Application of Groebner bases to the cup-length of oriented Grassmann manifolds

0.11 MB·English

by  Tomohiro Fukaya

APPLICATION OF GRO¨BNER BASES TO THE CUP-LENGTH OF
ORIENTED GRASSMANN MANIFOLDS
8
0 TOMOHIROFUKAYA
0
2
Abstract. For n = 2m+1 −4(m ≥ 2), wedetermine the cup-length of H∗(Gn,3;Z/2)
n
byfindingaGro¨bnerbasisassociatedwithacertainsubring,whereG istheoriented
a n,3 e
J Grassmann manifold SO(n+3)/SO(n)×SO(3). As its applications, we provide not
e
5 onlyalowerbutalsoanupper bound fortheLS-categoryofGn,3. Wealsostudythe
1 immersionproblemofG .
n,3 e

e
]
T
A
1. Introduction
.
h
t LetR be a commutative ring. The cup-length of R is defined by the greatest number
a
m n such that there exist x1,...,xn ∈ R\R× with x1···xn , 0. Wedenote the cup-length
of R by cup(R). In particular, for a space X and a commutative ring A, the cup-length
[
of X with the coefficient A,is defined by cup(H˜∗(X;A)). Wedenote itby cup (X). Itis
2 A
v well-knownthatcupA(X)isalowerboundfortheLS-categoryofX.
3 Theaim of this paper isto study cup (G ), whereG isthe oriented Grassmann
3 Z/2 n,3 n,k
0 manifoldSO(n+k)/SO(n)×SO(k). NotethaetGn,k is(nk)e-dimensional. Whilethecoho-
4 mology ofG is well-known, that ofG is in vague. However, Korbasˇ [Kor06] gave
. n,2 n,3 e
0 rough estimations for cup (G ) by considering the height of w ∈ H∗(G ;Z/2),
1 e Z/2 n,3 e 2 n,3
wherew isthesecondStiefel-Whitney class.
7 2 e e
0 Theauthorstudies H∗(G ;Z/2)byconsideringGro¨bnerbasesassociatedwithacer-
n,3
v: tain subring of H∗(Gn,3;Ze/2). It seems that, in principle, the method of Gro¨bner bases
i worksbetterinsuchcomplicatedcalculations thanthatofusualalgebraictopology. The
X
e
author employs a computer and carries a huge amount of calculations for finding the
r
a
aboveGro¨bnerbasesandthenhedarestoconjecture:
Conjecture1.1.
2m+1 −3 when2m+1−4 ≤ n ≤ 2m+1+2m −6,
cupZ/2(Gn,3)= 22mm++11 +−23m++k... wwhheennnn == 22mm++11++22mm +−·5·+·+k,20j−1≤−k2≤+2k,,
e  +2j+1+2j−1 +k 0≤ k ≤ 2j−1.
2000MathematicsSubjectClassification. Primary55M30,Secondary57T15,13P10.
Keywordsandphrases. Cup-length;LS-category;Gro¨bnerbases;Immersion.
TheauthorissupportedbyGrant-in-AidforJSPSFellows(193177)fromJapanSocietyforthePromo-
tionofScience.
1
2 TOMOHIROFUKAYA
Whenn= 2m+1 −4(m ≥ 2),ourmethodworksverywellandweobtain:
TheoremA. cup (G ) = n+1whenn = 2m+1−4(m ≥ 2).
Z/2 n,3
Byadimensional reeason, wehave
3
(1) cat(X) ≤ n,
2
where cat(X)denotes the LS-category of aspace X normalized as cat(∗) = 0. Theorem
Agivesnotonlylowerboundsforcat(G ),butalsorefinestheinequality (1). Actually
n,3
weobtain:
e
Corollary . n+1 ≤ cat(G ) < 3nwhen n = 2m+1 −4(m ≥ 2). In particular, wehave
n,3 2
cat(G ) = 5.
