Mixing
Normal Fuchsian subgroups
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I've been working with Fuchsian groups and from geometrical motivations finding a cocompact normal Fuchsian subgroups of $PSL(2,mathbbR)$ would have intresting properties for my research.
It is known that $SL(2,mathbbR)$ has no connected normal subgroups other than the its centre Id,-Id , however it might have discrete normal subgroups. I been working with quaternion generated cocompact Fuchsian subgroups which are in addition purely hyperbolic, all its elements have Trace bigger 2, and I couln't find any normal subgroups of $SL(2,mathbbR)$ there, I suspect that I have to include some elliptic elements to improve my chances.
To add up: it possible for $PSL(2,mathbbR)$ to admit discrete normal subgroups? how about cocompact normal subgroups?
Note: Notice that I'm not talking about normal subgroups of Fuchsian subgroups.
lie-groups fuchsian-groups normal-subgroups
asked 1 hour ago
P. W. Maunt
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up vote
2
down vote
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I've been working with Fuchsian groups and from geometrical motivations finding a cocompact normal Fuchsian subgroups of $PSL(2,mathbbR)$ would have intresting properties for my research.
It is known that $SL(2,mathbbR)$ has no connected normal subgroups other than the its centre Id,-Id , however it might have discrete normal subgroups. I been working with quaternion generated cocompact Fuchsian subgroups which are in addition purely hyperbolic, all its elements have Trace bigger 2, and I couln't find any normal subgroups of $SL(2,mathbbR)$ there, I suspect that I have to include some elliptic elements to improve my chances.
To add up: it possible for $PSL(2,mathbbR)$ to admit discrete normal subgroups? how about cocompact normal subgroups?
Note: Notice that I'm not talking about normal subgroups of Fuchsian subgroups.
lie-groups fuchsian-groups normal-subgroups
asked 1 hour ago
P. W. Maunt
133
New contributor
P. W. Maunt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I've been working with Fuchsian groups and from geometrical motivations finding a cocompact normal Fuchsian subgroups of $PSL(2,mathbbR)$ would have intresting properties for my research.
It is known that $SL(2,mathbbR)$ has no connected normal subgroups other than the its centre Id,-Id , however it might have discrete normal subgroups. I been working with quaternion generated cocompact Fuchsian subgroups which are in addition purely hyperbolic, all its elements have Trace bigger 2, and I couln't find any normal subgroups of $SL(2,mathbbR)$ there, I suspect that I have to include some elliptic elements to improve my chances.
To add up: it possible for $PSL(2,mathbbR)$ to admit discrete normal subgroups? how about cocompact normal subgroups?
Note: Notice that I'm not talking about normal subgroups of Fuchsian subgroups.
lie-groups fuchsian-groups normal-subgroups
asked 1 hour ago
P. W. Maunt
133
New contributor
P. W. Maunt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I've been working with Fuchsian groups and from geometrical motivations finding a cocompact normal Fuchsian subgroups of $PSL(2,mathbbR)$ would have intresting properties for my research.
It is known that $SL(2,mathbbR)$ has no connected normal subgroups other than the its centre Id,-Id , however it might have discrete normal subgroups. I been working with quaternion generated cocompact Fuchsian subgroups which are in addition purely hyperbolic, all its elements have Trace bigger 2, and I couln't find any normal subgroups of $SL(2,mathbbR)$ there, I suspect that I have to include some elliptic elements to improve my chances.
To add up: it possible for $PSL(2,mathbbR)$ to admit discrete normal subgroups? how about cocompact normal subgroups?
Note: Notice that I'm not talking about normal subgroups of Fuchsian subgroups.
lie-groups fuchsian-groups normal-subgroups
lie-groups fuchsian-groups normal-subgroups
asked 1 hour ago
P. W. Maunt
133
New contributor
P. W. Maunt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
P. W. Maunt
133
New contributor
P. W. Maunt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
P. W. Maunt
133
New contributor
P. W. Maunt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
P. W. Maunt
133
asked 1 hour ago
P. W. Maunt
133
133
New contributor
P. W. Maunt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
P. W. Maunt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
P. W. Maunt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
1 Answer
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accepted
Let $Gamma$ be a discrete subgroup of a connected Lie group $G$.
