Table Of ContentAN OPTIMAL BOUNDEDNESS ON WEAK Q-FANO
THREEFOLDS
JUNGKAI A. CHEN AND MENG CHEN
Abstract. LetX beaterminalweakQ-Fanothreefold. Weprove
8 that P−6(X) > 0 and P−8(X) > 1. We also prove that the anti-
0 canonicalvolumehasauniversallowerbound−K3 ≥1/330. This
X
0
lower bound is optimal.
2
n
a
J
8 1. Introduction
1
A threefold X is said to be a terminal (resp. canonical) Q-Fano
] threefold if X has at worst terminal (resp. canonical) singularities and
G
−K isample, whereK isacanonicalWeildivisoronX. X iscalleda
A X X
terminal weak Q-Fano threefold if X has at worst terminal singularities
.
h
and −K is nef and big.
t X
a We are interested in a conjecture of Miles Reid [8, Section 4.3] which
m
says that P (X) > 0 for almost all Q-Fano 3-folds. There are already
−2
[
several known examples with P = 0 by Iano-Fletcher [4] and Altinok
−2
2
and Reid [1]. Another question that we are interested in is the bound-
v
6 edness of Q-Fano 3-folds, which is equivalent to the boundedness of the
5 anti-canonical volume −K3 . Kawamata [5] first showed the bounded-
3 X
ness of −K3 for terminal Q-Fano 3-folds with Picard number 1. Then
4
. Koll´ar, Miyaoka, Mori and Takagi [7] gave the boundedness for all
2
1 canonical Q-Fano 3-folds. Recently Brown and Susuki [2] proved a
7 sharp lower bound of −K3 for certain Q-Fano 3-folds. However a prac-
0
tical lower bound of −K3 for all Q-Fano 3-folds is still unknown, which
:
v
is another motivation of our paper.
i
X Our main results are the following:
r
a Theorem 1.1. Let X be a terminal weak Q-Fano 3-fold. Then
(i) P > 0 with possibly one exception of basket of singularities;
−4
(ii) P > 0 and P > 1;
−6 −8
(iii) −K3 ≥ 1 . Furthermore −K3 = 1 if and only if the virtual
X 330 X 330
basket of singularities is
1 1 1 1
{ (1,−1,1), (1,−1,2), (1,−1,1), (1,−1,2)}.
2 5 3 11
The first author was partially supported by TIMS, NCTS/TPE and National
Science Council of Taiwan. The second author was supported by both the Pro-
gram for New Century Excellent Talents in University (#NCET-05-0358) and the
National Outstanding Young Scientist Foundation (#10625103).
1
2 JUNGKAIA. CHEN AND MENG CHEN
The lower bound 1 is optimal due to the following example by
330
Iano-Fletcher:
Example 1.2. ([4, Page 158]) The general hypersurface
X ⊂ P(1,5,6,22,33)
66
has −K3 = 1 .
X 330
We now sketch our method of baskets and explain the idea of the
proofs. Recall that Reid’s Riemann-Roch formula describes the Euler
characteristic by counting the contribution fromvirtual quotient singu-
larities, which he calls basket. We remark that when either K or−K
X X
is nef and big, then Euler characteristic is nothing but plurigenus or
anti-plurigenus. Our method in [3] provides a synthetic way to recover
baskets in terms of plurigenera. Even though one can not expect to
recover baskets completely with limited information from plurigenera.
However the possibility of baskets is finite when P is small for small
−m
m.
ThebehaviorofbasketsinQ-Fanocaseissomehownicer. Onereason
is that χ(O ) = 1. And thanks to many effective inequalities derived
X
from the basket trick, we can prove that there are only finitely many
baskets with given P and P (see 3.3). Furthermore we can give a
−1 −2
complete list of those small anti-plurigenera formal baskets satisfying
geometric constrains (2.1),(2.2) and (2.3). This allows us to prove our
statements.
We would like to thank J´anos Koll´ar for his comment on the possi-
bility of using our basket consideration in [3] to classify Q-Fano 3-folds
with small anti-volume. We are grateful to Miles Reid for telling us
their interesting examples with P = 0. Thanks are also due to De-Qi
−2
Zhang for the effective discussions during the preparation of this note.
