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The codimension of a parabolic subalgebra of a semisimple Lie algebra
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Given a complex semisimple Lie algebra $mathfrak g$ and a subalgebra $mathfrak h$. If we are given that the complex vector space $mathfrak g/mathfrak h$ has dimension $1$ over $mathbb C$. Is $mathfrak h$ a parabolic subalgebra. i.e., contains a Borel subalgebra?
Is the statement above still true when $dim_{mathbb C}mathfrak g/mathfrak h=2$?
lie-algebras
asked 21 hours ago
Amrat A
23617
add a comment |
up vote
6
down vote
favorite
Given a complex semisimple Lie algebra $mathfrak g$ and a subalgebra $mathfrak h$. If we are given that the complex vector space $mathfrak g/mathfrak h$ has dimension $1$ over $mathbb C$. Is $mathfrak h$ a parabolic subalgebra. i.e., contains a Borel subalgebra?
Is the statement above still true when $dim_{mathbb C}mathfrak g/mathfrak h=2$?
lie-algebras
asked 21 hours ago
Amrat A
23617
1
For your second question, for any $0 neq x in mathfrak{g}:=mathfrak{sl}_2(Bbb C)$, $mathfrak{h}:= Bbb Cx$ is a counterexample. (I suspect it is essentially the only one though, i.e. if $mathfrak{g}$ contains no direct summand $simeq mathfrak{sl}_2$, the statement might be true; but I'm not at all sure about this.)
– Torsten Schoeneberg
16 hours ago
1
The first assertion is true and actually the only possibility (up to direct product of both $mathfrak{g}$ and $mathfrak{h}$ by some other semisimple algebra) is when $mathfrak{g}$ is $mathfrak{sl}_2$.
– YCor
15 hours ago
add a comment |
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Given a complex semisimple Lie algebra $mathfrak g$ and a subalgebra $mathfrak h$. If we are given that the complex vector space $mathfrak g/mathfrak h$ has dimension $1$ over $mathbb C$. Is $mathfrak h$ a parabolic subalgebra. i.e., contains a Borel subalgebra?
Is the statement above still true when $dim_{mathbb C}mathfrak g/mathfrak h=2$?
lie-algebras
asked 21 hours ago
Amrat A
23617
Given a complex semisimple Lie algebra $mathfrak g$ and a subalgebra $mathfrak h$. If we are given that the complex vector space $mathfrak g/mathfrak h$ has dimension $1$ over $mathbb C$. Is $mathfrak h$ a parabolic subalgebra. i.e., contains a Borel subalgebra?
Is the statement above still true when $dim_{mathbb C}mathfrak g/mathfrak h=2$?
lie-algebras
lie-algebras
asked 21 hours ago
Amrat A
23617
asked 21 hours ago
Amrat A
23617
asked 21 hours ago
Amrat A
23617
asked 21 hours ago
Amrat A
23617
asked 21 hours ago
Amrat A
23617
23617
1
For your second question, for any $0 neq x in mathfrak{g}:=mathfrak{sl}_2(Bbb C)$, $mathfrak{h}:= Bbb Cx$ is a counterexample. (I suspect it is essentially the only one though, i.e. if $mathfrak{g}$ contains no direct summand $simeq mathfrak{sl}_2$, the statement might be true; but I'm not at all sure about this.)
– Torsten Schoeneberg
16 hours ago
1
The first assertion is true and actually the only possibility (up to direct product of both $mathfrak{g}$ and $mathfrak{h}$ by some other semisimple algebra) is when $mathfrak{g}$ is $mathfrak{sl}_2$.
– YCor
15 hours ago
add a comment |
1
For your second question, for any $0 neq x in mathfrak{g}:=mathfrak{sl}_2(Bbb C)$, $mathfrak{h}:= Bbb Cx$ is a counterexample. (I suspect it is essentially the only one though, i.e. if $mathfrak{g}$ contains no direct summand $simeq mathfrak{sl}_2$, the statement might be true; but I'm not at all sure about this.)
