Rational Curves in Calabi-Yau Threefolds

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