Descent in Buildings (AM-190), Volume I (Annals of Mathematics Studies) by Bernhard Mühlherr & Holger P. Petersson & Richard M. Weiss

Author:Bernhard Mühlherr & Holger P. Petersson & Richard M. Weiss [Mühlherr, Bernhard]
Language: eng
Format: azw3
Publisher: Princeton University Press
Published: 2015-09-21T16:00:00+00:00

Now suppose that R and T are parallel. By 21.17, R ∩ Σ and T ∩ Σ are parallel residues of Σ. Hence projR(T ∩ Σ) = R ∩ Σ (by 21.16). By (21.20), therefore, ΩR = ΩT.

Suppose, conversely, that ΩR = ΩT. If x, y ∈ Σ ∩ T are distinct, then there exists a root of ΩT containing x but not y. By (21.20), it follows that the restriction of projR to Σ ∩ T is injective. By symmetry, the restriction of projT to Σ ∩ R is injective. By 21.9(i), therefore, Σ ∩ R and Σ ∩ T are parallel residues of Σ. Hence by 21.17, R and T are parallel residues. Thus (i) holds.

We continue to assume that ΩR = ΩT. Let c ∈ Σ ∩ T. By (21.20) again, it follows that the set of roots in ΩT containing c is the same as the set of roots of ΩR containing projR(c). The chamber projR(c) is the only chamber of R contained in all these roots. Hence (ii) holds.

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Corollary 21.21. If Δ is thin, then parallelism is an equivalence relation on the set of all residues of Δ.

Proof. If Δ is thin, it consists of a single apartment. The claim is thus an immediate consequence of 21.19(i).

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Proposition 21.22. If R and T are parallel residues, then either R = T or .

Proof. Suppose that R and T are parallel residues and that c is a chamber in R ∩ T. Then projT(c) = c. By 21.11, it follows that δ(R, T) = 1 and hence dist(d, projT(d)) = ℓ(1) = 0 for all d ∈ R. Therefore T = projT(R) = R.

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Proposition 21.23. Let R, T, T′ be residues such that R is parallel to both T and T′ and suppose that . Then T = T′.

Proof. Let J = Typ(R), let c be a chamber in T ∩ T′, let d = projR(c) and let w = δ(c, d). By 21.10(iii), Typ(T) = wJw−1 = Typ(T′). Since T and T′ have the same type and both contain c, they are equal.

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We now want to make a number of observations (many of which can be found in [55]) concerning the notion of opposite residues defined in 19.25. We emphasize that only spherical buildings have opposite residues.

Proposition 21.24. Opposite residues in a spherical building are parallel.

Proof. Suppose that Δ is spherical and let R and T be opposite residues, let Σ be an apartment containing chambers of both R and T and let opΣ be the map which sends each chamber of Σ to its unique opposite in Σ. By [62, 9.8], R ∩ Σ and T ∩ Σ are opposite residues of Σ. By [62, 5.3], therefore, opΣ(T ∩ Σ) = R ∩ Σ and opΣ(R ∩ Σ) = T ∩ Σ. By 19.25, finally, a chamber of Σ is contained in a given root of Σ if and only if its image under opΣ is not contained in this root. We conclude that a root of Σ cuts R ∩ Σ if and only if it cuts T ∩ Σ.

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