Bourbaki

Proposition 6

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

Algebra, chapter IX, § 6. Propriétés spéciales aux formes hermitiennes

Let L be a finite-dimensional hermitian space over A, whose metric form has index 0. Any similitude u of L, with multiplier μ≠1\mu \neq 1, then admits one and only one fixed point.

Indeed, let aa be a point of L. There exists a similitude φ\varphi of L leaving aa fixed and a translation tt of L such that u=tφu = t \varphi. Saying that bb is a fixed point of uu amounts to saying that φ(b)−b=t\varphi(b) - b = t. To show that this equation admits one and only one solution bb, identify L with its space of translations T by taking aa as origin. It is then enough to prove that the endomorphism φ−1\varphi - 1 of T is invertible, in other words that the relation φ(x)−x=0\varphi(x) - x = 0 (x∈Tx \in T) implies x=0x = 0. Now, if φ(x)−x=0\varphi(x) - x = 0, we have Φ(x,x)=Φ(φ(x),φ(x))=μΦ(x,x)\Phi(x, x) = \Phi(\varphi(x), \varphi(x)) = \mu \Phi(x, x), hence Φ(x,x)=0\Phi(x, x) = 0 since μ≠1\mu \neq 1; this implies x=0x = 0 since Φ\Phi has index 0. QED.

Suppose that A is a field of characteristic ≠2\neq 2. Every displacement uu of L such that u2=1u^2 = 1 admits at least one fixed point, for example the midpoint 12(x+u(x))\frac{1}{2}(x + u(x)) of two homologous points; by taking this point as origin, one sees that the unitary automorphism of T associated with uu is a symmetry (No. 3). Let V be a non-isotropic linear variety in L; a displacement uu is said to be the symmetry with respect to V if, by taking an origin in V, uu is identified with the symmetry with respect to V of T. This amounts to saying that u(x)u(x) is obtained in the following way: denoting by x1x_1 the orthogonal projection of xx on V, one has u(x)−x=2(x1−x)u(x) - x = 2(x_1 - x).

Exercises. — 1) Suppose that A is a commutative field. Given a hermitian matrix RR of order nn over A, the principal minors of order rr of RR are called the minors obtained by deleting in RR n−rn - r rows and the n−rn - r columns of the same indices.

a) If a principal minor of order rr of RR is not zero, but if all the principal minors of orders r+1r + 1 and r+2r + 2 which contain this minor of order rr are zero, now prove that RR is of rank rr (cf. Chapter III, § 7, Exercise 1 and § 8, Exercise 11 and Chapter IV, § 2, Exercise 10). Deduce that, in order that RR be of rank rr, it is necessary and sufficient that there exist a principal minor of order rr which is ≠0\neq 0, and that all the principal minors of orders r+1r + 1 and r+2r + 2 be zero.

b) Deduce from a) that if RR is of rank rr, there exists a permutation σ∈Sn\sigma \in \mathfrak{S}_n such that, if the same permutation σ\sigma is performed on the rows and columns of RR, and if SS denotes the matrix obtained, and Δk\Delta_k the principal minor of order kk of SS obtained by suppressing in SS the rows and columns of index >k> k, the following two properties hold: 1∘Δr≠01^\circ \Delta_r \neq 0; 2∘2^\circ there is no index k<rk < r such that Δk=Δk+1=0\Delta_k = \Delta_{k+1} = 0.

2) Suppose that AA is a commutative field, and that EE is of finite dimension nn. Let Φ\Phi be a hermitian sesquilinear form on EE, satisfying condition (T) of § 4, No. 2, R=(αij)R = (\alpha_{ij}) the matrix of Φ\Phi with respect to a basis (ei)(e_i) of EE.

a) If Φ\Phi is of rank rr, and if the principal minor (Exercise 1) obtained by suppressing in RR the rows and columns of indices >r> r is not zero, show that there exists a new basis (fi)(f_i) of EE such that ei=fie_i = f_i for 1≤i≤r1 \leq i \leq r and that the matrix of Φ\Phi with respect to (fi)(f_i) is obtained by replacing by 0 in RR all the αij\alpha_{ij} such that i>ri > r or j>rj > r (consider the subspace E0E^0 orthogonal to EE).

b) Deduce from a) that if Φ\Phi is of rank nn, and if the cofactor Δn−1\Delta_{n-1} of αnn\alpha_{nn} in the determinant Δ=det⁡R\Delta = \det R is not zero, there exists a new basis (fi)(f_i) of EE such that fi=eif_i = e_i for 1≤i≤n−11 \leq i \leq n-1, and such that

Φ(x,y)=Φ(∑i=1nξifi,∑i=1nηifi)=∑i=1n−1∑j=1n−1αijξiηj‾+ΔΔn−1ξnηn‾ \Phi(x, y) = \Phi\left( \sum_{i=1}^n \xi_i f_i, \sum_{i=1}^n \eta_i f_i \right) = \sum_{i=1}^{n-1} \sum_{j=1}^{n-1} \alpha_{ij} \xi_i \overline{\eta_j} + \frac{\Delta}{\Delta_{n-1}} \xi_n \overline{\eta_n}

(consider the hermitian form whose matrix with respect to (ei)(e_i) is obtained by replacing αnn\alpha_{nn} by αnn−ΔΔn−1\alpha_{nn} - \frac{\Delta}{\Delta_{n-1}} in RR).

c) Suppose that Φ\Phi is of rank nn, that Δn−1=0\Delta_{n-1} = 0, but that the principal minor Δn−2\Delta_{n-2} of RR obtained by deleting the rows and columns with indices n−1n-1 and nn in RR is not zero. Now prove that there exists a new basis (fi)(f_i) of EE such that fi=eif_i = e_i for 1≤i≤n−21 \leq i \leq n-2, and such that

Φ(x,y)=(∑i=1nξifi,∑i=1nηifi)=∑i=1n−2∑j=1n−2αijξiηj‾+ξn−1ηn‾+ξnηn−1‾. \Phi(x, y) = \left( \sum_{i=1}^n \xi_i f_i, \sum_{i=1}^n \eta_i f_i \right) = \sum_{i=1}^{n-2} \sum_{j=1}^{n-2} \alpha_{ij} \xi_i \overline{\eta_j} + \xi_{n-1} \overline{\eta_n} + \xi_n \overline{\eta_{n-1}}.

