Quotient map from $GL_2(Bbb R)$

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
0
down vote

favorite
1












Is the following map a quotient map?
$$f:GL_2(Bbb R) to Bbb C setminus text0$$
$$
beginpmatrix
a & b\
c & d\
end pmatrix
mapsto fracaiota+bciota+d $$



I wanted a quotient map from $SL_2(Bbb R)$ to $mathscr H$(the upper half plane), if I could show $f$ is a quotient map then somehow the restriction to $SL_2(Bbb R)$ could be shown to be a quotient map.










share|cite|improve this question























  • @jykri What's IIRC?
    – Reznick
    Sep 2 at 5:11














up vote
0
down vote

favorite
1












Is the following map a quotient map?
$$f:GL_2(Bbb R) to Bbb C setminus text0$$
$$
beginpmatrix
a & b\
c & d\
end pmatrix
mapsto fracaiota+bciota+d $$



I wanted a quotient map from $SL_2(Bbb R)$ to $mathscr H$(the upper half plane), if I could show $f$ is a quotient map then somehow the restriction to $SL_2(Bbb R)$ could be shown to be a quotient map.










share|cite|improve this question























  • @jykri What's IIRC?
    – Reznick
    Sep 2 at 5:11












up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





Is the following map a quotient map?
$$f:GL_2(Bbb R) to Bbb C setminus text0$$
$$
beginpmatrix
a & b\
c & d\
end pmatrix
mapsto fracaiota+bciota+d $$



I wanted a quotient map from $SL_2(Bbb R)$ to $mathscr H$(the upper half plane), if I could show $f$ is a quotient map then somehow the restriction to $SL_2(Bbb R)$ could be shown to be a quotient map.










share|cite|improve this question















Is the following map a quotient map?
$$f:GL_2(Bbb R) to Bbb C setminus text0$$
$$
beginpmatrix
a & b\
c & d\
end pmatrix
mapsto fracaiota+bciota+d $$



I wanted a quotient map from $SL_2(Bbb R)$ to $mathscr H$(the upper half plane), if I could show $f$ is a quotient map then somehow the restriction to $SL_2(Bbb R)$ could be shown to be a quotient map.







general-topology






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 2 at 4:11

























asked Sep 2 at 4:05









Reznick

13710




13710











  • @jykri What's IIRC?
    – Reznick
    Sep 2 at 5:11
















  • @jykri What's IIRC?
    – Reznick
    Sep 2 at 5:11















@jykri What's IIRC?
– Reznick
Sep 2 at 5:11




@jykri What's IIRC?
– Reznick
Sep 2 at 5:11










1 Answer
1






active

oldest

votes

















up vote
1
down vote













Expanding my (possibly somewhat misguided) comment to an argument showing that $f:gammamapsto gammacdotiota$ is a quotient map from $SL_2(BbbR)$ to the upper half plane $H$.
First let's look at the following subgroup
$$
G=leftleft(beginarrayccy^1/2&xy^-1/2\0&y^-1/2endarrayright)mid x,yinBbbR,y>0rightle SL_2(BbbR)
$$
consisting of all the upper triangular matrices in $SL_2(BbbR)$ with positive diagonal entries.
Call the described matrix $gamma(x,y)$. We immediately see that $f(gamma(x,y))=x+iyin H$. In other words the restriction of $f$ to the subgroup $G$ is a bijection.



We also see easily that the stabilizer of $iota$ in $SL_2(BbbR)$ consists of the matrices of plane rotations w.r.t. the usual basis:
$$
Stab_SL_2(iota)=K=leftleft(beginarraycccostheta&-sintheta\ sintheta&costhetaendarrayright)mid thetainBbbRright.
$$



We also see that $G$ is a closed subgroup of $SL_2(BbbR)$. Futhermore, if we identify the elements of $G$ with the corresponding points in $H$, we get a homeomorphism.
Putting all the pieces together we see that:



  • If $Csubseteq H$ is closed, then the set $tildeC:=f^-1(C)cap
    G$ is closed as a subset of $SL_2(BbbR)$.

