Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 4 / 4

The Lie bracket is bilinear, so $\mathrm{ad}x$ is a linear map. Rearranging the Jacobi identity givesThuswhich is the defining Leibniz rule for a derivation of a Lie algebra. In particular $\mathrm{ad}x$ is an inner derivation of a Lie algebra.

Write $D=\mathrm{ad}x$, and choose $N$ with $D^N=0$. The exponential of a nilpotent Lie algebra derivation is the finite linear operatorFor a derivation of a Lie algebra, induction on $k$ using the Leibniz rule and Pascal's identity provesExpanding $[Ey,Ez]$ and grouping by $k=j+l$ consequently givesTerms with an exponent at least $N$ vanish, which justifies extending each inner sum; and $D^k=0$ for $k\ge N$. Finally the finite product of the two exponentials has coefficientso $\exp (-D)$ is the inverse. We have proved bracket preservation and invertibility, hence $\exp (\mathrm{ad}x)$ is a Lie algebra automorphism.

Because $\theta$ is a Lie algebra automorphism, for every $y$,Conjugation therefore also proves that $\mathrm{ad}(\theta x)$ is a nilpotent operator, and sends every power of $\mathrm{ad}x$ to the corresponding power of $\mathrm{ad}(\theta x)$. Applying this to the defining finite exponential of a nilpotent Lie algebra derivation yields

Apply the Jacobi identity to an arbitrary $z$:Thus the Adjoint representation of a Lie algebra preserves the Lie bracket:

The root-space decomposition gives $[L_r,L_s]\subseteq L_{r+s}$ when $r+s\ne 0$. The two roots are independent, so $r+s\ne 0$, and the missing root sum means that this weight space is zero. Hence $[e_r,e_s]=0$. These adjoint root operators are nilpotent: successive brackets shift root weights by the same root, so finite root strings force eventual vanishing; on the Cartan subalgebra the first bracket lies in its root space and the next is zero. By part (iv), the nilpotent operators $A=t\mathrm{ad}{e}_r$ and $B=u\mathrm{ad}{e}_s$ commute. Every power of $A$ therefore commutes with every power of $B$, so their finite matrix exponentials commute too. Using the inverses proved in part (ii),

Let every root have squared length $L$. Since $r+s$ is a root, $L=\Vert r+s{\Vert }^2=2L+2(r,s)$, hence $(r,s)=-L/2$. In particular the roots are independent. The potential higher sums satisfyso neither is a root. The root-space decomposition therefore gives $[e_r,e_{r+s}]=[e_s,e_{r+s}]=0$.

Set $A=t\mathrm{ad}{e}_r$, $B=u\mathrm{ad}{e}_s$ and $C=[A,B]=tu{N}_{r,s}\mathrm{ad}{e}_{r+s}$. Part (iv) shows $[A,C]=[B,C]=0$. To compute the sign, differentiate $F(v)=e^{-vB}A{e}^{vB}$:Thus $e^{-B}A{e}^B=A+C$. Conjugating the matrix exponential, and then using the commutation of $A,C$, givesThis proves the central-commutator exponential identity in the required order, and consequently the simply laced Chevalley commutator formula is