Lie Algebras

2 Semisimple Lie Algebras

Given a representation \(V\) of a Lie Algebra \(\mg \), the trace form is \((a,b)_V = \tr _V ab\). One easily checks it is symmetric \((a,b) = (b,a)\) and invariant \(([a,b],c) = (a,[b,c])\). The trace form on the adjoint representation is called the Killing form, denoted \(\kappa \).

Lemma 2.1. If \(\ma \) is an ideal of \(\mg \), then \(\kappa _{\mg }|_{\ma } =\kappa _{\ma }\).

Proof. \(\ad b\ad a, a\in \ma \) has image in \(\ma \), so its trace is equal to its trace on \(\ma \). □

Weights are related to roots by rational multiples on certain parts of the Cartan subalgebra.

Lemma 2.2 (Cartan’s Lemma). Let \(e \in \mg _a, f \in \mg _{-a}, [e,f]=h\), and suppose that \(V_{\lambda } \neq 0\). Then \(\lambda (h) = r \alpha (h)\), where \(r \in \QQ \) doesn’t depend on \(h\).

Proof. Consider \(W=\oplus _n V_{\lambda +n\alpha }\), upon which \(e,f,h\) act. \(h\)’s trace is \(0\) on this as it is a commutator. The trace is \(\sum _n (\lambda +n\alpha )(h)\dim V_{\lambda +n\alpha }=0\), giving the relation. □

Theorem 2.3 (Cartan’s Criterion). For a subalgebra of \(\gl _V\), the following are equivalent:
- 1. \((\mg ,[\mg ,\mg ])_V= 0\)
- 2. \(([\mg ,\mg ],[\mg ,\mg ])_V = 0\)
- 3. \(\mg \) is solvable.

Proof. \((1)\implies (2)\) is clear. \((3) \implies (1)\) follows from Lie’s theorem, since the product of two elements of that form is strictly uppertriangular.

\((2) \implies (3)\): If not, then there is a point in the derived series giving a subalgebra \(\mg '\) such that \([\mg ',\mg ']= \mg '\). If \(\mh \) is its Cartan subalgebra, then \(\mh = \sum _i [\mg '_{\alpha },\mg '_{-\alpha }]\). Choose an \(h = [e,f]\), and observe \(0 = (h,h)_V = \tr h^2 = \sum _{\lambda } \lambda (h)^2 \dim V_{\lambda }\). By the previous lemma, this is a nonnegative rational multiple of \(\alpha (h)\), so we must have \(\lambda (h)=0\) and \(V = V_0\). But then everything acts nilpotently, so by Engel’s theorem \(\mg '\) is nilpotent, a contradiction. □

Corollary 2.4. \(\mg \) is solvable iff \(\kappa (\mg ,[\mg ,\mg ]) =0\).

Definition 2.5. The radical of \(\mg \), \(\rad (\mg )\) is the largest solvable ideal.

Definition 2.6. \(\mg \) is semisimple if \(\rad (\mg ) = 0\).

Equivalently, there are no nontrivial abelian ideals. Note that a simple Lie algebra is clearly semisimple.

Corollary 2.7. If \(\mg \) is semisimple the trace form of any or every faithful representation \(V\) is nondegenerate.

Proof. By Cartan’s criterion, the nullspace of \((a,b)_V\) is a solvable ideal, so is \(0\). □

Corollary 2.8. \(\mg \) is semisimple iff its Killing form is nondegenerate.

Proof. If \(\mg \) is not semisimple, it has a nontrivial abelian ideal, which is in the kernel of \(\kappa \). □

Proposition 2.9. The Killing form of a semisimple Lie algebra \(\mg \) is nondegenerate when restricted to any ideal \(\ma \). A Lie algebra splits as a sum of simple Lie algebras.

