The structure theory of Lie algebras parallels finite group theory:
Definition 17.16.1 (Solvable and semisimple Lie algebras). The derived series of a Lie algebra is ,. The algebra is solvable if for some , and semisimple if it has no nonzero solvable ideals. Every Lie algebra decomposes as where is the radical (maximal solvable ideal) and is semisimple (Levi decomposition).
Note. Every semisimple Lie algebra decomposes (uniquely) as a direct sum of simple Lie algebras (those with no nontrivial ideals). The simple Lie algebras over are completely classified by Dynkin diagrams:
The classification proceeds via root systems: given a semisimple and a Cartan subalgebra (maximal abelian subalgebra of semisimple elements), the adjoint action of on decomposes into root spaces. The resulting set of roots is a root system, which is encoded by its Dynkin diagram. This is one of the great classification theorems in mathematics (Killing 1888, Cartan 1894).
Example 17.16.1 ( — the simplest semisimple Lie algebra). The Lie algebra has dimension 3 with standard basis satisfying ,,. This is the type Lie algebra. Its finite-dimensional representations are completely classified: for each there is a unique irreducible representation of dimension , realized on . The representation theory of is the foundation for the general theory.
Exercise 17.16.1 (Exercises).
Show that is not a PID by finding an ideal that is not principal. (Hint: consider .)
Prove that is a Euclidean domain with norm . (The key step: for any with , the quotient lies within distance of some Gaussian integer.)
Compute and draw its lattice of subgroups alongside the lattice of intermediate fields.
Show that the Galois group of over is the dihedral group of order 8 (not ).
Verify the Lie bracket relations ,, for the standard basis of .
Show that (real skew-symmetric matrices) is isomorphic as a Lie algebra to where is the cross product.
Let . Show that the Killing form equals .
Show solution
(1) The ideal is not principal. If , then and , so and . Thus . But no element of has norm 2 (since has no integer solutions). So , meaning is a unit and . But (checking: for leads to a contradiction modulo 2), so is proper and not principal.
(2) Given with , write with . Choose integers closest to , so and . Set and . Then .
(3) The extension has degree 4 (since is irreducible over ). The splitting field of is normal and separable, hence Galois. The Galois group has order 4 and is generated by and . Since and , we get . The three subgroups of order 2 correspond to the three intermediate fields ,,.
(4) The splitting field of over is . We have ( and ). The Galois group has order 8 and is determined by the actions on and . Let , (a rotation of the four roots) and , (complex conjugation). Then ,, and , giving . This is a proper subgroup of (order 8 vs 24).
(5) Direct computation: . Similarly . And .
(6) A basis for is ,,. Computing: ,,. Under the identification (standard basis of ), these bracket relations are exactly , etc. So is a Lie algebra isomorphism .
(7) For ,. So . Taking the trace over (summing over the basis with ) requires careful index manipulation. The result follows from the identities and the combinatorial sums. In particular, restricted to (where ) is just , which is nondegenerate—confirming is semisimple (Cartan’s criterion).