Why is the Levi-decomposition not available in positive characteristic?

Question

I am learning about Lie-Algebras and I came across a proof of the Levi-decomposition. Can somebody tell why the Lie-Algebra has to be over a field of characteristic $0$? Why does it fail for positive characteristic?

Thank you!

Answer 1

The short answer is, that there are counterexamples in characteristic $p$, see for example this MO-post. Moreover, most results of classical Lie algebra theory fail in characteristic $p>0$, such as Lie's Theorem, Cartan's criterion for semisimplicity (non-degenerate Killing form), almost all results for representations of semisimple Lie algebras, the classification of semisimple Lie algebras and so on.

It is more difficult to give a sufficient answer, why. Of course, many arguments involving linear algebra fail for characteristic $p$, e.g., arguments using the trace. A summary can be found, for example, in this post:

If the field of a vector space weren't characteristic zero, then what would change in the theory?

Whitehead's lemma, saying that the $H^1(L,M)=0$ for semisimple Lie algebras of characteristic zero also fails in characteristic $p$.

In characteristic zero, the short exact sequence

$$ {\displaystyle 0\rightarrow \mathrm {rad} ({\mathfrak {g}})\rightarrow {\mathfrak {g}}\,{\xrightarrow {\varphi }}\,{\mathfrak {g}}/\mathrm {rad} ({\mathfrak {g}})\rightarrow 0} $$

splits, because $\mathfrak{g}/{\rm rad}(\mathfrak{g})$ is semisimple and Whitehead's Lemma. Levi's Theorem is a simple consequence of this fact. As said, this fact fails in characteristic $p>0$.