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The Hodge index theorem on a blowup of the projective plane

Example

Assume the Axiom of Choice. Let $k$ be a field, let $X = P_{k}^{2}$ , let $p \in X (k)$ be a $k$ -rational point, let $X^{'} = {Bl}_{p} X$ be the blowup with exceptional curve $E$ and put $ℓ = π^{*} O_{X} (1)$ as in The Picard group of a point blowup of the projective plane, so that $Pic (X^{'}) = Z ℓ \oplus Z E$ with $ℓ^{2} = 1$ , $ℓ \cdot E = 0$ , $E^{2} = - 1$ and the strict transform $m = ℓ - E$ of a line through $p$ satisfies $m \cdot E = 1$ , $m^{2} = 0$ . Then:

The class $H : = ℓ$ has $H \cdot H = 1 > 0$ , so the Hodge index theorem The Hodge index theorem for smooth projective surfaces applies to $H$ even though $H$ is not ample ( $H \cdot E = 0$ ).
$H^{⊥} = R E$ and $E \cdot E = - 1 < 0$ ; thus the primitive part is negative definite of rank one, in agreement with Negative definiteness of the primitive part of the Neron-Severi space.
The intersection form on $N_{R}^{1} (X^{'})$ has matrix $(\begin{array}{cc} 1 & 0 \\ 0 & - 1 \end{array})$ in the basis $(ℓ, E)$ , of signature $0$ and index $(1, 1)$ ; the class $E$ is not numerically trivial because $E \cdot m = 1$ , so the negative direction in the Hodge index theorem is strict.

Facts & Assumptions

Given: a field $k$ , the plane $X = P_{k}^{2}$ , a $k$ -rational point $p$ , the blowup $X^{'} = {Bl}_{p} X$ with exceptional curve $E$ , the class $ℓ = π^{*} O_{X} (1)$ , and the strict transform $m$ of a line through $p$ .

[F1]

Blowup data: $X^{'}$ is an integral smooth projective surface over $k$ ; $Pic (X^{'}) = Z ℓ \oplus Z E$ ; $ℓ^{2} = 1$ , $ℓ \cdot E = 0$ , $E^{2} = - 1$ ; and the strict transform $m$ of a line through $p$ satisfies $π^{*} L = m + E$ for such a line $L$ , so $m = ℓ - E$ in the Picard group, $m \cdot E = 1$ and $m^{2} = 0$ by bilinearity (The Picard group of a point blowup of the projective plane, The intersection matrix of a point blowup of a regular surface, Total transform equals strict transform plus multiplicity times the exceptional divisor, The surface intersection product is symmetric and bilinear, Effective cartier divisor).

[F2]

Numerical space and definiteness data: $N_{R}^{1} (X^{'})$ is the real extension of $Pic (X^{'})$ modulo numerical equivalence; the parametrization $(a, b) \mapsto a ℓ + b E$ is an isomorphism at the level of Picard groups, and the intersection form on the real space is the bilinear extension, with inertia, rank and signature as in Positive and negative definiteness, the inertia $(p, q, r)$ , rank $p + q$ , and signature $p - q$ of a real symmetric bilinear or quadratic form (Numerical equivalence and the Neron-Severi space of a surface, Intersection numbers of Cartier divisors on a smooth projective surface).

[F3]

Hodge index theorem and its corollary: if $H \cdot H > 0$ and $L \cdot H = 0$ for invertible sheaves, then $L \cdot L \leq 0$ with equality exactly for numerically trivial $L$ ; and for $H \cdot H > 0$ the form is negative definite on the primitive part $h^{⊥}$ of the real Neron-Severi space (The Hodge index theorem for smooth projective surfaces, Negative definiteness of the primitive part of the Neron-Severi space).

Verification

Given: the data of the statement, with $ℓ$ , $E$ , $m$ as in [F1].

1.1F1F3

Positive self-intersection of $H = ℓ$ . By [F1], $ℓ^{2} = 1 > 0$ ; in particular $ℓ$ is not numerically trivial, since a numerically trivial class would have square $0$ by definition (Numerical equivalence and the Neron-Severi space of a surface). The Hodge index theorem [F3] therefore applies with $H = ℓ$ . The class $ℓ$ is not ample: $ℓ \cdot E = 0$ with $E$ a nonzero effective Cartier divisor, while an ample class meets every nonzero effective divisor positively (The intersection matrix of a point blowup of a regular surface for effectivity of $E$ ; compare Ample divisors meet nonzero effective divisors positively).

2.1F1F3step 1.1

The primitive part. Let $x = a ℓ + b E$ be a real class. By bilinearity and [F1], $x \cdot H = x \cdot ℓ = a (ℓ \cdot ℓ) + b (E \cdot ℓ) = a$ , so $x \cdot H = 0$ if and only if $a = 0$ ; hence $H^{⊥} = R E$ . On this line $E \cdot E = - 1 < 0$ , so the form is negative definite of rank one, in agreement with [F3]; in particular $E \cdot E = 0$ does not occur and the equality case of the Hodge index theorem is not met by a nonzero class of $H^{⊥}$ .

3.1F1F2step 2.1

The intersection matrix and its signature. In the basis $(ℓ, E)$ of the real vector space $N_{R}^{1} (X^{'})$ , which spans by [F1] and is independent because pairing a relation $a ℓ + b E = 0$ with $ℓ$ and $E$ gives $a = 0$ and $- b = 0$ , the Gram matrix is $(\begin{array}{cc} ℓ \cdot ℓ & ℓ \cdot E \\ E \cdot ℓ & E \cdot E \end{array}) = (\begin{array}{cc} 1 & 0 \\ 0 & - 1 \end{array})$ by [F1]; it is nondegenerate with eigenvalues $\pm 1$ , hence of inertia $(1, 1, 0)$ , index $(1, 1)$ and signature $0$ (Positive and negative definiteness, the inertia $(p, q, r)$ , rank $p + q$ , and signature $p - q$ of a real symmetric bilinear or quadratic form). The class $E$ is not numerically trivial: $m$ is an invertible sheaf class with $E \cdot m = 1 \neq 0$ [F1], so $E$ does not pair to zero against every class.

4.1step 1.1step 2.1step 3.1∎

Conclusion. Step 1.1 gives $H \cdot H > 0$ for the non-ample class $H = ℓ$ ; step 2.1 identifies $H^{⊥} = R E$ with negative definite form $E^{2} = - 1$ ; and step 3.1 computes the matrix, signature $0$ and index $(1, 1)$ and shows $E$ is numerically nontrivial, so the negative direction is strict. This verifies all three claims.

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Sources

Ravi Vakil, The Rising Sea: Foundations of Algebraic Geometry, pre-publication version 2025-10-21 (standard reference, not scraped)