2. Baskets of pairs and geometric inequalities
In this section, we would like to recall our method developed in [3],
together with some geometrical inequalities which will be the core of
our proof.
A basket B is a collection of pairs of integers (permitting weights)
{(b ,r )|i = 1,··· ,t;b coprime to r }.1 For simplicity, we will fre-
i i i i
quently write a basket in another way, say
B = {(1,2),(1,2),(2,5)}= {2×(1,2),(2,5)}.
1We may drop the assumption of coprime if we simply consider {(db,dr),∗} as
{d×(r,b),∗}. These two baskets share all the same numerical properties in our
discussion.
AN OPTIMAL BOUNDEDNESS ON WEAK Q-FANO THREEFOLDS 3
2.1. Reid’s formula. Let X be a terminal weak Q-Fano 3-fold. Ac-
cording to Reid [8], there is a basket of pairs
r
B = {(b ,r )|i = 1,··· ,t;0 < b ≤ i;b is coprime to r }.
X i i i i i
2
such that, for all integer n > 0,
1
P (X) = n(n+1)(2n+1)(−K3 )+(2n+1)−l(−n)
−n 12 X
where l(−n) = l(n+1) = n jbi(ri−jbi) and · means the smallest
PiPj=1 2ri
residue mod r .
i
The above formula can be rewritten as:
1 b2 1
P = (−K3 + i )− b +3,
−1 2 X X r 2 X i
i
i i
m2 b2 m
P −P = (−K3 + i )− b +2−∆m
−m −(m−1) 2 X X r 2 X i
i
i i
where ∆m = (bim(ri−bim) − bim(ri−bim)) and m ≥ 2.
Pi 2ri 2ri
Notice that all the anti-plurigenera P can be determined by the
−n
basket B and P (X). This leads us to set the following definitions
X −1
for formal baskets.
We recall some definitions and properties of baskets. Especially, we
introduce the notion of packing. All details can be found in Section 4
of [3].
SupposethatB = {(b ,r )|i = 1,··· ,t;0 < b ≤ ri;b is coprime to r }
i i i 2 i i
is a basket. Let n > 1 be an integer. For each i, set l := ⌊nbi⌋ and
i ri
define
1
∆n := l b n− (l2 +l )r ,
i i i 2 i i i
which can be shown to be a non-negative integer. Define ∆n(B) =
t ∆n. One can verify that ∆n(B) = (bin(ri−bin) − bin(ri−bin)).
Pi=1 i Pi 2ri 2ri
We set σ(B) := b and σ′(B) := b2i.
Pi i Pi ri
Given a basket B = {(b ,r )|i = 1,··· ,t} and assume that b + b
i i 1 2
is coprime to r + r , then we say that the new basket B′ := {(b +
1 2 1
b ,r +r ),(b ,r ),··· ,(b ,r )} is a packing of B, denoted as B ≻ B′.
2 1 2 3 3 t t
We call B ≻ B′ a prime packing if b r −b r = 1. A composition of
1 2 2 1
finite packings is also called a packing. So the relation “(cid:23)” is a partial
ordering on the set of baskets.
2.2. Properties of packings. As we have proved in [3], a packing
has the following properties:
Assume B (cid:23) B′. Then
i. σ(B) = σ(B′) and σ′(B) ≥ σ′(B′);
4 JUNGKAIA. CHEN AND MENG CHEN
ii. For all integer n > 1, ∆n(B) ≥ ∆n(B′);
2.3. Formal baskets. We call a pair (B,P˜ ) a formal basket if B is
−1
a basket and P˜ is a non-negative integer. We write
−1
(B,P˜ ) ≻ (B′,P˜ )
−1 −1
if B ≻ B′.
We define some invariants of formal baskets. Considering a formal
basket B = (B,P˜ ), define P˜ (B) := P˜ , the volume
−1 −1 −1
−K3(B) := 2P˜ +σ(B)−σ′(B)−6
−1
and P˜ (B) := 5P˜ +σ(B)−10. So one has
−2 −1
P˜ (B)−P˜ (B) = 2(−K3(B)+σ′(B))+2−σ(B).
−2 −1
For all m ≥ 2, we define the anti-plurigenus in an inductive way:
1 m+1
P˜ −P˜ = (m+1)2(−K3(B)+σ′(B))+2− σ−∆m+1(B).