4,3 e
Wee will give applications of Theorem A for the immersion problem of G . By the
n,3
classical result of Whitney [Whi44], we know that G immerses into R6n−1. We will
n,3 e
show:
e
Theorem B. The oriented Grassmann manifold G immerses into R6n−3 but not into
n,3
R3n+8 whenn = 2m+1−4(m ≥ 3)andG immerseintoR21 butnotintoR17.
4,3 e
Remark: Walgenbach [Wal01]obtaeined better results onthenon-immersion ofG :
n,3
G does not immerses into R4n−2m+3. On the other hand, due to R. Cohen [Coh85],
n,3 e
G is known tobe immersed into R6n−m+1. Then Theorem B gives a better estimation
en,3
whenm = 2,3.
e
The organization of this paper is as follows. In section 2, we consider the double
coveringmap p : G →G ,whereG istheunorientedGrassmannmanifoldO(n+
n n,3 n,3 n,3
3)/O(n)×O(3). We identify the subring Imp∗ of H∗(G ;Z/2) with a certain algebra
e n n,3
Z/2[w¯ ,w¯ ]/J ,wheregeneratorsof J aregiven. Insection3,settingn = 2m+1−4(m ≥
2 3 n n e
2), wewillgive anexplicit description of generators of the ideal J by using the binary
n
expansion. In section 4, we compute a Gro¨bner basis of J and obtain cup(Imp∗). In
n n
section5,weshowcup(Imp∗)determinescup (G )andobtainit. Asitsapplications,
n Z/2 n,3
wegivesomeestimations forcat(G )andstudytheimmersionproblem ofG .
n,3 e n,3
e e
2. CohomologyofG
n,3
Weconsider thedoublecovering e
(2) p : G →G .
n n,3 n,3
It will be shown that cup (G ) caen be determined by cup(Imp∗). Then we shall
Z/2 n,3 n
investigate cup(Imp∗).
n e
Themod2cohomology of BO(3)isgivenby
H∗(BO(3);Z/2) = Z/2[w ,w ,w ],
1 2 3
GROO¨BNERBASESOFORIENTEDGRASSMANNMANIFOLDS 3
where w is the i-th universal Stiefel-Whitney class. It is well-known that the canonical
i
map i: G → BO(3) induces an epimorphism i∗: H∗(BO(3);Z/2) → H∗(G ;Z/2).
n,3 n,3
Hereafterwedenotei∗(w)bythesamesymbolw ambiguously.
i i
Onecaneasilyseethattheabovedouble covering(2)induces theWangsequence as:
··· −→ Hq−1(G ;Z/2) −·w→1 Hq(G ;Z/2) −p→∗n Hq(G ;Z/2) −→ ··· .
n,3 n,3 n,3
Thenwehave e
Imp∗ (cid:27) Z/2[w ,w ,w ] (w ,Keri∗).
n 1 2 3 1
Letπ: Z/2[w ,w ,w ] → Z/2[w ,w ]betheab.stractringhomomorphismdefinedby
1 2 3 2 3
π(w ) = 0,π(w )= w andπ(w )= w . Thenitinduces theisomorphism
1 2 2 3 3
Imp∗ (cid:27) Z/2[w¯ ,w¯ ] J ,
n 2 3 n
.
whereπ(Keri∗)= J andwedenotew inH∗(G ;Z/2)byw¯ . Notethatthecommutative
n i n,3 i
diagram
e
Gn,3 pn // Gn,3
eı˜ i
(cid:15)(cid:15) (cid:15)(cid:15)
BSO(3) // BO(3)
p∞
yields that ı˜∗(w) = w¯ for i = 2,3 and p∗ : H∗(BO(3);Z/2) → H∗(BSO(3);Z/2) is
i i ∞
expressed byπ: Z/2[w ,w ,w ]→ Z/2[w ,w ].
