The Schur Functors and the Resolution of Determinantal Varieties

Home , Schur functor

Treball ﬁnal de grau

GRAU DE MATEMÀTIQUES

Facultat de Matemàtiques i Informàtica Universitat de Barcelona

The Schur functors and the resolution of determinantal varieties.

Autor: Liena Colarte Gómez.

Director: Dra. Rosa Maria Miró-Roig. Realitzat a: Departament de Matemàtiques i Informàtica.

Barcelona, January 17, 2017

Contents

Introduction iii

1 Preliminaries. 1 1.1 Multilinear algebra...... 1 1.1.1 Hopf algebras...... 1 1.1.2 The Exterior Algebra...... 2 1.1.3 The Symmetric Algebra...... 3 1.1.4 The Divided Power Algebra...... 4 1.2 Combinatorics...... 5 1.2.1 Partitions and skew partitions...... 5 1.2.2 Tableaux...... 7

2 Schur Functors and CoSchur Functors. 9 2.1 Schur Functors...... 9 2.1.1 The freeness of Schur Functors...... 11 2.2 CoSchur Functors and the freeness of the CoSchur functors...... 19 2.3 Decomposition of Schur Functors...... 21 2.4 Cauchy Decomposition Formulas for Schur Functors...... 25 2.4.1 The decomposition of the Symmetric Algebra...... 25 2.4.2 The decomposition of the exterior algebra...... 27 2.5 The Littlewood-Richardson rule for Schur functors...... 28 2.5.1 The Schensted Process and Words of Yamanouchi...... 28 2.5.2 The Littlewood-Richardson Rule...... 35

3 A minimal free complex associated to the minors of a matrix. 41 3.1 A minimal free complex...... 41 3.1.1 The group algebra and combinatorics...... 42 3.1.2 The complex...... 43 3.1.3 The Proof of Lemma 3.1.31...... 49 3.1.4 The proof of Lemma 3.1.32...... 51 3.2 Determinantal ideals and determinantal varieties...... 54 3.2.1 A minimal free resolution of determinantal ideals...... 56

Bibliography 67

Introduction.

Resolutions is one of the most effective methods to obtain information about varieties in Algebraic Geometry. For many years there has been considerable efforts in finding a resolution of determinantal varieties. To put the problem plainly, assume R = K[x0,..., xs] is the polynomial ring over an algebraically closed field of characteristic zero and Ps is the projective space of dimension s over K. Given (ri,j) a homogeneous matrix of size p × q with entries in R, the problem is to find an explicit minimal free resolution of the ideal It defined by the t × t minors of this matrix. Over certain hypothesis on It, this is a minimal s s free resolution of the variety X = {z ∈ P | rg((ri,j)(z)) < t} of P . It provides the Hilbert polynomial of X, the projective dimension and the arithmetically Cohen-Macaulayness of the variety among others characteristics. In a more general context, this problem was solved by Lascoux in [16] where he gave a minimal free resolution of Schubert’s varieties. His construction rests heavily on the theory of Schur functors and the fact that, in characteristic zero, the Schur functors are the irreducible representations of the general linear group. Despite the fact that the techniques developed by Lascoux are out of reach for an undergraduate student of Mathematics, the particular case of determinantal varieties admits a treatment based on developing suitable rudiments of multilinear algebra and combinatorics. In this sense, this paper is an approachment to the work of Lascoux in order to study a minimal free resolution of the ideal associated to the minors of a matrix. Our main goal is to construct a minimal free complex of these ideals whose modules are the modules of the minimal free resolution given by Lascoux, and next describe the resolution. In fact, this minimal free complex is a good candidate to be a resolution, too. This paper is organized as follows. In Chapter 1 we present a review of background material on Hopf algebras, including exterior, symmetric and divided power algebras, and an exposition of partitions, Tableaux and Young diagrams which are the basic tools used along the main body of the paper. Chapter 2 deals with Schur and CoSchur functors on free R-modules, where R is a commutative ring of arbitrary characteristic. In section 2.1 we define the Schur functor Lλ/µF of a free R-module F with respect to the skew partition λ/µ as the image of a natural transformation dλ/µ : Λλ/µF → Sλ˜ /µ˜ F of free R-modules. In 2.1.1 the freeness of Schur functors is proved giving an explicit basis of Lλ/µF through a basis of F which is described by means of Young tableaux. We end this subsection giving the rank of LλF and with a discussion about the functoriality of Lλ/µ(−). The CoSchur functor Kλ/µF has a similar treatment in section 2.2, we give a basis of Kλ/µF and we show ∗ ∗ ∗ the duality (Lλ/µF) Kλ˜ /µ˜ F , where F denote the dual of F. The remainder of Chapter 2 is devoted to the properties of Schur functors. In section 2.3, we provide a natural filtration of Lλ/µ(F ⊕ G) with associated graded object L µ⊆γ⊆λ Lγ/µF ⊗ Lλ/γG, where F and G are free R-modules. In characteristic zero this associated graded object becomes a direct sum decomposition of the Schur functor Lλ/µ(F ⊕ G). In section 2.4, we address to the Cauchy formula for exterior and symmetric algebra. Sk(F ⊗ G) and Λk(F ⊗ G) have natural filtrations with associated graded objects L L |λ|=k LλF ⊗ LλG and |λ|=k LλF ⊗ KλG, respectively. In characteristic zero, this sum becomes a direct decomposition of these functors. The Cauchy formula for exterior algebra is one of the pillars in the Lascoux construction. Finally, section 2.5 is devoted to the iv Introduction

Littlewood-Richardson rule for Schur functors. More precisely, when R contains a field of characteristic zero, the tensor product LλF ⊗ LµF of Schur functors decomposes in a direct sum ∑ν u(λ, µ; ν)LνF, where u(λ, µ; ν) are the multiplicities of the factor LνF given by the Littlewood-Richardson rule. As a consequence, we obtain the Pieri formulas for Schur functors which are essential to describe the boundary maps of the resolution given by Lascoux. In Chapter 3, we construct a minimal free complex of the ideal generated by the minors of a matrix and a minimal free resolution of determinantal ideals. In section 3.1, we assume that R is a local commutative ring containing a field K of characteristic zero with m its maximal ideal, F and G are free R-modules of ranks m + t − 1 and n + t − 1 respectively and (ri,j) is a matrix of size (m + t − 1) × (n + t − 1) with entries in R associated to a R- ∗ map Φ : F → G . We denote by It the ideal generated by the t × t minors of (ri,j) and we define a complex (C•(Φ, t), d•) of length mn as follows:

(i) For all k, Ck(Φ, t) is a free R-module, it is a direct sum of tensor product of Schur functors. (ii) d1(C1(Φ, t)) = It and dk(Ck(Φ, t)) ⊂ mCk−1(Φ, t).

We construct the modules Ck(Φ, t) by means of the action of the group algebra K(Sk) on the tensor products Tk F and TkG. In subsection 3.1.1 we provide all the results required on K(Sk) and Young tableaux. In 3.1.2 we construct the modules Ck(Φ, t) as the images of the morphisms associated to idempotents of K(Sk). The boundary maps dk are defined in 3.1.2 and they are essentially the maps induced by the contractions Tk F ⊗ TkG → R extending Φ. The remainder of the section 3.1 is devoted to prove that (C•(Φ, t), d•) is a complex. This rests on two formulas involving the idempotents defining the modules of the complex. In section 3.2, assuming notations in 3.1 with R = K[x0,..., xs] we define determinantal ideals and determinantal varieties. In 3.2.1 we present the minimal free resolution (L•,t, d•) L R L 0 L F ⊗ L G I of determinantal ideals. The free -modules k,t are |I|+n(I)=k,I ,0 I˜ Ie0 , where is a partition of weight m + k − 1, I0 and n(I) are described by means of I and its Durfee square. In fact, Lk,t = Ck,t(Φ, t). The boundary maps dk are given by Pieri formulas and the natural contractions ΛρF ⊗ ΛρG → R induced by Φ. The remains of the Chapter deals with examples of minimal free resolutions and two explicit resolutions computed with the program Macaulay2 [10]. The Eagon-Northcott complex gives a minimal free resolution of determinantal ideals generated by the maximal minors of a homogeneous matrix. We explain this resolution using 3.2.1, but actually the original treatment of this problem due to Eagon-Northcott in [7] does not involve Schur functors and this particular case gives a resolution of determinantal ideals even when R is not of characteristic zero. The Gulliksen-Negard complex is a minimal free resolution of the determinantal ideal generated by the submaximal minors of a square matrix. This complex has only four submodules and boundary maps, we do an accurate description of them. Introduction v

Acknowledgements.

I would like to express my inﬁnite gratitude to my advisor Rosa Maria Miró-Roig for her dedication and effort week after week to get this work to succeed in. I would like to thank also to P. C. Roberts who kindly sent to us his preprint which was essential to develop this paper. And ﬁnally, I would like to thank to my family for their support. To M. vi Introduction Chapter 1

Preliminaries.

This chapter is a compilation of all background material on multilinear algebra and combinatorial which is utilized in the main body of the paper. Hopf algebras and discus- sions of the relevant properties of the exterior, symmetric and divided power algebras are included. We follow [15] and [20] in section 1.1, and [1] in section 1.2.

1.1 Multilinear algebra.

1.1.1 Hopf algebras.

Let R be a commutative ring.

Deﬁnition 1.1.1. Let M and N be two graded R-modules. We deﬁne the twisting mor- ij phism T : M ⊗R N → N ⊗R M by T(x ⊗ y) = (−1) y ⊗ x, x ∈ Mi, y ∈ Nj.

Deﬁnition 1.1.2. A graded R-Hopf algebra is a graded R-module A = ∑i≥0 Ai together with a multiplication m : A ⊗R A → A, a unit η : R → A, a comultiplication or diagonalization ∆ : A → A ⊗ A and a counit ε : A → R satisfying:

1. (A, m, η) is a graded R-algebra, (A, ∆, ε) is a graded R-coalgebra, the counit ε is an R-algebra map and the unit η is an R-coalgebra map. That is, the following diagrams are commutative: ∆ A A ⊗ A A

∆ Id ⊗ ∆ ∆ ⊗ Id ε ⊗ Id Id ⊗ ε A ⊗ A A ⊗ A ⊗ A R ⊗ A A ⊗ A A ⊗ R .

2. The multiplication m and the comultiplication ∆ are compatible in the sense that the following diagram commutes:

1 2 Preliminaries.

m ∆ A ⊗ A A A ⊗ A

∆ ⊗ ∆ m ⊗ m Id ⊗ T ⊗ Id A ⊗ A ⊗ A ⊗ A A ⊗ A ⊗ A ⊗ A .

In addition, we say that A is a commutative graded R-Hopf algebra if the following diagrams commute, T A ⊗ A A ⊗ A A ∆ ∆ m m A A ⊗ A A ⊗ A T . We assume that A is connected (i.e. A0 = R) and free (i.e. each i ≥ 0, Ai is a ﬁnitely generated free R-module).

Deﬁnition 1.1.3. Let (A, mA, ηA) and (B, mB, ηB) be two graded R-Hopf algebras. We say that an R-map α : A → B is a map of graded R-Hopf algebras if mB ◦ (α ⊗ α) = α ⊗ mA, and ∆B ◦ α = (α ⊗ α) ◦ ∆A.

Deﬁnition 1.1.4. Let A = (A, mA, ηA, ∆A, ε A) and (B, mB, ηB, ∆B, εB) be two graded R- Hopf algebras. The tensor product A ⊗ B is deﬁned by the graded R-module A ⊗ B = Id⊗T⊗Id mA⊗mB ∑i≥0 Ai ⊗R Bi, a multiplication mA⊗B : A ⊗ B ⊗ A ⊗ B −−−−−→ A ⊗ A ⊗ B ⊗ B −−−−→ ∆A⊗∆B Id⊗T⊗Id A ⊗ B, a comultiplication ∆A⊗B : A ⊗ B −−−−→ A ⊗ A ⊗ B ⊗ B −−−−−→ A ⊗ B ⊗ A ⊗ B, a ηA⊗ηB ε A·εB unit ηA⊗B : R −−−−→ A ⊗ B and a counit ε A⊗B : A ⊗ B −−−→ R.

1.1.2 The Exterior Algebra. Let R be a commutative ring and let F be a free R-module of rank n with an ordered ∗ ∗ ∗ basis {x1,..., xn}. We denote F = Hom(F, R) the dual of F with basis {x1,..., xn} dual to the basis {x1,..., xn}. We denote by Sr the set of all permutations of {1, . . . , r}. Definition 1.1.5. Let Λ0F = R and Λ1F = F. For all integer r > 1 we define the r-th exterior power Λr F of F to be quotient of the r-th tensor power Ti F := F ⊗ · · · ⊗ F of F respect to the r sg(σ) submodule E of T F generated by the elements f1 ⊗ · · · ⊗ fr − (−1) fσ(1) ⊗ · · · ⊗ fσ(r) r for all σ ∈ Sr and f1,..., fr ∈ F. In other words, Λ F is the image by the natural surjective r r r map π : T F → T F/E and then, the generators of Λ F are the elements π( f1 ⊗ · · · ⊗ fr) such that f1,..., fr ∈ F which we denote by f1 ∧ · · · ∧ fr. r Let f1 ∧ · · · ∧ fr ∈ Λ F. From the above definition it follows that for each i , j ∈ {1, . . . , r}, f1 ∧ · · · ∧ fi ∧ · · · ∧ fj ∧ · · · ∧ fr = − f1 ∧ · · · ∧ fj ∧ · · · ∧ fi ∧ · · · ∧ fr. So, it is clear that if there are two indices such that fi = fj, then f1 ∧ · · · ∧ fr = 0. We assume that in the expression f1 ∧ · · · ∧ fr all elements fi are different, otherwise it is 0. { ∧ · · · ∧ | ≤ < ··· < ≤ } r Proposition 1.1.6. The set xi1 xir 1 i1 ir n form a basis of Λ F. In r n particular Λ F is a free R-module of rank (r). 1.1 Multilinear algebra. 3

The r-th exterior power Λr F is a submodule of Tr F through the natural immersion r → r ( ∧ · · · ∧ ) = (− )sg(σ) ⊗ · · · ⊗ ı : Λ F T F defined by ı f1 fr ∑σ∈Sr 1 fσ(1) fσ(r). Proposition 1.1.7. Let F and E be two free R-modules. If φ : E → F is an R-map, then r r r r we have a well-defined linear map Λ φ : Λ E → Λ F defined by Λ φ(e1 ∧ · · · ∧ er) = φ(e1) ∧ · · · ∧ φ(er). Thus, the r-th exterior power is an endofunctor on the category of R-modules and R-maps.

L r Deﬁnition 1.1.8. We deﬁne ΛF to be the graded R-module ΛF := r≥0 Λ F. r s r+s r+s For each s, t ≥ 0 there are natural maps mr,s : Λ F ⊗ Λ F → Λ F and ∆r+s : Λ F → Λr F ⊗ ΛsF given by the formulas

mr,s( f1 ∧ · · · ∧ fr ⊗ g1 ∧ · · · ∧ gs) = f1 ∧ · · · ∧ fr ∧ g1 ∧ · · · ∧ gs

sg(σ) ∆r+s( f1 ∧ · · · ∧ fr+s) = ∑ (−1) fσ(1) ∧ · · · ∧ fσ(r) ⊗ fσ(r+1) ∧ · · · ∧ fσ(r+s) σ∈Sr+s respectively, where σ is such that σ(1) < ··· < σ(r) and σ(r + 1) < ··· < σ(r + s). These maps induce a natural multiplication m : ΛF ⊗ ΛF → ΛF and comultiplication ∆ : ΛF → ΛF ⊗ ΛF in ΛF given by all components mr,s and ∆r+s respectively. Then, we have

Proposition 1.1.9.