– Torsten Schoeneberg
16 hours ago
1
The first assertion is true and actually the only possibility (up to direct product of both $mathfrak{g}$ and $mathfrak{h}$ by some other semisimple algebra) is when $mathfrak{g}$ is $mathfrak{sl}_2$.
– YCor
15 hours ago
1
1
For your second question, for any $0 neq x in mathfrak{g}:=mathfrak{sl}_2(Bbb C)$, $mathfrak{h}:= Bbb Cx$ is a counterexample. (I suspect it is essentially the only one though, i.e. if $mathfrak{g}$ contains no direct summand $simeq mathfrak{sl}_2$, the statement might be true; but I'm not at all sure about this.)
– Torsten Schoeneberg
16 hours ago
For your second question, for any $0 neq x in mathfrak{g}:=mathfrak{sl}_2(Bbb C)$, $mathfrak{h}:= Bbb Cx$ is a counterexample. (I suspect it is essentially the only one though, i.e. if $mathfrak{g}$ contains no direct summand $simeq mathfrak{sl}_2$, the statement might be true; but I'm not at all sure about this.)
– Torsten Schoeneberg
16 hours ago
1
1
The first assertion is true and actually the only possibility (up to direct product of both $mathfrak{g}$ and $mathfrak{h}$ by some other semisimple algebra) is when $mathfrak{g}$ is $mathfrak{sl}_2$.
– YCor
15 hours ago
The first assertion is true and actually the only possibility (up to direct product of both $mathfrak{g}$ and $mathfrak{h}$ by some other semisimple algebra) is when $mathfrak{g}$ is $mathfrak{sl}_2$.
– YCor
15 hours ago
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
The first assertion is true and actually the only possibility (up to direct product of both $mathfrak{g}$ and $mathfrak{h}$ by some other semisimple algebra) is when $mathfrak{g}$ is $mathfrak{sl}_2$.
(Edit: switched to an algebraic proof)
In $mathfrak{sl}_2$ it is easy to check that all codimension 1 subalgebras are conjugate to the Borel (parabolic) one.
It is enough to prove that if $mathbf{g}$ is simple of rank $ge 2$, then it has no codimension 1 subalgebra $mathfrak{h}$.
Choose a Cartan subalgebra $mathfrak{h}_0$ of $mathfrak{h}$. It induces a grading $(mathfrak{g}'_alpha)$ of $mathfrak{g}$, which has to be a quotient of its own Cartan grading.
If $mathfrak{h}_0=mathfrak{g}_0$, then $mathfrak{h}_0$ is a Cartan subalgebra of $mathfrak{g}$. In this case, it follows that $mathfrak{h}$ is a graded subalgebra for the Cartan grading $(mathfrak{g}_alpha)$ of $mathfrak{g}$, so there exists a nonzero root $alpha$ such that $mathfrak{h}=bigoplus_{betaneqalpha}mathfrak{g}_beta$.
Using that $mathfrak{g}$ has rank $ge 2$ and is simple, there exist two nonzero roots summing to $alpha$, and this implies that such $mathfrak{h}$ is not a subalgebra.
Next, if $mathfrak{h}_0neqmathfrak{g}_0$, then being a quotient of the Cartan grading, we have $mathfrak{g}_0$ semisimple and containing a Cartan subalgebra of $mathfrak{g}$. If $mathfrak{g}neqmathfrak{g}_0$, we can argue by induction and get a contradiction.
Otherwise, we have $mathfrak{g}=mathfrak{g}_0$: this means that $mathrm{ad}(h)$ is nilpotent for every $hinmathfrak{h}$. Letting $mathfrak{c}$ be a Cartan subalgebra of $mathfrak{g}$, this implies that every element of $mathfrak{h}capmathfrak{c}$ is in the kernel of every root. Since the intersection of kernels of roots is zero in a semisimple Lie algebra, this forces $mathfrak{c}$ to have dimension $le 1$. Hence $mathfrak{g}$ has rank $le 1$, a contradiction.
answered 15 hours ago
YCor
6,953827
1
Why is $G/H$ compact? (For a linear algebraic group, $G/H$ proper is equivalent to $H$ being a parabolic subgroup, so you seem to be starting with something pretty close to what you want to conclude).