1 2 3 2 3
Let us give explicit generators of J . Borel [Bor53] showed that Keri∗ is generated
n
bythehomogeneous components ofdegreesn+1,n+2andn+3in
1
.
1+w +w +w
1 2 3
Then it follows that J is generated by the homogeneous components of degrees n+1,
n
n+2andn+3in
1
.
1+w¯ +w¯
2 3
LetN betheuniqueintegerwhichsatisfies2N < n ≤ 2N+1. SincedimG < 4n ≤ 2N+3,
n,3
wehave
e
(1+w¯ +w¯ )2N+3 = 1
2 3
in H∗(G ;Z/2). Thenitfollowsthat
n,3
e 1 = (1+w¯ +w¯ )2N+3−1
1+w¯ +w¯ 2 3
2 3
andhence J isgenerated by
n
s
(3) g = w¯3s−rw¯r−2s
r 3s−r! 2 3
rX≤s≤r
3 2
forr = n+1,n+2,n+3.
4 TOMOHIROFUKAYA
3. Investigatinggeneratorsof J
n
In this section, we investigate generators gn+1, gn+2 and gn+3 of Jn by exploiting the
binaryexpansion.
Let us prepare notation for the binary expansion. To a non-negative integer x with
0 ≤ x< 2k,weassignasequence
ǫ (x) = (x ,...,x )∈ {0,1}k
k k−1 0
suchthat
k−1
(4) x= x2i.
i
Xi=0
(4) is, of course, the binary expansion of x. We denote 1−a by a with a ∈ {0,1}. For
example,wehave
ǫ (2k −1) = (1,...,1)
k
and
ǫ (2k −1− x) = (x ,...,x )
k k−1 0
forǫ(x) = (x ,...,x ). Weoftendenote(x ,...,x )∈ {0,1}k byx .
k−1 0 k 0 k
Tocalculate s modulo2,weusethefollowingwell-knownresultfromelementary
3s−r
numbertheory.(cid:16) (cid:17)
Lemma3.1. Letnand k be non-negative integers such that k ≤ n ≤ 2l −1 and ǫ(n) =
l
(n ,...,n ),ǫ(k) = (k ,...,k ). Then wehave n ≡ 1 (mod 2)if and only if k = 1
l−1 0 l l−1 0 k i
impliesn = 1foreachi. (cid:16) (cid:17)
i
Intherestofthispaper,weassumethat
n = 2m+1−4 (m ≥ 2).
ApplyingLemma3.1tothecoefficients ofgn+1,wehave:
Proposition 3.2. 3s−(sn+1) isevenforallinteger swith n+31 ≤ s≤ n+21,thatisgn+1 = 0.
(cid:16) (cid:17)
Proof. Letǫm(s) = (sm−1,...,s0)for n+31 ≤ s≤ n+21 andletǫm+1(n+1−2s) = (tm,...,t0).
Since s ≤ 2m − 2, there exists an integer i such that s = 0. Let i be the least integer
i
satisfying si = 0,thatis,ǫm(s) = (sm−1,...,si+1,0,1,...,1). Thenitiseasytoshowthat
t = 1. HenceitfollowsfromLemma3.1that s = s ≡ 0 (mod 2). (cid:3)
i 3s−(n+1) n+1−2s
(cid:16) (cid:17) (cid:16) (cid:17)
Next we investigate gn+2. Coefficients of gn+2 are well understood by considering
theirbinaryexpansion asintheabovecaseofgn+1. Let
Sk = s ∈ Z n(k3)+2 ≤ s≤ n(k2)+2,ǫk(s)= (sk−1,...,s0)satisfiesthatif sj = 0,then sj+1 = 1 ,
(cid:26) (cid:12) (cid:27)
(cid:12)
here n(k) =(cid:12)(cid:12)2k+1 − 4. Note that n(k)+2 ≤ s ≤ n(k)+2 implies that s is always equal
3 2 k−1
to 1 for each s ∈ S with ǫ (s) = (s ,...,s ). There is a one-to-one correspondence
k k k−1 0
betweennon-zero coefficients ofgn+2 andSm as:
GROO¨BNERBASESOFORIENTEDGRASSMANNMANIFOLDS 5
Lemma3.3.
s
≡ 1 (mod 2) ifandonlyif s∈ S .