Proof. \(\ma \cap \ma ^{\perp }\) is a solvable ideal by Cartan’s criterion, so is \(0\). Thus \(\mg = \ma \oplus \ma ^{\perp }\) as a Lie algebra. Iterating gives the last assertion. □

Example 2.10. \(\gl _n\) is \(k I \oplus \msl _n\) where \(\msl _n\) is semisimple. To see this, let \(E_{ij}\) be the matrix \((E_{ij})_{kl} = \delta _{ik}\delta _{jl}\). We can compute \(\ad (E_{ij})E_{kl} = E_{il}\delta _{jk} -E_{kj}\delta _{il}\). Thus \(\ad (E_{mn})\ad (E_{ij})E_{kl} = E_{ml}\delta _{jk}\delta _{ni} - E_{mj} \delta _{il}\delta _{nk} - E_{in} \delta _{jk}\delta _{lm} + E_{kn}\delta _{il}\delta _{jm}\), so taking trace, we get \(2n\delta _{ni}\delta _{mj} - 2\delta _{mn}\delta _{ij}\). By linearity, the Killing form is \((a,b) = 2n\tr ab-2\tr a\tr b\). From this we can see that it is nondegenerate on \(\msl _n\), and null on \(k I\).

Theorem 2.11. Given a faithful irreducible representation of \(\mg \) into \(V\) either \(\mg \) is semisimple, or it is \(kI\) summed with something semisimple.

Proof. If it isn’t semisimple, by Lie’s theorem, the radical has an nonzero weight space, which is invariant by Lie’s lemma. By irreducibility, it must be all of \(V\). Since the representation is faithful, the radical must be \(kI\). □

Definition 2.12. A Lie algebra is reductive if its adjoint representation is completely reducible.

Lemma 2.13. A Lie algebra is reductive iff it is a sum of a semisimple Lie algebra and an abelian one.

Proof. The irreducible summands of a reductive Lie algebra are ideals, so the Lie algebra splits into a sum of Lie algebras where the adjoint representation is both completely reducible and irreducible. If the center of such an algebra is trivial, by the above theorem, it is semisimple (in fact simple). Otherwise, since the center is a summand of the adjoint representation, it is everything, and the algebra is abelian. The other direction has already been proven. □

We will now see that semisimple Lie algebras are the ones having a semisimple representation theory. To see it is a necessary condition, observe that the adjoint representation must be irreducible so it is reductive, and it can’t have an abelian part since abelian Lie algebras aren’t semisimple.

Recall that representations of \(\mg \) are the same as representations of \(U(\mg )\). \(U(\mg )\) is a filtered cocommutative Hopf algebra, where the coproduct is \(x \mapsto 1 \otimes x +x\otimes 1\), and the antipode is \(x \mapsto -x\) on \(\mg \subset U(\mg )\). It is the algebra of differential operators on the associated formal group. \(U\) of a free Lie algebra is a tensor algebra, \(U(\mg \oplus \mh ) = U(\mg ) \otimes _k U(\mh )\), and \(U(k) = k[x]\).

The argument that representations are semisimple is similar to the argument for compact Lie group. There, via averaging over a bi-invariant metric, one makes any representation unitary, after which one can produce orthogonal complements to subrepresentations. Here the bi-invariant metric becomes the Killing form, and it hard to average, but we still have an infinitesmal version of averaging, namely the Laplacian operator, which here is called the Casimir element.

Any symmetric invariant form \(B\) on \(\mg \) can be thought of as a map of \(\mg \)-modules \(\mg \otimes \mg \to k\). Nondegeneracy means the adjoint \(\mg \to \mg ^*\) is an isomorphism. Via this identification, the identity \(\mg \to \mg \) is adjoint to a map \(k \to \mg \otimes \mg \), the image of \(1\) being the Casimir element \(L_B\). Alternatively it is adjoint to the map \(B\). Note that symmetry of the bilinear form implies that \(L_B\) is invariant under swapping the two terms. We can also view \(L\) in \(U(\mg )\) via the multiplication map \(\mg \otimes \mg \to U(\mg )\). It is central in \(U(\mg )\) since the map \(k \to \mg \otimes \mg \) is a map of \(\mg \)-modules.

Lemma 2.14. The category of f.d representations of \(\mg \) is semisimple iff \(\Ext ^1(k,V)=0\) for all irreducible \(V\), where \(k\) is the tensor unit.

Proof. We can identify \(\Hom _{\mg }(V,W)\) with \(\Hom _{\mg }(k,W\otimes V^*)\). \((-)\otimes V^*\) is exact and preserves injectives, so we can derive the isomorphism to \(\Ext \). Semisimplicity is equivalent to \(\Ext ^1(V,W)=0\). So we only need \(\Ext ^1(k,V) = 0\) for all \(V\), but the long exact sequence on \(\Ext \) shows it suffices to prove it for irreducible \(V\). □

Theorem 2.15 (Weyl Complete Reducability). For a semisimple \(\mg \), any finite-dimensional representation of \(U(\mg )\) is semisimple.

