For $X = \mathbb{P}^2$: $K_{\mathbb{P}^2} = -3H$, so $K^2 = 9$. The Betti numbers are $b_0 = 1, b_1 = 0, b_2 = 1, b_3 = 0, b_4 = 1$, giving $e(\mathbb{P}^2) = 3$.

Noether: $\chi(\mathcal{O}_{\mathbb{P}^2}) = \frac{1}{12}(9 + 3) = \frac{12}{12} = 1$.

Direct computation: $h^0(\mathcal{O}) = 1$, $h^1(\mathcal{O}) = 0$, $h^2(\mathcal{O}) = 0$, so $\chi(\mathcal{O}) = 1$.

For $X = \mathbb{P}^1 \times \mathbb{P}^1$: $K = -2F_1 - 2F_2$, so $K^2 = 2 \cdot (-2)(-2) \cdot F_1 \cdot F_2 = 8$. The Betti numbers are $b_0 = 1, b_1 = 0, b_2 = 2, b_3 = 0, b_4 = 1$, giving $e = 4$.

Noether: $\chi(\mathcal{O}) = \frac{1}{12}(8 + 4) = 1$.

Direct: $q = 0$, $p_g = 0$, so $\chi(\mathcal{O}) = 1$.

A K3 surface has $K_X = 0$ (so $K^2 = 0$), $q = 0$, and $p_g = 1$.

Noether: $\chi(\mathcal{O}_X) = 1 - 0 + 1 = 2 = \frac{1}{12}(0 + e(X))$, hence $e(X) = 24$.

Betti numbers: $b_0 = 1, b_1 = 0, b_2 = 22, b_3 = 0, b_4 = 1$. Check: $e = 1 + 22 + 1 = 24$.

Hodge diamond: $h^{0,0} = 1$, $h^{1,0} = h^{0,1} = 0$, $h^{2,0} = h^{0,2} = 1$, $h^{1,1} = 20$.

From the formula: $h^{1,1} = 10(2) - 0 + 2(1 - 0) = 22$... let us recheck: $h^{1,1} = 10\chi - K^2 + 2(1 - q) = 20 - 0 + 2 = 22$. But $b_2 = h^{2,0} + h^{1,1} + h^{0,2} = 1 + 20 + 1 = 22$. So indeed $h^{1,1} = 20$.

An abelian surface $A$ has $K_A = 0$ (so $K^2 = 0$), $q = 2$, $p_g = 1$.

Noether: $\chi(\mathcal{O}_A) = 1 - 2 + 1 = 0 = \frac{1}{12}(0 + e(A))$, hence $e(A) = 0$.

Betti numbers: $b_0 = 1, b_1 = 4, b_2 = 6, b_3 = 4, b_4 = 1$. Check: $e = 1 - 4 + 6 - 4 + 1 = 0$.

Hodge diamond: $h^{0,0} = 1$, $h^{1,0} = h^{0,1} = 2$, $h^{2,0} = h^{0,2} = 1$, $h^{1,1} = 4$.

An Enriques surface has $2K_X \sim 0$ but $K_X \not\sim 0$, $q = 0$, $p_g = 0$.

$K^2 = 0$ (since $2K \sim 0$ implies $4K^2 = (2K)^2 = 0$).

Noether: $\chi(\mathcal{O}) = 1 - 0 + 0 = 1 = \frac{1}{12}(0 + e)$, hence $e = 12$.

Betti numbers: $b_0 = 1, b_1 = 0, b_2 = 10, b_3 = 0, b_4 = 1$. Check: $e = 1 + 10 + 1 = 12$.

Let $X_d \subset \mathbb{P}^3$ be a smooth hypersurface of degree $d$. Then $K_{X_d} = (d - 4)H|_{X_d}$ by adjunction, so $K^2 = d(d-4)^2$.

The Euler characteristic of $X_d$ is computed from the Chern classes of $\mathbb{P}^3$ restricted to $X_d$. Using the exact sequence $0 \to T_{X_d} \to T_{\mathbb{P}^3}|_{X_d} \to N_{X_d/\mathbb{P}^3} \to 0$:

$e(X_d) = d^3 - 4d^2 + 6d.$

Noether: $\chi(\mathcal{O}_{X_d}) = \frac{1}{12}(d(d-4)^2 + d^3 - 4d^2 + 6d) = \frac{1}{12}(2d^3 - 12d^2 + 22d) = \frac{d(d-1)(d-2)}{6} + 1 - d + \frac{d(d-1)}{2}$... let us compute directly:

$\frac{1}{12}(d(d-4)^2 + d^3 - 4d^2 + 6d) = \frac{1}{12}(d^3 - 8d^2 + 16d + d^3 - 4d^2 + 6d) = \frac{1}{12}(2d^3 - 12d^2 + 22d) = \frac{d}{6}(d^2 - 6d + 11)$.

