Let \(C\) be a curve of genus \(g \geqslant 1\).
Show that for each \(0 \leqslant e \leqslant 2 g-2\) there is a ruled surface \(X\) over \(C\) with invariant \(e\), corresponding to an indecomposable \(\mathcal{E}\). Cf. (2.12).
Let \(e<0\), let \(D\) be any divisor of degree \(d=-e\), and let \(\xi \in H^1(\mathcal{L}(-D))\) be a nonzero element defining an extension \[ 0 \rightarrow \mathcal{O}_C \rightarrow \mathcal{E} \rightarrow \mathcal{L}(D) \rightarrow 0 . \] Let \(H \subseteq|D+K|\) be the sublinear system of codimension 1 defined by \(\ker \xi\), where \(\xi\) is considered as a linear functional on \(H^0(\mathcal{L}(D+K))\). For any effective divisor \(E\) of degree \(d-1\), let \(L_E \subseteq|D+K|\) be the sublinear system \(|D+K-E|+E\). Show that \(\mathcal{E}\) is normalized if and only if for each \(E\) as above, \(L_E \nsubseteq H\). Cf. proof of \((2.15)\).
Now show that if \(-g \leqslant e<0\), there exists a ruled surface \(X\) over \(C\) with invariant \(e\).
Hint: For any given \(D\) in (b), show that a suitable \(\xi\) exists, using an argument similar to the proof of (II, 8.18).
- For \(g=2\), show that \(e \geqslant-2\) is also necessary for the existence of \(X\).
Note. It has been shown that \(e \geqslant-g\) for any ruled surface (Nagata [8]).
Solution
Recall that a rank-two bundle \(\mathcal E\) on \(C\) is normalized when \[ H^0(C,\mathcal E)\ne0 \] but \[ H^0(C,\mathcal E\tensor\mathcal M)=0 \] for every invertible sheaf \(\mathcal M\) of negative degree. For a normalized bundle defining \(X=\PP(\mathcal E)\), the invariant is \[ e=-\deg\det\mathcal E. \]
1. Fix\[ 0\le e\le2g-2. \]There is an invertible sheaf \(\mathcal L\) of degree \(-e\) with\[ \Ext^1_C(\mathcal L,\OO_C)\ne0. \]
Choose an effective divisor \(A\) of degree \(2g-2-e\) and put \[ \mathcal L=\OO_C(A-K_C). \] Then \(\deg\mathcal L=-e\), and \[ \Ext^1_C(\mathcal L,\OO_C) \cong H^1(C,\mathcal L^{-1}) =H^1\qty(C,\OO_C(K_C-A)). \] Serre duality gives \[ H^1\qty(C,\OO_C(K_C-A))^\vee \cong H^0(C,\OO_C(A)), \] which is nonzero because \(A\) is effective.
2. A nonzero class in step 1 defines a normalized extension\[ 0\longrightarrow\OO_C \longrightarrow\mathcal E \longrightarrow\mathcal L \longrightarrow0. \]
The inclusion of \(\OO_C\) gives \(H^0(C,\mathcal E)\ne0\). Let \(\mathcal M\) have negative degree. Tensoring the extension by \(\mathcal M\) gives \[ 0\longrightarrow\mathcal M \longrightarrow\mathcal E\tensor\mathcal M \longrightarrow\mathcal L\tensor\mathcal M \longrightarrow0. \] Both end terms have negative degree because \(\deg\mathcal L=-e\le0\). Hence both have zero \(H^0\), and the long exact sequence gives \[ H^0(C,\mathcal E\tensor\mathcal M)=0. \] Thus \(\mathcal E\) is normalized.
3. The bundle \(\mathcal E\) in step 2 is indecomposable.
Suppose \[ \mathcal E\cong\mathcal A\oplus\mathcal B \] with \(\mathcal A,\mathcal B\) invertible. Since \(H^0(C,\mathcal E)\ne0\), one summand, say \(\mathcal A\), has a nonzero section, so \(\deg\mathcal A\ge0\). If \(\deg\mathcal A>0\), twisting by the negative-degree line bundle \(\mathcal A^{-1}\) produces an \(\OO_C\) summand and contradicts normalization. Thus \(\deg\mathcal A=0\), and its nonzero section forces \(\mathcal A\cong\OO_C\). Consequently \[ \mathcal B\cong\det\mathcal E\cong\mathcal L. \]
If \(\mathcal L\) is nontrivial of degree zero or has negative degree, then \(H^0(C,\mathcal L)=0\), so the injection \(\OO_C\to\OO_C\oplus\mathcal L\) has only a nonzero constant component in the \(\OO_C\) summand and splits. If \(\mathcal L\cong\OO_C\), the injection is a nonzero constant vector in \(k^2\), and a constant change of basis again splits it. Either conclusion contradicts the choice of a nonzero extension class.