– Stephen
14 hours ago
@Stephen you're right, it was just late in my time zone.
– YCor
6 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The first assertion is true and actually the only possibility (up to direct product of both $mathfrak{g}$ and $mathfrak{h}$ by some other semisimple algebra) is when $mathfrak{g}$ is $mathfrak{sl}_2$.
(Edit: switched to an algebraic proof)
In $mathfrak{sl}_2$ it is easy to check that all codimension 1 subalgebras are conjugate to the Borel (parabolic) one.
It is enough to prove that if $mathbf{g}$ is simple of rank $ge 2$, then it has no codimension 1 subalgebra $mathfrak{h}$.
Choose a Cartan subalgebra $mathfrak{h}_0$ of $mathfrak{h}$. It induces a grading $(mathfrak{g}'_alpha)$ of $mathfrak{g}$, which has to be a quotient of its own Cartan grading.
If $mathfrak{h}_0=mathfrak{g}_0$, then $mathfrak{h}_0$ is a Cartan subalgebra of $mathfrak{g}$. In this case, it follows that $mathfrak{h}$ is a graded subalgebra for the Cartan grading $(mathfrak{g}_alpha)$ of $mathfrak{g}$, so there exists a nonzero root $alpha$ such that $mathfrak{h}=bigoplus_{betaneqalpha}mathfrak{g}_beta$.
Using that $mathfrak{g}$ has rank $ge 2$ and is simple, there exist two nonzero roots summing to $alpha$, and this implies that such $mathfrak{h}$ is not a subalgebra.
Next, if $mathfrak{h}_0neqmathfrak{g}_0$, then being a quotient of the Cartan grading, we have $mathfrak{g}_0$ semisimple and containing a Cartan subalgebra of $mathfrak{g}$. If $mathfrak{g}neqmathfrak{g}_0$, we can argue by induction and get a contradiction.
Otherwise, we have $mathfrak{g}=mathfrak{g}_0$: this means that $mathrm{ad}(h)$ is nilpotent for every $hinmathfrak{h}$. Letting $mathfrak{c}$ be a Cartan subalgebra of $mathfrak{g}$, this implies that every element of $mathfrak{h}capmathfrak{c}$ is in the kernel of every root. Since the intersection of kernels of roots is zero in a semisimple Lie algebra, this forces $mathfrak{c}$ to have dimension $le 1$. Hence $mathfrak{g}$ has rank $le 1$, a contradiction.
answered 15 hours ago
YCor
6,953827
1
Why is $G/H$ compact? (For a linear algebraic group, $G/H$ proper is equivalent to $H$ being a parabolic subgroup, so you seem to be starting with something pretty close to what you want to conclude).
– Stephen
14 hours ago
@Stephen you're right, it was just late in my time zone.
– YCor
6 hours ago
add a comment |
up vote
2
down vote
accepted
The first assertion is true and actually the only possibility (up to direct product of both $mathfrak{g}$ and $mathfrak{h}$ by some other semisimple algebra) is when $mathfrak{g}$ is $mathfrak{sl}_2$.
(Edit: switched to an algebraic proof)
In $mathfrak{sl}_2$ it is easy to check that all codimension 1 subalgebras are conjugate to the Borel (parabolic) one.
It is enough to prove that if $mathbf{g}$ is simple of rank $ge 2$, then it has no codimension 1 subalgebra $mathfrak{h}$.
Choose a Cartan subalgebra $mathfrak{h}_0$ of $mathfrak{h}$. It induces a grading $(mathfrak{g}'_alpha)$ of $mathfrak{g}$, which has to be a quotient of its own Cartan grading.
If $mathfrak{h}_0=mathfrak{g}_0$, then $mathfrak{h}_0$ is a Cartan subalgebra of $mathfrak{g}$. In this case, it follows that $mathfrak{h}$ is a graded subalgebra for the Cartan grading $(mathfrak{g}_alpha)$ of $mathfrak{g}$, so there exists a nonzero root $alpha$ such that $mathfrak{h}=bigoplus_{betaneqalpha}mathfrak{g}_beta$.