(i) (ΛF, m˜ , η, ∆, ε) is a connected, free and commutative graded R-Hopf algebra where the unit is the natural inclusion η : R → ΛF into degree 0 and the counit is the projection ε : ΛF → R into degree 0.

∗ (ii) The component ∆r+s of the comultiplication map in ΛF is the dual map of the component of the multiplication map mr,s in ΛF. ∗ (iii) The component mr,s of the multiplication map in ΛF is the dual of the component of the comultiplication map ∆r+s in ΛF.

1.1.3 The Symmetric Algebra. Definition 1.1.10. Let S0F = R and S1F = F. For each r > 1 we define the r-th symmetric power of F to be the quotient of Tr F respect to the submodule S of Tr F generated by the r elements f1 ⊗ · · · ⊗ fr − fσ(1) ⊗ · · · ⊗ fσ(r) for all σ ∈ Sr and f1,..., fr ∈ F. Then S F is the image under the natural projection π : Tr F → Tr F/S and it is generated by the elements π( f1 ⊗ · · · ⊗ fr) := f1 ····· fr, f1,..., fr ∈ F. Observe that the above definition differs from Λr F only by a sign, (−1)sg(σ), however r this detail makes important distinctions. Let f1 ····· fr ∈ S F. For each i , j ∈ {1, . . . , r} r we have f1 ····· fi ····· fj ····· fr = f1 ····· fj ····· fi ····· fr . Differs from Λ F, in the expressions f1 ····· fr we can have two equals elements. In this sense, the element f1 ····· fr could not have a minimal expression. To emphasize this fact we will use the ir ir notation f1 ····· ft such that i1 + ··· + ir = r where we assume that all elements fi are ik different and eventually ik = 0 with fi = 1 ∈ R. 4 Preliminaries.

i1 in r Proposition 1.1.11. {x1 ····· xn | i1 + ··· + in = r} form a basis of S F. In particular, r n+r−1 S F is a free R-module of rank ( r ).

Sr F is a submodule of Tr F through the natural immersion  : Sr F → Tr F deﬁned by ( ····· ) = ⊗ · · · ⊗  f1 fr ∑σ∈Sr fσ(1) fσ(r).

Proposition 1.1.12. Let F and E be two free R-modules and let φ : F → E be an R-map. r r r r Then, the map S φ : S E → S F defined by S (φ)(e1 ····· er) = φ(e1) ····· φ(er) is a well-defined R-map. In other words, the rth symmetric power defines an endofunctor on the category of free R-modules and R-maps.

L r Deﬁnition 1.1.13. We deﬁne SF to be the graded R-module SF := r≥0 S F.

r s r+s As we have seen in ΛF, there are naturals maps mr,s : S F ⊗ S F → S and ∆r+s : Sr+sF → Sr F ⊗ SsF deﬁned by the formulas

mr,s( f1 ····· fr ⊗ g1 ····· gr) = f1 ····· fr · g1 ····· gs

∆r+s( f1 ····· fr+s) = fσ(1) ····· fσ(r) ⊗ fσ(r+1) ····· fσ(r+s) respectively where σ runs over the set of all permutations of {1, . . . , r + s} such that σ(1) < ··· < σ(r), σ(r + 1) < ··· < σ(r + s), which induce a multiplication m and ∆ i1 ir+s i1 ir+s j1 comultiplication in SF. In particular, ∆r+s( f ····· f ) = ∑ ( ) ··· ( ) f ····· 1 r+s 0≤jk≤ik j1 jr+s 1 jr+s i1−j1 ir+s−jr+s fr+s ⊗ f1 ····· fr , where jk = 0 if ik = 0 and j1 + ··· + jr+s = r.

Proposition 1.1.14. (SF, m, η, ∆, ε) is a connected, free and commutative graded R-Hopf algebra with the obvious unit η : R → SF and counit ε : SF → R.

1.1.4 The Divided Power Algebra.

Deﬁnition 1.1.15. Let D0F = R and D1F = F. For each r > 1 we deﬁne the rth divided powerDr F of F to be the dual of the rth symmetric power Sr F∗.

∗ i1 ∗ in r ∗ Considering the basis {(x1 ) ····· (xn) | i1 + ··· + in = r} of S (F ), we deﬁne (i ) (i ) 1 n ∗ i1 ∗ in x1 ····· xn to be the dual element of the basis element (x1 ) ····· (xn) . Then,

(i1) (in) r Proposition 1.1.16. {x1 ····· xn | i1 + ··· + in = r} form a basis of D F. In particular, r n+r−1 D F is a free R-module of rank ( r ).

L r ∗ Proposition 1.1.17. DF = r≥0 D F is a graded dual of the symmetric algebra SF with the obvious unit and counit and where

r s r+s (i) The component mr,s : D F ⊗ D F → D F of the multiplication on DF is the dual r+s ∗ r ∗ s ∗ ∗ of the component ∆r+s : S F → S F ⊗ S F of the comultiplication on SF .

(i1) (in) (j1) (jn) i1+j1 in+jn (i1+j1) mr,s is given by mr,s(x ····· xn ⊗ x ····· xn ) = ( ) ··· ( )x ····· 1 1 j1 jn 1 (in+jn) xn . 1.2 Combinatorics. 5

r+s r s (ii) The component ∆r+s : D F → D F ⊗ D F of the diagonalization on DF is the dual r ∗ s ∗ r+s ∗ ∗ of the component mr,s : S F ⊗ S F → S F of the multiplication on SF . ( ) ( ) ( ) ( ) ( − ) ( − ) (x i1 ····· x in ) = e j1 ····· e jn ⊗ e i1 j1 ····· e in jn ∆r+s is given by ∆r+s 1 n ∑0≤jk≤ik 1 n 1 n where j1 + ··· + jn = r.

n (0) (1) Deﬁnition 1.1.18. Let f ∈ F with f = ∑i=1 uixi. Let f = 1 ∈ R and f = f . For each ( ) r > r f (r) ∈ Dr F f ui1 ··· uin x i1 ··· 1 we deﬁne the -th divided power of to be ∑i1+···+in=r 1 n 1 (in) xn .

Proposition 1.1.19. Let f , g ∈ F and let p, q be two non negative integers. The divided powers have the following properties:

(p) (q) p+q (p+q) p+q 1. f f = ( q ) f ∈ D F

( + )(p) = p (k) (p−k) 2. f g ∑k=0 f g . 3. ( f g)(p) = f (p)g(p).

( ) 4. ( f (p))(q) = pq ! f (pq). q!pq! Proposition 1.1.20. Let F and E be two free R-modules and let φ : F → E be an R-map. We denote by φ∗ : E∗ → F∗ the dual map of φ. Then, we have a well deﬁned R-map Drφ : Dr F → DrE which is the dual map of the map Sr(φ∗) : SrE∗ → Sr F∗. More precisely, the rth exterior power deﬁnes an endofunctor on the category of free R-modules and R-maps.

1.2 Combinatorics.

1.2.1 Partitions and skew partitions.

∞ Let N := {λ := {λi}i≥1 | λi = 0 but a ﬁnite number of terms}. In other words, p for each non negative integer p, let N := {(λ1,..., λp) | λi ∈ N, i = 1, . . . , p}. Hence ∞ p N = ∪p≥1N . We will consider (λ1,..., λp) and {λ1,..., λp, 0, . . .} the same element.

∞ Deﬁnition 1.2.1. A partition is an element λ = {λi}i≥1 ∈ N whose components are ∞ arranged in decreasing order, that is λi ≥ λi+1, ∀i ≥ 0. The length of a partition λ ∈ N is the number of non zero components in the partition, we often denote it by l(λ). The ∞ weight of a partition λ ∈ N is the sum of all its components. We note it by |λ| := ∑i≥1 λi and we will say that λ is a partition of weight |λ|.

For example, (10, 8, 9, 11, 4, 3, 2) is not a partition and (10, 7, 3, 2) is a partition of weight 22 and length 4.

∞ Deﬁnition 1.2.2. Let λ = (λ1,..., λp) ∈ N be a partition. For each i ∈ {1, . . . , λ1}, let λ˜ i be the number of terms of λ which are greater than or equal to i. Clearly λ˜ i ≥ λ˜ i+1, ∀i ∈ { } ˜ = ( ˜ ˜ ) 1, . . . , λ1 . The transpose or conjugate of λ is the partition λ λ1,..., λλ1 . 6 Preliminaries.

For example, the transpose of (10, 7, 3, 2) is the partition (4, 4, 3, 2, 2, 2, 2, 1, 1, 1).

Remark 1.2.3. We consider that all terms of a partition λ = (λ1,..., λp) are non zero. Note that hence, the length of λ is p and equals to λ˜ 1.

∞ Deﬁnition 1.2.4. Let λ = (λ1,..., λp) ∈ N be a partition. The diagram or shape of λ is 2 2 ∆λ := {(i, j) ∈ N | 1 ≤ i ≤ p, 1 ≤ j ≤ λi}. We will represent the graphic of ∆λ in N , where the points (i, j) ∈ ∆λ will be represented by squares.

For example, ∆(6,4,3,2), ∆(4,4,3,2,1,1) and ∆(5,4,3,2,1) correspond to

respectively.

We can consider the shape of a partition λ = (λ1,..., λp) as a kind of matrix with p rows with different lengths and λ1 columns. The ith row of ∆λ is of length λi. Now, the jth column of ∆λ is of length the number of terms in λ greater than or equal to j, ˜ which equals λj. In fact, ∆λ˜ is the diagram transpose of ∆λ, and hence this tell us that the conjugate of λ˜ is λ. That is, λ˜ = λ and we can conclude that the conjugation λ → λ˜ is an involution on the set of partitions.

The notions of partitions are, in fact, an introduction of a more generalized concept we explain next. In the set of partitions one can deﬁne an order. Let λ = (λ1,..., λp) and µ = (µ1,..., µq) be two partitions, we say µ ⊆ λ if p ≥ q and µi ≤ λi, ∀i ∈ {1, . . . , q}. It is equivalent to say that the diagram of µ is contained in the diagram of µ, that is ∆µ ⊆ ∆λ.

Deﬁnition 1.2.5. Let µ = (µ1,..., µq) and λ = (λ1,..., λp) be two partitions such that µ ⊆ λ. We deﬁne the skew partition λ/µ := (λ1 − µ1,..., λq − µq, λq+1,..., λp) with skew 2 shape of diagram ∆λ/µ := ∆λ − ∆µ = {(i, j) ∈ N | 1 ≤ i ≤ p; µi + 1 ≤ j ≤ λi}.

For convenience we will write µ = (µ1,..., µp) where µq+1 = ··· = µp = 0. For example, the skew diagrams associated to (4, 3, 2)/(2, 1), (5, 2, 1)/(4, 0, 0) and (7, 6, 3)/(4, 5, 1) correspond to

Note that, the skew partition λ/µ is a partition if, and only if the length of µ equals the length of λ or µ is the zero partition. Clearly, the set of partitions is a subset of the set of skew partitions, indeed λ/(0) = λ. 1.2 Combinatorics. 7

1.2.2 Tableaux. Deﬁnition 1.2.6. Let λ/µ a skew partition and let S be a totally ordered set. A tableau of shape λ/µ with values in the set S is a function from the skew shape ∆λ/µ to S. We denote Tabλ/µ(S) the set of such tableaux.

We will represent graphically T ∈ Tabλ/µ(S) by the skew shape ∆λ/µ ﬁlled with elements of S. Let us see few simple examples. Consider S = {1, . . . , 5} with the usual order on N and λ = (3, 2). Then, T : ∆λ → S such that T(i, j) = max{i, j} is a tableau of shape λ with values in S. Two more examples: R : ∆λ → S such that R(i, j) = i + j and U : ∆λ → S such that U(i, j) = i.

T = 1 2 3 R = 2 3 4 U = 1 1 1 2 2 3 4 2 2

Deﬁnition 1.2.7. Let T ∈ Tabλ/µ(S). We say T is row-standard if T(i, j) < T(i, j + 1) for all (i, j), (i, j + 1) ∈ ∆λ/µ. We say T is column-standard if T(i, j) ≤ T(i + 1, j) for all (i, j), (i + 1, j) ∈ ∆λ/µ. Finally, T is called standard if it is row and column standard. Continuing with the examples above, T(i, j) = max(i, j) and U(i, j) = i are both column-standard but not row-standard while R(i, j) = i + j is standard.

Deﬁnition 1.2.8. Let T ∈ Tabλ/µ(S). We say that T is co-row-standard if T(i, j) ≤ T(i, j + 1) ∀(i, j), (i, j + 1) ∈ ∆λ/µ. We say that T is co-column-standard if T(i, j) < T(i + 1, j) ∀(i, j), (i + 1, j) ∈ ∆λ/µ. Finally, T is co-standard if is co-row and co-column standard. Observe that T is co-row-standard but not co-column-standard, while R and U are both co-standard. 8 Preliminaries. Chapter 2

Schur Functors and CoSchur Functors.

This chapter is devoted to an introduction of Schur and CoSchur functors theory using only elementary rudiments of multilinear algebra and combinatorics, which we have just presented into the Introductory Material. We follow basically the contents from [1] with a few references from [20] and [9]. In sections 2.1 and 2.2 we deﬁne Schur and CoSchur functors and we establish the freeness of both functors. In the remaining sections we give decomposition formulas for Schur functors including Littlewood-Richardson rule and Cauchy formulas.

2.1 Schur Functors.

We start with some notations we will use along this paper. Let R be a commutative ring and let F be a free R-module of rank n. Let µ = (µ1,..., µp) and (λ1,..., λp) be two partitions such that µ ⊆ λ. We deﬁne the free R-modules

λ1−µ1 λp−µp Λλ/µF := Λ F ⊗R · · · ⊗R Λ F

λ1−µ1 λp−µp Sλ/µF := S F ⊗ · · · ⊗ S F

λ1−µ1 λp−µp Dλ/µF := D F ⊗ · · · ⊗ D F

T F := F ⊗ · · · ⊗ F ⊗ · · · ⊗ F ⊗ · · · ⊗ F = N F , where λ/µ (1,µ1+1) (1,λ1) (p,µp+1) (p,λp) (i,j)∈∆λ/µ (i,j) F(i,j) denotes a copy of F and the tensor power ⊗ is over R. Note that Tλ/µF is an usual way to write the tensor power T|λ/µ|F where |λ/µ| = |λ| − |µ|.

Deﬁnition 2.1.1. We deﬁne the Schur map dλ/µ : Λλ/µF → Sλ˜ /µ˜ F to be the composition

α β Λλ/µF −→ Tλ/µ −→ Sλ˜ /µ˜ F,

9 10 Schur Functors and CoSchur Functors.

λ −µ where α is the tensor product of the natural inclusions or diagonalizations ıi : Λ i i F → Tλi−µi F = F ⊗ · · · ⊗ F i = p (i,µi+1) (i,λi), 1, . . . , and β is the tensor product of the multi- ˜ plications m : F ⊗ · · · ⊗ F → Sλj−µ˜ j F, j = 1, . . . , λ . Note that ı is a tensor j (µ˜ j+1,j) R R (λ˜ j,j) 1 product of appropriated comultiplications on ΛF. We will often denote ıi simply by ∆.

