Title Minimal free resolution of the third Veronese subring of threevariables

Author(s) Maeda, Takashi

Citation Ryukyu mathematical journal, 12: 9-30

Issue Date 1999-12-30

URL http://hdl.handle.net/20.500.12000/16091

Rights

Ryukyu Math. J .. 12(1999).9-30

MINIMAL FREE RESOLUTION

OF THE THIRD VERONESE SUBRING

OF THREE VARIABLES

TAKASHI MAEDA

ABSTRACT. The third component of the graded minimal free resolu­tion of the third Veronese subring of three variables over a field k ofcharacteristic zero, is proved to be decomposed into {741} EB {732} EB{651} EB {642} EB {633} EB {552} EB {543} 0 S( -4) as GL3(k)-module,where {n, m, l} is the irreducible representation of GL3(k) with thehighest weight (n, m, l) and S is the polynomial ring of ten variables.They are expressed applying the symbolic method of ternary cubicforms in the classical invariant theory.

Introduction

The purpose of this note is to describe the graded minimal free res­olution of the homogeneous coordinate ring of the projective plane JP>2embedded into JP>9 by the third Veronese embedding V3 over a field k ofcharacteristic zero. Our result is explicit because they are expressed byusing the symbolic method of ternary cubic forms in classical invarianttheory[G,GY]. Let R = k[Zll Z2, Z3](3) be the third Veronese subring ofthree variables, i.e. the homogeneous k-algebra generated by all themonomials in Zl, Z2, Z3 of degree three and let S be the homogeneouscoordinate ring of JP>9 with ten variables {aijk; 1 ::; i ::; j ::; k ::; 3},which form the basis of the third symmetric tensor representation ofGL3(k). It is known that the defining ideal of V3(JP>2) in JP>9 is generatedby all the 2 x 2 minors of the catalecticant matrix

C"I a1l2 a1l3 a122 a123 a,~ )a1l2 a122 a123 a222 a223 a223

a1l3 a123 a 133 a223 a233 a333

Received November 30 1999

(1)

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[cf.Pu] and that the graded ring R is the invariant subring of thepolynomial ring k[Zl, Z2, Z3] under the linear action of the cubic rootof unity which is an element of SL3 so that R is Gorenstein tWa] ofhomological dimension seven. Moreover the regularity of R is equal totwo implies that R is an extremal Gorenstein algebra [S,p93]. Fromthese facts and [S,p94,Th.B] we see the graded minimal free resolutionof R is of the form

0- 8(-9) _ 8(_7)27 _ 8(_6)105 _ 8(_5)189 (2)

-8( _4)189 _ 8(_3)105 _ S( _2)27 _ S _ R - 0

The action of GL3 on the polynomial ring 8 = k[aijk] induces a canon­ical action on the exact sequence (2). Let {n, m, l} be the irreduciblerepresentation of GL3 with the dominant heighest weight (n 2 m 2 1)[FH,p76], and the components in (2) denoted by Ui Q9 8(-i) for 2 ::::;i ::::; 7 and i = 9. Then the GL3-modules U2 and U3 are decomposedby

U2 = {4,2}, U3 = {621} EB {54} EB {531} EB {432}

which follow from [G](see Section 2). We shall show in this note

Theorem. U4 = {741}EB{732}EB{651}EB{642}EB{633}EB{552}EB{543}.

The differentials in (2) are explicitly described in the course of cal­culations except from the fourth to the third component S( _5)189 ­S( _4)189. The graded ring R = k[Zl, Z2, Z3](3) is Gorenstein impliesthat the exact sequence (2) is self-dual so Theorem above leads to theirreducible decompositions of the remaining components

Us = {852} EB {843} EB {762} EB {753} EB {744} EB {663} EB {654}

U6 = {954} EB {873} EB {864} EB {765}

U7 = {975}, U9 = {999}

As a Corollary the irreducible decomposition of the degree n com­ponent Rn = SnS3 V of R = k[Zl' Z2, Z3](3) for dim V = 3 and forany n is calculated by taking the degree n part of the exact sequence(2) and by decomposing each component by Littlewood-Richardsonrule[ABW,p249]. However the author does not know whether there isa rule for the decomposition of SnS3V as that of SnS2V [JPW,p117].

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Let V be the vector space of dimension n over k and let R = k[V](d)be the d-th Veronese subring of k[V], which is a module over the poly­nomial ring S = k[SdV] associated with the d-th symmetric represen­tation SdV of V. The graded minimal free resolutions of the S-moduleR = k[V](d) for the pair of integers (n = dim V, d) are known, includingdifferentials and GL(V)-module structures besides graded Betti num­bers, only in the cases (i) (n,2) for any n, (ii) (2, d) for any d (Thesimplest not contained in (i) and (ii) is (3,3), which is the case in thisnote). The case (i) is a special case of [JPW,p139,Th.3.19], Le. thedeterminantal variety of 2 x 2 minors of the generic symmetric matrixof degree n, while the case (ii) is a special case of [Po,p706,PropA.2.3]where she proves that the homogeneous coordinate ring of the n - 1­secant variety in the d-th Veronese embedding has the minimal freeresolution which is the Eagon-Northcott complex [EN] associated witha canonical map Sd-I V 0 S --t V* 0 S(l), where S = k[SdV], In par­ticular, [Po,p707,ExA.2.5] is the case n = d = 3, i.e. the determinantalvariety of the maximal minors of the 3 x 6 matrix (1). However theminimal free resolution of 2 x 2 minors of the generic 3 x 6 matrix [L],which has length ten, dose not yield the resolution (2).

