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If the radical of a Lie algebra is zero, we call it semi-simple. In the lecture notes that I'm following its stated that for any arbitrary Lie algebra (over a field with characteristic zero and finite dimensional) $\mathfrak{g}$ we have, $\mathfrak{h}:=\mathfrak{g}/\mathrm{rad}(\mathfrak{g})$ semi-simple. I have trouble seeing why.

I need to show that $\mathrm{rad}(\mathfrak{h})=0$. I think we can use the fact that if $\mathfrak{I}$ is a solvable ideal of $\mathfrak{g}$ such that $\mathfrak{g}/\mathfrak{I}$ is solvable, then $\mathfrak{g}$ is solvable, but I'm not really sure how...

Marius Jaeger
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  • Are there any conditions on your Lie algebra? For example, over a field of characteristic zero? Finite dimensional? – Matt Feb 01 '20 at 14:14
  • @Matt Yes, I forgot to mention that its over a field with characteristic zero, finite dimensional! – Marius Jaeger Feb 01 '20 at 14:16
  • So that you know, in this (rather nice) case, you actually have that this quotient $\mathfrak{h}$ is isomorphic to an actual subalgebra of $\mathfrak{g}$. This follows from something called Levi's theorem. This is a useful theorem, because if you think about it, there is no reason to think that $\mathfrak{h}$, as you've defined it, should be a subalgebra of $\mathfrak{g}$ at all! – Matt Feb 01 '20 at 14:21
  • @Matt if semisimple is defined as "trivial radical", this works over arbitrary fields (and finite-dim Lie algebras). There's no need to use Levi factors. – YCor Feb 01 '20 at 22:35

2 Answers2

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$\mathfrak{g}$ is semisimple if it has no non-zero abelian ideals.

Let $\pi: \mathfrak{g} \rightarrow \mathfrak{h}$ be the quotient map.

if $\mathfrak{h}$ is not semisimple, then $(0)\subsetneq rad(\mathfrak{h})$

therefore $\pi^{-1}(0) = rad(\mathfrak{g}) \subsetneq \pi^{-1}(rad(\mathfrak{h}))$

But $\pi^{-1}(rad(\mathfrak{h}))$ is an abelian ideal, this is a contradiction of the maximality of $rad(\mathfrak{g}).$

Alessandro
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HINT:

Suppose for contradiction that $\mathfrak{h}$ is not semi-simple. Write down exactly what this means.

You can now utilize a particular fact, and show that $\mathfrak{g}$ would have to therefore have a non-zero solvable ideal. This gives a contradiction.

Matt
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