Duality for Ideals in the Grassmann Algebra

Duality for Ideals in the Grassmann Algebra

I. Dibag

Department of Mathematics, Bilkent Uni_¨ersity, 06533 Bilkent, Ankara, Turkey Communicated by Walter Feit

Received May 4, 1994

A duality is established between left and right ideals of a finite dimensional Grassmann algebra such that if under the duality a left idealI and a right ideal J correspond then I is the left annihilator of J and J the right annihilator of I.

Another duality is established for two-sided ideals of the Grassmann algebra where two ideals that correspond are annihilators of each other. The dual of the principal ideal generated by an exterior 2-form is completely determined. Q 1996 Academic Press, Inc.

INTRODUCTION

Ž .

In a finite dimensional Grassmann algebra we define the left right Ž .

annihilator of a right left ideal as the set of elements whose products on

Ž . Ž .

the left right with elements of the ideal are zero and it is a left right ideal. Theorem 1.1.5 establishes a duality between left and right ideals of the Grassmann algebra such that if under the duality a left ideal I and a right ideal J correspond then I is the left annihilator of J and J is the right annihilator of I. We define the annihilator of a two-sided ideal to be the intersection of its left and right annihilators and it is a two-sided ideal.

Theorem 1.2.3 establishes a duality between two-sided ideals of the Grass- mann algebra such that if under the duality two ideals correspond then they are annihilators of one another. We define an ideal to be proper if its left and right annihilators coincide and are equal to its annihilator and prove that a homogeneous ideal is proper.

Ž .

A knowledge of the annihilator K I of an ideal I enables us to characterize I by means of exterior equations. If k . . . k are generators1 r

Ž .  < 4

for K I then I s i g HV i n k s ??? s i n k s 0 . For this reason1 r

we compute the annihilator of the ideal generated by a linearly indepen-

24 0021-8693r96 $18.00

CopyrightQ 1996 by Academic Press, Inc.

All rights of reproduction in any form reserved.

dent set of vectors in the underlying vector space and relate it to the factorization problem of an exterior form into a wedge product of k vectors and an exterior form. The second section of the paper is devoted to the determination of the annihilator of the principal ideal generated by an exterior 2-form. As a consequence given a 2-form m and a form v, we obtain a system of exterior equations whose satisfaction byv is a neces- sary and sufficient condition for v to factor into v s t n m for some exterior formt. The global version of this factorization problem is briefly discussed.

1. ANNIHILATORS OF IDEALS IN THE GRASSMANN ALGEBRA

1.1. Left and Right Annihilators

Let V be an n-dimensional vector space over a ground field k with dual space V * and HV s

[

_pⁿ_s0H^pV and H*V s HV * s

[

_pⁿ_s0H^pV * the corresponding Grassmann algebras.

Ž .  <

1.1.1. DEFINITION. Let I be a right ideal. Then K I s k g HVL

kni s 0 ;i g I is defined to be the left annihilator of I. It is readily4

Ž . Ž .

verified that KL I is a left ideal. The right annihilator K J of a leftL

ideal J is analogously defined and is a right ideal.

1.1.2. Left and right duality operators. Let xg HV and L : HV ª HVx

be left multiplication by x and dx^L: H*V ª H*V be its dual and let V g HⁿV * be a fixed dual volume element. We then define the left

L LŽ . ^LŽ . Ž .

duality operator ) : HV ª H*V by ) x sd V . If , is the dualx

Ž ^LŽ .. Ž .

pairing betweenHV and H*V then u, ) x s x n u, V ; x, u g HV.

 4  ^{U U}4

If e , . . . , e1 n is the basis for V, e , . . . , e1 n the dual basis for H*V,

ei₁n ??? n e N 0 F i - i - ??? - i F n the induced basis for HV andi_p 1 2 p 4

e^Ui₁n ??? n e N 0 F i - i - ??? - i F n the corresponding dual basis^Ui_n 1 2 p 4

U U LŽ

for H*V and if V s e n ??? n e , it can be verified that ) e n1 n i₁

. Ž .^{sgns U U} Ž .

e_i n ??? n e s y1_i e_j n ??? n e_j , where j j . . . j_{1 2 nyp} is the set

2 p 1 nyp

Ž .

of complementary indices and s s i . . . i j . . . j_{1 p 1 nyp} is a permutation of Ž1 2 . . . n .. ) is thus an isomorphism. We can also define the right^L duality operator )^R using the right multiplication R instead of L and_{x x}

Ž ^RŽ .. Ž . ^RŽ

obtain analogous equations: u,) x s u n x, V and ) e n ??? ni₁

. Ž .^{sgnt U U} Ž .

e_i s y1 e_j n ??? n e_j , wheret s j . . . j_{1 nyp 1}i . . . i_p is a permuta-

p 1 nyp

Ž . Ž .^{pŽ nyp.}

tion of 1 2 . . . n . We note that sgnt s y1 sgns and hence if e:

< ^p Ž .^{pŽ nyp.}

HV ª HV is the automorphism of HV defined bye H Vs y1 then)^R s )^L(e and )^L s )^R(e.