(If HH is the hyperplane generated by e1,…,en−1e_1, \ldots, e_{n-1}, which is isotropic, observe that the right orthogonal to HH is not in the subspace generated by e1,…,en−2e_1, \ldots, e_{n-2}, and use Prop. 2 of § 4, No. 2).

3) Let AA be a finite field, EE a finite-dimensional vector space over AA, Φ\Phi a non-degenerate hermitian sesquilinear form on EE, relative to an automorphism J≠1J \neq 1 of AA. Now prove that EE admits an orthonormal basis for Φ\Phi (cf. Chap. V, § 11, No. 5, Cor. of Th. 3).

4) Let AA be a finite field of characteristic ≠2\neq 2, EE a finite-dimensional vector space of dimension nn over AA.

a) Prove that for every non-degenerate symmetric bilinear form Φ\Phi on EE, there exists an orthogonal basis (ei)(e_i) of EE such that Φ(ei,ei)=1\Phi(e_i, e_i) = 1 for 1≤i≤n−11 \leq i \leq n-1, Φ(en,en)=Δ\Phi(e_n, e_n) = \Delta (discriminant of Φ\Phi with respect to (ei)(e_i)). (Remark that if αβ≠0\alpha \beta \neq 0, the equation αξ2+βη2=γ\alpha \xi^2 + \beta \eta^2 = \gamma always admits solutions (ξ,η)(\xi, \eta) in AA if γ≠0\gamma \neq 0 (chap. V, § 11, exerc. 4)).

b) In order that two non-degenerate symmetric bilinear forms on E be equivalent, it is necessary and sufficient that the quotient of their discriminants (with respect to the same basis of E) be a square in A. Deduce that, if n is odd, for every non-degenerate symmetric bilinear form Φ\Phi on E, there exists an orthogonal basis with respect to which the matrix of Φ\Phi is of the form λIn (λ∈A)\lambda I_n \ (\lambda \in A); the index of Φ\Phi is then (n−1)/2(n-1)/2.

c) If n=2mn = 2m is even, prove that the index of a non-degenerate symmetric bilinear form Φ\Phi on E is m if (−1)mΔ(-1)^m \Delta is a square in A, m−1m-1 otherwise.

5) Let A be a commutative field of characteristic ≠2\neq 2. Let I be a polynomial with coefficients in A, with respect to n(n+1)/2n(n+1)/2 indeterminates Xij (1≤i≤j≤n)X_{ij} \ (1 \leq i \leq j \leq n); for every symmetric matrix R=(αij)R = (\alpha_{ij}) over a commutative overfield A′A' of A, denote by I(R)I(R) the element of A′A' obtained by substituting αij\alpha_{ij} for the indeterminate Xij (i≤j)X_{ij} \ (i \leq j) in I.

Suppose that I is such that, for the matrix U=(uij)U = (u_{ij}) with uij=Xiju_{ij} = X_{ij} for i≤ji \leq j, uij=Xjiu_{ij} = X_{ji} for i>ji > j and the square matrix P=(Yij)P = (Y_{ij}) of order n (where the YijY_{ij} are n2n^2 other indeterminates), one has

I(PUP)=(det⁡P)hI(U) I(PUP) = (\det P)^h I(U)

where h is an integer >0> 0. Prove that h is even and that I(U)=γ(det⁡U)kI(U) = \gamma (\det U)^k, where h=2kh = 2k and γ∈A\gamma \in A. (Using th. 1, prove that for every symmetric matrix RR over the algebraic closure Ω\Omega of A, one has (I(R))2=λ(det⁡R)h(I(R))^2 = \lambda (\det R)^h, where λ∈Ω\lambda \in \Omega, and use the fact that the polynomial det⁡U\det U with respect to the XijX_{ij} is not a square, by considering the terms of this polynomial containing an XiiX_{ii}.

Soient A un corps valué complet non discret, commutatif et de caractéristique ≠2\neq 2 (Top. gén., chap. IX, § 3, no 2), Φ\Phi une forme hermitienne non dégénérée sur un espace vectoriel E de dimension finie n sur A, R=(αij)R = (\alpha_{ij}) la matrice de Φ\Phi par rapport à une base (ei)(e_i) de E. Show that there exists ε>0\varepsilon > 0 such that, for every hermitian matrix R′=(χij′)R' = (\chi'_{ij}) satisfying the conditions ∣αij′−αij∣≤ε|\alpha'_{ij} - \alpha_{ij}| \leq \varepsilon for every pair (i,j)(i, j), the form Φ′\Phi' having R′R' as matrix with respect to the basis (ei)(e_i) is equivalent to Φ\Phi. (Reduce to the case where RR is diagonal; use exerc. 2 b) by relying on the following lemma: there exists a number a>0a > 0 such that for ∣η∣≤a|\eta| \leq a, there exists in A an element ξ\xi such that ξ2=1−η\xi^2 = 1 - \eta. To prove this lemma, the binomial series for (1−x)1/2(1-x)^{1/2} will be used.)

Soient A un corps commutatif non ordonnable (chap. VI, § 2, exerc. 8) de caractéristique ≠2\neq 2, E un espace vectoriel de dimension finie n>0n > 0 sur A, Q une forme quadratique non dégénérée sur E, (ei)(e_i) une base orthogonale pour Q, de sorte que Q(∑i=1nξiei)=∑i=1nαiξi2Q \left( \sum_{i=1}^n \xi_i e_i \right) = \sum_{i=1}^n \alpha_i \xi_i^2. Pour 1≤r≤n1 \leq r \leq n, on pose Qr(ξ1,…,ξr)=∑i=1rαiξi2Q_r(\xi_1, \ldots, \xi_r) = \sum_{i=1}^r \alpha_i \xi_i^2, et on désigne par MrM_r l’ensemble des valeurs de Q_r lorsque les ξi (1≤i≤r)\xi_i \ (1 \leq i \leq r) parcourent A.

(a) Show that if, for an index r, one has Mr=Mr+1M_r = M_{r+1}, it follows that Mr=AM_r = A (observe that every element of A is a sum of squares (chap. VI, § 2, exerc. 7)).

