What is $T^1(mathbb H^2/PSL_2(mathbb Z))$?












2












$begingroup$


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 $T^1(mathbb H^2)$.










share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















2












$begingroup$


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 $T^1(mathbb H^2)$.










share|cite|improve this question











$endgroup$












  • $begingroup$
    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














2












2








2


1



$begingroup$


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 $T^1(mathbb H^2)$.










share|cite|improve this question











$endgroup$




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






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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


















  • $begingroup$
    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










2 Answers
2






active

oldest

votes


















0












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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





















1












$begingroup$

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 $93$).






share|cite|improve this answer











$endgroup$













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    2 Answers
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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      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


















    0












    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      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
















    0












    0








    0





    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$



    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.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    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




















    • $begingroup$
      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













    1












    $begingroup$

    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 $93$).






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      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 $93$).






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        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 $93$).






        share|cite|improve this answer











        $endgroup$



        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 $93$).







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 31 at 16:01

























        answered Jan 31 at 15:48









        Sir Wilfred Lucas-DockerySir Wilfred Lucas-Dockery

        369419




        369419






























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