$\operatorname{NS}(\mathbb{P}^2) = \mathbb{Z} \cdot H$ with $H^2 = 1$. The intersection form is the $1 \times 1$ matrix $(1)$, which has signature $(1, 0)$ since $\rho = 1$. The Hodge Index Theorem is vacuous: $H^{\perp} = 0$.

The Hodge inequality gives: for any curve $C$ of degree $d$ and any curve $D$ of degree $e$, $(de)^2 \geq d^2 \cdot e^2$. This is just $d^2 e^2 \geq d^2 e^2$, an equality. Since $\rho = 1$, every divisor is proportional to $H$.

$\operatorname{NS}(\mathbb{P}^1 \times \mathbb{P}^1) = \mathbb{Z} F_1 \oplus \mathbb{Z} F_2$ with $F_1^2 = F_2^2 = 0$ and $F_1 \cdot F_2 = 1$. The intersection matrix in the basis $\{F_1, F_2\}$ has entries $Q_{11} = Q_{22} = 0$ and $Q_{12} = Q_{21} = 1$.

The eigenvalues are $+1$ and $-1$ (eigenvectors $F_1 + F_2$ and $F_1 - F_2$), confirming signature $(1, 1)$.

Hodge inequality check: Take $C = F_1 + F_2$ (the diagonal, with $C^2 = 2 > 0$) and $D = aF_1 + bF_2$. Then $(C \cdot D)^2 = (a + b)^2$ and $C^2 \cdot D^2 = 2 \cdot 2ab = 4ab$. The inequality $(a+b)^2 \geq 4ab$ is equivalent to $(a - b)^2 \geq 0$ ✓.

$\operatorname{NS}(\operatorname{Bl}_P(\mathbb{P}^2)) = \mathbb{Z} H \oplus \mathbb{Z} E$ with $H^2 = 1$, $E^2 = -1$, $H \cdot E = 0$. The intersection matrix is $\operatorname{diag}(1, -1)$, which has signature $(1, 1)$ directly.

Let $D = aH - bE$. Then $D \cdot H = a$ and $D^2 = a^2 - b^2$. If $D \cdot H = 0$, then $a = 0$ and $D^2 = -b^2 \leq 0$, with equality iff $b = 0$ ✓.

Blowing up $\mathbb{P}^2$ at $n$ points in general position gives $\operatorname{NS}(X) = \mathbb{Z} H \oplus \mathbb{Z} E_1 \oplus \cdots \oplus \mathbb{Z} E_n$ with $H^2 = 1$, $E_i^2 = -1$, $H \cdot E_i = 0$, $E_i \cdot E_j = 0$ for $i \neq j$. The intersection matrix is $\operatorname{diag}(1, -1, -1, \ldots, -1)$, directly exhibiting signature $(1, n)$.

For $n = 6$: the del Pezzo surface of degree $3$ (a cubic surface in $\mathbb{P}^3$) has $\rho = 7$, intersection form $\operatorname{diag}(1, -1, -1, -1, -1, -1, -1)$, and contains the famous $27$ lines.

For $n = 8$: the del Pezzo surface of degree $1$ has $\rho = 9$, and the anti-canonical class $-K = 3H - E_1 - \cdots - E_8$ satisfies $(-K)^2 = 9 - 8 = 1$.

A K3 surface over $\mathbb{C}$ has $H^2(X, \mathbb{Z}) \cong \Lambda_{K3}$, the K3 lattice, which is isometric to $U^3 \oplus E_8(-1)^2$, a unimodular lattice of rank $22$ and signature $(3, 19)$.

The Neron--Severi group $\operatorname{NS}(X) = H^{1,1}(X) \cap H^2(X, \mathbb{Z})$ has rank $\rho$ with $1 \leq \rho \leq 20$. By the Hodge Index Theorem, the intersection form on $\operatorname{NS}(X)$ has signature $(1, \rho - 1)$.

• Generic K3 ($\rho = 1$): $\operatorname{NS}(X) = \mathbb{Z} H$ with $H^2 = 2d$ for some $d \geq 1$.
• $\rho = 2$ example: The Kummer surface of a generic abelian surface has $\rho = 17$ (not $2$!). A K3 with $\rho = 2$ could have NS lattice with Gram matrix having entries $2d, a$ on the first row and $a, -2e$ on the second, for appropriate $d, e, a$.
• $\rho = 20$ (singular K3): The Fermat quartic $x^4 + y^4 + z^4 + w^4 = 0$ over $\overline{\mathbb{F}_3}$ achieves $\rho = 22$ (the Tate conjecture), but over $\mathbb{C}$ the maximum is $\rho = 20$.