  • $f^-1(C)=tildeCK=ckmid cintildeC, kin K$ is a closed subset of $SL_2(BbbR)$ as a product of a closed and a compact subset.

  • If $D$ is not a closed subset of $H$ then $f^-1(D)cap G$ is not a closed subset of $G$ either (homeomorphism!). Consequently $f^-1(D)$ cannot be a closed subset of $SL_2$.

  • So $Csubseteq H$ is closed if and only if $f^-1(C)subseteq SL_2(BbbR)$ is closed proving that $f$ is a (topological) quotient map.


I'm sure that there exists a suitable more general result you can use. Say, about a topological group acting transitively on a space such that all the point stabilizers are compact? Or some such. I plead ignorance at this point, and look forward to be educated.






share|cite|improve this answer






















  • Wonder whether this has been handled in a number theory/modular form context? I didn't find a dupe with my first attempt, but I should search more carefully.
    – Jyrki Lahtonen
    Sep 2 at 17:05










Your Answer




StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);













 

draft saved


draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2902330%2fquotient-map-from-gl-2-bbb-r%23new-answer', 'question_page');

);

Post as a guest






























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote













Expanding my (possibly somewhat misguided) comment to an argument showing that $f:gammamapsto gammacdotiota$ is a quotient map from $SL_2(BbbR)$ to the upper half plane $H$.
First let's look at the following subgroup
$$
G=leftleft(beginarrayccy^1/2&xy^-1/2\0&y^-1/2endarrayright)mid x,yinBbbR,y>0rightle SL_2(BbbR)
$$
consisting of all the upper triangular matrices in $SL_2(BbbR)$ with positive diagonal entries.
Call the described matrix $gamma(x,y)$. We immediately see that $f(gamma(x,y))=x+iyin H$. In other words the restriction of $f$ to the subgroup $G$ is a bijection.



We also see easily that the stabilizer of $iota$ in $SL_2(BbbR)$ consists of the matrices of plane rotations w.r.t. the usual basis:
$$
Stab_SL_2(iota)=K=leftleft(beginarraycccostheta&-sintheta\ sintheta&costhetaendarrayright)mid thetainBbbRright.
$$



We also see that $G$ is a closed subgroup of $SL_2(BbbR)$. Futhermore, if we identify the elements of $G$ with the corresponding points in $H$, we get a homeomorphism.
Putting all the pieces together we see that:



  • If $Csubseteq H$ is closed, then the set $tildeC:=f^-1(C)cap
    G$ is closed as a subset of $SL_2(BbbR)$.

  • $f^-1(C)=tildeCK=ckmid cintildeC, kin K$ is a closed subset of $SL_2(BbbR)$ as a product of a closed and a compact subset.

  • If $D$ is not a closed subset of $H$ then $f^-1(D)cap G$ is not a closed subset of $G$ either (homeomorphism!). Consequently $f^-1(D)$ cannot be a closed subset of $SL_2$.

  • So $Csubseteq H$ is closed if and only if $f^-1(C)subseteq SL_2(BbbR)$ is closed proving that $f$ is a (topological) quotient map.


I'm sure that there exists a suitable more general result you can use. Say, about a topological group acting transitively on a space such that all the point stabilizers are compact? Or some such. I plead ignorance at this point, and look forward to be educated.






share|cite|improve this answer






















  • Wonder whether this has been handled in a number theory/modular form context? I didn't find a dupe with my first attempt, but I should search more carefully.
    – Jyrki Lahtonen
    Sep 2 at 17:05














up vote
1
down vote













Expanding my (possibly somewhat misguided) comment to an argument showing that $f:gammamapsto gammacdotiota$ is a quotient map from $SL_2(BbbR)$ to the upper half plane $H$.
First let's look at the following subgroup
$$
G=leftleft(beginarrayccy^1/2&xy^-1/2\0&y^-1/2endarrayright)mid x,yinBbbR,y>0rightle SL_2(BbbR)
$$
consisting of all the upper triangular matrices in $SL_2(BbbR)$ with positive diagonal entries.
Call the described matrix $gamma(x,y)$. We immediately see that $f(gamma(x,y))=x+iyin H$. In other words the restriction of $f$ to the subgroup $G$ is a bijection.



