Remark

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

Lie Groups and Lie Algebras, chapter VI, § 1. Root systems

We retain the notation above. Then:

1) If the length of S is 0, then $(\alpha|\gamma) = 0$ .

2) If the length of S is 1, then $n(\gamma, \alpha) = -1$ , and there are three cases:

n(\alpha, \gamma) = -1, \quad (\alpha|\alpha) = (\gamma|\gamma), \quad (\alpha|\gamma) = -\frac{1}{2}(\alpha|\alpha), \quad (\overline{\alpha, \gamma}) = \frac{2\pi}{3}

n(\alpha, \gamma) = -2, \quad (\alpha|\alpha) = 2(\gamma|\gamma), \quad (\alpha|\gamma) = -\frac{1}{2}(\alpha|\alpha), \quad (\overline{\alpha, \gamma}) = \frac{3\pi}{4}

n(\alpha, \gamma) = -3, \quad (\alpha|\alpha) = 3(\gamma|\gamma), \quad (\alpha|\gamma) = -\frac{1}{2}(\alpha|\alpha), \quad (\overline{\alpha, \gamma}) = \frac{5\pi}{6}

3) If the length of S is 2, then $n(\gamma, \alpha) = -2$ , so

n(\alpha, \gamma) = -1, \quad (\alpha|\alpha) = \frac{1}{2}(\gamma|\gamma), \quad (\alpha|\gamma) = -(\alpha|\alpha), \quad (\overline{\alpha, \gamma}) = \frac{3\pi}{4}.

4) If the length of S is 3, then $n(\gamma, \alpha) = -3$ , so

n(\alpha, \gamma) = -1, \quad (\alpha|\alpha) = \frac{1}{3}(\gamma|\gamma), \quad (\alpha|\gamma) = -\frac{3}{2}(\alpha|\alpha), \quad (\overline{\alpha, \gamma}) = \frac{5\pi}{6}.

We shall see (Plate X, Systems A₂, B₂, G₂) that all these cases are actually realised.

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

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