Bourbaki

Proposition 5

Tag 0AP8, permanent. Cite it as [Bourbaki, Tag 0AP8].

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

Let CC be a non-empty subset of EE. Assume that there exists a subset H′\mathfrak{H}' of H\mathfrak{H} with the following properties:

a) For any H∈H′H \in \mathfrak{H}', there exists an open half-space DHD_H bounded by HH such that C=⋂H∈H′DHC = \bigcap_{H \in \mathfrak{H}'} D_H.

b) *The set C does not meet any hyperplane belonging to H−H′\mathfrak{H} - \mathfrak{H}'. Under these conditions, C is a chamber defined by H\mathfrak{H} in E, and DH=DH(C)D_H = D_H(C) for all H∈HH \in \mathfrak{H}.

Properties a) and b) show that C is a convex subset of U; hence there is a chamber C′C' with C⊂C′C \subset C'. Since C⊂DHC \subset D_H, we have DH=DH(C)D_H = D_H(C) for all H in H′\mathfrak{H}', hence C=DH′(C)⊃DH(C)C = D_{\mathfrak{H}'}(C) \supset D_{\mathfrak{H}}(C) since H′⊂H\mathfrak{H}' \subset \mathfrak{H}; we have DH(C)=C′D_{\mathfrak{H}}(C) = C' by (6), hence C⊃C′C \supset C'. Finally therefore, C=C′C = C'.

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

Cited by

Cites

Nothing.

Transcribed from the printed volume. Not translated.