We also see easily that the stabilizer of $iota$ in $SL_2(BbbR)$ consists of the matrices of plane rotations w.r.t. the usual basis:
$$
Stab_SL_2(iota)=K=leftleft(beginarraycccostheta&-sintheta\ sintheta&costhetaendarrayright)mid thetainBbbRright.
$$



We also see that $G$ is a closed subgroup of $SL_2(BbbR)$. Futhermore, if we identify the elements of $G$ with the corresponding points in $H$, we get a homeomorphism.
Putting all the pieces together we see that:



  • If $Csubseteq H$ is closed, then the set $tildeC:=f^-1(C)cap
    G$ is closed as a subset of $SL_2(BbbR)$.

  • $f^-1(C)=tildeCK=ckmid cintildeC, kin K$ is a closed subset of $SL_2(BbbR)$ as a product of a closed and a compact subset.

  • If $D$ is not a closed subset of $H$ then $f^-1(D)cap G$ is not a closed subset of $G$ either (homeomorphism!). Consequently $f^-1(D)$ cannot be a closed subset of $SL_2$.

  • So $Csubseteq H$ is closed if and only if $f^-1(C)subseteq SL_2(BbbR)$ is closed proving that $f$ is a (topological) quotient map.


I'm sure that there exists a suitable more general result you can use. Say, about a topological group acting transitively on a space such that all the point stabilizers are compact? Or some such. I plead ignorance at this point, and look forward to be educated.






share|cite|improve this answer






















  • Wonder whether this has been handled in a number theory/modular form context? I didn't find a dupe with my first attempt, but I should search more carefully.
    – Jyrki Lahtonen
    Sep 2 at 17:05












up vote
1
down vote










up vote
1
down vote









Expanding my (possibly somewhat misguided) comment to an argument showing that $f:gammamapsto gammacdotiota$ is a quotient map from $SL_2(BbbR)$ to the upper half plane $H$.
First let's look at the following subgroup
$$
G=leftleft(beginarrayccy^1/2&xy^-1/2\0&y^-1/2endarrayright)mid x,yinBbbR,y>0rightle SL_2(BbbR)
$$
consisting of all the upper triangular matrices in $SL_2(BbbR)$ with positive diagonal entries.
Call the described matrix $gamma(x,y)$. We immediately see that $f(gamma(x,y))=x+iyin H$. In other words the restriction of $f$ to the subgroup $G$ is a bijection.



We also see easily that the stabilizer of $iota$ in $SL_2(BbbR)$ consists of the matrices of plane rotations w.r.t. the usual basis:
$$
Stab_SL_2(iota)=K=leftleft(beginarraycccostheta&-sintheta\ sintheta&costhetaendarrayright)mid thetainBbbRright.
$$



We also see that $G$ is a closed subgroup of $SL_2(BbbR)$. Futhermore, if we identify the elements of $G$ with the corresponding points in $H$, we get a homeomorphism.
Putting all the pieces together we see that:



  • If $Csubseteq H$ is closed, then the set $tildeC:=f^-1(C)cap
    G$ is closed as a subset of $SL_2(BbbR)$.

  • $f^-1(C)=tildeCK=ckmid cintildeC, kin K$ is a closed subset of $SL_2(BbbR)$ as a product of a closed and a compact subset.

