Exercise 125 Show that for the Lie algebra gln(k) with eithe matriix withjust one non-zero entry, a 1 in position i on the diagonal, we have (ei, ej) =2n − 2 if i = j and −2 if i 6= j. Deduce that the Killing form on sln(k) is 2ntimes the symmetric bilinear form associated to the standard representation,but the Killing form on gln(R) is not a multiple of the form of the standardrepresentation and has a non-trivial kernel.Exercise 126 If k is a field of characteristic 2 then the semidirect productgl2(k).k2is solvable. The Killing form is not identically zero on its derivedalgebra sl2(k).k2.Exercise 127 Let G be the Lie algebra slp(Fp) of 2 by 2 matrices over the fieldof p elements. Show that G is simple if p > 2. Show that the Killing form ofG is identically 0. Show that Trace(AB) is a non-degenerate invariant bilinearform on G.Exercise 128 Suppose G is the Lie algebra over a field of characteristic p > 0with basis aifor i ∈ Z/pZ and bracket [ai, aj] = (i − j)ai+j. Show that G is asimple Lie algebra but has no non-zero invariant bilinear form.Lemma 129 (Dieudonne) Suppose that a finite dimensional Lie algebra overany field of any characteristic has a non-degenerate bilinear form, and no abelianideals. Then it is a direct sum of simple subalgebras.Proof Fix a minimal ideal M. The derived ideal [M, M] is contained in M andcannot be 0 as M is non-abelian, so M = [M, M] is perfect. The orthogonalcomplement N of M is also an ideal as the bilinear form is invariant. It cannotcontain M as otherwise we would have (x, m) = (x,P[ai, bi]) =P([x, ai], bi) =0 so M would be in the kernel of (, ) which is not possible. So N ∩ M = 0 asit is a proper ideal of the minimal ideal M . So G splits as the direct sum of Mand N , so M is simple, and continuing by induction so is N. Exercise 130 Show that if L is a Lie algebra with an invariant symmetricbilinear form (, ) then L[t]/(tn) has an invariant symmetric bilinear form givenby the coefficient of tn−1of the bilinear form on L[t]/(tn) with values that aretruncated power series. If the form on L is non-degenerate show that the formon L[t]/(tn) is also non-degenerate. Find an example of a finite-dimensionalcomplex Lie algebra with a non-degenerate symmetric bilinear form that is nota sum of abelian and simple Lie algebras.Exercise 131 What is wrong with the following “proof” of the false result thata finite-dimensional complex Lie algebra L with a non-degenerate symmetricbilinear form is a sum of abelian and simple Lie algebras: take any ideal of L,and write L as the sum of the ideal and its orthogonal complement (which isalso an ideal). By repeating this we can write L as a sum of ideals with noproper subideals, so L is a sum of abelian and simple Lie algebras.Corollary 132 (Cartan’s criterion for semisimplicity) For a complex Lie alge-bra G the following conditions are equivalent:501. G has no nonzero solvable ideals2. G has no non-zero abelian ideals3. G has non-degenerate Killing form4. G is a direct sum of simple Lie algebras (in other words G is semisimple)Proof If the Killing form is degenerate, then its kernel is an ideal, and issolvable by one form of Cartan’s criterion. So if the algebra has no nonzerosolvable ideals then the Killing form is non-degenerate. Conversely if the Killingform is nondegenerate and A is an abelian ideal, then for any a ∈ A and g ∈ G,Ad(a)Ad(g) has square 0 so has trace 0 and therefore a = 0 as the Killing formis non-degenerate. So all Abelian ideals are 0,and therefore all solvable idealsare 0.We have seen that if the Killing form is non-degenerate then there are noabelian ideals, so if the Killing form is non-degenerate then by the previouslemma the Lie algebra is a sum of simple Lie algebras.If the algebra is a sum of simple subalgebras, it is obvious that it has nononzero solvable ideals. 11 Cartan subalgebras, Cartan subgroups andmaximal toriThe Lie algebra glnhas a subalgebra H of diagonal matrices, and under the ac-tion of this subalgebra glnsplits as the sum of eigenspaces. The zero eigenspaceis just H, while the other eigenspaces just correspond to the off-diagonal entriesof gln. The subalgebra H is an example of a Cartan subalgebra, and we wantto find a similar subalgebra for any Lie algebra. The first guess is to take amaximal abelian subalgebra, but this does not work: the algebra of matricesof block form0 A0 0is abelian but does not act nicely on the rest of the Liealgebra (and has dimension much larger than that of the diagonal matrices).Definition 133 A toral subalgebra of a Lie algebra is an abelian subalgebra thatacts semisimply on the adjoint representation.Definition 134 A Cartan of a Lie algebra is a self-centralizing nilpotent sub-algebra. (Self-centralizing means that it contains its centralizer.)For semisimple Lie algebras, maximal toral subalgebras and Cartan subalgebraswill turn out to be the same. In general it is really the maximal toral subalgebrasthat are important. It seems to be a historical accident that Cartan subalgebrashave this rather unintuitive definition. The properties of being nilpotent or selfnormalizing are not really that important or easy to use. The really impor-tant property is the semisimplicity, which means that one can decomp ose the(complex) Lie algebra into eigenspaces.51Exercise 135 Find maximal toral subalgebras for the algebra of all matrices,the algebra of upper triangular matrices, and the algebra of strictly upper tri-angular matrices (0’s on the diagonal).Theorem 136 The centralizer of a maximal toral subalgebra of G is a Cartansubalgebra, in other words it is self normalizing and