Both cases (i) and (ii) above are proved by a method of collapsinghomogeneous vector bundles [Ke,We], which we now explain briefly.Let G(r, V) be the Grassmannian of r-subspaces of V with the univer­sal rank r subbundle R of V ® 0G(r,V)' Denote by SAV and SAR theSchur modules associated with a partition A = (AI ~ ... ~ Ar ~ 0)and define 'H by the exact sequence

(3)

Then the vector bundle SARover G(r, V) is locally complete intere­section in the product SAV x G(r, V) and the minimal resolution ofthe structure sheaf OS>.R over Os>. VxG(r,V) is given by the Koszulcomplex A*p*'H, where p : SAR C SA V x G(r, V) --t G(r, V) is thesecond projection. On the other hand the image of the first projec­tion 1f : SAR C SA V x G(r, V) --t SA V is the rank variety X r definedin [Po,p689], which is a determinantal variety of r + 1 minors of anS := k[SA V] = S*(SAV*)-homomorphism SA/1 V* ® S --t V ® S( -1).From a spectral sequence of hypercohomology we see the minimal freeresolution of the affine coordinate ring of the rank variety X r overS = S* (SA V*) is given by ... --t Fk --t Fk-I --t ... --t F I --t Fa with

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the k-th component Tk written by

where S = k[S,x V] = S* (s'x V*) and 'H.* is the dual of 'H. [Po,p699,Th.3.3.4]. In the case>. = (d) the rank variety X r is equal to the cone of theT-secant variety Secr(vd(IP[V)) C IP[SdV], and if T = 1 then the exactsequence (3) is simply written by 0 ~ O(-d) ~ SdV 0 0 ~ 'H. ~ 0on G(l, V) = P[V]. Hence Ak+i'H.* fits into the exact sequence

o~ Ak+i 'H.* ~ (Ak+i SdV*) 0 0 ~ (Ak+i-1SdV*) 00(d)

~ (Ak+i-2SdV*) 0 O(2d) ~ ...

The spectral sequence ET(l = Hq(Ak+i-PSdV* 00(pd)) => Hp+q

(Ak+i'H.*) leads to HP(Ak+i'H.*) = Er = HP(E;'o), so the directsummand Hi (Ak+i'H.*) in (4) is given by the homology of the complex

(Ak+1SdV*) 0 SU-l)dV* ~(AkSdV*)0 SidV* (5)

~ (Ak-1SdV *) 0 S(j+l)dV*

In the case (i) d = 2 we can extract the components of :Fk from (5)explicitly [JPWj by the plethysm formula of A*S2V* combined withother collapsing[JPW,p140]. In the case (ii) n = 2 it turns out thatthe complex T* is the Eagon-Nothcott complex [Po]. However, exceptthese two cases, it seems difficUlt, up to now, to understand the co­homology module Hi(Ak+i'H.*) together with their differentials fromthe exact sequence above, and to determine the irreducible decompo­sition as well as graded Betti numbers of the minimal free resulutionof general Veronese subrings, in part for the lack of decomposition for­mula of Al SdV for I, d 2: 3. In this note we study the minimal freeresolution of the title starting with the exact sequence (1) and carryout the calculations from the view of the symbolic method of classi­cal invariant theory [GY]. Our calculations are somewhat lengthy, butthe arguments are quite elementary in which we are only using Pieri'sformula, invariant differential operators and the second fundamentaltheorem of ternary forms.

The paper is organized as follows. Section one contains preliminaryfacts we need from representation theory of GL3 including symbolic

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method on invariant theory. Section two (resp. three) is devoted tothe calculation for the irreducible decomposition of the second (resp.the third) component in the minimal free resolution of the title. Thebase field is assumed to be of characteristic zero. The author is gratefulto Prof. C. Miyazaki for providing the papers [Ka,Pu], as well as foruseful discussions.

1. Preliminaries

Let {Ul, U2, U3} be the basis of the standard representation of GL3

with its dual basis {Xl,X2,X3} :

(1)

for (5 E GL3 (k). For a ternary d-form and its dual form

where i, j, k ~ 0 in the summations run over i +j + k = d and C1k) =d!/i!j!k!, we define actions p and p* of GL3 (k) on the coefficients aijkand (}:ijkl respectively, by the porperty of GL3-invariance :

f = (5(f) = L; (i:k) PCT(aijk) . (xn i(x2Y (xg)k

r = (5(f*) = L; (i:k)P~((}:ijk)' (uni(u~nj(ug)k

for (5 E GL3 (k). If {aijd is a basis of a vector space V then V isthe d-th symmetric tensor representation of GL3 denoted by (d,O).Similarly, if {(}:ijd is a basis of a vector space then it is the dual of thed-th symmetric tensor representation of GL3 denoted by (0, d), and{(}:ijd is the dual basis with respect to {aijk} in the sense that theform L; aijk(}:ijk is GL3-invariant. Let

g* = L;

