Prove that $[frak{g}$ $,cen(I)] neq cen(I)$

0

$begingroup$

Let $frak{g}$ be a finite dimensional Lie algebra over an algebraically closed field, $k$, of characteristic $0$.

Suppose that $I triangleleft frak{g}$ is an ideal of co-dimension $1$, and that ${0} neq I cap Z(frak{g})$.

Moreover, define $cen(I) = {g in frak{g} : forall$ $x in I, [x,g] =0}$.

Prove that $[frak{g}$$,cen(I)] neq cen(I)$.

Where I'm at:

First, it can be easily shown that $cen(I)$ is an ideal itself, and hence we get that $[frak{g}$$,cen(I)] subset cen(I)$. So we must show that the inclusion is strict.

Since $I$ is of co-dimension $1$ we have that $[frak{g},g] subset$ $I$. Therefore $[frak{g}$$,cen(I)] subset I$.

Clearly, if $Z(frak{g})$ isn't contained in $I$ we can write that $frak{g} =$ $kx_0 oplus I$ for $x_0 in Z(frak{g})$. This gives $[frak{g}$$,cen(I)] = [kx_0 oplus I, cen(I)] = 0$ and we'd be done, since $I cap Z(frak{g})$ is non-zero and is contained in $cen(I)$.

Thus we can assume: ${0} neq Z(frak{g}) subset$ $I$.

Here I got stuck; I was trying to use Lie's lemma, and find a functional that is non zero on $cen(I)$, and has a common eigenvector. This would give the required inequality. However it is not clear to me how I may do so.

Moreover, I'm having trouble interpreting the assumption that ${0} neq I cap Z(frak{g})$

Any hints?

abstract-algebra lie-algebras

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edited Jan 2 at 16:14

Mariah

asked Jan 2 at 15:49

MariahMariah

1,3831518

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0

$begingroup$

Let $frak{g}$ be a finite dimensional Lie algebra over an algebraically closed field, $k$, of characteristic $0$.

Suppose that $I triangleleft frak{g}$ is an ideal of co-dimension $1$, and that ${0} neq I cap Z(frak{g})$.

Moreover, define $cen(I) = {g in frak{g} : forall$ $x in I, [x,g] =0}$.

Prove that $[frak{g}$$,cen(I)] neq cen(I)$.

Where I'm at:

First, it can be easily shown that $cen(I)$ is an ideal itself, and hence we get that $[frak{g}$$,cen(I)] subset cen(I)$. So we must show that the inclusion is strict.

Since $I$ is of co-dimension $1$ we have that $[frak{g},g] subset$ $I$. Therefore $[frak{g}$$,cen(I)] subset I$.

Clearly, if $Z(frak{g})$ isn't contained in $I$ we can write that $frak{g} =$ $kx_0 oplus I$ for $x_0 in Z(frak{g})$. This gives $[frak{g}$$,cen(I)] = [kx_0 oplus I, cen(I)] = 0$ and we'd be done, since $I cap Z(frak{g})$ is non-zero and is contained in $cen(I)$.

Thus we can assume: ${0} neq Z(frak{g}) subset$ $I$.

Here I got stuck; I was trying to use Lie's lemma, and find a functional that is non zero on $cen(I)$, and has a common eigenvector. This would give the required inequality. However it is not clear to me how I may do so.

Moreover, I'm having trouble interpreting the assumption that ${0} neq I cap Z(frak{g})$

Any hints?

abstract-algebra lie-algebras

share|cite|improve this question

edited Jan 2 at 16:14

Mariah

asked Jan 2 at 15:49

MariahMariah

1,3831518

$endgroup$

add a comment | 

0

0

0

$begingroup$

Let $frak{g}$ be a finite dimensional Lie algebra over an algebraically closed field, $k$, of characteristic $0$.

Suppose that $I triangleleft frak{g}$ is an ideal of co-dimension $1$, and that ${0} neq I cap Z(frak{g})$.

Moreover, define $cen(I) = {g in frak{g} : forall$ $x in I, [x,g] =0}$.

Prove that $[frak{g}$$,cen(I)] neq cen(I)$.

Where I'm at:

First, it can be easily shown that $cen(I)$ is an ideal itself, and hence we get that $[frak{g}$$,cen(I)] subset cen(I)$. So we must show that the inclusion is strict.

Since $I$ is of co-dimension $1$ we have that $[frak{g},g] subset$ $I$. Therefore $[frak{g}$$,cen(I)] subset I$.

Clearly, if $Z(frak{g})$ isn't contained in $I$ we can write that $frak{g} =$ $kx_0 oplus I$ for $x_0 in Z(frak{g})$. This gives $[frak{g}$$,cen(I)] = [kx_0 oplus I, cen(I)] = 0$ and we'd be done, since $I cap Z(frak{g})$ is non-zero and is contained in $cen(I)$.

Thus we can assume: ${0} neq Z(frak{g}) subset$ $I$.

Here I got stuck; I was trying to use Lie's lemma, and find a functional that is non zero on $cen(I)$, and has a common eigenvector. This would give the required inequality. However it is not clear to me how I may do so.

Moreover, I'm having trouble interpreting the assumption that ${0} neq I cap Z(frak{g})$

Any hints?

abstract-algebra lie-algebras

share|cite|improve this question

edited Jan 2 at 16:14

Mariah

asked Jan 2 at 15:49

MariahMariah

1,3831518

$endgroup$

Let $frak{g}$ be a finite dimensional Lie algebra over an algebraically closed field, $k$, of characteristic $0$.

Suppose that $I triangleleft frak{g}$ is an ideal of co-dimension $1$, and that ${0} neq I cap Z(frak{g})$.

