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Show that every compact Lie group contains a finitely generated dense subgroup.
Clash Royale CLAN TAG#URR8PPP
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I'm trying to show that the connected component has a finite number of distinct maximal tori. So the group generated by its generators must be dense. But I don't know if it is really true.
lie-groups
asked Jul 18 at 18:36
André Gomes
854516
add a comment |Â
up vote
4
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favorite
I'm trying to show that the connected component has a finite number of distinct maximal tori. So the group generated by its generators must be dense. But I don't know if it is really true.
lie-groups
asked Jul 18 at 18:36
André Gomes
854516
1
This proof strategy won't work: if a compact Lie group $G$ has non-abelian identity component, then it has uncountable many distinct maximal tori. This follows, e.g., from the Baire category theorem since each maximal torus is closed nowhere dense subset of the identity component of $G$.
– Jason DeVito
Jul 18 at 19:04
Do you have any hint for another strategy?
– André Gomes
Jul 18 at 19:12
2
Here is the approach I am trying, but I don't know if it works: Lemma (which I don't know how to prove, but I think it true): If $Hsubseteq G$ are connected Lie groups, with $Hneq G$, then there is an element $gin Gsetminus H$ for which the group generated by $g$ only intersects $H$ at $e$. If this lemma is true, I think I can prove your result by induction on $dim G$.
– Jason DeVito
Jul 18 at 19:15
I think I've got it, writing an answer now....
– Jason DeVito
Jul 18 at 19:20
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I'm trying to show that the connected component has a finite number of distinct maximal tori. So the group generated by its generators must be dense. But I don't know if it is really true.
lie-groups
asked Jul 18 at 18:36
André Gomes
854516
I'm trying to show that the connected component has a finite number of distinct maximal tori. So the group generated by its generators must be dense. But I don't know if it is really true.
lie-groups
asked Jul 18 at 18:36
André Gomes
854516
asked Jul 18 at 18:36
André Gomes
854516
asked Jul 18 at 18:36
André Gomes
854516
asked Jul 18 at 18:36
André Gomes
854516
854516
1
This proof strategy won't work: if a compact Lie group $G$ has non-abelian identity component, then it has uncountable many distinct maximal tori. This follows, e.g., from the Baire category theorem since each maximal torus is closed nowhere dense subset of the identity component of $G$.
– Jason DeVito
Jul 18 at 19:04
Do you have any hint for another strategy?
– André Gomes
Jul 18 at 19:12
2
Here is the approach I am trying, but I don't know if it works: Lemma (which I don't know how to prove, but I think it true): If $Hsubseteq G$ are connected Lie groups, with $Hneq G$, then there is an element $gin Gsetminus H$ for which the group generated by $g$ only intersects $H$ at $e$. If this lemma is true, I think I can prove your result by induction on $dim G$.
– Jason DeVito
Jul 18 at 19:15
I think I've got it, writing an answer now....
– Jason DeVito
Jul 18 at 19:20
add a comment |Â
1
This proof strategy won't work: if a compact Lie group $G$ has non-abelian identity component, then it has uncountable many distinct maximal tori. This follows, e.g., from the Baire category theorem since each maximal torus is closed nowhere dense subset of the identity component of $G$.
– Jason DeVito
Jul 18 at 19:04
Do you have any hint for another strategy?
– André Gomes
Jul 18 at 19:12
2
Here is the approach I am trying, but I don't know if it works: Lemma (which I don't know how to prove, but I think it true): If $Hsubseteq G$ are connected Lie groups, with $Hneq G$, then there is an element $gin Gsetminus H$ for which the group generated by $g$ only intersects $H$ at $e$. If this lemma is true, I think I can prove your result by induction on $dim G$.
– Jason DeVito
Jul 18 at 19:15
I think I've got it, writing an answer now....
– Jason DeVito
Jul 18 at 19:20
1
1
This proof strategy won't work: if a compact Lie group $G$ has non-abelian identity component, then it has uncountable many distinct maximal tori. This follows, e.g., from the Baire category theorem since each maximal torus is closed nowhere dense subset of the identity component of $G$.