The Hirzebruch surface $\mathbb{F}_n$ has $\operatorname{NS}(\mathbb{F}_n) = \mathbb{Z} C_0 \oplus \mathbb{Z} F$ with $C_0^2 = -n$, $F^2 = 0$, $C_0 \cdot F = 1$.

The intersection matrix has $Q_{11} = -n$, $Q_{12} = Q_{21} = 1$, $Q_{22} = 0$, with $\det(Q) = -1$ and eigenvalues $\frac{-n \pm \sqrt{n^2 + 4}}{2}$. One eigenvalue is positive, one is negative, confirming signature $(1, 1)$.

In the basis $H = C_0 + nF$ (which is nef with $H^2 = n$) and $F$: the matrix entries become $n, 1, 1, 0$ respectively, still with one positive and one negative eigenvalue.

Hodge inequality check: $C_0^2 = -n < 0$, so we cannot apply the Hodge inequality with $C = C_0$. Taking $C = C_0 + nF$ (with $C^2 = n > 0$ for $n \geq 1$) and $D = F$: $(C \cdot D)^2 = 1$ and $C^2 \cdot D^2 = n \cdot 0 = 0$, so $1 \geq 0$ ✓.

Let $f: X \to \mathbb{P}^1$ be an elliptic fibration with a singular fiber of Kodaira type $I_n$ (a cycle of $n$ rational curves $C_1, \ldots, C_n$, each with $C_i^2 = -2$, meeting in a cycle: $C_i \cdot C_{i+1} = 1$ for indices mod $n$).

The fiber is $F = C_1 + C_2 + \cdots + C_n$ with $F^2 = \sum_i C_i^2 + 2\sum_{i} C_i \cdot C_{i+1} = -2n + 2n = 0$ ✓.

The intersection matrix of the $C_i$ is the negative of the Cartan matrix of type $\tilde{A}_{n-1}$, which is negative semi-definite with $1$-dimensional kernel spanned by $(1, 1, \ldots, 1)$, confirming Zariski's lemma.

For a type $I_1$ fiber (a nodal rational curve $C$ with $C^2 = -1$... actually, $C^2 = 0$ since it is the entire fiber): a single irreducible fiber has $F = C$ with $C^2 = 0$.

Consider a Kodaira type $IV^*$ fiber ($\tilde{E}_6$): the fiber $F = C_0 + 2C_1 + 3C_2 + 2C_3 + C_4 + 2C_5 + C_6$ where all $C_i^2 = -2$. The intersection matrix of $C_0, \ldots, C_6$ is the negative of the extended Cartan matrix of $E_6$, which is negative semi-definite with kernel spanned by the coefficients $(1, 2, 3, 2, 1, 2, 1)$. One verifies $F^2 = 0$, and any proper sub-combination $D = \sum d_i C_i$ not proportional to $F$ satisfies $D^2 < 0$.

Let $f: \operatorname{Bl}_P(X) \to X$ with exceptional divisor $E$. Then $E^2 = -1 < 0$. Any effective divisor $D = aE$ ($a > 0$) supported on the exceptional locus has $D^2 = -a^2 < 0$ ✓.

After blowing up two points, the exceptional locus is $E_1 \cup E_2$ with $E_1^2 = E_2^2 = -1$ and $E_1 \cdot E_2 = 0$. The intersection matrix $\operatorname{diag}(-1, -1)$ is negative definite ✓.

Resolving an $A_n$ surface singularity (locally $xy = z^{n+1}$) produces a chain of $n$ exceptional curves $E_1, \ldots, E_n$ with $E_i^2 = -2$ and $E_i \cdot E_{i+1} = 1$. The intersection matrix is the negative of the Cartan matrix of $A_n$, which has eigenvalues $-2 + 2\cos(\pi k/(n+1))$ for $k = 1, \ldots, n$. All eigenvalues are negative, confirming negative definiteness.

For $n = 1$: a single $(-2)$-curve. For $n = 2$: the matrix has diagonal entries $-2, -2$ and off-diagonal entries $1, 1$, with eigenvalues $-1$ and $-3$.