Moreover, define $cen(I) = {g in frak{g} : forall$ $x in I, [x,g] =0}$.

Prove that $[frak{g}$$,cen(I)] neq cen(I)$.

Where I'm at:

First, it can be easily shown that $cen(I)$ is an ideal itself, and hence we get that $[frak{g}$$,cen(I)] subset cen(I)$. So we must show that the inclusion is strict.

Since $I$ is of co-dimension $1$ we have that $[frak{g},g] subset$ $I$. Therefore $[frak{g}$$,cen(I)] subset I$.

Clearly, if $Z(frak{g})$ isn't contained in $I$ we can write that $frak{g} =$ $kx_0 oplus I$ for $x_0 in Z(frak{g})$. This gives $[frak{g}$$,cen(I)] = [kx_0 oplus I, cen(I)] = 0$ and we'd be done, since $I cap Z(frak{g})$ is non-zero and is contained in $cen(I)$.

Thus we can assume: ${0} neq Z(frak{g}) subset$ $I$.

Here I got stuck; I was trying to use Lie's lemma, and find a functional that is non zero on $cen(I)$, and has a common eigenvector. This would give the required inequality. However it is not clear to me how I may do so.

Moreover, I'm having trouble interpreting the assumption that ${0} neq I cap Z(frak{g})$

Any hints?

abstract-algebra lie-algebras

abstract-algebra lie-algebras

share|cite|improve this question

edited Jan 2 at 16:14

Mariah

asked Jan 2 at 15:49

MariahMariah

1,3831518

share|cite|improve this question

edited Jan 2 at 16:14

Mariah

asked Jan 2 at 15:49

MariahMariah

1,3831518

share|cite|improve this question

share|cite|improve this question

edited Jan 2 at 16:14

Mariah

edited Jan 2 at 16:14

Mariah

edited Jan 2 at 16:14

Mariah

asked Jan 2 at 15:49

MariahMariah

1,3831518

asked Jan 2 at 15:49

MariahMariah

1,3831518

asked Jan 2 at 15:49

MariahMariah

1,3831518

1,3831518

add a comment | 

add a comment | 

1 Answer
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$begingroup$

Let $x$ an element which is not in $I$, $[g,cen(I)]=[kxoplus I,cen(I)]=[kx,cen(I)]$, this implies that $[g,cen(I)]$ is the image of the restriction of $ad_x$ defined by $ad_x(y)=[x,y]$ to $cen(I)$, but the kernel of the restriction of $ad_x$ to $cen(I)$ is not trivial since it contains $Z(I)$, thus $ad_x$ is not surjective and $[g,cen(I)]neq cen(I)$.

share|cite|improve this answer

answered Jan 2 at 16:16

Tsemo AristideTsemo Aristide

56.8k11444

$endgroup$

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$begingroup$

Let $x$ an element which is not in $I$, $[g,cen(I)]=[kxoplus I,cen(I)]=[kx,cen(I)]$, this implies that $[g,cen(I)]$ is the image of the restriction of $ad_x$ defined by $ad_x(y)=[x,y]$ to $cen(I)$, but the kernel of the restriction of $ad_x$ to $cen(I)$ is not trivial since it contains $Z(I)$, thus $ad_x$ is not surjective and $[g,cen(I)]neq cen(I)$.

share|cite|improve this answer

answered Jan 2 at 16:16

Tsemo AristideTsemo Aristide

56.8k11444

$endgroup$

add a comment | 

1

$begingroup$

Let $x$ an element which is not in $I$, $[g,cen(I)]=[kxoplus I,cen(I)]=[kx,cen(I)]$, this implies that $[g,cen(I)]$ is the image of the restriction of $ad_x$ defined by $ad_x(y)=[x,y]$ to $cen(I)$, but the kernel of the restriction of $ad_x$ to $cen(I)$ is not trivial since it contains $Z(I)$, thus $ad_x$ is not surjective and $[g,cen(I)]neq cen(I)$.

share|cite|improve this answer

answered Jan 2 at 16:16

Tsemo AristideTsemo Aristide

56.8k11444

$endgroup$

add a comment | 

1

1

1

$begingroup$

Let $x$ an element which is not in $I$, $[g,cen(I)]=[kxoplus I,cen(I)]=[kx,cen(I)]$, this implies that $[g,cen(I)]$ is the image of the restriction of $ad_x$ defined by $ad_x(y)=[x,y]$ to $cen(I)$, but the kernel of the restriction of $ad_x$ to $cen(I)$ is not trivial since it contains $Z(I)$, thus $ad_x$ is not surjective and $[g,cen(I)]neq cen(I)$.

share|cite|improve this answer

answered Jan 2 at 16:16

Tsemo AristideTsemo Aristide

56.8k11444

$endgroup$

Let $x$ an element which is not in $I$, $[g,cen(I)]=[kxoplus I,cen(I)]=[kx,cen(I)]$, this implies that $[g,cen(I)]$ is the image of the restriction of $ad_x$ defined by $ad_x(y)=[x,y]$ to $cen(I)$, but the kernel of the restriction of $ad_x$ to $cen(I)$ is not trivial since it contains $Z(I)$, thus $ad_x$ is not surjective and $[g,cen(I)]neq cen(I)$.

share|cite|improve this answer

answered Jan 2 at 16:16

Tsemo AristideTsemo Aristide

56.8k11444

share|cite|improve this answer

share|cite|improve this answer

answered Jan 2 at 16:16

Tsemo AristideTsemo Aristide

56.8k11444

answered Jan 2 at 16:16

Tsemo AristideTsemo Aristide

56.8k11444

answered Jan 2 at 16:16

Tsemo AristideTsemo Aristide

56.8k11444

56.8k11444

add a comment | 

add a comment | 

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