the local invariants.
I'm not going to write a treatise on the local invariants of singular fibers of Seifert fiber spaces, but I'll tell you how to proceed once you learn that theory.
In the case of $T^1(mathbb H^2 / PSL_2(mathbb Z))$, the base orbifold $mathbb H^2 / PSL_2(mathbb Z)$ is an open disc with an order 2 and an order 3 singularities. You can calculate the values of the local invariant of the singular fibers of $T^1(mathbb H^2 / PSL_2(mathbb Z))$ over those two singularities, by examining, for each $p in mathbb H^2$ with an order 2 or 3 stabilizer subgroup in $PSL_2(mathbb Z)$, how that subgroup acts on $T^1_p(mathbb H^2)$. That way you will have completely worked out the classification of $T^1(mathbb H^2 / PSL_2(mathbb Z))$ in Siefert fiber space language.
But, if you want to know some familiar example which is homeomorphic to $T^1(mathbb H^2 / PSL_2(mathbb Z))$, I think that's also possible: I believe it may be homeomorphic to the complement of the trefoil knot in $S^3$. That complement has the same base orbifold as $T^1(mathbb H^2 / PSL_2(mathbb Z))$ --- the open disc with an order 2 and order 3 singularity --- but I do not know offhand the local invariants of two singular fibers. One might also be able to directly compare the fundamental groups of those two 3-manifolds.
$endgroup$
$begingroup$
Isn't $mathbb{H}/PSL_2(mathbb{Z})$ a smooth manifold? Near the fixed points, isn't the action isomorphic to the quotient of the disc by some finite rotation group? So near a point with stabilizer of order $2$, wouldn't a coordinate chart be realised by a map of the form $zmapsto z^2$? Similarly, for points with stabilizer of order $3$?
$endgroup$
– Sir Wilfred Lucas-Dockery
Jan 31 at 12:36
$begingroup$
I should say, the coordinate chart is given by the homeomorphism $D/(mathbb{Z}/2mathbb{Z}) to D$ induced by $zmapsto z^2$.
$endgroup$
– Sir Wilfred Lucas-Dockery
Jan 31 at 13:39
$begingroup$
It is indeed true, as you say, that the smooth structure which $mathbb H / PSL_2(mathbb Z) - {text{the two cone points}}$ inherits from $mathbb H$ extends to a smooth structure on the whole of $mathbb H / PSL_2(mathbb Z)$. But that's a different thing from saying $mathbb H / PSL_2(mathbb Z)$ inherits a smooth structure from $mathbb H$.
$endgroup$
– Lee Mosher
Jan 31 at 14:50
$begingroup$
Isn’t there a unique complex structure making the quotient holomorphic?
$endgroup$
– Sir Wilfred Lucas-Dockery
Jan 31 at 14:52
$begingroup$
On the other hand, the orbifold structure on $mathbb H / PSL_2(mathbb Z)$ is naturally inherited from $mathbb H$. The conformal structure that $mathbb H / PSL_2(mathbb Z) - {text{the two cone points}}$ inherits from $mathbb H$ also extends naturally over $mathbb H / PSL_2(mathbb Z)$, because the singularities where those two cone points were removed are, indeed, removable singularities.
$endgroup$
– Lee Mosher
Jan 31 at 14:53
|
show 1 more comment
$begingroup$
If $Gamma$ is a torsion free lattice in $G:=PSL_2(mathbb{R})$, thus acting freely and discontinuously on $mathbb{H}$, one can unambiguously make sense of $T^1(Gammabackslashmathbb{H})$ since $Gammabackslashmathbb{H}$ is a nice Riemannian manifold with the projection preserving the Riemannian metric. Moreover, one can identify $Gammabackslash G$ with $T^1(Gammabackslashmathbb{H})$ and one can see that the $mathbb{R}$-action of diagonal matrices with entries $lbrace e^{t/2},e^{-t/2}rbrace$ is intertwined with the geodesic action on $T^1(Gammabackslashmathbb{H})$ under this identification.
But in general when $Gamma$ is has some fixed points, the notation $T^1(Gammabackslashmathbb{H})$ becomes ambiguousinnacurate as you have noticed. However, there is still the space $Gammabackslash G$ with a nice action by diagonal matrices. Most authors continue to work the the diagonal matrix action on this space and simply call it the 'unit tangent bundle over $Gammabackslashmathbb{H}$' with the 'geodesic action' as an abuse of notation (cf. the book by Bekka-Mayer, page $59$).
