Let $X = \mathbb{A}^2$ with coordinates $(x, y)$ and $P = (0, 0)$. The blowup is:

$\operatorname{Bl}_0(\mathbb{A}^2) = \{((x, y), [s:t]) \in \mathbb{A}^2 \times \mathbb{P}^1 \mid xt = ys\}.$

In the chart $s \neq 0$ (set $s = 1$, so $t = y/x$): the blowup is $\mathbb{A}^2$ with coordinates $(x, t)$ where $y = xt$. The map $\pi$ sends $(x, t) \mapsto (x, xt)$. The exceptional divisor in this chart is $x = 0$ (a copy of $\mathbb{A}^1$ with coordinate $t$).

In the chart $t \neq 0$ (set $t = 1$, so $s = x/y$): coordinates $(s, y)$ with $x = sy$. The exceptional divisor is $y = 0$.

The two charts glue to give $E \cong \mathbb{P}^1$, and $\operatorname{Bl}_0(\mathbb{A}^2)$ is again a smooth surface.

On $\operatorname{Bl}_0(\mathbb{A}^2)$, in the chart $(x, t)$ with $y = xt$, the exceptional divisor is $E = V(x)$. To compute $E^2$, we use the adjunction formula. Since $E \cong \mathbb{P}^1$ has genus $0$:

$2g(E) - 2 = E^2 + K_{\widetilde{X}} \cdot E.$

We will show $K_{\widetilde{X}} \cdot E = -1$ (from $K_{\widetilde{X}} = \pi^*K_X + E$), giving $-2 = E^2 + (-1)$, hence $E^2 = -1$.

Alternatively, compute directly: $\mathcal{O}_E(E) = N_{E/\widetilde{X}}$, the normal bundle. The normal bundle of $E \cong \mathbb{P}^1$ embedded via the tautological sub-bundle in the trivial rank-$2$ bundle is $\mathcal{O}_{\mathbb{P}^1}(-1)$. Therefore $E^2 = \deg \mathcal{O}_{\mathbb{P}^1}(-1) = -1$.

Consider $C = V(y) \subset \mathbb{A}^2$ (the $x$-axis) and the blowup $\pi: \operatorname{Bl}_0(\mathbb{A}^2) \to \mathbb{A}^2$. In the chart $(x, t)$ with $y = xt$:

$\pi^*C : y = 0 \iff xt = 0 \iff (x = 0) \cup (t = 0).$

So $\pi^*C = E + \widetilde{C}$ where $E = V(x)$ and $\widetilde{C} = V(t)$. Here $\operatorname{mult}_0(C) = 1$, consistent with $\pi^*C = \widetilde{C} + 1 \cdot E$.

The strict transform $\widetilde{C} = V(t)$ meets $E = V(x)$ at the single point $(x, t) = (0, 0)$, which corresponds to the tangent direction $[1:0]$ of the line $y = 0$ at the origin.

Let $C = V(y^2 - x^2 - x^3)$, a node at the origin with two branches tangent to $y = \pm x$. In the chart $(x, t)$ with $y = xt$:

$\pi^*C : (xt)^2 - x^2 - x^3 = 0 \iff x^2(t^2 - 1 - x) = 0.$

So $\pi^*C = 2E + \widetilde{C}$ where $\widetilde{C} = V(t^2 - 1 - x)$. Here $\operatorname{mult}_0(C) = 2$.

The strict transform $\widetilde{C}$ meets $E$ at $x = 0, t^2 = 1$, i.e., at $t = \pm 1$. These correspond to the two tangent directions $y = \pm x$ at the node. The strict transform $\widetilde{C}$ is smooth: the node has been resolved by one blowup.

Let $C = V(y^2 - x^3)$, a cusp at the origin. In the chart $(x, t)$ with $y = xt$:

$\pi^*C: x^2 t^2 - x^3 = 0 \iff x^2(t^2 - x) = 0.$

So $\pi^*C = 2E + \widetilde{C}$ with $\widetilde{C} = V(t^2 - x)$. The strict transform meets $E$ at $(0, 0)$ only (a single point with $t = 0$). Moreover, $\widetilde{C}$ is a smooth parabola: the cusp has been resolved by a single blowup.

Let $\widetilde{X} = \operatorname{Bl}_P(\mathbb{P}^2)$, $H = \pi^*\ell$ (pullback of a line), $E$ the exceptional divisor.

Basis: $\operatorname{Pic}(\widetilde{X}) = \mathbb{Z} H \oplus \mathbb{Z} E$ with $H^2 = 1$, $E^2 = -1$, $H \cdot E = 0$.