3s−(n+2)! m
Proof. Letǫ (s) = (s ,...,s )for n(m)+2 ≤ s ≤ n(m)+2. Thenwehave
m m−1 0 3 2
ǫm+1(n+2−2s) = (sm−1,...,s0,0)
andhenceLemma3.3followsfromLemma3.1. (cid:3)
Itisconvenientforcalculationsinsection4toindexcoefficientsofgn+2 byexponents
ofw¯ in(3),thatis,3s−(n+2),notby s∈ S . ThenwedefineasetP by
2 m k
P = {p∈ Z|p = 3s−(n(k)+2), s ∈ S }.
k k
P isexpressed bythebinaryexpansion as:
m
Proposition 3.4. Let
∆k = (pk−1,...,p0)∈ {0,1}k|If pl−1 = 1and pl = pl+1 = ··· = pl+2t = 0, then pl+2t+1 = 0 ,
n o
hereweassumethat p = 1. Thenwehave
−1
P = {p∈ Z|ǫ (p) ∈ ∆ }.
m m m
We list some properties of ∆ which will be useful in the following discussion. The
k
proofisstraightforward.
Proposition 3.5. Theset∆ hasthefollowingproperties.
k
(a) pk ∈ ∆k implies(1,pk) ∈ ∆k+1.
(b) pk ∈ ∆k implies(pk,1) ∈ ∆k+1.
(c) ∆ = {(1,p ) ∈ {0,1}m|p ∈ ∆ }⊔{(0,0,p ) ∈ {0,1}m|p ∈ ∆ }.
m m−1 m−1 m−1 m−2 m−2 m−2
ProofofProposition 3.4. Let s ∈ S with ǫ (s) = (s ,...,s ). If s = 0, then
m m m−1 0 m−2
one has s = 1 and s = 1 by definition of S . Then one can easily see that
m−1 m−3 m
(s ,...,s ) ∈ S . If s = 1, then onecan see that(s ,...,s ) ∈ S aswell.
m−2 0 m−3 m−2 m−2 0 m−1
Henceonehasobtained
(5) S = s+2m−1 s∈ S ⊔ s+2m−1 s∈ S .
m m−1 m−2
n (cid:12) o n (cid:12) o
We will show Proposition 3.4 b(cid:12)y induction. We suppo(cid:12)se that it is true for m − 1 and
(cid:12) (cid:12)
m−2. Let s ∈ S and p = 3(s+2m−1)−(n+2). Bythehypothesis oftheinduction,
m−1
ǫ (3s−(n′+2)) = p ∈ ∆ ,wheren′ = 2m −4. Since
m−1 m−1 m−1
p = 3(s+2m−1)−(n+2) = 3s−2m +2+2m−1 = 3s−(n′+2)+2m−1,
wehave
ǫ (p) = (1,p ) ∈ ∆ .
m m−1 m
Similarly,let s ∈ S and p = 3(s+2m−1)−(n+2). Bythehypothesisoftheinduction,
m−2
ǫ (3s−(n′′+2)) = p ∈ ∆ ,wheren′′ = 2m−1−4. Since
m−2 m−2 m−2
p = 3(s+2m−1)−(n+2) = 3s−2m−1+2 = 3s−(n′′+2),
6 TOMOHIROFUKAYA
wehave
ǫ (p) = (0,0,p ) ∈ ∆ .
m m−2 m
Thus,by(5),weobtain
P = p+2m−1 p ∈ P ⊔P
m m−1 m−2
n (cid:12) o
and,by(c)ofProposition 3.5,wehaveesta(cid:12)blished Proposition 3.4.