For $\mathbb{P}^1 \times \mathbb{P}^1$, $V = \mathbb{R}^2$ with the form $Q(a,b) = 2ab$. The positive cone is $\{(a,b) : ab > 0\}$, which has two components; the one containing $(1,1)$ is $\{a > 0, b > 0\}$. The projectivization $\mathbb{H}^1$ is parametrized by $t = a/b \in (0, \infty)$, a copy of the real line (the hyperbolic line).

The ample cone is $\{a > 0, b > 0\}$ (divisors of bidegree $(a,b)$ with $a, b > 0$), and the nef cone boundary consists of the rays $\mathbb{R}_{\geq 0} F_1$ and $\mathbb{R}_{\geq 0} F_2$ (the two rulings), which lie on the light cone.

For a K3 surface with $\rho = 2$ and NS lattice generated by $H, C$ with $H^2 = 4$, $C^2 = -2$, $H \cdot C = 1$: the positive cone in $\mathbb{R}^2$ is given by $(aH + bC)^2 = 4a^2 - 2b^2 + 2ab > 0$.

The ample cone is cut out by requiring additionally $D \cdot C > 0$ (i.e., $-2b + a > 0$) for the $(-2)$-curve $C$. So the ample cone is $\{4a^2 - 2b^2 + 2ab > 0\} \cap \{a - 2b > 0\}$.

In hyperbolic $1$-space $\mathbb{H}^1$, the ample cone corresponds to a half-line, bounded by the "wall" defined by the $(-2)$-curve $C$. Reflection across this wall is the Picard--Lefschetz reflection $s_C: D \mapsto D + (D \cdot C) C$.

For an elliptic curve $E/\mathbb{Q}$, the Neron--Tate height $\hat{h}: E(\mathbb{Q}) \otimes \mathbb{R} \to \mathbb{R}$ is a positive-definite quadratic form (after accounting for torsion). The rank of $E(\mathbb{Q})$ is the dimension of this space.

The arithmetic Hodge Index Theorem ensures that the height pairing matrix on a basis $P_1, \ldots, P_r$ of $E(\mathbb{Q})/\text{torsion}$ has $\det(\langle P_i, P_j \rangle) > 0$, which is the regulator of $E$. This appears in the BSD conjecture:

$L^{(r)}(E, 1) / r! = \Omega_E \cdot \operatorname{Reg}(E) \cdot \prod c_p \cdot |\text{III}| / |E(\mathbb{Q})_{\text{tors}}|^2.$

A K3 surface has $b_2 = 22$, $p_g = 1$, so the topological intersection form has signature $(3, 19)$ on $H^2(X, \mathbb{Z}) \cong U^3 \oplus E_8(-1)^2$.

The algebraic part $\operatorname{NS}(X)$ has signature $(1, \rho - 1)$, and the transcendental lattice $T(X)$ has signature $(2, 20 - \rho)$.

• For $\rho = 1$: $\operatorname{NS}(X) \cong \mathbb{Z}$ (signature $(1, 0)$) and $T(X)$ has signature $(2, 19)$.
• For $\rho = 20$: $\operatorname{NS}(X)$ has signature $(1, 19)$ and $T(X)$ has signature $(2, 0)$, meaning $T(X)$ is a positive-definite lattice of rank $2$.

For a minimal surface of general type, Noether's formula gives $\chi(\mathcal{O}_X) = \frac{1}{12}(K^2 + e(X))$ where $e(X) = 2 - 4q + b_2$ is the topological Euler characteristic. The Bogomolov--Miyaoka--Yau inequality gives $K^2 \leq 3 e(X)$ (equivalently $K^2 \leq 9 \chi(\mathcal{O}_X)$).

The topological intersection form on $H^2$ has signature $(2p_g + 1, b_2 - 2p_g - 1)$. By Freedman's theorem, the homeomorphism type of the underlying $4$-manifold is determined by this form (plus $b_1$). But the smooth structure carries more information (Donaldson theory), and the algebraic structure even more.

An Enriques surface $Y$ has $K_Y \not\equiv 0$ but $2K_Y \sim 0$. Its Neron--Severi group has rank $\rho = 10$, and the intersection form is the even unimodular lattice $U \oplus E_8(-1)$ of signature $(1, 9)$.

The Hodge Index Theorem gives: for $H$ ample on $Y$ and $D \cdot H = 0$, we have $D^2 \leq 0$. The lattice $U \oplus E_8(-1)$ is the unique even unimodular lattice of this signature, mirroring the uniqueness of the $E_8$ lattice.

