Rational Curves in Calabi-Yau Threefolds

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3944 Johnsen and Knutsen g xL þ yD þ zG, for z ¼ ± 1. Using g.L ¼ 1 ¼ 10x þ 3y þ 3z and g.D ¼ 1 ¼ 3x þ 2z, we find the integer solution (x, y, z) ¼ (1, 2, 1), which yields D 2 ¼ 6, where D :¼ G g as usual. However D.L ¼ 2, whence D has at most two components, contradicting its self-intersection number. We next consider m ¼ 6. If d 0 ¼ 1, then G is irreducible, so we must have (d0, a) ¼ (3, 2) and G.B ¼ 3. This yields d ¼ 6. If G is reducible, there exists a smooth rational curve g < G such that g.B < 0. Since g.L < G.L ¼ 3, we get by looking at the list in Lemma 5.7 that (g.L, g.D, g.B) ¼ (1, 1, 3) or (2, 1, 0), yielding respectively disc(L, D, g) ¼ 12 or 18, contradicting Lemma 5.6, which gives z 2 ¼ 2 or 3 respectively. Case III(b). G > R. We have 9 a < 3d0 ma ¼ G:B < 0; and (G R).D ¼ a 3 0, whence 3 a 8 and ma 3 3 a < d0 < ma : ð27Þ 3 We leave it to the reader to verify that there are no integer solutions to (27) for m ¼ 6 and that the only solutions for m ¼ 4 and 5 are m ¼ 4 : ðd0; aÞ ¼ð5; 4Þ; ð9; 7Þ; m ¼ 5 : ðd0; aÞ ¼ð6; 4Þ; ð8; 5Þ; ð13; 8Þ: The cases with m ¼ 5 belong to case (b) above. We now show that we can rule out the cases with m ¼ 4. Assume that G is reducible. Then there has to exist a smooth rational curve g < G such that g.B < 0, and we can use Lemma 5.7 again. Assume first that (d 0, a) ¼ (5, 4), which gives d ¼ 10. Then g.L < 5, whence by Lemma 5.7 we get the possibilities (g.L, g.D, g.B) ¼ (1, 1, 1) or (4, 4, 4), yielding, respectively, disc(L, D, g) ¼ 16 or 14, none of which are divisible by d ¼ 10, a contradiction. Assume now that (d0, a) ¼ (9, 7), which gives d ¼ 4. Then g.L < 9 and by Lemma 5.7 we get the possibilities (g.L, g.D, g.B) ¼ (1, 1, 1), (5, 4, 1), (6, 5, 2) or (4, 4, 4) yielding, respectively, disc(L, D, g) ¼ 16, 10, 2 or 14. By Lemma 5.6 the only possibility is therefore (g.L, g.D, g.B) ¼ (1, 1, 1) with g xL þ yD þ zG, for z ¼ ±2.Ifz ¼ 2 we get the absurdity 1 ¼ g.D ¼ 3x þ 14. If z ¼ 2, we get from the two equations 1 ¼ g.D ¼ 3x 14 and 1 ¼ g.L ¼ 8x þ 3y 18 the solution (x, y, z) ¼ (5, 7, 2), so g 5L 7D 2G, which yields g.G ¼ 0. Since we have just shown that
Rational Curves in Calabi-Yau Threefolds 3945 g is the only smooth rational curve satisfying g < G and g.B < 0, we can write: G kg þ D for an integer k 1 and D > 0 satisfying g 0 :B 0 for any smooth rational curve g 0 < D; ð28Þ D 2 ¼ 2ðk 2 þ 1Þ; ð29Þ D:L ¼ 9 k; whence k 8; ð30Þ D:B ¼ k 1: ð31Þ Now we claim that there has to exist a smooth rational curve g0 < D such that g0.B ¼ 0. Indeed, write D ¼ D0 þ D1, where D0 is the (possibly zero) moving part of jDj, and D1 its fixed part. (Note that D1 6¼ 0 by (29).) Then D 2 0 0 and D0.D1 0, whence D 2 1 D2 ¼ 2(k 2 þ 1). Now D1 is a finite sum of smooth rational curves, and let l denote the number of such curves, counted with multiplicities. One easily finds that D 2 1 2l 2 (29) ffiffiffiffiffiffiffiffiffiffiffiffiffi , whence by l k2 p þ 1 > D:B ¼ k 1; and it follows that there is a smooth rational curve g 0 < D such that g 0.B ¼ 0, as claimed. But then disc(L, D, g 0) ¼ 18, which is not divisible by d ¼ 4, a contradiction. & Now we summarize the numerical conditions obtained in Lemmas 5.5 and 6.1. We need d0 > ma 3 3 a and want L to be very ample with Cliff L ¼ 1 and such that the rational normal scroll T S defined by the pencil jDj is smooth and of maximally balanced scroll type. Moreover, we need G to be smooth and irreducible. If m ¼ 4 this is satisfied if d0 > 4a 3 3 a ,(d0, a) 6¼ (2, 2), (5, 4), (9, 7) and 3d0 6¼ 4a when a 9. The latter means that the tuples (d0, a) ¼ (4, 3) and (8, 6) are allowed. We therefore obtain the following values: For m ¼ 4 : ðd0; aÞ2fð4; 3Þ; ð8; 6Þg; or d0 > 4a 3 3 a ; ðd0; aÞ 62fð2; 2Þ; ð5; 4Þ; ð9; 7Þg and 3d0 6¼ 4a: If m ¼ 5 this is satisfied if (d0, a) 6¼ (2, 2) and either d0 4ord0 2a (since the cases (d0, a) ¼ (6, 4) and (13, 8) from Lemma 5.5 satisfy 4 < d0 < 2a and since also d0 2a implies d0 > 5a 3 3 a ). We also find that
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