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Existence of a non essentially real ideal in a semisimple Lie algebra
$begingroup$
Let $mathfrak{g}$ be the Lie algebra of a semisimple compact Lie group $G$. Denote by $mathbb{C} mathfrak g = mathbb{C} otimes mathfrak g$ the complexification of $mathfrak g$. I am assuming that $mathfrak g$ is not simple, then there exist an ideal $mathfrak i subsetmathfrak g$ that is different from ${0}$ and different from the entire Lie algebra $mathfrak g$.
A subalgebra $mathfrak h subset mathbb C mathfrak g$ is called essentially real if $overline {mathfrak h} = mathfrak h$. This basically means that there exist $mathfrak h_{mathbb R} subset mathfrak g$ such that $mathfrak h = mathbb{C} otimes mathfrak h_{mathbb R}$.
So, my question is: does it exists an ideal $mathfrak i subset mathbb C mathfrak g$ that is not essentially real?
In other words, I want an ideal $mathfrak i$ that is not just the complexification of an ideal from $mathfrak g$.
Thank you very much for any help.
lie-groups lie-algebras
asked Jan 10 at 1:32
Max Reinhold JahnkeMax Reinhold Jahnke
353213
$endgroup$
add a comment |
$begingroup$
Let $mathfrak{g}$ be the Lie algebra of a semisimple compact Lie group $G$. Denote by $mathbb{C} mathfrak g = mathbb{C} otimes mathfrak g$ the complexification of $mathfrak g$. I am assuming that $mathfrak g$ is not simple, then there exist an ideal $mathfrak i subsetmathfrak g$ that is different from ${0}$ and different from the entire Lie algebra $mathfrak g$.
A subalgebra $mathfrak h subset mathbb C mathfrak g$ is called essentially real if $overline {mathfrak h} = mathfrak h$. This basically means that there exist $mathfrak h_{mathbb R} subset mathfrak g$ such that $mathfrak h = mathbb{C} otimes mathfrak h_{mathbb R}$.
So, my question is: does it exists an ideal $mathfrak i subset mathbb C mathfrak g$ that is not essentially real?
In other words, I want an ideal $mathfrak i$ that is not just the complexification of an ideal from $mathfrak g$.
Thank you very much for any help.
lie-groups lie-algebras
asked Jan 10 at 1:32
Max Reinhold JahnkeMax Reinhold Jahnke
353213
$endgroup$
add a comment |
$begingroup$
Let $mathfrak{g}$ be the Lie algebra of a semisimple compact Lie group $G$. Denote by $mathbb{C} mathfrak g = mathbb{C} otimes mathfrak g$ the complexification of $mathfrak g$. I am assuming that $mathfrak g$ is not simple, then there exist an ideal $mathfrak i subsetmathfrak g$ that is different from ${0}$ and different from the entire Lie algebra $mathfrak g$.
A subalgebra $mathfrak h subset mathbb C mathfrak g$ is called essentially real if $overline {mathfrak h} = mathfrak h$. This basically means that there exist $mathfrak h_{mathbb R} subset mathfrak g$ such that $mathfrak h = mathbb{C} otimes mathfrak h_{mathbb R}$.
So, my question is: does it exists an ideal $mathfrak i subset mathbb C mathfrak g$ that is not essentially real?
In other words, I want an ideal $mathfrak i$ that is not just the complexification of an ideal from $mathfrak g$.
Thank you very much for any help.
lie-groups lie-algebras
asked Jan 10 at 1:32
Max Reinhold JahnkeMax Reinhold Jahnke
353213
$endgroup$
Let $mathfrak{g}$ be the Lie algebra of a semisimple compact Lie group $G$. Denote by $mathbb{C} mathfrak g = mathbb{C} otimes mathfrak g$ the complexification of $mathfrak g$. I am assuming that $mathfrak g$ is not simple, then there exist an ideal $mathfrak i subsetmathfrak g$ that is different from ${0}$ and different from the entire Lie algebra $mathfrak g$.
A subalgebra $mathfrak h subset mathbb C mathfrak g$ is called essentially real if $overline {mathfrak h} = mathfrak h$. This basically means that there exist $mathfrak h_{mathbb R} subset mathfrak g$ such that $mathfrak h = mathbb{C} otimes mathfrak h_{mathbb R}$.
So, my question is: does it exists an ideal $mathfrak i subset mathbb C mathfrak g$ that is not essentially real?
In other words, I want an ideal $mathfrak i$ that is not just the complexification of an ideal from $mathfrak g$.
