Linear Diff. Eqns. and Group Theory From Riemann to Poincare by J. Gray

By J. Gray

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Each such root vector has T -weight the label associated to the fundamental root βi . Now Z(CL (s)) is a torus of rank k which acts as scalars on each of the modules Vi . 7. CENTRALIZERS OF NILPOTENT ELEMENTS 33 if Ji = j=i Vj , then there is a 1-dimensional torus acting trivially on Ji . On the other hand, e ∈ L(CQ (s))2 ⊆ L(CQ (s))(1) and e is distinguished. Therefore e is not centralized by a nontrivial torus, and so e ∈ Ji for any i. It follows that each βi has label 2, completing the proof of (ii).

Then S is a non-empty open set of L(Q). Fix l ∈ S and suppose that l ∈ L(Q)g for some g ∈ G. By the above, there is a distinguished coset representative dJ such that g = ydJ u with y ∈ P, u ∈ UdJ . Then l ∈ L(Q)dJ u . Hence, l = au with a ∈ L(Q) ∩ L(Q)dJ . 5. NILPOTENT AND UNIPOTENT ELEMENTS 23 is possible. There are also only finitely many choices for dJ , so this completes the proof of the lemma. In the next lemma, we write L(Q)G for the set {lg : l ∈ L(Q), g ∈ G}. 18. Let P = QL be a parabolic subgroup of G.

Orbit O such that O Now O ∩ L(Q) is non-empty, as otherwise O ∩ L(Q)G would be empty. If e ∈ O ∩ L(Q), then dim eP = dim P ≥ dim P = dim P = dim P = dim P − dim CP (e) − dim CG (e) − dim G + dim O − dim G + dim G − dim L − dim L = dim L(Q). It follows that eP is open dense in L(Q). This holds for any element of O ∩ L(Q), so if e is another such element, then the open orbits eP and e P must intersect and e and e are in the same P -orbit. Therefore O ∩ L(Q) = eP , completing the proof. 36 (another result of Richardson) to follow.

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