Indeed, since the restriction $\Phi'$ of $\Phi$ to F is non-degenerate by assumption, the mapping $d_{\Phi'}$ from F into its dual $F^*$ associated on the right with $\Phi'$ is injective, hence bijective since F and $F^*$ are two spaces of the same finite dimension. Consequently, for every $y \in E$, there exists one and only one element $y_0$ of F such that $\Phi(x, y) = \Phi(x, y_0)$ for every $x \in F$, that is, $y - y_0 \in F^0$; this proves that E is the direct sum of F and $F^0$.

a) F is non-isotropic. b) F^0 is non-isotropic. c) E is the direct sum of F and F^0.

Proposition 1 shows indeed that a) implies c), and that c) implies a) and b). Finally, if F^0 is non-isotropic, we have F ∩ F^0 ⊂ F^0 ∩ F^{00} = {0}; hence F is non-isotropic, which shows that b) implies a).

Put $\Phi(x, x) = \beta + \varepsilon \bar{\beta}$ (resp. $Q(x) = \beta$). For $y \in F$ we then have $\Phi(x + y, x + y) = (\beta + \Phi(x, y)) + \varepsilon (\beta + \Phi(x, y))$ since $\Phi(y, y) = 0$ (resp. $Q(x + y) = \beta + \Phi(x, y)$ since $Q(y) = 0$). Since $x$ is not orthogonal to F, the affine linear function $y \to \Phi(x, y) + \beta$ on F is not constant; it therefore takes the value $\alpha$ for some element $y$ of F, which thus answers the question.

On appelle décomposition de Witt de E toute décomposition de E en somme directe de trois sous-espaces F, F', G tels que F et F' soient totalement isotropes (resp. totalement singuliers) et que G soit non isotrope et soit orthogonal à $F + F'$; si E est de dimension finie, la matrice de $\Phi$ par rapport à une base de E adaptée à une décomposition de Witt de E se met sous la forme

(1)

On dit que $\Phi$ est une forme neutre si elle est non dégénérée et si E est de dimension finie et est somme directe de deux sous-espaces totalement isotropes (resp. totalement singuliers). La somme directe de deux formes neutres est une forme neutre.

a) Si $F'$ est un sous-espace totalement isotrope de dimension r tel que $F' \cap F^0 = \{0\}$, alors $F + F'$ est non isotrope et, pour toute base $(f_i)$ de F, il existe une base $(f'_i)$ de $F'$ telle que $\Phi(f_i, f'_j) = \delta_{ij}$ (indice de Kronecker) pour $i, j = 1, \ldots, r$.

b) Si G est un sous-espace totalement isotrope (resp. totalement singulier) de dimension $\leq r$ tel que $G \cap F^0 = \{0\}$, il existe un sous-espace totalement isotrope (resp. totalement singulier) $F' \supset G$ de dimension r tel que $F' \cap F^0 = \{0\}$.

Soit $\Psi$ la restriction de $\Phi$ à $F \times F'$; pour $x' \in F'$, la relation «$\Phi(x, x') = 0$ pour tout $x \in F$ » entraîne $x = 0$ puisque $F' \cap F^0 = \{0\}$. L’assertion a) résulte alors du cor. de la prop. 6 du § 1, No. 6, à l’exception du fait que $F + F'$ est non isotrope. Or le sous-espace $H = (F + F') \cap (F + F')^0$ est égal à $(F + F') \cap F^0 \cap {F'}^0$. Comme $F \subset F^0$, on a $(F + F') \cap F^0 = F + (F' \cap F^0) = F$, d’où $H = F \cap {F'}^0$; donc $H = \{0\}$ puisqu’on a vu que $\Psi$ est non dégénérée. Ceci prouve bien que $F + F'$ est non isotrope.