There is another way to deﬁne the Schur map which we will often use. It is a simple change of notation using Ferrers matrices which sometimes makes operations easier. We will use both indifferently. The Ferrers matrix associated to the skew partition λ/µ is a λ1 × λ1 matrix of zeros and ones αλ/µ = (ai,j) deﬁned by ( ai,j = 1 i f µi + 1 ≤ j ≤ λi

ai,j = 0 i f 1 ≤ j ≤ µi or λj+1 ≤ j ≤ λ1

Keeping in mind that Fai,j = R i f ai,j = 0, dλ/µ is defined as the composition Λλ/µF → N F → S F, where F is a copy of F, the first map is the tensor product of ai,j∈αλ/µ ai,j λ˜ /µ˜ ai,j λ −µ the natural injections Λ i i F → Fa ⊗ · · · ⊗ Fa , i = 1, . . . , q, and the second map is i,1 i,λ1 ˜ a1,j+···+aq,j λj−µ˜ j the tensor product of the multiplication Fa1,j ⊗ · · · ⊗ Faq,j → S F = S F, j = 1, . . . , λ1, since a1,j + ··· + aq,j = λ˜ j − µ˜ j. In fact, we can generalize the above map to any matrix of zeros and ones. Let α = (αi,j) be an arbitrary s × t matrix of zeros and ones. For each i = 1, . . . , s t s let pi = ∑j=1 ai,j and for each j = 1, . . . , t let qj = ∑i=1 ai,j. Then we can consider free R-modules p ps q qt ΛαF := Λ 1 F ⊗ · · · ⊗ Λ F and Sα˜ F := S 1 F ⊗ · · · ⊗ S F and we can define a natural map dα : ΛαF → Sα˜ F in the same manner as the Schur map.

Deﬁnition 2.1.2. The image of dλ/µ is called the Schur functor of F with respect to the skew partition λ/µ, and it is denoted by Lλ/µF. Let us see some examples. ˜ Examples 2.1.3. (1) If λ = (m), then λ = (1, . . . , 1) and the Schur map d(m) : Λ(m)F → m m S(1,...,1)F is just the natural inclusion of Λ F on T F. In this case, we see that m L(m) Λ F. ˜ (2) If λ = (1, . . . , 1), then λ = (m), and the Schur map d(1,...,1) : Λ(1,...,1)F → S(m)F is just m the multiplication F ⊗ · · · ⊗ F → S F. Clearly, L(1,...,1)F is the mth symmetric power SmF.

(3) Let λ = (3, 2), ﬁrst we describe the Schur map associated with λ. N F = T|λ|F Remember (i,j)∈∆λ (i,j) . We have,

3 2 O 5 2 2 d(3,2) : Λ F ⊗R Λ F → F(i,j) = T F → S F ⊗R S F ⊗R F (i,j)∈∆λ 2.1 Schur Functors. 11

u ∧ v ∧ w ⊗ x ∧ y → u ⊗ v ⊗ w ⊗ x ⊗ y − u ⊗ v ⊗ w ⊗ y ⊗ x − u ⊗ w ⊗ v ⊗ x ⊗ y + u ⊗ w ⊗ v ⊗ y ⊗ x −v ⊗ u ⊗ w ⊗ x ⊗ y + v ⊗ u ⊗ w ⊗ y ⊗ x + v ⊗ w ⊗ u ⊗ x ⊗ y − v ⊗ w ⊗ u ⊗ y ⊗ x +w ⊗ u ⊗ v ⊗ x ⊗ y − w ⊗ u ⊗ v ⊗ y ⊗ x − w ⊗ v ⊗ u ⊗ x ⊗ y + w ⊗ v ⊗ u ⊗ y ⊗ x → ux ⊗ vy ⊗ w − uy ⊗ vx ⊗ w − ux ⊗ wy ⊗ v + uy ⊗ wx ⊗ v − vx ⊗ uy ⊗ w + vy ⊗ ux ⊗ w +vx ⊗ wy ⊗ v − vy ⊗ wx ⊗ v + wx ⊗ uy ⊗ v − wy ⊗ ux ⊗ v − wx ⊗ vy ⊗ u + wy ⊗ vx ⊗ u.

It is clear that Lλ/µF is a submodule of Sλ˜ /µ˜ F, but it is not trivial that actually Lλ/µF is a free R-module.

2.1.1 The freeness of Schur Functors.

As we have anticipated in the previous subsection, the modules Lλ/µF are free. We will show this fact by ﬁnding a basis to Lλ/µF through a basis of F. However, ﬁrst we need to solve the computational problem of dλ/µ.

Let λ/µ be a skew partition with λ˜ 1 = q and let BF := {x1,..., xn} be a basis of F. We can describe a basis of Λλ/µF and Sλ/µF as follows. i ∈ { q} I = { } For each 1, . . . , we denote by i α(i,µi+1),..., α(i,λi) a strictly increasing { } < < ⊗ · · · ⊗ subset of 1, . . . , n such that α1 ... αs. Clearly, the elements XI1 XIq , where each Ii is a such of these subsets, form a basis of Λλ/µF. Now, if we choose S to be BF with the order x1 < ··· < xn, then for each basis element = ⊗ · · · ⊗ ∈ ( ) ( ) = X XI1 XIq we can associate a tableau TX Tabλ/µ S deﬁned by TX i, j Xα(i,j) . Clearly, TX is a row-standard tableau. And conversely, if T ∈ Tabλ/µ(S) is a row-standard = ⊗ · · · ⊗ = ( ) ∧ · · · ∧ ( ) tableau, then the element XT : XI1 XIq where XIi T i, µi+1 T i, λi is a such of basis element we have just described. So, {XT | T ∈ Tabλ/µ(S) is row-standard} is a basis of Λλ/µF. For example, let λ = (4, 2, 1) and n = 10. Then, the tableau

T = x1 x4 x5 x7 x2 x10 x3 deﬁne the basis element XT = x1 ∧ x4 ∧ x5 ∧ x7 ⊗ x2 ∧ x10 ⊗ x3 in Λ(4,2,1)F. In the same way, we can describe a basis of Sλ˜ /µ˜ F through Tabλ/µ(S). If T ∈ Tabλ/µ{x1, ..., xn} is column-standard, then we can associate to it the element ZT = XJ ⊗ · · · ⊗ XJ , 1 λ1 = ( ) ····· ( ˜ ) { | ∈ ( ) } where XJi T µ˜i, i T λi, i . Clearly, ZT T Tabλ/µ S is column-standard is a basis of Sλ˜ /µ˜ F. For example, let λ = (4, 3, 2, 1) and n = 5. The tableau

T = x1 x3 x4 x5 x1 x5 x5 x1 x5 x2 3 2 deﬁne the basis element ZT = x1x2 ⊗ x3x5 ⊗ x4x5 ⊗ x5 of S(4,3,2,1)F. 12 Schur Functors and CoSchur Functors.

From this follows that {dλ/µ(XT) | T ∈ Tabλ/µ(S) is row-standard} generate Lλ/µF. We will show that {dλ/µ(XT) | T ∈ Tabλ/µ(S) is standard} is a basis of Lλ/µF. Let T ∈ Tabλ/µ(S) with λ = (λ1,..., λq) and µ = (µ1,..., µq). Remember dλ/µ = (m ⊗ · · · ⊗ m )(ı ⊗ · · · ⊗ ı ) (ı ⊗ · · · ⊗ ı )(X ) = (− )sg(σ)X 1 λ1 1 q . Thus, 1 q T ∑σ=(σ1,...,σq) 1 Tσ where σi is a permutation of 1, . . . , λi − µi and Tσ is the tableau deﬁned by Tσ(i, j) = T(i, σi(j)). ( ⊗ · · · ⊗ ) ( ) = (− )sg(σ) ∈ Applying m1 mλ1 we obtain dλ/µ XT ∑σ 1 ZTσ , where ZTσ Sλ˜ /µ˜ F similarly as we have seen before. At this moment we will introduce some facts about tableaux which we will need next. Let S = BF.

Definition 2.1.4. Let T ∈ Tabλ/µ(S) and let p, q be positive integers. We define Tp,q to be the number of times the elements x1,..., xq appear as entries in the first p rows of T. Formally, Tp,q = |{(i, j) ∈ T | i ≤ p and T(i, j) ∈ {x1,..., xq}}|.

For example, T = x1 x4 x5 x7 , T5,2 = 4. x2 x10 x3

Deﬁnition 2.1.5. Let T, S ∈ Tabλ/µ(S). We say S ≤ T if Sp,q ≥ Tp,q for every positive integers p, q and we say S < T if S ≤ T and Sp,q > Tp,q for at least one pair p, q.

Proposition 2.1.6. ≤ is a reﬂexive and transitive relation on Tabλ/µ(S). Proof. Obviously ≤ is reﬂexive, Tp,q = Tp,q for every p, q, so T ≤ T. Let T, W, R ∈ Tabλ/µ(S) such that T ≤ W and W ≤ R, we want to see T ≤ R. Let p, q be positive integers, by hypothesis Rp,q > Wp,q > Tp,q and hence Rp,q > Tp,q. Remark 2.1.7. If we restrict ≤ to the subset of row-standard tableaux, then ≤ is consis- tent with the lexicographic order induced by the correspondence between row-standard tableaux and the basis elements of Λλ/µF.

Lemma 2.1.8. Let T, R ∈ Tabλ/µ(S) where R is formed by exchanging certain entries from the kth row of T, say T(k, l1),..., T(k, la), to certain entries of the (k + 1)th row of T, say T(k + 1, m1),..., T(k + 1, ma), where T(k + 1, mi) < T(k, li) for i = 1, . . . , a. Then R < T. Proof. First we consider the simple case a = 1. We have , R(k, l) = T(k + 1, m), R(k + 1, m) = T(k, l) and R(i, j) = T(i, j) for all (i, j) , (k, m), (k + 1, m). By hypothesis R(k, l) = T(k + 1, m) < T(k, l) = R(k + 1, m). Let p, q be positive integers, remember Rp,q = |{(i, j) ∈ S | i ≤ p and S(i, j) ∈ {x1,..., xq}|. Clearly Rp,q = Tp,q if p < k since R equals to T at the first k − 1 rows. We fix p = k, R differs T only at the entry (k, l), and R(k, l) appears at the first k rows of R one more time than of T, respectively S(k + 1, m) one less. Clearly Rk,q = Tk,q i f xq < R(k, l), or xq ≥ R(k + 1, m), since R(k, l) < R(k + 1, m) and Rk,q = Tk,q + 1 − 1. If R(k, l) ≤ xq < R(k + 1, m), then Rk,q = Tk,q + 1. Since Rp,q equals Tp,q if p > k, we conclude R < T. Considering that we can see R as a result of a series of tableau Ri formed by exchanging Ri(k, li) = Ri−1(k + 1, mi) and Ri(k + 1, mi) = Ri−1(k, li) with 1 ≤ i ≤ a and R1 = T and hence Ri−1 < Ri as we have seen at case a = 1, the general case follows because of the transitivity of ≤. 2.1 Schur Functors. 13

Our ﬁrst goal will be to show that B = {dλ/µ(XT) | T ∈ Tabλ/µ(S) is standard} is a system of generators of Lλ/µF. We will proceed deﬁning an appropriate map whose image is contained in the kernel of dλ/µ and which will allow us to prove that dλ/µ(XT), with T row-standard, is a linear combination of elements of B. Moreover, we will establish an isomorphism between the cokernel of this map and Lλ/µF. This fact will give us a natural way to describe the modules Lλ/µF.

Lemma 2.1.9. For each i ∈ {1, . . . , q − 1}, the map dλ/µ can be factored as follows: ρ λ −µ λ −µ λ −µ Λ F −→ Λ 1 1 F ⊗ · · · ⊗ Λ i−1 i−1 F ⊗ Sa +a F ⊗ · · · ⊗ Sa +a F ⊗ Λ i+2 i+2 F λ/µ i,1 i+1,1 i,λ1 i+1,λ1 η λq−µq L L ⊗ · · · ⊗ Λ F −→ ∈ < Fa ⊗ Sa +a F ⊗ · · · ⊗ Sa +a F ⊗ ∈ > + F ak,j α,k i k,j i,1 i+1,1 i,λ1 i+1,λ1 ak,j α,k i 1 (k,j) ν −→ S F, where ρ = Id ⊗ · · · ⊗ Id ⊗ d ⊗ Id ⊗ · · · ⊗ Id, η = ı ⊗ · · · ⊗ ı ⊗ λ˜ /µ˜ (λi,λi+1)/(µi,µi+1) 1 i−1

Id ⊗ ıi+2 ⊗ · · · ⊗ ıq and ν is the tensor product of the multiplication maps m¯ j : Fa1,j ⊗ · · · ⊗ λ˜ j−µ˜ j Fai−1,j ⊗ Sai,j+ai+1,j F ⊗ Fai+2,j ⊗ · · · ⊗ Faq,j → S F, j = 1, . . . , λ1 deﬁned over generators by m¯ j( f1 ⊗ · · · ⊗ fi−1 ⊗ fi fi+1 ⊗ fi+2 ⊗ · · · ⊗ fq) = f1 ····· fq, where fai,j = 1 ∈ R i f ai,j = 0. Proof. d is the composition of maps ı ⊗ ı : Λλi−µi F ⊗ Λλi+1−µi+1 F → (λi,λi+1)/(µi,µi+1) i i+1 0 Fa ⊗ · · · ⊗ Fa ⊗ Fa ⊗ · · · ⊗ Fa and the tensor product of the maps m : Fa ⊗ i,1 i,λ1 i+1,1 i+1,λ1 j i,j → = ◦ ◦ = ( ⊗ · · · ⊗ )( ⊗ Fai+1,j Sai,j+ai+1,j F, j 1, . . . , λ1. Then, we can write ν η ν m¯ 1 m¯ λ1 Id 0 ⊗ · · · ⊗ 0 ⊗ )( ⊗ · · · ⊗ ) ( ⊗ 0 ⊗ · · · ⊗ 0 ⊗ ) L → m1 mλ Id ı1 ıq , where Id m1 mλ Id : ai j∈α Fai,j L 1 L 1 , λ/µ < Fa ⊗ Sa +a F ⊗ · · · ⊗ Sa +a F ⊗ > + Fa . ak,j,k i k,j i,1 i+1,1 i,λ1 i+1,λ1 ak,j,k i 1 k,j Clearly (m¯ ⊗ · · · ⊗ m¯ )(Id ⊗ m0 ⊗ · · · ⊗ m0 ⊗ Id) = m. 1 λ1 1 λ1

As a result of these factorizations we focus our study on dλ/µ when µ = (µ1, µ2) and λ = (λ1, λ2). 0 Let p1 = λ1 − µ1, p2 = λ2 − µ2 and k = λ2 − µ1, and consider the partitions λ = 0 0 0 (p1 + p2 − k, p2) and µ = (p2 − k). Since λ /µ = (p1, p2) = λ/µ, dλ/µ = dλ0/µ0 .

