Proposition 3

*Let $\Phi$ be a non-degenerate $\varepsilon$-hermitian form satisfying (T) (resp. the bilinear form associated with a non-degenerate quadratic form Q). Let $F_1$ and $F_2$ be two maximal totally isotropic (resp. totally singular) subspaces of E, one of them being finite-dimensional. Put $F = F_1 \cap F_2$. Let $S_i \ (i = 1, 2)$ be a supplementary subspace to $F$ in $F_i$; put $S = S_1 + S_2$. Then there exist two subspaces G and H of E such that a) The subspaces $G + F, S$ and H are non-isotropic and pairwise orthogonal; b) E is the direct sum of F, S, G and H; c) There is no nonzero isotropic (resp. singular) vector in H; d) G is totally isotropic (resp. totally singular). Moreover $F_1$ and $F_2$ are both finite-dimensional and we have dim $F_1 = \dim F_2, \dim G = \dim F, \dim S_1 = \dim S_2, \operatorname{codim} H = 2 \dim F_1$.

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.