(cid:12)
(cid:3)
For the last of this section, we investigate gn+3. Coefficients of gn+3 can be well
understood byusingthebinaryexpansion aswellasabove. Let
S′k = s′ ∈ Z n(k3)+3 ≤ s′ ≤ n(k2)+3,ǫk(s′) = (sk−1,...,s1,1)satisfiesthatif sj = 0,then sj+1 = 1 .
(cid:26) (cid:12) (cid:27)
(cid:12)
Quitesimilarl(cid:12)ytoLemma3.3,wecansee:
(cid:12)
Lemma3.6.
s′
≡ 1 (mod 2)ifandonlyif s′ ∈ S′ .
3s′−(n+3)! m
Wegiveanexplicitdescription oftheset
P′ = p′ ∈ Z p′ = 3s′−(n(k)+3), s′ ∈ S′
k k
n (cid:12) o
aswell. Defineamap (cid:12)
(cid:12)
ι : S → S′
m−1 m
byι(s)= 2s+1. Then,obviously, itisbijective. Notethat,for s′ = ι(s),
3s′−(n+3) = 3ι(s)−2m+1+1 = 6s−2m+1+4= 2(3s−(n′+2))
where n′ = 2m −4. Then we have p ∈ P if and only if p′ ∈ P′ such that ǫ (p′) =
m−1 m m
(p ,0)forǫ (p) = p . Hencewehaveobtained:
m−1 m−1 m−1
Proposition 3.7. P′ = {p ∈ Z|ǫ (p) = (p ,0), p ∈ ∆ }.
m m m−1 m−1 m−1
4. Gro¨bnerbasisandcup-length
In this section, by using the result of the previous section, we search for a Gro¨bner
basisof J inordertodeterminecup(Imp∗).
n n
4.1. Gro¨bner bases. Wefirst recall the definition and some facts of Gro¨bner bases by
restrictingtoourspecificcase. Inordertoclarifyourdiscussionandtosimplifynotation,
weshallmakeaconvention ofidentifying atwovariablepolynomial ringwithacertain
set as follows. Let X = {(p,q) ∈ Z2|p ≥ 0,q ≥ 0} and let P[X] denote the set of
finite subset of X. By assigning F ∈ P[X] to w¯pw¯q, we can identify P[X]
(p,q)∈F 2 3
withapolynomialringZ/2[w¯2,w¯3]andweshallmPakethisidentificationthroughoutthis
section. This identification translates the operations in Z/2[w¯ ,w¯ ] into P[X] as: For
2 3
F,G ∈ P[X],
F +G = F ∪G\F ∩G,
GROO¨BNERBASESOFORIENTEDGRASSMANNMANIFOLDS 7
F ·G = (p+r,q+ s).
(p,q)∈XF,(r,s)∈G
Thistranslationofoperationsenablesustohandlethefollowingpolynomialcalculations
easily.
TheorderofXisgivenbytheusuallexicographicorder. Namely,for(p,q),(r,s) ∈ X,
(p,q) ≥ (r,s)ifandonlyif p> r or p = r,q ≥ s.
Byemploying thisorder,wesearchforaGro¨bnerbasisoftheideal J ⊂ P[X].
n
In order to define Gro¨bner bases, we prepare some notation and terminology. The
leadingtermofapolynomial F ∈ P[X]isthemonomial
LT(F) = max{(p,q) ∈ F}.
If there is a monomial (p,q) ∈ X such that (p,q) · LT(G) ∈ F, then the polynomial
F−(p,q)·LT(G)iscalledtheremainderofF ondivisionbyG. Wedenotetheremainder
R = F −(p,q)·LT(G)of F ondivision byG,by
F −G−→∗ R.
