Let \(Y \cong \PP^1\) be a curve in a surface \(X\), with \(Y^2<0\). Show that \(Y\) is contractible (5.7.2) to a point on a projective variety \(X_0\) (in general singular).
Solution
Under the standing hypotheses for this section, \(X\) is a nonsingular projective surface. Put \[ m=-Y^2>0. \]
1. There is a very ample divisor \(H\) on \(X\) such that, with\[ k=H\cdot Y, \]one has\[ \boxed{ k\ge2, \qquad H^1\bigl(X,\OO_X(mH)\bigr)=0.} \]
Choose any very ample divisor \(A\). Since \(A\cdot Y>0\), for all sufficiently large integers \(n\) the divisor \[ H=nA \] is very ample and satisfies \[ H\cdot Y=n(A\cdot Y)\ge2. \] Serre vanishing applied to the ample line bundle \(\OO_X(A)\) gives \[ H^1\bigl(X,\OO_X(mnA)\bigr)=0 \] for all sufficiently large \(n\). Choosing one \(n\) satisfying both conditions gives the required \(H\).
Define \[ D=mH+kY \] and \[ \mathcal M=\OO_X(D). \]
2. The divisor \(D\) has degree zero on \(Y\), and hence\[ \boxed{\mathcal M|_Y\cong\OO_{\PP^1}.} \]
By the definitions of \(m\) and \(k\), \[ \begin{aligned} D\cdot Y &=(mH+kY)\cdot Y\\ &=m(H\cdot Y)+kY^2\\ &=mk-km\\ &=0. \end{aligned} \] Since \(Y\cong\PP^1\), every line bundle of degree zero on \(Y\) is trivial. Therefore \[ \mathcal M|_Y\cong\OO_Y\cong\OO_{\PP^1}. \]
3. For every integer\[ 0\le i\le k, \]one has\[ \boxed{ H^1\bigl(X,\OO_X(mH+iY)\bigr)=0.} \]
We argue by induction on \(i\). The case \(i=0\) is step 1.
Assume the assertion for \(i-1\), where \(1\le i\le k\). The divisor sequence for \(Y\) gives \[ 0 \longrightarrow \OO_X(mH+(i-1)Y) \longrightarrow \OO_X(mH+iY) \longrightarrow \OO_Y(mH+iY) \longrightarrow 0. \] Its restriction term has degree \[ \begin{aligned} (mH+iY)\cdot Y &=mk-im\\ &=m(k-i)\ge0. \end{aligned} \] Thus, because \(Y\cong\PP^1\), \[ \OO_Y(mH+iY) \cong \OO_{\PP^1}\bigl(m(k-i)\bigr) \] and \[ H^1\!\left( Y,\OO_Y(mH+iY) \right)=0. \] The long exact cohomology sequence, together with the induction hypothesis, therefore gives \[ H^1\bigl(X,\OO_X(mH+iY)\bigr)=0. \]
Let \[ s_Y\in H^0(X,\OO_X(Y)) \] be the canonical section whose zero divisor is \(Y\).
4. The line bundle \(\mathcal M\) is generated by global sections at everypoint of\[ X\setminus Y. \]
Since \(H\) is very ample and \(m\ge1\), the line bundle \[ \OO_X(mH) \] is very ample, hence globally generated. Multiplication by \(s_Y^k\) gives a linear map \[ H^0\bigl(X,\OO_X(mH)\bigr) \longrightarrow H^0(X,\mathcal M). \] On \(X\setminus Y\), the section \(s_Y\) is nowhere zero. Therefore at every point \(x\notin Y\), a section of \(\OO_X(mH)\) nonzero at \(x\) remains nonzero after multiplication by \(s_Y^k\). Hence these multiplied sections generate \(\mathcal M\) at every point outside \(Y\).
5. The restriction map\[ H^0(X,\mathcal M) \longrightarrow H^0(Y,\mathcal M|_Y) \]is surjective.
Tensor the divisor sequence of \(Y\) by \(\mathcal M\). Since \[ \mathcal M(-Y) = \OO_X(mH+(k-1)Y), \] we obtain \[ 0 \longrightarrow \OO_X(mH+(k-1)Y) \longrightarrow \mathcal M \longrightarrow \mathcal M|_Y \longrightarrow 0. \] Step 3 gives \[ H^1\bigl(X,\OO_X(mH+(k-1)Y)\bigr)=0. \] The associated long exact sequence therefore makes the displayed restriction map on \(H^0\) surjective.
6. The line bundle \(\mathcal M\) is globally generated on all of \(X\).
Step 4 gives generation away from \(Y\). By step 2, \[ \mathcal M|_Y\cong\OO_Y. \] Choose the nowhere-zero constant section \[ 1\in H^0(Y,\mathcal M|_Y). \] By step 5 it lifts to a section \[ \tau\in H^0(X,\mathcal M). \] For every \(y\in Y\), the value \(\tau(y)\) is nonzero in the fibre \(\mathcal M_y\otimes k(y)\), because its restriction is the constant section \(1\). Thus \(\tau\) generates the fibre at every point of \(Y\). Together with step 4 this proves global generation.