Using that $mathfrak{g}$ has rank $ge 2$ and is simple, there exist two nonzero roots summing to $alpha$, and this implies that such $mathfrak{h}$ is not a subalgebra.
Next, if $mathfrak{h}_0neqmathfrak{g}_0$, then being a quotient of the Cartan grading, we have $mathfrak{g}_0$ semisimple and containing a Cartan subalgebra of $mathfrak{g}$. If $mathfrak{g}neqmathfrak{g}_0$, we can argue by induction and get a contradiction.
Otherwise, we have $mathfrak{g}=mathfrak{g}_0$: this means that $mathrm{ad}(h)$ is nilpotent for every $hinmathfrak{h}$. Letting $mathfrak{c}$ be a Cartan subalgebra of $mathfrak{g}$, this implies that every element of $mathfrak{h}capmathfrak{c}$ is in the kernel of every root. Since the intersection of kernels of roots is zero in a semisimple Lie algebra, this forces $mathfrak{c}$ to have dimension $le 1$. Hence $mathfrak{g}$ has rank $le 1$, a contradiction.
answered 15 hours ago
YCor
6,953827
1
Why is $G/H$ compact? (For a linear algebraic group, $G/H$ proper is equivalent to $H$ being a parabolic subgroup, so you seem to be starting with something pretty close to what you want to conclude).
– Stephen
14 hours ago
@Stephen you're right, it was just late in my time zone.
– YCor
6 hours ago
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The first assertion is true and actually the only possibility (up to direct product of both $mathfrak{g}$ and $mathfrak{h}$ by some other semisimple algebra) is when $mathfrak{g}$ is $mathfrak{sl}_2$.
(Edit: switched to an algebraic proof)
In $mathfrak{sl}_2$ it is easy to check that all codimension 1 subalgebras are conjugate to the Borel (parabolic) one.
It is enough to prove that if $mathbf{g}$ is simple of rank $ge 2$, then it has no codimension 1 subalgebra $mathfrak{h}$.
Choose a Cartan subalgebra $mathfrak{h}_0$ of $mathfrak{h}$. It induces a grading $(mathfrak{g}'_alpha)$ of $mathfrak{g}$, which has to be a quotient of its own Cartan grading.
If $mathfrak{h}_0=mathfrak{g}_0$, then $mathfrak{h}_0$ is a Cartan subalgebra of $mathfrak{g}$. In this case, it follows that $mathfrak{h}$ is a graded subalgebra for the Cartan grading $(mathfrak{g}_alpha)$ of $mathfrak{g}$, so there exists a nonzero root $alpha$ such that $mathfrak{h}=bigoplus_{betaneqalpha}mathfrak{g}_beta$.
Using that $mathfrak{g}$ has rank $ge 2$ and is simple, there exist two nonzero roots summing to $alpha$, and this implies that such $mathfrak{h}$ is not a subalgebra.
Next, if $mathfrak{h}_0neqmathfrak{g}_0$, then being a quotient of the Cartan grading, we have $mathfrak{g}_0$ semisimple and containing a Cartan subalgebra of $mathfrak{g}$. If $mathfrak{g}neqmathfrak{g}_0$, we can argue by induction and get a contradiction.
Otherwise, we have $mathfrak{g}=mathfrak{g}_0$: this means that $mathrm{ad}(h)$ is nilpotent for every $hinmathfrak{h}$. Letting $mathfrak{c}$ be a Cartan subalgebra of $mathfrak{g}$, this implies that every element of $mathfrak{h}capmathfrak{c}$ is in the kernel of every root. Since the intersection of kernels of roots is zero in a semisimple Lie algebra, this forces $mathfrak{c}$ to have dimension $le 1$. Hence $mathfrak{g}$ has rank $le 1$, a contradiction.
answered 15 hours ago
YCor
6,953827
The first assertion is true and actually the only possibility (up to direct product of both $mathfrak{g}$ and $mathfrak{h}$ by some other semisimple algebra) is when $mathfrak{g}$ is $mathfrak{sl}_2$.
(Edit: switched to an algebraic proof)
In $mathfrak{sl}_2$ it is easy to check that all codimension 1 subalgebras are conjugate to the Borel (parabolic) one.
