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

Solution

Write

D = ad x

, and choose

N

with

D^{N} = 0

E = exp D = \sum_{j = 0}^{N - 1} \frac{D ^{j}}{j !} .

k

using the Leibniz rule and Pascal's identity proves

D^{k} [y, z] = \sum_{j = 0}^{k} (j k) [D^{j} y, D^{k - j} z] .

Expanding

[E y, E z]

and grouping by

k = j + l

consequently gives

[E y, E z] = \sum_{k = 0}^{2 N - 2} \frac{1}{k !} \sum_{j = 0}^{k} (j k) [D^{j} y, D^{k - j} z] = \sum_{k = 0}^{2 N - 2} \frac{D ^{k} [ y , z ]}{k !} = E [y, z] .

Terms with an exponent at least

N

vanish, which justifies extending each inner sum; and

D^{k} = 0

for

k \geq N

. Finally the finite product of the two exponentials has coefficient

[D^{k}] (exp D exp (- D)) = \sum_{j = 0}^{k} \frac{( - 1 ) ^{k - j}}{j ! ( k - j )!} = \frac{( 1 - 1 ) ^{k}}{k !},

exp (- D)

is the inverse. We have proved bracket preservation and invertibility, hence $exp (ad x)$ is a Lie algebra automorphism.

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