Bourbaki

Exercise 2, § 1

Lie Groups and Lie Algebras, VI

Let R be a root system, α\alpha and β\beta two roots. If t is a scalar such that β+tα∈R\beta + t\alpha \in R, then 2t∈Z2t \in \mathbf{Z}. (For n(β+tα,α)=n(β,α)+2t.n(\beta + t\alpha, \alpha) = n(\beta, \alpha) + 2t.) If α\alpha is indivisible, then t∈Zt \in \mathbf{Z}. (If not, show, by using Prop. 9, that there exists a root γ\gamma orthogonal to α\alpha such that γ+12α∈R\gamma + \frac{1}{2}\alpha \in R: then (γ+12α∣α)>0(\gamma + \frac{1}{2}\alpha|\alpha) > 0, so 12α∈R.\frac{1}{2}\alpha \in R.)

Solution

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