Ideals of a characteristic ideal
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We say $frak{h}$ is a characteristic ideal of a lie algebra $frak{g}$ if $[frak{h},frak{g}]subsetfrak{h}$, and $D(frak{h})subsetfrak{h}$ for every derivation $Din Der(frak{g})$.
The theorem states that if $frak{h}$ is a characteristic ideal of a lie algebra $frak{g}$, then every ideal of $frak{h}$ is also an ideal in $frak{g}$. Essentially, if we let $frak{a}$ be an ideal in $frak{h}$, we want to show that $[frak{a},frak{g}]subset frak{a}$, but I don't know what derivation $Din Der(frak{g})$ should be used?
lie-algebras
asked Nov 18 at 5:58
Sid Caroline
1,5382514
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We say $frak{h}$ is a characteristic ideal of a lie algebra $frak{g}$ if $[frak{h},frak{g}]subsetfrak{h}$, and $D(frak{h})subsetfrak{h}$ for every derivation $Din Der(frak{g})$.
The theorem states that if $frak{h}$ is a characteristic ideal of a lie algebra $frak{g}$, then every ideal of $frak{h}$ is also an ideal in $frak{g}$. Essentially, if we let $frak{a}$ be an ideal in $frak{h}$, we want to show that $[frak{a},frak{g}]subset frak{a}$, but I don't know what derivation $Din Der(frak{g})$ should be used?
lie-algebras
asked Nov 18 at 5:58
Sid Caroline
1,5382514
I don't think this is true--at least, the corresponding statement for groups is definitely not true.
– Eric Wofsey
Nov 18 at 6:11
1
It is not to prove that if $asubset hsubset g$ and $a$ is a characteristic ideal of $h$ then it is an ideal of $g$?
– Tsemo Aristide
Nov 18 at 6:13
@TsemoAristide Yes I think you are right.
– Sid Caroline
Nov 18 at 7:47
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
We say $frak{h}$ is a characteristic ideal of a lie algebra $frak{g}$ if $[frak{h},frak{g}]subsetfrak{h}$, and $D(frak{h})subsetfrak{h}$ for every derivation $Din Der(frak{g})$.
The theorem states that if $frak{h}$ is a characteristic ideal of a lie algebra $frak{g}$, then every ideal of $frak{h}$ is also an ideal in $frak{g}$. Essentially, if we let $frak{a}$ be an ideal in $frak{h}$, we want to show that $[frak{a},frak{g}]subset frak{a}$, but I don't know what derivation $Din Der(frak{g})$ should be used?
lie-algebras
asked Nov 18 at 5:58
Sid Caroline
1,5382514
We say $frak{h}$ is a characteristic ideal of a lie algebra $frak{g}$ if $[frak{h},frak{g}]subsetfrak{h}$, and $D(frak{h})subsetfrak{h}$ for every derivation $Din Der(frak{g})$.
The theorem states that if $frak{h}$ is a characteristic ideal of a lie algebra $frak{g}$, then every ideal of $frak{h}$ is also an ideal in $frak{g}$. Essentially, if we let $frak{a}$ be an ideal in $frak{h}$, we want to show that $[frak{a},frak{g}]subset frak{a}$, but I don't know what derivation $Din Der(frak{g})$ should be used?
lie-algebras
lie-algebras
asked Nov 18 at 5:58
Sid Caroline
1,5382514
asked Nov 18 at 5:58
Sid Caroline
1,5382514
asked Nov 18 at 5:58
Sid Caroline
1,5382514
asked Nov 18 at 5:58
Sid Caroline
1,5382514
asked Nov 18 at 5:58
Sid Caroline
1,5382514
1,5382514
I don't think this is true--at least, the corresponding statement for groups is definitely not true.
– Eric Wofsey
Nov 18 at 6:11
1
It is not to prove that if $asubset hsubset g$ and $a$ is a characteristic ideal of $h$ then it is an ideal of $g$?
– Tsemo Aristide
Nov 18 at 6:13
@TsemoAristide Yes I think you are right.
– Sid Caroline
Nov 18 at 7:47
add a comment |
I don't think this is true--at least, the corresponding statement for groups is definitely not true.
