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Caleb Proctor 2022-07-03 Answered
Let Δ be an indecomposable root system in a real inner product space E, and suppose that Φ is a simple system of roots in Δ, with respect to an ordering of E. If Φ = { α 1 , , α l }, prove that
α 1 + + α l Δ
I know that any positive root γ may be written as a sum of simple roots, and furthermore that every partial sum is itself a root, but I am unsure if that will help me or not. Any hints to get me started?
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Answers (2)

gozaderaradiox5
Answered 2022-07-04 Author has 19 answers
First, I would let Φ be a base for an indecomposable root system, and choose labels for the elements of Φ according to the following algorithm:
Suppose you have chosen Φ k := { α 1 , , α k } such that the corresponding root system Δ k is indecomposable (when k = 1 this is automatic). Then, there exists α k + 1 Φ such that Φ k + 1 is a base for an indecomposable root system Δ k + 1 . Indeed, otherwise
s p a n ( Φ k ) s p a n ( Φ Φ k ) ,
so Δ is not indecomposable, a contradiction.
This is enough to obtain strict inequality since, by construction, we must have ( α j , α l ) < 0 for some j < l.
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Lucian Maddox
Answered 2022-07-05 Author has 8 answers

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