Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 3 / 1

The derived series of a Lie algebra is defined recursively by ${\mathfrak{g}}^{(0)}=\mathfrak{g}$ and ${\mathfrak{g}}^{(r+1)}=[{\mathfrak{g}}^{(r)},{\mathfrak{g}}^{(r)}]$, where the bracket denotes the linear span of all brackets of pairs of elements. A solvable Lie algebra is one whose derived series of a Lie algebra terminates at zero:The solvable radical $\mathrm{rad}\mathfrak{g}$ of a finite-dimensional Lie algebra is its largest solvable ideal of a Lie algebra. A semisimple Lie algebra has no nonzero solvable ideals of a Lie algebra, equivalentlyThese are definitions; identifying semisimplicity with a direct sum of simple ideals additionally uses the structure theorem in characteristic zero. The calculations and representation theorems below are over $\mathbb{C}$, as in the complex representation setting of this paper.

Let $L_A$ and $R_A$ be left and right multiplication by $A$ on the vector space $M_n(\mathbb{C})$. Then $\mathrm{ad}A=L_A-R_A$. On matrix units, or from $M_n(\mathbb{C})={\mathbb{C}}^n\otimes ({\mathbb{C}}^n{)}^{\ast }$, one obtainsFor example, the coefficient of $E_{ij}$ in $A{E}_{ij}B$ is $A_{ii}{B}_{jj}$; summing over $i,j$ proves the mixed-trace identity. Expanding the product of the two adjoint operators therefore gives the Killing form of the general linear Lie algebra:For $A,B$ in the special linear Lie algebra, both individual traces vanish. Furthermore $M_n(\mathbb{C})={\mathfrak{sl}}_n\oplus \mathbb{C}I$, and every adjoint operator kills the scalar summand, so its product has the same trace on $M_n$ and on ${\mathfrak{sl}}_n$. Consequently the Killing form of the special linear Lie algebra isNo uniqueness theorem for invariant bilinear forms on a Lie algebra is needed: this calculation establishes the constant directly. For the degenerate case $n=1$, the Lie algebra is zero, both forms vanish, and proportionality does not determine a unique constant.

For this diamond Lie algebra, every bracket lies in $⟨c,p,q⟩$, and each of these three vectors occurs as a bracket. Thus its derived algebra is exactly ${\mathfrak{g}}^{(1)}=⟨c,p,q⟩$. Inside that subalgebra the only nonzero bracket is a multiple of $[p,q]=c$, so ${\mathfrak{g}}^{(2)}=\mathbb{C}c$. Centrality of $c$ then gives ${\mathfrak{g}}^{(3)}=0$. HenceIn particular it is a solvable Lie algebra, although its lower central series of a Lie algebra does not terminate: $[\mathfrak{g},⟨c,p,q⟩]=⟨c,p,q⟩$. Solvability does not imply nilpotence.

Define a symmetric bilinear form $B$ by pairing $c$ with $d$ and $p$ with $q$, each with value one, and setting all other pairings of basis vectors to zero. Its Gram matrix in the order $(c,d,p,q)$ isThe determinant is one, so this is a nondegenerate bilinear form.

For invariance, put $T(x,y,z)=B([x,y],z)$. The nonzero values on distinct basis triples are generated byand the reversed triples have value minus one. Every value involving $c$, or a repeated basis vector, vanishes. Thus $T$ is an alternating trilinear form, explicitly $d^{\ast }\wedge p^{\ast }\wedge q^{\ast }$. In particular $T(x,y,z)=T(y,z,x)$. By symmetry of $B$, this reads $B([x,y],z)=B(x,[y,z])$, the defining identity for an invariant bilinear form on a Lie algebra.

Suppose a finite-dimensional Lie algebra representation $\rho$ realized this form as $B(x,y)=\mathrm{tr}(\rho (x)\rho (y))$. The Lie theorem says that every finite-dimensional representation of a complex solvable Lie algebra admits a basis in which all representing matrices are upper triangular. A commutator of upper triangular matrices has zero diagonal, hence is strictly upper triangular. Therefore $\rho ([\mathfrak{g},\mathfrak{g}])$ is strictly upper triangular, and its product with any representing matrix still has zero diagonal. It follows thatThis is the general fact that trace forms of solvable Lie algebras annihilate the derived algebra. Here $c\in [\mathfrak{g},\mathfrak{g}]$, so such a trace form would have $B(c,d)=0$, contradicting the constructed value one. Thus the nondegenerate invariant form exists, but no finite-dimensional representation has it as its trace form. Finite dimensionality is implicit in the ordinary representation trace; no trace on arbitrary infinite-dimensional endomorphisms is being assumed.