Strict transform of a line through $P$: $\widetilde{\ell} = H - E$ (since $\operatorname{mult}_P(\ell) = 1$). Then $\widetilde{\ell}^2 = 1 - 1 = 0$. This makes sense: two lines through $P$ become disjoint after blowup (their intersection point has been separated).

Strict transform of a conic through $P$: $\widetilde{C} = 2H - E$. Then $\widetilde{C}^2 = 4 - 1 = 3$.

Strict transform of a conic tangent to a line at $P$: the conic has $\operatorname{mult}_P = 1$, so $\widetilde{C} = 2H - E$. The line has $\widetilde{\ell} = H - E$. Then $\widetilde{C} \cdot \widetilde{\ell} = 2 - 1 = 1$. The tangency has been "separated" by the blowup.

For $X = \mathbb{P}^2$, $K_X = -3H$. So:

$K_{\operatorname{Bl}_P(\mathbb{P}^2)} = -3H + E.$

Check with adjunction on $E$: $K \cdot E = (-3H + E) \cdot E = 0 + (-1) = -1$. Then $2g(E) - 2 = E^2 + K \cdot E = -1 + (-1) = -2$, giving $g(E) = 0$.

Check $K^2$: $(-3H + E)^2 = 9 - 1 = 8$. Since $K_{\mathbb{P}^2}^2 = 9$, blowing up one point decreases $K^2$ by $1$.

For the blowup of $\mathbb{P}^2$ at $n$ points: $K^2 = 9 - n$.

Let $S = V(xy - z^2) \subset \mathbb{A}^3$, which has an $A_1$ (node) singularity at the origin. Blow up the origin in $\mathbb{A}^3$ and take the strict transform.

In the chart $[1:u:v]$ (i.e., $y = ux, z = vx$): the equation becomes $x^2(u - v^2) = 0$. The strict transform $\widetilde{S} = V(u - v^2)$ is smooth, isomorphic to $\mathbb{A}^2$ with coordinates $(x, v)$.

The exceptional curve on $\widetilde{S}$ is $E = \widetilde{S} \cap V(x) = V(x, u - v^2)$, parametrized by $v$. So $E \cong \mathbb{P}^1$. One computes $E^2 = -2$ on $\widetilde{S}$: this is the minimal resolution of the $A_1$ singularity.

The $A_n$ singularity is (locally) $x^2 + y^2 + z^{n+1} = 0$ (over $\mathbb{C}$, equivalently $uv = z^{n+1}$). The minimal resolution has exceptional divisor consisting of a chain of $n$ smooth rational curves $E_1, E_2, \ldots, E_n$, each with $E_i^2 = -2$, and:

• $E_i \cdot E_{i+1} = 1$ for $i = 1, \ldots, n - 1$,
• $E_i \cdot E_j = 0$ for $|i - j| \geq 2$.

The intersection matrix is minus the Cartan matrix of the $A_n$ Dynkin diagram. For $A_2$ ($uv = z^3$): two $(-2)$-curves meeting in one point. For $A_3$: three $(-2)$-curves in a chain.

The blowup $\operatorname{Bl}_P(\mathbb{P}^2)$ is isomorphic to the Hirzebruch surface $\mathbb{F}_1 = \mathbb{P}(\mathcal{O}_{\mathbb{P}^1} \oplus \mathcal{O}_{\mathbb{P}^1}(1))$.

To see this, the linear system $|H - E|$ (strict transforms of lines through $P$) defines a morphism $\operatorname{Bl}_P(\mathbb{P}^2) \to \mathbb{P}^1$. The fibers are strict transforms of lines through $P$, each isomorphic to $\mathbb{P}^1$ with self-intersection $0$. This exhibits $\operatorname{Bl}_P(\mathbb{P}^2)$ as a ruled surface over $\mathbb{P}^1$.

The exceptional divisor $E$ is a section with $E^2 = -1$, and the fibers $F = H - E$ satisfy $F^2 = 0$, $F \cdot E = 1$. This matches $\mathbb{F}_1$: the unique Hirzebruch surface with a section of self-intersection $-1$.

Numerical invariants:

• $\operatorname{Pic} = \mathbb{Z} H \oplus \mathbb{Z} E$, with $H^2 = 1, E^2 = -1, H \cdot E = 0$.
• $K = -3H + E$, $K^2 = 8$.
• $\chi(\mathcal{O}) = 1$, $p_g = q = 0$.

Let $\widetilde{X} = \operatorname{Bl}_{P_1, \ldots, P_n}(\mathbb{P}^2)$ be the blowup of $\mathbb{P}^2$ at $n$ points in general position.

