Quotient map from $GL_2(Bbb R)$

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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
add a comment |Â
up vote
0
down vote
favorite
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
@jykri What's IIRC?
â Reznick
Sep 2 at 5:11
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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
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
general-topology
edited Sep 2 at 4:11
asked Sep 2 at 4:05
Reznick
13710
13710
@jykri What's IIRC?
â Reznick
Sep 2 at 5:11
add a comment |Â
@jykri What's IIRC?
â Reznick
Sep 2 at 5:11
@jykri What's IIRC?
â Reznick
Sep 2 at 5:11
@jykri What's IIRC?
â Reznick
Sep 2 at 5:11
add a comment |Â
1 Answer
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1
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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.
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.
â Jyrki Lahtonen
Sep 2 at 17:05
add a comment |Â
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.
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.
â Jyrki Lahtonen
Sep 2 at 17:05
add a comment |Â
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.
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.
â Jyrki Lahtonen
Sep 2 at 17:05
add a comment |Â
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.
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.
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.
â Jyrki Lahtonen
Sep 2 at 17:05
add a comment |Â
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.
â 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.
â 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.
â Jyrki Lahtonen
Sep 2 at 17:05
add a comment |Â
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@jykri What's IIRC?
â Reznick
Sep 2 at 5:11