Choose F ,...,F ∈ P[X] and give them an arbitrary order. Then it is known that
1 s
thereisanalgorithm toprovidethedecomposition of F ∈ P[X]as
F = A F +···+A F +R
1 1 s s
suchthat A ,...,A ∈ P[X]andRisalinearcombination ofmonomials, noneofwhich
1 s
is divisible by each LT(F ),...,LT(F ). The above R is called the remainder of F on
1 s
division by{F ,...,F }aswell. However,thisdecomposition depends onthechoiceof
1 s
an order of F ,...,F and F ∈ (F ,...,F ) does not imply the remainder R = 0. We
1 s 1 s
canovercomethisdifficultyofremainders bychoosing aGro¨bnerbasisdefinedas:
Definition4.1. LetI beanidealof P[X]. AfinitesubsetG = {G ,...,G }isaGro¨bner
1 s
basisofI if
({LT(F)|F ∈ I})= (LT(G ),...,LT(G )).
1 s
Theorem 4.2. Let I be an ideal of P[X] and let {G ,...,G } be a Gro¨bner basis of I.
1 s
Thentheremainder ofF ∈ I ondivision by{G ,...,G }iszero.
1 s
Buchberger [CLO97]gave acriterion foraset ofpolynomials being aGro¨bner basis
oftheidealgenerated byitasfollows. ForF,G ∈ P[X],theleastcommonmultipleofF
andGisthemonomial
LCM(F,G) = (max{p,r},max{q,s}),
whereLT(F) = (p,q)andLT(G) = (r,s). TheS-polynomial ofF andG ∈ P[X]is
LCM(F,G) LCM(F,G)
S(F,G)= F + G.
LT(F) LT(G)
8 TOMOHIROFUKAYA
Theorem 4.3 ([CLO97]). The set of polynomials {G ,...,G } ⊂ P[X] is a Gro¨bner
1 s
basis of the ideal (G ,...,G ) if and only if the remainder of S(G,G ) on division by
1 s i j
{G ,...,G }iszeroforeachi , j.
1 s
4.2. Search for a Gro¨bner basis of J . The author found the following polynomials
n
experimentally byacomputer calculation. Fornon-negative integersi,t witht−2(2m −
2i) ≡ 0 (mod 3),wedefineapolynomial P(t,i)by
i
P(t,i) = p, t−2p ∈ Xǫ (p) = (p ,0,...,0), p ∈ ∆ ,
( 3 (cid:12) m m−i m−i m−i)
(cid:16) (cid:17) (cid:12)(cid:12) z }| {
P = P(2i+n+1,i)(cid:12)(cid:12).
i (cid:12)
Weshallprovethat{P ,...,P }isaGro¨bnerbasisof J .
0 m n
In order to investigate P, we define the following sets which will be useful for ex-
i
pression. Let∆(i, j,l)and∆¯(i,l)be
l i
∆(i, j,l) = (p ,p ,1,...,1,0,...,0) ∈ {0,1}m (p ,p )∈ ∆ , p , (1,...,1) ,
m−j j−i−l m−j j−i−l m−i−l j−i−l
n z }| { z }| { (cid:12)(cid:12) o
l i (cid:12)
(cid:12)
∆¯(i,l) = (p ,0,0,1,...,1,0,...,0)∈ {0,1}m p ∈ ∆ .
m−i−l−2 m−i−l−2 m−i−l
n z }| { z }| { (cid:12)(cid:12) o
Itiseasytocheck: (cid:12)
(cid:12)
Lemma4.4.
∆(i, j,l) = ∆¯(i,l)⊔∆(i, j,l+1).
Letusbegininvestigating P. Itiseasytoverifythat
i
(6) LT(P)= (2m −2i,2i−1).
i
Proposition 4.5. Wehave P0,...,Pm ∈ Jn. Inparticular P0 = gn+2,P1 = gn+3.