−(m+1) −m
2 2
Notice that P˜ −P˜ is an integer because −K3(B)+σ′(B) =
−(m+1) −m
2P˜ +σ(B)−6 has the same parity as that of σ(B).
−1
Now if B = B for a terminal weak Q-Fano 3-fold X and P˜ =
X −1
P (X), then one can verify that −K3(B) = −K3 and P˜ (B) =
−1 X −m
P (X) for all m ≥ 2.
−m
2.4. Properties of packings (of formal baskets). By 2.2 and
the above formulae, one can see the following immediate properties of
formal baskets:
Assume B := (B,P˜ ) (cid:23) B′ := (B′,P˜ ). Then
−1 −1
iii. −K3(B)+σ′(B) = −K3(B′)+σ′(B′);
iv. −K3(B) ≤ −K3(B′);
v. P˜ (B) ≤ P˜ (B′) for all m ≥ 2.
−m −m
2.5. Canonial sequence of baskets. Next we recall the “canonical”
sequence of a basket B. Set S(0) := {1|n ≥ 2}, S(5) := S(0) ∪{2} and
n 5
inductively for all n ≥ 5,
b n
S(n) := S(n−1) ∪{ | 0 < b < , b coprime to n}.
n 2
Defined in this way, then each set S(n) gives a division of the interval
(0, 1] = [ω(n),ω(n)] with ω(n),ω(n) ∈ S(n). Let ω(n) = qi+1 and ω(n) =
2 S i+1 i i i+1 i+1 pi+1 i
i
qi with g.c.d(q ,p ) = 1 for l = i,i + 1. Then it is easy to see that
pi l l
q p −p q = 1 for all n and i (cf. [3, Claim A]).
i i+1 i i+1
Now given a basket B = {(b ,r )|r = 1,··· ,t}, we would like to
i i
define new baskets B(n)(B). For each B = (b ,r ) ∈ B, if bi ∈
i i i ri
S(n), then we set B(n) := {(b ,r )}. If bi 6∈ S(n), then ω(n) < bi <
i i i ri l+1 ri
ω(n) for some l. We write ω(n) = ql and ω(n) = ql+1 respectively.
l l pl l+1 pl+1
AN OPTIMAL BOUNDEDNESS ON WEAK Q-FANO THREEFOLDS 5
In this situation, we can unpack (b ,r ) to B(n) := {(r q − b p ) ×
i i i i l i l
(q ,p ),(−r q +b p )×(q ,p )}. Adding up those B(n), we get
l+1 l+1 i l+1 i l+1 l l i
a new basket B(n)(B). B(n)(B) is uniquely defined according to our
construction and B(n)(B) ≻ B for all n. Notice that B = B(n)(B) for
n sufficiently large, e.g. for n ≥ max{r }.
i
In fact, we have
B(n−1)(B) = B(n−1)(B(n)(B)) ≻ B(n)(B)
for all n ≥ 1 (cf. [3, Claim B]). Therefore we have a chain of baskets:
B(0)(B) ≻ B(5)(B) ≻ ... ≻ B(n)(B) ≻ ... ≻ B.
The step B(n−1)(B) ≻ B(n)(B) can be achieved by a number of
successive prime packings. Let ǫ (B) be the number of such prime
n
packings.
We recall the following easy but essential properties.
Lemma 2.6. ([3, Lemma 4.15]) For the sequence {B(n)(B)}, the fol-
lowing statements are true:
(i) ∆j(B(0)(B)) = ∆j(B) for j = 3,4;
(ii) ∆j(B(n−1)(B)) = ∆j(B(n)(B)) for all j < n;
(iii) ∆n(B(n−1)(B)) = ∆n(B(n)(B))+ǫ (B).
n
It follows that ∆j(B(n)(B)) = ∆j(B) for all j ≤ n and
ǫ (B) = ∆n(B(n−1)(B))−∆n(B(n)(B)) = ∆n(B(n−1)(B))−∆n(B).
n
Moreover, given a formal basket B = (B,P˜ ), we can similarly
−1
consider B(n)(B) := (B(n)(B),P˜ ). It follows that
−1
P˜ (B(n)(B)) = P˜ (B) for all j ≤ n.
−j −j
Therefore we can realize the canonical sequence of formal baskets as
an approximation of formal baskets via anti-plurigenera.