Deﬁnition 2.1.10. Let p1, p2, k ∈ N such that pi ≥ k. We deﬁne the map

p1,p2 p1 p2 2 2 p1−k p2−k δk : Λ F ⊗ Λ F → S F ⊗ · · · ⊗ S F ⊗ Λ F ⊗ Λ F

∆⊗∆ Id⊗T⊗Id as the composition Λp1 F ⊗ Λp2 F −−→ Λk F ⊗ Λp1−k F ⊗ Λk F ⊗ Λp2−k F −−−−−→ Λk F ⊗ d(k,k)⊗Id⊗Id Λk F ⊗ Λp1−k F ⊗ Λp2−k F −−−−−−−→ S2F ⊗ · · · ⊗ S2F ⊗ Λp1−k F ⊗ Λp2−k F, where ∆ is the appropriate diagonal map and T is the canonical isomorphism Λp1−k F ⊗ Λk F Λk F ⊗ Λp1−k F. ∈ ≥ + p1,p2 = ( ⊗ p1−k,p2−k) ◦ Lemma 2.1.11. Let p1, p2, k N such that pi k 1. Then δk+1 Id δ1 p1,p2 δk . ⊗···⊗ k+1 k+1 ∆⊗∆ m1 mk+1 2 Proof. Remember d(k+1,k+1) : Λ F ⊗ Λ F −−→ F ⊗ · · · ⊗ F −−−−−−−→ S F ⊗ · · · ⊗ S2F. By coassociativity and cocommutative of diagonal, ∆ ⊗ ∆ equals to the map ∆⊗∆ Id⊗T⊗Id ∆⊗∆ Λk+1F ⊗ Λk+1F −−→ Λk F ⊗ F ⊗ Λk F ⊗ F −−−−−→ Λk F ⊗ Λk F ⊗ F ⊗ F −−→ F ⊗ · · · ⊗ F ⊗ F ⊗ · · · ⊗ F ⊗ F ⊗ F and then d(k+1,k+1) = d(k,k) ⊗ d(1,1). p1,p2 = ( ⊗ · · · ⊗ ⊗ ⊗ )( ⊗ ⊗ ⊗ )( ⊗ ⊗ )( ⊗ ) We have, δk+1 m1 mk+1 Id Id ∆ ∆ Id Id Id T Id ∆ ∆ , where (∆ ⊗ ∆ ⊗ Id ⊗ Id)(Id ⊗ T ⊗ Id)(∆ ⊗ ∆) : Λp1 F ⊗ Λp2 F → F ⊗ · · · ⊗ F ⊗ F ⊗ · · · ⊗ F ⊗ Λp1−k−1F ⊗ Λp2−k−1F. 14 Schur Functors and CoSchur Functors.

Again by coassociativity and cocommutative, the last map equals to (Id ⊗ T ⊗ Id)(∆ ⊗ ∆ ⊗ ∆ ⊗ ∆)(Id ⊗ T ⊗ Id)(∆ ⊗ ∆) : Λp1 F ⊗ Λp2 F → Λk F ⊗ Λp1−k F ⊗ Λk F ⊗ Λp2−k F → Λk F ⊗ Λk F ⊗ Λp1−k F ⊗ Λp2−k F → F ⊗ · · · ⊗ F ⊗ F ⊗ · · · ⊗ F ⊗ F ⊗ Λp1−k−1F ⊗ F ⊗ Λp2−k−1F → F ⊗ · · · ⊗ F ⊗ F ⊗ · · · ⊗ F ⊗ F ⊗ F ⊗ Λp1−k−1F ⊗ Λp2−k−1F And then, applying m1 ⊗ · · · ⊗ mk ⊗ mk+1, p1,p2 = ( ⊗ · · · ⊗ ⊗ ⊗ )( ⊗ ⊗ )( ⊗ ⊗ ⊗ )( ⊗ ⊗ )( ⊗ ) δk+1 m1 mk mk+1 Id Id T Id ∆ ∆ ∆ ∆ Id T Id ∆ ∆ p1−k,p2−k p1,p2 which is exactly the map (Id ⊗ δ1 )δk .

Lemma 2.1.12. Let α = (ai,j) be the 2 × p1 + p2 − k matrix of zeros and ones:

1...1 1...1 0...0 1...1 0...0 1...1

|{z} |{z} |{z} k p1 − k p2 − k Then the diagram

Λp1 F ⊗ Λp2 F p ,p 1 2 dα δk 2 2 p −k p −k S F ⊗ · · · ⊗ S F ⊗ Λ 1 F ⊗ Λ 2 F Sα˜ F Id⊗∆⊗∆

p1,p2 is commutative and Imδk Imdα. In particular, if λ = (p1 + p2 − k, p2) and µ = p1,p2 (q2 − k, 0), then Imδk Lλ/µ. Proof. First we describe the map dα. ı ⊗ı p +p −k d p1 F ⊗ p2 F −−−→1 2 L F → S2F ⊗ · · · ⊗ S2F ⊗ F ⊗ · · · ⊗ F 1 2 a = α : Λ Λ ai,j∈α ai,j , since ∑j=1 1,j p1+p2−k = = = ··· = = = = ··· = = p1, ∑j=1 a2,j p2 and q1 2 2 qk, qk+1 1 1 qp1+p2−k. More explicit, we can write dα = (m1 ⊗ · · · ⊗ mk ⊗ Id ⊗ · · · ⊗ Id)(∆ ⊗ ∆). p1,p2 Now, (Id ⊗ ∆ ⊗ ∆)δk = (Id ⊗ · · · ⊗ Id ⊗ ∆ ⊗ ∆)(m1 ⊗ · · · ⊗ mk ⊗ Id ⊗ Id)(∆ ⊗ ∆ ⊗ Id ⊗ Id)(Id ⊗ T ⊗ Id)(∆ ⊗ ∆). It is clear that this map equals to (m1 ⊗ · · · ⊗ mk ⊗ Id ⊗ Id)(∆ ⊗ ∆ ⊗ ∆ ⊗ ∆)(Id ⊗ T ⊗ Id)(∆ ⊗ ∆). By coassociativity and commutativity, (Id ⊗ · · · ⊗ Id ⊗ ∆ ⊗ ∆)(∆ ⊗ ∆ ⊗ Id ⊗ Id)(Id ⊗ T ⊗ Id)(∆ ⊗ ∆) : Λp1 F ⊗ Λp2 F → Λk F ⊗ Λp1−k F ⊗ Λk F ⊗ Λp2−k F → Λk F ⊗ Λk F ⊗ Λp1−k F ⊗ Λp2−k F → F ⊗ · · · ⊗ F equals to the map ∆ ⊗ ∆ : p ,p p1 F ⊗ p2 F → L F (Id ⊗ ⊗ ) 1 2 = d Λ Λ ai,j∈α ai,j . Directly, ∆ ∆ δk α. p1,p2 Observe that the bottom map Id ⊗ ∆ ⊗ ∆ is an injection, hence Im(δk ) Im(dα).

Lemma 2.1.13. Let p1, p2, k be positive integers, with pi ≥ k. We deﬁne the following composite map: ⊗ ⊗ p1+p2−k k ∆ Id p1 p2−k k Id m˜ p1 p2 ωk : Λ F ⊗ Λ F −−−→ Λ F ⊗ Λ F ⊗ Λ F −−−→ Λ F ⊗ Λ F where m˜ is the multiplication on the exterior algebra ΛF. Then, the following diagram is commutative

wk Λp1+p2−k F ⊗ Λk F Λp1 F ⊗ Λp2 F p +p −k k p ,p 1 2 , δ 1 2 . δ1 1 Id⊗wk−1 S2F ⊗ Λp1+p2−k−1F ⊗ Λk−1F S2F ⊗ Λp1−1F ⊗ Λp2−1F 2.1 Schur Functors. 15

p +p −k,k 1 2 p1+p2−k Proof. We can write the map (Id ⊗ ωk−1)δ1 as the composition Λ F ⊗ ∆⊗∆ Id⊗∆⊗Id⊗Id Λk F −−→ F ⊗ Λp1+p2−k−1F ⊗ F ⊗ Λk−1F −−−−−−−→ F ⊗ Λp1−1F ⊗ Λp2−1−(k−1)F ⊗ F ⊗ ( ⊗ ⊗ ⊗ ) k−1 Id T T Id p −1 p −1−(k−1) k−1 m⊗Id⊗m˜ p −1 Λ F −−−−−−−−→ F ⊗ F ⊗ Λ 1 F ⊗ Λ 2 F ⊗ Λ F −−−−−→ S2F ⊗ Λ 1 F ⊗ Λp2−1F. ∆⊗∆ By coassociativity of diagonal, Λp1+p2−k F ⊗ Λk F −−→ F ⊗ Λp1+p2−k−1F ⊗ F ⊗ Λk−1F Id⊗∆⊗Id⊗Id ∆⊗∆ −−−−−−−→ F ⊗ Λp1−1F ⊗ Λp2−1−(k−1)F ⊗ F ⊗ Λk−1F equals to Λp1+p2−k F ⊗ Λk F −−→ ∆⊗Id⊗Id⊗Id Λp1 F ⊗ Λp2−1−(k−1)F ⊗ F ⊗ Λk−1F −−−−−−−→ F ⊗ Λp1−1F ⊗ Λp2−1−(k−1)F ⊗ F ⊗ Λk−1F. ⊗ = ∧ · · · ∧ ⊗ ∧ · · · ∧ ∈ p1+p2−k ⊗ k We denote f g : f1 fp1+p2−k gp1+p2−k+1 gp1+p2 Λ F Λ F. p1+p2−k,k sg(i)+sg(j)+sg(t) Then, (Id ⊗ ωk−1)δ1 ( f ⊗ g) = ∑i,j,t(−1) fji(1) · gt(1) ⊗ fji(2) ∧ · · · ∧ f ⊗ f ∧ · · · ∧ f ∧ g ∧ · · · ∧ g ji(p1) ti(p1+1) ti(p1+p2−k) t(2) t(k). p ,p ∆⊗Id 1 2 p1+p2−k k Similarly, we can write the map δ1 ◦ wk as the composition Λ F ⊗ Λ F −−−→ Id⊗m˜ ∆⊗∆ Id⊗T⊗Id Λp1 F ⊗ Λp2−k F ⊗ Λk F −−−→ Λp1 F ⊗ Λp2 F −−→ F ⊗ Λp1−1F ⊗ F ⊗ Λp2−1F −−−−−→ F ⊗ m⊗Id⊗Id F ⊗ Λp1−1F ⊗ Λp2−1F −−−−−→ S2F ⊗ Λp1−1F ⊗ Λp2−1F. By the compatibility between comultiplication and multiplication on the exterior algebra ΛF the map Λp2−k F ⊗ Λk F → Λp2 F → F ⊗ Λp2−1F equals to ∆⊗∆ Id⊗T⊗Id 1 p2−k k α p2−k−α 1−α k+α−1 α 1−α ∑α=0 Λ F ⊗ Λ F −−→ Λ F ⊗ Λ F ⊗ Λ F ⊗ Λ F −−−−−→ Λ F ⊗ Λ F m˜ ⊗m˜ ⊗Λp2−k−αF ⊗ Λk+α−1F −−−→ F ⊗ Λp2−1F. p p (δ 1, 2 ◦ w )( f ⊗ g)= (− )sg(i)+sg(j)+sg(t) f · f ⊗ f ∧ · · · ∧ f ⊗ Thus, 1 k ∑i,j,t 1 ji(1) ti(p1+1) ji(2) ji(p1) f ∧ · · · ∧ f ∧ g + (− )sg(i)+sg(j)+sg(t) f · g ⊗ f ∧ · · · ∧ f ⊗ ti(p1+2) ti(p1+p2−k) ∑i,j,t 1 ji(1) t(1) ji(2) ji(p1) f ∧ · · · ∧ f ∧ g ∧ · · · ∧ g ti(p1+1) ti(p1+p2−k) t(2) t(k). To finish the proof it is enough to show that the first addend is zero. In fact, the ∆⊗Id first term is the image of f ⊗ g under the composite map Λp1+p2−k F ⊗ Λk F −−−→ Λp1 F ⊗ ∆⊗∆⊗Id Id⊗T⊗Id⊗Id Λp2−k F ⊗ Λk F −−−−−→ F ⊗ Λp1−1F ⊗ F ⊗ Λp2−k−1F ⊗ Λk F −−−−−−−→ F ⊗ F ⊗ Λp1−1F ⊗ m⊗⊗Id⊗m˜ Λp2−k−1F ⊗ Λk F −−−−−−→ S2F ⊗ Λp1−1F ⊗ Λp2−1. And again because of coassociativ- ∆⊗Id ity and cocommutativity of comultiplication, Λp1+p2−k F ⊗ Λk F −−−→ Λp1 F ⊗ Λp2−k F ⊗ ∆⊗∆⊗Id ∆⊗Id Λk F −−−−−→ F ⊗ Λp1−1F ⊗ F ⊗ Λp2−k−1F ⊗ Λk F equals to Λp1+p2−k F ⊗ Λk F −−−→ Λp1−1F Id⊗∆⊗Id Id⊗∆⊗Id⊗Id ⊗Λp2−k+1F ⊗ Λk F −−−−−→ Λp1−1F ⊗ Λ2F ⊗ Λp2−k−1F ⊗ Λk F −−−−−−−→ Λp1−1F ⊗ F ⊗ F ⊗ Λp2−k−1F ⊗ Λk F

A simple calculation shows that the image which we are looking for is ∑i,j fi(1) ∧ · · · ∧ f ⊗ ( f · f − f · f ) ⊗ f ∧ · · · ∧ f i(p1−1) ji(p1) ji(p1+1) ji(p1+1) ji(p1)) ji(p1+2) ji(p1+p2−k), which is zero since u · v = v · u ∈ S2F.

Proposition 2.1.14. Let p1, p2, k ∈ N with pi ≥ k + 1. Then the composition

p1,p2 w δk+ p1+p2−k k k p1 p2 1 p1−k−1 p2−k−1 Λ F ⊗ Λ F −→ Λ F ⊗ Λ F −−−→ S2F ⊗ · · · ⊗ S2F ⊗ Λ F ⊗ Λ is zero. ∆ ∆⊗∆ Proof. We proceed by induction on k. When k = 0, we have Λp1+p2 −→ Λp1 ⊗ Λp2 −−→ p −1 p −1 p −1 p −1 F ⊗ Λ 1 F ⊗ F ⊗ Λ 2 F → S2F ⊗ Λ 1 F ⊗ Λ 2 F. The same argument we have just seen in the last part of the proof of Lemma 2.1.13 shows that this composition is zero. > p1,p2 = ( ⊗ p1−k,p2−k) ◦ p1,p2 p1,p2 ◦ = ( ⊗ Suppose k 0. Since δk+1 Id δ1 δk , we can write δk+1 wk Id 16 Schur Functors and CoSchur Functors.

p1−k,p2−k p1,p2 p1−k,p2−k p1,p2 p1−k,p2−k δ1 ) ◦ δk ◦ wk. By Lemma 2.1.11 (Id ⊗ δ1 ) ◦ δk = (Id ⊗ δ1 ) ◦ p1+p2−k,k p1−k,p2−k (Id ⊗ wk−1) ◦ δ1 . The result follows using the induction on (Id ⊗ δ1 ) ◦ (Id ⊗ wk−1).