– Jason DeVito
Jul 18 at 19:04
This proof strategy won't work: if a compact Lie group $G$ has non-abelian identity component, then it has uncountable many distinct maximal tori. This follows, e.g., from the Baire category theorem since each maximal torus is closed nowhere dense subset of the identity component of $G$.
– Jason DeVito
Jul 18 at 19:04
Do you have any hint for another strategy?
– André Gomes
Jul 18 at 19:12
Do you have any hint for another strategy?
– André Gomes
Jul 18 at 19:12
2
2
Here is the approach I am trying, but I don't know if it works: Lemma (which I don't know how to prove, but I think it true): If $Hsubseteq G$ are connected Lie groups, with $Hneq G$, then there is an element $gin Gsetminus H$ for which the group generated by $g$ only intersects $H$ at $e$. If this lemma is true, I think I can prove your result by induction on $dim G$.
– Jason DeVito
Jul 18 at 19:15
Here is the approach I am trying, but I don't know if it works: Lemma (which I don't know how to prove, but I think it true): If $Hsubseteq G$ are connected Lie groups, with $Hneq G$, then there is an element $gin Gsetminus H$ for which the group generated by $g$ only intersects $H$ at $e$. If this lemma is true, I think I can prove your result by induction on $dim G$.
– Jason DeVito
Jul 18 at 19:15
I think I've got it, writing an answer now....
– Jason DeVito
Jul 18 at 19:20
I think I've got it, writing an answer now....
– Jason DeVito
Jul 18 at 19:20
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
We begin with a Lemma.
Lemma: Suppose $Hsubseteq G$ are compact connnected Lie groups with $Hneq G$. Then there is a subgroup $S^1subseteq G$ for which $S^1cap H$ is finite.
Proof: Look on the Lie algebra level. We have $mathfrakhsubseteq mathfrakg$, and $mathfrakhneq mathfrakg$ because $H$ and $G$ are connected and $Hneq G$. Then $mathfrakgsetminus mathfrakh$ is an open dense subset of $mathfrakg$, so contains a vector $v$ for which $exp(tv)$ closes up. Then $exp(tv) = S^1$ is the desired $S^1$.
To see this, note that if $S^1cap H$ is infinite, then it has an accumulation point (since $G$ is compact). Now, by using the group multiplication, we may assume this accumulation point is the identity. It follows that $exp(t_n v) in H$ for a decreasing sequence $t_nrightarrow 0$. This, then, implies that $vinmathfrakh$, a contradiction. $square$
Now, we prove the theorem. Suppose $G$ is any connected Lie group. Let $H = Tsubseteq G$ be a maximal torus. Under the identification $Tcong mathbbR^n/mathbbZ^n$, if we pick an element $x=(x_1,...,x_n)in T$ for which $operatornamespan_mathbbQ1,x_1, x_2,...,x_n$ has dimension $n+1$, it follows that $x$ generates a dense subgroup of $T$.
If $T = G$, we are done. Otherwise, using the lemma, pick $S^1subseteq G$ with $S^1cap T$ finite. We pick $yin S^1$ which generates a dense subgroup of $S^1$.
Let $langlelangle x,yranglerangle$ denote the closure of the subgroup generated by $x$ and $y$. Then clearly $langlelangle x,yranglerangle$ contains $H$ and $S^1$. On the Lie algebra level, the Lie algebra of $langle langle x,yranglerangle$ is a subspace containing $mathfrakt$ and $v$, so has dimension at least that of $dim T + 1$.
Now, we induct. If $langle langle x,yranglerangle neq G$, we use the lemma to pick $zin Gsetminus langlelangle x,yrangle rangle$. By the same argument as above, $langle langle x,y,zranglerangle$ has larger dimension that $langle langle x,y,rangle rangle$. Continuing in this way, since the dimension of the generated subset increases at every stage, the process must stop after finitely many steps.
answered Jul 18 at 19:40
Jason DeVito
29.5k473129
Why is $mathfrakgsetminusmathfrakh$ open and dense in $mathfrakg$? And what do you mean by "closes up"?
