Correspondence between complex and real subalgebras

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Let $mathfrak g$ be a real Lie algebra and let $mathfrak g^mathbb C$ be its complexification. Is every complex subalgebra of $mathfrak g^mathbb C$ a complexification of some subalgebra of $mathfrak g$? If not, what is a counterexample?

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asked Jul 17 at 22:27

Ronald

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Let $mathfrak g$ be a real Lie algebra and let $mathfrak g^mathbb C$ be its complexification. Is every complex subalgebra of $mathfrak g^mathbb C$ a complexification of some subalgebra of $mathfrak g$? If not, what is a counterexample?

lie-algebras

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asked Jul 17 at 22:27

Ronald

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up vote
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Let $mathfrak g$ be a real Lie algebra and let $mathfrak g^mathbb C$ be its complexification. Is every complex subalgebra of $mathfrak g^mathbb C$ a complexification of some subalgebra of $mathfrak g$? If not, what is a counterexample?

lie-algebras

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asked Jul 17 at 22:27

Ronald

1,5391821

Let $mathfrak g$ be a real Lie algebra and let $mathfrak g^mathbb C$ be its complexification. Is every complex subalgebra of $mathfrak g^mathbb C$ a complexification of some subalgebra of $mathfrak g$? If not, what is a counterexample?

lie-algebras

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asked Jul 17 at 22:27

Ronald

1,5391821

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share|cite|improve this question

asked Jul 17 at 22:27

Ronald

1,5391821

asked Jul 17 at 22:27

Ronald

1,5391821

asked Jul 17 at 22:27

Ronald

1,5391821

1,5391821

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add a comment | 

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Counterexample:
$$mathfrakg = mathfraksu_2 =lbrace pmatrixai & b+ci\ -b+ci & -ai : a,b,c in Bbb R rbrace$$
$$mathfrakg_Bbb C simeq mathfraksl_2(Bbb C) = lbrace pmatrixx & y\ z & -x : x,y,z in Bbb C rbrace$$
Then the subalgebra $mathfrakb :=lbrace pmatrixx & y\ 0 & -x : x,y,z in Bbb C rbrace subsetneq mathfrakg_Bbb C$ does not come from a subalgebra of $mathfrakg$. (In general, many real forms of semisimple Lie algebras do not contain subalgebras which become Borel subalgebras after complexification; only the so-called quasi-split ones do. There's certainly tons of other examples for other classes of Lie algebras.)

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answered Jul 17 at 23:59

Torsten Schoeneberg

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

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active

oldest

votes

up vote
1
down vote



accepted

Counterexample:
$$mathfrakg = mathfraksu_2 =lbrace pmatrixai & b+ci\ -b+ci & -ai : a,b,c in Bbb R rbrace$$
$$mathfrakg_Bbb C simeq mathfraksl_2(Bbb C) = lbrace pmatrixx & y\ z & -x : x,y,z in Bbb C rbrace$$
Then the subalgebra $mathfrakb :=lbrace pmatrixx & y\ 0 & -x : x,y,z in Bbb C rbrace subsetneq mathfrakg_Bbb C$ does not come from a subalgebra of $mathfrakg$. (In general, many real forms of semisimple Lie algebras do not contain subalgebras which become Borel subalgebras after complexification; only the so-called quasi-split ones do. There's certainly tons of other examples for other classes of Lie algebras.)

share|cite|improve this answer

answered Jul 17 at 23:59

Torsten Schoeneberg

2,6021732

add a comment | 

up vote
1
down vote



accepted

Counterexample:
$$mathfrakg = mathfraksu_2 =lbrace pmatrixai & b+ci\ -b+ci & -ai : a,b,c in Bbb R rbrace$$
$$mathfrakg_Bbb C simeq mathfraksl_2(Bbb C) = lbrace pmatrixx & y\ z & -x : x,y,z in Bbb C rbrace$$
Then the subalgebra $mathfrakb :=lbrace pmatrixx & y\ 0 & -x : x,y,z in Bbb C rbrace subsetneq mathfrakg_Bbb C$ does not come from a subalgebra of $mathfrakg$. (In general, many real forms of semisimple Lie algebras do not contain subalgebras which become Borel subalgebras after complexification; only the so-called quasi-split ones do. There's certainly tons of other examples for other classes of Lie algebras.)

share|cite|improve this answer

answered Jul 17 at 23:59

Torsten Schoeneberg

2,6021732

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Counterexample:
$$mathfrakg = mathfraksu_2 =lbrace pmatrixai & b+ci\ -b+ci & -ai : a,b,c in Bbb R rbrace$$
$$mathfrakg_Bbb C simeq mathfraksl_2(Bbb C) = lbrace pmatrixx & y\ z & -x : x,y,z in Bbb C rbrace$$
Then the subalgebra $mathfrakb :=lbrace pmatrixx & y\ 0 & -x : x,y,z in Bbb C rbrace subsetneq mathfrakg_Bbb C$ does not come from a subalgebra of $mathfrakg$. (In general, many real forms of semisimple Lie algebras do not contain subalgebras which become Borel subalgebras after complexification; only the so-called quasi-split ones do. There's certainly tons of other examples for other classes of Lie algebras.)

share|cite|improve this answer

answered Jul 17 at 23:59

Torsten Schoeneberg

2,6021732

Counterexample:
$$mathfrakg = mathfraksu_2 =lbrace pmatrixai & b+ci\ -b+ci & -ai : a,b,c in Bbb R rbrace$$
$$mathfrakg_Bbb C simeq mathfraksl_2(Bbb C) = lbrace pmatrixx & y\ z & -x : x,y,z in Bbb C rbrace$$
Then the subalgebra $mathfrakb :=lbrace pmatrixx & y\ 0 & -x : x,y,z in Bbb C rbrace subsetneq mathfrakg_Bbb C$ does not come from a subalgebra of $mathfrakg$. (In general, many real forms of semisimple Lie algebras do not contain subalgebras which become Borel subalgebras after complexification; only the so-called quasi-split ones do. There's certainly tons of other examples for other classes of Lie algebras.)

share|cite|improve this answer

answered Jul 17 at 23:59

Torsten Schoeneberg

2,6021732

share|cite|improve this answer

share|cite|improve this answer

answered Jul 17 at 23:59

Torsten Schoeneberg

2,6021732

answered Jul 17 at 23:59

Torsten Schoeneberg

2,6021732

answered Jul 17 at 23:59

Torsten Schoeneberg

2,6021732

2,6021732

add a comment | 

add a comment | 

 

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