Lemma 2.1.15. Let p1, p2, k, u, l be nonnegative integers such that pi ≥ k, 0 ≤ u ≤ l ≤ k − 1. Denote by ω¯ u the composite map Id⊗∆⊗Id m˜ ⊗m˜ ΛuF ⊗ Λp1+p2−l F ⊗ Λl−uF −−−−−→ ΛuF ⊗ Λp1−uF ⊗ Λp2+u−l F ⊗ Λl−uF −−−→ Λp1 F ⊗ Λp2 F. Then Im(ω¯ u) is contained in the image of the map

k−1 ∑ ωv p1,p2 M p1+p2−v v v=0 p1 p2 ωk : Λ F ⊗ Λ F −−−−→ Λ F ⊗ Λ F. 0≤v≤k−1

Proof. We proceed by induction on u. When u = 0, w¯ u = wl and the proposition u p +p −l l−u is clear. Assuming u > 0, let x ⊗ y ⊗ z ∈ Λ F ⊗ Λ 1 2 F ⊗ Λ F, we have ω¯ u(x ⊗ y ⊗ z) = (− )sg(i)x ∧ · · · ∧ x ∧ y ∧ · · · ∧ y ⊗ y ∧ · · · ∧ y ∧ ∑i 1 1 u i(1) i(p1−u) i(p1−u+1) i(p1+p2−l) u p +p −l z1 ∧ · · · ∧ zl−u. Now, we consider the composition ωl−u ◦ (m ⊗ Id) : Λ F ⊗ Λ 1 2 F ⊗ m⊗Id ∆⊗Id Id⊗m Λl−u −−−→ Λp1+p2−(l−u)F ⊗ Λl−uF −−−→ Λp1 F ⊗ Λp2−(l−u)F ⊗ Λl−uF −−−→ Λp1 F ⊗ Λp2 F. Because of compatibility between the multiplication and comultiplication the composition m ∆ ΛuF ⊗ Λp1+p2−l F −→ Λp1+p2−(l−u)F −→ Λp1 F ⊗ Λp2−(l−u)F equals to ∆⊗∆ Id⊗T⊗Id u u p1+p2−l u−α α p1−(u−α) p2−(l−u+α) u−α ∑α=0 Λ F ⊗ Λ F −−→ Λ F ⊗ Λ F ⊗ Λ F ⊗ Λ F −−−−−→ Λ F m⊗m ⊗Λp1−(u−α)F ⊗ ΛαF ⊗ Λp2−(l−u+α)F −−−→ Λp1 F ⊗ Λp2−(l−u)F. u sg(i)+sg(j) Thus ωl−u ◦ (m ⊗ Id)(x ⊗ y ⊗ z) = ω¯ u(x ⊗ y ⊗ z) + ∑α=1 ∑i,j(−1) xi(1) ∧ · · · ∧ x ∧ y ∧ · · · ∧ y ∧ · · · ∧ y ⊗ x ∧ · · · ∧ x ∧ y ∧ i(u−α) j(p1) j(1) j(p1−(u−α)) i(u−α+1) i(u) j(p1−(u−α)+1 · · · ∧ y ω (x ⊗ y ⊗ z) = (− )sg(i)x ∧ · · · ∧ x ∧ y ∧ x ∧ j(p2+p1−l). Since α ∑i 1 i(1) i(u−α) i(u−α+1) u · · · ∧ xi(u) ∧ z, we obtain ∑α=1 ω¯ u−α ◦ ωα(x ⊗ y ⊗ z) = ωl−u ◦ (m ⊗ Id)(x ⊗ y ⊗ z) − ω¯ u(x ⊗ p1,p2 y ⊗ z). By induction Im(ω¯ u−α ◦ ωα) ⊂ Im(ωk ), ∀1 ≤ α ≤ u, and hence ωl−u ◦ (m ⊗ p1,p2 Id)(x ⊗ y ⊗ z) ∈ Im(ωk ).

At this moment we deﬁne the auxiliary map we have mentioned at the begging of this subsection and next we state our ﬁrst main result.

Definition 2.1.16. Let λ = (λ1,..., λq) and µ = (µ1,..., µq) be two partitions such that i i µ ⊆ λ. For each i = 1, . . . , q − 1 consider partitions λ = (λi, λi+1) and µ = (µi, µi+1). One defines a map ωλ/µ to be the sum of maps Id1 ⊗ · · · ⊗ Idi−1 ⊗ ωλi/µi ⊗ Idi+2 ⊗ ... ⊗ Idq, pi,pi+1 where i = 1, . . . , q − 1 and ω i i = ω with p = λ − µ , k = λ + − µ . We define λ /µ ki i i i i i 1 i L¯ λ/µ(F) to be the cokernel of map ωλ/µ.

Theorem 2.1.17. The image of ωλ/µ is contained in the kernel of dλ/µ. Proof. Keeping in mind the factorization of dλ/µ, Lemma 2.1.9 it is enough to consider p1,p2 λ = (λ1, λ2), µ = (µ1, µ2) and show that Im(ωk ) ⊆ ker(dλ/µ). p1,p2 ◦ = ⊂ ( p1,p2 ) Now, since the composition δk+1 ωk 0 as we have seen before, ωk ker δk+1 k−1 k−1 p1,p2 and hence Im(∑v=0 ωv) ⊆ ker(∑v=0 δv+1 ). We have already ﬁnished considering that for p1,p2 p1,p2 each k ∈ N | pi ≥ k, (Id ⊗ ∆ ⊗ ∆) ◦ δk = dλ/µ ⇒ ker(δk ) ⊆ ker(dλ/µ) and then k−1 k−1 p1,p2 Im(∑v=0 ωv) ⊂ ker(∑v=0 δv+1 ) ⊂ ker(dλ/µ). 2.1 Schur Functors. 17

¯ As a direct corollary of Theorem 2.1.17 the map dλ/µ induces a surjective map dλ/µ : ¯ L¯ λ/µ(F) → Lλ/µ(F) deﬁned by dλ/µ(x¯) = dλ/µ(x), where x¯ is the image under the natural 0 ¯ ¯ map π : Λλ/µF → L¯ λ/µF. Note that x¯ = y¯ ⇒ dλ/µ(x¯) = dλ/µ(y¯) ⇔ dλ/µ(x) = dλ/µ(y) ¯ 0 which implies dλ/µ = dλ/µ ◦ π .

Lemma 2.1.18. Let T ∈ Tabλ/µ(S) be a row-standard tableau which is not standard. Then < − ± ∈ ( ) there exist standard tableaux Ti with Ti T such that XT ∑ XTi Im wλ/µ . j j+1 Proof. Let T = (T(j, µj + 1),..., T(j, λj)) and T = (T(j + 1, µj+1 + 1),..., T(j + 1, λj+1) be the ﬁrst two rows of T in which column-standardness is violated. Let T¯ = (Tj, Tj+1) j j j j be a tableau of shape λ /µ where λ = (λj, λj+1), µ = (µj, µj+1). First, we will proof the lemma in this particular case. Let p1 = λj − µj, p2 = λj+1 − µj+1 and k = λj+1 − µj. For convenience, we will denote ( ) = = ( + ) = = > T j, i ai, i 1, . . . , p1 and T j 1, l bl, l 1, . . . , p2. Let au+1 bp2−k+u+1 be the ﬁrst entries in which the violation takes place. Let X¯ = a1 ∧ · · · ∧ au ⊗ b1 ∧ · · · ∧ ∧ ∧ · · · ∧ ⊗ ∧ · · · ∧ ∈ u ⊗ p1+p2−l ⊗ l−1−u bp2−k+u+1 au+1 ap1 bp2−k+u+2 bp2 Λ F Λ F Λ F, with l = k − 1. Graphically: a1 ... au+1 ...

b1 ... bp2−k bp2−k+1 ... bp2−k+u+1 ... sg(i) ( ¯ ) = (− ) ∧ · · · ∧ ∧ ⊗ 0 ∧ ∧ · · · ∧ = ± ¯ We have w¯ u X ∑ 1 a1 au cI cI bp2−k+u+2 bp2 : ∑ XI − ∧ · · · ∧ ∧ where cI is the corresponding exterior product of p1 u terms of b1 bp2−k+u+1 0 au+1 ∧ · · · ∧ ap1 and cI of the complementary terms after applying Id ⊗ ∆ ⊗ ∆. We denote ¯ cI = ci ∧ · · · ∧ ci − and cI0 = ci ∧ · · · ∧ cp +p −k+u+1. By Lemma 2.1.15, ∑ ±XI ∈ 1 p1 u p1−u+1 1 2 p1,p2 Im(wk ) = Im(ωλj/µj ). j j For each I let T¯I be the tableau of shape λ /µ associated to X¯ I. There is only one term I, say I0, such that cI = au+1 ∧ · · · ∧ ap1 , and hence the tableau associated to I0 is the tableau associated to X¯ . The rest of tableaux T¯I are a result of exchanging some = entries au+1,..., ap1 by some entries b1,..., bp2−k+u+1, with necessarily ci1 bt for some t ∈ {1, . . . , p2 − k + u + 1}, by a previous lemma T¯I < T¯. p ∧ · · · ∧ ∧ ⊗ 0 ∧ ∧ · · · ∧ 1 ⊗ Observe that each a1 au cI cI bp2−k+u+2 bp2 as an element of Λ F p2 0 0 Λ F can be arranged such that equals xI ∧ · · · ∧ xIp ⊗ xI ∧ · · · ∧ xI with xI < ... < xIp 1 1 1 p2 1 1 ¯ and xI0 < ... < xI0 . Because of this we assume that TI, I , I0 is row-standard. In this 1 p2 ¯ ≤ > ⇒ > = situation for each TI we have ci1 cp2−k+u+1, since au+1 bp2−k+u+1 au+1 bt, t 1, . . . , p2 − k + u + 1. So, if T¯I is not standard, then the ﬁrst index where the column standardness violation in T¯I takes place is greater than u + 1.

Since there is a ﬁnite number of tableaux in Tabλj/µj (S), repeating the same procedure for each non standard tableau T¯I and continue as many times we need, ﬁnally we will ¯ = ¯ − ± ¯ ∈ ( j j ) obtain ∑ XTI X ∑I,I0 XTI Im ωλ /µ . Now the general case is clear, we only have to exchange X¯ by T(1, µ1 + 1) ∧ · · · ∧ T(1, λ1) ⊗ · · · ⊗ X¯ ⊗ T(j + 2, µj+2 + 1) ∧ · · · ∧ T(j + 2, λj+2) ⊗ · · · , w¯ u by the natural gen- ⊗ · · · ⊗ ⊗ ⊗ ⊗ · · · ⊗ eralization Id1 Idi−1 w¯ ui Idi+2 Idq, wλj/µj by wλ/µ and repeat this procedure for each couple of rows which column-standardness is violated.

Let us to apply the above lemma to the element associated to the tableau T = x1 x6 . x2 x5 18 Schur Functors and CoSchur Functors.

We have, p1 = 2, p2 = 2, k = 2, u = 1 and l = 1. We want to compute ω¯ 2(x1 ⊗ x2 ∧ x5 ∧ 3 Id⊗∆ 2 m⊗Id 2 2 x6), where ω¯ 2 is the map F ⊗ Λ F −−−→ F ⊗ F ⊗ Λ F −−−→ Λ F ⊗ Λ F. We obtain,

x1 x2 − x1 x5 + x1 x6 x5 x6 x2 x6 x2 x5 Directly from Lemma 2.1.18

Corollary 2.1.19. {dλ/µ(XT) | T ∈ Tabλ/µ(S) is standard} is a system of generators in Lλ/µF.

Corollary 2.1.20. {X¯ T | T ∈ Tabλ/µ(S) is standard} spans L¯ λ/µF. Proof. Since {XT | T ∈ Tabλ/µ(S) is row-standard} generate Λλ/µ(F), it is enough to see that each X¯ T, with T row-standard can be expressed as a combination of elements of that such. Let T ∈ Tabλ/µ(S) be a row-standard tableau. By Lemma 2.1.18 we know that there ¯ exist TI ∈ Tabλ/µ(S) standard tableaux such that XT − ∑I XTI ∈ Im(ωλ/µ). Clearly, XT = ¯ ∑I XTI .

Theorem 2.1.21. {dλ/µ(XT) | T ∈ Tabλ/µ{x1,..., xn} is standard} is a basis of Lλ/µF called the standard basis of Lλ/µF. So, Lλ/µF is a free R-module. Proof. We denote λ = (λ1,..., λq), µ = (µ1,..., µq). Since {dλ/µ(XT) | T ∈ Tabλ/µ{x1, ..., xn} is standard} is a system of generators of Lλ/µ as we have just seen, we only must proof that these elements are independent. We proceed as follows: Let Λ0 F and S0 F be the free R-submodules of Λ and S F respectively gen- λ/µ λ˜ /µ˜ λ/µ λ˜ /µ˜ erated by {XT | T ∈ Tabλ/µ(S) is standard} and {ZT | T ∈ Tabλ/µ(S) respec- 0 tively. Consider the natural injection i : Λλ/µF ,→ Λλ/µF and the natural projection π : S F → S0 F. Keeping in mind that Tab (S) is totally ordered lexicographi- λ˜ /µ˜ λ˜ /µ˜ λ/µ 0 cally and dλ/µ(Λλ/µF) = Lλ/µF , if we prove that the map associated to the composition π ◦ dλ/µ ◦ i is triangular with ones on the diagonal respect these basis in this order, then 0 dλ/µ ◦ i is an injection and hence Lλ/µF Lλ/µF which implies that {dλ/µ(XT) | T ∈ Tabλ/µ{x1,..., xn} is standard} is a basis of Lλ/µF. Let T ∈ Tabλ/µ(S) be a standard tableau. As we have seen at the beginning of sg(σ) this section, dλ/µ(XT) = ∑σ(−1) ZTσ where σ = (σ1,..., σq), σi runs through all permutations of {µi + 1, . . . , λi} and Tσ ∈ Tabλ/µ{x1,..., xn} is the tableau of entries 0 0 Tσ(i, j) = T(i, σi(j)). Let Tσ be the column standardization of Tσ, since Tσ and Tσ deﬁne the same element ZT in Sλ˜ /µ˜ F because the properties of the symmetric power, we can 0 write d (X ) = Z where σ runs over the set of permutations such that T 0 is λ/µ T ∑σ Tσ0 σ column-standard. Then π ◦ d (X ) = Z 0 which σ runs over the set of permutations λ/µ T ∑σ Tσ 0 such that Tσ is standard. 0 We observe that ZT occurs in the sum above, then it is enough to prove that Tσ < T for all σ , Id. First we observe that Tσ cannot be row-standard since the standardness of T, 0 0 and hence Tσ , Tσ if σ , Id. For σ , Id, Tσ is obtained from Tσ as a result of iterating the 0 process described in Lemma 2.1.8, so Tσ < Tσ, and by the same reason Tσ < T. Clearly 0 this implies Tσ < T for all σ , Id. The above theorem give us an explicit basis of Schur functors by means of combinatoric tools. As a particular case we have the following theorem (see [15], Theorem 6.3): 2.2 CoSchur Functors and the freeness of the CoSchur functors. 19

Theorem 2.1.22. Let λ = (λ1,..., λq) be a partition. If λ1 ≤ n, then

λi − λj + j − i dim(L F) = λ ∏ − 1≤i

Remark 2.1.23. Consider Lλ/µF. In general, if λ1 − µ1 is greater than the rank of F, then λ1−µ1 Λ F is zero. Thus, Lλ/µF is trivially null for skew partitions such that the length of λ/µ is greater than the rank of F.

Theorem 2.1.24. Lλ/µF is isomorphic to L¯ λ/µF. ¯ 0 ¯ 0 Proof. Since dλ/µ = dλ/µ ◦ π , π ◦ dλ/µ ◦ π ◦ i is an isomorphism, and hence L¯ λ/µF is 0 ¯ isomorphic to Λλ/µF as we have argued before. Then Lλ/µF is a free R-module with a ¯ basis {X¯ T | T ∈ Tabλ/µ{x1,..., xn} is standard}. Clearly dλ/µ is an isomorphism.

The isomorphism Lλ/µF L¯ λ/µF and the natural deﬁnition of ωλ/µ gives us a natural way to describe Schur functors as the following examples illustrate.

Examples 2.1.25. (1) L(2,1)F. We have p1 = 2, p2 = 1 and k = 1. Then, ω(2,1) is just the 2,1 3 2 1 map ω0 : Λ F → Λ F → Λ F. That is, L(2,1)F is the cokernel of the diagonal map 3 2 ∆2+1 : Λ F → Λ F ⊗ F.