1.1.3. PROPOSITION. Let I be a right ideal. Then Ž .i KRŽKLŽI s I...

Ž .ii dim KLŽI q dim I s dim HV s 2 .. ⁿ Žiii. There exist commutati_¨e diagrams

L rest.^{n R} rest.ⁿ

) 6 6 ) 6 6 UŽ .

HV H*V I* HV H*V KL I

6 ^f 6 ^f

6 6

HVrI Ž .

HVrK IL

Ž .

Proof. Let i: I ; HV and j: K I ; HV be the inclusions and i*:L

UŽ . ^UŽ . ^L

H*V ª K I and j*: H*V ª K I be the restriction maps. Since )L L

is an isomorphism and i* is onto it follows that the composite i*()^L is

Ž ^L. Ž ^LŽ .. Ž ..

onto. ug Ker i*() iff x,) u s u n x, V s 0 ; x g I, which

Ž . Ž ^L. Ž .

clearly shows that KL I ; Ker i*() and if uf K I then thereL

 4

exists xg I such that u n x / 0. Let us choose a basis e , . . . , e for V1 n

and express un x s Ýli . . . i_{1 p}ei₁n ??? n e . Since u n x / 0 it follows thati_p

Ž . Ž .

l_{i . . . i} / 0 for some i , . . . , i . Let j , . . . , j_{1 p 1 nyp} be the set of comple-

1 p

mentary indices andt s e n ??? n ej₁ j_nyp. Then ys x nt g I since I is a

Ž ^LŽ .. Ž .

right ideal and y,) u s u n x nt , V s "li . . . i_{1 p}/ 0 and this gives

Ž ^L . Ž .

a contradiction. Hence Ker) (i* s K I and this proves the first partL

Ž .

of iii . If we take dimensions of both sides under the isomorphism,

Ž . f Ž .

HVrK I ª I* we obtain ii . By reasoning analogous to that used toL

Ž ^R. Ž Ž ..

prove the first diagram we deduce that I : Ker j*() s K K IR L

f U

Ž Ž .. Ž .

and we obtain an isomorphism HVrK K I ª K I . HenceR L L

Ž Ž . Ž . Ž . Ž .

dim KR KL I q dim K I s dim HV s dim K I q dim I by ii .L L

Ž Ž .. Ž Ž ..

Thus dim KR KL I s dim I and this gives equality, i.e., K K I sR L

Ž . Ž .

I, which proves both i and the second diagram in iii . 1.1.4. PROPOSITION. Let I be a left ideal. Then

Ž .i KLŽKRŽI s I...

Ž .ii dim KRŽI q dim I s dim HV s 2 .. ⁿ Žiii. There exist commutati_¨e diagrams

R rest.^{n L} rest.ⁿ

) 6 6 ) 6 6 U

Ž .

HV H*V I* HV H*V KR I

6 ^f 6 ^f

6 6

Ž . HVrI

HVrK IR

Proof. Identical with that of Proposition 1.1.3.

1.1.5. THEOREM. There exists a duality between left and right ideals of the Grassmann algebra such that if under the duality a left ideal I and a right

Ž . Ž .

ideal J correspond then I s K J and J s K I .L R

Proof. Let A and B be the set of left and right ideals of the Grass- mann algebra, respectively. We define mappings T : Aª B and U: B ª A

Ž . Ž . Ž . Ž .

by T I s K I ;I g A and U J s K J ;J g B. Then TU s 1 byR R

Proposition 1.1.3 and UTs 1 by Proposition 1.1.4.

1.1.6. COROLLARY. Let I be a two-sided ideal. Then the automorphism

Ž . Ž .

e: HV ª HV interchanges K I and K I .L R

Proof. Let i: I ; HV be the inclusion and i*: H*V ª I* be the Ž . restriction maps. Then by Propositions 1.1.3 and 1.1.4, KL I s

Ž ^L . w Ž ^Rx Ž Ž ..

Ker i*() (e s e Ker i*() s e K I .L

1.2. The Annihilator of a Two-Sided Ideal

From now onward when we say ideal we shall mean a two-sided ideal.

Ž .

1.2.1. DEFINITION. The annihilator K I of an ideal I is defined by

Ž . Ž . Ž .

K I s K I l K I .L R

1.2.2. PROPOSITION.

Ž .i KŽI is an ideal left in. ¨ariant under the automorphisme.

Ž .ii K KŽ ŽI s I...

Proof.

Ž .i Let kg K I ,Ž . v g HV, and x g I. Then v n k n x s v nŽ . Žkn x s 0 since k g K I and hence. Ž . v n k g K I . Also, x n v nLŽ . Ž

. Ž . Ž .

k s x nv n k s 0 since x n v g I and k g K I and thus v n kR

Ž . Ž . Ž . Ž . Ž .

g K I . HenceR v n k g K I l K I s K I . This shows that K IL R

Ž .

is a left ideal. Similar argument shows that K I is also a right ideal.

Ž . Ž .

Hence K I is an ideal. The fact that e leaves K I invariant follows from Corollary 1.1.6.

