Suppose we have no non-trivial solvable ideals, and as mathfrak{g} is semisimple, it is not solvable, thus the maximum solvable ideal is {0}, hence Rad(mathfrak{g})=0. 4 Rightarrow 3 Suppose we have a Lie algebra with solvable ideals, so there exists (mathfrak{i} subset mathfrak{g}) such that [mathfrak{i}supset mathfrak{i}^{(1)} supset cdots supset mathfrak{i}^{(n)}=0] then mathfrak{i}^{(n-1)} must be a one-dimensional subspace, and so must be Abelian, so Abelian Ideals imply solvability. end{proof} Now we have defined semisimplicity for...
Suppose we have no non-trivial solvable ideals, and as mathfrak{g} is semisimple, it is not solvable, thus the maximum solvable ideal is {0}, hence Rad(mathfrak{g})=0.
4 Rightarrow 3 Suppose we have a Lie algebra with solvable ideals, so there exists (mathfrak{i} subset mathfrak{g}) such that
[mathfrak{i}supset mathfrak{i}^{(1)} supset cdots supset mathfrak{i}^{(n)}=0]
then mathfrak{i}^{(n-1)} must be a one-dimensional subspace, and so must be Abelian, so Abelian Ideals imply solvability. end{proof}
Now we have defined semisimplicity for Lie Algebras, we can show an interesting fact for mathfrak{sl}_n. The problem with dealing in Killing forms is that they are often difficult to calculate. It is a useful fact that any two symmetric bilinear forms on semisimple lie algebras are equivalent up to a scalar.
The map (x,y) rightarrow Tr(x,y) is an invariant, symmetric bilinear form on mathfrak{sl}_n
We can use this fact to consider the Killing form on mathfrak{sl}_n, so consider a special subalgebra (that will re-appear later) on mathfrak{sl}_n, namely the diagonal matrices which are traceless.