Specific values:

• $d = 1$ (plane): $\chi = \frac{1}{6}(1 - 6 + 11) = 1$, $K^2 = 9$, $e = 3$.
• $d = 2$ (quadric): $\chi = \frac{2}{6}(4 - 12 + 11) = 1$, $K^2 = 8$, $e = 4$.
• $d = 3$ (cubic): $\chi = \frac{3}{6}(9 - 18 + 11) = 1$, $K^2 = 3$, $e = 9$.
• $d = 4$ (quartic = K3): $\chi = \frac{4}{6}(16 - 24 + 11) = 2$, $K^2 = 0$, $e = 24$.
• $d = 5$ (quintic): $\chi = \frac{5}{6}(25 - 30 + 11) = 5$, $K^2 = 5$, $e = 55$.

The cubic surface $X_3 \subset \mathbb{P}^3$ is rational: it is the blowup of $\mathbb{P}^2$ at $6$ points. So $q = 0$, $p_g = 0$, $\chi(\mathcal{O}) = 1$.

$K_{X_3} = -H|_{X_3}$, so $K^2 = H^2 \cdot X_3 = 3$ (the degree). But $K^2 = 3$ for the cubic surface.

From the blowup description: $K_{\operatorname{Bl}_6(\mathbb{P}^2)} = -3H + E_1 + \cdots + E_6$, and $K^2 = 9 - 6 = 3$.

$e = 12\chi - K^2 = 12 - 3 = 9$. Alternatively: $b_2 = 1 + 6 = 7$ (the class of the line plus $6$ exceptional divisors), so $e = 1 + 7 + 1 = 9$.

Let $X = C_1 \times C_2$ where $C_i$ has genus $g_i$. Then:

• $K_X = \pi_1^* K_{C_1} + \pi_2^* K_{C_2}$, so $K^2 = 2(2g_1 - 2)(2g_2 - 2) = 8(g_1 - 1)(g_2 - 1)$.
• $e(X) = e(C_1) \cdot e(C_2) = (2 - 2g_1)(2 - 2g_2) = 4(g_1 - 1)(g_2 - 1)$.
• $q = g_1 + g_2$, $p_g = g_1 g_2$.
• $\chi(\mathcal{O}_X) = 1 - (g_1 + g_2) + g_1 g_2 = (1 - g_1)(1 - g_2)$.

Noether check: $\frac{1}{12}(8(g_1-1)(g_2-1) + 4(g_1-1)(g_2-1)) = \frac{12(g_1-1)(g_2-1)}{12} = (g_1-1)(g_2-1) = (1-g_1)(1-g_2)$.

Specific cases:

• $C_1 = C_2 = \mathbb{P}^1$ ($g = 0$): $K^2 = 8$, $e = 4$, $\chi = 1$. This is $\mathbb{P}^1 \times \mathbb{P}^1$.
• $C_1 = E$, $C_2 = E'$ (elliptic, $g = 1$): $K^2 = 0$, $e = 0$, $\chi = 0$. This is an abelian surface.
• $C_1 = C_2$ genus $2$: $K^2 = 8$, $e = 4$, $q = 4$, $p_g = 4$, $\chi = 1$.
• $C_1 = C_2$ genus $3$: $K^2 = 32$, $e = 16$, $q = 6$, $p_g = 9$, $\chi = 4$.

Let $X = X_{d_1, d_2} \subset \mathbb{P}^4$ be a smooth complete intersection of hypersurfaces of degrees $d_1, d_2$. By adjunction:

$K_X = (d_1 + d_2 - 5)H|_X$, $\deg(X) = d_1 d_2$, $K^2 = d_1 d_2(d_1 + d_2 - 5)^2$.

The Euler characteristic is $e(X) = d_1 d_2(d_1^2 + d_2^2 + 3d_1 d_2 - 5d_1 - 5d_2 + 10) / 1$... more precisely, from the Chern class computation:

$e(X) = d_1 d_2(d_1^2 - 5d_1 + d_2^2 - 5d_2 + 3d_1 d_2 + 10)$.

Specific cases:

• $(2,2)$ in $\mathbb{P}^4$: $K = -H$, $K^2 = 4$, $e = 4 \cdot (4 - 10 + 4 - 10 + 12 + 10) = 4 \cdot 10 = 40$... let me compute more carefully. The degree is $4$. $e = c_2(T_X) = 4(4+4+12-10-10+10) = 4 \cdot 10 = 40$. But $\chi = \frac{1}{12}(4 + 40) = \frac{44}{12}$ which is not an integer. Let me recalculate.

For $(2,2)$ in $\mathbb{P}^4$: this is a del Pezzo surface of degree $4$. It is the blowup of $\mathbb{P}^2$ at $5$ points. So $K^2 = 9 - 5 = 4$, $q = 0$, $p_g = 0$, $\chi = 1$, $e = 12 - 4 = 8$.

