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Non-trivial open connected subgroup of a connected Lie Group
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Let $G$ be a connected Lie group. Then, I would like to show that there does not
exist a open connected Lie subgroup $H$ such that $esubsetneq H subsetneq G$.
Any help or hint would be very helpful!
lie-groups
asked Jul 28 at 8:38
Shubham Namdeo
321111
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up vote
0
down vote
favorite
Let $G$ be a connected Lie group. Then, I would like to show that there does not
exist a open connected Lie subgroup $H$ such that $esubsetneq H subsetneq G$.
Any help or hint would be very helpful!
lie-groups
asked Jul 28 at 8:38
Shubham Namdeo
321111
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $G$ be a connected Lie group. Then, I would like to show that there does not
exist a open connected Lie subgroup $H$ such that $esubsetneq H subsetneq G$.
Any help or hint would be very helpful!
lie-groups
asked Jul 28 at 8:38
Shubham Namdeo
321111
Let $G$ be a connected Lie group. Then, I would like to show that there does not
exist a open connected Lie subgroup $H$ such that $esubsetneq H subsetneq G$.
Any help or hint would be very helpful!
lie-groups
asked Jul 28 at 8:38
Shubham Namdeo
321111
asked Jul 28 at 8:38
Shubham Namdeo
321111
asked Jul 28 at 8:38
Shubham Namdeo
321111
asked Jul 28 at 8:38
Shubham Namdeo
321111
321111
add a comment |Â
add a comment |Â
2 Answers
2
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1
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accepted
That's true because, on any connected Lie group, any neighborhood of $e$ spans the whole group. So, if $H$ is open, it is a neighborhhod of $e$ and therefore the subgroup spanned by $H$ (which is $H$ itself) is the whole group $G$.
answered Jul 28 at 8:44
José Carlos Santos
112k1696173
add a comment |Â
up vote
1
down vote
It is not hard to show that open subgroups are also topologically closed. Now your intermediate subgroup and its complement separate $G$, contradicting connectivity. You do not need the connectedness assumption on $H$.
answered Jul 28 at 17:50
Randall
7,2221825
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
That's true because, on any connected Lie group, any neighborhood of $e$ spans the whole group. So, if $H$ is open, it is a neighborhhod of $e$ and therefore the subgroup spanned by $H$ (which is $H$ itself) is the whole group $G$.
answered Jul 28 at 8:44
José Carlos Santos
112k1696173
add a comment |Â
up vote
1
down vote
accepted
That's true because, on any connected Lie group, any neighborhood of $e$ spans the whole group. So, if $H$ is open, it is a neighborhhod of $e$ and therefore the subgroup spanned by $H$ (which is $H$ itself) is the whole group $G$.
answered Jul 28 at 8:44
José Carlos Santos
112k1696173
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
That's true because, on any connected Lie group, any neighborhood of $e$ spans the whole group. So, if $H$ is open, it is a neighborhhod of $e$ and therefore the subgroup spanned by $H$ (which is $H$ itself) is the whole group $G$.
answered Jul 28 at 8:44
José Carlos Santos
112k1696173
That's true because, on any connected Lie group, any neighborhood of $e$ spans the whole group. So, if $H$ is open, it is a neighborhhod of $e$ and therefore the subgroup spanned by $H$ (which is $H$ itself) is the whole group $G$.
answered Jul 28 at 8:44
José Carlos Santos
112k1696173
answered Jul 28 at 8:44
José Carlos Santos
112k1696173
answered Jul 28 at 8:44
José Carlos Santos
112k1696173
answered Jul 28 at 8:44
José Carlos Santos
112k1696173
112k1696173
add a comment |Â
add a comment |Â
up vote
1
down vote
It is not hard to show that open subgroups are also topologically closed. Now your intermediate subgroup and its complement separate $G$, contradicting connectivity. You do not need the connectedness assumption on $H$.
answered Jul 28 at 17:50
Randall
7,2221825
add a comment |Â
up vote
1
down vote
It is not hard to show that open subgroups are also topologically closed. Now your intermediate subgroup and its complement separate $G$, contradicting connectivity. You do not need the connectedness assumption on $H$.
answered Jul 28 at 17:50
Randall
7,2221825
add a comment |Â
up vote
1
down vote
up vote
1
down vote
It is not hard to show that open subgroups are also topologically closed. Now your intermediate subgroup and its complement separate $G$, contradicting connectivity. You do not need the connectedness assumption on $H$.
answered Jul 28 at 17:50
Randall
7,2221825
It is not hard to show that open subgroups are also topologically closed. Now your intermediate subgroup and its complement separate $G$, contradicting connectivity. You do not need the connectedness assumption on $H$.
answered Jul 28 at 17:50
Randall
7,2221825
answered Jul 28 at 17:50
Randall
7,2221825
answered Jul 28 at 17:50
Randall
7,2221825
answered Jul 28 at 17:50
Randall
7,2221825
7,2221825
add a comment |Â
add a comment |Â
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