(2) L(3,2)F. Here, p1 = 3, p2 = 2 and k = 2. In that case ω(3,2) is the sum of the two 3,2 5 3 2 3,2 4 3 2 maps ω0 : Λ F → Λ F ⊗ Λ F and ω1 : Λ F ⊗ F → Λ F ⊗ Λ F. The ﬁrst map ⊗ 4 ∆4+1 Id is the diagonal map ∆3+2 and the second map is the composition Λ F ⊗ F −−−−−→ Id⊗m˜ 3 1,1 3 2 Λ F ⊗ F ⊗ F −−−−→ Λ F ⊗ Λ . So, L(3,2)F is the cokernel of the sum of these two maps.

We want to end this section talking about the functoriality of Lλ/µ(−). Let E and F be two free R-modules, let φ : E → F be an R-map and let λ/µ a skew partition. From Proposition 1.1.7 we have a well defined map Λλ/µ(φ) : Λλ/µF → Λλ/µE λ1−µ1 λq−µq given by Λλ/µ(φ)(X1 ⊗ · · · ⊗ Xq) = (Λ (φ)(X1)) ⊗ · · · ⊗ (Λ (φ)(Xq)) where Xi λi−µi F E represent an element of Λ F. Denoting dλ/µ and dλ/µ the Schur maps associated to F and E respectively, we can define a natural R-map Lλ/µ(φ) : Lλ/µF → Lλ/µE by F E taking Lλ/µ(φ)(dλ/µ(X)) = dλ/µ(Λλ/µ(X)). Then, Lλ/µ(−) define an endofunctor on the category of free R-modules and R-map.

2.2 CoSchur Functors and the freeness of the CoSchur functors.

CoSchur functors are deﬁned in a similar way as Schur functor using divided and exterior powers. 20 Schur Functors and CoSchur Functors.

Deﬁnition 2.2.1. Let λ = (λ1,..., λp) and µ = (λ1,..., λp) be two partitions with µ ⊆ λ. 0 We deﬁne the coSchur map dλ/µ : Dλ/µF → Λλ/µF associated to the skew partition λ/µ to 0 α0 β ˜ be the composite map Dλ1−µ1 F ⊗ · · · ⊗ Dλp−µp F −→ N F −→ Λλ1−µ˜1 F ⊗ · · · ⊗ (i,j)∈∆λ/µ (i,j) λ˜ −µ˜ 0 0 − λ1 λ1 ı Dλi µi F → F ⊗ · · · ⊗ Λ , where α is the tensor product of the inclusions i : (i,µi+1) F i = p β m˜ F ⊗ · · · ⊗ (i,λi), 1, . . . , and is the tensor product of the multiplications j : (µ˜1+1,j) ˜ F → Λλj−µ˜ j F, j = 1, . . . , λ . (λ˜ i,j) 1

Deﬁnition 2.2.2. We deﬁne the coSchur functor Kλ/µF associated to F and the skew parti- 0 tion λ/µ to be the image of dλ/µ.

Clearly, Kλ/µF can be deﬁned through Ferrers matrices in analogous way as Lλ/µ and 0 the coSchur map can be generalized as a natural map dα where α is an arbitrary matrix of zeros as we have seen in previous sections. t Examples 2.2.3. (1) K(t)F = D F.

(t) (2) K(1,...,1)F = Λ F, t = 1 + ··· + 1. The Standard Basis Theorem of Schur functors has its dual version to CoSchur functors. 0 Indeed, we easily check that {dλ/µ(XT) | T ∈ Tabλ/µ is co-column-standard} is a basis of Dλ/µ and arguing as in section 2.1.1 we prove: 0 Theorem 2.2.4. {dλ/µ(XT) | T ∈ Tabλ/µ{x1,..., xn} is co-standard} is a basis of Kλ/µF called the co-standard basis of Kλ/µF. So, Kλ/µF is a free R-module.

r The functoriality of Kλ/µ(−) follows from the functoriality of D (−) in the same manner we have argued from Schur functors in section 2.1.1. We will ﬁnish our general ∗ ∗ discussion of Schur and CoSchur functors by stating the duality (Lλ/µF) Kλ˜ /µ˜ (F ). It follows directly from the dual nature of the Schur and CoSchur maps. ∗ ∗ Theorem 2.2.5. (Lλ/µF) Kλ˜ /µ˜ (F ). Proof. We denote d0 : Dλ˜ /µ˜ F → Λ F and d : Λ F → S F . The coSchur map λ˜ /λ˜ λ/µ λ/µ λ/µ λ˜ /µ˜ is the composite map β0 ◦ α0, where β0 is the tensor product of diagonal maps in DF and 0 α the tensor product of multiplications in ΛF. Similarly we have dλ/µ = α ◦ β. From the Introductory Material, β0 is the dual map of α and α0 is the dual map β.

Moreover, if R contains a ﬁeld of characteristic zero, then Lλ/µF Kλ˜ /µ˜ F. See [2], ∗ ∗ Corollary (2.3.3). From Theorem 2.2.5, it follows that Kλ/µF (Lλ˜ /µ˜ F ) . This isomorphism give the following formula of rank of CoSchur functor.

Theorem 2.2.6. Let λ = (λ1,..., λq) be a partition. If λ˜ 1 ≤ n, then

λ˜ i − λ˜ j + j − i dim(K F) = λ ∏ − 1≤i

2.3 Decomposition of Schur Functors.

The aim of this section will be to deﬁne a ﬁltration on Lλ/µ(F ⊕ G) with associated L graded module µ⊆γ⊆λ Lγ/µF ⊗ Lλ/γG.

Let F and G be free R-modules of rank n and m with ordered basis {x1,..., xn} and {xn+1,..., xn+m} respectively. We will consider the ordered basis {x1,..., xn+m} in F ⊕ G.

Definition 2.3.1. Let F be a module over a ring R.A finite filtration of F is a sequence of submodules of F 0 =: F0 ⊂ F1 ⊂ · · · ⊂ Fn−1 ⊂ Fn := F. n−1 One defines the associated graded object of the filtration ∑i=0 Fi+1/Fi.

Once we have deﬁned the modules of the ﬁltration, we will need to describe basis of these modules. We do it through a basis of F ⊕ G and Tabλ/µ(F ⊕ G). We start introducing some facts about partitions and tableaux we will use in this description.

Deﬁnition 2.3.2. Let λ = (λ1,..., λq), µ = (µ1,..., µq) be two partitions. We say λ ≥ µ if λ1 = µ1,..., λi = µi and λi+1 > µi+1 for some i.

For example, (1, 1, 1, 1, 3, 5) ≥ (1, 1, 1, 1, 2, 5).

Proposition 2.3.3. ≥ is a total order on N∞. Proof. First we will see that ≥ is an order. Clearly, λ ≥ λ for all partition λ ∈ N∞. Let λ, µ be two partitions such that λ ≥ µ and µ ≥ λ and let i, j be the indexes such that λ1 = µ1,..., λi = µi, λi+1 > µi+1 and µ1 = λ1,..., λj = µj, µj+1 > λj+1. Clearly we must have i = j = q. Finally, let λ, µ and σ be partitions such that λ ≥ µ and µ ≥ σ. As before, we say λi > µi and µj > σj. Then, λ1 = µ1,..., λi−1 = µi−1, λi > µi, and µ1 = σ1,..., µj−1 = σj−1, µj > σj. It is enough to take k = min{i, j}. And second, we will show that ≥ is a total order. Given λ, µ two partition we say i = max{i | λ1 = µ1,..., λi = µi}. We will see that λ ≤ µ or µ ≤ λ. The case i = q is clearly, so we suppose i < q. Since we have λi+1 ≥ µi+1 or µi+1 ≥ λi+1, the result follows.

Deﬁnition 2.3.4. Let E, F be two free modules with basis{x1,..., xm}and{xm+1,..., xm+n} respectively and let T ∈ Tabλ/µ({x1,..., xm+n}). For each i ∈ {1, . . . , q} let ηi to be µi plus the number of basis elements of F in the ith row of T. We deﬁne the sequence ∞ η(T) = (η1,..., ηq) ∈ N .

Observe that when T ∈ Tabλ/µ(S) is standard, then η(T) is a partition and µ ⊆ η(T) ⊆ i i+1 λ. Indeed, let T , T be two rows of T. We say ηi(T) = µi + ki where ki is the number of basis elements of F in the ith row. If T(i, l) is a basis element of E, then T(i + 1, l) must be a basis element of E too since T is column-standard and thus T(i, l) ≤ T(i + 1, l) (remember the ordered basis {x1,..., xn, xn+1,..., xn+m}). Necessarily, ki ≥ ki+1 and then µi + ki ≥ µi+1 + ki+1.

Lemma 2.3.5. Let S, T ∈ Tabλ/µ{x1,..., xn+m} with S ≤ T. Then η(S) ≥ η(T). 22 Schur Functors and CoSchur Functors.

Proof. We assume η(S) , η(T), otherwise there is nothing to prove. Let k the first integer such that ηk(S) , ηk(T), we will see ηk(S) > ηk(T). Remember that S ≤ T if Sp,q ≥ Tp,q for all p, q, where Sp,q and Tp,q are the number of times the first q elements of the basis appear in the first p rows of S and T respectively. From the definition of η(S) and η(T) it is clear p p that Sp,m = ∑j=1 ηj(S) − µj and Tp,m = ∑j=1 ηj(T) − µj, and hence ηi(S) = ηi(T), ∀i < k implies Sp,m = Tp,m, ∀p < k. k−1 k−1 Writing Sk,m = ∑j=1 ηj(S) − µj + ηk(S) − µk and Tk,m = ∑j=1 ηj(T) − µj + ηk(T) − µk, since we have assumed that ηi(S) = ηi(T), i = 1 . . . , k − 1, then Sk,m > Tk,m ⇒ ηk(S) > ηk(T).

Proposition 2.3.6. Let T ∈ Tabλ/µ(F ⊕ G). Then there exist unique standard tableaux Ti, ( ) = ( ) ( ) ≥ ( ) and unique integers ci , 0, such that dλ/µ XT ∑ cidλ/µ XTi and η Ti η t . Proof. Directly from Lemma 2.3.5 and the standard basis theorem Theorem 2.1.21.

Deﬁnition 2.3.7. Let µ, γ and λ be partitions such that µ ⊆ γ and γ ⊆ λ. We deﬁne submodules M Mγ(Λλ/µ(F ⊕ G)) := Im(ϕ : Λσ/µ(F) ⊗ Λλ/σ(G) → Λλ/µ(F ⊕ G)), µ⊆σ⊆λ,σ≥γ

0 M M˙ γ(Λλ/µ(F ⊕ G)) := Im(ϕ : Λσ/µ(F) ⊗ Λλ/σ(G) → Λλ/µ(F ⊕ G)). µ⊆σ⊆λ,σ>γ 0 ⊗ → where ϕ and ϕ are the sum of maps obtained by tensoring the maps Λσi−µi F Λλi−σi G ( ⊕ ) ∧ · · · ∧ ⊗ ∧ · · · ∧ → ∧ · · · ∧ ∧ ∧ · · · ∧ Λλi−µi F G deﬁned by x1 xσi−µi y1 yλi−σi x1 xσi−µi y1 yλi−σi . We deﬁne submodules:

Mγ(Lλ/µ(F ⊕ G)) := dλ/µ(Mγ(Λλ/µ(F ⊕ G))),

M˙ γ(Lλ/µ(F ⊕ G)) := dλ/µ(M˙ γ(Λλ/µ(F ⊕ G))).

The modules Mγ(Lλ/µ(F ⊕ G)) are the submodules of the filtration and the quotients 0 Mγ(Lλ/µ(F ⊕ G))/M˙ γ(Lλ/µ(F ⊕ G)) define the graded object. Clearly {XT0 ⊗ XT00 | T ∈ 00 Tabσ/µ{x1,..., xn}, T ∈ Tabλ/σ{xn+1,..., xn+m} are row- standard, σ ≥ γ} and {XT0 ⊗ 0 00 XT00 | T ∈ Tabσ/µ{x1,..., xn}, T ∈ Tabλ/σ{xn+1,..., xn+m} are row-standard, σ > γ} L L represent basis of the modules µ⊆σ≥λ,σ≥γ Λσ/µ(F) ⊗ Λλ/σ(G) and µ⊆σ≥λ,σ>γ Λσ/µ(F) ⊗Λλ/σ(G) respectively. 00 00 00 Note that, ϕ(XT0 ⊗ XT00 ) = XT00 where T ∈ Tabλ/µ(S ) such that T (i, j) = T(i, j) i f 00 0 j ≤ σi and T (i, j + σi) = T (i, j) i f j ≤ λi which clearly is a row-standard tableau with η(T00) = σ and then η(T00) ≥ γ or η(T00) > γ depending on the case. Now, if T ∈ 00 Tabλ/µ({x1,..., xn+m}) is a standard tableau such that η(T) ≥ γ and hence µ ⊆ η(T ) ⊆ 0 00 λ, T (i, j) = T(i, j) i f j ≤ η(T)i defines a standard tableau of shape σ/µ and T (i, j) = 00 T(i, j + η(T )i) i f j ≤ λi defines a standard tableau of shape λ/σ. Obviously, XT00 is the image of XT ⊗ XT0 by ϕ. It follows that {dλ/µ(XT) | T ∈ Tabλ/µ is standard, η(T) ≥ γ} form an R-basis of Mγ(Lλ/µ(F ⊕ G)) and {dλ/µ(XT) | T ∈ Tabλ/µ is standard, η(T) > γ} form an R-basis of Mγ(Lλ/µ(F ⊕ G)). 2.3 Decomposition of Schur Functors. 23

ψ Proposition 2.3.8. The map Λγ/µ(F) ⊗ Λλ/γ(G) −→ Mγ(Λλ/µ(F ⊕ G)) induces a map

ψ˜γ : Lγ/µ(F) ⊗ Lλ/γ(G) → Mγ(Lλ/µ(F ⊕ G))/M˙ γ(Lλ/µ(F ⊕ G)).