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where i + j + k = e, be e-forms with linearly independent coefficients{bijk } and {{3ijk}, respectively. Expand the products fg and J*g* as

f - 't""' (d + e) ,L_i k f* .. _ 't""' (d + e) i j k9 - U ijk CijkXl X-2 X 3, 9 - U ijk l'ijkUl u2u3

where Cijk (resp. I'ijk) is a bilinear form of {aijd and {bijk } (resp.{aijd and {{3ijd). Then {cijd (resp. {I'ijd) is a basis of the (d+e)­th symmetric tensor representation (d + e, 0) (resp. the dual of the(d + e)-th symmetric tensor representation (O,d + e)). Next let usconsider the product

f .. - l: Xpqr i x!, k p q r ,pqr (d) ( e) r.l (2). 9 - ijkXl 2 x 3 u l U2 1la, /\ijk = ijk pqT aijkfJpqr

with i + j + k = d and p +q +T = e. In general we call a homogeneouspolynomial of degree d in {Xi} and of degree e in {Ui} a (d, e)- farm.The linearly independent coefficients {AfJ;} of the (d, e)-form fg* in(2) form a basis of the tensor product (d, 0) 0 (0, e). We see from thedefinition above

Lemma 1. Let F = l: J.lfJ;xix~x~ufu~U3 be a (d, e)-form whosecoefficients J.lfj

q; are vectors in (d, 0) 0 (0, e). Then the subspace of

(d,O) 0 (0, e) generated by {J.lfjq;} is GL3 -subspace if and only if F is

an SLa -invariant polynomial.

In view of Lemma 1 we define

Definition. A (d,e)-form F is associated to a GL3 -subspace W of(d, 0) ® (0, e) if the coefficients of F generate W.

We remark that if a form F associated to an irreducible subspaceW in the definition above, then it is uniquely determined by W up toscalar multiples because the action ofGL3 on {xd and {Ui} are definedby (1) and W is irreducible. Now the tensor product (d, 0) 0 (0, e) isdecomposed by

(d,O) 0 (O,e) = (d,e) EB ((d -1,0) 0 (O,e -1)) (3)

where (d, e) is the irreducible representation of SL3 of the highestweight (d, e) [FH,p223]. We shall express in Lemma 2 below the (d, e)­form associated to the irreducible subspace (d, e) of (d, 0) 0 (0, e) ex­plicitly. Let us consider the differential operator

{)2

6. = l:f=18 8Ui Xi

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which is GL3-invariant because {xd is dual to {Ui} with respect tothe action of GL3 . The GL3-homomorphism l!!. : (d,O) 0 (0, e) ~(d - 1,0) 0 (0, e - 1) is the contraction map [FH,p182] so that the(d, e)-form 1Lx l!!.(Jg*) (where Ux = ~1::;i::;3UiXi) is associated to thesubspace (d - 1,0) 0 (0, e - 1) of (d,O) 0 (0, e) in the decomposition(3). Therefore if a nonzero (d, e)-form F whose coefficients are vectorsin (d, 0)0(0, e) is annihilated by l!!. then F is associated to the subspace(d, e) because (d, e) is irreducible.

Lemma 2. Assume d 2 e and let F = fg* in (2). Then the (d,e)­form associated to (d, e) is equal to

F(1") = 1" + ),luxl!!.(1") + ),2u;l!!.2(F') + ... + ),eu~l!!.(1") (4)

where Ak = (-I)k(k!)-2 . (d +:+I) -1

Proof We have only to prove l!!.F = O. First we show

l!!.(u~l!!.kF) = k(d + e + 2 - k)u~-ll!!.kF + u~l!!.k+l 1".

We see from Leibniz rule

(82/8UJJxi)U~l!!.kF

= (8/&xi){kxiu~-1l!!.kF + u~(8l!!.Fk/&ui}

= k{ u~-ll!!.kF + Xi (k - 1)UiU~-2l!!.kF + XiU~-l (8l!!.kF / 8Xi)}

+ {kUiU~-l (8l!!. Ie F / 8Ui) + u~ (82l!!. Ie F / 8Ui8xi)}'

Since l!!.kF is a (d - k, e - k)-form we see from Euler's identity

~ Xi(&l!!.kF/8xi) = (d - k)6.kF, ~ ui(&l!!.k1"/8ui) = (e - k)6. le 1".

Hence l!!.(u~l!!.kF) is equal to

k{3 + (k - 1) + (d - k) + (e - k)}U~-ll!!.k1" + u~l!!.k+l F

= k(d + e + 2 - k)U~-ll!!.k1" + u~l!!.k+l 1".

Now we shall show l!!.F(F) = O. If F(1") is set as in (4) then

l!!.F(F) = l!!.F + ),ll!!.(ux l!!.F) + ),2 (u;l!!.2 F) + ...= l!!.F + ~le21),dk(d + e + 2 - k)U~-l6.kP + u~6.k+lF}

= ~k21 {k(d + e + 2 - k),k + ),k_l}U~-ll!!.kF (5)

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with >'0 = 1. Hence if >'k = (-1)k(k!)-2 . (d+~+l) -1 then the coeffici­ennt of u~-l~kF in (5) is equal to zero and ~F(F) = D. 0

Remark. FF(F) = F(F) because F(F(F) - F) is equal to

(F(F) - F) + >'lux~(F(F) - F) + >'2U;~2(F(F) - F) + ...= F(F) - F - >'lux~(F) - >'2U~~2F - ... = D.

