Exercise 3, § 6

Lie Groups and Lie Algebras, II, 201

Let $M(U, V) = H(-U, U + V)$ . Then

\exp(M(U, V)) = \exp(-U) \cdot \exp(U + V).

Let

M_1(U, V)

(resp.

H_1(U, V)

) denote the sum of the bihomogeneous terms of the series

M

(resp.

H

) whose degree in

V

is 1.

(a) Show that

M_1(U, V) = \sum_{n=0}^\infty \frac{(-1)^n}{(n+1)!} (\mathrm{ad}\, U)^n(V) = (f(\mathrm{ad}\, U))(V)

where

f(T) = (1 - e^{-T})/T

(Same argument as for the proof of Proposition 5. It can also be reduced to Proposition 5 by using the identity

\exp(U) \cdot \exp(M(U, V)) \cdot \exp(-U) = \exp(H(U + V, -U)),

together with Exercise 1.)

(b) Show that $H(U, M(U, V)) = U + V$ . Deduce that

H_1(U, M_1(U, V)) = V.

g(T) = 1/f(T) = 1 + T/2 + \sum_{n=1}^\infty \frac{1}{(2n)!} b_{2n} T^{2n}

, where the

b_{2n}

are the Bernoulli numbers (Functions of a Real Variable, Chapter VI, § 1, no. 4). Show that

H_1(U, V) = (g(\operatorname{ad} U))(V) = V + \frac{1}{2}[U, V] + \sum_{n=1}^\infty \frac{1}{(2n)!} b_{2n} (\operatorname{ad} U)^{2n}(V).

Derive the first few terms of the expansion of

H_1(U, V)

H_1(U, V) = V + \frac{1}{2}(\operatorname{ad} U)(V) + \frac{1}{12} (\operatorname{ad} U)^2(V) - \frac{1}{720} (\operatorname{ad} U)^4(V)

+ \frac{1}{30240} (\operatorname{ad} U)^6(V) - \cdots

(d) Show that the sum of the bihomogeneous terms of

H(U, V)

whose degree in

U

is 1 is the series:

U + \frac{1}{2}[U, V] + \sum_{n=1}^\infty \frac{1}{(2n)!} b_{2n} (\operatorname{ad} V)^{2n}(U).

(Apply (c) to

H(V, U)

and use Exercise 2 to pass from

H(V, U)

H(U, V)

.) (e) Let

W

be another indeterminate and let

X(U, V, W)

be the sum of the trihomogeneous terms of

H(U, V + W)

whose degree in

W

is 1. Show that

X(U, H_1(V, W)) = H_1(H(U, V), W).

(Use the identity

H(U, H(V, W)) = H(H(U, V), W)

.) Deduce that

X(U, V, W) = (g(\operatorname{ad} H(U, V)) \circ f(\operatorname{ad} V))(W).

Solution

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