Pour démontrer b), nous procéderons par récurrence descendante sur $s = \dim G$. Il nous suffit ainsi de prouver que, si $s < r$, il existe un sous-espace totalement isotrope (resp. totalement singulier) $G'$ contenant G, de dimension $s + 1$, et tel que $G' \cap F^0 = \{0\}$. Comme $\dim G < \dim F$, la restriction de $\Phi$ à $F \times G$ est dégénérée, et comme $G \cap F^0$ est nul, $F \cap G^0$ est non nul. Si l’on avait alors $G + F^0 \supset G^0$, on en déduirait, en prenant les sous-espaces orthogonaux et en remarquant que $F = F^{00}$ et que $G = G^{00}$ ($§ 1$, No. 6, cor. 1 de la prop. 4), que $G^0 \cap F \subset G$, d’où

ce qui est impossible. Il existe alors un élément $x$ de $G^0$ tel que $x \notin G + F^0$; comme $F \subset F^0$, on peut ajouter à $x$ un vecteur de $G^0 \cap F$ sans modifier ces propriétés; comme $G^0 \cap F$ est totalement isotrope (resp. totalement singulier) et $\neq \{0\}$, le lemme 1 montre qu’on peut choisir $x$ isotrope (resp. singulier).

Then the subspace $G' = G + Ax$ is of dimension $s + 1$, and is totally isotropic (resp. totally singular); moreover we have $G' \cap F^0 = \{0\}$ since, if $y = z + ax$ ($z \in G,\ a \in A$) is in $F^0$, we have $a = 0$, for otherwise $x \in F^0 + G$, contrary to the choice of $x$, whence $y = z \in G \cap F^0 = \{0\}$ and $y = 0$. Consequently the subspace $G'$ answers the question.

It suffices to take $G = \{0\}$ in prop. 2, b).

First observe that, if $N$ is a maximal totally isotropic (resp. totally singular) subspace, then every isotropic (resp. singular) vector $x$ orthogonal to $N$ is an element of $N$, since otherwise $N + Ax$ would contradict the maximal character of $N$. Hence if, for $i = 1,\ 2,\ x_i$ is an isotropic (resp. singular) vector of $F_i^0$, we have $x_i \in F_i$. On the other hand, if $y$ is an element of $S_1$ orthogonal to $S_2$, it is orthogonal to $F_1$ since $F_1$ is totally isotropic, hence to $F$, and consequently to $F_2 = S_2 + F$. Since $y$ is isotropic (resp. singular) and is orthogonal to $F_2$, we have

One therefore has $S_1 \cap S_2^0 = \{0\}$, and analogously $S_2 \cap S_1^0 = \{0\}$. Since one of the two subspaces $F_1, F_2$, for example $F_1$, is finite-dimensional, $S_1$ is finite-dimensional, and hence $S_1^0$ is of finite codimension ($§ 1, \mathrm{n}^\circ 6, \mathrm{cor.\ 1\ de\ la\ prop.\ 4}$), and consequently $S_2$ is finite-dimensional since $S_2 \cap S_1^0 = \{0\}$; moreover this shows that $\dim S_2 \leqslant \operatorname{codim} S_1^0 = \dim S_1$; analogously $\dim S_1 \leqslant \dim S_2$, whence $\dim S_1 = \dim S_2$. Prop. 2 a) then shows that $S = S_1 + S_2$ is non-isotropic.

Now the orthogonal $N$ of $S$ is non-isotropic ($\mathrm{n}^\circ 1, \mathrm{cor.\ de\ la\ prop.\ 1}$) and contains $F$; Cor. 1 of Prop. 2 therefore shows that there exists a subspace $G$ totally isotropic (resp. totally singular) of $N$ such that $\dim G = \dim F$, that $G \cap F = \{0\}$ and that $G + F$ is non-isotropic. Thus d) is satisfied by $G$. We shall then satisfy a) and b) by taking for $H$ the orthogonal of $G + F$ in $N$. As for c), we note that, since $H$ is orthogonal to $F_1 = S_1 + F$, there is no nonzero isotropic (resp. singular) vector in $H$ by virtue of what was noted at the beginning of the proof and of the fact that $H \cap F_1 = \{0\}$. Finally some of the assertions concerning dimensions have been proved along the way; the others follow from them trivially.