The Lemma 2 implies the following usefull criterion which makes thecalculations in Section 3 considerably simple.

Corollary 3. If a (d,e)-form F satisfies F - 0 mod ux then thepolynomial F(F) is equal to zero.

Let ax = ~t=lalXl be the (1, D)-form associated to the standardrepresentation (1,0) and let

be the (d, e)-form associated to the irreducible representation (d, e). ByPieri's formula [FH,p79] the tensor product (d, e)0(1, D) is decomposedas

(d, e) 0 (1,0) = (d + 1, e) EB (d - 1, e + 1) EB (d, e - 1) (6)

where we assume d, e 2 1. We shall express in Lemma 4 below thebasis of the three irreducible representation of the right hand-side of(6) in terms of bilinear forms of {>'fjq;} and {at}.

(i) Case (d + 1, e). For the (d + 1, e)-form F· ax, Lemma 2 impliesF(J"'ax ) is associated to the irreducible subspace (d+ 1, e) of (d+ 1, 0) ®(D, e).

(ii) Case (d - 1, e + 1). Consider the (d - 1, e + I)-form

a1 U1 OXl

(auox)F = a2 'U2 OX2 Fa3 'U3 OX3

whereiJax; = -a .x·t

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we seeal OXl FXl

~(a1tox)F= (a1Lox)~F - a2 OX2 FX2 = Da3 OX3 FX3

because ~F = D. Therefore the (d-l, e+l)-form (avlJx)F is associatedto the irreducible subspace (d - 1, e + 1) of (d - 1, D) 0 (D, e + 1).

(iii) Case (d, e - 1). The (d, e - I)-form

(a8j8x)F = ('E~=lai8joxi)F

satisfies ~(a8j8x)F = (a8jfJx)~F = Dso (a8jfJx)F is associated tothe irreducible subspace (d, e - 1) of (d, D) 0 (D, e - 1).

Summing up we obtain

Lemma 4. The basis of the three irreducible GL3-subspaces in thedecomposition (6) are given by the coefficients of the forms

F(Fax), (au8x)F, (a8j8x)F

respectively.

Next we recall briefly the symbolic method in classical invarianttheory[G,G-Y]. Let

f = E C;k) aijkXix~x~be an (n, D)-form associated to (n, D). If we set aijk = aiajak then fbecomes the n-th power of the (1, D)-form ax = aIxI + a2x2 + a3x3 :f = a~. Let F be a (d, e)-form associated to (d, e) as in (2). For eachthree integers ml, m2, m3 ~ Dsuch that

ml + m2 + m3 = n, m2 :S d, m3 :s: e

the tensor product (d, e) 0 (n, D) = {e + d, e, D} 0 {n, D, D} contains theirreducible representation

{e+d+ml,e+m2,m3} = (d+ml-m2,e+m2 -m3) (7)

with multiplicity one. Lemma 4 implies

Corollary 5. The basis of the irreducible subspace (7) in (d,e)0(n,D)is given by the coefficients of the form substituted aiajak = aijk in the(d + ml - m2, e + m2 - m3)-form

F((a8j8u)m3 (au8x )m2 Fa;"!).

We use Corollary 5 only in the case n = 3.

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2. The second component

Let R = k[Z1' Z2, Z3](3) be the third Veronese subring of three variables{Z1' Z2, Z3} and let 8 be the polynomial ring of ten variables {aij k I i +j + k = 3} with the surjection cPo : 8 ---7 R, cPO(aijk) = ZiZjZk. The groupGL3 acts on R and on 8 canonically such that {aijd is the basis of thethird symmtric tensor representation (3,0), which means the (3,0)-form

i+j+k=3

is associated to (3,0). The kernel of cPo : 8 ---7 R is generated by all thequadrics contained in the subspace {420} = (2,2) of

82 = 8ymm2 (3, 0) = (6,0) EB (2,2).

From Corollary 5 the (2,2)-form associated to the subspace (2,2) of 82is written by

() = (1/3!)(qu8xfp~qx = (pqu)2pxqx

Wl'th p3x = qx3 = f l' e p p p - q q q - a.. i j k - i j k - ijk·

Let

0---7 8( -9) ---7 8( _7)27 ---7 8(_6)105 ---7 8( _5)189 (1)

---78(_4)189 ti S( _3)105 ~ 8(_2)27 ti S ---7 R ---70

be the graded minimal free resolution of the Veronese subring R =k[zI, Z2, z3F3) (see Introduction) and denote by L i = Ui 0 S(-i) com­ponents in (1) for 2 ~ i ~ 7 and i = 9 with a GLa-module Ui . Wesaw U2 = {4, 2} = (2,2) as above. Next we calculate Ua and cP2

Ua 0 S(-3) --t lIz 0 S( -2). The third graded part of (1) is given by

By Pieri's formula we see U2 081 = {420} 0 {300} = (2,2) 0 (3,0) isdecomposed as

{72}EB{63} EB {522} EB {441}

EB{621} EB {54} EB {531} EB {432}.

Typeset by AMS-TEX-18-

Lemma 6. U3 = {521} EB {54} EB {531} EB {432}.