(b) Suppose that the subgroup S of the multiplicative group A∗A^*, formed by the squares of elements of A, is of finite index s in A∗A^*. Deduce from (a) that if n>sn > s, every non-degenerate quadratic form on E has index >0> 0 (observe that every set MrM_r is the union of 0 and of classes mod. S). Application to the case where A is a pp-adic field Qp\mathbf{Q}_p (Top. gén., chap. III, § 5, exerc. 35).

Soient A un corps commutatif de caractéristique ≠2\neq 2, E un espace vectoriel de dimension finie n sur A, Q une forme quadratique non dégénérée d’indice 0 sur E. Soient A′A' une extension algébrique de A, de degré fini et impair, E′E' l’espace vectoriel sur A′A' obtenu par extension à A′A' du corps des scalaires de E. Show that the extension Q′Q' of Q to E′E' (§3§ 3, no 4, prop. 3) is still of index 0. (Reduce to the case where A′=A[X]/(f)A' = A[X]/(f), f being an irreducible polynomial of odd degree m over A. Let (ei)(e_i) be an orthogonal basis of E for Q, and let ρi=Q(ei)\rho_i = Q(e_i); show that, in A[X]A[X], a relation of the form ∑iρi(gi(X))2=f(X)h(X)\sum_i \rho_i(g_i(X))^2 = f(X)h(X), where the gig_i are non-zero polynomials not all zero, of degree ≤m−1\leq m - 1, is impossible; observe for this that h would necessarily be of odd degree, and consider an irreducible factor of h, of odd degree).

Soient A un corps, E un espace vectoriel sur A admettant une base dénombrable (en)n≥1(e_n)_{n \geq 1}, Φ\Phi une forme sesquilinéaire hermitienne non dégénérée sur E, satisfaisant à la condition (T) (§4§ 4, no 2).

(a) Show that if the conditions (C) of th. 1 are not simultaneously verified, there exists in E an orthogonal basis for Φ\Phi (reason as in exerc. 4 of the §5§ 5).

b) Suppose in addition that A is commutative, and that there exists an integer s such that on every finite-dimensional vector space of dimension >s> s over A, every non-degenerate Hermitian sesquilinear form has index >0> 0 (cf. exerc. 7). Now prove that there then exists in E an orthonormal basis for Φ\Phi. (Reason as in a), observing that for every element of A of the form α=λ+λ‾\alpha = \lambda + \overline{\lambda}, and every non-degenerate Hermitian form Ψ\Psi on a space F of dimension >s> s, there exists z∈Fz \in F such that Ψ(z,z)=α\Psi(z, z) = \alpha (cf. §4§ 4, no 2, prop. 4).)

¶ 10) a) Let A be a principal ideal domain in which there is only one maximal ideal AπA\pi, such that 2 is not divisible by π\pi (chap. VII, § 1, exerc. 4). Let E be a free module over A, of dimension n. Now prove that every symmetric bilinear form Φ\Phi on E admits an orthogonal basis. (Let r be the greatest exponent such that πr\pi^r divides all the elements Φ(x,y)\Phi(x, y); show that there exists a∈Ea \in E such that Φ(a,a)=απr\Phi(a, a) = \alpha \pi^r, where α\alpha is invertible in A; deduce that E is the direct sum of F=AaF = Aa and the submodule F0F^0 orthogonal to F.)

b) Give an example (for n=2n = 2) in which Φ\Phi is non-degenerate and in which there exists a non-isotropic submodule F of E, of rank 1, admitting a complement in E but such that F0F^0 is not a complement of F.

c) Let (ei)(e_i) be an orthogonal basis for Φ\Phi, and αi=Φ(ei,ei)\alpha_i = \Phi(e_i, e_i). Now prove that the ideals AαiA\alpha_i are, up to order, independent of the orthogonal basis considered (cf. §5§ 5, th. 1).

These ideals are called the invariant factors of the form Φ\Phi. Give an example of two forms having the same invariant factors and not equivalent (take two forms for which the quotient of the discriminants is not a square).

d) Soient FF un sous-module de EE, ΦF\Phi_F la restriction de Φ\Phi à F×FF \times F, AαiA\alpha_i (1≤i≤r1 \leq i \leq r) les facteurs invariants non nuls de Φ\Phi, rangés de sorte que αi\alpha_i divise αi+1\alpha_{i+1}, AβiA\beta_i (1≤i≤s1 \leq i \leq s) les facteurs invariants non nuls de ΦF\Phi_F, rangés de sorte que βi\beta_i divise βi+1\beta_{i+1}. Montrer que l’on a s≤rs \leq r et que βi\beta_i est multiple de αi\alpha_i pour 1≤i≤s1 \leq i \leq s (même méthode que dans l’exerc. 1 a) du § 5).

e) Suppose Φ\Phi is non-degenerate; let FF, GG be two non-isotropic submodules of EE such that F0F^0 (resp. G0G^0) is supplementary to FF (resp. GG). Suppose that the restrictions of Φ\Phi to FF and to GG are equivalent; now prove that there exists an automorphism uu of EE, leaving Φ\Phi invariant, and such that u(F)=Gu(F) = G. (Using aa), reduce to the case where F=Aa,G=Ab,Φ(a,a)=Φ(b,b)F = Aa, G = Ab, \Phi(a, a) = \Phi(b, b). Let (cj)(c_j) be a base of G0G^0, and let b′,cj′b', c'_j (1≤j≤n−11 \leq j \leq n-1) be the components of bb and cjc_j respectively in F0F^0; now prove that there exist scalars μj\mu_j (1≤j≤n−11 \leq j \leq n-1) such that the elements dj=cj′+μjb′d_j = c'_j + \mu_j b' satisfy the relations Φ(dj,dk)=Φ(cj,ck)\Phi(d_j, d_k) = \Phi(c_j, c_k) for every pair of indices; one will note for this that for every λ∈A\lambda \in A, one of the elements 1±λ1 \pm \lambda is invertible in AA.

11) Let AA be a principal ideal domain of characteristic 0, in which there is only one principal ideal π\pi, such that 2 is divisible by π\pi. If (e1,e2)(e_1, e_2) is the canonical basis of E=A2E = A^2, Φ\Phi the symmetric bilinear form on EE defined by Φ(ξ1e1+ξ2e2,η1e1+η2e2)=ξ1η2+ξ2η1\Phi(\xi_1 e_1 + \xi_2 e_2, \eta_1 e_1 + \eta_2 e_2) = \xi_1 \eta_2 + \xi_2 \eta_1, now prove that there is no orthogonal base of EE for Φ\Phi.