If F ∩ F' = {0}, we shall have G = {0} with the notation of Prop. 3, and F + F' will be non-isotropic. The other assertions follow trivially from Prop. 3 and Cor. 1 of Prop. 2.

(2)

(3)

Indeed, let F_1 and F_2 be two maximal totally singular subspaces such that F_1 ∩ F_2 = {0} (Cor. 1) and let q be their dimension. We then have G = {0} with the notation of Prop. 3. Taking a basis (e_i)_{1 \leq i \leq q} of F_1 and a basis (e_i)_{q+1 \leq i \leq 2q} of F_2 such that Φ(e_i, e_{j+q}) = δ_{ij} for i, j = 1, ..., q (Prop. 2 a)), we see that it suffices to show that dim H ≤ 1. Now, if x ∈ H, y ∈ H and if x ≠ 0, the equation Q(y - ax) = Q(y) - aΦ(x, y) + a^2 Q(x) = 0 has at least one solution a_0 since Q(x) ≠ 0, and we have y = a_0 x since every singular vector of H is zero.

Indeed, according to Cor. 1 of Prop. 2, there exists a subspace $G = F + F'$ ($F, F'$: totally singular subspaces of dimension 1) of E, of dimension 2, such that the restriction of Q to G is neutral. If $\{ e, e' \}$ ($e \in F, e' \in F'$) is a basis of G, we have

It suffices therefore to take for $y$ the vector $ae + b^{-1}e'$.

Indeed, let u be a metric isomorphism of E'_1 onto E'_2. By Theorem 1, u extends to a metric isomorphism ϕ of E_1 onto E_2. Since Φ_i is non-degenerate, E''_i is the orthogonal of E'_i in E_i, hence ϕ maps E''_1 onto E''_2. QED.

This follows immediately from Theorem 1 and Corollary 2 of Proposition 3.

Exercises. — 1) a) Let K be a field of characteristic 2, J : ξ → ¯ξ an involutory antiautomorphism of K, Z the center of K. Show that if the restriction of J to Z is not the identity, every element μ of K such that ¯μ = μ is of the form λ + ¯λ (observe that there is an element ρ ≠ 0 in Z which can be written in the form ζ + ¯ζ, with ζ ∈ Z); every hermitian form on a vector space over K then satisfies condition (T).

b) Give examples of fields of characteristic 2, admitting an involutory antiautomorphism ξ → ¯ξ distinct from the identity map, and for which there are elements $\mu = \bar{\mu}$ which are not of the form $\lambda + \bar{\lambda}$ (cf. Chapter VIII, § 11, Exercise 4).

2) Let A be a field, E a vector space over A, $\Phi$ (resp. Q) a non-degenerate $\varepsilon$-hermitian form on E, satisfying condition (T) (resp. a non-degenerate quadratic form on E, Φ then denoting the symmetric bilinear form associated with Q).

a) Show that in order that a plane P \subset E be isotropic (resp. singular) and not totally isotropic (resp. not totally singular), it is necessary and sufficient that it contain only one isotropic (resp. singular) right (cf. Exercise 14 e)).

b) Suppose that dim $E \geqslant 3$, and that there exist in E isotropic vectors $\neq 0$. Show that if P is a non-totally isotropic plane in E, there exists a non-isotropic vector subspace V \subset E, of dimension 3, containing isotropic vectors $\neq 0$, and such that $P \subset V$.

3) The hypotheses being the same as in Exercise 2, show that if dim $E \geqslant 3$, every isotropic right in E is the intersection of two non-isotropic planes.

