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Component roots intersect the complementary parabolic exactly
Statement
Use the arbitrary-rank geometric representation of Coxeter-system, roots, length and weak-order conventions and the orbit definition from Standard parabolic subgroups and parabolic roots. Let , , , , and , with connected-component notation from Connected component notation [I,a]. Vectors have finite support, even if the vertex set is infinite. Then . In particular . No rank-finiteness assumption is needed.
Facts & Assumptions
Given: the diagram and reflection formula from Coxeter-system, roots, length and weak-order conventions, the root orbits from Standard parabolic subgroups and parabolic roots, and the component from Connected component notation [I,a]. Put . [F1] There is no edge between and ; hence and for . [F2] The subspaces and have zero intersection, and every root is nonzero because it is the invertible image of a basis vector.
Proof
Goal: Establish the assertion in the Statement.
Proof technique: direct.
The two disjoint parts of are and . By [F1] every finite word in can be reordered to a word in followed by a word in . Generators in act identically on , and generators in act identically on , by the reflection formula. Consequently the orbit under of a simple root indexed by is exactly its orbit under ; the orbit of a simple root indexed by stays in . Thus , with and .
Every root in lies in , because its generators preserve . By [F2] and step 1.1, a root belonging also to must lie in . Conversely gives directly from the orbit definitions. This proves equality. Finally , so a root in is outside by the equality and hence lies in .
Depends on
Dependency tree · 2 levels
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