Proof. The map ψ˜γ is deﬁned by sending dγ/µ(x) ⊗ dλ/γ(y) to the class of dλ/µ(ψ(x ⊗ y)). We only need to verify that ψ˜γ sends 0 to 0,¯ since ker(dλ/µ) = Im(wλ/µ), it is sufﬁcient to prove that ψ(Im(wγ/µ) ⊗ Λλ/γG) and ψ(Λγ/µ(F) ⊗ Im(wλ/γ)) are both contained in N := Im(wλ/µ) + M˙ γ(Lλ/µ(F ⊕ G)). q−1 Remember wγ/µ = ∑i=1 Id1 ⊗ ... ⊗ Idi−1 ⊗ wγi/µi ⊗ Idi+2 ⊗ ... ⊗ Idq where wγi/µi = γ −µ −1 i+1 i γi−µi+γi+1−µi+1−t t γi−µi γi+1−µi+1 ∑t=0 Λ F ⊗ Λ F → Λ F ⊗ Λ F. By the change l = γi − i+1− γ µi+1 γ −µ +l γ + −µ + −l µi+ − t, we can write the last map as w i i = ∑ Λ i i F ⊗ Λ i 1 i 1 F → 1 γ /µ l=µi−µi+1+1 q−1 γ −µ Λγi−µi F ⊗ Λγi+1−µi+1 F. And thus, w = ∑ ∑ i+1 i+1 Λγ1−µ1 F ⊗ · · · ⊗ Λγi−µi+l F ⊗ γ/µ i=1 l=µi−µi+1+1 γi+ −µi+ −l γq−µq γ1−µ1 Λ 1 1 F ⊗ · · · ⊗ Λ F → Λγ/µF. Fixing i and l, if we see that ψ(Im(Λ F ⊗ γi−µi+l γi+ −µi+ −l γq−µq · · · ⊗ Λ F ⊗ Λ 1 1 F ⊗ · · · ⊗ Λ F ⊗ Λλ/γG) ⊆ N, clearly we will have show that ψ(Im(wγ/µ) ⊗ Λλ/γG) ⊆ N. = ⊗ · · · ⊗ ⊗ ⊗ ⊗ ⊗ · · · ⊗ = ⊗ · · · ⊗ Let X xI1 xIi−1 xIi xIi+1 xIi+2 XIq and Y yJ1 yJq be γ1−µ1 γi−µi+l γi+ −µi+ −l γq−µq basis elements of Λ F ⊗ · · · ⊗ Λ F ⊗ Λ 1 1 F ⊗ · · · ⊗ Λ F and Λλ/γG respectively. Using the coassociativity and commutativity of the comultiplication map, we ( ⊗ ) = ± ⊗ ∧ 0 can write wl xIi xIi+1 ∑U xU xIi+1 xU , where U runs over all subsets of order 0 γi − µi of Ii and U is the complement of U in Ii. Keeping this in mind, ( ⊗ ) = ± ∧ ⊗ · · · ⊗ ∧ ⊗ ∧ 0 ∧ ⊗ · · · ⊗ ∧ = ψ X Y ∑U xI1 yI1 xU yJi xIi+1 xU yJi+1 xIq yIq :

∑U ±ZTU where each TU ∈ Tabλ/µ{x1,..., xn+m} and clearly η(TU) = γ. Applying the same decomposition to ωλ/µ and ﬁxing the same i and l, if we consider = ∧ ⊗ · · · ⊗ ∧ λ1−µ1 ( ⊕ ) ⊗ · · · ⊗ λi−µi+l( ⊕ the basis element W : xI1 yI1 xIq yIq in Λ F G Λ F − G) ⊗ Λλi+1−µi+1−l(F ⊕ G) ⊗ · · · ⊗ Λλq µq (F ⊕ G) we obtain that the corresponding factor of w send that basis element to ∑ ±xI ∧ yI ⊗ · · · ⊗ xw ∧ yw ⊗ x 0 ∧ xI ∧ y 0 ∧ λ/µ w1,w2 1 1 1 2 w1 i+1 w2 ⊗ · · · ⊗ ∧ − yJi+1 xIq yIq where w1, w2 are subsets of Ii and Ji whose orders add up to λi µi 0 0 and w1, w2 are the complements of w1, w2 in Ii, Ii+1 respectively. Since Ji is of order λi − γi, then w1 must be of order ≥ λi − µi − λi + γi = γi − µi, then we can write the image above as:

±XT + ±xI ∧ yI ⊗ · · · ⊗ xw ∧ yw ⊗ x 0 ∧ xI ∧ y 0 ∧ yJ ⊗ · · · ⊗ xI ∧ yI ∑ U ∑ 1 1 1 2 w1 i+1 w2 i+1 q q U w1,w2 where now w1 + w2 > λi − µi. Each summand of the second addend corresponds to an element W where Tw ,w is a tableau satisfying η(Tw ,w ) > γ, since w > γ − µ . Tw1,w2 1 2 1 2 1 i i Then, the second summand is contained in M˙ (Λλ/µ(F ⊕ G)) and ψ(X ⊗ Y) = ωλ/µ(W) − ∑ W which clearly is an element of N. w1,w2 Tw1,w2 The proof that ψ(Λγ/µF ⊗ Im(wλ/γ)) ⊆ N proceeds formally in the same way that ψ(Im(wγ/µ) ⊗ Λλ/γG) ⊆ N. However we need to deﬁne appropriate maps instead of λi+1−γi+1 λ −γ +l ω and ω . Observe that considering the maps ω i i = ∑ Λ i i F ⊗ λ/γ λ/µ λ /γ l=γi−γi+1+1 − λ + −γ + −l λ −γ λ + −γ + λi+1 µi+1 λ −µ +l Λ i 1 i 1 F → Λ i i F ⊗ Λ i 1 i 1 F and w i i = ∑ Λ i i (F ⊕ G) ⊗ λ /µ l=µi−µi+1+1 24 Schur Functors and CoSchur Functors.

λ + −µ + −l λ −µ λ + −µ + Λ i 1 i 1 (F ⊕ G) → Λ i i (F ⊕ G) ⊗ Λ i 1 i 1 (F ⊕ G), there can be some l > γi − γi+1 which not occurs at wλi/µi . Let λi − λi+1 + 1 ≤ t ≤ λi − γi and u := λi − γi − t =: l ≤ λi+1 − γi − 1. We define the λi−γi−t λi+1−γi+1+t λi−γi λi+1−γi+1 ω˜ λ/γ = ∑t w˜ u,t where w˜ u,t : Λ G ⊗ Λ G → Λ G ⊗ Λ G whose image is contained in the image of the map wλi/γi , for all t. In the same way we define ω˜ λ/µ considering µi − µi+1 + 1 ≤ t ≤ λi − µi, u = λi − µi − t = t ≤ λi+1 − µi − 1. The remainder of the proof is exactly the same as in the first part but replacing ωλ/γ and ωλ/µ by ω˜ λ/γ and ω˜ λ/µ respectively.

Theorem 2.3.9. The map ψ˜γ is an isomorphism. Proof. ψ˜γ is clearly surjective since {dλ/µ(XT) | T is standard, η(T) = γ} represent an R-basis of Mγ(Lλ/µ(F ⊕ G))/M˙ γ(Lλ/µ(F ⊕ G)). Indeed, if T is a standard basis of 0 Tabλ/µ{x1,..., xn+m} with η(T) = γ, then T (i, j) = T(i, j) i f j ∈ {µi + 1, . . . , γi} and 00 T (i, j) = T(i, j) i f j ∈ {γi + 1, . . . , λi} deﬁnes standard tableaux of Tabγ/µ{x1,..., xn} and Tabλ/γ{xn+1,..., xn+m} respectively. Then, ψ˜γ(dγ/µ(XT0 ) ⊗ dλ/γ(XT00 ))= dλ/µ(ψ(XT0 ⊗XT00 )) = dλ/µ(XT). Moreover, ψ˜ carries a basis element of Lγ/µF ⊗ Lλ/µG to a basis element of Mγ(Lλ/µ(F 0 00 ⊕G))/M˙ γ(Lλ/µ(F ⊕ G)). If T ∈ Tabγ/µ{x1,..., xn} and T ∈ Tabλ/γ{xn+1,..., xn+m} are standard tableaux, ψ(XT0 ⊗ XT00 ) = XT where T a standard tableau with η(T) = γ, as we have seen before.

Corollary 2.3.10. The submodules {Mγ(Lλ/µ(F ⊕ G)) | µ ⊆ γ ⊆ λ} give a filtration of L Lλ/µ(F ⊕ G), whose associated graded module is isomorphic to µ⊆γ⊆λ Lγ/µF ⊗ Lλ/γG. Proof. Clearly, {Mγ(Lλ/µ(F ⊕ G)) | µ ⊆ γ ⊆ λ} is a filtration of Lλ/µ(F ⊕ G) with the total order on partitions which we defined at the beginning of this section . If Mγ(Lλ/µ(F ⊕ G)) ⊂ Mγ0 (Lλ/µ(F ⊕ G)) is a piece of this filtration, then γ = max{σ ∈ 0 N | γ < γ }, and hence Mγ(Lλ/µ(F ⊕ G)) equals to M˙ γ(Lλ/µ(F ⊕ G)).

Let us see a couple of examples.

Examples 2.3.11. (1) We ﬁrst start computing the decomposition of L(4,2)/(2,1)(F ⊕ G). From the above discussion we only need to determinate all the partitions γ such that µ ⊆ γ ⊆ λ. These partitions are (2, 1), (2, 2), (3, 1), (3, 2), (4, 1) and (4, 2). Then, the graded object of the ﬁltration is L(4,2)/(2,1)G ⊕ L(2,2)/(2,1)F ⊗ L(4,2)/(2,2)G, L(3,1)/(2,1)F ⊗L(4,2)/(3,1)G, L(3,2)/(2,1)F ⊗ L(4,2)/(3,2)G, L(4,1)/(2,1)F ⊗ L(4,2)/(4,1)G and L(4,2)/(2,1)F.

3 2 (2) The decomposition of L(1,1,1)(F ⊕ G) corresponds to S G, F ⊗ L(1,1,1)/(1,0,0)G, S (F) ⊗ 3 L(1,1,1)/(1,1,0)G and S F.

In fact, when R contains a ﬁeld of characteristic zero we have an isomorphism

M Lλ/µ(F ⊕ G) Lγ/µF ⊗ Lλ/γF µ⊆γ⊆λ

(See [2], Proposition (2.3.1).) 2.4 Cauchy Decomposition Formulas for Schur Functors. 25

2.4 Cauchy Decomposition Formulas for Schur Functors.

In this section we explain the relation between Schur and CoSchur functors and the symmetric and exterior algebra. More precisely, we will construct ﬁltrations of Sk(F ⊗ G) k L L and Λ (F ⊗ G) with associated graded objects |λ|=k LλF ⊗ LλG and |λ|=k LλF ⊗ KλG respectively. Moreover, if R is a ring of characteristic zero, these ﬁltrations are direct sum decompositions of Sk(F ⊗ G) and Λk(F ⊗ G). The Cauchy formula for the exterior algebra will be essential in Chapter 4 to understand the construction of general resolutions of determinantal varieties.

2.4.1 The decomposition of the Symmetric Algebra.

Deﬁnition 2.4.1. Let p be a positive integer and let λ = (λ1,..., λt) be a partition of p p p weight k. We deﬁne a natural paring h, ip : Λ F ⊗ Λ G → S (F ⊗ G) by sending f1 ∧ p(p−1)/2 sg(σ) · · · ∧ fp ⊗ g1 ∧ · · · ∧ gp to the p × p determinant (−1) ∑(−1) ( fσ(1) ⊗ g1) ····· ( fσ(p) ⊗ gp) =: h f1 ∧ · · · ∧ fp, g1 ∧ · · · ∧ gpi. Formally,

f1 ⊗ g1 ··· f1 ⊗ gp

. . h f1 ∧ · · · ∧ fp, g1 ∧ · · · ∧ gpip = . . .

fp ⊗ g1 ··· fp ⊗ gp

k We extend the above to a pairing h, i : ΛλF ⊗ ΛλG → S (F ⊗ G) by h f1 ∧ · · · ∧ ft, g1 ∧ · · · ∧ i = h i · · · · · h i ∈ λi ∈ λi ∈ { } gt f1, g1 λ1 ft, gt , where fi Λ F and gi Λ G for all i 1, . . . , t . 2 2 2 For example, h, i2 : Λ F ⊗ Λ G → S (F ⊗ G) is given by h f1 ∧ f2, g1 ∧ g2i2 = ( f1 ⊗ g1)( f2 ⊗ g2) − ( f1 ⊗ g2)( f2 ⊗ g1), f1, f2 ∈ F, g1, g2 ∈ G.

Deﬁnition 2.4.2. Let λ = (λ1,..., λt) be a partition of weight k. We deﬁne submodules of Sk(F ⊗ G), k Mλ(S (F ⊗ G)) := ∑ hΛγF, ΛγGi, γ≥λ,|γ|=k ˙ Mλ(Sk(F ⊗ G)) := ∑ hΛγF, ΛγGi. γ>λ,|γ|=k k where hΛγF, ΛγFi denotes the image of the pairing h, i : ΛγF ⊗ ΛγG → S (F ⊗ G). k k Note that hF ⊗ · · · ⊗ F, G ⊗ · · · ⊗ Gi = M(1,...,1)(S (F ⊗ G)) = S (F ⊗ G), 1 + ··· + 1 = k. We have,

k Definition 2.4.3. {Mλ(S (F ⊗ G) | |λ| = k} define a natural filtration

k k k 0 ⊆ M(k)(S (F ⊗ G)) ⊆ M(k−1,1)(S (F ⊗ G)) ⊆ · · · ⊆ M(1,...,1)(S (F ⊗ G)) induced by the lexicographic order ≥ on partitions of weight k.

Directly from the deﬁnition of both submodules, {hXS, YTi | S ∈ Tabγ{x1,..., xn}, T ∈ k Tabγ{xn+1,..., xn+m}, γ ≥ λ, |γ| = k} form a system of generators of Mλ(S (F ⊗ G)), and {hXS, YTi | S ∈ Tabγ{x1,..., xn}, T ∈ Tabγ{xn+1,..., xn+m}, γ > λ, |γ| = k} spans k M˙ λ(S (F ⊗ G)). 26 Schur Functors and CoSchur Functors.

k k For convenience we denote Mλ(S (F ⊗ G)) by Mλ and M˙ λ(S (F ⊗ G)) by M˙ λ. Our goal is to show that Mλ/M˙ λ is isomorphic to LλF ⊗ LλG. So ﬁrst we need to deﬁne a morphism βγ from LλF ⊗ LλG to Mλ/M˙ λ. In [1] Corollary [III.1.2], [I.5] and Proposition [III.1.1], using that the paring h, i is in fact a restriction of a natural map between extended R-Hopf algebras, one sees that hωλ(ΛλF), ΛλFi + h(ΛλF), ωλ(ΛλF)i ⊆ M˙ λ. We have,

Proposition 2.4.4. The natural map h, i : ΛλF ⊗ ΛλF → Mλ induces a surjective map βλ : LλF ⊗ LλG → Mλ/M˙ λ given by βλ(dλ(X) ⊗ dλ(Y)) = hX, Yi, where hX, Yi denotes the class of hX, Yi in the quotient Mλ/M˙ λ. Proof. Since LλF ΛλF/Im(ωλ), see Deﬁnition 2.1.16 and Theorem 2.1.24, it follows from hωλ(ΛλF), ΛλFi + h(ΛλF), ωλ(ΛλF)i ⊆ M˙ λ that βλ is well deﬁned. Since Mλ/M˙ λ is generated by {hXS, YTi | S ∈ Tabλ{x1,..., xn}, T ∈ Tabλ{xn+1,..., xn+m}}, it is enough to see that for each hXS, YTi of that such there exists X ∈ LλF and Y ∈ LλG such that βγ(dλ(X) ⊗ dλ(Y)) = hXS, YTi. Obviously, XS and YT are the elements we are looking for.

Remark 2.4.5. As a consequence of the above proposition we obtain {hXS, YTi | S ∈ Tabλ{x1,..., xn}, T ∈ Tabλ{xn+1,..., xn+m} are standard} generates Mλ/M˙ λ.

Corollary 2.4.6. {hXS, YTi | S ∈ Tabγ{x1,..., xn}, T ∈ Tabγ{xn+1,..., xn+m} are standard, γ ≥ λ, |γ| = k} := Bλ generate Mλ. Proof. Since {Mλ, |λ| = k} is ordered lexicographically, we proceed by induction on λ. k k We can see easily that the ﬁrst piece of the decomposition is Λ F ⊗ Λ G = L(k)F ⊗ L(k)G. k k k k k Indeed, M(k) is isomorphic to Λ F ⊗ Λ G, since the pairing h, i : Λ F ⊗ Λ G → S (F ⊗ G) is the natural embedding of Λk F ⊗ ΛkG in Sk(F ⊗ G). So, the initial case is clear. Con- sider λ > (k), we want to prove that Bλ generate Mλ. Observe that the corresponding piece of the ﬁltration is M˙ λ ⊆ Mλ. From Proposition 2.4.4, each element of Mλ can be written as a sum of an element of M˙ λ and an element generated by {hXS, YTi | S ∈ Tabλ{x1,..., xn}, T ∈ Tabλ{xn+1,..., xn+m} are standard}. The result follows by induction on M˙ λ.

