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Student Seminar: Classical and Quantum Integrable Systems

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It is clear that<br />

( ) 1 0<br />

α + β<br />

0 1<br />

( ) 0 1<br />

, α<br />

−1 0<br />

( ) 1 0<br />

+ (β + 2πk)<br />

0 1<br />

( ) 0 1<br />

−1 0<br />

has the the same image under the exponential map. In the sufficiently small neighbourhood<br />

of 0 in M(n, R) the map exp A is a diffeomorphism. The inverse map is<br />

constructed by means of series<br />

for x sufficiently close to the identity.<br />

ln x = (x − I) − 1 2 (x − I)2 + 1 3 (x − I)3 − · · ·<br />

Linear representation of a Lie algebra. Adjoint representation. Let J be a<br />

Lie algebra. We say that a map<br />

ρ : J → M(n, R)<br />

defines a representation of the Lie algebra J is the following equality is satisfied<br />

for any two vectors ζ, η ∈ J .<br />

ρ[ζ, η] = [ρ(η), ρ(ζ)]<br />

Let F (t) be a one-parameter subgroup in G. Then g → F gF −1 generates a oneparameter<br />

group of transformations in the Lie algebra<br />

AdF (t) : T → T .<br />

The vector d AdF (t)| dt t=0 lies in the Lie algebra. Let a ∈ T <strong>and</strong> let F (t) = exp(bt)<br />

then<br />

d<br />

dt AdF (t)| t=0 a = d (<br />

)<br />

exp(bt)a exp(−bt) | t=0 = [b, a]<br />

dt<br />

Thus to any element b ∈ J we associate an operator ad b which acts on the Lie<br />

algebra:<br />

ad b : J → J , ad b a = [b, a] .<br />

This action defines a representation of the Lie algebra on itself. This representation<br />

is called adjoint. To see that this is indeed representation we have to show that it<br />

preserves the commutation relations, i.e. that from [x, y] = z it follows that<br />

We compute<br />

[adx, ady] = adz .<br />

[adx, ady]w = adx adyw − ady adxw = [x, [y, w]] − [y, [x, w]] = [x, [y, w]] + [y, [w, x]] =<br />

− [w, [x, y]] = [[x, y], w] = [z, w] = adzw .<br />

– 89 –

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