Thank you very much for any help.
lie-groups lie-algebras
lie-groups lie-algebras
asked Jan 10 at 1:32
Max Reinhold JahnkeMax Reinhold Jahnke
353213
asked Jan 10 at 1:32
Max Reinhold JahnkeMax Reinhold Jahnke
353213
edited Jan 10 at 18:15
Max Reinhold Jahnke
asked Jan 10 at 1:32
Max Reinhold JahnkeMax Reinhold Jahnke
353213
asked Jan 10 at 1:32
Max Reinhold JahnkeMax Reinhold Jahnke
353213
asked Jan 10 at 1:32
Max Reinhold JahnkeMax Reinhold Jahnke
353213
353213
add a comment |
add a comment |
1 Answer
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$begingroup$
No such ideal exists.
A simple Lie algebra $mathfrak{g}$ over a field $k$ is called absolutely simple if for every algebraic extension $Kvert k$ (or equivalently: for an algebraic closure $Kvert k$) , the scalar extension $Kotimes mathfrak{g}$ is also simple (note that it necessarily is semisimple; for an example where it is not simple, see below).
One can show that if a Lie algebra over a field $k$ is simple but not absolutely simple, it is the scalar restriction of an absolutely simple Lie algebra over some algebraic extension $K vert k$. Actually, one can compute $K$ as the (associative) subalgebra of $End_k(mathfrak{g})$ consisting of those elements that commute with all $ad_mathfrak{g}(x), x in mathfrak{g}$. As far as I know, this was first shown by Jacobson in Duke Math. J., Volume 3, Number 3 (1937), 544-548, doi:10.1215/S0012-7094-37-00343-0, and holds for more general kinds of algebras. I wrote a little overview of that in section 4.1 of my thesis.
Now to your question: Since scalar extension commutes with direct sums, and ideals of semisimple Lie algebras are direct summands, your question is equivalent to asking whether there exists a simple compact real Lie algebra $mathfrak{g}$ which is not absolutely simple. But by the above theory, and the fact that the only proper algebraic extension of $Bbb R$ is $Bbb C$, the only simple but not absolutely simple real Lie algebras are: the simple complex Lie algebras considered as $Bbb R$-algebras. The first example maybe being $mathfrak{sl}_2(Bbb C)$ viewed as a Lie algebra over $Bbb R$ (six-dimensional); it is simple, but not absolutely simple, as its scalar extension $Bbb C otimes_{Bbb R} mathfrak{sl}_2(Bbb C)$ actually is isomorphic to the sum of two copies of $mathfrak{sl}_2(Bbb C)$.
However, none of these scalar restrictions of simple complex Lie algebras correspond to compact Lie groups, e.g. because they obviously contain nilpotent elements.
answered Jan 11 at 0:17
Torsten SchoenebergTorsten Schoeneberg
4,0262833
$endgroup$
$begingroup$
Thank you very much for your detailed answer. I was seeking such simple ideals as a way to construct a semisimple subalgebra of $mathbb C mathfrak g$ having a specific property. Now it looks that my idea is not going to work. Maybe you can help me with it. If you not mind, please, take a look at this question: math.stackexchange.com/questions/3069692/…
$endgroup$
– Max Reinhold Jahnke
Jan 11 at 10:40
add a comment |
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$begingroup$
No such ideal exists.
A simple Lie algebra $mathfrak{g}$ over a field $k$ is called absolutely simple if for every algebraic extension $Kvert k$ (or equivalently: for an algebraic closure $Kvert k$) , the scalar extension $Kotimes mathfrak{g}$ is also simple (note that it necessarily is semisimple; for an example where it is not simple, see below).
One can show that if a Lie algebra over a field $k$ is simple but not absolutely simple, it is the scalar restriction of an absolutely simple Lie algebra over some algebraic extension $K vert k$. Actually, one can compute $K$ as the (associative) subalgebra of $End_k(mathfrak{g})$ consisting of those elements that commute with all $ad_mathfrak{g}(x), x in mathfrak{g}$. As far as I know, this was first shown by Jacobson in Duke Math. J., Volume 3, Number 3 (1937), 544-548, doi:10.1215/S0012-7094-37-00343-0, and holds for more general kinds of algebras. I wrote a little overview of that in section 4.1 of my thesis.
Now to your question: Since scalar extension commutes with direct sums, and ideals of semisimple Lie algebras are direct summands, your question is equivalent to asking whether there exists a simple compact real Lie algebra $mathfrak{g}$ which is not absolutely simple. But by the above theory, and the fact that the only proper algebraic extension of $Bbb R$ is $Bbb C$, the only simple but not absolutely simple real Lie algebras are: the simple complex Lie algebras considered as $Bbb R$-algebras. The first example maybe being $mathfrak{sl}_2(Bbb C)$ viewed as a Lie algebra over $Bbb R$ (six-dimensional); it is simple, but not absolutely simple, as its scalar extension $Bbb C otimes_{Bbb R} mathfrak{sl}_2(Bbb C)$ actually is isomorphic to the sum of two copies of $mathfrak{sl}_2(Bbb C)$.