Proof In general the dimension of the irreducible representation {m, n,I} = (n - m, m-l) is equal to (rn- n + l)(n -I + l)(m-1 +2)/2 [FH,p77]so we see the sum of the dimensions of the right hand-side is equal to 105.Hence we have only to show the four subspaces {521}, {54}, {531}, {432}are mapped to zero by (p! : U2 0 S( -2(~ S. By Corollary 5 the formsassociated to these four subspaces are given by

A = {521} = (4,1) = F((a8/8'U)()a'~J

B = {54} = (1,4) = F((au8x)2()ax )

C = {531} = (2,2) = F((a8/8u)(au8x)Oax)

D = {432} = (1,1) = (a8/8u)2(au8x)O

respectively. Since () = (pqu )2pxqx these are equal to

A = (pqa)(pqu)pxqxa~ - 3-1ux(pqa)2pxqxax

B = (pqu)2(pau)(qau)ax - 3-1ux(pqa)(pqu) (pau) (qau)

C = (pqa)(pqu)(pau)qxax - 5- 1ux(pqa)2(pau)qx

D = (pqa)2(pau)qx

In order to prove that the kernel of (p! is equal to the four subspaces inLemma 5 we have to show that the associated forms A, B, C, D above areequal to zero after identifying P~ = q~ with a~, Le. Pi = qi = ai'

(i) A. Since the symbolic product (pqa)pxqxax are invariant under thecyclic permutation a ~ p ~ q ~ a we see the first term (pqa) (pqu)Pxqxa;of A is equal to

hence

(pqa)(pqu)pxqxa; = 3-1(pqa)pxqx ax{ax(pqu) - px(aqu) + qx(apu)}

= 3- 1(pqa)pxqxax . ux(apq) = 3-1ux(pqa)2pxqxax.

Thus A = O.

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(ii) B. Similarly the symbolic product (pqu)(qau)(apu) is invariantunder the cyclic permutaion p -+ q -+ a -+ p we see the first term(pqu)2(pau)(qau)a:r; = -(pqu)(qau)(apu) . a:r;(pqu) of B is equal to

3-1 (pqu)(qau) (apu) . u:r;(apq) = 3-1(pqa) (pqu) (pau) (qau)

Thus B = O.

(iii) C. The first term (pqa)(pqu)(pau)q:r;a:r; is equal to zero by inter­changing q with a, and the second term (pqa)2(pau)q:r; is equal to zero byinterchanging p with a.

(iv) D = (pqa)2(pau)q:r; = 0 by interchanging p with a. 0

3. The third component

In this section we describe the kernel of ¢2 in the fourth graded partof (1) in Section 2 :

It was determined in the proof of Lemma 5 that the forms A, B, C, Dassociated to the irreducible representation in the second componentU3 = {521}EB{54}EB{531}EB{432}. By Pieri's formula the tensor productU3 0 51 is decomposed as

{521} 0 {3} = {921} EB {831} EB {822} EB {741} EB {732} EB {551} EB {542}

{54} 0 {3} = {840} EB {750} EB {741} EB {551} EB {542} EB {652} EB {543}

{531} 0 {3} = {831} EB {741} EB {732} EB {651} EB {642} EB {533} EB {552}

EB {543}

{432} 0 {3} = {732} EB {542} EB {533} EB {543} (1)

By Corollary 5 the forms assoicated to the irrerducible subspaces in thedecomposition in (1) above are written down by using A, B, C, D togetherwith the (1, O)-form b~ associated to 51 = (1,0). On the other hand thehomomorphism ¢2 above factors through

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with the canonical inclusion t : U2 C U1 081 and the multiplicationm : 8 1 (2) 8 1 -4 8 2 , Hence an irreducible subspace W in the decompostion(1) is mapped to zero under th if and only if the form associated toW is identically zero after identifying ax with bx. Moreover Corollary3 guarantees that we may carry out the calculations modulo u x , hencewe need not to use the operator F in Lemma 2. In the proof of thenext Theorem we will use without mention the following identities ofdeterminants (the second fundamental theorem) repeatedly.

ax(bcd) - bx(acd) + cx(abd) - dx(abc) = 0 (2)

(abc)(deu) - (dbc)(aeu) + (ebc) (adu) - (ubc)(ade) = 0 (3)

Theorem. U4 = {741}EB{732}EB{651}EB{642}EB{633}EB{552}EB{543}.

Proof The sum of the dimension of the right hand-side is equal to 189, sowe have only to find forms associated to the seven irreducible subspacesin the right hand-side which are equal to zero after identifying ax (resp.Px) with bx (resp. qx).

Case 1. {741} = (3,3). {741} appears with multiplicity three inU2 081 , which are given by

mod U x ' In what follows, =means modulo U x '

A' - (bu8x)2{(pqa)(pqu)pxqxa; - 3-1ux . (pqa)2pxqxax}bx

=(pqa)(pqu){(pbu)(qbu)a; + pxqx(abu)2

+ 2(pbu)qx(abu)ax + 2px(qbu)(abu)ax}bx

Since Px = qx we see

A' == (pqa) (pqu)(pbu) (qbu)a;bx + 4(pqa)(pqu)(pbu) qx (abu)axbx

+ (pqa )(pqu)(abu)2pxqxbx

:= h +4h+h

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Here we see from (2)

h - 12 = (pqa)(pqu)(pbu){(qbu)ax - (abu)qx}axbx

- (pqa)(pqu)(pbu)(qau)axb; := 1412 - !J = (pqa)(pqu)(abu){(pbu)ax - (abu)px}qxbx

- (pqa) (pqu) (pau) (abu )qxb;

(pqa) (pqu) (qau )(abu )Pxb;

(II - 12)- (12 -!J) - (pqa)(pqu)(qau){(pbu)ax - (abu)px}b;

- (pqa)(pqu)(qau)(pau)b~ := Is.