– André Gomes
Jul 23 at 20:50
1
Up to isomorphism, $mathfrakhsubseteq mathfrakg$ is just $mathbbR^ksubseteq mathbbR^n$ for some $k < n$. View $mathbbR^k$ as spanned by the first $k$ of the standard basis vectors of $mathbbR^n$. Considering $f:=d(cdot, mathbbR^k):mathbbR^nrightarrow mathbbR$, $mathbbR^nsetminus mathbbR^k$ is $f^-1( mathbbRsetminus 0)$, so is open. For density, just note that every open set in $mathbbR^n$ contains points whose $k+1$st coordinate is non-zero. "Closes up" means $exp(tv):tinmathbbR$ is a circle.
– Jason DeVito
Jul 23 at 21:01
Can you recomend me any reference?
– André Gomes
Jul 25 at 18:29
1
The sources I've used for most of my knowledge are: 1. Lecture notes by my advisor, Wolfgang Ziller math.upenn.edu/~wziller/math650/LieGroupsReps.pdf. 2. Fulton and Harris - Representation theory 3. Brocker and tom Dieck - Representations of Compact Lie groups
– Jason DeVito
Jul 25 at 19:57
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
We begin with a Lemma.
Lemma: Suppose $Hsubseteq G$ are compact connnected Lie groups with $Hneq G$. Then there is a subgroup $S^1subseteq G$ for which $S^1cap H$ is finite.
Proof: Look on the Lie algebra level. We have $mathfrakhsubseteq mathfrakg$, and $mathfrakhneq mathfrakg$ because $H$ and $G$ are connected and $Hneq G$. Then $mathfrakgsetminus mathfrakh$ is an open dense subset of $mathfrakg$, so contains a vector $v$ for which $exp(tv)$ closes up. Then $exp(tv) = S^1$ is the desired $S^1$.
To see this, note that if $S^1cap H$ is infinite, then it has an accumulation point (since $G$ is compact). Now, by using the group multiplication, we may assume this accumulation point is the identity. It follows that $exp(t_n v) in H$ for a decreasing sequence $t_nrightarrow 0$. This, then, implies that $vinmathfrakh$, a contradiction. $square$
Now, we prove the theorem. Suppose $G$ is any connected Lie group. Let $H = Tsubseteq G$ be a maximal torus. Under the identification $Tcong mathbbR^n/mathbbZ^n$, if we pick an element $x=(x_1,...,x_n)in T$ for which $operatornamespan_mathbbQ1,x_1, x_2,...,x_n$ has dimension $n+1$, it follows that $x$ generates a dense subgroup of $T$.
If $T = G$, we are done. Otherwise, using the lemma, pick $S^1subseteq G$ with $S^1cap T$ finite. We pick $yin S^1$ which generates a dense subgroup of $S^1$.
Let $langlelangle x,yranglerangle$ denote the closure of the subgroup generated by $x$ and $y$. Then clearly $langlelangle x,yranglerangle$ contains $H$ and $S^1$. On the Lie algebra level, the Lie algebra of $langle langle x,yranglerangle$ is a subspace containing $mathfrakt$ and $v$, so has dimension at least that of $dim T + 1$.
Now, we induct. If $langle langle x,yranglerangle neq G$, we use the lemma to pick $zin Gsetminus langlelangle x,yrangle rangle$. By the same argument as above, $langle langle x,y,zranglerangle$ has larger dimension that $langle langle x,y,rangle rangle$. Continuing in this way, since the dimension of the generated subset increases at every stage, the process must stop after finitely many steps.
answered Jul 18 at 19:40
Jason DeVito
29.5k473129
Why is $mathfrakgsetminusmathfrakh$ open and dense in $mathfrakg$? And what do you mean by "closes up"?