As a beginner in ergodic theory, I have noticed that working with torsion lattices in $G$ is quite a headache. In Bekka-Mayer's book, they assume torsion-free-ness for a number of their results simply for convenience (cf. section III$.3$, page $93$).
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
The key word here is Seifert fiber spaces (don't be fooled by the Wikipedia link, Seifert fiber spaces do not have to be closed manifolds).
Oriented 3-manifolds $M$ that fiber over a connected 2-manifold $S$ with circle fiber $C$ have a pretty simple classification theory. That theory has a few intricacies in the case that $M$ (and $S$) is closed, centered on the concept of the Euler number of the fibration. However, if $M$ (and $S$) are not closed then the classification is very simple: each is fiber isomorphic to the product $C times S$.
More generally, an oriented Seifert fibered 3-manifold $M$ has the structure of a fibration in the category of orbifolds, where the base 2-orbifold $O$ is oriented and hence is a surface with a collection of singularities each determined by its order (a natural number $ge 2$), and where the generic fiber is a circle. The Siefert fiber structure has a local invariant over each singularity of $O$, which describes how the generic circles over points near $O$ wind around the singular circle wind over $O$; this invariant is a rational number modulo $1$, whose denominator is equal to the order of the singularity. Again, the full classification theory is complicated by Euler number if $M$ is closed, but when $M$ is not closed (equivalently $S$ is not closed), the Siefert fiber structure is completely determined by the local invariants.
I'm not going to write a treatise on the local invariants of singular fibers of Seifert fiber spaces, but I'll tell you how to proceed once you learn that theory.
In the case of $T^1(mathbb H^2 / PSL_2(mathbb Z))$, the base orbifold $mathbb H^2 / PSL_2(mathbb Z)$ is an open disc with an order 2 and an order 3 singularities. You can calculate the values of the local invariant of the singular fibers of $T^1(mathbb H^2 / PSL_2(mathbb Z))$ over those two singularities, by examining, for each $p in mathbb H^2$ with an order 2 or 3 stabilizer subgroup in $PSL_2(mathbb Z)$, how that subgroup acts on $T^1_p(mathbb H^2)$. That way you will have completely worked out the classification of $T^1(mathbb H^2 / PSL_2(mathbb Z))$ in Siefert fiber space language.
But, if you want to know some familiar example which is homeomorphic to $T^1(mathbb H^2 / PSL_2(mathbb Z))$, I think that's also possible: I believe it may be homeomorphic to the complement of the trefoil knot in $S^3$. That complement has the same base orbifold as $T^1(mathbb H^2 / PSL_2(mathbb Z))$ --- the open disc with an order 2 and order 3 singularity --- but I do not know offhand the local invariants of two singular fibers. One might also be able to directly compare the fundamental groups of those two 3-manifolds.
$endgroup$
$begingroup$
Isn't $mathbb{H}/PSL_2(mathbb{Z})$ a smooth manifold? Near the fixed points, isn't the action isomorphic to the quotient of the disc by some finite rotation group? So near a point with stabilizer of order $2$, wouldn't a coordinate chart be realised by a map of the form $zmapsto z^2$? Similarly, for points with stabilizer of order $3$?
$endgroup$
– Sir Wilfred Lucas-Dockery
Jan 31 at 12:36
$begingroup$
I should say, the coordinate chart is given by the homeomorphism $D/(mathbb{Z}/2mathbb{Z}) to D$ induced by $zmapsto z^2$.
$endgroup$
– Sir Wilfred Lucas-Dockery
Jan 31 at 13:39
$begingroup$
It is indeed true, as you say, that the smooth structure which $mathbb H / PSL_2(mathbb Z) - {text{the two cone points}}$ inherits from $mathbb H$ extends to a smooth structure on the whole of $mathbb H / PSL_2(mathbb Z)$. But that's a different thing from saying $mathbb H / PSL_2(mathbb Z)$ inherits a smooth structure from $mathbb H$.
