Rational Curves in Calabi-Yau Threefolds

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3936 Johnsen and Knutsen Inserting into 1 2 E2 ¼ 0 ¼ mx 2 z 2 þ 3xy þ d0xz þ ayz, we find ½aðma 3d0Þ 9Šz 2 ¼ m 6: ð8Þ If m ¼ 4, we get from (8) that z ¼ ±1 and a(4a 3d0) ¼ 7. Since d0 > 0, we must have (d0, a) ¼ (9, 7), which is present in case (b). (From (7) we get the integer solution (x, y, z) ¼ ( 2, 3, 1).) If m ¼ 5, we get from (8) that z ¼ ±1 and a(5a 3d0) ¼ 8. Since d0 > 0, we must have (d 0, a) ¼ (2, 2), (6, 4) or (13, 8), which are present in case (c). (We can however check from (7) that (2, 2) and (13, 8) do not give any integer solutions for x and y, whereas (6, 4) gives the integer solution (x, y, z) ¼ ( 1, 2, 1).) If m ¼ 6, we get from (8) that either z ¼ 0ora(2a d0) ¼ 3. In the first case, we get the absurdity x ¼ 1=3 from (7), and in the latter we get the only solution (d0, a) ¼ (5, 3), which inserted in (7) gives the absurdity x ¼ 1=3 z. We have therefore shown that L is base point free and that Cliff L ¼ 1. To show that L is ample we have to show by the Nakai criterion that there is no smooth curve E satisfying E 2 ¼ 2 and E.L ¼ 0. By the Hodge index theorem again we have 2L 2 ð E:D 1Þ ¼L 2 ðD EÞ 2 ðL:ðD EÞÞ 2 ¼ 9; giving 1 E.D 1. The cases E.D ¼ ±1 are symmetric by interchanging E and E, so we can restrict to treating the cases E.D ¼ 0 and E.D ¼ 1. We can write E xL þ yD þ zG. We get x ¼ a E:D z þ : ð9Þ 3 3 Combining this with E.L ¼ 0 ¼ 2mx þ 3y þ d0z, we get y ¼ 2ma 3d0 z 9 2mðE:DÞ : 9 Now we use 1 2 E2 ¼ 1 ¼ mx 2 z 2 þ 3xy þ d0xz þ ayz, and find ½aðma 3d0Þ 9Šz 2 ¼ mðE:DÞ 2 We first treat the case E.D ¼ 0. We get ð10Þ 9: ð11Þ ½aðma 3d0Þ 9Šz 2 ¼ 9; ð12Þ which means that z ¼ ±1 or z ¼ ±3.
Rational Curves in Calabi-Yau Threefolds 3937 If z ¼ ±1, we find from (12) that ma 3d0 ¼ 0, and from (9) we find x ¼ a 3 , whence 3ja. If in addition 9ja or m ¼ 6, then (x, y, z) ¼ ±( a=3, ma=9, 1) defines an effective divisor E with E 2 ¼ 2andE.L¼E.D¼0. If z ¼ ±3, we find from (10) that 3j2ma. But since a(ma 3d0) ¼ 8, we get the absurdity 3j16. We now treat the case E.D ¼ 1. Then we have ½aðma 3d0Þ 9Šz 2 ¼ m 9: ð13Þ We now divide into the three cases m ¼ 4, 5, and 6. If m ¼ 4, then (13) reads ½að4a 3d0Þ 9Šz 2 ¼ 5; ð14Þ which means that z ¼ ±1 and a(4a 3d0) ¼ 4, in particular a ¼ 1, 2 or 4. Since d 0 > 0 we only get the solutions ða; d0Þ ¼ð2; 2Þ or ða; d0Þ ¼ð4; 5Þ: ð15Þ From (9) we get x ¼ 1 3 ð1 azÞ ¼1 3 ð1 aÞ, which means that we only have the possibilities (x, z, a, d0) ¼ (1, 1, 2, 2) and (x, z, a, d0) ¼ ( 1, 1, 4, 5). Inserting into (10) we get y ¼ 1 9 ½ð8a 3d0Þz 8Š ¼ 2 and 1; ð16Þ respectively, whence (x, y, z, a, d0) ¼ (1, are the only solutions. If m ¼ 5, then (13) reads 2, 1, 2, 2) and ( 1, 1, 1, 4, 5) ½að5a 3d0Þ 9Šz 2 ¼ 4; ð17Þ which means that z ¼ ±1 or z ¼ ±2. If z ¼ ±1, we have a(5a 3d0) ¼ 5, and since d0 > 0, there is no solution. So z ¼ ±2 and a(5a 3d0) ¼ 8. Again, since d 0 > 0, we only get the solutions ða; d0Þ ¼ð2; 2Þ; ða; d0Þ ¼ð4; 6Þ or ða; d0Þ ¼ð8; 13Þ: ð18Þ From (9) we get x ¼ 1 3 ð1 azÞ ¼1 3 ð1 2aÞ, which means that we only have the possibilities (x, z, a, d0) ¼ ( 1, 2, 2, 2), (3, 2, 4, 6) or ( 5, 2, 8, 13). Inserting into (10) we get y ¼ 1 9 ½ð10a 3d0Þz 10Š ¼2; 6or8; ð19Þ respectively, whence (x, y, z, a, d 0) ¼ ( 1, 2, 2, 2, 2), (3, 6, 2, 4, 6) and ( 5, 8, 2, 8, 13) are the only solutions.
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