Points in general position and exceptional curves on blowups of the plane

The common resolution of a quadratic transformation is described in P-AGH542QUADTRANSFORM. The pullback of a target line is \[ 2h-e_1-e_2-e_3, \] and the target exceptional divisors are the strict transforms of the three fundamental lines, namely \[ h-e_2-e_3,\qquad h-e_1-e_3,\qquad h-e_1-e_2. \] Because no other \(P_i\) lies on a fundamental line, its exceptional divisor is unchanged. This gives the displayed basis formulas. The quadratic transformation is an involution, so solving for the coefficients of \(D\) in the target basis gives the same formulas with primed and unprimed coefficients interchanged, which is the boxed transformation law.

It suffices to consider a transformation centered at \(P_1,P_2,P_3\). Denote the target base points by \(Q_1,Q_2,Q_3\) and the images of \(P_4,P_5,P_6\) by \(R_4,R_5,R_6\).

A line through \(R_4,R_5,R_6\) has class \[ h'-e_4'-e_5'-e_6' = 2h-e_1-e_2-e_3-e_4-e_5-e_6. \] Thus these three target points are collinear exactly when the six source points lie on a conic.

A line through, say, \(Q_1,R_4,R_5\) has class \[ h'-e_1'-e_4'-e_5' = h-e_1-e_4-e_5. \] Hence such a target collinearity would give three collinear source points. The same calculation applies after permuting the labels.

A line through two of \(Q_1,Q_2,Q_3\) and one \(R_j\) would pull back, for example, to the class \[ h'-e_1'-e_2'-e_j'=e_3-e_j. \] No effective curve has this class: its plane degree is zero, so an effective divisor of this degree is supported on exceptional curves with nonnegative coefficients. Finally \(Q_1,Q_2,Q_3\) themselves are the three base points of the inverse quadratic transformation and are noncollinear.

It remains to exclude a conic through all six target points. Its class pulls back to \[ 2h'-\sum_{i=1}^6e_i' = h-e_4-e_5-e_6. \] Such a conic would therefore force \(P_4,P_5,P_6\) to be collinear. Thus the displayed source condition implies the same condition on the target. Since the quadratic transformation is an involution, the converse holds as well.

If the six points satisfy the two stated conditions, step 2 shows that every successive admissible transformation preserves them. In particular no three points become collinear after any finite sequence, so the six points are in general position.

Conversely, general position includes the requirement that the original six points have no collinear triple. Suppose all six nevertheless lie on a conic. Choose any three as centers of an admissible transformation. By the first calculation in step 2, the images of the other three points are collinear. This contradicts general position. This proves (a).

Let \(\mathbf Q\) be obtained from a general-position configuration \(\mathbf P\) by a finite admissible sequence \(w\). Any further finite admissible sequence \(v\) starting from \(\mathbf Q\) gives the concatenated finite sequence \[ v\circ w \] starting from \(\mathbf P\). By general position of \(\mathbf P\), the terminal configuration has no three collinear. Since this holds for every \(v\), \(\mathbf Q\) is itself in general position. This proves (b).

Let \(U\subseteq\PP^2\) be nonempty open and let \[ Z_1,Z_2,\ldots \] be proper closed subsets. Each \(Z_n\) has only finitely many one-dimensional irreducible components. Since \(k\) is uncountable, there are uncountably many lines in \(\PP^2\), so one can choose a line \(\ell\) which is not a component of any \(Z_n\) and which meets \(U\).

For every \(n\), the intersection \[ \ell\cap Z_n \] is finite. Hence \[ \ell(k)\cap\bigcup_nZ_n \] is countable. But \(\ell(k)\cong\PP^1(k)\) is uncountable, and \(U\cap\ell\) is the complement of finitely many points in \(\ell\). Therefore \(U\cap\ell\) contains a point outside every \(Z_n\).

Work on the open subset on which the successive centers occurring in \(w\) are noncollinear and all quadratic transformations are defined at the remaining labeled points. The standard quadratic formula P-AGH46CREMONA shows inductively that every terminal point has homogeneous coordinates which are rational functions of the coordinates of \(P\). Collinearity of a specified terminal triple is therefore the vanishing of a rational determinant. Clearing denominators gives a closed condition on the initial point \(P\).

This determinant is not identically zero. Indeed, cancel the transformations of \(w\) in reverse order, using that each quadratic transformation is birational and involutive. At the level of function fields, pullback by a birational map is injective, so a nonzero incidence equation cannot become the zero equation. Continuing to the initial configuration, an identically vanishing condition would force either a collinearity among the fixed \(r\)-tuple after a finite admissible sequence, contrary to its general position, or an incidence condition requiring the free point \(P\) to lie on a fixed proper plane curve. The latter is also not an identity on \(\PP^2\). Thus the cleared determinant defines a proper closed subset.

