Bourbaki

Proposition 7

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

Lie Groups and Lie Algebras, chapter V, § 1. Hyperplanes, chambers and facets

Let L be an affine subspace of E and Ω\Omega a non-empty open subset of L.

(i) There exists a point a in Ω\Omega that does not belong to any of the hyperplanes of H\mathfrak{H} that do not contain L.

(ii) If L is a hyperplane and L∉HL \notin \mathfrak{H}, there exists a chamber that meets Ω\Omega.

(iii) If L is a hyperplane and L∈HL \in \mathfrak{H}, there exists a point a in Ω\Omega that does not belong to any hyperplane H≠LH \neq L of H\mathfrak{H}.

Denote by N\mathfrak{N} the set of hyperplanes H with H∈HH \in \mathfrak{H} and L⊄HL \not\subset H, and by L\mathscr{L} the set of hyperplanes of the affine space L of the form L∩HL \cap H with H∈NH \in \mathfrak{N}. It is clear that L\mathscr{L} is a locally finite set of hyperplanes in L, and Prop. 6 shows that Ω\Omega meets a chamber Γ\Gamma defined by L\mathscr{L} in L. If aa is a point of Γ∩Ω\Gamma \cap \Omega, then a∉Ha \notin H for all H∈NH \in \mathfrak{N}, hence (i).

Assume now that L is a hyperplane; any hyperplane containing L is then equal to it, so we may distinguish two cases:

a) L∉HL \notin \mathfrak{H}: then N=H\mathfrak{N} = \mathfrak{H}, and we have a∉Ha \notin H for all H∈HH \in \mathfrak{H}; thus aa belongs to a chamber defined by H\mathfrak{H} in E, hence (ii).

b) L∈HL \in \mathfrak{H}: then N=H−{L}\mathfrak{N} = \mathfrak{H} - \{L\}, hence (iii).

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.