However, none of these scalar restrictions of simple complex Lie algebras correspond to compact Lie groups, e.g. because they obviously contain nilpotent elements.
answered Jan 11 at 0:17
Torsten SchoenebergTorsten Schoeneberg
4,0262833
$endgroup$
$begingroup$
Thank you very much for your detailed answer. I was seeking such simple ideals as a way to construct a semisimple subalgebra of $mathbb C mathfrak g$ having a specific property. Now it looks that my idea is not going to work. Maybe you can help me with it. If you not mind, please, take a look at this question: math.stackexchange.com/questions/3069692/…
$endgroup$
– Max Reinhold Jahnke
Jan 11 at 10:40
add a comment |
$begingroup$
No such ideal exists.
A simple Lie algebra $mathfrak{g}$ over a field $k$ is called absolutely simple if for every algebraic extension $Kvert k$ (or equivalently: for an algebraic closure $Kvert k$) , the scalar extension $Kotimes mathfrak{g}$ is also simple (note that it necessarily is semisimple; for an example where it is not simple, see below).
One can show that if a Lie algebra over a field $k$ is simple but not absolutely simple, it is the scalar restriction of an absolutely simple Lie algebra over some algebraic extension $K vert k$. Actually, one can compute $K$ as the (associative) subalgebra of $End_k(mathfrak{g})$ consisting of those elements that commute with all $ad_mathfrak{g}(x), x in mathfrak{g}$. As far as I know, this was first shown by Jacobson in Duke Math. J., Volume 3, Number 3 (1937), 544-548, doi:10.1215/S0012-7094-37-00343-0, and holds for more general kinds of algebras. I wrote a little overview of that in section 4.1 of my thesis.
Now to your question: Since scalar extension commutes with direct sums, and ideals of semisimple Lie algebras are direct summands, your question is equivalent to asking whether there exists a simple compact real Lie algebra $mathfrak{g}$ which is not absolutely simple. But by the above theory, and the fact that the only proper algebraic extension of $Bbb R$ is $Bbb C$, the only simple but not absolutely simple real Lie algebras are: the simple complex Lie algebras considered as $Bbb R$-algebras. The first example maybe being $mathfrak{sl}_2(Bbb C)$ viewed as a Lie algebra over $Bbb R$ (six-dimensional); it is simple, but not absolutely simple, as its scalar extension $Bbb C otimes_{Bbb R} mathfrak{sl}_2(Bbb C)$ actually is isomorphic to the sum of two copies of $mathfrak{sl}_2(Bbb C)$.
However, none of these scalar restrictions of simple complex Lie algebras correspond to compact Lie groups, e.g. because they obviously contain nilpotent elements.
answered Jan 11 at 0:17
Torsten SchoenebergTorsten Schoeneberg
4,0262833
$endgroup$
$begingroup$
Thank you very much for your detailed answer. I was seeking such simple ideals as a way to construct a semisimple subalgebra of $mathbb C mathfrak g$ having a specific property. Now it looks that my idea is not going to work. Maybe you can help me with it. If you not mind, please, take a look at this question: math.stackexchange.com/questions/3069692/…
$endgroup$
– Max Reinhold Jahnke
Jan 11 at 10:40
add a comment |
$begingroup$
No such ideal exists.
A simple Lie algebra $mathfrak{g}$ over a field $k$ is called absolutely simple if for every algebraic extension $Kvert k$ (or equivalently: for an algebraic closure $Kvert k$) , the scalar extension $Kotimes mathfrak{g}$ is also simple (note that it necessarily is semisimple; for an example where it is not simple, see below).
One can show that if a Lie algebra over a field $k$ is simple but not absolutely simple, it is the scalar restriction of an absolutely simple Lie algebra over some algebraic extension $K vert k$. Actually, one can compute $K$ as the (associative) subalgebra of $End_k(mathfrak{g})$ consisting of those elements that commute with all $ad_mathfrak{g}(x), x in mathfrak{g}$. As far as I know, this was first shown by Jacobson in Duke Math. J., Volume 3, Number 3 (1937), 544-548, doi:10.1215/S0012-7094-37-00343-0, and holds for more general kinds of algebras. I wrote a little overview of that in section 4.1 of my thesis.