Žii. K KŽ ŽI.. s K K ILŽ Ž .. l KRŽKŽI.. : K K ILŽ RŽ .. l Ž Ž ..

KR KL I s I l I s I, by Propositions 1.1.3 and 1.1.4 the other inclu- Ž Ž ..

sion trivially holds, and hence we have equality, i.e., K K I s I.

1.2.3. THEOREM. There exists a duality among ideals of the Grassmann algebra. Two ideals that correspond under this duality are annihilators of each other.

Proof. Let S be the set of ideals of the Grassmann algebra and define

Ž . Ž . ²

a mappingf: S ª S by f I s K I ;I g S. Then, f s 1 by Proposi- tion 1.2.2.

Ž .

The annihilator K I of an ideal I will be called the dual ideal of I.

This is not to be confused with the vector space dual I* of I.

Ž . Ž .

1.2.4. DEFINITION. An ideal I is called proper iff K I s K I sL

Ž . Ž .

KR I . Note that if an ideal is proper then dim K I q dim I s dim H Vs 2ⁿ.

1.2.5. OBSERVATION. A necessary and sufficient condition for an ideal

Ž . Ž .

I to be proper is that the automorphism e leave either K I or K IL R

invariant.

Proof. It follows from Corollary 1.1.6.

Ž . 1.2.6. OBSERVATION. If an ideal I is proper so is K I .

Ž Ž .. Ž Ž .. Ž Ž .. Ž Ž ..

Proof. KL K I s K K I s I and K K I s K K I sL R R R L

Ž Ž .. Ž Ž ..

I by Propositions 1.1.3 and 1.1.4 and thus K K I s K K I s I.L R

1.2.7. LEMMA. Let I and I be proper ideals. Then I l I and1 2 1 2

I q I are also proper and1 2

Ž .i KŽI l I s K I q K I1 2. Ž 1. Ž 2. Ž .ii KŽI q I s K I l K I .1 2. Ž 1. Ž 2.

Ž . Ž .

Proof. Let kig K I i s 1, 2 , k s k q k ,1 2 i g I l I . Then i g1 2

I and k n1 1 i s 0 and similarly k n i s 0. Hence k n i s 0 and thus2

Ž . Ž . Ž . Ž .

kg K I l I . Hence 1. K I q K I : K I l I . ReplacingL 1 2 1 2 L 1 2

Ž . Ž .

I by K Ii i is 1, 2 and using Proposition 1.2.2 we obtain I q I :1 2

Ž Ž . Ž ..

KL K I l K I . Taking dimensions of both sides gives dim I q1 2 1

Ž . ⁿ w Ž . Ž .x ⁿ

dim I y dim I l I F 2 y dim K I l K I , i.e., 2 y2 1 2 1 2

Ž . Ž ⁿ . Ž ⁿ . w Ž .

dim I l I F 2 y dim I q 2 y dim I y dim K I l1 2 1 2 1

Ž .x Ž . Ž . Ž . w Ž .

K I , dim K I l I F dim K I q dim K I y dim K I l2 L 1 2 1 2 1

Ž .x Ž . w Ž . Ž .x

K I , and dim K I l I F dim K I q K I . We deduce from2 L 1 2 1 2

Ž . Ž . Ž . Ž .

this and inclusion 1 that K I q K I s K I l I . Similarly K I1 2 L 1 2 1

Ž . Ž . Ž .

q K I s K I l I2 R 1 2 and hence I l I is proper and K I q1 2 1

Ž . Ž . Ž . Ž . Ž .

K I s K I l I , which proves i . Replacing I by K I2 1 2 i i is 1, 2 and taking K of both sides yield ii .Ž .

1.2.8. OBSERVATION. If an ideal I is multiplicatively generated by

evŽ . ^{w n r2x 2 i}

generators which lie inH V s

[

_is0 H V then I is proper.

evŽ . Ž .

Proof. Let gig H V be the multiplicative generators 1 F i F r . If

Ž . Ž .

kg K I , g n k s k n g s 0 1 F i F n since g lies in the center ofL i i i

Ž . Ž . Ž .

HV and thus k g K I . Hence K I : K I and the reverse inclu-R L R

sion symmetrically follows.

1.2.9. LEMMA. A homogeneous ideal is proper and its dual ideal is also homogeneous and proper.

Proof. Let I be a homogeneous ideal. Then I s

[

pⁿs0I , wherep

p n Ž . ^p

I s I l H V. Let k s Ýp ps0 pk g K I for k g H V and x g I .L p q q

Then 0s k n x s Ý_q ⁿ_{ps0 p}k n x and since HV s_q

[

_pⁿ_s0H^pV is a direct sum decomposition, it follows that kpn x s 0. Varying x over I forq q q

Ž . Ž .

all 0F q F n we see that k g K I and K I is thus a homogeneousp L L

Ž .

ideal. The automorphism e hence leaves K I invariant and hence I isL

Ž . Ž .

proper by Observation 1.2.5. K I s K I is homogeneous and henceL

also proper.

1.2.10. Remark. Proposition 1.2.2 enables us to characterize proper ideals of the Grassmann algebra by means of exterior equations. Suppose

Ž .