There are only finitely many choices of a triple of labels at each stage, so there are only countably many finite words of admissible transformations. For each word there are finitely many terminal triples. Step 6 therefore produces only countably many proper closed bad subsets of \(\PP^2\).

Also exclude the finitely many original lines through pairs of the fixed points and the finitely many points \(P_i\) themselves. Let \(B\) be the union of all these bad subsets and put \[ V=\PP^2\setminus B. \] Step 5, applied inside every nonempty open subset of \(\PP^2\), shows that \(V\) is dense. For \(P_{r+1}\in V\), no collinear triple occurs initially or after any finite admissible word. This is exactly the definition of general position, proving (c).

If \(a=0\), the hypothesis forces every \(b_i=0\), so the conclusion is immediate. Assume \(a>0\).

Put \(c=b_3\). Since \(r\le8\) and \(b_i\le c\) for \(i\ge3\), \[ \sum_{i=1}^r b_i^2 \le b_1^2+b_2^2+6c^2. \] Also \[ b_1+b_2\le a-c \] and \(b_2\ge c\). For fixed sum, the square sum is maximized by making the two entries as unequal as the constraint \(b_2\ge c\) permits. Hence \[ b_1^2+b_2^2 \le (a-2c)^2+c^2. \] Because \(b_1,b_2\ge c\) and \(b_1+b_2+c\le a\), one has \(3c\le a\). Therefore \[ \begin{aligned} \sum_i b_i^2 &\le (a-2c)^2+7c^2\\ &= a^2-4ac+11c^2\\ &\le a^2. \end{aligned} \] The last inequality follows from \(c\le a/3\), since then \(11c\le 11a/3<4a\). Thus \[ D^2=a^2-\sum_i b_i^2\ge0. \]

Let \(C\subseteq X(\mathbf P)\) be irreducible with \(C^2<0\). If \(C\) is an exceptional divisor, the assertion is immediate. Otherwise its plane image has positive degree, so after relabeling its class is \[ C\sim ah-\sum_{i=1}^r b_i e_i, \qquad a>0, \qquad b_1\ge\cdots\ge b_r\ge0. \] By step 8, negativity forces \[ b_1+b_2+b_3>a. \] Perform the admissible quadratic transformation centered at the three points with these largest multiplicities. Step 1 gives the new plane degree \[ a'=2a-b_1-b_2-b_3<a. \] The marked blowup itself is unchanged up to isomorphism, so the transformed curve remains irreducible and has the same negative self-intersection. Step 4 says the new point configuration is again in general position.

Repeat. The nonnegative plane degree decreases strictly, so after finitely many steps it is zero. An irreducible curve of plane degree zero on a point-blowup is one of the exceptional divisors. Reversing the sequence of isomorphisms shows that the original \(C\) is isomorphic to an exceptional divisor. Hence \[ C\cong\PP^1, \qquad C^2=-1. \]

Step 9 gives \[ C^2=-1 \] and \(p_a(C)=0\). Adjunction gives \[ -2=C\cdot(C+K_X)=-1+K_X\cdot C, \] so \[ K_X\cdot C=-1. \] Using \[ K_X=-3h+\sum_i e_i \] and the intersection form gives \[ -3a+\sum_i b_i=-1, \] which is the first equality. The equation \(C^2=-1\) gives the second.

For \(a>0\), step 10 and Cauchy–Schwarz give \[ (3a-1)^2 \le 7(a^2+1), \] hence \[ a^2-3a-3\le0. \] Thus \(a\le3\).

For \(a=1\) or \(2\), subtracting the two equations in step 10 gives \[ \sum_i b_i(b_i-1)=(a-1)(a-2)=0. \] Thus every \(b_i\) is \(0\) or \(1\), and their sums are respectively \(2\) and \(5\), giving the first two nonexceptional rows.

For \(a=3\), put \(x_i=b_i-1\). The equations become \[ \sum_i x_i=1, \qquad \sum_i x_i^2=1. \] Hence exactly one \(x_i\) is \(1\) and the others are zero. This gives \((2,1,1,1,1,1,1)\).

Together with the seven exceptional divisors, the total number of classes is \[ 7+21+21+7=\boxed{56}. \]

There cannot be two distinct irreducible curves in the same listed class: their intersection number would be the self-intersection \(-1\), whereas distinct irreducible curves on a smooth surface have nonnegative intersection. Hence each class has exactly one representative. Step 9 says every irreducible negative curve must be one of these \((-1)\)-curves. This proves the \(r=7\) assertion in (d).