12) Let AA be the finite field Fq2\mathbf{F}_{q^2}, JJ the involutive automorphism ξ→ξq\xi \to \xi^q of AA, whose field of invariants is Fq\mathbf{F}_q. If EE is a vector space of dimension nn over AA, Φ\Phi a non-degenerate hermitian sesquilinear form (for JJ) on EE, now prove that the order of the unitary group U(Φ)\mathbf{U}(\Phi) is equal to

(qn−(−1)n)qn−1(qn−1−(−1)n−1)qn−2…(q2−1)q(q+1) (q^n - (-1)^n) q^{n-1}(q^{n-1} - (-1)^{n-1}) q^{n-2} \ldots (q^2 - 1) q(q + 1)
(method analogous to that of Exercise 10 of § 5, using Exercise 3).

13) Let AA be the finite field Fq\mathbf{F}_q (qq not a multiple of 2), EE a vector space of dimension nn over AA, QQ a non-degenerate quadratic form on EE. Now prove that: a) If nn is odd, the order of the group SO(Q)\mathbf{SO}(Q) is

(qn−1−1)qn−2(qn−3−1)qn−4…(q2−1)q. (q^{n-1} - 1) q^{n-2}(q^{n-3} - 1) q^{n-4} \ldots (q^2 - 1) q.
b) If n=2mn = 2m is even, the order of the group SO(Q)\mathbf{SO}(Q) is equal to
(q2m−1−εqm−1)(q2m−2−1)q2m−3…(q2−1)q (q^{2m-1} - \varepsilon q^{m-1}) (q^{2m-2} - 1) q^{2m-3} \ldots (q^2 - 1) q
where ε=1\varepsilon = 1 if (−1)mΔ(-1)^m \Delta is a square in AA, ε=−1\varepsilon = -1 otherwise, Δ\Delta denoting the discriminant of QQ with respect to any basis of EE. (Method analogous to that of Exerc. 12, using Exerc. 3 of §6§ 6 and Exerc. 5 of Chap. V, § 11.)

14) Suppose that AA is a commutative field, EE a vector space of finite dimension n⩾2n \geqslant 2 over AA, Φ\Phi a non-degenerate hermitian sesquilinear form on EE, satisfying condition (T) (§4§ 4, No. 2). Now prove that the only endomorphisms ω\omega of EE commuting with all the automorphisms uu belonging to the special unitary group SU(Φ)\mathbf{SU}(\Phi) are the homotheties, except when one has simultaneously n=2n = 2, J=1J = 1, AA being of characteristic ≠2\neq 2. (If n⩾3n \geqslant 3, write that ω\omega commutes with the involutions u∈SU(Φ)u \in \mathbf{SU}(\Phi), and use Exerc. 3 of the §4§ 4; if n=2n = 2 and J≠1J \neq 1, write that ω\omega commutes with the elements of SU(Φ)\mathbf{SU}(\Phi) whose matrix is of the form (λ00λ−1)\begin{pmatrix} \lambda & 0 \\ 0 & \lambda^{-1} \end{pmatrix} with respect to an orthogonal basis of EE.)

§15§ 15) Let AA be a commutative field of characteristic ≠2\neq 2, EE a vector space of dimension n⩾1n \geqslant 1 over AA, QQ a non-degenerate quadratic form on EE. For every automorphism u∈O(Q)u \in \mathbf{O}(Q), let ω=u−1\omega = u - 1, and let rr be the rank of ω\omega, and W=ω‾(0)W = \overline{\omega}(0).

a) Show that ω(E)\omega(E) is the subspace W0W^0 orthogonal to WW.

b) Show that if n=2,r=2n = 2, r = 2, uu is product of two symmetries with respect to lines of EE. (Establish that if ω(x)\omega(x) is isotropic for every non-isotropic vector x∈Ex \in E, ω(x)\omega(x) is isotropic for every x∈Ex \in E; one will consider separately the case where AA has at least 5 elements and the case A=F3A = \mathbf{F}_3.)

c) Suppose that nn and rr are arbitrary. Show that if ω(E)\omega(E) is not totally isotropic, uu is product of rr symmetries with respect to hyperplanes of EE, and cannot be product of a smaller number of symmetries. (Reduce to the case where WW is totally isotropic, and proceed by induction on nn and rr. If W≠{0}W \neq \{0\}, show that there exists a vector a∈W0a \in W^0 such that ω(a)\omega(a) is not isotropic, by reasoning by contradiction and using the fact that a plane of which all the lines except at most one are isotropic is necessarily totally isotropic; take then the symmetry ss with respect to the hyperplane orthogonal to ω(a)\omega(a), and consider the automorphism susu. If W={0}W = \{0\}, take a∈Ea \in E such that ω(a)\omega(a) is not isotropic, and, with the same meaning for ss, consider again the automorphism susu, and use b).)

d) Suppose that ω(E)\omega(E) is totally isotropic. If ss is a symmetry with respect to a non-isotropic hyperplane HH, show that the subspace of vectors invariant under susu is H∩WH \cap W, hence is of dimension n−r−1n - r - 1, and deduce that susu cannot be product of fewer than r+1r + 1 symmetries with respect to hyperplanes. Deduce then from c) that uu is product of r+2r + 2 symmetries with respect to hyperplanes, but cannot be product of a smaller number of symmetries.

e) Deduce from c) and d) that every orthogonal automorphism is product of at most nn symmetries with respect to hyperplanes.

f) Show that if nn is odd (resp. even), for every automorphism u∈O(Q)u \in \mathbf{O}(Q) of determinant 1 (resp. – 1), there exists a vector x≠0x \neq 0 invariant under uu (use e)).

16) The hypotheses being the same as in Exerc. 15, show that, if n⩾3n \geqslant 3, the group SO(Q)\mathbf{SO}(Q) is generated by the symmetries with respect to the non-isotropic subspaces of EE of dimension n−2n - 2 (reason as in Prop. 5 of No. 4).