  • If $D$ is not a closed subset of $H$ then $f^-1(D)cap G$ is not a closed subset of $G$ either (homeomorphism!). Consequently $f^-1(D)$ cannot be a closed subset of $SL_2$.

  • So $Csubseteq H$ is closed if and only if $f^-1(C)subseteq SL_2(BbbR)$ is closed proving that $f$ is a (topological) quotient map.


I'm sure that there exists a suitable more general result you can use. Say, about a topological group acting transitively on a space such that all the point stabilizers are compact? Or some such. I plead ignorance at this point, and look forward to be educated.






share|cite|improve this answer














Expanding my (possibly somewhat misguided) comment to an argument showing that $f:gammamapsto gammacdotiota$ is a quotient map from $SL_2(BbbR)$ to the upper half plane $H$.
First let's look at the following subgroup
$$
G=leftleft(beginarrayccy^1/2&xy^-1/2\0&y^-1/2endarrayright)mid x,yinBbbR,y>0rightle SL_2(BbbR)
$$
consisting of all the upper triangular matrices in $SL_2(BbbR)$ with positive diagonal entries.
Call the described matrix $gamma(x,y)$. We immediately see that $f(gamma(x,y))=x+iyin H$. In other words the restriction of $f$ to the subgroup $G$ is a bijection.



We also see easily that the stabilizer of $iota$ in $SL_2(BbbR)$ consists of the matrices of plane rotations w.r.t. the usual basis:
$$
Stab_SL_2(iota)=K=leftleft(beginarraycccostheta&-sintheta\ sintheta&costhetaendarrayright)mid thetainBbbRright.
$$



We also see that $G$ is a closed subgroup of $SL_2(BbbR)$. Futhermore, if we identify the elements of $G$ with the corresponding points in $H$, we get a homeomorphism.
Putting all the pieces together we see that:



  • If $Csubseteq H$ is closed, then the set $tildeC:=f^-1(C)cap
    G$ is closed as a subset of $SL_2(BbbR)$.

  • $f^-1(C)=tildeCK=ckmid cintildeC, kin K$ is a closed subset of $SL_2(BbbR)$ as a product of a closed and a compact subset.

  • If $D$ is not a closed subset of $H$ then $f^-1(D)cap G$ is not a closed subset of $G$ either (homeomorphism!). Consequently $f^-1(D)$ cannot be a closed subset of $SL_2$.

  • So $Csubseteq H$ is closed if and only if $f^-1(C)subseteq SL_2(BbbR)$ is closed proving that $f$ is a (topological) quotient map.


I'm sure that there exists a suitable more general result you can use. Say, about a topological group acting transitively on a space such that all the point stabilizers are compact? Or some such. I plead ignorance at this point, and look forward to be educated.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Sep 4 at 8:07

























answered Sep 2 at 17:03









Jyrki Lahtonen

105k12163358




105k12163358











  • Wonder whether this has been handled in a number theory/modular form context? I didn't find a dupe with my first attempt, but I should search more carefully.
    – Jyrki Lahtonen
    Sep 2 at 17:05
















  • Wonder whether this has been handled in a number theory/modular form context? I didn't find a dupe with my first attempt, but I should search more carefully.
    – Jyrki Lahtonen
    Sep 2 at 17:05















Wonder whether this has been handled in a number theory/modular form context? I didn't find a dupe with my first attempt, but I should search more carefully.
– Jyrki Lahtonen
Sep 2 at 17:05




Wonder whether this has been handled in a number theory/modular form context? I didn't find a dupe with my first attempt, but I should search more carefully.
– Jyrki Lahtonen
Sep 2 at 17:05

















 

draft saved


draft discarded















































 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2902330%2fquotient-map-from-gl-2-bbb-r%23new-answer', 'question_page');

);

Post as a guest













































































這個網誌中的熱門文章

tkz-euclide: tkzDrawCircle[R] not working

How to combine Bézier curves to a surface?

1st Magritte Awards