Rational Curves in Calabi-Yau Threefolds

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3938 Johnsen and Knutsen If m ¼ 6, then (13) reads ½að6a 3d0Þ 9Šz 2 ¼ 3z 2 ½að2a d0Þ 3Š ¼ 3: ð20Þ We obtain z 2 [a(2a d0) 3] ¼ 1, which gives z ¼ ±1 and a(2a d0) ¼ 2. Since d0 > 0, the latter yields (a, d0) ¼ (2, 3). If z ¼ 1, then (9) gives the absurdity x ¼ 1 3 . For z ¼ 1 we obtain (x, y) ¼ (1, 3) from (9) and (10). Hence (x, y, z, a, d0) ¼ (1, 3, 1, 2, 3) is the only solution for m ¼ 6. So we have proved that L is ample except for the cases (a)–(d). It is well-known (see e.g., Saint-Donat, 1974 or Knutsen, 2001a) that an ample line bundle with Cliff L ¼ 1 is very ample. If L is ample, it follows from Proposition 5.1 that jDj and jL Dj are base point free and h 1 (L D) ¼ h 1 (L 2D) ¼ 0. & We get the corresponding statement for H: Lemma 5.5. Assume n, d and g do not satisfy any of the following conditions: (i) na ¼ 3d, with 9ja ifn 1, 2 (mod 3), and 3ja ifn 0(mod 3). (ii) n 0(mod 3), a ¼ 2 and d ¼ 3 þ 2 (iii) 3 ðn 6Þ. n 1(mod 3) and d ¼ d0 þ 2 3 ðn 4Þ, for (d0, a)¼ (2, 2), (5, 4) or (9, 7). (iv) n 2(mod 3) and d ¼ d0 þ 2 3 ðn 5Þ, for (d0, (13, 8). a)¼ (2, 2), (6, 4) or Then H is very ample and Cliff H ¼ 1. Moreover, h 1 (H iD) ¼ 0 for all i 0 such that H iD 0. Denote by T the scroll defined by D. Then T is smooth and of maximally balanced scroll type. Proof. The first two statements are clear since D.H ¼ 3andH¼Lþ n 4 b 3 cD, with Cliff L ¼ 1 and D nef, since the cases (i), (ii), (iii) and (iv) are direct translations of the cases (a), (d), (b) and (c), respectively, in Lemma 5.4 above. We now prove that h 1 (H iD) ¼ 0 for all i 0 such that H iD 0. By Lemma 5.4 we have that h 1 (L D) ¼ h 1 (L 2D) ¼ 0. Since D is nef we have h 1 (L þ iD) ¼ 0 for all i 0. Now let R :¼ L 2D. Then R 2 ¼ 4, 2 or 0, corresponding to L 2 ¼ 8, 10 or 12. Since we have h 1 (R) ¼ 0weget h 0 (R) ¼ 0, 1 or 2 respectively. In the first case we are therefore done, and in the second, we clearly have h 0 (R D) ¼ 0, and now we also want
Rational Curves in Calabi-Yau Threefolds 3939 to show this for L 2 ¼ 12. Assume that h 0 (R D) > 0. Then, we have R ¼ D þ D; where jDj is the moving part of jRj and D is the fixed part. Since R.D ¼ 3, we get D.D ¼ 3, and since D 2 ¼ 0 we get D 2 ¼ 6. Moreover D.L ¼ (L 3D).L ¼ 3. This yields D ¼ G1 þ G2 þ G3; where the Gi’s are smooth non-intersecting rational curves satisfying Gi.L ¼ Gi.D ¼ 1 for i ¼ 1, 2, 3. Writing Gi ¼ xiL þ yiD þ ziG, the three equations Gi.L ¼ 1, Gi.D ¼ 1 and G 2 i ¼ 2 yield at most two integer solutions (xi, yi, zi), whence at least two of the Gis have to be equal, a contradiction. So we have proved that h 1 (H iD) ¼ 0 for all i 0 such that H iD 0. By Lemma 5.2 it follows that the scroll T is of maximally balanced type, whence smooth. & In the next section we will describe under what conditions G is a smooth rational curve. We end this section with two helpful lemmas. Lemma 5.6. Assume L is ample. Let D xL þ yD þ zG be a divisor on S such that D 2 ¼ 2 and set d 0 :¼jdisc(L, D, D)j. Then d 0 ¼ z 2 d, and is zero if and only if D L 2D and m ¼ 5. Proof. It is an easy computation to show that d 0 ¼ z 2 d. Hence it is zero if and only if z ¼ 0. It is then an easy exercise to find that D L 2D and m ¼ 5. & Lemma 5.7. Let B :¼ 3L mD. (We have B 2 ¼ 0 and B.D ¼ 9, whence by Riemann-Roch B > 0.) If D is a smooth rational curve satisfying D 2 ¼ 2 and D.B 0, then we only have the following possibilities: m ¼ 4 and ðD:L; D:D; D:BÞ ¼ ð1; 1; 1Þ; ð4; 3; 0Þ; ð4; 4; 4Þ; ð5; 4; 1Þ; ð6; 5; 2Þ; ð8; 6; 0Þ; ð9; 7; 1Þ m ¼ 5 and ðD:L; D:D; D:BÞ ¼ ð1; 1; 2Þ; ð3; 2; 1Þ; ð4; 3; 3Þ m ¼ 6 and ðD:L; D:D; D:BÞ ¼ ð1; 1; 3Þ; ð2; 1; 0Þ; ð3; 2; 3Þ; ð4; 2; 0Þ:
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Rational Curves in Calabi-Yau Threefolds

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