Step 10 and Cauchy–Schwarz give \[ (3a-1)^2 \le 8(a^2+1), \] so \[ a^2-6a-7\le0 \] and hence \(a\le7\). Equality in Cauchy–Schwarz would be necessary when \(a=7\), because then \[ \frac{(3a-1)^2}{8}=a^2+1=50. \] It would force all eight integers \(b_i\) to equal \(20/8=5/2\), impossible. Thus \(a\le6\).

For \(a=1,2\), the argument of step 11 gives respectively two and five entries equal to \(1\).

For \(a=3\), writing \(x_i=b_i-1\) gives \[ \sum_i x_i=0, \qquad \sum_i x_i^2=2. \] Hence one \(x_i=1\), one \(x_j=-1\), and all others are zero, giving \((2,1,1,1,1,1,1,0)\).

For \(a=4\), again put \(x_i=b_i-1\). Then \[ \sum_i x_i=3, \qquad \sum_i x_i^2=3. \] Since \[ \sum_i x_i(x_i-1)=0 \] and every integer \(x_i\ge-1\) has \(x_i(x_i-1)\ge0\), each \(x_i\) is \(0\) or \(1\). Exactly three are \(1\), giving \((2,2,2,1,1,1,1,1)\).

For \(a=5\), put \(y_i=b_i-2\). Then \[ \sum_i y_i=-2, \qquad \sum_i y_i^2=2. \] Thus exactly two \(y_i\) equal \(-1\) and the rest are zero, giving \((2,2,2,2,2,2,1,1)\).

For \(a=6\), the same substitution gives \[ \sum_i y_i=1, \qquad \sum_i y_i^2=1. \] Hence exactly one \(y_i=1\), giving \((3,2,2,2,2,2,2,2)\).

The numbers of permutations are \[ 8,\ 28,\ 56,\ 56,\ 56,\ 28,\ 8, \] whose sum is \[ \boxed{240}. \]

As in step 12, apply an admissible transformation at three points with largest multiplicities. Direct substitution in step 1 gives the degree reductions \[ 6\longmapsto5\longmapsto4\longmapsto2\longmapsto1\longmapsto0 \] for the corresponding rows, while the degree-\(3\) row maps to the degree-\(2\) row. Thus every listed class reduces to an exceptional divisor. Reversing the admissible sequence produces a smooth rational representative of that class on the original surface.

Uniqueness again follows from negative self-intersection: two distinct irreducible curves in the same class would have intersection \(-1\). Finally step 9 excludes every other irreducible curve of negative self-intersection. This completes (d).

Start with the line \[ L=P_1P_2, \] whose strict-transform class is \[ C_0=h-e_1-e_2. \] It is a nonsingular rational \((-1)\)-curve. Suppose at some stage its image has class \[ C=dh-\sum_{i=1}^9m_i e_i, \qquad d>0, \qquad m_i\ge0. \] Because admissible transformations identify the marked blowups, \(C\) remains a rational \((-1)\)-curve. Hence adjunction gives \[ K_X\cdot C=-1, \] or equivalently \[ \sum_{i=1}^9m_i=3d-1. \]

Relabel so that \[ m_1\le m_2\le m_3\le\cdots\le m_9. \] The sum of the three smallest multiplicities is at most one third of the total: \[ m_1+m_2+m_3 \le \frac{3d-1}{3} < d. \] Since it is an integer, \[ m_1+m_2+m_3\le d-1. \] Transform at these three points. Step 1 gives the new degree \[ d' = 2d-m_1-m_2-m_3 \ge d+1. \] The new multiplicities at the three target base points are \[ d-m_2-m_3,\qquad d-m_1-m_3,\qquad d-m_1-m_2. \] Each is positive because each pair sum is at most \(m_1+m_2+m_3\le d-1\); the other six multiplicities are unchanged. Thus the same argument applies again.

Step 4 guarantees that every successive nine-point configuration remains in general position. Iterating therefore gives finite admissible sequences for which the plane degree of the transform of \(L\) tends to infinity.

This line class lies in the same orbit as an exceptional class: a quadratic transformation centered at \(P_1,P_2,P_3\) contracts the line \(P_1P_2\) to the target exceptional point \(Q_3\), so on the blowups its class becomes \(e_3'\). Hence the orbit of an exceptional class is infinite.

For any finite admissible word, take an exceptional divisor on the terminal marked blowup and pull it back through the induced isomorphism to the original surface \(X\). The result is an irreducible nonsingular rational curve of self-intersection \(-1\), and its Picard class is the corresponding orbit element. Infinitely many orbit classes therefore give infinitely many distinct curves on \(X\). This proves (e).

Steps <1>2–<1>3 prove (a), step 4 proves (b), and steps <1>5–<1>7 prove (c). Steps <1>8–<1>14 classify all negative curves for \(r=7,8\) and give the counts \(56\) and \(240\), proving (d). Steps <1>15–<1>16 prove the starred \(r=9\) assertion (e).