Theorem 1 (« law of inertia »)

Suppose that $A$ satisfies the hypotheses at the beginning of this paragraph, and that $E$ is of finite dimension $n$. Let $\Phi$ be a hermitian form on $E$. Then:

a) There exists a decomposition of $E$ into a direct sum of the subspace $E^0$ orthogonal to $E$, and of two subspaces $E^+$ and $E^-$ such that the restriction of $\Phi$ to $E^+$ (resp. $E^-$) is positive (resp. negative) and non-degenerate.

b) There exists an orthogonal basis $(e_i)_{1 \leq i \leq n}$ of $E$ such that

c) The dimensions $s$ of $E^+$ and $t$ of $E^-$ are the same for all the decompositions into a direct sum satisfying the conditions stated in a) ; the integer $s$ (resp. $t$) is the maximum of the dimensions of the subspaces $F$ of $E$ such that the restriction of $\Phi$ to $F$ is positive (resp. negative) and non-degenerate.

d) The rank of $\Phi$ is $s + t$.

e) If $\Phi$ is non-degenerate, its index is equal to $\inf(s, t)$ (§ 4, no 2, def. 2).

Indeed, let $(x_i)$ ($i = 1, \ldots, n$) be an orthogonal basis of $E$ (§ 6, No. 1, Theorem 1); recall that $\Phi(x, x) \in K$ for every $x \in E$. We may suppose that $\Phi(x_i, x_i) > 0$ for $i = 1, \ldots, s$, $\Phi(x_i, x_i) < 0$ for $i = s + 1, \ldots, s + t$ and $\Phi(x_i, x_i) = 0$ for $i = s + t + 1, \ldots, n$. This proves a), since we take for $E^+$ (resp. $E^-$) the subspace generated by $x_1, \ldots, x_s$ (resp. by $x_{s+1}, \ldots, x_{s+t}$), and $E^0$ is then generated by $x_{s+t+1}, \ldots, x_n$. We deduce b) from this by observing that $\Phi(x_i, x_i)$ is of the form $\rho \overline{\rho}$ (resp.

- $\rho \bar{\rho} )\ (\rho \in A^*)$ for $i = 1, \ldots, s$ (resp. $i = s + 1, \ldots, s + t$). To prove c), consider a subspace P of E such that the restriction of $\Phi$ to P is positive and non-degenerate; we then have $P \cap (E^- + E^0) = \{0\}$, and the sum $P + E^- + E^0$ is therefore direct; we conclude that $\dim P \leq \dim E^+ = s$, and this proves c). Assertion d) follows at once from a).

Suppose finally that $\Phi$ is non-degenerate, put $q = \inf (s, t)$, and show that q is the index of $\Phi$. With the notations of b), the vectors $e_i + e_{s+i}$ (resp. $e_i - e_{s+i}$) ($i = 1, \ldots, q$) generate a totally isotropic subspace F (resp. $F'$). Since K is of characteristic 0, $F + F'$ is generated by $e_1, \ldots, e_q, e_{s+1}, \ldots, e_{s+q}$ and is therefore non-isotropic. The restriction of $\Phi$ to the subspace $H = (F + F')^0$ is then positive (or negative) and non-degenerate, and H contains therefore no isotropic vector $\neq 0$. Since $F^0$ contains F and H, and since $\dim(F + H) = \operatorname{codim} F' = \operatorname{codim} F$, we have $F^0 = F + H$. Thus an isotropic vector z orthogonal to F is necessarily in F, because, in the direct sum $F + H$, the component of z in H is isotropic. Consequently F is a maximal totally isotropic subspace, and this proves e).