– Eric Wofsey
Nov 18 at 6:11
1
It is not to prove that if $asubset hsubset g$ and $a$ is a characteristic ideal of $h$ then it is an ideal of $g$?
– Tsemo Aristide
Nov 18 at 6:13
@TsemoAristide Yes I think you are right.
– Sid Caroline
Nov 18 at 7:47
I don't think this is true--at least, the corresponding statement for groups is definitely not true.
– Eric Wofsey
Nov 18 at 6:11
I don't think this is true--at least, the corresponding statement for groups is definitely not true.
– Eric Wofsey
Nov 18 at 6:11
1
1
It is not to prove that if $asubset hsubset g$ and $a$ is a characteristic ideal of $h$ then it is an ideal of $g$?
– Tsemo Aristide
Nov 18 at 6:13
It is not to prove that if $asubset hsubset g$ and $a$ is a characteristic ideal of $h$ then it is an ideal of $g$?
– Tsemo Aristide
Nov 18 at 6:13
@TsemoAristide Yes I think you are right.
– Sid Caroline
Nov 18 at 7:47
@TsemoAristide Yes I think you are right.
– Sid Caroline
Nov 18 at 7:47
add a comment |
1 Answer
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If I am right, then let $in frak{g}, ainfrak{a}$, $ad_x:frak{h}rightarrowfrak{h}$ defined by $ad_x(h)=[x,h]$ is a derivation (Jacobi). This implies that $ad_x(a)=[x,a]in frak{a}$ since $frak{a}$ is a characteristic ideal of $frak{h}$. This implies that $frak{a}$ is an ideal of $frak{g}$.
answered Nov 18 at 7:53
Tsemo Aristide
54.5k11444
I wonder if there exists a counter example to my original statement.
– Sid Caroline
Nov 18 at 8:03
@SidCaroline yes you have a counterexample of the form $K^2rtimes_D K$ where $K$ is the ground field; assuming that the derived subalgebra is $K^2$ (i.e. $D$ is an invertible $2times 2$ matrix), it is characteristic, and all lines in this $K^2$ are ideals of $K^2$. But if $D$ is not a scalar matrix, there are lines that are not $D$-invariant, and this yields ideals of $K^2$ that are not ideals of the whole Lie algebra.
– YCor
Nov 18 at 11:24
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If I am right, then let $in frak{g}, ainfrak{a}$, $ad_x:frak{h}rightarrowfrak{h}$ defined by $ad_x(h)=[x,h]$ is a derivation (Jacobi). This implies that $ad_x(a)=[x,a]in frak{a}$ since $frak{a}$ is a characteristic ideal of $frak{h}$. This implies that $frak{a}$ is an ideal of $frak{g}$.
answered Nov 18 at 7:53
Tsemo Aristide
54.5k11444
I wonder if there exists a counter example to my original statement.
– Sid Caroline
Nov 18 at 8:03
@SidCaroline yes you have a counterexample of the form $K^2rtimes_D K$ where $K$ is the ground field; assuming that the derived subalgebra is $K^2$ (i.e. $D$ is an invertible $2times 2$ matrix), it is characteristic, and all lines in this $K^2$ are ideals of $K^2$. But if $D$ is not a scalar matrix, there are lines that are not $D$-invariant, and this yields ideals of $K^2$ that are not ideals of the whole Lie algebra.
– YCor
Nov 18 at 11:24
add a comment |
up vote
1
down vote
accepted
If I am right, then let $in frak{g}, ainfrak{a}$, $ad_x:frak{h}rightarrowfrak{h}$ defined by $ad_x(h)=[x,h]$ is a derivation (Jacobi). This implies that $ad_x(a)=[x,a]in frak{a}$ since $frak{a}$ is a characteristic ideal of $frak{h}$. This implies that $frak{a}$ is an ideal of $frak{g}$.
answered Nov 18 at 7:53
Tsemo Aristide
54.5k11444
I wonder if there exists a counter example to my original statement.
– Sid Caroline
Nov 18 at 8:03
@SidCaroline yes you have a counterexample of the form $K^2rtimes_D K$ where $K$ is the ground field; assuming that the derived subalgebra is $K^2$ (i.e. $D$ is an invertible $2times 2$ matrix), it is characteristic, and all lines in this $K^2$ are ideals of $K^2$. But if $D$ is not a scalar matrix, there are lines that are not $D$-invariant, and this yields ideals of $K^2$ that are not ideals of the whole Lie algebra.