2.7. Solving formal baskets by anti-plurigenera.
Wenowstudytherelationbetweenformalbasketsandanti-plurigenera
more closely. For a given formal basket B = (B,P˜ ), we begin by
−1
computing the non-negative number ǫ and B(0),B(5) in terms of P˜ .
n −m
From the definition of P˜ we get:
−m
σ(B) = 10−5P˜ +P˜ ,
−1 −2
1
∆m+1 = (2−5(m+1)+2(m+1)2)+ (m+1)(2−3m)P˜
−1
2
1
+ m(m+1)P˜ +P˜ −P˜ .
−2 −m −(m+1)
2
6 JUNGKAIA. CHEN AND MENG CHEN
In particular, we have:
∆3 = 5−6P˜ +4P˜ −P˜ ;
−1 −2 −3
∆4 = 14−14P˜ +6P˜ +P˜ −P˜ ;
−1 −2 −3 −4
Assume B(0)(B) = {n0 ×(1,r)|r ≥ 2}. By Lemma 2.6, we have
1,r
σ(B) = σ(B(0)(B)) = n0 ;
X 1,r
∆3(B) = ∆3(B(0)(B)) = n0 ;
1,2
∆4(B) = ∆4(B(0)(B)) = 2n0 +n0 .
1,2 1,3
Thus one gets B(0) as follows:
n0 = 5−6P˜ +4P˜ −P˜
1,2 −1 −2 −3
n01,3 = 4−2P˜−1 −2P˜−2 +3P˜−3 −P˜−4
n0 = 1+3P˜ −P˜ −2P˜ +P˜ −σ
1,4 −1 −2 −3 −4 5
n0 = n0 ,r ≥ 5,
1,r 1,r
where σ := n0 . A computation gives:
5 r≥5 1,r
P
ǫ = 2+P˜ −2P˜ +P˜ −σ .
5 −2 −4 −5 5
Therefore we get B(5) as follows:
n5 = 3−6P˜ +3P˜ −P˜ +2P˜ −P˜ +σ
1,2 −1 −2 −3 −4 −5 5
n5 = 2+P˜ −2P˜ +P˜ −σ
2,5 −2 −4 −5 5
n5 = 2−2P˜ −3P˜ +3P˜ +P˜ −P˜ +σ
1,3 −1 −2 −3 −4 −5 5
n5 = 1+3P˜ −P˜ −2P˜ +P˜ −σ
1,4 −1 −2 −3 −4 5
n5 = n0 ,r ≥ 5
1,r 1,r
Because B(5) = B(6), we see ǫ = 0 and on the other hand
6
ǫ = 3P˜ +P˜ −P˜ −P˜ −P˜ +P˜ −ǫ = 0
6 −1 −2 −3 −4 −5 −6
where ǫ := 2σ −n0 ≥ 0.
5 1,5
Going on a similar calculation, one gets:
ǫ = 1+P˜ +P˜ −P˜ −P˜ +P˜ −2σ +2n0 +n0
7 −1 −2 −5 −6 −7 5 1,5 1,6
ǫ = 2P˜ +P˜ +P˜ −P˜ −P˜ −P˜ +P˜
8 −1 −2 −3 −4 −5 −7 −8
−3σ +3n0 +2n0 +n0
5 1,5 1,6 1,7
2.8. Geometric inequalities. We say that a formal basket B =
(B,P˜ ) is geometric if B = (B ,P (X)) for a terminal weak Q-Fano
−1 X −1
3-fold X. By [7], one has that −K · c (X) ≥ 0. Therefore [8, 10.3]
X 2
gives the following inequality:
t t
1
γ(B) := − r +24 ≥ 0 (2.1)
X r X i
i
i=1 i=1
AN OPTIMAL BOUNDEDNESS ON WEAK Q-FANO THREEFOLDS 7
Furthermore −K3(B) = −K3 > 0 gives the inequality:
X
σ′(B) < 2P˜ +σ(B)−6. (2.2)
−1
Moreover by [6, Lemma 15.6.2], whenever P > 0 and P > 0,
−m −n
one has
P ≥ P +P −1. (2.3)
−m−n −m −n
3. Plurigenus
We begin with the following observation, which follows immediately
from the definition of packing and γ:
Lemma 3.1. Given a packing of baskets B ≻ B′, we have γ(B) >
γ(B′). In particular, if inequality (2.1) doesn’t hold for B, then it
doesn’t hold for B′.