– André Gomes
Jul 23 at 20:50
1
Up to isomorphism, $mathfrakhsubseteq mathfrakg$ is just $mathbbR^ksubseteq mathbbR^n$ for some $k < n$. View $mathbbR^k$ as spanned by the first $k$ of the standard basis vectors of $mathbbR^n$. Considering $f:=d(cdot, mathbbR^k):mathbbR^nrightarrow mathbbR$, $mathbbR^nsetminus mathbbR^k$ is $f^-1( mathbbRsetminus 0)$, so is open. For density, just note that every open set in $mathbbR^n$ contains points whose $k+1$st coordinate is non-zero. "Closes up" means $exp(tv):tinmathbbR$ is a circle.
– Jason DeVito
Jul 23 at 21:01
Can you recomend me any reference?
– André Gomes
Jul 25 at 18:29
1
The sources I've used for most of my knowledge are: 1. Lecture notes by my advisor, Wolfgang Ziller math.upenn.edu/~wziller/math650/LieGroupsReps.pdf. 2. Fulton and Harris - Representation theory 3. Brocker and tom Dieck - Representations of Compact Lie groups
– Jason DeVito
Jul 25 at 19:57
add a comment |Â
up vote
2
down vote
We begin with a Lemma.
Lemma: Suppose $Hsubseteq G$ are compact connnected Lie groups with $Hneq G$. Then there is a subgroup $S^1subseteq G$ for which $S^1cap H$ is finite.
Proof: Look on the Lie algebra level. We have $mathfrakhsubseteq mathfrakg$, and $mathfrakhneq mathfrakg$ because $H$ and $G$ are connected and $Hneq G$. Then $mathfrakgsetminus mathfrakh$ is an open dense subset of $mathfrakg$, so contains a vector $v$ for which $exp(tv)$ closes up. Then $exp(tv) = S^1$ is the desired $S^1$.
To see this, note that if $S^1cap H$ is infinite, then it has an accumulation point (since $G$ is compact). Now, by using the group multiplication, we may assume this accumulation point is the identity. It follows that $exp(t_n v) in H$ for a decreasing sequence $t_nrightarrow 0$. This, then, implies that $vinmathfrakh$, a contradiction. $square$
Now, we prove the theorem. Suppose $G$ is any connected Lie group. Let $H = Tsubseteq G$ be a maximal torus. Under the identification $Tcong mathbbR^n/mathbbZ^n$, if we pick an element $x=(x_1,...,x_n)in T$ for which $operatornamespan_mathbbQ1,x_1, x_2,...,x_n$ has dimension $n+1$, it follows that $x$ generates a dense subgroup of $T$.
If $T = G$, we are done. Otherwise, using the lemma, pick $S^1subseteq G$ with $S^1cap T$ finite. We pick $yin S^1$ which generates a dense subgroup of $S^1$.
Let $langlelangle x,yranglerangle$ denote the closure of the subgroup generated by $x$ and $y$. Then clearly $langlelangle x,yranglerangle$ contains $H$ and $S^1$. On the Lie algebra level, the Lie algebra of $langle langle x,yranglerangle$ is a subspace containing $mathfrakt$ and $v$, so has dimension at least that of $dim T + 1$.
Now, we induct. If $langle langle x,yranglerangle neq G$, we use the lemma to pick $zin Gsetminus langlelangle x,yrangle rangle$. By the same argument as above, $langle langle x,y,zranglerangle$ has larger dimension that $langle langle x,y,rangle rangle$. Continuing in this way, since the dimension of the generated subset increases at every stage, the process must stop after finitely many steps.
answered Jul 18 at 19:40
Jason DeVito
29.5k473129
Why is $mathfrakgsetminusmathfrakh$ open and dense in $mathfrakg$? And what do you mean by "closes up"?
– André Gomes
Jul 23 at 20:50
1
Up to isomorphism, $mathfrakhsubseteq mathfrakg$ is just $mathbbR^ksubseteq mathbbR^n$ for some $k < n$. View $mathbbR^k$ as spanned by the first $k$ of the standard basis vectors of $mathbbR^n$. Considering $f:=d(cdot, mathbbR^k):mathbbR^nrightarrow mathbbR$, $mathbbR^nsetminus mathbbR^k$ is $f^-1( mathbbRsetminus 0)$, so is open. For density, just note that every open set in $mathbbR^n$ contains points whose $k+1$st coordinate is non-zero. "Closes up" means $exp(tv):tinmathbbR$ is a circle.