Picard group: $\operatorname{Pic}(\widetilde{X}) = \mathbb{Z} H \oplus \mathbb{Z} E_1 \oplus \cdots \oplus \mathbb{Z} E_n$ with:

• $H^2 = 1$, $E_i^2 = -1$, $H \cdot E_i = 0$, $E_i \cdot E_j = 0$ for $i \neq j$.

Canonical class: $K = -3H + E_1 + \cdots + E_n$, $K^2 = 9 - n$.

Anticanonical class: $-K = 3H - E_1 - \cdots - E_n$, $(-K)^2 = 9 - n$.

For the anticanonical class to be effective, we need $n \leq 9$ (curves of degree $3$ through $n$ points exist iff $n \leq 9$).

Key classes of divisors:

• Lines through $P_i$: class $H - E_i$, self-intersection $0$.
• Lines through $P_i, P_j$: class $H - E_i - E_j$, self-intersection $-1$ (a $(-1)$-curve!).
• Conics through $P_1, \ldots, P_5$: class $2H - E_1 - \cdots - E_5$, self-intersection $-1$.
• Cubics through $P_1, \ldots, P_6$ with a double point at $P_1$: class $3H - 2E_1 - E_2 - \cdots - E_6$, self-intersection $-1$.

On $\mathbb{F}_1 = \operatorname{Bl}_P(\mathbb{P}^2)$, the exceptional curve $E$ with $E^2 = -1$ can be blown down to recover $\mathbb{P}^2$. But $\mathbb{F}_1$ has another family of $(-1)$-curves: the strict transforms of lines through $P$, which have class $H - E$ and self-intersection $0$... wait, $(H-E)^2 = 0$, not $-1$. Those are fibers, not $(-1)$-curves.

Actually, on $\mathbb{F}_1$ the only $(-1)$-curve is $E$ itself. Blowing down $E$ recovers $\mathbb{P}^2$. On $\mathbb{F}_1$ there is no other contraction to a smooth surface.

On $S = \operatorname{Bl}_{P_1, \ldots, P_n}(\mathbb{P}^2)$ with $n \leq 8$: $-K = 3H - E_1 - \cdots - E_n$.

Check ampleness via Nakai--Moishezon:

• $(-K)^2 = 9 - n > 0$ for $n \leq 8$. Good.
• $(-K) \cdot E_i = 1 > 0$. Good.
• $(-K) \cdot (H - E_i) = 2 > 0$ (strict transform of a line through one point). Good.
• $(-K) \cdot (H - E_i - E_j) = 1 > 0$ (strict transform of a line through two points). Good.
• $(-K) \cdot (2H - E_1 - \cdots - E_5) = 1 > 0$ (conic through five points). Good.

For all irreducible curves $C$ on $S$, one verifies $(-K) \cdot C > 0$ provided the points are in general position. Hence $-K$ is ample.

A smooth cubic surface $S \subset \mathbb{P}^3$ is isomorphic to $\operatorname{Bl}_{P_1, \ldots, P_6}(\mathbb{P}^2)$ where $P_1, \ldots, P_6$ are in general position.

The anticanonical embedding $|-K_S| = |3H - E_1 - \cdots - E_6|$ maps $S$ into $\mathbb{P}^3$ as a cubic surface. Here $(-K)^2 = 3$, confirming the degree.

The 27 lines on $S$: A line $\ell \subset S \subset \mathbb{P}^3$ is a curve with $\deg \ell = 1$, i.e., $(-K) \cdot \ell = 1$, and $\ell \cong \mathbb{P}^1$. The 27 lines are:

• The $6$ exceptional divisors: $E_1, \ldots, E_6$. Each has $(-K) \cdot E_i = 1$.
• The $15$ strict transforms of lines $\overline{P_iP_j}$: class $H - E_i - E_j$ with $(-K) \cdot (H - E_i - E_j) = 3 - 1 - 1 = 1$. There are $\binom{6}{2} = 15$ of these.
• The $6$ strict transforms of conics through $5$ of the $6$ points: class $2H - E_1 - \cdots - \widehat{E_k} - \cdots - E_6 = 2H - \sum_{i \neq k} E_i$. Each has $(-K) \cdot (2H - \sum_{i \neq k} E_i) = 6 - 5 = 1$. There are $\binom{6}{5} = 6$ of these.

Total: $6 + 15 + 6 = 27$ lines. Each is a $(-1)$-curve on $S$.