¶ 17) The hypotheses are the same as in Exerc. 15.

a) Show that, for n⩾2n \geqslant 2, the commutator group Ω(Q)\Omega(Q) of the orthogonal group O(Q)O(Q) is generated by the elements (st)2(st)^2, where ss and tt range over the set of symmetries with respect to hyperplanes (use Prop. 5 of No. 4, and observe that for every group Γ\Gamma, the subgroup generated by the squares of the elements of Γ\Gamma contains the commutator group of Γ\Gamma).

b) Show that if n⩾3n \geqslant 3, the commutator group of SO(Q)SO(Q) is generated by the squares of the elements of SO(Q)SO(Q) (use Exerc. 16); deduce that this group is identical with Ω(Q)\Omega(Q), and that the quotient group SO(Q)/Ω(Q)SO(Q)/\Omega(Q) is a commutative group all of whose elements are of order 2.

c) A plane P⊂EP \subset E is said to be hyperbolic if it is non-isotropic and if it contains isotropic lines (necessarily 2 in number). An automorphism u∈O(Q)u \in O(Q) is said to be hyperbolic if there exists a hyperbolic plane PP such that u(x)=xu(x) = x for all x∈P0x \in P^0; uu is then said to be a hyperbolic transform associated with PP. Show that if QQ is of index ⩾1\geqslant 1, every u∈O(Q)u \in O(Q) is a product of hyperbolic transforms (use Prop. 5 of No. 4 and Exerc. 4 a) of § 4). Deduce that if PP is a hyperbolic plane, every u∈O(Q)u \in O(Q) can be written u=tφu = t \varphi, where tt is a hyperbolic transform associated with PP and φ∈Ω(Q)\varphi \in \Omega(Q).

**¶ 18) Let A be a commutative field, E a vector space of dimension nn over A, Φ\Phi a non-degenerate hermitian sesquilinear form on E, satisfying condition (T) (§4§ 4, No. 2). Let V be a vector subspace of E, HvH_v the subgroup of the unitary group U(Φ)U(\Phi) formed by the unitary automorphisms uu such that u(V)=Vu(V) = V.

a) Show that, when V is not a totally isotropic subspace of dimension n/2n/2, the image of HvH_v by the mapping u→det⁡uu \to \det u is the subgroup of A∗A^* formed by the ρ∈A\rho \in A such that ρρ‾=1\rho \overline{\rho} = 1.

b) If nn is even and if V is a totally isotropic subspace of dimension n/2n/2, show that the image of HvH_v by the mapping u→det⁡uu \to \det u is the subgroup of A∗A^* formed by the elements of the form λˉ/λ\bar{\lambda}/\lambda (use Prop. 2 of the §4§ 4, No. 2).

c) Let V, W be two vector subspaces of E such that the restrictions of Φ\Phi to V and W are equivalent. Show that there exists u∈SU(Φ)u \in SU(\Phi) such that u(V)=Wu(V) = W in the following cases: 1° J is distinct from the identity (use Th. 3 of Chap. V, § 11, No. 5). 2° J=1J = 1, A is of characteristic ≠2\neq 2, V and W are not totally isotropic subspaces of dimension n/2n/2.

d) Suppose that J=1J = 1, that A is of characteristic ≠2\neq 2, that n=2mn = 2m is even, and that Φ\Phi is a nondegenerate symmetric bilinear form of index mm. Let V, W be two totally isotropic subspaces of dimension mm in E; prove that if dim⁡(V∩W)=q\dim(V \cap W) = q, then, for every orthogonal automorphism uu such that u(V)=Wu(V) = W, one has det⁡u=(−1)m−q\det u = (-1)^{m-q} (use b) and prop. 2 of the §4§ 4, no 2). Deduce that the set of totally isotropic subspaces of dimension mm is the union of two classes of intransitivity N1,N2N_1, N_2 for the group SU(Φ)SU(\Phi); if V and W belong to the same class (resp. to different classes), the dimension of V ∩ W has the same parity as m (resp. does not have the same parity as m). For a similitude u (for Φ) to be direct, it is necessary and sufficient that u(N1)=N1u(N_1) = N_1 (use exerc. 4 c) of § 4.

19) Let A be a field, E a finite-dimensional vector space and > 0 over A, Φ an ε-hermitian sesquilinear form on E, nondegenerate and nonalternating. Let u be an automorphism of E such that one has

Φ(u(x),u(x))=αΦ(x,x) \Phi(u(x), u(x)) = \alpha \Phi(x, x)

for all x∈Ex \in E, with α∈A\alpha \in A. Prove that u is a similitude of multiplier α\alpha except when the following conditions are simultaneously satisfied: A is commutative and of characteristic 2, J is the identity (use exerc. 8 of § 1).

20) Let A be a field, L a finite-dimensional hermitian space over A; suppose that the metric form Φ of L satisfies condition (T) (§ 4, no 2). Prove that if the index of Φ is > 0, there may be similitudes of L, of multiplier ≠1\neq 1, and which admit no fixed point (use the reasoning of prop. 6 of no 6, and prop. 2 of § 4, no 2).

21) Let A be a commutative field of characteristic ≠2\neq 2, L a finite-dimensional euclidean space over A, Φ the metric form of L.

a) Prove that every bijection u of L onto itself, such that

Φ(u(x)−u(y),u(x)−u(y))=Φ(x−y,x−y) \Phi(u(x) - u(y), u(x) - u(y)) = \Phi(x - y, x - y)

whatever x, y in L may be, is a displacement (use exerc. 7 of § 1).

b) Prove that the group of displacements is generated by the symmetries with respect to the nonisotropic hyperplanes of the affine space L (using prop. 5 of no 4, reduce oneself to proving that every nonisotropic translation is the product of two such symmetries).

22) In a hermitian space L, two linear varieties are said to be perpendicular if their directions are weakly orthogonal subspaces (§ 3, exerc. 11). Suppose L finite-dimensional; let V₁, V₂ be two linear varieties, W₁, W₂ their respective directions. Suppose that p=dim⁡(W1+W2)<np = \dim(W_1 + W_2) < n; prove that if W1+W2W_1 + W_2 is not isotropic, there exists at least one linear variety U of dimension n−pn - p, perpendicular to V₁ and to V₂, and meeting each of the linear varieties V₁, V₂ in a single point; moreover, if q=dim⁡(W1∩W2)q = \dim(W_1 \cap W_2), the union of all the linear varieties U having the preceding properties is a linear variety of dimension n−p+qn - p + q.

