$\chi(\mathcal{O}_X(0)) = \frac{1}{2} \cdot 0 \cdot (0 - K_X) + \chi(\mathcal{O}_X) = \chi(\mathcal{O}_X)$ ✓. The formula is consistent.

$\chi(\mathcal{O}(K_X)) = \frac{1}{2}K_X \cdot (K_X - K_X) + \chi(\mathcal{O}_X) = 0 + \chi(\mathcal{O}_X) = \chi(\mathcal{O}_X)$.

By Serre duality, $\chi(\omega_X) = \chi(\mathcal{O}_X)$ (since $h^i(\omega_X) = h^{2-i}(\mathcal{O}_X)$, the alternating sum is preserved). ✓

The Serre duality pairing gives $h^i(\mathcal{O}(D)) = h^{2-i}(\mathcal{O}(K - D))$, so $\chi(\mathcal{O}(D)) = \chi(\mathcal{O}(K - D))$.

From the formula: $\frac{1}{2}D(D - K) + \chi(\mathcal{O}_X)$ vs. $\frac{1}{2}(K-D)(K-D-K) + \chi(\mathcal{O}_X) = \frac{1}{2}(K-D)(-D) + \chi(\mathcal{O}_X) = -\frac{1}{2}KD + \frac{1}{2}D^2 + \chi(\mathcal{O}_X)$.

Both equal $\frac{1}{2}(D^2 - KD) + \chi(\mathcal{O}_X)$ ✓. The formula is automatically compatible with Serre duality.

On $\mathbb{P}^2$: $K = -3H$, $\chi(\mathcal{O}) = 1$ (since $q = p_g = 0$). For $D = dH$:

$\chi(\mathcal{O}(d)) = \frac{1}{2}d(d + 3) + 1 = \frac{d^2 + 3d + 2}{2} = \frac{(d+1)(d+2)}{2} = \binom{d+2}{2}.$

This recovers:

• $d = 0$: $\chi = 1 = h^0(\mathcal{O}) = 1$ ✓.
• $d = 1$: $\chi = 3 = h^0(\mathcal{O}(1)) = 3$ (linear forms $x, y, z$) ✓.
• $d = 2$: $\chi = 6$ (monomials $x^2, xy, xz, y^2, yz, z^2$) ✓.
• $d = -1$: $\chi = 0$. Indeed $h^0(\mathcal{O}(-1)) = 0$, $h^1(\mathcal{O}(-1)) = 0$, $h^2(\mathcal{O}(-1)) = 0$ ✓.
• $d = -2$: $\chi = 0$. Same vanishing ✓.
• $d = -3$: $\chi = 1$. We have $\mathcal{O}(-3) = \omega_{\mathbb{P}^2}$, so $h^2(\mathcal{O}(-3)) = h^0(\mathcal{O}) = 1$, $h^0 = h^1 = 0$, giving $\chi = 0 - 0 + 1 = 1$ ✓.
• $d = -4$: $\chi = 3$. $h^2(\mathcal{O}(-4)) = h^0(\mathcal{O}(1)) = 3$ ✓.

The Hirzebruch surface $\mathbb{F}_n = \mathbb{P}(\mathcal{O}_{\mathbb{P}^1} \oplus \mathcal{O}_{\mathbb{P}^1}(-n))$ has:

• $\operatorname{Pic}(\mathbb{F}_n) = \mathbb{Z} C_0 \oplus \mathbb{Z} F$ where $C_0$ is the negative section ($C_0^2 = -n$) and $F$ is a fiber ($F^2 = 0$, $C_0 \cdot F = 1$).
• $K_{\mathbb{F}_n} \sim -2C_0 - (n + 2)F$.
• $\chi(\mathcal{O}_{\mathbb{F}_n}) = 1$ (rational surface, so $q = p_g = 0$).

For $D \sim aC_0 + bF$: $D^2 = -na^2 + 2ab$, $D \cdot K = 2na - 2b - (n+2)a = (n-2)a - 2b$. So:

$\chi(\mathcal{O}(D)) = \frac{1}{2}(-na^2 + 2ab - (n-2)a + 2b) + 1.$

Check: $D = C_0$ ($a = 1, b = 0$): $\chi = \frac{1}{2}(-n - n + 2) + 1 = \frac{1}{2}(2 - 2n) + 1 = 2 - n$. For $n = 0$ ($\mathbb{P}^1 \times \mathbb{P}^1$): $\chi(\mathcal{O}(C_0)) = 2$, and $h^0(\mathcal{O}(1,0)) = 2$ ✓.

A K3 surface has $K_X = 0$, $\chi(\mathcal{O}_X) = 2$ (since $h^0 = 1$, $h^1 = 0$, $h^2 = 1$). Riemann–Roch gives:

$\chi(\mathcal{O}(D)) = \frac{1}{2}D^2 + 2.$

• If $D$ is a primitive ample class with $D^2 = 2d$, then $\chi(\mathcal{O}(D)) = d + 2$. By Kodaira vanishing, $h^1 = h^2 = 0$ for ample $D$, so $h^0(D) = d + 2$. The linear system $|D|$ has dimension $d + 1$.