Suppose that $Gamma$ is normal. For given $gammainGamma$ the conjugacy class is the image of $G$ under the continuous map $xmapsto xgamma x^-1$, therefore it is connected. Since it lies in $Gamma$, it consists of one point only, and as $x=1$ occurs, this point is $gamma$. This means that $Gamma$ lies in the center of $G$.
In the case $G=SL_2(mathbb R)$ the center is $ pm 1$. So $Gamma $ is either trivial or $pm 1$.
answered 1 hour ago
Corbennick
6,0171835
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Let $Gamma$ be a discrete subgroup of a connected Lie group $G$.
Suppose that $Gamma$ is normal. For given $gammainGamma$ the conjugacy class is the image of $G$ under the continuous map $xmapsto xgamma x^-1$, therefore it is connected. Since it lies in $Gamma$, it consists of one point only, and as $x=1$ occurs, this point is $gamma$. This means that $Gamma$ lies in the center of $G$.
In the case $G=SL_2(mathbb R)$ the center is $ pm 1$. So $Gamma $ is either trivial or $pm 1$.
answered 1 hour ago
Corbennick
6,0171835
add a comment |Â
up vote
3
down vote
accepted
Let $Gamma$ be a discrete subgroup of a connected Lie group $G$.
Suppose that $Gamma$ is normal. For given $gammainGamma$ the conjugacy class is the image of $G$ under the continuous map $xmapsto xgamma x^-1$, therefore it is connected. Since it lies in $Gamma$, it consists of one point only, and as $x=1$ occurs, this point is $gamma$. This means that $Gamma$ lies in the center of $G$.
In the case $G=SL_2(mathbb R)$ the center is $ pm 1$. So $Gamma $ is either trivial or $pm 1$.
answered 1 hour ago
Corbennick
6,0171835
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Let $Gamma$ be a discrete subgroup of a connected Lie group $G$.
Suppose that $Gamma$ is normal. For given $gammainGamma$ the conjugacy class is the image of $G$ under the continuous map $xmapsto xgamma x^-1$, therefore it is connected. Since it lies in $Gamma$, it consists of one point only, and as $x=1$ occurs, this point is $gamma$. This means that $Gamma$ lies in the center of $G$.
In the case $G=SL_2(mathbb R)$ the center is $ pm 1$. So $Gamma $ is either trivial or $pm 1$.
answered 1 hour ago
Corbennick
6,0171835
Let $Gamma$ be a discrete subgroup of a connected Lie group $G$.
Suppose that $Gamma$ is normal. For given $gammainGamma$ the conjugacy class is the image of $G$ under the continuous map $xmapsto xgamma x^-1$, therefore it is connected. Since it lies in $Gamma$, it consists of one point only, and as $x=1$ occurs, this point is $gamma$. This means that $Gamma$ lies in the center of $G$.
In the case $G=SL_2(mathbb R)$ the center is $ pm 1$. So $Gamma $ is either trivial or $pm 1$.
answered 1 hour ago
Corbennick
6,0171835
answered 1 hour ago
Corbennick
6,0171835
answered 1 hour ago
Corbennick
6,0171835
answered 1 hour ago
Corbennick
6,0171835
6,0171835
add a comment |Â
add a comment |Â
P. W. Maunt is a new contributor. Be nice, and check out our Code of Conduct.
P. W. Maunt is a new contributor. Be nice, and check out our Code of Conduct.
P. W. Maunt is a new contributor. Be nice, and check out our Code of Conduct.
P. W. Maunt is a new contributor. Be nice, and check out our Code of Conduct.
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