7. The complete linear system \(|\mathcal M|\) defines a morphism\[ \varphi:X\longrightarrow\PP^N \]which maps all of \(Y\) to a single point\[ \boxed{P=\varphi(Y).} \]
Global generation from step 6 gives the morphism associated with the complete linear system.
By step 2, \[ \mathcal M|_Y\cong\OO_Y, \] so every global section of \(\mathcal M\), when restricted to \(Y\), is a constant multiple of one fixed nowhere-zero trivializing section. Therefore all homogeneous coordinate ratios of \(\varphi|_Y\) are constant. Hence \(\varphi|_Y\) is constant, with image a single point \(P\).
8. One has\[ \boxed{\varphi^{-1}(P)=Y.} \]
Choose a basis \[ \sigma_0,\ldots,\sigma_r \] of \[ H^0\bigl(X,\OO_X(mH)\bigr). \] The sections \[ s_Y^k\sigma_0,\ldots,s_Y^k\sigma_r \in H^0(X,\mathcal M) \] all restrict to zero on \(Y\), so their corresponding homogeneous coordinates vanish at \(P\).
Now let \(x\in X\setminus Y\). Since \(\OO_X(mH)\) is globally generated, some \(\sigma_j(x)\) is nonzero. Also \(s_Y(x)\ne0\), so \[ (s_Y^k\sigma_j)(x)\ne0. \] Thus the image \(\varphi(x)\) has at least one of these coordinates nonzero and cannot equal \(P\). Hence no point outside \(Y\) maps to \(P\), proving the assertion.
9. The restriction\[ \boxed{ \varphi|_{X\setminus Y}: X\setminus Y \xrightarrow{\sim} \varphi(X)\setminus\{P\}} \]is an isomorphism.
Let \[ \iota_m:X\hookrightarrow\PP^r \] be the closed immersion defined by the very ample complete linear system \(|mH|\) and the basis \(\sigma_0,\ldots,\sigma_r\) from step 8.
Choose homogeneous coordinates on \(\PP^N\) for \(\varphi\) so that the first \(r+1\) coordinates are the sections \[ s_Y^k\sigma_0,\ldots,s_Y^k\sigma_r. \] Let \[ L=V(z_0,\ldots,z_r)\subseteq\PP^N. \] Step 8 shows \[ \varphi(X)\cap L=\{P\}. \] Hence the linear projection \[ \rho:\PP^N\setminus L\longrightarrow\PP^r, \qquad [z_0:\ldots:z_N]\longmapsto[z_0:\ldots:z_r] \] is defined on \(\varphi(X)\setminus\{P\}\).
For \(x\in X\setminus Y\), \[ \begin{aligned} \rho(\varphi(x)) &= [s_Y(x)^k\sigma_0(x):\cdots:s_Y(x)^k\sigma_r(x)]\\ &= [\sigma_0(x):\cdots:\sigma_r(x)]\\ &= \iota_m(x). \end{aligned} \] Thus \[ \rho\circ\varphi|_{X\setminus Y} = \iota_m|_{X\setminus Y}. \] Because \(\iota_m\) is a closed immersion, its inverse on its image is a morphism. Therefore \[ \bigl(\iota_m|_{X\setminus Y}\bigr)^{-1} \circ \rho \] is a morphism on \(\varphi(X)\setminus\{P\}\) and is inverse to \(\varphi|_{X\setminus Y}\). Step 8 ensures that every point of \(\varphi(X)\setminus\{P\}\) comes from \(X\setminus Y\). Hence the restriction is an isomorphism.
10. Let\[ X_0=\varphi(X)\subseteq\PP^N. \]Then \(X_0\) is a projective variety, and\[ \boxed{ \varphi:X\longrightarrow X_0 } \]contracts precisely \(Y\) to the point \(P\) and is an isomorphism away from\(Y\).
Since \(X\) is projective, its image under the projective morphism \(\varphi\) is closed in \(\PP^N\). Since \(X\) is irreducible, \(X_0\) is irreducible. Thus \(X_0\) is a projective variety.
Step 7 maps \(Y\) to \(P\), step 8 says the fibre over \(P\) is exactly \(Y\), and step 9 says the morphism is an isomorphism \[ X\setminus Y \xrightarrow{\sim} X_0\setminus\{P\}. \] Moreover \(X_0\) has dimension two because it contains the open subset isomorphic to \(X\setminus Y\). This is precisely the required contraction in the sense of (5.7.2). The point \(P\) need not be nonsingular when \(m>1\), which is why the target is allowed to be singular.
11. Q.E.D.
Steps <1>1–<1>6 construct the globally generated line bundle \(\OO_X(mH+kY)\). Steps <1>7–<1>8 show that its morphism contracts exactly \(Y\), while step 9 proves that nothing else changes. Step 10 identifies the image as the required projective surface \(X_0\).