4) The hypotheses are those of Exercise 2, and suppose in addition that E is finite-dimensional.

a) If the index $\nu$ of $\Phi$ (resp. Q) is $\geqslant 1$, show that for every isotropic (resp. singular) vector $a \neq 0$ in E, there exists a basis $(e_i)$ of E consisting of isotropic (resp. singular) vectors, such that $e_1 = a$ (cf. Exercise 14 e)).

b) Let V, W be two totally isotropic (resp. totally singular) subspaces of the same dimension $r \leqslant \nu$; show that there exist two maximal totally isotropic (resp. totally singular) subspaces V_1, W_1, such that V \subset V_1, W \subset W_1, and $V_1 \cap W_1 = V \cap W$. (If $U = V \cap W$, reason in $U^0/U$).

c) Let V, W, $V_1, W_1$ be four totally isotropic (resp. totally singular) subspaces of the same dimension, such that $V + W$ and $V_1 + W_1$ are non-isotropic. Show that there exists a metric automorphism $u$ of E such that $u(V) = V_1$ and $u(W) = W_1$.

d) Let $f$ be a linear form on E, $\alpha$ an element of A of the form $\lambda + \varepsilon \bar{\lambda}$ (resp. an element of $\Lambda$). Consider the sesquilinear form on E

(resp. the quadratic form

Show that if $\Phi_1$ (resp. $Q_1$) is non-degenerate and if $\nu_1$ denotes its index, one has $|\nu_1 - \nu| \leqslant 1$.

5) a) Let B be a ring, $\xi \to \bar{\xi}$ an involutory antiautomorphism of B, $\varepsilon$ an element of the center of B such that $\varepsilon \bar{\varepsilon} = 1$. Show that if $\beta$ is an invertible element of B such that $\beta + \varepsilon \bar{\beta} \neq 0$, there exists an invertible element $\mu \neq 1$ in B, such that $\mu(\beta + \varepsilon \bar{\beta})\mu = \beta + \varepsilon \bar{\beta}$. (Show that one can take $\mu$ such that $\mu \beta \bar{\mu} = \beta$).

b) Let A be a field, E a vector space over A, $\Phi$ a non-degenerate $\varepsilon$-hermitian sesquilinear form on E, satisfying (T). Show that if $\Phi$ is not alternating, for every non-isotropic hyperplane $H$ of E, there exists a metric automorphism of E, distinct from the identity, leaving every element of $H$ invariant (use a).

6) The hypotheses being those of Exercise 2, let $a$ be an isotropic vector $\neq 0$ in $E$ (resp. a nonsingular isotropic vector (note that such vectors exist only if $A$ is of characteristic 2)). Let $\lambda \in A$ such that $\lambda + \varepsilon \bar{\lambda} = 0$ (resp. $\lambda = (Q(a))^{-1}$); show that the transvection $x \to x + \Phi(x, a)\lambda a$ (Chapter II, § 6, Exercise 7) is a metric automorphism of $E$; converse.

7) The hypotheses being those of Exercise 2, let $G$ be the group of metric automorphisms of $E$. Show that the only semilinear bijections of $E$ onto itself which commute with all the elements of $G$ are the homotheties of $E$, except in the following three cases: dim $E = 2$, $G$ is the group of metric automorphisms corresponding to a quadratic form of index 1 on $E$, and $A$ is one of the three fields $\mathbf{F}_2, \mathbf{F}_3$ or $\mathbf{F}_4$. (Use Exercises 5, 6 and 3; examine separately the case of a quadratic form on a vector space of dimension 2).

*8) Let $A$ be a field, $E$ a finite-dimensional vector space $> 0$ over $A$, $\Phi$ a non-degenerate $\varepsilon$-hermitian sesquilinear form on $E$, satisfying (T). Let $M(\Phi)$ be the group of multipliers of the similarities of $E$ for $\Phi$ ($§ 6, \mathrm{n}^o\ 5$).

a) Let $V_1, V_2$ be two vector subspaces of $E$, of the same dimension, and let $\Phi_1, \Phi_2$ be the restrictions of $\Phi$ to $V_1, V_2$ respectively. For there to exist a similarity $u$ such that $u(V_1) = V_2$, it is necessary and sufficient that there exist $\alpha \in M(\Phi)$ such that $\Phi_2$ is equivalent to $\alpha \Phi_1$ (use Witt's theorem).

