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Let $L$ be a Lie algebra. It is known that $L$ admits a Levi decomposition (possibly non unique):

$L=S\oplus rad(L) $,

where $rad(L)$ is the solvable radical and $S$ is a semisimple subalgebra.

If $L$ is reductive, this is, if

$rad(L)=Z(L)$,

where $Z(L)$ is the center, then $S$ is an ideal.

Is the converse true? Or is it possible to exhibit a non reductive Lie algebra that admits a Levi decomposition such that $S$ is an ideal?

Remarks: $L$ is finite dimensional and the underlying field is algebraically closed of characteristic $0$ . The direct sum sign means direct sum of vector subspaces.

Let $L$ be a Lie algebra. It is known that $L$ admits a Levi decomposition (possibly non unique):

$L=S\oplus rad(L) $,

where $rad(L)$ is the solvable radical and $S$ is a semisimple subalgebra.

If $L$ is reductive, this is, if

$rad(L)=Z(L)$,

where $Z(L)$ is the center, then $S$ is an ideal.

Is the converse true? Or is it possible to exhibit a non reductive Lie algebra that admits a Levi decomposition such that $S$ is an ideal?

(edited) Is there any method to determine whether is possible to have $S$ as an ideal or not?

Remarks: $L$ is finite dimensional and the underlying field is algebraically closed of characteristic $0$ . The direct sum sign means direct sum of vector subspaces.

Let $L$ be a Lie algebra. It is known that $L$ admits a Levi decomposition (possibly non unique):

$L=S\oplus rad(L) $,

where $rad(L)$ is the solvable radical and $S$ is a semisimple subalgebra.

If $L$ is reductive, this is, if

$rad(L)=Z(L)$,

where $Z(L)$ is the center, then $S$ is an ideal.

Is the converse true? Or is it possible to exhibit a non reductive Lie algebra that admits a Levi decomposition such that $S$ is an ideal?

Remark: $L$ is finite dimensional and the underlying field is algebraically closed of characteristic $0$ .

Let $L$ be a Lie algebra. It is known that $L$ admits a Levi decomposition (possibly non unique):

$L=S\oplus rad(L) $,

where $rad(L)$ is the solvable radical and $S$ is a semisimple subalgebra.

If $L$ is reductive, this is, if

$rad(L)=Z(L)$,

where $Z(L)$ is the center, then $S$ is an ideal.

Is the converse true? Or is it possible to exhibit a non reductive Lie algebra that admits a Levi decomposition such that $S$ is an ideal?

Remarks: $L$ is finite dimensional and the underlying field is algebraically closed of characteristic $0$ . The direct sum sign means direct sum of vector subspaces.

Let $L$ be a Lie algebra. It is known that $L$ admits a Levi decomposition (possibly non unique):

$L=S\oplus rad(L) $,

where $rad(L)$ is the solvable radical and $S$ is a semisimple subalgebra.

If $L$ is reductive, this is, if

$rad(L)=Z(L)$,

where $Z(L)$ is the center, then $S$ is an ideal.

Is the converse true? Or is it possible to exhibit a non reductive Lie algebra that admits a Levi decomposition such that $S$ is an ideal?

Remark: L is finite dimensional and the underlying field is algebraically closed of characteristic $0$ .

Let $L$ be a Lie algebra. It is known that $L$ admits a Levi decomposition (possibly non unique):

$L=S\oplus rad(L) $,

where $rad(L)$ is the solvable radical and $S$ is a semisimple subalgebra.

If $L$ is reductive, this is, if

$rad(L)=Z(L)$,

where $Z(L)$ is the center, then $S$ is an ideal.

Is the converse true? Or is it possible to exhibit a non reductive Lie algebra that admits a Levi decomposition such that $S$ is an ideal?

Remark: $L$ is finite dimensional and the underlying field is algebraically closed of characteristic $0$ .