Proof. We need to show that a short exact sequence \(0 \to V \to W \to k \to 0\) splits when \(V\) is irreducible. WLOG \(V\) is faithful by passing to a quotient of \(\mg \). \(L_V\) is \(0\) on \(k\), and \(\tr _{L_V}(V) = \dim \mg \neq 0\). So \(L_V\) is nonzero, and thus an isomorphism by Schur’s Lemma. An element in the kernel of \(L_W\) gives a splitting. □

A remark about the proof: we should expect \(k\) to be the only irreducible representation for which \(L\) acts as \(0\) since every irrep embeds in \(L^2\), where \(L\) is literally the Laplacian, and the only harmonic functions are constant.

This has an interpretation in terms of derivations. A derivation \(d:\mg \to M\) is a map such that \(d[a,b] = adb-bda\) It is the same as a \(1\)-cocycle in the Chevalley-Eilenberg complex (i.e the de Rham complex) of \(M\). A \(1\)-boundary is an inner derivation or a map of the form \(dx = xa, a \in M\). We can write these groups as \(\Der (\mg ,M)\) and \(\Ider (\mg ,M)\).

Corollary 2.16. Every derivation of a semisimple Lie into a finite-dimensional module is inner.

A special case of this is the Lie algebra of derivations of \(\mg \) into \(\mg \). Here is an alternate proof of the fact in that case:

Proposition 2.17. If \(\mg \) is semisimple, the map \(\ad : \mg \to \Der (\mg )\) is an isomorphism.

Proof. The restriction of the tautological representation \(\Der (\mg )\) to \(\mg \) is the adjoint representation. Since \(\kappa \) is nondegenerate, \(\Der (\mg )\) splits as \(\ad \mg \oplus \ad \mg ^{\perp }\). If \(D \in \ad \mg ^{\perp }\) then \(0 = [D,\ad a] = \ad Da\), so \(Da \in Z(\mg ) =0\) showing \(D = 0\). □

Definition 2.18. An abstract Jordan decomposition of \(a\) is a decomposition \(a=a_s+a_n\) where \([a_s,a_n]=0\) and \(\ad a_s,\ad a_n\) are semisimple and nilpotent.

They are unique iff \(Z(\mg )=0\), because that is the kernel of map \(\ad \), and we know Jordan decompositions to be unique in \(\gl _V\).

Lemma 2.19. Abstract Jordan decompositions exist for reductive Lie algebras.

Proof. It suffices to assume \(\mg \) semisimple. \(A=\ad g\) has a Jordan decomposition as an endomorphism, so it suffices to show that \(A_s\) is a derivation. By linearity it suffices to do check this for elements in a root space \(x,y\), with eigenvalues \(\lambda ,\mu \) for \(A\). Indeed, \(A_s[x,y] = (\mu +\lambda )[x,y] =[\lambda x,y] + [x,\mu y] = [A_sx,y] + [x,A_sy]\). □

Now we can better understand the Killing forms of a semisimple Lie algebra.

Theorem 2.20. Let \(\kappa \) be the killing form on \(\mg \), a semisimple Lie algebra, and let \(\mh \) be a Cartan subalgebra.
- 1. \(\kappa (\mg _{\alpha },\mg _{\beta })=0\) unless \(\alpha +\beta =0\).
- 2. \(\kappa _{\mg _{\alpha }+ \mg _{-\alpha }}\) is nondegenerate.
- 3. \(\mh \) is abelian
- 4. \(\mh \) consists of semisimple elements

Proof. \((1)\): This is immediate from Theorem 1.15 by looking at how \(\ad \) acts on the root spaces.

\((2)\): This follows from \((1)\) and the fact that \(\kappa \) is nondegenerate.

\((3)\): Cartan’s criterion shows that \(\kappa (\mh ,[\mh ,\mh ])= 0\), but along with \((2)\) on \(\mg _0=\mh \) this means \([\mh ,\mh ]=0\).

\((4)\): \(\mh _s \subset \mh \) because of how \(\ad \mh _s\) behaves, but note that by \((3)\), \(\kappa (h,h') = \kappa (h_s,h')\) for \(h,h' \in \mh \). Thus by \((2)\) on \(\mg _0=\mh \), \(h = h_s\). □