23) Let A be a commutative field of characteristic ≠2\neq 2, E a finite-dimensional vector space of dimension n+1≥2n + 1 \geq 2 over A, Q a quadratic form on E, Φ the symmetric bilinear form associated with Q. The set C of the x∈Ex \in E such that Q(x)=0Q(x) = 0 is called the isotropic cone with vertex 0 and equation Q(x)=0Q(x) = 0. If it is not reduced to 0, the image S of C – {0} in the projective space P(E)\mathbf{P}(E), by the canonical mapping π\pi of E−{0}E - \{0\} onto P(E)\mathbf{P}(E) (chap. II, 2e éd., App. III), is called projective quadric (resp. projective conic if n=2n = 2) of homogeneous equation Q(x)=0Q(x) = 0. One says that S is degenerate if Q is degenerate. One says that two projective linear varieties V1,V2V_1, V_2 of P(E)\mathbf{P}(E) are conjugate with respect to S if π‾(V1)\overline{\pi}(V_1) and π‾(V2)\overline{\pi}(V_2) are orthogonal (for Φ\Phi). The polar V0V^0 of a projective linear variety V⊂P(E)V \subset \mathbf{P}(E) with respect to S is the variety such that π‾(V0)∪{0}\overline{\pi}(V^0) \cup \{0\} is the totally orthogonal subspace (for Φ\Phi) to π‾(V)∪{0}\overline{\pi}(V) \cup \{0\}; if V is a hyperplane and if S is nondegenerate, V0V^0 is reduced to a point, called the pole of V. A projective linear variety V is said to be tangent to S if π‾(V)∪{0}\overline{\pi}(V) \cup \{0\} is an isotropic subspace (for Φ\Phi).

Suppose in what follows that S is nonempty and nondegenerate.

a) Prove that the intersection of S and a projective linear variety V is empty or is a quadric in V; for this quadric to be degenerate, it is necessary and sufficient that V be tangent to S.

b) Prove that the tangent hyperplane to S at a point z∈Sz \in S is the union of the lines passing through z and tangent to S.

c) Suppose z∉Sz \notin S. For every line D passing through z and meeting S at two points a,ba, b (distinct or not), let z′z' be the harmonic conjugate of z with respect to aa and bb, that is to say, the point of D such that [abz′z]=−1\begin{bmatrix} a & b \\ z' & z \end{bmatrix} = -1 (chap. II, 2e éd., App. III, exerc. 4); prove that z′z' belongs to the polar hyperplane of z with respect to S, and that there exist n of these points forming a projectively free family in P(E)\mathbf{P}(E) and belonging to S (cf. § 4, exerc. 4 a)).

d) Suppose that n=3n = 3 and that Φ\Phi is of maximum index v=2v = 2. The set of lines contained in S is then the union of two sets N1,N2N_1, N_2 such that every line of N1N_1 meets every line of N2N_2, but two distinct lines of N1N_1 (resp. N2N_2) do not meet (exerc. 18 d)). Let D, D’ be two distinct lines belonging to N1N_1; for every z∈Dz \in D there exists one and only one line Δ∈N2\Delta \in N_2 passing through z; if u(z)u(z) is the point where Δ\Delta meets D’, now prove that uu is a projective linear mapping of D onto D’.

e) Supposing still n=3n = 3, let D, D’, D’’ be three lines of P(E)\mathbf{P}(E) of which any two do not meet. Now prove that the union of the lines meeting D, D’ and D’’ is a nondegenerate quadric.

24) The hypotheses and notations are those of exerc. 23, the quadric S being supposed nonempty and nondegenerate.

a) Now prove that the subgroup Γ\Gamma of the projective group PGL(E)\mathbf{PGL}(E) formed by the projective linear bijections transforming S into itself, is the canonical image of the group of similitudes relative to Q. (Use exerc. 23 c) above, exerc. 2 a) of § 4 and exerc. 8 of § 1.)

b) Let aa be a point of P(E)\mathbf{P}(E) not belonging to S, and let Φ1\Phi_1 be the restriction of Φ\Phi to the orthogonal hyperplane to π‾(a)\overline{\pi}(a) in E. Now prove that the subgroup of Γ\Gamma leaving aa invariant is isomorphic to the quotient of the orthogonal group U(Φ1)\mathbf{U}(\Phi_1) by its center.

c) Let bb be a point of SS, FF the (isotropic) hyperplane orthogonal to π‾(b)\overline{\pi}(b) in EE, MM a nonisotropic supplementary space of π‾(b)\overline{\pi}(b) with respect to FF, and Φ2\Phi_2 the restriction of Φ\Phi to MM. Now prove that the subgroup of Γ\Gamma leaving bb invariant is isomorphic to the group of similitudes of a Euclidean space LL of dimension n−1n - 1, having as metric form the inverse form (§1§ 1, no 7) of Φ2\Phi_2. (Observe that if a similitude for Φ\Phi transforms the line π‾(b)\overline{\pi}(b) into itself, it transforms FF into itself, and is entirely determined by its restriction to FF).

25) Let AA be a commutative field of characteristic ≠2\neq 2, LL a finite-dimensional affine space of dimension n⩾2n \geqslant 2 over AA. We identify LL with the complement of a projective hyperplane H0H_0 (“hyperplane at infinity”) of a projective space P(E)\mathbf{P}(E) of dimension nn (chap. II, 2nd ed., App. III, No. 4). We say that a nonempty set S⊂LS \subset L is an affine quadric (resp. affine conic if n=2n = 2) if SS is the intersection of LL and a projective quadric (resp. conic) in P(E)\mathbf{P}(E) (exerc. 23).

(a) Show that if there exists a non-degenerate projective quadric S‾⊂P(E)\overline{S} \subset \mathbf{P}(E) such that S=L∩S‾S = L \cap \overline{S}, this quadric is the only one having these properties, unless n=2n = 2, A=F3A = \mathbf{F}_3 and SS is reduced to 2 elements (observe that outside this exceptional case, for every point z∈H0z \in H_0 not belonging to S‾\overline{S}, there exists a right passing through zz and meeting SS in two distinct points). We then say that SS is a non-degenerate affine quadric. We say that two affine linear varieties V1,V2V_1, V_2 contained in LL are conjugate with respect to SS if the projective linear varieties V‾1,V‾2\overline{V}_1, \overline{V}_2 such that Vi=L∩V‾i (i=1,2)V_i = L \cap \overline{V}_i \ (i = 1, 2) are conjugate with respect to S‾\overline{S}; we define analogously the polar (when it is not contained in H0H_0) or the pole of an affine linear variety with respect to SS, and the affine linear varieties tangent to SS.