• Degree-$2$ K3 ($D^2 = 2$): $h^0(D) = 3$, so $|D|$ is a pencil in $\mathbb{P}^2$... rather, $|D|$ gives a $2:1$ map to $\mathbb{P}^2$.

• Degree-$4$ K3 ($D^2 = 4$): $h^0(D) = 4$, embedding as quartic in $\mathbb{P}^3$.

• Degree-$6$ K3 ($D^2 = 6$): $h^0(D) = 5$, complete intersection of a quadric and cubic in $\mathbb{P}^4$.

• Degree-$8$ K3 ($D^2 = 8$): $h^0(D) = 6$, complete intersection of three quadrics in $\mathbb{P}^5$.

For a minimal surface of general type ($K_X$ nef and big, $K_X^2 > 0$):

$\chi(\mathcal{O}(nK_X)) = \frac{n(n-1)}{2}K_X^2 + \chi(\mathcal{O}_X).$

Pluricanonical dimensions:

• $n = 1$: $\chi(\omega_X) = \chi(\mathcal{O}_X) = 1 + p_g - q$.
• $n = 2$: $\chi(\mathcal{O}(2K)) = K^2 + \chi(\mathcal{O}_X)$. By Kawamata–Viehweg vanishing ($2K - K = K$ is nef and big), $h^i(\mathcal{O}(2K)) = 0$ for $i > 0$, so $P_2 = h^0(2K) = K^2 + \chi(\mathcal{O}_X)$.
• $n = 3$: $P_3 = 3K^2 + \chi(\mathcal{O}_X)$.
• $n \to \infty$: $P_n \sim \frac{1}{2}n^2 K^2$ (quadratic growth — this is the hallmark of a surface of general type).

Bombieri's theorem: $|5K_X|$ gives a birational map for all surfaces of general type.

An abelian surface $A$ has $K_A = 0$, $\chi(\mathcal{O}_A) = 0$ (since $h^0 = 1$, $h^1 = 2$, $h^2 = 1$). So:

$\chi(\mathcal{O}(D)) = \frac{1}{2}D^2.$

For an ample $D$ with $D^2 = 2d$: $\chi(\mathcal{O}(D)) = d$. By the Mumford vanishing theorem, $h^i(D) = 0$ for $i > 0$ when $D$ is ample, so $h^0(D) = d$.

• Principal polarization ($D^2 = 2$): $h^0(D) = 1$. The theta divisor $\Theta$ has a unique section.
• $(1,2)$-polarization ($D^2 = 4$): $h^0(D) = 2$, giving a $2:1$ map $A \to \mathbb{P}^1$... actually, $|D|$ maps to $\mathbb{P}^1$ — but abelian surfaces don't admit maps to $\mathbb{P}^1$ (no rational curves). More precisely, $|D|$ has base locus and $\phi_D$ maps $A$ to $\mathbb{P}^1$ as a genus-$2$ fibration after resolving.
• $(1,3)$-polarization ($D^2 = 6$): $h^0(D) = 3$. Maps $A \to \mathbb{P}^2$ as $3:1$.

Complete list of common vanishing situations:

• $D$ ample on $\mathbb{P}^2$: $h^i(\mathcal{O}(d)) = 0$ for $0 < i < 2$ and $d \geq -2$ (from the explicit computation of cohomology of $\mathbb{P}^n$).
• $D$ ample on K3: $h^1 = h^2 = 0$ by Kodaira vanishing (since $D = (D - K) + K = D + 0$, and $D$ ample implies vanishing). So $h^0(D) = \frac{1}{2}D^2 + 2$.
• $D$ ample on an Enriques surface ($K^2 = 0$, $\chi(\mathcal{O}) = 1$, $2K \sim 0$): $h^0(D) = \frac{1}{2}D^2 + 1$.
• $nK_X$ with $n \geq 2$ on a surface of general type: $h^0(nK) = \frac{n(n-1)}{2}K^2 + \chi(\mathcal{O}_X)$.

Riemann–Roch determines the expected dimension of families of curves:

• Curves of degree $d$ in $\mathbb{P}^2$: The linear system $|dH|$ has dimension $\binom{d+2}{2} - 1$. A curve passes through $n$ general points iff $\binom{d+2}{2} - 1 \geq n$. Through $\binom{d+2}{2} - 1$ general points, there is a unique degree-$d$ curve.

• Lines through 2 points: $\binom{3}{2} - 1 = 2$; imposing $2$ points gives $0$-dimensional family, i.e., a unique line. ✓

• Conics through 5 points: $\binom{4}{2} - 1 = 5$; imposing $5$ conditions gives a unique conic. ✓

• Cubics through 9 points: $\binom{5}{2} - 1 = 9$; but $9$ points may not impose independent conditions. Cayley–Bacharach: if $8$ of the $9$ base points of two cubics are specified, the $9$th is determined.