I is a proper ideal and K I is multiplicatively generated by generators

Ž .  <

k , . . . , k . Then by ii of Proposition 1.2.2,1 r Is vgHV vnk s???s1

4 Ž .

v n k s 0 . This if course presupposes that K I is known. The determi-r

Ž .

nation of K I may be a formidable problem, as we shall see in Section 2.

wŽ .x Ž . 1.2.11. LEMMA. Let 0/ x g V. Then K x s x .

Proof. Let U be a complementary subspace in V to the 1-dimensional

f Ž .

subspace generated by x. There is an isomorphism, HU ª x given by

Ž . ^ny1 Ž . wŽ .x

D ª x n D. Thus dim x s dim HU s 2 . Clearly x : K x and

wŽ .x ⁿ Ž . ^{n ny1 ny1} wŽ .x

dim K x s 2 y dim x s 2 y 2 s 2 . Hence dim K x s

Ž . wŽ .x Ž .

dim x and thus K x s x .

1.2.12. COROLLARY. Let 0/ x g V and v g HV. Then v s t n x for somet g HV iff v n x s 0.

 4

1.2.13. PROPOSITION. Let x , . . . , x_{1 k} be a linearly independent set of

wŽ .x Ž .

¨ectors in V. Then K x , . . . , x1 k s x n ??? n x .1 k

Proof. Let U be a complementary subspace in V to the k-dimensio- nal subspace generated by x , . . . , x . Then there is an isomorphism,_{1 k}

f Ž . Ž

HU ª x n ??? n x given by D ª x n ??? n x n D and thus dim x n1 k 1 k 1

. ^nyk

??? n x s dim HU s 2_k . Also we have a direct-sum decomposition,

Ž . Ž . ^{n nyk}

HV s HU [ x n ??? n x and hence dim x n ??? n x s 2 y 2_{1 k 1 k} .

Ž . wŽ .x wŽ .x

Clearly x1n ??? n x : K x n ??? n xk 1 k and dim K x1n ??? n xk s

n Ž . ^{n k} Ž . wŽ

2 y dim x n ??? n x s 2 y 2 s dim x n ??? n x . Thus K x n1 k 1 k 1

.x Ž .

??? n xk s x n ??? n x .1 k

 4

1.2.14. COROLLARY. Let x , . . . , x_{1 k} be a linearly independent set of

¨ectors in V and v g HV. Then v s t n x n ??? n x for some t g HV1 k

iff x1nv s . . . s x n v s 0.k

w x

As a further corollary we recover 1, Theorem 1 , e.g.,

1.2.15. THEOREM. Let v g HV and R : HV ª HV be right multipli-_v cation by v. Then v factors into the wedge product of k 1-¨ectors and an exterior form iff dim Ker RvG k.

1.3. Abelian Ideals

1.3.1. DEFINITION. A proper ideal I is called an abelian ideal iff Ž .

I : K I .

1.3.2. DEFINITION. An abelian ideal I is called a maximal abelian Ž .

ideal iff I s K I . Note that the dimension of a maximal abelian ideal is

1dimHV s 2ny1.

2

1.3.3. Types of maximal ideals

 4 ^{n p}

Let e , . . . , e1 n be a basis for V and I s

[

ps1H V the unique

Ž .

maximal ideal of HV. Let I s e , . . . , ek 1 2 ky1 be the ideal in HV

Ž wŽ . x.

multiplicatively generated by e , . . . , e1 2 ky1 1F k F n q 1 r2 . Then its k th-power, I^k_k is a maximal abelian ideal. I^k_k and its images under automorphisms of V are called maximal abelian ideals of type k. Two maximal ideals of types k and l are isomorphisms iff ks l. Thus classes of different types of maximal abelian ideals form the non-isomorphic classes

k Ž ^k.

of maximal abelian ideals inHV. Also, I s Ds g AutŽV .s I .k

2. THE DUAL OF THE PRINCIPAL IDEAL GENERATED BY AN EXTERIOR 2-FORM

Ž . 2.1. The Principal Ideal m

evŽ .

Let m be an exterior 2-form in V. Since m g H V , it follows from Ž .

Observation 1.2.8 that the principal ideal m generated by m is a proper ideal.

Ž . 2.2. The Idealu m

Ž .

Suppose rank m s 2 s. Then there exists a linearly independent set

x , . . . , x , y , . . . , y of vectors in V such that1 s 1 s4 m s x n y q ??? qx n y .1 1 s s

Ž .

Define m s x n yj j j 1F j F n so that m s m q ??? qm . Then1 s

Žm y m n m q m s m q m n m y m s 0 andi j. Ž i j. Ž i j. Ž i j. Žm y m s y2m n m si j.² y i j 2 xin y n x n y . Take all possible partitionsi j j

Ži j1 1.??? i j k k ??? kŽ r r.Ž 1 2 sy2 r., ikF j 1 F k F r , i - ??? - i , k - ???k Ž . 1 r 1

w x Ž .