Hence A' is equal to

A' 6h - 6(1I - h) + (II - h) - (12 - h) = 6!I - 6f4 + Is. (4)

Next B' is equal to

B' _ (b8j8u){(pqu)2(pau)(qau)ax - 3-1ux ' (pqa)(pqu)(pau)(qau)}b;

- {2(pqb)(pqu)(pau)(qau) + (pqu)2(pab)(qau)

+ (pqu)2(pau)(qab)}axb; - 3-1(pqa)(pqu)(pau)(qau)b~

= 2(pqu)(pqu)(pau)(qau)axb; + 2(pab) (pqu)2 qau)axb; - 3-1fs

= 2h + 216 - 3- 1Is

where the third equality follows from ax = bx and Px = qx. Since

h - Ie = {(pqb)(pau) - (pab)(qau)}(pqu)(qau)axb;

= (pqa)(pbu)(pqu)(qa)axb; = f4

we see

Lastly C' is equal to

C' _ (bu8x){ (pqa) (pqu) (pau)qxax - 5- 1ux . (pqa)2(pau)qx}b;

(pqa) (pqu) (pau) {(qbu )ax + qx (aim) }b;

= (pqa)(pqu)(pau)(qbu)axb; + (pqa)(pqu) (pau)(abu) qxb;

:= f4 + 17 since Px = qx

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Here f4 - h is equal to

f4 -- h = (pqa)(pqu)(pau){(qbu)ax - (abu)qx}b;

(pqa)(pqu)(pau)(qau)b~ = f5

Hence C' is equal to

c' - 2f4 - (j4 - h) = 2f4 - f5.

We see from (4), (5), (6) that 2A' - 3B' + 3C' is equal to

2(6h - 6f4 + f5) - 3(4h - 214 - 3- 115) + 3(2f4 - f5) = o.

(6)

Case 2. {732} = (1,4). {732} appears with multipilcity three inU2 0 81 ,

A' (b%u)(buox)Abx, C' _ (b%u)Cb;, D' = Db~.

First we calculate A' :

A' - (b%u)(buox){(pqa)(pqu)pxqxa; - 3- 1ux . (pqa)2pxqxax}bx

= (buoxH (pqa) (pqb)px qxa; - 3- 1 (pqa)2pxqxaxbx}bx

= (pqa)(pqb){(pbu)qxa; + px(qbu)ax2+ 2pxqx(abu)ax}bx

- 3-1 (pqa)2{(pbu)qxaxbx + px(qbu)axbx + pxqx(abu)bx}bx

The first (resp. the fourth) term is equal to the second (resp. the fifth)term since Px = qx, and the third term is equal to zero since ax = bx,hence

A' _ 2(pqa) (pqb) (pbu)qxa;bx - (2/3)(pqa)2(pbu)qxaxb;

-- 3- 1(pqa)2 pxqx (aJm)b; := 2f - (2/3)h - 3- 11

Here we have

.f - h = (pqa)(pqb)(pau)qxaxb; - (pqa)2(pbu)qxaxb;

= (pqa)(pab)(pqu)qxax b; := 9

h -1 = (pqa) 2(pbu)qxaxb; - (pqa)2(abu)pxqxb;

= (pqa)2(pau)qx . b~ := D'

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Hence A' is equal to

2f - (2/3)(f - g) - 3- 1{(f - g) - D' } = f + 9 + 3- 1D' . (7)

On the other hand C' is equal to

C' (b%u){(pqa)(pqu)(pau)qxax - 5- 1ux . (pqa)2(pau)qx}b;

=(pqa){(pqb)(pau) + (pqu)(pab)}qxaxb; - 5-1(pqa)2(pau)qxb~

= f + 9 - 5- 1D' . (8)

Thus we see from (7) and (8) that A' - C ' - (8/15)D' mod u x .

Case 3. {651} = (1,4). {651} appears with multipilcity three inU2 0 Sl'

A' = (buox)3A, B ' - (b%u)(buox)Bbx1 C' (buox)2C.

First A' is equal to

A' = (buox)3{(pqa)(pqu)pxqxa; - 3-1ux . (pqa)2pxqxax}

= (pqa)(pqu){2(pbu) (qbu) (abu)ax + (pbu)qx(abu)2

+ px(qbu)(abu)2}

= 2(pqa)(pqu) (pbu)(qbu) (abu )ax + 2(pqa) (pqu)(pbu)(abu)2qx

:= 2ft +212·

where the third equality follows from Px = qx.

ft - 12 = (pqa) (pqu )(pbu)(abu){ax(qbu) - qx (abu)}

(pqa)(pqu)(pbu)(abu)(qau)bx := f4'

Hence A' - 4ft - 2(ft - h) = 4ft - !4.

