Normal subgroup in a matrix Lie group
$begingroup$
Prop: If $G$ is a Matrix Lie group, then the connected component that contains the identity $I$ is a normal subgroup of $G$.
I have problem in the proof of this. Suppose that $A$ and $B$ belong to the connected component that contains $I$, then exist two continuos function ($A(t), B(t)$) in $G$ such that $A(0)=B(0)=I$ and $A(1)=A$, $B(1)=B$.
My question is: if we consider $A(t)B(t)$ why this path is contained in the connected component of $G$?
lie-groups lie-algebras
asked Dec 23 '18 at 12:50
Domenico VuonoDomenico Vuono
2,3161623
$endgroup$
add a comment |
$begingroup$
Prop: If $G$ is a Matrix Lie group, then the connected component that contains the identity $I$ is a normal subgroup of $G$.
I have problem in the proof of this. Suppose that $A$ and $B$ belong to the connected component that contains $I$, then exist two continuos function ($A(t), B(t)$) in $G$ such that $A(0)=B(0)=I$ and $A(1)=A$, $B(1)=B$.
My question is: if we consider $A(t)B(t)$ why this path is contained in the connected component of $G$?
lie-groups lie-algebras
asked Dec 23 '18 at 12:50
Domenico VuonoDomenico Vuono
2,3161623
$endgroup$
add a comment |
$begingroup$
Prop: If $G$ is a Matrix Lie group, then the connected component that contains the identity $I$ is a normal subgroup of $G$.
I have problem in the proof of this. Suppose that $A$ and $B$ belong to the connected component that contains $I$, then exist two continuos function ($A(t), B(t)$) in $G$ such that $A(0)=B(0)=I$ and $A(1)=A$, $B(1)=B$.
My question is: if we consider $A(t)B(t)$ why this path is contained in the connected component of $G$?
lie-groups lie-algebras
asked Dec 23 '18 at 12:50
Domenico VuonoDomenico Vuono
2,3161623
$endgroup$
Prop: If $G$ is a Matrix Lie group, then the connected component that contains the identity $I$ is a normal subgroup of $G$.
I have problem in the proof of this. Suppose that $A$ and $B$ belong to the connected component that contains $I$, then exist two continuos function ($A(t), B(t)$) in $G$ such that $A(0)=B(0)=I$ and $A(1)=A$, $B(1)=B$.
My question is: if we consider $A(t)B(t)$ why this path is contained in the connected component of $G$?
lie-groups lie-algebras
lie-groups lie-algebras
asked Dec 23 '18 at 12:50
Domenico VuonoDomenico Vuono
2,3161623
asked Dec 23 '18 at 12:50
Domenico VuonoDomenico Vuono
2,3161623
asked Dec 23 '18 at 12:50
Domenico VuonoDomenico Vuono
2,3161623
asked Dec 23 '18 at 12:50
Domenico VuonoDomenico Vuono
2,3161623
asked Dec 23 '18 at 12:50
Domenico VuonoDomenico Vuono
2,3161623
2,3161623
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
Because the mapt $tmapsto A(t)B(t)$ is continuous and its domain is conneced (it is an interval). Therefore, its range is connected too. Since, furtheremore, $operatorname{Id}$ belongs to the range, the range is a subset of the connected component of $operatorname{Id}$.
answered Dec 23 '18 at 13:04
José Carlos SantosJosé Carlos Santos
176k24136245
$endgroup$
$begingroup$
thanks, very easy
$endgroup$
– Domenico Vuono
Dec 23 '18 at 13:07
add a comment |
$begingroup$
If $gammacolon[0,1]$ is a path from $a$ to $b$, then $a^{-1}gamma$ is a path from $1$ to $a^{-1}b$. Hence in any topological group, the path-connected component of $1$ is a subgroup.
Moreover, for $gin G$, if $gamma$ is a path from $1$ to $a$, then $g^{-1}gamma g$ is a path from $1$ to $g^{-1}ag$, hence that subroup is normal.
answered Dec 23 '18 at 14:09
Hagen von EitzenHagen von Eitzen
2943
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Because the mapt $tmapsto A(t)B(t)$ is continuous and its domain is conneced (it is an interval). Therefore, its range is connected too. Since, furtheremore, $operatorname{Id}$ belongs to the range, the range is a subset of the connected component of $operatorname{Id}$.
answered Dec 23 '18 at 13:04
José Carlos SantosJosé Carlos Santos
176k24136245
$endgroup$
$begingroup$
thanks, very easy
$endgroup$
– Domenico Vuono
Dec 23 '18 at 13:07
add a comment |
$begingroup$
Because the mapt $tmapsto A(t)B(t)$ is continuous and its domain is conneced (it is an interval). Therefore, its range is connected too. Since, furtheremore, $operatorname{Id}$ belongs to the range, the range is a subset of the connected component of $operatorname{Id}$.
