Do a complex Lie algebra and its realification have the same radical?

Question

Let $\mathfrak{g}$ be a finite-dimensional Lie algebra. The radical of a Lie algebra $\mathfrak{g}$, denoted as $\mathrm{rad}(\mathfrak{g})$, is the unique maximal solvable ideal of $\mathfrak{g}$.

Let $\mathfrak{g}$ be a finite-dimensional complex Lie algebra and $\mathfrak{g}_{\mathbb{R}}$ its realification, which means that $\mathfrak{g}_{\mathbb{R}}$ is $\mathfrak{g}$ considered as a real Lie algebra. Of course we have $\mathrm{rad}(\mathfrak{g})\subset \mathrm{rad}(\mathfrak{g}_{\mathbb{R}})$ since solvability does not rely on the ground field. But can this inclusion be strict?

Thank you in advance for any help.

(Note: From this question we know that for finite-dimensional complex Lie algebra $\mathfrak{g}$, if $\mathrm{rad}(\mathfrak{g}) = \{0\}$, then $\mathrm{rad}(\mathfrak{g}_{\mathbb{R}}) = \{0\}$.)

Answer 1

$\DeclareMathOperator\g{\mathfrak{g}}\DeclareMathOperator\h{\mathfrak{h}}\DeclareMathOperator\rad{rad}$Let $K\subset L$ be fields. Let $\g$ be a finite-dimensional Lie algebra over $L$. I claim that $\rad(\g)=\rad(\g|_K)$. The latter means the sum of all solvable ideals ($\g$ might be infinite-dimensional over $K$).

As already mentioned, $\rad(\g)\subset\rad(\g|_K)$ is clear (since $\rad(\g)$ is an ideal of $\g|_K$). Assuming by contradiction the inclusion proper, there exists a solvable ideal of $\g|_K$ not contained in $\rad(\g)$. Projecting, we deduce that the Lie $K$-algebra $\g/\rad(\g)$ contains a nonzero solvable ideal. Hence passing to an element in its derived series, we deduce that the Lie $K$-algebra $\g/\rad(\g)$ contains a nonzero abelian ideal. By the next lemma, $\g/\rad(\g)$ also contains a nonzero abelian ideal (as $L$-algebra).

Lemma: Let $\h$ be a Lie algebra over $L$ and $V$ an abelian ideal of $\h|_K$. Then the $L$-span $W$ of $\h$ is an abelian ideal of $\h$.

Proof: we have $[\sum_i t_iv_i,\sum_i u_jv'_j]=\sum_{i,j}t_iu_i[v_i,v'_j]=0$, so $W$ is abelian. For $h\in\h$ we have $[h,\sum_i t_iv_i]=\sum_i t_i[h,v_i]$ so $W$ is an ideal.

Answer 2

I'm trying to provide an answer, which is based on the claim that $\mathfrak{g}$ is semisimple if and only if $\mathfrak{g}_{\mathbb{R}}$ is semisimple here if $\mathfrak{g}$ has finite dimension. (I'm not sure that this result is correct when $\mathfrak{g}$ has infinite dimension).

Lemma. Let $\mathfrak{g}$ be a finite-dimensional Lie algebra over a field for characteristic $0$ and $\mathrm{rad}(\mathfrak{g})$ its radical. Then for an ideal $\mathfrak{h}\subset \mathfrak{g}$, $\mathfrak{g}/\mathfrak{h}$ is semisimple if and only if $\mathrm{rad}(\mathfrak{g})\subset \mathfrak{h}$.

$\Leftarrow$: $\mathfrak{g}/\mathfrak{h} \cong (\mathfrak{g}/\mathrm{rad}(\mathfrak{g}))/(\mathfrak{h}/\mathrm{rad}(\mathfrak{g}))$ is a quotient of $\mathfrak{g}/\mathrm{rad}(\mathfrak{g})$, hence being semisimple.

$\Rightarrow$: $\mathrm{rad}(\mathfrak{g})/(\mathrm{rad}(\mathfrak{g})\cap \mathfrak{h})\cong (\mathrm{rad}(\mathfrak{g}) + \mathfrak{h})/\mathfrak{h}$ is an ideal of $\mathfrak{g}/\mathfrak{h}$, hence being semisimple. But it's also solvable, being quotient of the solvable algebra $\mathrm{rad}(\mathfrak{g})$. So $\mathrm{rad}(\mathfrak{g})/(\mathrm{rad}(\mathfrak{g})\cap \mathfrak{h}) = \{0\}$, $\mathrm{rad}(\mathfrak{g})\subset \mathfrak{h}$.

(I doubt that this proof works only for $\mathfrak{g}$ being over a field for characteristic $0$. There is also another approach also works only for such $\mathfrak{g}$: from this question, $\mathrm{rad}(\mathfrak{h})$ is an ideal of $\mathrm{rad}(\mathfrak{g})$ and $\mathrm{rad}(\mathfrak{g}/\mathfrak{h})\cong \mathrm{rad}(\mathfrak{g})/\mathrm{rad}(\mathfrak{h})$, so $\mathfrak{g}/\mathfrak{h}$ semisimple $\Leftrightarrow \mathrm{rad}(\mathfrak{h}) = \mathrm{rad}(\mathfrak{g}) \Leftrightarrow \mathrm{rad}(\mathfrak{g})\subset \mathfrak{h}$.)

Now for the original question, we have $\mathfrak{g}/\mathrm{rad}(\mathfrak{g})$ being semisimple, so $\mathfrak{g}_{\mathbb{R}}/\mathrm{rad}(\mathfrak{g})_{\mathbb{R}} = (\mathfrak{g}/\mathrm{rad}(\mathfrak{g}))_{\mathbb{R}}$ is semisimple. From the lemma above we know that $\mathrm{rad}(\mathfrak{g}_\mathbb{R})\subset \mathrm{rad}(\mathfrak{g})_{\mathbb{R}}$. Hence these two are equal.