– YCor
Nov 18 at 11:24
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If I am right, then let $in frak{g}, ainfrak{a}$, $ad_x:frak{h}rightarrowfrak{h}$ defined by $ad_x(h)=[x,h]$ is a derivation (Jacobi). This implies that $ad_x(a)=[x,a]in frak{a}$ since $frak{a}$ is a characteristic ideal of $frak{h}$. This implies that $frak{a}$ is an ideal of $frak{g}$.
answered Nov 18 at 7:53
Tsemo Aristide
54.5k11444
If I am right, then let $in frak{g}, ainfrak{a}$, $ad_x:frak{h}rightarrowfrak{h}$ defined by $ad_x(h)=[x,h]$ is a derivation (Jacobi). This implies that $ad_x(a)=[x,a]in frak{a}$ since $frak{a}$ is a characteristic ideal of $frak{h}$. This implies that $frak{a}$ is an ideal of $frak{g}$.
answered Nov 18 at 7:53
Tsemo Aristide
54.5k11444
answered Nov 18 at 7:53
Tsemo Aristide
54.5k11444
answered Nov 18 at 7:53
Tsemo Aristide
54.5k11444
answered Nov 18 at 7:53
Tsemo Aristide
54.5k11444
54.5k11444
I wonder if there exists a counter example to my original statement.
– Sid Caroline
Nov 18 at 8:03
@SidCaroline yes you have a counterexample of the form $K^2rtimes_D K$ where $K$ is the ground field; assuming that the derived subalgebra is $K^2$ (i.e. $D$ is an invertible $2times 2$ matrix), it is characteristic, and all lines in this $K^2$ are ideals of $K^2$. But if $D$ is not a scalar matrix, there are lines that are not $D$-invariant, and this yields ideals of $K^2$ that are not ideals of the whole Lie algebra.
– YCor
Nov 18 at 11:24
add a comment |
I wonder if there exists a counter example to my original statement.
– Sid Caroline
Nov 18 at 8:03
@SidCaroline yes you have a counterexample of the form $K^2rtimes_D K$ where $K$ is the ground field; assuming that the derived subalgebra is $K^2$ (i.e. $D$ is an invertible $2times 2$ matrix), it is characteristic, and all lines in this $K^2$ are ideals of $K^2$. But if $D$ is not a scalar matrix, there are lines that are not $D$-invariant, and this yields ideals of $K^2$ that are not ideals of the whole Lie algebra.
– YCor
Nov 18 at 11:24
I wonder if there exists a counter example to my original statement.
– Sid Caroline
Nov 18 at 8:03
I wonder if there exists a counter example to my original statement.
– Sid Caroline
Nov 18 at 8:03
@SidCaroline yes you have a counterexample of the form $K^2rtimes_D K$ where $K$ is the ground field; assuming that the derived subalgebra is $K^2$ (i.e. $D$ is an invertible $2times 2$ matrix), it is characteristic, and all lines in this $K^2$ are ideals of $K^2$. But if $D$ is not a scalar matrix, there are lines that are not $D$-invariant, and this yields ideals of $K^2$ that are not ideals of the whole Lie algebra.
– YCor
Nov 18 at 11:24
@SidCaroline yes you have a counterexample of the form $K^2rtimes_D K$ where $K$ is the ground field; assuming that the derived subalgebra is $K^2$ (i.e. $D$ is an invertible $2times 2$ matrix), it is characteristic, and all lines in this $K^2$ are ideals of $K^2$. But if $D$ is not a scalar matrix, there are lines that are not $D$-invariant, and this yields ideals of $K^2$ that are not ideals of the whole Lie algebra.
– YCor
Nov 18 at 11:24
add a comment |
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I don't think this is true--at least, the corresponding statement for groups is definitely not true.
– Eric Wofsey
Nov 18 at 6:11
1
It is not to prove that if $asubset hsubset g$ and $a$ is a characteristic ideal of $h$ then it is an ideal of $g$?
– Tsemo Aristide
Nov 18 at 6:13
@TsemoAristide Yes I think you are right.
– Sid Caroline
Nov 18 at 7:47