Rational Curves in Calabi-Yau Threefolds

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3940 Johnsen and Knutsen Proof. Set as before R :¼ L 2D. We have 3D.R ¼ D.(3L 6D) D.(3L mD) 0, whence D.R 0, with equality only if m ¼ 6. If equality occurs, we have D < R (since R 2 ¼ 0andR > 0 by Proposition 5.1), whence we have a nontrivial effective decomposition R D þ D0. Since R.L ¼ 6 and L is ample, we have D.L 5, whence (D.L, D.D) ¼ (4, 4) and (2, 1) are the only possibilities. If D ¼ R, then R 2 ¼ 2, whence m ¼ 5and(D.L, D.D, D.B) ¼ (4, 3, 3). So for the rest of the proof, we can assume that D.R < 0 with D 6¼ R. If R 2 ¼ 2 or 0 (i.e., m ¼ 5 or 6), then R > 0 by Proposition 5.1, whence D < R. IfD.R 2, we get a nontrivial effective decomposition R D þ D0 with D.D0 0. But this contradicts Proposition 5.1. So D.R ¼ 1. Since R.L ¼ 4 and 6 for m ¼ 5 and 6 respectively, and L is ample, we have D.L 3and5 respectively. If m ¼ 6 and D.L ¼ 5, we get D.D ¼ 3 and we calculate jdisc(L, D, D)j¼0, contradicting Lemma 5.6. This leaves us with the possibilities listed above for m ¼ 5 and 6. Now we treat the case R 2 ¼ 4, i.e., m ¼ 4. Then we have h 0 (R) ¼ h 1 (R) ¼ 0. If D.R ¼ 1, then, since D.B ¼ 3D.R þ 2D.D ¼ 3 þ 2D.D and D is nef, we get the only possibility (D.L, D.D) ¼ (1, 1). If D.R 2, we get (R D) 2 2, whence by Riemann-Roch either R D > 0 or D R > 0. In the first case we get the contradiction R > D > 0, so we must have L 2D < D < 3L 4D (the latter inequality due to the fact that B 2 ¼ 0, B > 0 and D.B 0). Since L is ample, we therefore get 3 D:L 11; ð21Þ and from the Hodge index theorem that is, 16ð D:B 1Þ ¼ðB DÞ 2 L 2 ððB DÞ:LÞ 2 ¼ð12 D:LÞ 2 ; ð12 D:LÞ2 ð12 3Þ2 D:B b þ 1c b þ 1c ¼6: ð22Þ 16 16 If D.B ¼ 0, then D:D ¼ 3D:L 4 , which means by (21) that (D.L, D.D) ¼ (4, 3) or (8, 6). If D.B ¼ 1, then D:D ¼ 3D:Lþ1 4 , which means by (21) that (D.L, D.D) ¼ (5, 4) or (9, 7). If D.B ¼ 2, then (21) and (22) gives 3 D.L 8 and D:D ¼ 3D:Lþ2 4 , which means that (D.L, D.D) ¼ (6, 5).
Rational Curves in Calabi-Yau Threefolds 3941 Continuing this way up to D.B ¼ 6, one ends up with the choices listed in the lemma. & 6. PROOF OF THEOREM 4.3 We will now complete the four steps of the proof of Theorem 4.3. 6.1. Proof of Step (I) We start with some further investigations of the K3 surface with the Picard lattice introduced in Lemma 5.3: Lemma 6.1. Assume L is ample. Then G is a smooth rational curve if and only if none of these special cases occurs: (a) m ¼ 4, 3d0 ¼ 4a and a > 9, in which case G (3L 4D) þ (G 3L þ 4D) is a nontrivial effective decomposition. (b) m ¼ 5 and 4 < d 0 < 2a, in which case G (L 2D) þ (G L þ 2D) is a nontrivial effective decomposition. (c) m ¼ 6, d0 ¼ 2a and a > 3, in which case G (L 2D) þ (G L þ 2D) is a nontrivial effective decomposition. Proof. Since G 2 ¼ 2andG.H > 0 we only need to show that G is irreducible. Consider B ¼ 3L mD as defined above. Then d :¼ jdiscðL; D; GÞj ¼ j2ðG:DÞðG:BÞþ18j: ð23Þ Case I. G.B > 0. Then d > 18. Assume that G is not irreducible. Then there has to exist a smooth rational curve g < G such that g.B G.B. We also have g.D G.D by nefness of D. Set D :¼ G g > 0. If g.B ¼ G.B, then D.B ¼ 3D.L mD.D ¼ 0, whence D.D > 0, since L is ample, and hence g.D < G.D. In other words we always have (g.D)(g.B) < (G.D)(G.B), whence discðL; D; gÞ ¼2ðg:DÞðg:BÞþ18 < d: ð24Þ If now g.B < 0, we get from Lemma 5.7 that (g.D)(g.B) 16, whence discðL; D; gÞ ¼2ðg:DÞðg:BÞþ18 32 þ 18 14 > d: ð25Þ So we must have disc(L, D, g) ¼ 0, whence by Lemma 5.6 we have m ¼ 5 and g L 2D ¼: R. By ampleness of L we must have
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