Rational Curves in Calabi-Yau Threefolds

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3942 Johnsen and Knutsen 0 < D.L ¼ (G L þ 2D).L ¼ d0 10 þ 6 ¼ d0 4, whence d0 > 4. We will now show that d0 < 2a as well, so that we end up in case (b) above. To get a contradiction, assume that d0 2a. Write G kR þ Dk, for an integer k 1 such that Dk > 0 and R ä Dk. By our assumption we have D 2 k ¼ 2(k2 þ 1 þ k(d0 2a)) 2, so Dk must have at least one smooth rational curve in its support. Since we have just shown that the only smooth rational curve g such that g.B 0isR, we have g0.B > 0 for any smooth rational curve g0 Dk. Pick one such g0 Dk such that g0.B Dk.B ¼ G.B þ 3k. Then, since also 0 g0.D Dk.D ¼ G.D 3k, and G.B ¼ 3d0 tion 5a a ¼ G.D by our assumptions, we get the contradic- 0 < discðL; D; g0Þ¼2ðg0:DÞðg0:BÞþ18 2ðG:D 3ÞðG:B þ 3Þ þ 18 < 2ðG:DÞðG:BÞþ18 ¼ d: So we are in case (b). Conversely, if m ¼ 5 and 4 < d0 < 2a, one sees that (G L þ 2D) 2 2 and (G L þ 2D).L > 0, whence by Riemann- Roch (G L þ 2D) > 0andG (L 2D) þ (G L þ 2D) is a nontrivial effective decomposition. Case II. G.B ¼ 0. Then d ¼ 18 and 3d0 ¼ ma. Since we assume that L is ample, we have that 9 does not divide a if m ¼ 4 or 5 and 3 does not divide a if m ¼ 6 by Lemma 5.4. Assume that G is not irreducible. Then there has to exist a smooth rational curve g < G such that g.B 0. If g.B < 0, then g.D > 0 by Lemma 5.7, and we can argue as in Case I above. We end up in the case m ¼ 5andg¼L 2D, and since d0 ¼ 5a 3 < 2a this is a special case of (b). So we can assume that g.B ¼ 0 for any smooth rational curve in the support of B. By Lemma 5.7 again, for any such g we have the possibilities ðm; g:L; g:DÞ ¼ð4; 4; 3Þ; ð4; 8; 6Þ; ð6; 2; 1Þ or ð6; 4; 2Þ; ð26Þ whence the case m ¼ 5 is ruled out. To prove that we end up in the cases (a) and (c) above, we have to show that a 6¼ 3, 6 when m ¼ 4 and a 6¼ 1, 2 when m ¼ 6. So assume m ¼ 4and(d0, a) ¼ (4, 3) or (8, 6). Since g.L < G.L ¼ d, we see from (26) that (d0, a) ¼ (8, 6) and (g.L, g.D) ¼ (4, 3) for any g < G. For any such g, consider D :¼ G g > 0. Then (D.L, D.D) ¼ (4, 3). If D 2 2, we get the contradiction from the Hodge index theorem: 64 ¼ 8L 2 L 2 ðD þ DÞ 2 ðL:ðD þ DÞÞ 2 ¼ 49:
Rational Curves in Calabi-Yau Threefolds 3943 If D 2 ¼ 0, we write D xL þ yD þ zG and use the three equations D:L ¼ 8x þ 3y þ 8z ¼ 4; D:D ¼ 3x þ 6z ¼ 3; D 2 ¼ 8x 2 2z 2 þ 6xy þ 16xz þ 12yz ¼ 0: to find the absurdity (x, y, z) ¼ (1, 4=3, 0). So D 2 2, which means that D 2 ¼ 2, since for any smooth rational curve g0 in its support we must have (g0.L, g0.D) ¼ (4, 3). But then D.G ¼ g.G ¼ 1, and D and g have the same intersection numbers with all three generators of Pic S. But then g ¼ D and G would be divisible, a contradiction. Assume now that m ¼ 6 and (d0, a) ¼ (2, 1) or (4, 4). As above we end up with the only possibility (d0, a) ¼ (4, 4) and (g.L, g.D) ¼ (2, 1). Now also (D.L, D.D) ¼ (2, 1), and D 2 ¼ as above. 2 and we reach the same contradiction So we have proved that we end up in cases (a) and (c) above. Conversely, if m ¼ 4 and d0 and a satisfy the conditions in (a), one easily checks that G (3L 4D) þ (G 3L þ 4D) is a nontrivial effective decomposition, since both the components have self-intersection 2 and positive intersection with L. The same holds for the decomposition G (L 2D) þ (G L þ 2D) ifm¼6andd0and a satisfy the conditions in (c). Case III. G.B < 0. As in the proof of Lemma 5.7 we have G.R < 0. If m ¼ 4 and G.R ¼ 1, we again end up with the possibility (G.L, G.D) ¼ (1, 1), in which case G is irreducible. In all other cases, we have (R G) 2 2, whence by Riemann-Roch either R G > 0orG R > 0. Case III(a). R > G. We must have m ¼ 5 or 6, since h 0 (R) ¼ 0ifm ¼ 4by Proposition 5.1. We proceed as in the proof of Lemma 5.7, with G in the place of D, and show that G.R ¼ 1, which gives the cases: ðm; d0; aÞ ¼ð5; 1; 1Þ; ð5; 3; 2Þ; ð5; 4; 3Þ; ð6; 1; 1Þ or ð6; 3; 2Þ: We consider m ¼ 5 first. If d0 ¼ 1, then G is irreducible, and if (d0, a) ¼ (4, 3), we get d ¼ 0, whence the absurdity G L 2D by Lemma 5.6. So we must have (d0, a) ¼ (3, 2) and G.B ¼ 1. 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, 2). Since then disc(L, D, g) ¼ disc(L, D, G) ¼ 14, we get from Lemma 5.6 that
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