b) Let $(F, F', G)$ be a Witt decomposition of $E$ ($\mathrm{n}^o\ 2$), and let $\Phi_0$ be the restriction of $\Phi$ to the non-isotropic subspace $G$. Show that $M(\Phi) = M(\Phi_0)$ if $G \neq \{0\}$. (Use Witt's theorem and Proposition 2 of $\mathrm{n}^o\ 2$).

c) Now prove that if the index $v$ of $\Phi$ is such that dim $E = 2v$, $M(\Phi)$ is the group of elements $\zeta \neq 0$ of the center of $A$ such that $\bar{\zeta} = \zeta$. If dim $E = 2v + 1$, $M(\Phi)$ is the group of elements of the form $\rho \bar{\rho}$, where $\rho$ runs through the multiplicative group of the elements $\neq 0$ of the center of $A$ (use Witt's theorem). (Cf. § 10, exerc. 18).*

9) Let $A$ be a commutative field, $E$ a vector space over $A$, $Q$ a non-degenerate quadratic form on $E$. A similarity for $Q$ is any automorphism $u$ of $E$ such that there exists an element $\alpha \neq 0$ of $A$ for which $Q(u(x)) = \alpha Q(x)$ for all $x \in E$; $u$ is then also a similarity for the bilinear form associated with $Q$. Assuming the dimension of $E$ to be finite and $> 0$, state and exercise prove, for the similarities relative to $Q$, the analogues of the results of exerc. 8.

*10) Let $A$ be a field, $E$ a vector space over $A$ of dimension $> 2$, $\Phi_1$ (resp. $\Phi_2$) a non-degenerate sesquilinear form $\varepsilon_1$-hermitian (resp. $\varepsilon_2$-hermitian) on $E$ for an involutory antiautomorphism $J_1$ (resp. $J_2$) of $A$, satisfying condition (T). Now prove that if the group of metric automorphisms of $E$ for $\Phi_1$ is a subgroup of the group of similarities for $\Phi_2$, there exists $\alpha \in A$ such that $\Phi_2 = \Phi_1 \alpha$ (use exerc. 5 b) and 6).

Exercise prove the analogous property when A is assumed commutative, and when $\Phi_1$ and $\Phi_2$ are replaced in the statement by two non-degenerate quadratic forms $Q_1, Q_2$ on E.*

11) Let A be a ring, J an involutory antiautomorphism of A, E an A-module admitting a finite basis $(e_i)$, $\Phi$ a non-degenerate $\varepsilon$-hermitian form on E, R the matrix of $\Phi$ with respect to $(e_i)$; the group of metric automorphisms of $\Phi$ is identified with the group G of invertible matrices U such that $^tU . R . U^j = R$.

a) Suppose that there exists a matrix P such that $R = ^tP + \varepsilon P^j$. Now prove that for every matrix S such that $^tS + \varepsilon S^j = 0$, and such that $P + S$ is invertible, $U = (^tP^j - \varepsilon^j S)^{-1}(P + S)$ belongs to G, and $\varepsilon I + U$ is invertible. Conversely (show that for every matrix $U \in G$ such that $\varepsilon I + U$ is invertible, we have

b) Show that the condition in a) is satisfied when $\Phi$ satisfies condition (T). Case where in A the equation $2\xi = \alpha$ has one and only one solution for every $\alpha \in A$.

¶ 12) Let A be a commutative field, E a finite-dimensional vector space of dimension n over A, $\Phi$ a nondegenerate $\varepsilon$-Hermitian sesquilinear form on E.

a) Let u be an endomorphism of E; show that if the $r_i (1 \leq i \leq m)$ are the similarity invariants of u (Chapter VII, § 5, No. 1, Def. 1), the similarity invariants of the adjoint $u^*$ of u with respect to $\Phi$ are the polynomials $\bar{r}_i (1 \leq i \leq m)$, where $\bar{r}_i$ is obtained from $r_i$ by applying the automorphism J to each coefficient (cf. Chapter VII, § 5, exerc. 2). For every monic irreducible polynomial $p \in A[X]$ dividing the minimal polynomial of u, let $F_k(u, p)$ be the kernel of $(p(u))^k$ in E, and let $F(u, p)$ be the union of the $F_k(u, p)$ for all integers $k > 0$. Show that if p and q are distinct monic irreducible polynomials dividing the minimal polynomial of u, the subspaces $F(u, p)$ and $F(u^*, \bar{q})$ are orthogonal (use Bezout's identity). Finally, if G is a vector subspace of E such that $u(G) \subset G$, we have $u^*(G^0) \subset G^0$.

