Bourbaki

Remark 27

Tag 0DN0, permanent. Cite it as [Bourbaki, Tag 0DN0].

Algebra, chapter IX, § 7. Formes hermitiennes et corps ordonnés

Let K be a maximal ordered field, E a vector space of dimension n over K, Φ\Phi a non-degenerate symmetric bilinear form on E, of signature (s, t) distinct from (n, 0) and from (0, n). Let (ei)(e_i) be an orthogonal basis for Φ\Phi, such that Φ(ei,ei)=1\Phi(e_i, e_i) = 1 for 1≤i≤s1 \leq i \leq s, Φ(ei,ei)=−1\Phi(e_i, e_i) = -1 for s+1≤i≤ns+1 \leq i \leq n. For every orthogonal transform u∈U(Φ)u \in \mathbf{U}(\Phi), let

U=(MNPQ) U = \begin{pmatrix} M & N \\ P & Q \end{pmatrix}

be the matrix of u with respect to (ei)(e_i), written in the form of a square tableau of matrices corresponding to the partition of (1,n)(1, n) into (1,s)(1, s) and (s+1,n)(s+1, n).

a) Prove the relations

tM⋅M−tP⋅P=IstQ⋅Q−tN⋅N=IttM⋅N−tP⋅Q=0. \begin{align*} {}^tM \cdot M - {}^tP \cdot P &= I_s \\ {}^tQ \cdot Q - {}^tN \cdot N &= I_t \\ {}^tM \cdot N - {}^tP \cdot Q &= 0. \end{align*}

b) Let R be a matrix over K with t rows and s columns, such that Is−tR⋅RI_s - {}^tR \cdot R is the matrix of a positive non-degenerate hermitian form (on KsK^s). Show that det⁡(M+λNR)\det(M + \lambda NR) does not change sign for −1≤λ≤1-1 \leq \lambda \leq 1 in K (show, using a), that t(M+λNR)(M+λNR){}^t(M + \lambda NR)(M + \lambda NR) is the matrix of a positive non-degenerate symmetric form).

c) Put σ(u)=σ(U)=sgn⁡(det⁡M)\sigma(u) = \sigma(U) = \operatorname{sgn}(\det M); show that, for any two elements u,vu, v of the orthogonal group U(Φ)\mathbf{U}(\Phi), one has σ(uv)=σ(u)σ(v)\sigma(uv) = \sigma(u)\sigma(v) (use a) and b)). Deduce that the commutator group of the special orthogonal group SU(Φ)\mathbf{SU}(\Phi) is distinct from SU(Φ)\mathbf{SU}(\Phi) (cf. § 10, exerc. 9).

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

Cited by

Nothing.

Cites

Nothing.

Translated with machine help, not yet reviewed by a person.