By Jean Renault

ISBN-10: 0387099778

ISBN-13: 9780387099774

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6 we have r = 0. 7 We have r ∩ g = 0. Proof Consider the ideal [rg] ⊂ g. If x ∈ g, a ∈ r, and s is any linear transformation of V which commutes with all transformations defined by the elements of r, then Tr [xa]s = Tr (xas − axs) = Tr (xas − xsa) = Tr x(as − sa) = 0. Since r is abelian, we can take s to be any power of [xa] ∈ r. Then the trace of any power of [xa] is zero, and this implies that [xa] is a nilpotent linear transformation. (The coefficients of the characteristic polynomial are up to sign the symmetric functions in the eigenvalues, and these can be expressed in terms of sums of powers.

8] J-P. Serre. Alg`ebres de Lie semi-simple complexes. Benjamin, 1966.

9 The centre of o(2l + 1) is 0. Proof Let z = ξ + tα Xα be an element of the centre, ξ ∈ h. We have [zh] = tα α(h)Xα = 0 for any h ∈ h, which implies that tα α(h) = 0 since the vectors Xα are linearly independent. This holds identically on h, thus tα = 0. Now [ξ, Xα ] = α(ξ)Xα = 0 for all α. Since the linear forms α : h → C span h∗ we conclude that ξ = 0. QED An argument similar to that we used for sl(n + 1) shows that h ⊂ o(2l + 1) is a Cartan subalgebra. , αl−1 = el−1 − el , αl = el form a basis of the root system of o(2l + 1).

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