– Jason DeVito
Jul 23 at 21:01
Can you recomend me any reference?
– André Gomes
Jul 25 at 18:29
1
The sources I've used for most of my knowledge are: 1. Lecture notes by my advisor, Wolfgang Ziller math.upenn.edu/~wziller/math650/LieGroupsReps.pdf. 2. Fulton and Harris - Representation theory 3. Brocker and tom Dieck - Representations of Compact Lie groups
– Jason DeVito
Jul 25 at 19:57
add a comment |Â
up vote
2
down vote
up vote
2
down vote
We begin with a Lemma.
Lemma: Suppose $Hsubseteq G$ are compact connnected Lie groups with $Hneq G$. Then there is a subgroup $S^1subseteq G$ for which $S^1cap H$ is finite.
Proof: Look on the Lie algebra level. We have $mathfrakhsubseteq mathfrakg$, and $mathfrakhneq mathfrakg$ because $H$ and $G$ are connected and $Hneq G$. Then $mathfrakgsetminus mathfrakh$ is an open dense subset of $mathfrakg$, so contains a vector $v$ for which $exp(tv)$ closes up. Then $exp(tv) = S^1$ is the desired $S^1$.
To see this, note that if $S^1cap H$ is infinite, then it has an accumulation point (since $G$ is compact). Now, by using the group multiplication, we may assume this accumulation point is the identity. It follows that $exp(t_n v) in H$ for a decreasing sequence $t_nrightarrow 0$. This, then, implies that $vinmathfrakh$, a contradiction. $square$
Now, we prove the theorem. Suppose $G$ is any connected Lie group. Let $H = Tsubseteq G$ be a maximal torus. Under the identification $Tcong mathbbR^n/mathbbZ^n$, if we pick an element $x=(x_1,...,x_n)in T$ for which $operatornamespan_mathbbQ1,x_1, x_2,...,x_n$ has dimension $n+1$, it follows that $x$ generates a dense subgroup of $T$.
If $T = G$, we are done. Otherwise, using the lemma, pick $S^1subseteq G$ with $S^1cap T$ finite. We pick $yin S^1$ which generates a dense subgroup of $S^1$.
Let $langlelangle x,yranglerangle$ denote the closure of the subgroup generated by $x$ and $y$. Then clearly $langlelangle x,yranglerangle$ contains $H$ and $S^1$. On the Lie algebra level, the Lie algebra of $langle langle x,yranglerangle$ is a subspace containing $mathfrakt$ and $v$, so has dimension at least that of $dim T + 1$.
Now, we induct. If $langle langle x,yranglerangle neq G$, we use the lemma to pick $zin Gsetminus langlelangle x,yrangle rangle$. By the same argument as above, $langle langle x,y,zranglerangle$ has larger dimension that $langle langle x,y,rangle rangle$. Continuing in this way, since the dimension of the generated subset increases at every stage, the process must stop after finitely many steps.
answered Jul 18 at 19:40
Jason DeVito
29.5k473129
We begin with a Lemma.
Lemma: Suppose $Hsubseteq G$ are compact connnected Lie groups with $Hneq G$. Then there is a subgroup $S^1subseteq G$ for which $S^1cap H$ is finite.
Proof: Look on the Lie algebra level. We have $mathfrakhsubseteq mathfrakg$, and $mathfrakhneq mathfrakg$ because $H$ and $G$ are connected and $Hneq G$. Then $mathfrakgsetminus mathfrakh$ is an open dense subset of $mathfrakg$, so contains a vector $v$ for which $exp(tv)$ closes up. Then $exp(tv) = S^1$ is the desired $S^1$.
