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By Antoni Zygmund
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Therefore we may assume that, for every i E I, the restriction to [N, S] of the representation 7ri ® 7T; contains the trivial representation. By a standard argument, this means that the restriction of 7r; to [N, S] contains a finite-dimensional subrepresentation. Since [N, S] is solvable, by Lie's theorem every finite-dimensional unitary representation is a direct sum of unitary characters. 7ri such that the one-dimensional subspace C~i is invariant under [N, Sl; the vector ~; is then fixed by [[N, S], [N, S]] (since a character of a group is trivial on the derived subgroup).
Classification of Lie Groups 44 Proof. 2, n = n S EB ns. 2) Indeed, Y is a sum of elements of the form Ad(s)Z - Z, with ZEn, and, since X = Ad(s)X, [X, Ad(s)Z - Z] = [Ad(s)X, Ad(s)Z]- [X, Z] = Ad(s)[X, Z] - [X, Z] = 0, where the last equality follows from our assumption that S centralizes [N, N]. We now prove the lemma itself. First, for X E n, write X = X S + Xs in the decomposition n = n S EB ns. 2), [X, Y] = [Xs, Y] E [i, i]. Hence [n, ns] C [i, i]. 2(1)), and moreover [i, i] is an ideal in n, we have [n, i] ~ [i, i]; the converse inclusion is obvious.
The analytic subgroup corresponding to the ideal i therefore contains [N, S], so i contains the Lie algebra of [N, S], proving (1). 1]). Therefore it is enough to check that i/[n, i] has no nonzero fixed points under the action of S. From the first part of the lemma, we deduce that i = ns + [n, i]. Since S is semisimple, the quotient module i/[n, i] may be identified with a submodule of ns; the result then follows from the remark preceding the lemma. 3. 2, assume also that S centralizes [N, N].
Trigonometric Series by Antoni Zygmund