answered Dec 23 '18 at 13:04
José Carlos SantosJosé Carlos Santos
176k24136245
$endgroup$
$begingroup$
thanks, very easy
$endgroup$
– Domenico Vuono
Dec 23 '18 at 13:07
add a comment |
$begingroup$
Because the mapt $tmapsto A(t)B(t)$ is continuous and its domain is conneced (it is an interval). Therefore, its range is connected too. Since, furtheremore, $operatorname{Id}$ belongs to the range, the range is a subset of the connected component of $operatorname{Id}$.
answered Dec 23 '18 at 13:04
José Carlos SantosJosé Carlos Santos
176k24136245
$endgroup$
Because the mapt $tmapsto A(t)B(t)$ is continuous and its domain is conneced (it is an interval). Therefore, its range is connected too. Since, furtheremore, $operatorname{Id}$ belongs to the range, the range is a subset of the connected component of $operatorname{Id}$.
answered Dec 23 '18 at 13:04
José Carlos SantosJosé Carlos Santos
176k24136245
answered Dec 23 '18 at 13:04
José Carlos SantosJosé Carlos Santos
176k24136245
answered Dec 23 '18 at 13:04
José Carlos SantosJosé Carlos Santos
176k24136245
answered Dec 23 '18 at 13:04
José Carlos SantosJosé Carlos Santos
176k24136245
176k24136245
$begingroup$
thanks, very easy
$endgroup$
– Domenico Vuono
Dec 23 '18 at 13:07
add a comment |
$begingroup$
thanks, very easy
$endgroup$
– Domenico Vuono
Dec 23 '18 at 13:07
$begingroup$
thanks, very easy
$endgroup$
– Domenico Vuono
Dec 23 '18 at 13:07
$begingroup$
thanks, very easy
$endgroup$
– Domenico Vuono
Dec 23 '18 at 13:07
add a comment |
$begingroup$
If $gammacolon[0,1]$ is a path from $a$ to $b$, then $a^{-1}gamma$ is a path from $1$ to $a^{-1}b$. Hence in any topological group, the path-connected component of $1$ is a subgroup.
Moreover, for $gin G$, if $gamma$ is a path from $1$ to $a$, then $g^{-1}gamma g$ is a path from $1$ to $g^{-1}ag$, hence that subroup is normal.
answered Dec 23 '18 at 14:09
Hagen von EitzenHagen von Eitzen
2943
$endgroup$
add a comment |
$begingroup$
If $gammacolon[0,1]$ is a path from $a$ to $b$, then $a^{-1}gamma$ is a path from $1$ to $a^{-1}b$. Hence in any topological group, the path-connected component of $1$ is a subgroup.
Moreover, for $gin G$, if $gamma$ is a path from $1$ to $a$, then $g^{-1}gamma g$ is a path from $1$ to $g^{-1}ag$, hence that subroup is normal.
answered Dec 23 '18 at 14:09
Hagen von EitzenHagen von Eitzen
2943
$endgroup$
add a comment |
$begingroup$
If $gammacolon[0,1]$ is a path from $a$ to $b$, then $a^{-1}gamma$ is a path from $1$ to $a^{-1}b$. Hence in any topological group, the path-connected component of $1$ is a subgroup.
Moreover, for $gin G$, if $gamma$ is a path from $1$ to $a$, then $g^{-1}gamma g$ is a path from $1$ to $g^{-1}ag$, hence that subroup is normal.
answered Dec 23 '18 at 14:09
Hagen von EitzenHagen von Eitzen
2943
$endgroup$
If $gammacolon[0,1]$ is a path from $a$ to $b$, then $a^{-1}gamma$ is a path from $1$ to $a^{-1}b$. Hence in any topological group, the path-connected component of $1$ is a subgroup.
Moreover, for $gin G$, if $gamma$ is a path from $1$ to $a$, then $g^{-1}gamma g$ is a path from $1$ to $g^{-1}ag$, hence that subroup is normal.
answered Dec 23 '18 at 14:09
Hagen von EitzenHagen von Eitzen
2943
answered Dec 23 '18 at 14:09
Hagen von EitzenHagen von Eitzen
2943
answered Dec 23 '18 at 14:09
Hagen von EitzenHagen von Eitzen
2943
answered Dec 23 '18 at 14:09
Hagen von EitzenHagen von Eitzen
2943
2943
add a comment |
add a comment |
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