3.2. Natation and Convention. For simplicity, we write P for
−m
P˜ in what follows.
−m
In this section, we mainly study those formal baskets (B,P ) satis-
−1
fying inequalities (2.1) and (2.2).
We may and often do abuse the notation of B with B when P˜ is
−1
given.
Thefollowingpropositionprovidesanevidenceabouthowourmethod
is going to work effectively.
Proposition 3.3. Given p ∈ Z+, there are only finitely many formal
i
baskets admitting (P ,P ) = (p ,p ) and satisfying (2.1).
−1 −2 1 2
Proof. The number of pairsin B is upper bounded by σ = 10−5p +p .
1 2
Assume B = {(b ,r )}. Then inequality (2.1) gives
i i
t
1 σ
r ≤ 24+ ≤ 24+ .
X i X r 2
i
i=1 i
Clearly B has finite number of possibilities. This completes the proof.
(cid:3)
3.4. Geometrically constrained baskets with P = 0. We now
−2
study formal baskets, satisfying (2.1) and (2.2), with P = P = 0
−1 −2
and will give a complete classification in this situation. In fact, some
other geometric constrains such as P ≥ P are tacitly employed in
−2 −1
our argument.
Given a formal basket B = (B,0) with P = P = 0. The initial
−1 −2
basket B(0)(B) has datum:
n0 = 5−P ,
1,2 −3
n0 = 4+3P −P ,
1,3 −3 −4
n0 = 1−2P +P −σ .
1,4 −3 −4 5
8 JUNGKAIA. CHEN AND MENG CHEN
By Lemma 3.1, B(0)(B) satisfies (2.1) and thus
1
0 ≤ γ(B(0)(B)) = ( −r)n0 +24
X r 1,r
r≥2
1 24
≤ ( −r)n0 − σ +24
X r 1,r 5 5
r=2,3,4
25 13 21
= +P − P − σ .
−3 −4 5
12 12 20
It follows that
P +σ ≤ P +1. (3.1)
−4 5 −3
We need a more refined inequality, due to the fact that B(5) satisfies
(2.1) again by Lemma 3.1. Because γ(B(5)(B)) = γ(B(0)(B)) − 19ǫ ,
30 5
one gets
25 13 21 19
0 ≤ +P − P − σ − ǫ . (3.2)
−3 −4 5 5
12 12 20 30
On the other hand, by n0 ≥ 0, we have
1,4
P ≥ 2P −1.
−4 −3
Thus we conclude that (P ,P ) = (0,0),(0,1),(1,1),(1,2),(2,3).
−3 −4
Here is our complete classification:
Theorem 3.5. Any geometric basket with P = 0 is among the fol-
−2
lowing list:
Table A
B −K3 P−3 P−4 P−5 P−6 P−7 P−8
No.1. {2×(1,2),3×(2,5),(1,3),(1,4)} 1/60 0 0 1 1 1 2
No.2. {5×(1,2),2×(1,3),(2,7),(1,4)} 1/84 0 1 0 1 1 2
No.3. {5×(1,2),2×(1,3),(3,11)} 1/66 0 1 0 1 1 2
No.4. {5×(1,2),(1,3),(3,10),(1,4)} 1/60 0 1 0 1 1 2
No.5. {5×(1,2),(1,3),2×(2,7)} 1/42 0 1 0 1 2 3
No.6. {4×(1,2),(2,5),2×(1,3),2×(1,4)} 1/30 0 1 1 2 2 4
No.7. {3×(1,2),(2,5),5×(1,3)} 1/30 1 1 1 3 3 4
No.8. {2×(1,2),(3,7),5×(1,3)} 1/21 1 1 1 3 4 5
No.9. {(1,2),(4,9),5×(1,3)} 1/18 1 1 1 3 4 5
No.10. {3×(1,2),(3,8),4×(1,3)} 1/24 1 1 1 3 3 5
No.11. {3×(1,2),(4,11),3×(1,3)} 1/22 1 1 1 3 3 5
No.12. {3×(1,2),(5,14),2×(1,3)} 1/21 1 1 1 3 3 5
No.13. {2×(1,2),2×(2,5),4×(1,3)} 1/15 1 1 2 4 5 7