It is enough to prove that if $mathbf{g}$ is simple of rank $ge 2$, then it has no codimension 1 subalgebra $mathfrak{h}$.
Choose a Cartan subalgebra $mathfrak{h}_0$ of $mathfrak{h}$. It induces a grading $(mathfrak{g}'_alpha)$ of $mathfrak{g}$, which has to be a quotient of its own Cartan grading.
If $mathfrak{h}_0=mathfrak{g}_0$, then $mathfrak{h}_0$ is a Cartan subalgebra of $mathfrak{g}$. In this case, it follows that $mathfrak{h}$ is a graded subalgebra for the Cartan grading $(mathfrak{g}_alpha)$ of $mathfrak{g}$, so there exists a nonzero root $alpha$ such that $mathfrak{h}=bigoplus_{betaneqalpha}mathfrak{g}_beta$.
Using that $mathfrak{g}$ has rank $ge 2$ and is simple, there exist two nonzero roots summing to $alpha$, and this implies that such $mathfrak{h}$ is not a subalgebra.
Next, if $mathfrak{h}_0neqmathfrak{g}_0$, then being a quotient of the Cartan grading, we have $mathfrak{g}_0$ semisimple and containing a Cartan subalgebra of $mathfrak{g}$. If $mathfrak{g}neqmathfrak{g}_0$, we can argue by induction and get a contradiction.
Otherwise, we have $mathfrak{g}=mathfrak{g}_0$: this means that $mathrm{ad}(h)$ is nilpotent for every $hinmathfrak{h}$. Letting $mathfrak{c}$ be a Cartan subalgebra of $mathfrak{g}$, this implies that every element of $mathfrak{h}capmathfrak{c}$ is in the kernel of every root. Since the intersection of kernels of roots is zero in a semisimple Lie algebra, this forces $mathfrak{c}$ to have dimension $le 1$. Hence $mathfrak{g}$ has rank $le 1$, a contradiction.
answered 15 hours ago
YCor
6,953827
edited 5 hours ago
answered 15 hours ago
YCor
6,953827
answered 15 hours ago
YCor
6,953827
answered 15 hours ago
YCor
6,953827
6,953827
1
Why is $G/H$ compact? (For a linear algebraic group, $G/H$ proper is equivalent to $H$ being a parabolic subgroup, so you seem to be starting with something pretty close to what you want to conclude).
– Stephen
14 hours ago
@Stephen you're right, it was just late in my time zone.
– YCor
6 hours ago
add a comment |
1
Why is $G/H$ compact? (For a linear algebraic group, $G/H$ proper is equivalent to $H$ being a parabolic subgroup, so you seem to be starting with something pretty close to what you want to conclude).
– Stephen
14 hours ago
@Stephen you're right, it was just late in my time zone.
– YCor
6 hours ago
1
1
Why is $G/H$ compact? (For a linear algebraic group, $G/H$ proper is equivalent to $H$ being a parabolic subgroup, so you seem to be starting with something pretty close to what you want to conclude).
– Stephen
14 hours ago
Why is $G/H$ compact? (For a linear algebraic group, $G/H$ proper is equivalent to $H$ being a parabolic subgroup, so you seem to be starting with something pretty close to what you want to conclude).
– Stephen
14 hours ago
@Stephen you're right, it was just late in my time zone.
– YCor
6 hours ago
@Stephen you're right, it was just late in my time zone.
– YCor
6 hours ago
add a comment |
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For your second question, for any $0 neq x in mathfrak{g}:=mathfrak{sl}_2(Bbb C)$, $mathfrak{h}:= Bbb Cx$ is a counterexample. (I suspect it is essentially the only one though, i.e. if $mathfrak{g}$ contains no direct summand $simeq mathfrak{sl}_2$, the statement might be true; but I'm not at all sure about this.)
– Torsten Schoeneberg
16 hours ago
1
The first assertion is true and actually the only possibility (up to direct product of both $mathfrak{g}$ and $mathfrak{h}$ by some other semisimple algebra) is when $mathfrak{g}$ is $mathfrak{sl}_2$.
– YCor
15 hours ago