Bourbaki

Lemma 8

Tag 0AY9, permanent. Cite it as [Bourbaki, Tag 0AY9].

Lie Groups and Lie Algebras, chapter VI, § 4. Classification of root systems

Let i1,i2,…,ipi_1, i_2, \ldots, i_p be vertices of X such that {i1,i2},{i2,i3},…,{ip−1,ip}\{i_1, i_2\}, \{i_2, i_3\}, \ldots, \{i_{p-1}, i_p\} are edges of order 3. Then q(∑r=1preir)=12p(p+1)q(\sum_{r=1}^p r e_{i_r}) = \frac{1}{2} p(p+1).

We have (eir∣eir)=1,(eir∣eir+1)=−12,(eir∣eis)=0(e_{i_r}|e_{i_r}) = 1, (e_{i_r}|e_{i_{r+1}}) = -\frac{1}{2}, (e_{i_r}|e_{i_s}) = 0 if s>r+1s > r + 1. Thus

q(∑r=1preir)=∑r=1pr2−2∑r=1p−112r(r+1)=p2−∑r=1p−1r. q(\sum_{r=1}^p r e_{i_r}) = \sum_{r=1}^p r^2 - 2 \sum_{r=1}^{p-1} \frac{1}{2} r(r+1) = p^2 - \sum_{r=1}^{p-1} r.

By Theory of Sets, Chap. III. § 5, no. 8. Cor. to Prop. 14, this is equal to

p2−12p(p−1)=12p(p+1). p^2 - \frac{1}{2} p(p-1) = \frac{1}{2} p(p+1).

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Transcribed from the printed volume. Not translated.