• $(2,3)$ in $\mathbb{P}^4$: degree $6$, $K = 0$. This is a K3 surface. $K^2 = 0$, $\chi = 2$, $e = 24$.

• $(3,3)$ in $\mathbb{P}^4$: degree $9$, $K = H|_X$, $K^2 = 9$. $\chi = \frac{1}{12}(9 + e)$. By explicit computation, $q = 0$ and $p_g = 4$, so $\chi = 5$. Then $e = 60 - 9 = 51$.

• $(2,4)$ in $\mathbb{P}^4$: degree $8$, $K = H|_X$, $K^2 = 8$. We have $q = 0$, $p_g = 3$, $\chi = 4$, $e = 48 - 8 = 40$.

• Quintic in $\mathbb{P}^3$ ($d = 5$): $K^2 = 5$, $p_g = 4$. Inequality: $5 \geq 2(4) - 4 = 4$.
• Product of two genus-$2$ curves: $K^2 = 8$, $p_g = 4$. Inequality: $8 \geq 4$.
• Godeaux surface ($\pi_1 = \mathbb{Z}/5$): $K^2 = 1$, $p_g = 0$. Inequality: $1 \geq -4$.
• Barlow surface: $K^2 = 1$, $p_g = 0$, $q = 0$. The simplest simply connected surface of general type.
• Horikawa surfaces: these lie on the Noether line $K^2 = 2p_g - 4$, achieving equality. They exist for all $p_g \geq 3$.

• $\mathbb{P}^2$: $c_1^2 = 9$, $c_2 = 3$, ratio $c_1^2/c_2 = 3$. This achieves equality! Indeed, $\mathbb{P}^2 = \mathbb{B}^2/\Gamma$... no, $\mathbb{P}^2$ is simply connected and rational, not of general type. The BMY inequality applies only to surfaces of general type.
• Quintic surface ($d = 5$): $K^2 = 5$, $e = 55$, $\chi = 5$. Check: $5 \leq 9 \cdot 5 = 45$ and $5 \leq 3 \cdot 55 = 165$.
• Ball quotients: the fake projective planes have $c_1^2 = 3c_2 = 9$, $\chi = 1$, $p_g = 0$, $q = 0$. There are exactly $100$ fake projective planes (50 conjugate pairs), classified by Prasad--Yeung and Cartwright--Steger.
• Product of two genus-$2$ curves: $K^2 = 8$, $e = 4$, $\chi = 1$. Check: $8 \leq 3 \cdot 4 = 12$ and $8 \leq 9 \cdot 1 = 9$.

Key surfaces and their positions in the $(K^2, \chi)$-plane:

On or near the Noether line ($K^2 = 2\chi - 6$):

• Horikawa surfaces: $K^2 = 2p_g - 4$, ranging over $K^2 = 2, 4, 6, \ldots$ with $\chi = (K^2 + 6)/2 = 4, 5, 6, \ldots$.

On the BMY line ($K^2 = 9\chi$):

• Fake projective planes: $(K^2, \chi) = (9, 1)$.
• Ball quotients with $\chi = 2$: $(K^2, \chi) = (18, 2)$.

Between the lines:

• Godeaux surface: $(K^2, \chi) = (1, 1)$.
• Campedelli surface: $(K^2, \chi) = (2, 1)$.
• Barlow surface: $(K^2, \chi) = (1, 1)$ (same invariants as Godeaux, but simply connected).
• Catanese surface: $(K^2, \chi) = (2, 1)$.
• Burniat surfaces: $(K^2, \chi) = (2, 1), (3, 1), (4, 1), (5, 1), (6, 1)$.
• Beauville surface: $(K^2, \chi) = (8, 1)$.
• Quintic in $\mathbb{P}^3$: $(K^2, \chi) = (5, 5)$.

Rational surfaces ($X$ birational to $\mathbb{P}^2$):

• $\chi(\mathcal{O}_X) = 1$, $q = 0$, $p_g = 0$.
• Minimal models: $\mathbb{P}^2$ ($K^2 = 9, e = 3$) and $\mathbb{F}_n$ for $n \neq 1$ ($K^2 = 8, e = 4$).
• Blowups: $\operatorname{Bl}_r(\mathbb{P}^2)$ has $K^2 = 9 - r$, $e = 3 + r$.

Ruled surfaces over a curve of genus $g$ (geometrically ruled: $X = \mathbb{P}(\mathcal{E}) \to C$):

• $q = g$, $p_g = 0$, $\chi(\mathcal{O}_X) = 1 - g$.
• $K^2 = 8(1 - g)$, $e = 4(1 - g)$.
• Noether: $12(1 - g) = 8(1 - g) + 4(1 - g)$.