$endgroup$
– Lee Mosher
Jan 31 at 14:50
$begingroup$
Isn’t there a unique complex structure making the quotient holomorphic?
$endgroup$
– Sir Wilfred Lucas-Dockery
Jan 31 at 14:52
$begingroup$
On the other hand, the orbifold structure on $mathbb H / PSL_2(mathbb Z)$ is naturally inherited from $mathbb H$. The conformal structure that $mathbb H / PSL_2(mathbb Z) - {text{the two cone points}}$ inherits from $mathbb H$ also extends naturally over $mathbb H / PSL_2(mathbb Z)$, because the singularities where those two cone points were removed are, indeed, removable singularities.
$endgroup$
– Lee Mosher
Jan 31 at 14:53
|
show 1 more comment
$begingroup$
The key word here is Seifert fiber spaces (don't be fooled by the Wikipedia link, Seifert fiber spaces do not have to be closed manifolds).
Oriented 3-manifolds $M$ that fiber over a connected 2-manifold $S$ with circle fiber $C$ have a pretty simple classification theory. That theory has a few intricacies in the case that $M$ (and $S$) is closed, centered on the concept of the Euler number of the fibration. However, if $M$ (and $S$) are not closed then the classification is very simple: each is fiber isomorphic to the product $C times S$.
More generally, an oriented Seifert fibered 3-manifold $M$ has the structure of a fibration in the category of orbifolds, where the base 2-orbifold $O$ is oriented and hence is a surface with a collection of singularities each determined by its order (a natural number $ge 2$), and where the generic fiber is a circle. The Siefert fiber structure has a local invariant over each singularity of $O$, which describes how the generic circles over points near $O$ wind around the singular circle wind over $O$; this invariant is a rational number modulo $1$, whose denominator is equal to the order of the singularity. Again, the full classification theory is complicated by Euler number if $M$ is closed, but when $M$ is not closed (equivalently $S$ is not closed), the Siefert fiber structure is completely determined by the local invariants.
I'm not going to write a treatise on the local invariants of singular fibers of Seifert fiber spaces, but I'll tell you how to proceed once you learn that theory.
In the case of $T^1(mathbb H^2 / PSL_2(mathbb Z))$, the base orbifold $mathbb H^2 / PSL_2(mathbb Z)$ is an open disc with an order 2 and an order 3 singularities. You can calculate the values of the local invariant of the singular fibers of $T^1(mathbb H^2 / PSL_2(mathbb Z))$ over those two singularities, by examining, for each $p in mathbb H^2$ with an order 2 or 3 stabilizer subgroup in $PSL_2(mathbb Z)$, how that subgroup acts on $T^1_p(mathbb H^2)$. That way you will have completely worked out the classification of $T^1(mathbb H^2 / PSL_2(mathbb Z))$ in Siefert fiber space language.
But, if you want to know some familiar example which is homeomorphic to $T^1(mathbb H^2 / PSL_2(mathbb Z))$, I think that's also possible: I believe it may be homeomorphic to the complement of the trefoil knot in $S^3$. That complement has the same base orbifold as $T^1(mathbb H^2 / PSL_2(mathbb Z))$ --- the open disc with an order 2 and order 3 singularity --- but I do not know offhand the local invariants of two singular fibers. One might also be able to directly compare the fundamental groups of those two 3-manifolds.
$endgroup$
$begingroup$
Isn't $mathbb{H}/PSL_2(mathbb{Z})$ a smooth manifold? Near the fixed points, isn't the action isomorphic to the quotient of the disc by some finite rotation group? So near a point with stabilizer of order $2$, wouldn't a coordinate chart be realised by a map of the form $zmapsto z^2$? Similarly, for points with stabilizer of order $3$?
$endgroup$
– Sir Wilfred Lucas-Dockery
Jan 31 at 12:36
$begingroup$
I should say, the coordinate chart is given by the homeomorphism $D/(mathbb{Z}/2mathbb{Z}) to D$ induced by $zmapsto z^2$.
$endgroup$
– Sir Wilfred Lucas-Dockery
Jan 31 at 13:39
$begingroup$
It is indeed true, as you say, that the smooth structure which $mathbb H / PSL_2(mathbb Z) - {text{the two cone points}}$ inherits from $mathbb H$ extends to a smooth structure on the whole of $mathbb H / PSL_2(mathbb Z)$. But that's a different thing from saying $mathbb H / PSL_2(mathbb Z)$ inherits a smooth structure from $mathbb H$.