B' (bo / ou)(buox ){ (pqu)2 (pau) (qau )ax

- 3- 1 (pqa)(pqu)(qau) (abu)}bx

(bo / ou) (pqu) 2 (pau) (qau) (abu )bx

= 2(pqb) (pqu)(pau) (qau) (abu)bx + (pqu)2(pab)(qau)(abu)bx

+ (pqu)2(pau)(qab)(abu)bx

= - 2(pqa)(pqu)(pbu)(qbu)(abu) + 2(pqu)2(pab)(qau)(abu)

:= - 2ft + 2f3

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The last equality follows from ax = bx and Px = qx.

- II - 13 = {pqb)(pau) - (pab)(pqu)}(pqu)(pau)abu)bx= (pqa)(pqu)(pbu)(qau)(abu)bx = 14

Hence B' = -4II + 2(1I + h) = -4II - 2/4'

C' == (buox)2{(pqa)(pqu)(pau)qxax - 5- 1ux ' (pqa)2(pau)qx}bx= (pqa)(pqu) (pau)(qlYu)(aim )bx = !4.

Thus A' + 8' - -4/5 = -4C'.

Case 4. {642} = (2,2). {642} appears with multipilcity four in U2 051'

A' = (buox)2(b% u)A,

C' - (b%u)(buox)Bbx,

Now A' is equal to

B' = (bO/ou)2Bbx

D' (buox)Bb;

A' = (buox)2(b%u){(pqa)(pqu)pxqxa; - 3- 1ux . (pqa)2pxqxax}

= (buox)2{(pqa)(pqb)pxqxa; - 3-1 (pqa)2pxqxaxbx}

= (pqa)(pqb){(pbu)(qbu)a; + 2(pbu) qx (abu)ax

+ 2px(qbu)(abu)ax +pxqx(abu)2}

- 3-1 (pqa) 2{(pbu) (qbu)ax + (pbu)qx(abu) + Px (qbu) (alru)}bx

Since Px = qx we see

A' = (pqa)(pqb)(pbu)(qbu)a; + 4(pqa)(pqb) (pbu)(abu)qxax

+ (pqa)(pqb)(abu)2 pxqx - 3-1 (pqa)2(pbu)(qbu)axbx

- (2/3)(pqa)2(pbu)(abu)qxbx

:= II + 412 + 13 - 3- 114 - (2/3)/5

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where the differences of {Ii} are given by

11 - 12 = (pqa)(pqb)(pbu){(qbu)ax - (abu)qx}ax(pqa)(pqb)(pbu)(qau)axbx := f6

12 - fa = (pqa)(pqb)(abu){(pbu)ax - (abu)Px}qx

_ (pqa) (pqb) (pau) (abu )qxbx = (pqa) (pqb) (qau) (abu )Pxbx

(11 - 12)- (12 - fa) = (pqa) (pqb) (qau){(pbu)ax - (abu)px}bx

- (pqa)(pqb)(pau)(qau)b; = 1114 - fs = (pqa)2(pbu){(qbu)ax - (abu)px}bx

(pqa)2(pbu)(qau)b; := 17

Hence A f is equal to

11 + 412 + fa - 3-1(f4 + 2fs)

= 611 - 6(11 - h) + (11 - h) - (12 - fa) - 3- 1{3/4 - 2(f4 - fs)}

= 611 - 6f6 + 11 - 3- 1 (3f4 - 217)

= 711 - f4 - 6f6 + (2/3)17·

Next B f is equal to

B' - (b% u )2{(pqu)2(pau)(qau)ax - 3- 1ux . (pqa)(pqu)(pau)(qau)}bx

= ((pqb) 2 (pau) (qau) + 2(pqb) (pqu )(pab)(qau)

+ 2(pqb)(pqu)(pau(qab) + (pqu)2(pab)(qab)}axbx

- 3-1b; (pqa){ (pqb) (pau)(qau)

+ (pqu)(pab) (qau) + (pqu) (pau) (qab)}

Since Px = qx we see

B f = (pqb)2(pau)(qau)axbx + 4(pqb) (pab) (pqu)(qau)axbx

+ (pab)(qab)(pq)2 axbx - 3- 1 (pqa)(pqb)(pau) (qau)b;

- (2/3) (pqa) (pab) (pqu )(qau)b;

:= f4 + 4fs + f9 - 3- 1 fl - (2/3)f1O

Hence B f 14 + 4fs + f9 - 3- 1 (11 + 2flO) is equal to

B f = 614 - 6(f4 - Is) + (J4 - fs) - (fs - 19) - 3-1(311 - 2(11 - flO)

- - 11 + 714 - 6f6 + (2/3)17

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Next C' is equal to

C' = (b%u)(buox){(pqa)(pqu)(pau)qxax - 51ux ' (pqa)2(pau)qx}bx= (pqa){(pqb)(pau) + (pqu)(pab)}{(qbu)ax + qx(abu)}bx

- 5-1 (pqa)2(qbu) (pau)b;

= (pqa) (pqb) (pau) (qbu )axbx + (pqa) (qab) (pqu) (qbu )axbx

(pqa)(pqb)(pau)(abu)qxbx + (pqa)(qab)(pqu)(abu)qxbx - 5- 117:= 16 + III + h2 + h3 - 5-117