- ksy2 r, for all 0F r F sr2 and let u m be the homogeneous ideal

Ž . Ž .

multiplicatively generated by generators g_as m y m n m y mi₁ j₁ i₂ j₂

Ž . ^s

n ??? n m y m n_{i j} ¨_k n ??? n¨_k in H V, where ¨_k is either x_k or

r r 1 sy2 r j j

Ž . Ž .

yk_j 1F j F s y 2r . u m is a homogeneous ideal and is thus proper by Ž . wŽ .x

Lemma 1.2.9. Also,u m : K m . Ž .

2.2.1. LEMMA. E_¨ery element v g u m has an expression of the form v s

Ý

l n ga a

a

for l g HV.a

Ž .

Proof. Every element v g u m by definition has an expression v s

Ž . Ž .

Ý_{a a}a n g n b , a , b g HV; g s_{a a a a a} m y m n m y m n ??? ni₁ j₁ i₂ j₂

Žm y m nir jr. ¨k1n ??? n¨ksy2 r. Let bas Ýⁿps0t , where ta , p a , pg H V.^p

Ž .^{pŽ sy2 r. X n} Ž .^{pŽ sy2 r.}

Then ganta , ps y1 ta , pn g . Define b s Ýa a ps0 y1 t .a , p

n Ž .^{pŽ sy2 r. X}

Then g_an b s Ý_{a ps0}g_ant_{a , p}s Ý y1 t_{a , p}n g s b n g_{a a a} and hence a_an g n b s a n b_{a a a a}^X n g . Putting_a l s a n b_{a a a}^X yields the re- sult.

2.2.2. Equi_¨alence classes of generators

We define an equivalence relation on the set of generators. Let gas Žm y m n m y m n ??? n m y m ni1 j1. Ž i2 j2. Ž ir jr. ¨k1n ??? n¨ksy2 rand gbs Žm y m n m y m_iX _jX. Ž _iX _jX. n ??? nŽm y m n u n ??? n u_iX _jX. _kX _kX be two

1 1 2 2 t t 1 sy2 t

Ž^{X X X X}.

generators. Then g_a; g iff r s t, i j ??? i j is a permutation of_b 1 1 t t

Ži j1 1 ??? i j , k s k , and u sr r. ^Xj j kX_j ¨k_j Ž1F j F s y 2r . The equivalence.

Ž . w x

classes are denoted by D ¨_k1,_¨k2, . . . ,_{¨ksy2 r} for 0F r F sr2 and ¨_kj is Ž .

either xk_j or y . If s is even,k_j D f denotes the equivalence class of

Ž . Ž . Ž .

generators, gas m y m n m y m n ??? n mi₁ j₁ i₂ j₂ i_sr2ymj_sr2 . Let

Ž .

D ¨_k,_¨k , . . . ,_¨k be an equivalence class and let u_k be the comple-

1 2 sy2 r j

 4  4  4 Ž .

ment of _¨k_j in the set x , yk_j k_j , i.e., _¨k_j, uk_j s x , yk_j k_j 1F j F s y 2r .

Ž .

Then D u , u , . . . , u_{k k k} is called the ‘‘dual’’ class of

1 2 sy 2 r

Ž .

D ¨_k,_¨k , . . . ,_¨k .

1 2 sy2 r

Ž .

2.2.3. LEMMA. Let g_a and g_b be two generators for u m . If their equi_¨alence classes are not dual then gan g s 0.b

Proof. Since g and g do not belong to dual equivalence classes there_{a b}

< <

exists k such that WLG x_{1 k} g_a but y_k ¦ g . If x_{b k} g_b then obviously

1 1 1

Ž .<

g_an g s 0. Suppose there exists k / k such that_{b 2 1} m y m_{k k} g . Thus_b

1 2

Ž .

g_an g contains x n_b k1 m y mk1 k2 s yx nk1 m as a factor. If g hask2 a

either xk2 or yk2 at the k th-place then g2 an g contains the wedgeb

product of this with xk1nm , which is zero and hence g n g s 0. So letk2 a b

us assume WLG that there exists k3/ k such that g contains2 a Žm yk₂

. Ž .

mk₃ as a factor. Then gan g contains yx nb k₁ m n m y m sk₂ k₂ k₃

xk₁nm n m as a factor, which would be zero if k s k . We thusk₂ k₃ 3 1

Ž .

assume that k , k , k_{1 2 3} are all distinct. Continuing in this manner, we find either that gan g s 0 or that there exist distinct integers k , k , k , . . .b 1 2 3

Ž . Ž . Ž .

such that gas x nk₁ m y m n m y m n ??? g s m y mk₂ k₃ k₄ k₅ b k₁ k₂ n

. Ž . Ž .

m y m n . . . and this is a contradiction since deg g / deg g .k3 k4 a b

 4

2.2.4. LEMMA. Let V be a_¨ector space with basis e , . . . , e . Define the_{1 2}

Ž . ^{U U}

isomorphismf: V ª V * by f e s e , where e is the dual of e . If U ; Vi i i i

is any subspace of V then the composite mapc : U ; V ª^f V *ª^rest.nU* is an isomorphism.

 4 Ž .