No.14. {(1,2),(3,7),(2,5),4×(1,3)} 17/210 1 1 2 4 6 8
No.15. {2×(1,2),(2,5),(3,8),3×(1,3)} 3/40 1 1 2 4 5 8
No.16. {2×(1,2),(5,13),3×(1,3)} 1/13 1 1 2 4 5 8
No.17. {(1,2),3×(2,5),3×(1,3)} 1/10 1 1 3 5 7 10
No.18. {4×(1,2),5×(1,3),(1,4)} 1/12 1 2 2 5 6 9
No.19. {4×(1,2),4×(1,3),(2,7)} 2/21 1 2 2 5 7 10
No.20. {4×(1,2),3×(1,3),(3,10)} 1/10 1 2 2 5 7 10
No.21. {3×(1,2),(2,5),4×(1,3),(1,4)} 7/60 1 2 3 6 8 12
No.22. {3×(1,2),7×(1,3)} 1/6 2 3 4 9 12 17
No.23. {2×(1,2),(2,5),6×(1,3)} 1/5 2 3 5 10 14 20
(cid:3)
Proof. This theorem follows from Propositions 3.6, 3.7.
AN OPTIMAL BOUNDEDNESS ON WEAK Q-FANO THREEFOLDS 9
Proposition 3.6. If (P ,P ) = (0,0), then B is of type No.1 in
−3 −4
Table A.
Proof. Now σ ≤ 1 and ǫ = 2+P −σ ≤ 3 by (3.2).
5 5 −5 5
Claim 1. The situation (σ ,P ) = (0,0) does not happen.
5 −5
Proof of the claim. We have
B(5)(B) = {3×(1,2),2×(2,5),2×(1,3),(1,4)}.
If B = B(5)(B), then −K3(B) = σ − σ′ − 6 = − 1 < 0, a contra-
60
diction. Thus B 6= B(5)(B). Assume that B has totally t pairs. Then
t < 8 since B(5) ≻ B. From B(5)(B) we know r = 26. Thus (2.1)
i i
becomes t 1 ≥ 2. Assume r ≤ r ≤ ··· ≤ rP. If t ≤ 4, then r > 2
Pi=1 ri 1 2 t t
and 1 ≤ 3 + 1 < 2. So (2.1) fails. If t = 5, we consider the value
P ri 2 rt
of r . Whenever r = 2, one has r ≥ 3 and 26 = 6+r +r ≤ 6+2r
3 3 4 4 5 5
gives r ≥ 10. Thus 1 ≤ 3 + 1 + 1 < 2. So (2.1) fails. Whenever
5 Pi ri 2 3 10
r ≥ 3, then r ≥ 3 and r ≥ 7. So again 1 ≤ 1 + 2 + 1 < 2, a
3 4 5 Pi ri 3 7
contradiction to (2.1). Therefore we have seen t = 6,7, which means
that B is exactly obtained by 1 or 2 prime packings from B(5)(B).
When t = 7, B must be one of the following cases:
(I). {2×(1,2),(3,7),(2,5),2×(1,3),(1,4)}; −K3 = − 1 < 0 (con-
60
tradiction);
(II). {3×(1,2),(2,5),(3,8),(1,3),(1,4)};−K3 = − 1 < 0 (contra-
120
diction);
(III). {3×(1,2),2×(2,5),(1,3),(2,7)}; −K3 = − 1 < 0 (contradic-
210
tion);
When t = 6, B is nothing but an extra prime packing from one of
I,II and III:
(I-1). {(1,2),2×(3,7),2×(1,3),(1,4)}; 1 < 2 (contradiction);
Pi ri
(I-2). {(1,2),(4,9),(2,5),2×(1,3),(1,4)}; 1 < 2 (contradiction);
Pi ri
(I-3). {2×(1,2),(5,12),2×(1,3),(1,4)}; −K3 = 0 (contradiction);
(I-4). {2×(1,2),(3,7),(3,8),(1,3),(1,4)}; 1 < 2 (contradiction);
Pi ri
(I-5). {2×(1,2),(3,7),(2,5),(1,3),(2,7)}; 1 < 2 (contradiction);
Pi ri
(II-1). {3 × (1,2),(5,13),(1,3),(1,4)}; −K3 = − 1 < 0 (contradic-
156
tion);
(II-2). {3 × (1,2),(2,5),(4,11),(1,4)}; −K3 = − 1 < 0 (contradic-
220
tion);
(II-3). {3×(1,2),(2,5),(3,8),(2,7)}; 1 < 2 (contradiction);
Pi ri
(III-1). {3×(1,2),2×(2,5),(3,10)}; −K3 = 0 (contradiction);
(cid:3)
We go on proving Proposition 3.6.