(b) We suppose that SS is non-degenerate; show that one can take an origin aa in LL such that, by identifying LL in this way with a vector space, there is a basis (ei)(e_i) of LL such that SS is the set of the x=∑i=1nξieix = \sum_{i=1}^n \xi_i e_i satisfying one of the two equations of the forms

α1ξ12+⋯+αnξn2=1α1ξ12+⋯+αn−1ξn−12+ξn=0. \alpha_1 \xi_1^2 + \cdots + \alpha_n \xi_n^2 = 1 \\ \alpha_1 \xi_1^2 + \cdots + \alpha_{n-1} \xi_{n-1}^2 + \xi_n = 0.

In the first case, the point aa is well determined and is the pole with respect to S‾\overline{S} of the hyperplane at infinity H0H_0 (called the center of SS). (Distinguish two cases according as H0H_0 is or is not tangent to S‾\overline{S}; use Theorem 1 of § 6, No. 1 and Proposition 2 of § 4, No. 2.)

26) Let AA be a commutative algebraically closed field of characteristic ≠2\neq 2, EE a finite-dimensional vector space over AA, QQ a non-degenerate quadratic form on EE. Let u∈O(Q)u \in \mathbf{O}(Q); with the notation of exerc. 12 of § 4, we have G(p,p)={0}G(p, p) = \{0\} unless p(X)=X−1p(X) = X - 1 and p(X)=X+1p(X) = X + 1. Let MM be a minimal element of the set of non-isotropic subspaces contained in G(p,p)G(p, p) and stable under uu, and let php^h be the minimal polynomial of the restriction of uu to MM. Now prove that if hh is odd, MM is an indecomposable submodule of EuE_u, and that if hh is even, MM is a direct sum of two isomorphic indecomposable submodules of EuE_u. (To see that if h=2kh = 2k is even, MM cannot be indecomposable, show that N=pk(u)(M)N = p^k(u)(M) would then be totally isotropic; if (ei)1≤i≤2k(e_i)_{1 \leq i \leq 2k} is a basis of MM such that u(ei)=εei+ei+1u(e_i) = \varepsilon e_i + e_{i+1} for i≤2k−1i \leq 2k - 1, u(e2k)=εe2ku(e_{2k}) = \varepsilon e_{2k} (with ε=±1\varepsilon = \pm 1), show that eke_k cannot be orthogonal to ek+1e_{k+1}, and deduce that the relation Q(u(ek))=Q(ek)Q(u(e_k)) = Q(e_k) leads to a contradiction).

27) Let AA be a commutative field of characteristic 2, EE a vector space over AA, of finite dimension nn, QQ a quadratic form on EE, Φ\Phi the associated bilinear form, which is alternating, hence of even rank 2m2m (§5§ 5, no 1, cor. 1 of th. 1).

a) Prove that if E0E^0 is the subspace of EE (of dimension n−2mn - 2m) orthogonal to EE for Φ\Phi, one has Q(λx+μy)=λ2Q(x)+μ2Q(y)Q(\lambda x + \mu y) = \lambda^2 Q(x) + \mu^2 Q(y) for all x,yx, y in E0E^0; in other words, the restriction Q0Q_0 of QQ to E0E^0 is a semilinear mapping from E0E^0 (considered as a vector space over AA) into AA (considered as a vector space over the subfield A2A^2), relative to the isomorphism ξ→ξ2\xi \to \xi^2 of AA onto A2A^2. Let qq be the dimension (over AA) of the kernel E0∩Q‾(0)E^0 \cap \overline{Q}(0) of QQ, and let E1E_1 be a supplementary subspace of E0∩Q‾(0)E^0 \cap \overline{Q}(0) with respect to E0E^0; one has n−2m−q≤[A:A2]n - 2m - q \leq [A : A^2].

b) Deduce from a) that there exists a base (ei)1≤i≤n(e_i)_{1 \leq i \leq n} of EE, the first 2m2m vectors of which form a base of a supplementary subspace E2E_2 of E0E^0 in EE, the next n−2m−qn - 2m - q vectors a base of E1E_1, such that, for x=∑i=1nξieix = \sum_{i=1}^n \xi_i e_i

Q(x)=∑i=1m(αiξi2+ξiξm+i+βiξm+i2)+∑i=2m+1n−qγiξi2 Q(x) = \sum_{i=1}^m (\alpha_i \xi_i^2 + \xi_i \xi_{m+i} + \beta_i \xi_{m+i}^2) + \sum_{i=2m+1}^{n-q} \gamma_i \xi_i^2

the γi\gamma_i (2m+1≤i≤n−q2m + 1 \leq i \leq n - q) being elements of AA linearly independent with respect to A2A^2.

c) The index of QQ is called the maximum dimension of the totally singular subspaces VV of EE such that V∩E0={0}V \cap E^0 = \{0\}. Prove that if vv is the index of QQ, one can take the base (ei)(e_i) of EE having the properties stated in b) in such a way that αi=βi=0\alpha_i = \beta_i = 0 for 1≤i≤v1 \leq i \leq v and that the restriction of QQ to the subspace of E2E_2 generated by ev+1,…,em,em+v+1,…,e2me_{v+1}, \ldots, e_m, e_{m+v+1}, \ldots, e_{2m} is a quadratic form (non-degenerate) of index 0.

d) We suppose q=0q = 0; let O(Q)O(Q) be the automorphism group of EE leaving QQ invariant. If u∈O(Q)u \in O(Q), now prove that u(x)=xu(x) = x for every x∈E0x \in E^0. For every x∈E2x \in E_2, let u(x)=u0(x)+u2(x)u(x) = u_0(x) + u_2(x), where u0(x)∈E0u_0(x) \in E^0 and u2(x)∈E2u_2(x) \in E_2; show that u2u_2 belongs to the symplectic group Sp(Φ2)Sp(\Phi_2) (where Φ2\Phi_2 is the restriction of Φ\Phi to E2E_2) and that Q(u2(x))+Q(x)∈Q(E0)Q(u_2(x)) + Q(x) \in Q(E^0). Conversely, for every automorphism u2∈Sp(Φ2)u_2 \in Sp(\Phi_2) such that Q(u2(x))+Q(x)∈Q(E0)Q(u_2(x)) + Q(x) \in Q(E^0) for every x∈E2x \in E_2, show that there exists one and only one linear mapping u0u_0 of E2E_2 into E0E^0 such that the linear mapping equal to u0+u2u_0 + u_2 in E2E_2, to the identity in E0E^0, belongs to O(Q)\mathbf{O}(Q).