$endgroup$
– Lee Mosher
Jan 31 at 14:50
$begingroup$
Isn’t there a unique complex structure making the quotient holomorphic?
$endgroup$
– Sir Wilfred Lucas-Dockery
Jan 31 at 14:52
$begingroup$
On the other hand, the orbifold structure on $mathbb H / PSL_2(mathbb Z)$ is naturally inherited from $mathbb H$. The conformal structure that $mathbb H / PSL_2(mathbb Z) - {text{the two cone points}}$ inherits from $mathbb H$ also extends naturally over $mathbb H / PSL_2(mathbb Z)$, because the singularities where those two cone points were removed are, indeed, removable singularities.
$endgroup$
– Lee Mosher
Jan 31 at 14:53
|
show 1 more comment
$begingroup$
The key word here is Seifert fiber spaces (don't be fooled by the Wikipedia link, Seifert fiber spaces do not have to be closed manifolds).
Oriented 3-manifolds $M$ that fiber over a connected 2-manifold $S$ with circle fiber $C$ have a pretty simple classification theory. That theory has a few intricacies in the case that $M$ (and $S$) is closed, centered on the concept of the Euler number of the fibration. However, if $M$ (and $S$) are not closed then the classification is very simple: each is fiber isomorphic to the product $C times S$.
More generally, an oriented Seifert fibered 3-manifold $M$ has the structure of a fibration in the category of orbifolds, where the base 2-orbifold $O$ is oriented and hence is a surface with a collection of singularities each determined by its order (a natural number $ge 2$), and where the generic fiber is a circle. The Siefert fiber structure has a local invariant over each singularity of $O$, which describes how the generic circles over points near $O$ wind around the singular circle wind over $O$; this invariant is a rational number modulo $1$, whose denominator is equal to the order of the singularity. Again, the full classification theory is complicated by Euler number if $M$ is closed, but when $M$ is not closed (equivalently $S$ is not closed), the Siefert fiber structure is completely determined by the local invariants.
I'm not going to write a treatise on the local invariants of singular fibers of Seifert fiber spaces, but I'll tell you how to proceed once you learn that theory.
In the case of $T^1(mathbb H^2 / PSL_2(mathbb Z))$, the base orbifold $mathbb H^2 / PSL_2(mathbb Z)$ is an open disc with an order 2 and an order 3 singularities. You can calculate the values of the local invariant of the singular fibers of $T^1(mathbb H^2 / PSL_2(mathbb Z))$ over those two singularities, by examining, for each $p in mathbb H^2$ with an order 2 or 3 stabilizer subgroup in $PSL_2(mathbb Z)$, how that subgroup acts on $T^1_p(mathbb H^2)$. That way you will have completely worked out the classification of $T^1(mathbb H^2 / PSL_2(mathbb Z))$ in Siefert fiber space language.
But, if you want to know some familiar example which is homeomorphic to $T^1(mathbb H^2 / PSL_2(mathbb Z))$, I think that's also possible: I believe it may be homeomorphic to the complement of the trefoil knot in $S^3$. That complement has the same base orbifold as $T^1(mathbb H^2 / PSL_2(mathbb Z))$ --- the open disc with an order 2 and order 3 singularity --- but I do not know offhand the local invariants of two singular fibers. One might also be able to directly compare the fundamental groups of those two 3-manifolds.
$endgroup$
The key word here is Seifert fiber spaces (don't be fooled by the Wikipedia link, Seifert fiber spaces do not have to be closed manifolds).
Oriented 3-manifolds $M$ that fiber over a connected 2-manifold $S$ with circle fiber $C$ have a pretty simple classification theory. That theory has a few intricacies in the case that $M$ (and $S$) is closed, centered on the concept of the Euler number of the fibration. However, if $M$ (and $S$) are not closed then the classification is very simple: each is fiber isomorphic to the product $C times S$.