Proof. Let u , . . . , u1 m be a basis for U mF n and express u si

n Ž . Ž . Ž .

Ýjs1 i j ja e 1F i F m , where A s a is an m = n -matrix with linearlyi j

independent rows and has rank m. Let A^t be the transpose of A. Then

t Ž . Ž . Ž .Ž .

Bs AA is a non-singular m = m -matrix. Let B s b . Theni j c u ui j

Ž ^{n U}.Ž ⁿ . ⁿ Ž . ^{m U}

s Ý_{ks1 ik k}a e Ý_{ls1 jl l}a e s Ý_{ks1 ik jk}a a s b , i.e.,_{i j} c u s Ý_{i js1 i j j}b u .

 Ž .4

Since B is nonsingular, it follows that c u is a linearly independent seti

and hence is a basis for U*. Thus c is an isomorphism.

 4

2.2.5. DEFINITION. Let Vs x , y , . . . , x , y_{1 1 2 n 2 n} be a 4 n-dimensional

Ž . ^{2 n}

vector space. m s x n y 1 F j F 2n . Let T be the subspace of H Vj j j

Ž . Ž . Ž .4 Ž

spanned by m y m n m y m n ??? n m y m , i F ji₁ j₁ i₂ j₂ i_n j_n k k 1F

. Ž .Ž . Ž .

kF n , i - ??? - i as i , j i , j ??? i , j_{1 n 1 1 2 2 n n} runs through the set of Ž2, 2, . . . , 2 partitions of 1, 2, . . . , 2 n . Define K. Ž . TŽT.s t g T t n t9 s <

0;t9 g T .4

Ž . 2.2.6. LEMMA. KT T s 0.

Ž . ^U Ž . ^U

Proof. Let f: V ª V * be defined by f x s x and f y s y andi i i i

Hf: HV ª H*V be the induced map which maps the induced basis for Ž . HV into the dual basis for H*V. Let ): HV ª H*V be the left duality operator. Then an easy computation shows that

)



Žm y m n ??? n m y mi₁ j₁. Ž i_n j_n.

4

s . HfŽ .



Žm y m n ??? n m y mi₁ j₁. Ž i_n j_n.

4

. Ž .1 Let i: T ; HV be the inclusion and i*: H*V ª T* be the restriction map. Definea, c : T ª T* by a s i*()(i and c s i*(f (i. Applying i*

to Eq. 1 yieldsŽ .

a



Žm y m n ??? n m y mi₁ j₁. Ž i_n j_n.

4

s .c



Žm y m n ??? n m y mi₁ j₁. Ž i_n j_n.

4

. Ž .2 Ž .

a maps K T to zero. However, c is an isomorphism by Lemma 2.2.4T

Ž . Ž .

and thus a is an isomorphism by Eq. 2 . Hence K T s 0.T

 4 Ž .

2.2.7. Remark. Let Ujs x , y 1 F j F s and let U be a complemen-j j

tary subspace to U1[ U [ ??? [ U in V. In the following proposition we2 s

2Ž .

shall regard m q m q ??? qm1 2 sy1g H U [ U [ ??? [ U1 2 sy1[ U and

Ž . Ž .

u m q m q ??? qm1 2 sy1 ; H U [ U [ ??? [ U1 2 sy1[ U and similarly for the other terms.

Ž . Ž Ž .. Ž . ²

2.2.8. PROPOSITION. u m l K u m s u m q ??? qm1 sy1 m H U qs

Ž . ² Ž . ²

u m q ??? qm1 sy2qm m H Us sy1q ??? qu m q ??? qm m H U .2 s 1

Ž . ²

Proof. Let ggu m q ??? qm1 sy1 m H U . Then g s g ns a m fors

Ž . Ž . Ž .

ga gu m q ??? qm1 sy1 . ga s m y m n ??? n m y m ni₁ j₁ i_r j_r ¨k₁

wŽ . x Ž .Ž . Ž .

n ??? n¨sy1y2 r, where 0F r F s y 1 r2 and i j i j ??? i j1 1 2 2 r r

Žk . . . k1 sy1y2 r.is a partition of 1 2 . . . sŽ y 1 . g s g n. a m s g n x ns a s

Ž . Ž .

ys s .x ns m y m n ??? n m y m ni₁ j₁ i_r j_r ¨k₁n ??? n¨sy2 rn y gs

Ž .

u m , hence RHS : LHS.

Ž . Ž Ž ..

Conversely, let ggu m l K u m . By Lemma 2.2.1 we can write

gs

Ý

aan g sa

Ý Ý

aan g .a

a DŽ¨k₁,¨k₂, . . . ,¨k_{sy2 r}. gagDŽ¨k₁,¨k₂, . . . ,¨k_{sy2 r}.

Ž .

Fix an equivalence class D ¨_k,_¨k , . . . ,_¨k of generators and let

1 2 sy2 r

Ž .

D u , u , . . . , uk1 k2 ksy2 r be the ‘‘dual’’ equivalence class and let gbg

Ž . Ž .