The minimal surfaces with $\kappa = 0$ fall into four classes:

K3 surfaces: $q = 0$, $p_g = 1$, $\chi = 2$, $K^2 = 0$, $e = 24$.

Enriques surfaces: $q = 0$, $p_g = 0$, $\chi = 1$, $K^2 = 0$, $e = 12$. (In characteristic $\neq 2$, $2K \sim 0$.)

Abelian surfaces: $q = 2$, $p_g = 1$, $\chi = 0$, $K^2 = 0$, $e = 0$.

Hyperelliptic (bielliptic) surfaces: $q = 1$, $p_g = 0$, $\chi = 0$, $K^2 = 0$, $e = 0$. (These are quotients $(E_1 \times E_2)/G$ where $G$ is a finite group acting on both elliptic curves.)

In all cases $K^2 = 0$. Noether gives $\chi = e/12$, so $e$ is a multiple of $12$ for $\chi \in \mathbb{Z}$. Indeed: $e \in \{0, 0, 12, 24\}$.

Minimal surfaces with $\kappa = 1$ are properly elliptic surfaces: they admit a fibration $f \colon X \to C$ whose general fiber is an elliptic curve, and $K_X$ is not numerically trivial.

For a relatively minimal elliptic fibration $f \colon X \to C$ with $g(C) = g$:

• $\chi(\mathcal{O}_X) = \chi(\mathcal{O}_C) \cdot \chi(\mathcal{O}_F) + \text{(singular fiber contribution)} = \deg f_* \omega_{X/C}$.
• $K^2 = 0$ (for minimal models).
• $e(X) = 12\chi(\mathcal{O}_X)$ (by Noether, since $K^2 = 0$).

The Euler characteristic $e(X)$ is computed from the singular fibers via Ogg's formula: $e(X) = \sum_v e(X_v)$ where the sum is over singular fibers.

Example: Elliptic surface over $\mathbb{P}^1$ with $\chi = 1$: $K^2 = 0$, $e = 12$. If it has $12$ nodal fibers (type $I_1$), each contributes $e = 1$ to the Euler characteristic: $e = 12 \cdot 1 = 12$.

Example: Elliptic K3 surface: $\chi = 2$, $e = 24$. Possible singular fiber configuration: $24$ nodal fibers (all $I_1$), or two fibers of type $I_0^*$ (each contributes $e = 6$, giving $12$) plus $12$ nodal fibers, etc.

• $\mathbb{P}^2$: $\sigma = 4(1) - 3 = 1$. Intersection form on $H^2$: the $1 \times 1$ matrix $(1)$, signature $+1$.
• $\mathbb{P}^1 \times \mathbb{P}^1$: $\sigma = 4(1) - 4 = 0$. Intersection form: the hyperbolic form $H$ with eigenvalues $\pm 1$, signature $0$.
• K3 surface: $\sigma = 4(2) - 24 = -16$. The intersection form is $3H \oplus (-E_8)^2$ (three hyperbolic planes plus two copies of the negative $E_8$ lattice), with $b_2^+ = 3, b_2^- = 19$, signature $3 - 19 = -16$.
• Abelian surface: $\sigma = 4(0) - 0 = 0$. The intersection form on $H^2(\mathbb{Z}) \cong \mathbb{Z}^6$ has $b_2^+ = 3, b_2^- = 3$, signature $0$.
• Enriques surface: $\sigma = 4(1) - 12 = -8$. The intersection form is $H \oplus (-E_8)$ (one hyperbolic plane plus the negative $E_8$ lattice), with $b_2^+ = 1, b_2^- = 9$, signature $-8$.
• Quintic surface: $\sigma = 4(5) - 55 = -35$.

A reference table of invariants for the most important surfaces.

Rational surfaces:

• $\mathbb{P}^2$: $K^2 = 9$, $e = 3$, $\chi = 1$, $q = 0$, $p_g = 0$, $\sigma = 1$, $b_2 = 1$.
• $\mathbb{P}^1 \times \mathbb{P}^1$: $K^2 = 8$, $e = 4$, $\chi = 1$, $q = 0$, $p_g = 0$, $\sigma = 0$, $b_2 = 2$.
• $\operatorname{Bl}_1(\mathbb{P}^2)$: $K^2 = 8$, $e = 4$, $\chi = 1$, $q = 0$, $p_g = 0$, $\sigma = 0$, $b_2 = 2$.
• $\operatorname{Bl}_6(\mathbb{P}^2)$ (cubic surface): $K^2 = 3$, $e = 9$, $\chi = 1$, $q = 0$, $p_g = 0$, $\sigma = -5$, $b_2 = 7$.
• $\operatorname{Bl}_8(\mathbb{P}^2)$: $K^2 = 1$, $e = 11$, $\chi = 1$, $\sigma = -7$, $b_2 = 9$.