Negative definiteness of the primitive part of the Neron-Severi space

The orthogonal decomposition. For $x\in {\mathrm{N}}_{\mathbb{R}}^1(X)$ put $a:=(x\cdot h)\mathrm{/}(h\cdot h)$, a real number because $h\cdot h>0$ by hypothesis (Positive and negative definiteness, the inertia $(p,q,r)$, rank $p+q$, and signature $p-q$ of a real symmetric bilinear or quadratic form). Then $(x-ah)\cdot h=x\cdot h-a(h\cdot h)=0$, so $x_0:=x-ah\in h^{\perp }$ and $x=ah+x_0$. If $x=a^{\mathrm{\prime }}h+x_0^{\mathrm{\prime }}$ is another such decomposition, then $(a-a^{\mathrm{\prime }})h=x_0^{\mathrm{\prime }}-x_0\in h^{\perp }$ and both summands are orthogonal to $h$, so $(a-a^{\mathrm{\prime }})(h\cdot h)=((a-a^{\mathrm{\prime }})h)\cdot h=0$ and $a=a^{\mathrm{\prime }}$, whence $x_0=x_0^{\mathrm{\prime }}$; orthogonality of $\mathbb{R}h$ and $h^{\perp }$ is the definition of the latter and $h\cdot h\mathrm{\ne }0$. This proves (1).

Negative semidefiniteness on the primitive part. Suppose $x\in h^{\perp }$ has $x\cdot x>0$. Write $x$ as a finite real combination of classes $[L_1],\dots ,[L_r]$ of invertible sheaves, consider the real span $V$ of $h,[L_1],\dots ,[L_r]$ and the rational span $W$ of the $[L_i]$ inside $V$; the map $p(y):=y-\frac{y\cdot h}{h\cdot h}h$ is $\mathbb{R}$-linear with rational coefficients on $W$, so $p(W)$ is a dense $\mathbb{Q}$-subspace of $p(V)=V\cap h^{\perp }$. The set $\{y\in V\cap h^{\perp }:y\cdot y>0\}$ is open in $V\cap h^{\perp }$ and nonempty because it contains $x$, hence contains some $y=p(w)\in p(W)$. Clearing denominators of the rational coefficients of $y$ produces an integer $N\ge 1$ and an element $z=Ny\in {\mathrm{N}}^1(X)$ of the form $\sum a_i[L_i]+a_{0}h$ with $a_i,a_0\in \mathbb{Z}$, which is the class of the invertible sheaf ${⨂}_i{L}_i^{\otimes a_i}\otimes H^{\otimes a_0}$; moreover $z\cdot h=N(y\cdot h)=0$ and $z\cdot z=N^2(y\cdot y)>0$. The Hodge index theorem [F2] applied to $H$ and this invertible sheaf gives $z\cdot z\le 0$, a contradiction. Hence $x\cdot x\le 0$ for every $x\in h^{\perp }$.

The equality case. Let $x\in h^{\perp }$ satisfy $x\cdot x=0$. If $x\mathrm{\ne }0$, nondegeneracy [F1] supplies $z$ with $z\cdot x\mathrm{\ne }0$. Put $z^{\mathrm{\prime }}:=z-\frac{z\cdot h}{h\cdot h}h\in h^{\perp }$, so $z^{\mathrm{\prime }}\cdot x=z\cdot x\mathrm{\ne }0$. For $t\in \mathbb{R}$, the class $z^{\mathrm{\prime }}+tx$ lies in $h^{\perp }$ and has square $z^{\mathrm{\prime }}\cdot z^{\mathrm{\prime }}+2t(z^{\mathrm{\prime }}\cdot x)$, which is positive for $t$ of a suitable sign and sufficiently large magnitude. This contradicts step 1.2. Hence $x\cdot x=0$ forces $x=0$; the converse is immediate by bilinearity. Together with step 1.2, this proves negative definiteness.

The signature statement. Assume $\rho (X)={\dim }_{\mathbb{R}}{\mathrm{N}}_{\mathbb{R}}^1(X)$ is finite. By (1) the form decomposes as the orthogonal direct sum of $\mathbb{R}h$, on which it is positive definite because $h\cdot h>0$, and of $h^{\perp }$, on which it is negative definite by step 2.1; the decomposition is nondegenerate with $\dim h^{\perp }=\rho (X)-1$ (if $\rho (X)=0$ the space is zero and the statement is vacuous, while $\rho (X)\ge 1$ here because $h\mathrm{\ne }0$). Hence the inertia is $(1,\rho (X)-1,0)$, the rank is $\rho (X)$ and the signature is $1-(\rho (X)-1)=2-\rho (X)$ by [F4]. The Axiom of Choice is inherited from [F5].