If σ = 0 and P > 0. Because 2+ P = ǫ ≤ 3, we see P = 1
5 −5 −5 5 −5
andB(5)(B) = {2×(1,2),3×(2,5),(1,3),(1,4)}. Acomputationshows
10 JUNGKAIA. CHEN AND MENG CHEN
that any non-trivial packing of B(5) has γ < 0. Hence B = B(5)(B).
So B corresponds to case No.1 in Table A.
If σ = 1 and P = 0, then we have B(5)(B) = {4×(1,2),(2,5),3×
5 −5
(1,3),(1,s)} with s ≥ 5, −K3 = 1 − 1 and γ = 5−s+ 1 + 1. When
5 s 5 s
s ≥ 6, we have γ < 0, a contradiction. Hence we must have s = 5.
Since −K3(B(5)) = 0, so B(5)(B) ≻ B is nontrivial. However, any non-
trivial packing of B(5)(B) has γ < 0, which still gives a contradiction.
Thus this case can not happen.
Finally if σ = 1 and P > 0, then we get a contradiction from
5 −5
(3.2).
(cid:3)
We have proved Proposition 3.6.
Proposition 3.7. (1) If (P ,P ) = (0,1), then B is of type No.2-
−3 −4
No.6 in Table A;
(2) If (P ,P ) = (1,1), then B is of type No.7-No.17 in Table A;
−3 −4
(3) If (P ,P ) = (1,2), then B is of type No.18-No.21 in Table A;
−3 −4
(4) If (P ,P ) = (2,3), then B is of type No. 22, No. 23 in Table
−3 −4
A.
Proof. (1) By (3.1), we have σ = 0, hence B(0)(B) = {5×(1,2),3×
5
(1,3),2×(1,4)}. By (3.2), we have P = ǫ ≤ 1.
−5 5
If P = 0, then we can easily compute all possible formal baskets B
−5
withB(5)(B) = {5×(1,2),3×(1,3),2×(1,4)},γ > 0and−K3(B) > 0.
In fact, by classifying all baskets with B(5)(B) as above, and verifying
inequalities (2.1),(2.2), the reader should have no difficulty to see that
B has 4 types which correspond to No.2 through No.5 in Table A.
If P = 1, then B(5)(B) = {4 × (1,2),(2,5),2× (1,3),2× (1,4)}.
−5
Because any basket dominated by B(5)(B) has γ < 0, we see B =
B(5)(B) which corresponds to No. 6 in Table A.
(2) In this case, 0 ≤ n0 = −σ gives σ = 0. By (3.2), we have
1,4 5 5
ǫ ≤ 3.
5
If ǫ = 0, then we get B = B(5)(B) = {4 × (1,2),6 × (1,3)} with
5
−K3(B) = 0, a contradiction.
If ǫ = 1, then we get B(5)(B) = {3 × (1,2),(2,5),5× (1,3)}. By
5
computation, we see that B corresponds to No. 7 through No. 12 in
Table A.
If ǫ = 2, then we get B(5)(B) = {2×(1,2),2×(2,5),4×(1,3)}. We
5
see that B corresponds to No. 13 through No. 16 in Table A.
If ǫ = 3, then we get B(5)(B) = {(1,2),3 × (2,5),3 × (1,3)}. We
5
see that B has only one possibility, which is B(5)(B) corresponding to
No. 17 in Table A.
(3) By (3.1), we must have σ = 0. Moreover, by (3.2), we have
5
ǫ ≤ 1.
5
If ǫ = 0, then we get B(5)(B) = {4×(1,2),5×(1,3),(1,4)} and B
5
corresponds to No. 18 through No. 20 in Table A.