D u , u , . . . , u_{k k k} . Then g_an g s 0 ;g f D_{b a} ¨_k,_¨k , . . . ,_¨k by

1 2 sy2 r 1 2 sy2 r

Lemma 2.2.3. Then

0s g n g s_b

Ý

a_an g n g_{a b} Ž .1

Ž .

g_agD ¨k₁,¨k₂, . . . ,¨k_{sy2 r}

Ž .

Let i , i , . . . , i1 2 2 r , 1F i - i - ??? - i F s, be the complementary in-1 2 2 r

Ž . Ž .

dices to k , k , . . . , k_{1 2 sy2 r} in 1 2 . . . s . Let W be the 4 r-dimensional

 4 ²

subspace of V spanned by x , y , . . . , x , yi₁ i₁ i_{2 r} i_{2 r} , m s x n y g H Wj i_j i_j

Ž1F j F 2r . Let T be the subspace of H W generated by. ^{2 r} Žm yl₁

. Ž .4 Ž . Ž .

m_m n ??? n m y m_{l m} as l m_{1 1} . . . l m_{r r} runs through the set

1 r r

Ž2, 2, . . . , 2 partitions of i , i , . . . , i. Ž1 2 2 r.. Then gas t na ¨k₁n ??? n¨k_{sy2 r}

and g_bs t n u n ??? n u_b k₁ k_{sy2 r} for some t , ta bg T. Define V9 s¨k₁

n ??? n¨k_{sy 2 r}n u n ??? n uk₁ k_{sy2 r} s .x n y n ??? n xk₁ k₁ k_{sy2 r}n yk_{sy2 r} s .m n m n ??? n mk1 k2 ksy2 r. Then gan g s t n t n V9 and define V0b a b

s .x n y n ??? n x n y s .m n m n ??? n m . Then t n t si1 i1 i2 r i2 r i1 i2 i2 r a b

Ž .

n_abV0 for unique integers n . The matrix n_{ab ab} is non-singular by Lemma 2.2.6. Define V s V9 n V0 s x n y n ??? n x n y s1 1 2 r 2 r m n1

m n ??? n m . Then g n g s n V. Substituting into Eq. 1 yields2 2 r a b ab Ž .

0s

Ý

nab aa n V.

Ž .

g_agD¨k₁,¨k₂, . . . ,¨k_{sy2 r}

Ž .

Since the matrix n_ab is non-singular, we deduce that a_an V s 0 ;g g_a

Ž .

D ¨k₁,_¨k₂, . . . ,_¨k_{sy2 r} , i.e., aan x n y n ??? n x n y s 0. It follows1 1 2 r 2 r

from Proposition 1.2.13 that aX _as aŽ _{a , 1}n x q a1 ^Xa , 1^X n y q ??? qa1 a , 2 rn x2 rq aa , 2 rX n y . Then a n g s a2 r. Ž a .a a , 1Žn x q a1 a , 1. n y q ??? qa1 a , 2 rn x_{2 r}q a_{a , 2 r}n y_{2 r} n m y m_{l m} n ??? n m y m_{l m} n¨_k n ??? n¨_k .

1 1 r r 1 sy2 r

Ž . Ž .

Take the term x1n m y ml₁ m₁ n ??? n m y ml_r m_r n¨k₁n ??? n

 4

¨k_{sy 2 r}. Suppose 1g k , k , . . . , k1 2 sy2 r and assume WLG that k1s 1. Ei-

ther _¨k₁s x , in which case the term is zero or1 ¨k₁s y and the term1

Ž . Ž . Ž

equals m y ml₁ m₁ n ??? n m y ml_r m_r n¨k₁n ??? n¨k_{sy2 r}nm g u m1 2

. ²  4

q ??? qm m H U . Now suppose that 1 f k , k , . . . , k_{s 1 1 2 sy2 r} and the

Ž . Ž .

term equals x1n m y ml₁ m₁ n ??? n m y ml_r m_r n¨k₁n ??? n¨k_{sy2 r}s Žml₂ y mm₂.n ??? nŽml_r y mm_r.n x n1 ¨k₁n ??? n¨k_{sy2 r}nmm₁ g

Ž . ²

u m q ??? qm q ??? qm m H U1 ˆm₁ s m₁ and similar for the other terms.

This shows that aan g g RHS and hence that g g RHS and this showsa

that LHS: RHS and hence equality, i.e., LHS s RHS.

Ž . Ž .

2.3. Duality between m and u m

Ž . Ž . w wŽ .x Ž

2.3.1. LEMMA. m l m s K m q ??? qms 1 sy1 q m1

.x ² q ??? qmsy1 m H U .s

Proof. Let U be a complementary subspace to U1[ ??? [ U in V ands

w wŽ .x Ž .x ²

let v g K m q ??? qm1 sy1 q m q ??? qm1 sy1 m H U . Thens v s

wŽ .x

v n m q v n m for v g K m q ??? qm1 s 2 s 1 1 sy1 and v s t n2

Žm q ??? qm1 sy1.fort g H U [ ??? [ UŽ 1 sy1[ U . Then. v n m s v n1 1

Žm q ??? qm1 sy 1. q v n m s v n m1 s 1 s and v n m s t n2 s

Žm q ??? q m1 sy1. n m s t n m n m and thus v s v n m q t n ms s 1

Ž . Ž . Ž . Ž .

nm s v q t n m n m g m . Also, v g m . Hence v g m ls 1 s s

Žm . Conversely, let v g m l m . Then, v s t n m for some t g HVs. Ž . Ž s.