Here we see

16 - III = (pqa){(pqb)(pau) - (pab)(pqu)}(qbu)axbx

= (pqa)2(pbu)(qbu)axbx = 14

16 - 112 = (pqa)(pqb)(pau){(qbu)ax - (abu)qx}bx

= (pqa)(pqb)(pau)(qau)qxb; = II

h2 - 113 = (pqa){(pqb)(pau) .- (pab)(qau)}(atru)qxbx

= (pqa)2(pbu)(abu)qxbx = (pqa)2(qbu)(abu)pxbx

(I6 - fll)- (f12 - h3) = (pqa)2(qbu){(pbu)ax - (abu)px}bx

- (pqa)2(qbu)(abu)b; = 17

Thus C' is equal to

C' = f6 + fll + h2 + f13 - 5-117= 4f6 - 2(J6 - hd - 2(f6 - h2) + (f6 - fll - f12 + h3) - 5- 117= - 2h - 2f4 + 4f6 + (4/5)17

Lastly D' is equal to D' (buox)(pqa)2(pau)(qbu)b'; = 17. Suming upwe have modulo ('u x , h)

A' 7!I - 14 - 6f6

B' - - fl + 7f4 - 6f6

C' = - 2h - 214 + 4/6

hence A' + H' + 3C' =0 mod (tLX1 D').

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Case 5. {633} = (3,0). {633} appears with multipilcity two in U20S1.

D = (bOlou)[)b~

We see

C' (bOlou)2{(pqa)(pqu)(pau)qxax - 5- 1ux . (pqa)2(pau)qx}bx

= (pqa) (pqb)(pab)qxaxbx - 5-1(pqa)2(pab)qxb~

where the first term is equal to zero since ax = bx. On the other hand

hence C' -5- 1D'.

Case 6. {552} = (0,3). {552} appears with multipilcity two in U2f2)Sl.

First B' is equal to

B' = (bOIOu)2(pqu)2(pau)(qau)(abu)

= (pqb)2(pau)(qau)(abu) + (pqu) 2 (pab)(qab) (abu)

+ 2(pqb)(pqu){(pab) (qau) + (pau)(qab)}(abu)

:= f+ 2g+2h

where (pqu)2(pab)(qab)(abu) in the middle is equal to zero since ax = bx.

f - 9 = {(pqb)(pau) - (pqu)(pab)}(pqb)(qau)(abu)

= (pqa )(pqb)(qau) (pbu )(abu) = 0

9 - h = (pqb)(pqu){pab)(qau) - (pau)(qab) }(abu)

= (pqa )(pqb)(pqu) (abu)2 = 0

because ax = bx and Px = qx, Thus H' = 5f.

C' = (bOIOu)(pqa)(pqu)(pau)(qbu)(abu)

= (pqa){(pqb)(pau) + (pqu)(pab)}(qbu)(abu)

= 2(pqa) (pqa) (pau) (qbu)(abu) - (pqa) 2 (pbu )(qbu)(abu) = -.f

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Thus 8' = 5f = -5C'.

Case 7. {543} = (1,1). {543} appears with multipilcity two in [12051 ,

H' = (bfJ/fJu)3 B, C' = (bufJx)(bfJ/fJu)2C, D' (bufJx)(bfJ/fJu)DbxFirst H' is equal to

(bfJ/fJu)3{(pqu)2(pau)(qau)ax - 3- 1ux . (pqa)(pqu)(pau)(qau))

= {(pqa)2(pab)(qau) + (pqb)2(pau)(qab) + 2(pqb)(pqu) (pab)(qab)}ax

- 3- 1bx(pqa){pqb) (pab)(qau) + (pqb)(pau)(qab)

+ (pqu)(pab)(qab)} - 3- 1ux . (pqa)(pqb)(pab)(qab)

The first and the second (resp. the fifth and the sixth) terms are equalsince Px = qx while the third and fourth terms are equal since ax = bx,hence

H' - 21 - (3/2)h + (5/3)g.

Here

f - 9 = (pqb)(pab){(pqb)(qau) - (pqu)(qab)}ax= (pqa)pqb)(pab)(qbu)ax = -(pqa)(pqb)(pab)(qau)bx = -h

Hence B' is equal to

B' _ 21 + (3/5)(f + h) - (2/3)h = (11/3)1 + h.

Next C' is equal to

C' (bufJx)(bfJ/fJu)2{(pqa)(pqu)(pau)qxax - 5- 1ux . (pqa)2(pau)qx}

= (bufJx){(pqa)(pqb)(pab)qxax - 5- 1(pqa)2 (pab)qxbx}

= (pqa)(pqb)(pab){(qbu)ax + qx(abu)} - 5- 1(pqa)2(pab)qxbx

= -h+1-5- 1fwhere -h - I is equal to

-h -l = (pqa)(pqb)(pab){(qbu)ax - qx(abu)}

_ - (pqa)(pqb)(pab)(qau)bx = (pqa)(pqb)(abu)(qbu)ax = h

Hence C' is equal to

C' - h + I - 5-1f = - 2h - (-h - 1) - 5-1f= - 2h - h - 5-1 f = -3h - 5- 11

Lastly D' is equal to

D'(bufJx)(bfJ/fJu)(pqa)2(pau)qxbx =(pqa)2(pab)(qbu)bx = f.Therefore we see C' = -3h = -3B' mod (ux , f).

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Department of Mathematical SciencesFaculty of ScienceUniversity of RyukyusNishihara-Cho, Okinawa 903-0213JAPAN

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