Proof. By Proposition 3.4 and Proposition 3.7, one has P0 = gn+2 and P1 = gn+3. For
i < j,itfollowsfrom(6)that
S(P,P ) =(0,2j −2i)·P +(2j−2i,0)·P
i j i j
i
= p,q ∈ X ǫ (p) = (p ,p ,0,...,0), (p ,p ) ∈ ∆
i,j m m−j j−i m−j j−i m−i
( )
(cid:16) (cid:17) (cid:12)(cid:12) z }| {
(cid:12)
(cid:12) j−i i
+ p,q ∈ X ǫ (p) = (p ,1,...,1,0,...,0), p ∈ ∆
i,j m m−j m−j m−j
( )
(cid:16) (cid:17) (cid:12)(cid:12) z }| { z }| {
(cid:12)
= p,q (p) ∈ X(cid:12) ǫ (p) ∈ ∆(i, j,0) ,
i,j m
(cid:26)(cid:16) (cid:17) (cid:12)(cid:12) (cid:27)
where (cid:12)
(cid:12)
3·2j−2·2i +n+1−2p
q (p) = .
i,j
3
GROO¨BNERBASESOFORIENTEDGRASSMANNMANIFOLDS 9
By the definition of ∆ , one can easily see that ∆(i,0,i+1) = ∆ . Then it follows
k m−i−2
thatS(Pi,Pi+1) = Pi+2 andhencewehaveestablished Proposition 4.5. (cid:3)
Wecalculate theremainders ofS(P,P )ondivisionby{P ,...,P }.
i j 0 m
Lemma4.6. TheremainderofQ = p,q (p) ∈ X ǫ (p) ∈ ∆(i, j,l) ondivisionby
i,j,l i,j m
Pi+l+2 is Qi,j,l+1. (cid:26)(cid:16) (cid:17) (cid:12)(cid:12)(cid:12) (cid:27)
(cid:12)
m − i−l− 2 l i
Proof. Let p(i,l)beǫ (p(i,l)) = (1,...,1,0,0,1,...,1,0,...,0). Thenitiseasytosee
m
z }| { z }| { z }| {
LT(Q(i, j,l)) = p(i,l),q (p(i,l))
i,j
(cid:16) (cid:17)
anditfollowsfrom(6)that
LT(Pi+l+2)= p(i+l,0),qi+j+2,i+j+2(p(i+l,0)) .
(cid:16) (cid:17)
Hencewehave
(2i+l −2i,2j −2i+l+1)·LT(Pi+l+2)= Q(i, j,l).
Ontheotherhand,onecaneasilycheckthat
(2i+l −2i,2j−2i+l+1)·Pi+l+2 = p,qi,j(p) ∈ X ǫm(p) ∈ ∆¯(i,l)
(cid:26)(cid:16) (cid:17) (cid:12)(cid:12) (cid:27)
andthenitfollowsfromLemma4.4that (cid:12)
(cid:12)
Q(i, j,l) −−P−i+−l+−→2∗ Q(i, j,l)+(2i+l −2i,2j−2i+l+1)·Pi+l+2 = Q(i, j,l+1).
(cid:3)
Theorem4.7. Theset{P ,...,P }isaGro¨bnerbasisof J .
0 m n
Proof. By Proposition 4.5, we have J = (P ,...,P ). As in the proof of Proposition
n 0 m
4.5,wehaveS(P,P )= Q(i, j,0)andthenitfollowsfromLemma4.6that,fori< j,
i j
S(P,P ) = Q(i, j,0) −−P−i+−→2∗ Q(i, j,1) −−P−i+−3→∗ ··· −P−→j∗ Q(i, j, j−i−1) −−P−j+−→1∗ 0.
i j
(cid:3)
4.3. Cup-lengthofImp∗. Inordertodeterminecup(Imp∗),letusintroduce newpoly-
n n
nomials. Fornon-negative integers i, j,swith s−2m+1 +2i+1 ≡ 0 (mod 3),wedefinea
polynomial Pˆ(s,i, j)by
Pˆ(s,i, j)= p, s−2p ∈ X ǫ(p) ∈ ∆¯(i, j) .