Now to your question: Since scalar extension commutes with direct sums, and ideals of semisimple Lie algebras are direct summands, your question is equivalent to asking whether there exists a simple compact real Lie algebra $mathfrak{g}$ which is not absolutely simple. But by the above theory, and the fact that the only proper algebraic extension of $Bbb R$ is $Bbb C$, the only simple but not absolutely simple real Lie algebras are: the simple complex Lie algebras considered as $Bbb R$-algebras. The first example maybe being $mathfrak{sl}_2(Bbb C)$ viewed as a Lie algebra over $Bbb R$ (six-dimensional); it is simple, but not absolutely simple, as its scalar extension $Bbb C otimes_{Bbb R} mathfrak{sl}_2(Bbb C)$ actually is isomorphic to the sum of two copies of $mathfrak{sl}_2(Bbb C)$.
However, none of these scalar restrictions of simple complex Lie algebras correspond to compact Lie groups, e.g. because they obviously contain nilpotent elements.
answered Jan 11 at 0:17
Torsten SchoenebergTorsten Schoeneberg
4,0262833
$endgroup$
No such ideal exists.
A simple Lie algebra $mathfrak{g}$ over a field $k$ is called absolutely simple if for every algebraic extension $Kvert k$ (or equivalently: for an algebraic closure $Kvert k$) , the scalar extension $Kotimes mathfrak{g}$ is also simple (note that it necessarily is semisimple; for an example where it is not simple, see below).
One can show that if a Lie algebra over a field $k$ is simple but not absolutely simple, it is the scalar restriction of an absolutely simple Lie algebra over some algebraic extension $K vert k$. Actually, one can compute $K$ as the (associative) subalgebra of $End_k(mathfrak{g})$ consisting of those elements that commute with all $ad_mathfrak{g}(x), x in mathfrak{g}$. As far as I know, this was first shown by Jacobson in Duke Math. J., Volume 3, Number 3 (1937), 544-548, doi:10.1215/S0012-7094-37-00343-0, and holds for more general kinds of algebras. I wrote a little overview of that in section 4.1 of my thesis.
Now to your question: Since scalar extension commutes with direct sums, and ideals of semisimple Lie algebras are direct summands, your question is equivalent to asking whether there exists a simple compact real Lie algebra $mathfrak{g}$ which is not absolutely simple. But by the above theory, and the fact that the only proper algebraic extension of $Bbb R$ is $Bbb C$, the only simple but not absolutely simple real Lie algebras are: the simple complex Lie algebras considered as $Bbb R$-algebras. The first example maybe being $mathfrak{sl}_2(Bbb C)$ viewed as a Lie algebra over $Bbb R$ (six-dimensional); it is simple, but not absolutely simple, as its scalar extension $Bbb C otimes_{Bbb R} mathfrak{sl}_2(Bbb C)$ actually is isomorphic to the sum of two copies of $mathfrak{sl}_2(Bbb C)$.
However, none of these scalar restrictions of simple complex Lie algebras correspond to compact Lie groups, e.g. because they obviously contain nilpotent elements.
answered Jan 11 at 0:17
Torsten SchoenebergTorsten Schoeneberg
4,0262833
answered Jan 11 at 0:17
Torsten SchoenebergTorsten Schoeneberg
4,0262833
answered Jan 11 at 0:17
Torsten SchoenebergTorsten Schoeneberg
4,0262833
answered Jan 11 at 0:17
Torsten SchoenebergTorsten Schoeneberg
4,0262833
4,0262833
$begingroup$
Thank you very much for your detailed answer. I was seeking such simple ideals as a way to construct a semisimple subalgebra of $mathbb C mathfrak g$ having a specific property. Now it looks that my idea is not going to work. Maybe you can help me with it. If you not mind, please, take a look at this question: math.stackexchange.com/questions/3069692/…
$endgroup$
– Max Reinhold Jahnke
Jan 11 at 10:40
add a comment |
$begingroup$
Thank you very much for your detailed answer. I was seeking such simple ideals as a way to construct a semisimple subalgebra of $mathbb C mathfrak g$ having a specific property. Now it looks that my idea is not going to work. Maybe you can help me with it. If you not mind, please, take a look at this question: math.stackexchange.com/questions/3069692/…
$endgroup$
– Max Reinhold Jahnke
Jan 11 at 10:40
$begingroup$
Thank you very much for your detailed answer. I was seeking such simple ideals as a way to construct a semisimple subalgebra of $mathbb C mathfrak g$ having a specific property. Now it looks that my idea is not going to work. Maybe you can help me with it. If you not mind, please, take a look at this question: math.stackexchange.com/questions/3069692/…
$endgroup$
– Max Reinhold Jahnke
Jan 11 at 10:40
$begingroup$
Thank you very much for your detailed answer. I was seeking such simple ideals as a way to construct a semisimple subalgebra of $mathbb C mathfrak g$ having a specific property. Now it looks that my idea is not going to work. Maybe you can help me with it. If you not mind, please, take a look at this question: math.stackexchange.com/questions/3069692/…
$endgroup$
– Max Reinhold Jahnke
Jan 11 at 10:40
add a comment |
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