Ideals of a characteristic ideal
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
1,5532514
add a comment |
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
1,5532514
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
1,5532514
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
1,5532514
asked Nov 18 at 5:58
Sid Caroline
1,5532514
asked Nov 18 at 5:58
Sid Caroline
1,5532514
asked Nov 18 at 5:58
Sid Caroline
1,5532514
asked Nov 18 at 5:58
Sid Caroline
1,5532514
1,5532514
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
1
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
54.6k11444
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
54.6k11444
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
54.6k11444
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
54.6k11444
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
54.6k11444
answered Nov 18 at 7:53
Tsemo Aristide
54.6k11444
answered Nov 18 at 7:53
Tsemo Aristide
54.6k11444
answered Nov 18 at 7:53
Tsemo Aristide
54.6k11444
54.6k11444
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3003181%2fideals-of-a-characteristic-ideal%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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