Each of the 27 lines on a cubic surface meets exactly $10$ of the other $26$ lines. The intersection pattern:

Type $E_i$ (6 lines): $E_i \cdot E_j = 0$ for $j \neq i$. $E_i \cdot (H - E_j - E_k) = 1$ if $i \in \{j, k\}$, $0$ otherwise. $E_i \cdot (2H - \sum_{j \neq k} E_j) = 1$ if $i = k$, $0$ otherwise. So $E_i$ meets: the $5$ lines $H - E_i - E_j$ ($j \neq i$) plus the $5$ conics $2H - \sum_{j \neq k} E_j$ ($k \neq i$). Total: $10$.

Type $H - E_i - E_j$ (15 lines): meets $E_i$, $E_j$, and $4$ lines of type $H - E_k - E_l$ with $\{k,l\} \cap \{i,j\} = \varnothing$ (since $(H-E_i-E_j)(H-E_k-E_l) = 1$ iff $\{i,j\} \cap \{k,l\} = \varnothing$, giving $\binom{4}{2} = 6$... let me recount). We get $(H - E_i - E_j) \cdot (H - E_k - E_l) = 1 - \delta_{ik} - \delta_{il} - \delta_{jk} - \delta_{jl}$, which is $1$ when the sets are disjoint and $0$ when they share one element. With $6$ points, the number of disjoint pairs is $\binom{4}{2} = 6$. Adding the $2$ exceptional curves $E_i, E_j$, and $2$ conics, the count reaches $10$.

This incidence structure is the Schläfli graph, which has deep connections to the $E_6$ root system and the Weyl group $W(E_6)$ of order $51840$, which acts as the symmetry group of the $27$ lines.

The standard Cremona transformation $\sigma: \mathbb{P}^2 \dashrightarrow \mathbb{P}^2$ is defined by $[x:y:z] \mapsto [1/x : 1/y : 1/z] = [yz : xz : xy]$. It is a birational involution, undefined at the three coordinate points $P_1 = [1:0:0]$, $P_2 = [0:1:0]$, $P_3 = [0:0:1]$.

The Cremona transformation factors through blowups:

$\mathbb{P}^2 \xleftarrow{\pi_1} \operatorname{Bl}_{P_1, P_2, P_3}(\mathbb{P}^2) \xrightarrow{\pi_2} \mathbb{P}^2.$

Here $\pi_1$ blows up $P_1, P_2, P_3$, and $\pi_2$ blows down the strict transforms of the three coordinate lines $\overline{P_1P_2}$, $\overline{P_1P_3}$, $\overline{P_2P_3}$ (each of which is a $(-1)$-curve on the blowup).

On the blowup, the map $\pi_2$ is the anticanonical map $|{-K}| = |3H - E_1 - E_2 - E_3|$, which has $(-K)^2 = 9 - 3 = 6$... actually $\sigma^*H' = 2H - E_1 - E_2 - E_3$ (pulling back a line in the target gives a conic through the $3$ points). So $\sigma$ acts on $\operatorname{Pic}$ by $H' \mapsto 2H - E_1 - E_2 - E_3$.

Over $\mathbb{C}$, the blowup $\operatorname{Bl}_P(X)$ has:

• $e(\operatorname{Bl}_P(X)) = e(X) + 1$, where $e$ denotes the topological Euler characteristic. This is because we replace a point ($e = 1$) by $\mathbb{P}^1$ ($e = 2$).

• $b_2(\operatorname{Bl}_P(X)) = b_2(X) + 1$ (one extra generator $[E]$ in $H_2$).

• $H^2(\operatorname{Bl}_P(X), \mathbb{Z}) \cong H^2(X, \mathbb{Z}) \oplus \mathbb{Z} \cdot [E]$, and the intersection form gains a $(-1)$ summand.

Concrete values for $\operatorname{Bl}_n(\mathbb{P}^2)$:

• $e(\mathbb{P}^2) = 3$, so $e(\operatorname{Bl}_n(\mathbb{P}^2)) = 3 + n$.
• $b_2 = 1 + n$, $b_1 = 0$, $b_0 = b_4 = 1$.
• Intersection form: $\operatorname{diag}(1, -1, -1, \ldots, -1)$ of signature $(1, n)$.
• $\chi(\mathcal{O}) = 1$ (birational invariant for rational surfaces).

The Kodaira dimension is a birational invariant, so blowing up does not change it:

• $\operatorname{Bl}_n(\mathbb{P}^2)$ has $\kappa = -\infty$ for all $n$: it is rational.
• Blowing up a point on a K3 surface gives $\kappa = 0$, but now the surface has a $(-1)$-curve, so it is not minimal. Blowing down recovers the K3.
• Blowing up a point on a surface of general type with $K^2 > 0$: the new surface has $K^2 - 1$, still $\kappa = 2$ as long as $K$ is still big (which it always is for surfaces of general type).