Theorem 1 (Witt)

Let E and $E'$ be two vector spaces of finite dimensions, provided respectively with two non-degenerate $\varepsilon$-hermitian forms $\Phi$ and $\Phi'$ satisfying condition (T) of No. 2 (resp. with two non-degenerate quadratic forms Q and Q'), and isomorphic for these structures. Given an arbitrary subspace F of E, every injective metric homomorphism from F into E' extends to a metric isomorphism from E onto E'.

Using the given isomorphism from E onto E', one sees that it suffices to show that every injective metric homomorphism u from F into E extends to a metric automorphism of E. Let us remark that if, for i = 1, 2, F_i is a subspace of E and u_i a metric homomorphism from F_i into E, such that F_1 ∩ F_2 = {0} and that Φ(u_1(x_1), u_2(x_2)) = Φ(x_1, x_2) for x_i ∈ F_i (i = 1, 2), then the homomorphism ψ : x_1 + x_2 → u_1(x_1) + u_2(x_2) from F_1 + F_2 into E which extends u_1 and u_2 is metric: indeed, whatever x_i, y_i in F_i (i = 1, 2), the expansion of each of the expressions Φ(x_1 + x_2, y_1 + y_2) and Φ(u_1(x_1) + u_2(x_2), u_1(y_1) + u_2(y_2)) (resp. Q(x_1 + x_2) and Q(u_1(x_1) + u_2(x_2))) contains four (resp. three) terms equal each to each according to the assumptions made. Moreover, if u_1 and u_2 are injective and if u_1(F_1) ∩ u_2(F_2) = {0}, then ψ is injective.

1) We first prove the Witt theorem in the case where the set of elements invariant by u is a hyperplane U of F. The set of vectors of the form u(x) − x with x ∈ F is then a right D. If F' is a subspace orthogonal to D such that F' ∩ F = F' ∩ u(F) = {0}, we shall have Φ(u(x), y) = Φ(x, y) for x ∈ F and y ∈ F' ; our initial remark therefore applies to u and to the identical mapping of F' into E, showing that u extends to F + F' by leaving the points of F' fixed; the set of vectors of the form u(x) − x (x ∈ F + F') is still the right D. Now we have, for x ∈ F, y ∈ F

(5) $\Phi(u(x), u(y) - y) = \Phi(u(x), u(y)) - \Phi(u(x), y) = \Phi(x - u(x), y)$,

which, when x ∈ U (that is to say when u(x) = x), shows that x ∈ D^0; in other words we have U ⊂ D^0. We shall distinguish two cases:

a) $F \not\subset D^0$. Formula (5) shows that $u(F)$ is not contained in $D^0$, hence $F \cap D^0 = u(F) \cap D^0 = U$. One may then take for F' a supplementary subspace of U in D^0; since F + F' contains the hyperplane D^0 and is distinct from it, we have F + F' = E, and in this case we have found the required extension of u to E.

b) F ⊂ D^0. Formula (5) shows that u(F) ⊂ D^0, and therefore that

D ⊂ D^0 ; the right D is therefore isotropic (resp. singular, since we have Q(u(x) − x) = Q(u(x)) − Φ(x, u(x)) + Q(x) = 2Q(x) − Φ(x, x) = 0 for x ∈ F). We shall show that, under these conditions, there exists a subspace F' of D^0 which is supplementary to F and u(F) in D^0. This is immediate if F = u(F). Otherwise, let x and y be vectors such that x ∈ F, x ∉ U, y ∈ u(F), y ∉ U; we then have F = U + Ax, u(F) = U + Ay, and F does not contain x + y otherwise y = (x + y) − x would belong to F ∩ u(F) = U; one sees analogously that x + y does not belong to u(F); thus the right A(x + y) is supplementary to F and u(F) in the subspace F + u(F); it suffices then to put F' = A(x + y) + G where G is supplementary to F + u(F) in D^0. This being so, we have F + F' = u(F) + F' = D^0, and, in this case, what was said at the beginning of 1) shows that there exists an extension of u to the hyperplane D^0 of E, and that D^0 is stable for this extension.

We are therefore reduced to the case where F is the hyperplane D^0 and where u is an automorphism of F. We prove that, for every z ∈ E, there exists z' ∈ E such that

(6)

for every x ∈ F; indeed the linear form x → Φ(u^{-1}(x), z) on F is the restriction of a linear form on E, a form which is of the type x → Φ(x, z') since Φ is non-degenerate; hence (6) holds. Moreover, if z ∉ F, there exists a vector z' ∈ E satisfying (6) and such that Φ(z', z') = Φ(z, z) (resp. Q(z') = Q(z)): indeed formula (6) remains valid if one adds to z' an element u(y) − y (y ∈ F) of D since F = D^0, and Lemma 1 of No. 2 allows one to conclude since z is not orthogonal to D. Our initial remark then shows that there exists a metric homomorphism ϕ from F + Az = E into E which extends u and which transforms z into z'. Since Φ is non-degenerate, ϕ is the required metric automorphism of E.

2) In the general case, we reason by induction on r = dim F. The case r = 0 is trivial. Let then r > 0, that is to say F ≠ {0}, and let U be a hyperplane of F. The restriction u_0 of u to U extends, by the induction hypothesis, to a metric automorphism ϕ_0 of E. If ϕ_0 extends u, the theorem is proved. Otherwise U is the set of elements invariant by v_0^{-1}u, and there exists, according to 1), a metric automorphism v_1 of E extending v_0^{-1}u. The automorphism v_0v_1 is then the required extension of u. QED.