To see this, note that if $S^1cap H$ is infinite, then it has an accumulation point (since $G$ is compact). Now, by using the group multiplication, we may assume this accumulation point is the identity. It follows that $exp(t_n v) in H$ for a decreasing sequence $t_nrightarrow 0$. This, then, implies that $vinmathfrakh$, a contradiction. $square$
Now, we prove the theorem. Suppose $G$ is any connected Lie group. Let $H = Tsubseteq G$ be a maximal torus. Under the identification $Tcong mathbbR^n/mathbbZ^n$, if we pick an element $x=(x_1,...,x_n)in T$ for which $operatornamespan_mathbbQ1,x_1, x_2,...,x_n$ has dimension $n+1$, it follows that $x$ generates a dense subgroup of $T$.
If $T = G$, we are done. Otherwise, using the lemma, pick $S^1subseteq G$ with $S^1cap T$ finite. We pick $yin S^1$ which generates a dense subgroup of $S^1$.
Let $langlelangle x,yranglerangle$ denote the closure of the subgroup generated by $x$ and $y$. Then clearly $langlelangle x,yranglerangle$ contains $H$ and $S^1$. On the Lie algebra level, the Lie algebra of $langle langle x,yranglerangle$ is a subspace containing $mathfrakt$ and $v$, so has dimension at least that of $dim T + 1$.
Now, we induct. If $langle langle x,yranglerangle neq G$, we use the lemma to pick $zin Gsetminus langlelangle x,yrangle rangle$. By the same argument as above, $langle langle x,y,zranglerangle$ has larger dimension that $langle langle x,y,rangle rangle$. Continuing in this way, since the dimension of the generated subset increases at every stage, the process must stop after finitely many steps.
answered Jul 18 at 19:40
Jason DeVito
29.5k473129
answered Jul 18 at 19:40
Jason DeVito
29.5k473129
answered Jul 18 at 19:40
Jason DeVito
29.5k473129
answered Jul 18 at 19:40
Jason DeVito
29.5k473129
29.5k473129
Why is $mathfrakgsetminusmathfrakh$ open and dense in $mathfrakg$? And what do you mean by "closes up"?
– André Gomes
Jul 23 at 20:50
1
Up to isomorphism, $mathfrakhsubseteq mathfrakg$ is just $mathbbR^ksubseteq mathbbR^n$ for some $k < n$. View $mathbbR^k$ as spanned by the first $k$ of the standard basis vectors of $mathbbR^n$. Considering $f:=d(cdot, mathbbR^k):mathbbR^nrightarrow mathbbR$, $mathbbR^nsetminus mathbbR^k$ is $f^-1( mathbbRsetminus 0)$, so is open. For density, just note that every open set in $mathbbR^n$ contains points whose $k+1$st coordinate is non-zero. "Closes up" means $exp(tv):tinmathbbR$ is a circle.
– Jason DeVito
Jul 23 at 21:01
Can you recomend me any reference?
– André Gomes
Jul 25 at 18:29
1
The sources I've used for most of my knowledge are: 1. Lecture notes by my advisor, Wolfgang Ziller math.upenn.edu/~wziller/math650/LieGroupsReps.pdf. 2. Fulton and Harris - Representation theory 3. Brocker and tom Dieck - Representations of Compact Lie groups
– Jason DeVito
Jul 25 at 19:57
add a comment |Â
Why is $mathfrakgsetminusmathfrakh$ open and dense in $mathfrakg$? And what do you mean by "closes up"?
– André Gomes
Jul 23 at 20:50
1
Up to isomorphism, $mathfrakhsubseteq mathfrakg$ is just $mathbbR^ksubseteq mathbbR^n$ for some $k < n$. View $mathbbR^k$ as spanned by the first $k$ of the standard basis vectors of $mathbbR^n$. Considering $f:=d(cdot, mathbbR^k):mathbbR^nrightarrow mathbbR$, $mathbbR^nsetminus mathbbR^k$ is $f^-1( mathbbRsetminus 0)$, so is open. For density, just note that every open set in $mathbbR^n$ contains points whose $k+1$st coordinate is non-zero. "Closes up" means $exp(tv):tinmathbbR$ is a circle.