b) Suppose that $uu^* = u^*u$ (the case where u is said to be a normal endomorphism for $\Phi$; cf. § 7, No. 3); show that we then have $u^*(F_k(u, p)) \subset F_k(u, p)$ for every k, and consequently $u^*(F(u, p)) \subset F(u, p)$. If we put $G(p, \bar{q}) = F(u, p) \cap F(u^*, \bar{q})$, show that E is the direct sum of the subspaces $G(p, \bar{q})$, and that $G(p, \bar{q})$ and $G(p_1, \bar{q}_1)$ are orthogonal if $p \neq q_1$ or if $p_1 \neq q$; in particular $G(p, \bar{q})$ is totally isotropic if $p \neq q$. Show that $G(p, \bar{p})$ is reduced to 0 or is nonisotropic, and that if $p \neq q$, no nonzero vector of $G(p, \bar{q})$ is orthogonal to $G(q, \bar{p})$ (use the fact that $\Phi$ is nondegenerate); deduce that if $p \neq q$, $G(p, \bar{q})$ and $G(q, \bar{p})$ are totally isotropic subspaces of the same dimension, and that $G(p, \bar{q}) + G(q, \bar{p})$ is nonisotropic.

c) Suppose that $J$ is not the identity or that $A$ is not of characteristic 2, and that $u^* = u$. Let $\mathscr{M}$ be the set of non-isotropic subspaces $M \subset G(p, \bar{p})$, stable under $u$ (hence submodules of the $A[X]$-module $E_u$ (Chapter VII, § 5, No. 1)). Now prove that if $M$ is a minimal element of $\mathfrak{M}$, $M$ is an indecomposable submodule of $E_u$ (Chapter VII, § 4, No. 7). (Suppose that $M$ is the direct sum of an indecomposable submodule $M_1$ and a submodule $M_2 \neq \{0\}$, the minimal polynomials $p^h$ and $p^k$ of the restrictions of $u$ to $M_1$ and $M_2$ respectively being such that $h \geq k$. Observe then that $M_1$ is necessarily isotropic and that every $z \neq 0$ in $M_1$ such that $p(u).z = 0$ is orthogonal to $M_1$ (use the fact that every submodule of $M_1$ is monogenic); write $z = (p(u))^{h-1}.x$ and that $z$ is not orthogonal to $M_2$, and deduce that necessarily $k = h$. Now prove that there exists an indecomposable submodule $N_2$ of $M_2$ such that $p^h$ is the minimal polynomial of the restriction of $u$ to $N_2$, and that $M_1 + N_2$ is non-isotropic; conclude that $M_2 = N_2$. Finally, if $y \in M_2$ is not orthogonal to $z$, consider the submodule $P$ of $M$ generated by $w = x + \lambda y$, where $\lambda \in \mathbf{A}$, and prove that $\lambda$ can be chosen so that $P$ is non-isotropic, by proving that $\Phi((p(u))^{h-1}.w, w) \neq 0$; which leads to a contradiction).

d) Deduce from c) that $G(p, \overline{p})$ is the direct sum of pairwise orthogonal indecomposable submodules $H_i$. If $p^h$ is the minimal polynomial of the restriction of $u$ to $H_i$, and if $d$ is the degree of $p$, prove that there exists in $H_i$ a totally isotropic subspace of dimension $d.[h/2]$. Case where $E$ contains no isotropic vector $\neq 0$ (cf. § 7, No. 3).

e) State and prove the analogous properties to those of c) and d) when $u^* = -u$ or $u^*u = 1$.

f) Give an example where $n = 4$, $\Phi$ is symmetric and of index 2, $p = \overline{p} = X - 1$, $u$ is normal, $E = G(p, \overline{p})$, but $E$ is not the direct sum of minimal submodules of $\mathfrak{M}$, and where there exists an eigenvector of $u$ which is not an eigenvector of $u^*$ (cf. § 7, No. 3).