Ž .

and sinceHV s H U [ ??? [ U1 sy1[ U m HU . We can writes t s t q0

Ž . Ž .

t n x q t n y q t n m for t g H U [ ??? [ U1 s 2 s 3 s i 1 sy1[ U 1 F i F 3 . Ž . Thus v s t n m q t n x n m q t n y n m q t n m n m. v g m0 1 s 2 s 3 s s

wŽ .x

s K x , ys s by Proposition 1.2.13 and thus 0sv n m s t n m n ms 0 s

w Ž .x Ž .

s t n m q ??? q m n m g H U [ ??? [ U0 1 s s 1 sy1[ U m HU .s

Ž . wŽ .x

Hence t n m q ??? qm_{0 1 sy1} s 0, i.e., t g K m q ??? qm_{0 1 sy1} . Also,

Ž .

t n m s t n m q ??? qm0 0 1 sy1qm s t n m . 0 s v n x q t n ys 0 s s 2 s

w Ž .x

nm n x s yt n m n m s y t n m q ??? qms 2 s 2 1 sy1 nm . Hence ts 2

Ž . Ž .

n m q ??? qm1 sy1 s 0 and t n y n m s t n m q ??? qm2 s 2 1 sy1 n ys

Ž .

s 0. Similarly,t n x n m s 0. t n m n m s t n m q ??? qm1 s 3 s 3 1 sy1 n

Ž . Ž .

m s a n m , where a g m q ??? qms s 1 sy1 . Thus v s t q a n m g0 s

w wŽK m q ??? qm1 sy1.xqŽm q ??? qm1 sy1.xm H U .² s

w Ž .x Ž . w Ž . ²

2.3.2. LEMMA. K u m l m s K u m q ??? qms 1 sy2 m H Usy1q

Ž . ² x ²

??? qu m q ??? qm2 sy1 m H U m H U .1 s

w Ž .x Ž . Ž

Proof. Let v g K u m l m . Then v s t n m for t g H U [s s 1

??? [ Usy1[ U , where U is a complementary subspace to U [ U [. 1 2

Ž . Ž .

??? [ U in V. Let g ss a m y m n ??? n m y m ni₁ j₁ i_r j_r ¨k₁n ??? n¨k_{sy2 r}

Ž . w x Ž . Ž .

be a generator for u m , where 0 F r F sr2 , i j_{1 1} . . . i j_{r r} Žk k . . . k_{1 2 sy2 r}. is a partition of 1 2 . . . s andŽ . _¨k is either x_k or y . If_k

j j j

 4

sg k , . . . , k1 sy2 r then ganv s 0 anyway; so the generators which

 4

contribute non-trivially are those for which sg i j . . . i j . Suppose g1 1 r r a

is such a generator and assume WLG that jrs s. Then

0s g n_a v s m y m n ??? n m y m n m nŽ _{i1 j1}. Ž _{ir jr}. _s ¨_k1n ??? n¨_{ksy2 r}nt sŽm y m n ??? n m_{i1 j1}. Ž _iry1ym_jry1.nm n m n_{ir s} ¨_k1n ??? n¨_{ksy2 r}nt s g n_b m n t , where g g u m q ??? qm q ??? qmir b

Ž

1 ˆir sy1

.

1F i F s y 1 .

Ž _r .

w Ž . ² x w Ž

Thus t g K u m q ??? qm q ??? qm1 ˆi_r sy1 m H U : Ki_r u m q1

. ² Ž .x ²

??? qmsy2 m H Usy1q ??? qu m q ??? qm2 sy1 m H U and hence1

w Ž . ² Ž .

v g K u m q ??? qm1 sy2 m H Usy1q ??? qu m q ??? qm2 sy1 m

2 x ²

H U m H U and the argument can be reversed so as to prove the1 s

converse.

w Ž .x Ž . 2.3.3. THEOREM. K u m s m .

By using the duality between and ideal and its annihilator, we can also state Theorem 2.3.3 in an equivalent form.

wŽ .x Ž . 2.3.3. THEOREM*. K m s u m .

Ž . w Ž .x Proof. Clearly m : K u m .

Ž .

The converse will be proved by induction on ss rank m . For s s 1,

Ž . Ž . wŽ .x wŽ .x

m s x n y and u m s x, y s K x n y s K m . Let s ) 1 and as-

Ž .

sume the induction hypothesis for sy 1 . Let v s x n y q ??? qx n y ,1 1 s s

 4 Ž .

Ujs x , yj j 1F j F s and let U be a complementary subspace to

Ž . Ž

U1[ ??? [ U in V and regards u m q ??? qm1 sy1 ; H U [ ??? [ U1 sy1[

. Ž .

U . Let IIs denote the restriction of x , ys s to HU . Thens