– Jason DeVito
Jul 23 at 21:01
Can you recomend me any reference?
– André Gomes
Jul 25 at 18:29
1
The sources I've used for most of my knowledge are: 1. Lecture notes by my advisor, Wolfgang Ziller math.upenn.edu/~wziller/math650/LieGroupsReps.pdf. 2. Fulton and Harris - Representation theory 3. Brocker and tom Dieck - Representations of Compact Lie groups
– Jason DeVito
Jul 25 at 19:57
Why is $mathfrakgsetminusmathfrakh$ open and dense in $mathfrakg$? And what do you mean by "closes up"?
– André Gomes
Jul 23 at 20:50
Why is $mathfrakgsetminusmathfrakh$ open and dense in $mathfrakg$? And what do you mean by "closes up"?
– André Gomes
Jul 23 at 20:50
1
1
Up to isomorphism, $mathfrakhsubseteq mathfrakg$ is just $mathbbR^ksubseteq mathbbR^n$ for some $k < n$. View $mathbbR^k$ as spanned by the first $k$ of the standard basis vectors of $mathbbR^n$. Considering $f:=d(cdot, mathbbR^k):mathbbR^nrightarrow mathbbR$, $mathbbR^nsetminus mathbbR^k$ is $f^-1( mathbbRsetminus 0)$, so is open. For density, just note that every open set in $mathbbR^n$ contains points whose $k+1$st coordinate is non-zero. "Closes up" means $exp(tv):tinmathbbR$ is a circle.
– Jason DeVito
Jul 23 at 21:01
Up to isomorphism, $mathfrakhsubseteq mathfrakg$ is just $mathbbR^ksubseteq mathbbR^n$ for some $k < n$. View $mathbbR^k$ as spanned by the first $k$ of the standard basis vectors of $mathbbR^n$. Considering $f:=d(cdot, mathbbR^k):mathbbR^nrightarrow mathbbR$, $mathbbR^nsetminus mathbbR^k$ is $f^-1( mathbbRsetminus 0)$, so is open. For density, just note that every open set in $mathbbR^n$ contains points whose $k+1$st coordinate is non-zero. "Closes up" means $exp(tv):tinmathbbR$ is a circle.
– Jason DeVito
Jul 23 at 21:01
Can you recomend me any reference?
– André Gomes
Jul 25 at 18:29
Can you recomend me any reference?
– André Gomes
Jul 25 at 18:29
1
1
The sources I've used for most of my knowledge are: 1. Lecture notes by my advisor, Wolfgang Ziller math.upenn.edu/~wziller/math650/LieGroupsReps.pdf. 2. Fulton and Harris - Representation theory 3. Brocker and tom Dieck - Representations of Compact Lie groups
– Jason DeVito
Jul 25 at 19:57
The sources I've used for most of my knowledge are: 1. Lecture notes by my advisor, Wolfgang Ziller math.upenn.edu/~wziller/math650/LieGroupsReps.pdf. 2. Fulton and Harris - Representation theory 3. Brocker and tom Dieck - Representations of Compact Lie groups
– Jason DeVito
Jul 25 at 19:57
add a comment |Â
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1
This proof strategy won't work: if a compact Lie group $G$ has non-abelian identity component, then it has uncountable many distinct maximal tori. This follows, e.g., from the Baire category theorem since each maximal torus is closed nowhere dense subset of the identity component of $G$.
– Jason DeVito
Jul 18 at 19:04
Do you have any hint for another strategy?
– André Gomes
Jul 18 at 19:12
2
Here is the approach I am trying, but I don't know if it works: Lemma (which I don't know how to prove, but I think it true): If $Hsubseteq G$ are connected Lie groups, with $Hneq G$, then there is an element $gin Gsetminus H$ for which the group generated by $g$ only intersects $H$ at $e$. If this lemma is true, I think I can prove your result by induction on $dim G$.
– Jason DeVito
Jul 18 at 19:15
I think I've got it, writing an answer now....
– Jason DeVito
Jul 18 at 19:20