13) The hypotheses are those of Exercise 2, and suppose in addition that $E$ admits a countable basis $(e_n)$. Let $F$ be a totally isotropic (resp. totally singular) subspace of $E$ such that $F^{00} = F$; now prove that there exists a totally isotropic (resp. totally singular) subspace $F'$ such that: $1^\circ \ F \cap F' = \{0\}$; $2^\circ$ there exists a basis $(a_m)_{m \in I}$ of $F$ and a basis $(a'_m)_{m \in I}$ of $F'$ (the interval of $\mathbf{N}$ with origin 0) such that $\Phi(a_i, a'_j) = \delta_{ij}$ for every pair of indices; $3^\circ$ $(F + F')^{00} = F + F'$ and $E$ is the direct sum of $F + F'$ and $G = (F + F')^0$. (Form by induction an increasing sequence $(L_n)$ of non-isotropic subspaces, whose union is $E$, such that $\dim L_{n+1} - \dim L_n = 2$, and apply Proposition 2 of No. 2 to each of the $L_n$; to form this sequence, consider, for each $n$, the smallest integer $k$ such that $e_k \notin L_n$, and use Exercise 9 b) of § 1).

14) Let $A$ be a field of characteristic 2, $E$ a finite-dimensional vector space of dimension $n$ over $A$, $\Phi$ a non-degenerate hermitian form on $E$, not necessarily satisfying condition (T).

a) Show that the set $V$ of the $x \in E$ such that $\Phi(x, x)$ is of the form $\alpha + \overline{\alpha}$ is a vector subspace of $E$.

b) Let $V_1 = V \cap V^0$, $q = \dim V_1$, $V_2$ a supplementary space to $V_1$ with respect to $V$, $V_3$ a supplementary space to $V_1$ with respect to $V^0$. Now prove that there exists a base $(e_i)_{1 \leq i \leq 2q}$ of $(V_2 + V_3)^0 = V_2^0 \cap V_3^0$ such that the vectors $e_1, \ldots, e_q$ form a base of $V_1$ and that $\Phi(e_i, e_{q+j}) = \delta_{ij}$ for $1 \leq i \leq q, 1 \leq j \leq q$.

c) Let $G(\Phi)$ be the group of metric automorphisms of $E$ (for $\Phi$). Now prove that for every $u \in G(\Phi)$, we have $u(x) = x$ for every $x \in V^0$.

d) For every $u \in G(\Phi)$, we have $u(V) = V$; let $u_v$ be the restriction of $u$ to $V$, and let $G_v$ be the group formed by the $u_v$. Now prove that: $1^o$ the kernel of the homomorphism $u \to u_v$ from $G(\Phi)$ onto $G_v$ is commutative; $2^o$ if $\Phi_2$ is the restriction of $\Phi$ to $V_2$ and $G(\Phi_2)$ the group of metric automorphisms of $V_2$ for $\Phi_2$, there exists a homomorphism from $G_v$ onto $G(\Phi_2)$ whose kernel is commutative (use $b$ and $c$).

e) Suppose that $A$ is commutative and $J$ is the identity; let $E$ be a vector space of dimension 3 over $A$, $(e_i)_{1 \leq i \leq 3}$ a base of $E$, $\Phi$ the non-degenerate symmetric form on $E$ whose matrix with respect to $(e_i)$ is

Show that all isotropic vectors in $E$ are contained in the hyperplane generated by $e_2$ and $e_3$ (cf. exerc. 4 a)). Give an example of a non-isotropic plane containing only one isotropic right (cf. exerc. 2 a)). Show that there is no automorphism $u \in G(\Phi)$ such that $u(e_1) = e_1 + e_2$, though we have $\Phi(e_1, e_1) = \Phi(e_1 + e_2, e_1 + e_2)$.