More generally, an oriented Seifert fibered 3-manifold $M$ has the structure of a fibration in the category of orbifolds, where the base 2-orbifold $O$ is oriented and hence is a surface with a collection of singularities each determined by its order (a natural number $ge 2$), and where the generic fiber is a circle. The Siefert fiber structure has a local invariant over each singularity of $O$, which describes how the generic circles over points near $O$ wind around the singular circle wind over $O$; this invariant is a rational number modulo $1$, whose denominator is equal to the order of the singularity. Again, the full classification theory is complicated by Euler number if $M$ is closed, but when $M$ is not closed (equivalently $S$ is not closed), the Siefert fiber structure is completely determined by the local invariants.
I'm not going to write a treatise on the local invariants of singular fibers of Seifert fiber spaces, but I'll tell you how to proceed once you learn that theory.
In the case of $T^1(mathbb H^2 / PSL_2(mathbb Z))$, the base orbifold $mathbb H^2 / PSL_2(mathbb Z)$ is an open disc with an order 2 and an order 3 singularities. You can calculate the values of the local invariant of the singular fibers of $T^1(mathbb H^2 / PSL_2(mathbb Z))$ over those two singularities, by examining, for each $p in mathbb H^2$ with an order 2 or 3 stabilizer subgroup in $PSL_2(mathbb Z)$, how that subgroup acts on $T^1_p(mathbb H^2)$. That way you will have completely worked out the classification of $T^1(mathbb H^2 / PSL_2(mathbb Z))$ in Siefert fiber space language.
But, if you want to know some familiar example which is homeomorphic to $T^1(mathbb H^2 / PSL_2(mathbb Z))$, I think that's also possible: I believe it may be homeomorphic to the complement of the trefoil knot in $S^3$. That complement has the same base orbifold as $T^1(mathbb H^2 / PSL_2(mathbb Z))$ --- the open disc with an order 2 and order 3 singularity --- but I do not know offhand the local invariants of two singular fibers. One might also be able to directly compare the fundamental groups of those two 3-manifolds.
edited Jan 26 at 17:21
answered Jan 26 at 15:44
Lee MosherLee Mosher
50.5k33787
50.5k33787
$begingroup$
Isn't $mathbb{H}/PSL_2(mathbb{Z})$ a smooth manifold? Near the fixed points, isn't the action isomorphic to the quotient of the disc by some finite rotation group? So near a point with stabilizer of order $2$, wouldn't a coordinate chart be realised by a map of the form $zmapsto z^2$? Similarly, for points with stabilizer of order $3$?
$endgroup$
– Sir Wilfred Lucas-Dockery
Jan 31 at 12:36
$begingroup$
I should say, the coordinate chart is given by the homeomorphism $D/(mathbb{Z}/2mathbb{Z}) to D$ induced by $zmapsto z^2$.
$endgroup$
– Sir Wilfred Lucas-Dockery
Jan 31 at 13:39
$begingroup$
It is indeed true, as you say, that the smooth structure which $mathbb H / PSL_2(mathbb Z) - {text{the two cone points}}$ inherits from $mathbb H$ extends to a smooth structure on the whole of $mathbb H / PSL_2(mathbb Z)$. But that's a different thing from saying $mathbb H / PSL_2(mathbb Z)$ inherits a smooth structure from $mathbb H$.
$endgroup$
– Lee Mosher
Jan 31 at 14:50
$begingroup$
Isn’t there a unique complex structure making the quotient holomorphic?
$endgroup$
– Sir Wilfred Lucas-Dockery
Jan 31 at 14:52
$begingroup$
On the other hand, the orbifold structure on $mathbb H / PSL_2(mathbb Z)$ is naturally inherited from $mathbb H$. The conformal structure that $mathbb H / PSL_2(mathbb Z) - {text{the two cone points}}$ inherits from $mathbb H$ also extends naturally over $mathbb H / PSL_2(mathbb Z)$, because the singularities where those two cone points were removed are, indeed, removable singularities.
$endgroup$
– Lee Mosher
Jan 31 at 14:53
|
show 1 more comment
$begingroup$
Isn't $mathbb{H}/PSL_2(mathbb{Z})$ a smooth manifold? Near the fixed points, isn't the action isomorphic to the quotient of the disc by some finite rotation group? So near a point with stabilizer of order $2$, wouldn't a coordinate chart be realised by a map of the form $zmapsto z^2$? Similarly, for points with stabilizer of order $3$?