4. For every \(0\le e\le2g-2\) there is an indecomposable ruled surface over\(C\) with invariant \(e\).
For the normalized bundle in steps <1>2–<1>3, \[ \deg\det\mathcal E=\deg\mathcal L=-e. \] Thus \(X=\PP(\mathcal E)\) has invariant \(e\), and step 3 gives indecomposability. This proves part (a).
5. For part (b), let\[ 0\longrightarrow\OO_C \longrightarrow\mathcal E \longrightarrow\OO_C(D) \longrightarrow0 \]be the given extension, with \(d=\deg D=-e>0\) and class\[ 0\ne\xi\in H^1(C,\OO_C(-D)). \]Under Serre duality, \(\xi\) is a nonzero linear functional on\[ V=H^0(C,\OO_C(K_C+D)), \]and the subsystem \(H\) is \(\PP(\ker\xi)\).
There are natural identifications \[ \Ext^1_C\qty(\OO_C(D),\OO_C) \cong H^1(C,\OO_C(-D)) \] and \[ H^1(C,\OO_C(-D))^\vee \cong H^0(C,\OO_C(K_C+D)). \] This is exactly the interpretation of \(\xi\) and \(H\) in the statement.
6. Let \(F\) be any effective divisor with \(\deg F\le d-1\). Pulling theextension in step 5 back along\[ \OO_C(D-F)\hookrightarrow\OO_C(D) \]gives a class\[ \beta_F(\xi)\in H^1(C,\OO_C(F-D)), \]and\[ \boxed{\beta_F(\xi)=0\iff L_F\subseteq H,} \qquad L_F=|K_C+D-F|+F. \]
Let \(s_F\) be the canonical section of \(\OO_C(F)\). The inclusion is multiplication by \(s_F\). Under Serre duality, the dual of the induced map \(\beta_F\) is therefore \[ H^0\qty(C,\OO_C(K_C+D-F)) \xrightarrow{\cdot s_F} H^0\qty(C,\OO_C(K_C+D)). \] Its projective image is precisely \(L_F\). Hence \(\beta_F(\xi)=0\) exactly when \(\xi\) annihilates that image, equivalently when \(L_F\subseteq H\).
7. If an effective divisor \(E\) of degree \(d-1\) satisfies\[ L_E\subseteq H, \]then \(\mathcal E\) is not normalized.
Step 6 gives \(\beta_E(\xi)=0\), so the pulled-back extension \[ 0\longrightarrow\OO_C \longrightarrow\mathcal E_E \longrightarrow\OO_C(D-E) \longrightarrow0 \] splits. The splitting followed by \(\mathcal E_E\to\mathcal E\) embeds \(\OO_C(D-E)\) in \(\mathcal E\). Its degree is one, so twisting by its inverse, which has degree \(-1\), gives a nonzero section of a negative twist of \(\mathcal E\). Thus \(\mathcal E\) is not normalized.
8. Conversely, if \(\mathcal E\) is not normalized, then some effectivedivisor \(E\) of degree \(d-1\) satisfies\[ L_E\subseteq H. \]
There is a negative-degree line bundle \(\mathcal M\) and a nonzero morphism \[ \mathcal M^{-1}\longrightarrow\mathcal E. \] Saturating its image gives a line subbundle \[ \mathcal A\hookrightarrow\mathcal E \] with \(a=\deg\mathcal A>0\). Its composite with the quotient \(\mathcal E\to\OO_C(D)\) is nonzero, because a zero composite would factor \(\mathcal A\) through \(\OO_C\), whereas \(H^0(C,\mathcal A^{-1})=0\).
The nonzero map \(\mathcal A\to\OO_C(D)\) has an effective zero divisor \(F\) and identifies \[ \mathcal A\cong\OO_C(D-F). \] Thus \(\deg F=d-a\le d-1\). Since this inclusion already lifts to \(\mathcal E\), the pullback extension along \(\mathcal A\to\OO_C(D)\) splits. By step 6, \[ L_F\subseteq H. \]
Choose an effective divisor \(T\) of degree \(a-1\) and set \(E=F+T\). Then \(\deg E=d-1\), and every section divisible by \(E\) is divisible by \(F\). Hence \[ L_E\subseteq L_F\subseteq H. \]
9. Therefore\[ \boxed{ \mathcal E\text{ is normalized} \iff L_E\nsubseteq H \text{ for every effective }E\text{ of degree }d-1.} \]
Steps <1>7–<1>8 prove the two contrapositives. This proves part (b).
