Let \(f: X \rightarrow X^{\prime}\) be a birational morphism of nonsingular surfaces.
If \(Y \subseteq X\) is an irreducible curve such that \(f(Y)\) is a point, then \(Y \cong \PP^1\) and \(Y^2<0\).
Let \(P^{\prime} \in X^{\prime}\) be a fundamental point of \(f^{-1}\), and let \(Y_1, \ldots, Y_r\) be the irreducible components of \(f^{-1}\left(P^{\prime}\right)\). Show that the matrix \(\left\|Y_i . Y_j\right\|\) is negative definite.
Solution
We use the standing hypotheses of the surfaces chapter, so \(X\) and \(X'\) are nonsingular projective surfaces.
1. The birational morphism \(f\) factors as a finite sequence of pointblowups:\[ X=X_n \xrightarrow{\pi_n} X_{n-1} \longrightarrow \cdots \longrightarrow X_1 \xrightarrow{\pi_1} X_0=X', \]where each\[ \pi_j:X_j\longrightarrow X_{j-1} \]is the blowup of a point and has exceptional curve\[ E_j\cong\PP^1. \]
This is the factorization theorem for birational morphisms of nonsingular projective surfaces T-SRFZMT. Each elementary factor is a monoidal transformation at a point.
2. Every irreducible curve \(Y\subseteq X\) contracted by \(f\) is the finalstrict transform of one of the exceptional curves \(E_j\). In particular,\[ \boxed{Y\cong\PP^1.} \]
For \[ 0\le i\le n, \] let \(Y_i\subseteq X_i\) be the image of \(Y\) under \[ X=X_n\longrightarrow X_i. \] We have \(Y_n=Y\), while \[ Y_0=f(Y) \] is a point. Choose the least index \(j\ge1\) for which \(Y_j\) is a curve. Then \(Y_{j-1}\) is a point and \[ \pi_j(Y_j)=Y_{j-1}. \] The only irreducible curve contracted by the blowup morphism \(\pi_j\) is its exceptional curve, so \[ Y_j=E_j. \]
For every \(i>j\), the morphism \[ \pi_i:Y_i\longrightarrow Y_{i-1} \] does not contract \(Y_i\), hence \(Y_i\) is the strict transform of \(Y_{i-1}\). Blowing up a nonsingular surface at a point preserves a nonsingular curve under strict transform: if the centre lies on the curve, the restriction of the blowup to its strict transform is the blowup of a nonsingular curve at a Cartier divisor, hence an isomorphism; if the centre does not lie on it, nothing changes.
Starting from \[ E_j\cong\PP^1 \] therefore gives \[ Y=Y_n\cong\PP^1. \]
3. The numerical classes of all irreducible curves contracted by \(f\) arelinearly independent in\[ \operatorname{NS}(X)_\RR. \]
At the \(j\)th blowup, the standard blowup decomposition gives \[ \operatorname{NS}(X_j)_\RR = \pi_j^*\operatorname{NS}(X_{j-1})_\RR \oplus \RR[E_j], \] with \[ E_j^2=-1 \] and \(E_j\) orthogonal to the pullback summand FE-SRFBLOW.
Pull \([E_j]\) back through the later blowups by total transform and denote its final class on \(X\) by \[ e_j. \] Iterating the displayed direct-sum decomposition shows that \[ e_1,\ldots,e_n \] are linearly independent.
Let \(\overline E_j\subseteq X\) be the final strict transform of \(E_j\). At any later blowup, a strict-transform class is either unchanged under pullback or is the pullback minus the new exceptional class. Consequently \[ [\overline E_j] = e_j+\sum_{\ell>j}a_{j\ell}e_\ell \] for some integers \(a_{j\ell}\). Thus the change-of-basis matrix from the \(e_j\) to the \([\overline E_j]\) is triangular with every diagonal entry equal to \(1\). Hence \[ [\overline E_1],\ldots,[\overline E_n] \] are linearly independent.
By step 2, every irreducible curve contracted by \(f\) is one of the \(\overline E_j\). Any subset of these classes is therefore linearly independent.
4. Let \(H\) be a very ample divisor on \(X'\) and put\[ D=f^*H. \]Then\[ \boxed{ D^2=H^2>0, \qquad D\cdot C=0 } \]for every curve \(C\) contracted by \(f\).
Because \(f\) is birational of degree one, the projection formula for intersection products gives \[ (f^*H)^2=H^2. \] Very ampleness implies \[ H^2>0. \]
If \(C\) is contracted, then \(f_*C=0\) as a one-cycle. Hence the projection formula gives \[ D\cdot C = f^*H\cdot C = H\cdot f_*C =0. \]
Notice that \(D\) is generally not ample: this zero intersection with the exceptional curves is exactly why the retained source’s statement “pullback of ample is ample” cannot be used.
5. The intersection form is negative definite on the orthogonalcomplement\[ D^\perp \subseteq \operatorname{NS}(X)_\RR. \]
The Hodge index theorem says that the intersection form on \[ \operatorname{NS}(X)_\RR \] has signature \[ (1,\rho(X)-1) \] T-SRFHODGE.
Step 4 gives \(D^2>0\). In a real vector space with a symmetric form of signature \((1,\rho-1)\), the orthogonal complement of any positive-square vector is negative definite: the positive index is already exhausted by the line \(\RR D\). Therefore every nonzero \[ \alpha\in D^\perp \] satisfies \[ \alpha^2<0. \]
6. If \(Y\subseteq X\) is an irreducible curve contracted by \(f\), then\[ \boxed{Y^2<0.} \]
By step 4, \[ D\cdot Y=0, \] so \[ [Y]\in D^\perp. \] The numerical class \([Y]\) is nonzero: if \(A\) is any ample divisor on \(X\), then \[ A\cdot Y>0. \] Step 5 therefore gives \[ Y^2<0. \] Together with step 2 this proves part (a).
7. For the components\[ Y_1,\ldots,Y_r \]of the fibre over the fundamental point \(P'\), every nonzero real vector\((a_1,\ldots,a_r)\) satisfies\[ \boxed{ \left(\sum_{i=1}^r a_iY_i\right)^2<0.} \]
Each \(Y_i\) is contracted by \(f\), so step 4 gives \[ D\cdot Y_i=0. \] Hence \[ Z=\sum_{i=1}^r a_iY_i \] lies in \(D^\perp\).
If the coefficient vector is nonzero, step 3 says that the numerical classes \([Y_i]\) are linearly independent. Thus \[ [Z]\ne0 \] in \(\operatorname{NS}(X)_\RR\). Step 5 now gives \[ Z^2<0. \]
8. The intersection matrix\[ \boxed{ \bigl(Y_i\cdot Y_j\bigr)_{1\le i,j\le r} \text{ is negative definite}.} \]
Let \[ A=(Y_i\cdot Y_j)_{i,j}. \] For a column vector \[ a=(a_1,\ldots,a_r)^T\in\RR^r, \] bilinearity of the intersection product gives \[ a^TAa = \left(\sum_{i=1}^r a_iY_i\right)^2. \] By step 7 this quantity is strictly negative for every nonzero \(a\). This is exactly negative definiteness of \(A\), proving part (b).
9. Q.E.D.
Steps <1>1–<1>3 identify contracted irreducible curves as independent strict transforms of exceptional \(\PP^1\)’s. Steps <1>4–<1>6 combine the positive-square pullback of a very ample divisor with Hodge index to prove part (a). Steps <1>7–<1>8 apply the same negative-definite orthogonal complement to the whole exceptional fibre and prove part (b).