$endgroup$
– Sir Wilfred Lucas-Dockery
Jan 31 at 12:36
$begingroup$
I should say, the coordinate chart is given by the homeomorphism $D/(mathbb{Z}/2mathbb{Z}) to D$ induced by $zmapsto z^2$.
$endgroup$
– Sir Wilfred Lucas-Dockery
Jan 31 at 13:39
$begingroup$
It is indeed true, as you say, that the smooth structure which $mathbb H / PSL_2(mathbb Z) - {text{the two cone points}}$ inherits from $mathbb H$ extends to a smooth structure on the whole of $mathbb H / PSL_2(mathbb Z)$. But that's a different thing from saying $mathbb H / PSL_2(mathbb Z)$ inherits a smooth structure from $mathbb H$.
$endgroup$
– Lee Mosher
Jan 31 at 14:50
$begingroup$
Isn’t there a unique complex structure making the quotient holomorphic?
$endgroup$
– Sir Wilfred Lucas-Dockery
Jan 31 at 14:52
$begingroup$
On the other hand, the orbifold structure on $mathbb H / PSL_2(mathbb Z)$ is naturally inherited from $mathbb H$. The conformal structure that $mathbb H / PSL_2(mathbb Z) - {text{the two cone points}}$ inherits from $mathbb H$ also extends naturally over $mathbb H / PSL_2(mathbb Z)$, because the singularities where those two cone points were removed are, indeed, removable singularities.
$endgroup$
– Lee Mosher
Jan 31 at 14:53
$begingroup$
Isn't $mathbb{H}/PSL_2(mathbb{Z})$ a smooth manifold? Near the fixed points, isn't the action isomorphic to the quotient of the disc by some finite rotation group? So near a point with stabilizer of order $2$, wouldn't a coordinate chart be realised by a map of the form $zmapsto z^2$? Similarly, for points with stabilizer of order $3$?
$endgroup$
– Sir Wilfred Lucas-Dockery
Jan 31 at 12:36
$begingroup$
Isn't $mathbb{H}/PSL_2(mathbb{Z})$ a smooth manifold? Near the fixed points, isn't the action isomorphic to the quotient of the disc by some finite rotation group? So near a point with stabilizer of order $2$, wouldn't a coordinate chart be realised by a map of the form $zmapsto z^2$? Similarly, for points with stabilizer of order $3$?
$endgroup$
– Sir Wilfred Lucas-Dockery
Jan 31 at 12:36
$begingroup$
I should say, the coordinate chart is given by the homeomorphism $D/(mathbb{Z}/2mathbb{Z}) to D$ induced by $zmapsto z^2$.
$endgroup$
– Sir Wilfred Lucas-Dockery
Jan 31 at 13:39
$begingroup$
I should say, the coordinate chart is given by the homeomorphism $D/(mathbb{Z}/2mathbb{Z}) to D$ induced by $zmapsto z^2$.
$endgroup$
– Sir Wilfred Lucas-Dockery
Jan 31 at 13:39
$begingroup$
It is indeed true, as you say, that the smooth structure which $mathbb H / PSL_2(mathbb Z) - {text{the two cone points}}$ inherits from $mathbb H$ extends to a smooth structure on the whole of $mathbb H / PSL_2(mathbb Z)$. But that's a different thing from saying $mathbb H / PSL_2(mathbb Z)$ inherits a smooth structure from $mathbb H$.
$endgroup$
– Lee Mosher
Jan 31 at 14:50
$begingroup$
It is indeed true, as you say, that the smooth structure which $mathbb H / PSL_2(mathbb Z) - {text{the two cone points}}$ inherits from $mathbb H$ extends to a smooth structure on the whole of $mathbb H / PSL_2(mathbb Z)$. But that's a different thing from saying $mathbb H / PSL_2(mathbb Z)$ inherits a smooth structure from $mathbb H$.
$endgroup$
– Lee Mosher
Jan 31 at 14:50
$begingroup$
Isn’t there a unique complex structure making the quotient holomorphic?
$endgroup$
– Sir Wilfred Lucas-Dockery
Jan 31 at 14:52
$begingroup$
Isn’t there a unique complex structure making the quotient holomorphic?