e) We suppose that AA is a perfect field (A2=AA^2 = A) and that q=0q = 0. Deduce from b) that every vector subspace of EE, of dimension ⩾3\geqslant 3, contains at least one vector xx such that Q(x)=0Q(x) = 0. If nn is odd, one necessarily has m=νm = \nu and n=2m+1n = 2m + 1, so that there exists a basis (ei)(e_i) of EE with respect to which one has

Q(∑i=1nξiei)=ξ1ξm+1+⋯+ξmξ2m+ξ2m+12, Q(\sum_{i=1}^n \xi_i e_i) = \xi_1 \xi_{m+1} + \cdots + \xi_m \xi_{2m} + \xi_{2m+1}^2,

and (with the notations of dd)) O(Q)\mathbf{O}(Q) is isomorphic to Sp(Φ2)\mathbf{Sp}(\Phi_2); all quadratic forms such that q=0q = 0 are then equivalent. If nn is even, one necessarily has n=2m,ν=mn = 2m, \nu = m or ν=m−1\nu = m - 1, and there exists a basis (ei)(e_i) of EE with respect to which one has

(1)Q(∑i=1nξiei)=ξ1ξm+1+⋯+ξm−1ξ2m−1+ξmξ2m+λ(ξm2+ξ2m2) (1) \quad Q(\sum_{i=1}^n \xi_i e_i) = \xi_1 \xi_{m+1} + \cdots + \xi_{m-1} \xi_{2m-1} + \xi_m \xi_{2m} + \lambda(\xi_m^2 + \xi_{2m}^2)

where λ∈A\lambda \in A. Let A1A_1 be the field obtained by adjoining to AA the roots of the polynomial λX2+X+λ\lambda X^2 + X + \lambda; prove that this field is independent of the basis (ei)(e_i) with respect to which QQ can be written in the form (1), and that in order that two quadratic forms (such that q=0q = 0) be equivalent, it is necessary and sufficient that the quadratic extensions of AA which correspond to them in this way be identical (use Witt's theorem). Case where AA is a finite field of characteristic 2.

¶ 28) Let AA be a commutative field of characteristic 2, distinct from F2\mathbf{F}_2, EE a vector space of dimension n=2mn = 2m over AA, QQ a non-degenerate quadratic form on EE.

a) Prove that the orthogonal group O(Q)\mathbf{O}(Q) is generated by the symmetries (which here are nothing other than the transvections belonging to O(Q)\mathbf{O}(Q) (§4§ 4, exercise 6)) (reason as in exercise 11 of §5§ 5). Deduce that the commutator group of O(Q)\mathbf{O}(Q) is generated by the squares of the elements of O(Q)\mathbf{O}(Q) (cf. exercise 17).

b) Suppose that QQ is of maximum index; let V,WV, W be two totally singular subspaces of EE (§4§ 4, No. 1) of dimension mm. Let uu be a symmetry x→x+Φ(x,a)Q(a)ax \to x + \frac{\Phi(x, a)}{Q(a)} a (§4§ 4, exercise 6) ; let kk be the dimension of V∩WV \cap W. Prove that the dimension of V∩u(W)V \cap u(W) is k+1k + 1 if aa is orthogonal to V∩W,k−1V \cap W, k - 1 otherwise (in the first case remark that a=x+ya = x + y, or x∈V,y∈Wx \in V, y \in W, and prove that u(y)=xu(y) = x; in the second, remark that uu cannot leave invariant any non-orthogonal singular vector to aa).

c) Suppose again that the index of QQ is arbitrary. Prove that the subgroup SO(Q)\mathbf{SO}(Q) of O(Q)\mathbf{O}(Q), formed by the automorphisms of EE which are products of an even number of symmetries, is a distinguished subgroup of index 2 of O(Q)\mathbf{O}(Q). (Prove that the product of an odd number of symmetries cannot be the identity, by considering the extension of QQ to the es-

Bourbaki XXIV.

pace vectoriel E' obtained by extension of the field of scalars of E to its algebraic closure ; use then b).) (Cf. § 9, exercise 9.)

d) If V₁, V₂ are two totally singular subspaces of E, of the same dimension < m, prove that there exists an automorphism u ∈ SO(Q) such that u(V₁) = V₂. On the contrary, if V₁ and V₂ are two totally singular subspaces of dimension m, in order that there exists an automorphism u ∈ SO(Q) such that u(V₁) = V₂, it is necessary and sufficient that the dimension of V₁ ∩ V₂ have the same parity as m (reason as in exercise 18, using b)).

e) A plane P ⊂ E is said to be hyperbolic if it is non-isotropic and contains singular lines (necessarily 2 in number). A transformation u ∈ O(Q) is said to be hyperbolic if there exists a hyperbolic plane P such that u(x) = x for every x ∈ P⁰ ; one then says that u is a hyperbolic transformation associated with P. Prove that if Q is of index > 0, every u ∈ O(Q) is a product of hyperbolic transformations (use a)). Deduce that if P is a hyperbolic plane, every transformation u ∈ O(Q) can be written u = sv, where s is a hyperbolic transformation associated with P and v belongs to the commutator group of O(Q).

29) The hypotheses being those of exercise 28, suppose in addition that Q is of maximum index m; let (eᵢ) be a symplectic basis of E (for the alternating form Φ associated with Q) formed by singular vectors (§ 4, No. 2, Proposition 2), so that the matrix of Φ with respect to this basis is the matrix denoted R in exercise 14 of § 5. With the notations of this last exercise, prove that, in order that a symplectic matrix (tD + S)⁻¹(D + S) be the matrix of an automorphism u ∈ O(Q), it is necessary and sufficient that S be alternating (write that every vector u(eᵢ) is singular, remarking that one has (tD + S).u(eᵢ) = (D + S).eᵢ).

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

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