# Unipotent and nilpotent classes in simple algebraic groups by Martin W. Liebeck, Gary M. Seitz

By Martin W. Liebeck, Gary M. Seitz

This publication issues the idea of unipotent components in uncomplicated algebraic teams over algebraically closed or finite fields, and nilpotent components within the corresponding uncomplicated Lie algebras. those subject matters were a tremendous quarter of analysis for many years, with purposes to illustration conception, personality concept, the subgroup constitution of algebraic teams and finite teams, and the category of the finite basic teams. the focus is on acquiring complete details on category representatives and centralizers of unipotent and nilpotent components. even though there's a vast literature in this subject, this ebook is the 1st unmarried resource the place such details is gifted thoroughly in all features. additionally, a few of the effects are new--for instance, these referring to centralizers of nilpotent parts in small features. certainly, the full method, whereas utilizing a few rules from the literature, is novel, and yields many new common and particular proof about the constitution and embeddings of centralizers

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**Example text**

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.