$endgroup$
– Sir Wilfred Lucas-Dockery
Jan 31 at 14:52
$begingroup$
On the other hand, the orbifold structure on $mathbb H / PSL_2(mathbb Z)$ is naturally inherited from $mathbb H$. The conformal structure that $mathbb H / PSL_2(mathbb Z) - {text{the two cone points}}$ inherits from $mathbb H$ also extends naturally over $mathbb H / PSL_2(mathbb Z)$, because the singularities where those two cone points were removed are, indeed, removable singularities.
$endgroup$
– Lee Mosher
Jan 31 at 14:53
$begingroup$
On the other hand, the orbifold structure on $mathbb H / PSL_2(mathbb Z)$ is naturally inherited from $mathbb H$. The conformal structure that $mathbb H / PSL_2(mathbb Z) - {text{the two cone points}}$ inherits from $mathbb H$ also extends naturally over $mathbb H / PSL_2(mathbb Z)$, because the singularities where those two cone points were removed are, indeed, removable singularities.
$endgroup$
– Lee Mosher
Jan 31 at 14:53
|
show 1 more comment
$begingroup$
If $Gamma$ is a torsion free lattice in $G:=PSL_2(mathbb{R})$, thus acting freely and discontinuously on $mathbb{H}$, one can unambiguously make sense of $T^1(Gammabackslashmathbb{H})$ since $Gammabackslashmathbb{H}$ is a nice Riemannian manifold with the projection preserving the Riemannian metric. Moreover, one can identify $Gammabackslash G$ with $T^1(Gammabackslashmathbb{H})$ and one can see that the $mathbb{R}$-action of diagonal matrices with entries $lbrace e^{t/2},e^{-t/2}rbrace$ is intertwined with the geodesic action on $T^1(Gammabackslashmathbb{H})$ under this identification.
But in general when $Gamma$ is has some fixed points, the notation $T^1(Gammabackslashmathbb{H})$ becomes ambiguousinnacurate as you have noticed. However, there is still the space $Gammabackslash G$ with a nice action by diagonal matrices. Most authors continue to work the the diagonal matrix action on this space and simply call it the 'unit tangent bundle over $Gammabackslashmathbb{H}$' with the 'geodesic action' as an abuse of notation (cf. the book by Bekka-Mayer, page $59$).
As a beginner in ergodic theory, I have noticed that working with torsion lattices in $G$ is quite a headache. In Bekka-Mayer's book, they assume torsion-free-ness for a number of their results simply for convenience (cf. section III$.3$, page $93$).
$endgroup$
add a comment |
$begingroup$
If $Gamma$ is a torsion free lattice in $G:=PSL_2(mathbb{R})$, thus acting freely and discontinuously on $mathbb{H}$, one can unambiguously make sense of $T^1(Gammabackslashmathbb{H})$ since $Gammabackslashmathbb{H}$ is a nice Riemannian manifold with the projection preserving the Riemannian metric. Moreover, one can identify $Gammabackslash G$ with $T^1(Gammabackslashmathbb{H})$ and one can see that the $mathbb{R}$-action of diagonal matrices with entries $lbrace e^{t/2},e^{-t/2}rbrace$ is intertwined with the geodesic action on $T^1(Gammabackslashmathbb{H})$ under this identification.
But in general when $Gamma$ is has some fixed points, the notation $T^1(Gammabackslashmathbb{H})$ becomes ambiguousinnacurate as you have noticed. However, there is still the space $Gammabackslash G$ with a nice action by diagonal matrices. Most authors continue to work the the diagonal matrix action on this space and simply call it the 'unit tangent bundle over $Gammabackslashmathbb{H}$' with the 'geodesic action' as an abuse of notation (cf. the book by Bekka-Mayer, page $59$).
As a beginner in ergodic theory, I have noticed that working with torsion lattices in $G$ is quite a headache. In Bekka-Mayer's book, they assume torsion-free-ness for a number of their results simply for convenience (cf. section III$.3$, page $93$).
