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On a connected compact Kähler manifold of complex dimension two, the intersection form restricted to real $(1,1)$ cohomology has signature $(1,h^{1,1}-1)$. The Kähler class spans a positive line, and its orthogonal complement consists of primitive classes, which are negative definite. Thus a totally isotropic subspace of real $(1,1)$ cohomology has dimension at most one.

For a real quadratic form on an $n$-dimensional vector space, with rank $r$ and signature $s$ defined as positive minus negative index, the maximum dimension of a totally isotropic subspace is $n-(r+|s|)/2$. Quotienting by the radical of a bilinear form leaves a nondegenerate form of indices $p,q$. Projection of an isotropic subspace to each sign-coordinate space is injective, bounding its dimension by $\min (p,q)$. The radical and vectors pairing one positive with one negative unit coordinate attain the bound $n-r+\min (p,q)$.

Associate to $f$ the symmetric bilinear form $B(u,v)=\frac{1}{2}(f(u+v)-f(u)-f(v))$. If $f$ is identically zero, every basis is diagonal. Otherwise choose $v$ with $f(v)\ne 0$. Every $w$ has the decompositionwhose second term lies in $v^{\perp }$. Thus $V=\mathbb{R}v\oplus v^{\perp }$, with no cross term in $f$. Induction on dimension gives a diagonal basis. Rescaling its nonzero entries reduces the form towith $k$ additional zero coordinates. Define the rank of a quadratic form $r=p+q$ and its signature $s=p-q$. Rank is intrinsic, since it is the rank of $v\mapsto B(v,\cdot )$, equivalently $n-\dim \mathrm{rad}B$. Also $p$ is the greatest dimension of a subspace on which the restriction of $f$ is a positive-definite quadratic form: any such subspace intersects the negative-plus-zero coordinate space only at zero, so its dimension is at most $p$, while the positive coordinate space attains $p$. The corresponding negative statement identifies $q$. This proves the basis independence of $r$ and $s$, the Sylvester's law of inertia conclusion.

For a totally isotropic subspace $W$ on which $f$ vanishes, polarization identity also makes $B$ vanish on $W\times W$. Quotient by the radical of a bilinear form. In the resulting nondegenerate form, projection of an isotropic subspace to either the positive or negative coordinate space is injective, so its dimension is at most $\min (p,q)$. Consequently $\dim W\le k+\min (p,q)$. The radical of a bilinear form together with the vectors $e_i^{+}+e_i^{-}$, $1\le i\le \min (p,q)$, attains this bound. The maximum dimension of a totally isotropic subspace is thereforeThe criterion for membership in a diagonal basis follows similarly: a nonzero vector can belong to a diagonal basis if it has nonzero quadratic value, by the same orthogonal splitting, or if it lies in the radical of a bilinear form, by choosing a radical basis containing it. An isotropic vector in a diagonal basis must be orthogonal to every basis vector, and hence must lie in that radical. For the specified form, its radical is the $z$-axis, giving exactlyThe zero vector never belongs to a basis.

Use the usual connected-manifold convention for this assertion. Suppose, for contradiction, that $H^2(M,{\mathcal{O}}_M)=0$. The Dolbeault theorem gives $H^{0,2}=0$, and the Hodge decomposition theorem for compact Kähler manifolds and its conjugation symmetry give $H^{2,0}=0$. Hence every real class in $H^2(M,\mathbb{R})$ has a real harmonic differential form of type $(1,1)$ as its representative.

The intersection form is $Q([u],[v])={\int }_{M}u\wedge v$. Its value on the Kähler class is positive: $Q([\omega ],[\omega ])={\int }_M{\omega }^2>0$. Every real harmonic $(1,1)$ form splits uniquely as $c\omega +u_0$, with $u_0$ primitive and $c$ constant. Indeed, $\Lambda u$ is a harmonic function, hence constant by connectedness, and $\Lambda \omega =2$; subtracting $c\omega$ kills $\Lambda u$, and part (b) makes $u_0$ primitive.

Part (c) gives $\ast u_0=-u_0$, while $\omega \wedge u_0=0$. ThusThis proves the Hodge index theorem for compact Kähler surfaces in this case: $Q$ has signature $(1,b_2-1)$.

On the alleged two-dimensional totally isotropic subspace $V$, the linear functional $v\mapsto Q(v,[\omega ])$ has a nonzero vector in its kernel. Its harmonic differential form representative is primitive, so the preceding negative-definiteness statement gives $Q(v,v)<0$, contradicting the vanishing of $Q$ on $V$. ThereforeConnectedness is essential if one's definition of a manifold allows disconnected spaces. For $M=Y\bigsqcup Y$, with $Y={\mathbb{CP}}^1\times {\mathbb{CP}}^1$, one has $H^2(M,{\mathcal{O}}_M)=0$. Take on each component the pullback of the degree-two class from its first factor. Each class has square zero, and their mutual pairing is zero because their supports lie on different components. Their span is then a two-dimensional totally isotropic subspace, disproving the unqualified disconnected version.