3
(cid:26)(cid:16) (cid:17) (cid:12)(cid:12) (cid:27)
Thenwehave (cid:12)
(cid:12)
(7) LT(P) = P(2i+n+1,i)+P(2i+n+1,i+2)+ Pˆ(2i+n+1,i, j).
i
1≤j≤Xm−i−2
In order to investigate cup(Imp∗), we shall calculate min{p|(p,0)·LT(P) ∈ J } for
n i n
eachiasfollows.
10 TOMOHIROFUKAYA
Lemma4.8. Letα = min{α|(α,0)·P(t,i)∈ J }fornon-negative integers i,twith
i n
2i−2 +n+1 ≤ t < 2i +n+1, t−2(2m −2i)≡ 0 (mod 3).
Thenwehaveα = 2m−2i−1.Inparticular, α isindependent fromtasabove.
i i
Proof. Notethat
(2i−1,0)·P(t,i) =((cid:18)p, t+2i3−2p(cid:19) ∈ X(cid:12)(cid:12)ǫm(p) = (pm−i1,0,...,0) ∈ {0,1}m, (pm−i,1) ∈ ∆m−i+1),
(cid:12)
(cid:12)
(cid:16)0, t+2i−13−n−1(cid:17)·Pi−1 =((cid:18)p, t+2i3−2p(cid:19) ∈ X(cid:12)(cid:12)(cid:12)ǫm(p) = (pm−i+1,0,...,0) ∈ {0,1}m, pm−i+1 ∈ ∆m−i+1).
(cid:12)
ByProposition 3.5,wehave (cid:12)(cid:12)
(2i−1,0)·P(t,i)+ 0, t+2i−1−n−1 ·P = P(t+2i,i+1)
3 i−1
(cid:16) (cid:17)
andhence
(2i−1,0)·P(t,i)−−P−i−−→1∗ P(t+2i,i+1).
Thenweobtain
(2i−1,0)·P(t,i) −−P−i−−→1∗ P(t+2i,i+1),
(2i,0)·P(t+2i,i+1) −P−→i∗ P(t+2i +2i+1,i+2),
.
.
.
(2m−1,0)·P(t+2i+···+2m−1,m) −P−m−−−1→∗ 0
andthiscompletes theproofofLemma4.8. (cid:3)
Lemma4.9. Letα beasinLemma4.8. Thenwehave(α,0)·Pˆ(s,i, j) ∈ J .
i i n
Proof. QuitesimilarlytotheproofofLemma4.8,onehas
(2j+i +2i,0)·Pˆ(s,i, j)+ 0, s+2i3+1+1 ·Pj+i+1 = P(s+2j+i+1 +2i+1, j+i+3).
(cid:16) (cid:17)
ByLemma4.8,wehave2j+i+2i+αj+i+3 < αiandthenLemma4.9isaccomplished. (cid:3)
ItfollowsfromLemma4.8andLemma4.9that:
Proposition 4.10. Let αi be as in Lemma 4.8. Then we have αi+1 = min{α|(α,0) ·
LT(P)∈ J }.
i n
Corollary4.11. Letχ beafixedintegersuchthat2m+1−2i+1−2i+2 ≤ χ < 2m+1−2i−
1 1
2i+1 andletχ = max{z|(χ ,z) < J }. Thenwehaveχ = 2i−1.
2 1 n 2
Proof. Let(p,q) = LT(P). Then,by(6)andLemma4.8,wehave
i i i
··· > pi−1 +αi+1 > pi+αi+2 > pi+1+αi+3··· ,
··· < qi−1 < qi < qi+1 < ··· .
Hence, by Theorem 4.2 and Theorem 4.7, we have established that, for pi+1 + αi+3 =
2m+1−2i+1−2i+2 ≤ χ1 < 2m+1−2i−2i+1 = pi+αi+2,onehasχ2 = qi+1−1 = 2i+1−2. (cid:3)