$endgroup$
add a comment |
$begingroup$
If $Gamma$ is a torsion free lattice in $G:=PSL_2(mathbb{R})$, thus acting freely and discontinuously on $mathbb{H}$, one can unambiguously make sense of $T^1(Gammabackslashmathbb{H})$ since $Gammabackslashmathbb{H}$ is a nice Riemannian manifold with the projection preserving the Riemannian metric. Moreover, one can identify $Gammabackslash G$ with $T^1(Gammabackslashmathbb{H})$ and one can see that the $mathbb{R}$-action of diagonal matrices with entries $lbrace e^{t/2},e^{-t/2}rbrace$ is intertwined with the geodesic action on $T^1(Gammabackslashmathbb{H})$ under this identification.
But in general when $Gamma$ is has some fixed points, the notation $T^1(Gammabackslashmathbb{H})$ becomes ambiguousinnacurate as you have noticed. However, there is still the space $Gammabackslash G$ with a nice action by diagonal matrices. Most authors continue to work the the diagonal matrix action on this space and simply call it the 'unit tangent bundle over $Gammabackslashmathbb{H}$' with the 'geodesic action' as an abuse of notation (cf. the book by Bekka-Mayer, page $59$).
As a beginner in ergodic theory, I have noticed that working with torsion lattices in $G$ is quite a headache. In Bekka-Mayer's book, they assume torsion-free-ness for a number of their results simply for convenience (cf. section III$.3$, page $93$).
$endgroup$
If $Gamma$ is a torsion free lattice in $G:=PSL_2(mathbb{R})$, thus acting freely and discontinuously on $mathbb{H}$, one can unambiguously make sense of $T^1(Gammabackslashmathbb{H})$ since $Gammabackslashmathbb{H}$ is a nice Riemannian manifold with the projection preserving the Riemannian metric. Moreover, one can identify $Gammabackslash G$ with $T^1(Gammabackslashmathbb{H})$ and one can see that the $mathbb{R}$-action of diagonal matrices with entries $lbrace e^{t/2},e^{-t/2}rbrace$ is intertwined with the geodesic action on $T^1(Gammabackslashmathbb{H})$ under this identification.
But in general when $Gamma$ is has some fixed points, the notation $T^1(Gammabackslashmathbb{H})$ becomes ambiguousinnacurate as you have noticed. However, there is still the space $Gammabackslash G$ with a nice action by diagonal matrices. Most authors continue to work the the diagonal matrix action on this space and simply call it the 'unit tangent bundle over $Gammabackslashmathbb{H}$' with the 'geodesic action' as an abuse of notation (cf. the book by Bekka-Mayer, page $59$).
As a beginner in ergodic theory, I have noticed that working with torsion lattices in $G$ is quite a headache. In Bekka-Mayer's book, they assume torsion-free-ness for a number of their results simply for convenience (cf. section III$.3$, page $93$).
edited Jan 31 at 16:01
answered Jan 31 at 15:48
Sir Wilfred Lucas-DockerySir Wilfred Lucas-Dockery
369419
369419
add a comment |
add a comment |
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$begingroup$
See page $5$ here, why the unit tangent bundle of the modular surface $M=Bbb H^2/PSL_2(Bbb Z)$ is just the quotient space $X=PSL_2(Bbb R)/PSL_2(Bbb Z)$.
$endgroup$
– Dietrich Burde
Jan 26 at 11:43
$begingroup$
The document mentions that $mathbb H^2/PSL_2(mathbb Z)$ is not a manifold but an "orbifold". In that case it is not clear what is the meaning of $T^1(mathbb H^2/PSL_2(mathbb Z))$. I looked up the definition of an orbifold on wikipedia but didn't help. Can you say what is going on here?
$endgroup$
– caffeinemachine
Jan 26 at 12:10
$begingroup$
Yes, the fibered bundle has singularities at the elliptic fixed points, but this can be fixed, see S. Katok, Fuchsian groups, $§ 3.6$.
$endgroup$
– Dietrich Burde
Jan 26 at 12:31
$begingroup$
$mathbb{H}to mathbb{H}/PSL_2(mathbb{Z})$ is a submersion? What is the differential at $i$ or $1/2+isqrt{3}/2$?
$endgroup$
– Sir Wilfred Lucas-Dockery
Jan 31 at 13:07
$begingroup$
@Unpud For $mathbb Hto mathbb H/PSL_2(mathbb Z)$ to be a submersion the target has to be a manifold. But it is not. Or were you trying to say something else?
$endgroup$
– caffeinemachine
Jan 31 at 13:13