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

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