Any $(-2)$-curve $C$ on $Y$ satisfies $C^2 = -2$ and defines a reflection $s_C: D \mapsto D + (D \cdot C)C$ preserving the lattice. The Weyl group generated by these reflections acts on the ample cone.

An abelian surface $A$ has $\rho(A)$ ranging from $1$ to $4$ (over $\mathbb{C}$). The Neron--Severi group embeds into $\operatorname{End}(A)^{\text{sym}}$ via the Appell--Humbert theorem. The intersection form on $\operatorname{NS}(A)$ has signature $(1, \rho - 1)$.

• $\rho = 1$: $\operatorname{NS}(A) = \mathbb{Z} L$ with $L^2 = 2d$ (a polarization of type $(1, d)$). Signature $(1, 0)$.
• $\rho = 2$ (generic principally polarized): $\operatorname{NS}(A) \cong \mathbb{Z}^2$ with signature $(1, 1)$.
• $\rho = 3$ ($A$ has real multiplication): the extra endomorphisms produce additional algebraic classes.
• $\rho = 4$ ($A \cong E_1 \times E_2$ with CM): $\operatorname{NS}(A)$ is rank $4$ with signature $(1, 3)$.

Let $X = \mathbb{P}^2$ blown up at one point, with $\operatorname{NS}(X) = \mathbb{Z} H \oplus \mathbb{Z} E$. Let $C$ be an irreducible curve of class $aH - bE$ (degree $a$, multiplicity $b$ at the blown-up point). By irreducibility, $C$ is effective and:

• $C \cdot H = a > 0$ and $C \cdot E = b \geq 0$.
• $C^2 = a^2 - b^2 \geq -1$ (genus formula: $g = \frac{(a-1)(a-2)}{2} - \frac{b(b-1)}{2} \geq 0$).

The Hodge inequality with $H$ (taking $H^2 = 1 > 0$, $D = C$): $(H \cdot C)^2 \geq H^2 \cdot C^2$, i.e., $a^2 \geq a^2 - b^2$, giving $b^2 \geq 0$, which is trivial. But applying with the ample class $A = 3H - E$ ($A^2 = 8$) and $C = aH - bE$: $(3a - b)^2 \cdot 1 \geq (a^2 - b^2) \cdot 1$... this requires more care.

A more useful application: if $C, D$ are two distinct irreducible curves both passing through the blown-up point with multiplicities $b, d$, then by the Hodge inequality on the original $\mathbb{P}^2$: their intersection number satisfies $C \cdot D = ae - bd \cdot ...$. The cleanest use is: $(C \cdot D)^2 \geq C^2 \cdot D^2$ when $C^2 > 0$.

Suppose a smooth projective surface $X$ has two divisors $F_1, F_2$ with $F_1^2 = F_2^2 = 0$ and $F_1 \cdot F_2 > 0$. Then $F_1$ and $F_2$ are independent in $\operatorname{NS}(X)$ and span a sublattice of signature $(1, 1)$ (the form $2(F_1 \cdot F_2) ab$ in coordinates $aF_1 + bF_2$ gives eigenvalues $\pm F_1 \cdot F_2$).

By the Hodge Index Theorem, $\rho \geq 2$. Moreover, if $F_1$ is the class of a fiber of a fibration $f_1: X \to B_1$, then $F_1^2 = 0$ is automatic. Two independent fibrations $f_1, f_2$ force $\rho \geq 2$ and the sublattice $\mathbb{Z} F_1 \oplus \mathbb{Z} F_2$ has signature $(1, 1)$.

This is realized by $\mathbb{P}^1 \times \mathbb{P}^1$: the two projections give fiber classes $F_1, F_2$ with $F_1 \cdot F_2 = 1$.

The cubic surface $X = \operatorname{Bl}_{p_1, \ldots, p_6}(\mathbb{P}^2)$ has $\rho = 7$ and NS lattice $\mathbb{Z}^{1,6}$ (one positive, six negative directions). The $27$ lines on the cubic surface are the $(-1)$-curves, and the effective cone is generated by these $27$ classes.

The ample cone is the interior of the dual of the effective cone of curves, cut out by the $27$ inequalities $D \cdot L_i > 0$ for each line $L_i$. In the hyperbolic $6$-space $\mathbb{H}^6$, this is a polytope bounded by $27$ hyperplanes.

The Weyl group $W(E_6)$ of order $51840$ acts on the NS lattice, permuting the $27$ lines and preserving the intersection form.