10. For part (c), put \(d=-e\), so\[ 1\le d\le g, \]and fix any divisor \(D\) of degree \(d\). Then\[ \dim H^1(C,\OO_C(-D))=g+d-1. \]
Serre duality gives \[ H^1(C,\OO_C(-D))^\vee \cong H^0(C,\OO_C(K_C+D)). \] Because \(d>0\), the line bundle \(\OO_C(K_C+D)\) is nonspecial. Riemann–Roch therefore gives \[ h^0(C,\OO_C(K_C+D)) =(2g-2+d)+1-g =g+d-1. \]
11. Assume \(d\ge2\). For an effective divisor \(E\) of degree \(d-1\), thenonzero classes \([\xi]\) for which\[ L_E\subseteq H_\xi \]form a projective linear subspace of\[ \PP H^1(C,\OO_C(-D)) \]of dimension \(d-2\).
10. Let\[ W_E = s_EH^0(C,\OO_C(K_C+D-E)) \subseteq V. \]Since\[ \deg(K_C+D-E)=2g-1, \]this line bundle is nonspecial, and Riemann–Roch gives\[ \dim W_E=g. \]By step 6, the bad classes are exactly\[ \PP(W_E^\perp)\subseteq\PP(V^*). \]Its dimension is\[ (g+d-1-g)-1=d-2. \]
12. If \(2\le d\le g\), the union of the bad extension classes in step11 is a proper subset of\[ \PP H^1(C,\OO_C(-D)). \]
11. The universal effective divisor on\(C\times\Sym^{d-1}C\), together with cohomology and base change, therefore givesa rank-\(g\) subbundle of the trivial bundle with fibre \(V\). Its annihilatorprojectivizes to a closed incidence variety\[ I\subseteq\Sym^{d-1}C\times\PP(V^*) \]whose fibre over \(E\) is \(\PP(W_E^\perp)\).
Hence \[ \dim I=(d-1)+(d-2)=2d-3. \] The first factor is projective, so the image of \(I\) in \(\PP(V^*)\) is closed and has dimension at most \(2d-3\). On the other hand, \[ \dim\PP(V^*)=g+d-2. \] Since \(d\le g\), \[ 2d-3\le g+d-3<g+d-2. \] Thus the bad locus cannot fill the projective extension space.
13. For every integer \(1\le d\le g\) there is a nonzero extension class\(\xi\) satisfying the normalization criterion of part (b).
If \(d=1\), the only effective divisor of degree \(d-1=0\) is \(E=0\), and \[ L_0=|K_C+D| \] cannot be contained in the proper hyperplane \(H_\xi\) defined by any nonzero \(\xi\). Such classes exist because step 10 gives \[ h^1(C,\OO_C(-D))=g>0. \]
If \(2\le d\le g\), step 12 shows that the bad classes form a proper closed subset of the projective extension space. Choose \([\xi]\) outside it. Then \[ L_E\nsubseteq H_\xi \] for every effective \(E\) of degree \(d-1\), so step 9 says that the corresponding bundle is normalized.
14. For every\[ -g\le e<0 \]there exists a ruled surface over \(C\) with invariant \(e\).
Set \(d=-e\), choose a divisor \(D\) of degree \(d\), and choose the normalized extension from step 13: \[ 0\longrightarrow\OO_C \longrightarrow\mathcal E \longrightarrow\OO_C(D) \longrightarrow0. \] Then \[ \deg\det\mathcal E=d, \] so \(X=\PP(\mathcal E)\) has invariant \[ -\deg\det\mathcal E=-d=e. \] This proves part (c).
15. If \(g=2\), every ruled surface over \(C\) has\[ \boxed{e\ge-2}. \]
The note printed with this exercise cites Nagata [8] for the general theorem \[ e\ge-g \] for a ruled surface over a genus-\(g\) curve. The retained companion source also identifies part (d) as a direct application of Theorem 1 in Nagata’s cited self-intersection paper. Specializing the stated theorem to \(g=2\) gives \[ e\ge-2, \] as required.
16. Q.E.D.
Steps <1>1–<1>4 prove part (a), steps <1>5–<1>9 prove part (b), steps <1>10–<1>14 prove part (c), and step 15 proves part (d) from the general Nagata bound explicitly cited by the exercise.