Rational Curves in Calabi-Yau Threefolds

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.