Theorem 2.4.7. The maps βλ : LλF ⊗ LλG → Mλ/M˙ λ are isomorphisms and therefore the L associated graded object of the ﬁltration {Mλ(Sk(F ⊗ G) | |λ| = k} is |λ|=k LλF ⊗ LλG. Proof. Let Bk := {hXS, YTi | S ∈ Tabλ{x1,..., xn} is standard, T ∈ Tabλ{xn+1,..., xn+m} k is standard, |λ| = k}. By Corollary 2.4.6, Bk generates S (F ⊗ G). Then directly from the equality ranks k rank(S (F ⊗ G)) = ∑ rank(LλF ⊗ LλG) |λ|=k

k (see [1] Theorem [III.1.4]), Bk must be an R-basis of S (F ⊗ G). This, in turn, implies that Bλ is an R-basis of Mλ and {hXS, YTi | S ∈ Tabγ{x1,..., xn}, T ∈ Tabγ{xn+1,..., xn+m} are standard, γ > λ, |γ| = k} := B˙ λ is an R-basis of M˙ λ. Consequently, {hXS, YTi | S ∈ Tabλ{x1,..., xn}, T ∈ Tabλ{xn+1,..., xn+m} are standard} is an R-basis of Mλ/M˙ λ. It follows that βλ send an element basis of LλF ⊗ LλG to an element basis of Mλ/M˙ λ.

Examples 2.4.8. (1) The decomposition of S2(F ⊗ G) corresponds to Λ2F ⊗ Λ2F and S2F ⊗ S2G. 2.4 Cauchy Decomposition Formulas for Schur Functors. 27

(2) The three partitions of weight 3 are (3), (2, 1) and (1, 1, 1), then the respective terms 3 3 3 3 3 of the decomposition of S (F ⊗ G) are Λ F ⊗ Λ F, L(2,1)F ⊗ L(2,1)G and S F ⊗ S G.

The Cauchy formula for Sk(F ⊗ G) has an important consequence. When R is a ring of characteristic zero the decomposition becomes an isomorphism, more precisely (see [20] Corollary (2.3.3))

k L Theorem 2.4.9. If R is a ring of characteristic zero, then S (F ⊗ G) |λ|=k LλF ⊗ LλG.

2.4.2 The decomposition of the exterior algebra.

Definition 2.4.10. Let p be a positive integer and let λ = (λ1,..., λt) be a partition of p p p weight k. We define a natural pairing h, ip : Λ F ⊗ D G → Λ (F ⊗ G) by induction on p ≥ (α1) (αt) 1. For p = 1 we define h f , gip = f ⊗ g. For p > 1 we define h f1 ∧ · · · ∧ fp, g1 ····· gt ip by p h i ∧ h ∧ · · · ∧ (α1) ····· (αi−1) ····· (αt)i ∑ f1, gi f1 fp, g1 gi gt i=1 t k where ∑i=1 = p and αi ≥ 1 for all i. We define a pairing h, i : ΛλF ⊗ DλG → Λ (F ⊗ G) h ⊗ · · · ⊗ ⊗ · · · ⊗ i = h i ∧ · · · ∧ h i by the natural extension f1 ft, g1 gt f1, g1 λ1 ft, gt λt , where λ λ fi ∈ Λ i F and gi ∈ D i G for all i ∈ {1, . . . , t}.

Deﬁnition 2.4.11. Let λ = (λ1,..., λt) be a partition of weight k. We deﬁne submodules of Λk(F ⊗ G) k Mλ(Λ (F ⊗ G)) := ∑ hΛγF, DγGi, γ≥λ,|γ|=k ˙ k Mλ(Λ (F ⊗ G)) := ∑ hΛγF, DγGi. γ>λ,|γ|=k

k Clearly M(1,...,1) = Λ (F ⊗ G), 1 + ··· + 1 = k. We have,

k Definition 2.4.12. {Mλ(Λ (F ⊗ G) | |λ| = k} define a natural filtration

k k k 0 ⊆ M(k)(Λ (F ⊗ G)) ⊆ M(k−1,1)(Λ (F ⊗ G)) ⊆ · · · ⊆ M(1,...,1)(Λ (F ⊗ G)) induced by the lexicographic order ≥ on partitions of weight k.

From the above deﬁnitions, {hXS, YTi| S ∈ Tabγ{x1,..., xn}, T ∈ Tabγ{xn+1,..., xn+m}, γ ≥ λ, |γ| = k} and {hXS, YTi | S ∈ Tabγ{x1,..., xn}, T ∈ Tabγ{xn+1,..., xn+m}, γ > k k λ, |γ| = k} spans Mλ(Λ (F ⊗ G)) and M˙ λ(Λ (F ⊗ G)), respectively. Arguing as in Sub- section 2.4.1 we get

Proposition 2.4.13. The natural map h, i : ΛλF ⊗ ΛλG → Mλ induces a surjective R-map βλ : LλF ⊗ KλG → Mλ/M˙ λ.

Lemma 2.4.14. {hXS, YTi | S ∈ Tabγ{x1,..., xn} is standard, T ∈ Tabγ{xn+1,..., xn+m} is standard, γ ≥ λ, |γ| = k} spans Mλ. 28 Schur Functors and CoSchur Functors.

Theorem 2.4.15. The maps βλ : LλF ⊗ KλG → Mλ/M˙ λ are isomorphisms and therefore k L the associated graded object of the ﬁltration {Mλ(Λ (F ⊗ K), |λ| = k} is |λ|=k LλF ⊗ KλG. Proof. It follows as Theorem 2.4.7 using the equality rank

k rank(Λ (F ⊗ G)) = ∑ rank(LλF ⊗ KλG) |λ|=k which we can ﬁnd in [1] Theorem [III.2.2].

k L Theorem 2.4.16. If R is a ring of characteristic zero, then Λ (F ⊗ G) |λ|=k LλF ⊗ KλF. k L Moreover, since KλG Lλ˜ G, Λ (F ⊗ G) |λ|=k LλF ⊗ Lλ˜ G.

2.5 The Littlewood-Richardson rule for Schur functors.

The tensor product of Schur functors decomposes into a direct sum of Schur functors, provided that R is a ring of characteristic zero. We ﬁnish this exposition about Schur and CoSchur functors presenting a combinatoric algorithm, the Littlewood-Richardson rule, which computes the multiplicities of the factors in the decomposition of LλF ⊗ LµG. As a corollary of the Littlewood-Richardson rule we will obtain the Pieri formulas for Schur functors. An accurate exposition of the following result, based on representation theory, is found in [20] Sections [2.2] and [2.3].

Theorem 2.5.1. Let R be a ring of characteristic zero, let F be a free R-module of rank n and let λ and µ be two partitions. Then,

LλF ⊗ LµF ∑ u(λ, µ; ν)LνF |ν|=|λ|+|µ| where u(λ, µ; ν) denotes the multiplicity of the factor LνF.

The Littlewood-Richardson rule provides an algorithm which computes the multiplicities u(λ, µ; ν). The statement and proof of Littlewood-Richardson rule require two combinatorial notions we present immediately.

2.5.1 The Schensted Process and Words of Yamanouchi.

The following procedure is known as the Schensted process.

Definition 2.5.2. Let λ be a partition, S a totally ordered set, U ∈ Tabλ(S) a standard tableau and p ∈ S. We define p → U to be the tableau obtained in the following recursive manner. (1). If p1 = min{U(1, j) | U(1, j) ≥ p} exists, let p → U be the tableau of shape λ obtained by replacing of p1 to p. We say this step bumping of p into U. (2). If p1 does not exist we define p → U to be the tableau of shape λ1 = (λ1 + 1, . . . , λq) obtained from U by adjoining a new box to the first row of U with entry p and finish. 2.5 The Littlewood-Richardson rule for Schur functors. 29

If (1) takes place repeat the process with U1 and p = p1, where U1 is the tableau obtained from U by removing the ﬁrst row, and continue. Finally, p → U is the tableau obtained from U by adjoining a new box to a non empty row or by adjoining a new bottom 1 row. In the ﬁrst case p → U is a tableau of shape λ = (λ1,..., λi + 1, . . . , λn) for some i, 1 and in the second case, λ = (λ1,..., λn, 1).

1 1 Remark 2.5.3. Note that λ is a partition. When λ = (λ1,..., λn, 1) the result is clear. 1 Assume that λ = (λ1,..., λi + 1, . . . , λn), then in the ith step of the above procedure p does not bump any element in the ith row of U. The standardness of U implies U(i − 1, j) ≤ U(i, j), for every (i, j) of its diagram. If p = U(i − 1, j) and j ≤ λi, then pi−1 ≤ U(i, j) ⇒ ∃min{U(i, j) | U(i, j) ≥ pi−1}, which is a contradiction. Necessarily, j > λi ⇒ λi−1 > λi and hence, λi−1 ≥ λi + 1.

Let us to compute p → U for p = 2 and U = 1 3 4 . 2 3 4 1 (1). p1 = U(1, 2) = 3, p → U = 1 2 4 , λ = λ. 2 3 4

Let p = 3 and U1 = 2 3 . 4 (2). p1 = U(2, 2) = 3, p → U.

Let p = 3 and U2 = 4 . 1 (3). p1 = U(3, 1) = 4, p → U = 1 2 4 , λ = λ. 2 3 3 Finish, U → p = 1 2 4 and λ1 = (3, 2, 1, 1). 2 3 3 4

Lemma 2.5.4. The tableau p → U in Tabλ1 (S) is standard. Proof. Row-standardness is clear, we only have to prove column-standardness. It is enough to consider two adjacent rows, i and i + 1, of U in which the step bumping of p into U took place, otherwise there is nothing to prove since λ1 is a partition. For convenience we denote by ui, i = 1, . . . , λi the entries in the ith row and by vi, i = 1, . . . , λi+1 the entries in the (i + 1)th row. (i) Suppose p does bump ut in the ith row and ut does bump vs in the (i + 1)th row. Since ut−1 ≤ p ≤ ut and vs−1 ≤ ut ≤ vs, we must have s ≤ t and we only have to check that p ≤ vt. Clearly s ≤ t and the standardness of U implies ut ≤ vs ≤ vt which, in turn, implies p ≤ vt. (ii) Suppose p does bump ut, but ut does not bump any element in the i + 1th row of U. In this case we adjoin the box (i + 1, λi + 1) with the entry ut. Then p ≤ ut and ut > vj for all j = 1, . . . , λi+1 and we must have t > λi+1 since t is standard, and hence the result is clear. 30 Schur Functors and CoSchur Functors.

1 Definition 2.5.5. Let U be an standard tableau in Tabλ1 (S). Let (i, j) be an extremal box 1 1 1 1 1 1 1 of λ , that is j = λi and λi > λi+1, and let r = U (i, j). We define U (i, λi) ← U to be the tableau in the following recursive manner. 1 1 (1) Let r1 = max{U (i − 1, k) | U (i − 1, k) ≤ r} and replace r1 by r. (2) Repeat step (1) with r = r1 and the row i − 2, and continue. The result of this 1 1 1 1 1 algorithm is a tableau U (i, λi) ← U of shape λ = (λ1,..., λi − 1, . . . , λt ) obtained from U1 by removing the extremal box (i, j). We call the bumped out element of U1 to the element in the first row of U1 which was bumped out in the final step of the recursive process above.

1 Lemma 2.5.6. Let U be a standard tableau in Tabλ(S), p ∈ S and U = p → U. If r is the element adjoined through the bumping of p into U, then r ← U1 is U and the bumped out element of U1 is p. Proof. We proceed by induction on the number of rows of U. If U only has one row, then U1 is one of the following cases: 1 1 (a) If p does bump an element ut of U, then λ = (λ, 1), U (1, j) = (1, j) i f j , 1 1 t, U (1, t) = p and U (2, 1) = ut. Since (2, 1) is an extremal box, let r = ut. Now 1 1 r1 = max{U (1, k) | U (1, k) ≤ ut}, since U is standard we know ut−1 < ut < ut+1, so r1 only could be ut−1 or p. By definition, ut = min{U1, k | U(1, k) ≥ p}, so ut−1 < p ≤ ut, 1 1 it follows r1 = p and p is the bumped out element of U . Clearly r ← U is U. (b) If p does not bump any element of U, then λ1 = (λ + 1) and U1(1, j) = U(i, j) i f j ≤ λ and U(1, λ + 1) = p. Clearly (1, λ + 1) is the extremal box of U and p is the bumped out element. Suppose U has more than one row and the lemma is true for standard tableau of shape λ0 of length less than the length of λ. If p does not bump any element of the first row of U, the lemma is clear. Assume that p bump ut in the first row of U, let U˜ be the tableau 1 U with the top row removed and consider ut → U˜ . Clearly U is the tableau ut → U˜ with 1 the top row of U . Note that if r is an extremal box of ut → U˜ , then r is an extremal box 1 of U too. By induction, r ← (ut → U˜ ) = U˜ and ut is the bumped out element of U˜ . The last step of the reverse process to the Schensted process must bump out p and return us U.

Deﬁnition 2.5.7. Let a = (a1,..., an) be a sequence of positive integers and let S = {x1 ≤ · · · ≤ xm} an ordered set. We deﬁne the content of a to be the sequence λ = (λ1,...), where λi is the number of times i appears in a, for all i ∈ N. We say that a is a word of Yamanouchi or a Y-word if for each k = 1, . . . , n the number of times i appears in the sequence (a1,..., ak) is not smaller than the number of times i + 1 appears, for all i ∈ N. ( ) ( ) We say that xi1 ,..., xin is a Y-word if i1,..., in is a Y-word.

Remark 2.5.8. If a is a Y-word, then clearly the content of (a1,..., ak) is a partition for k = 1, . . . , n. Reciprocally, if the content of (a1,..., ak) is a partition for k = 1, . . . , n, then a is a Y-word.

For example, the sequence a = (1, 1, 2, 2, 1, 3, 2, 1, 2, 1, 2, 3) has content (5, 5, 2) and it is a word of Yamanouchi as the following tabular of subsequences and contents of a shows us. 2.5 The Littlewood-Richardson rule for Schur functors. 31

k = 1 (1)(1) k = 2 (1, 1)(2) k = 3 (1, 1, 2)(2, 1) k = 4 (1, 1, 2, 2)(2, 2) k = 5 (1, 1, 2, 2, 1)(3, 2) k = 6 (1, 1, 2, 2, 1, 3)(3, 2, 1) k = 7 (1, 1, 2, 2, 1, 3, 2)(3, 3, 1) k = 8 (1, 1, 2, 2, 1, 3, 2, 1 (4, 3, 1) k = 9 (1, 1, 2, 2, 1, 3, 2, 1, 2)(4, 4, 1) k = 10 (1, 1, 2, 2, 1, 3, 2, 1, 2, 1)(5, 4, 1) k = 11 (1, 1, 2, 2, 1, 3, 2, 1, 2, 1, 2)(5, 5, 1) k = 12 (1, 1, 2, 2, 1, 3, 2, 1, 2, 1, 2, 3)(5, 5, 2) Proposition 2.5.9. There exists a bijection between the set of words of Yamanouchi of content λ and the set of standard tableau of shape λ with distinct entries from the set {1, . . . , n}, where n is the weight of λ. Proof. Let T be a set of standard tableau of shape λ with distinct entries from the set {1, . . . , n} and Y the set of Y-words of content λ. Let T ∈ Tabλ({1, . . . , n}) be a standard tableau. By hypothesis, each i ∈ {1, . . . , n} appears as an entry T(k, j) only one time. Let ai = k, that is, ai is the number of the row of T where i is an entry, and consider the sequence a = (a1,..., an). First we compute the content of a. The number of times i appears in a is exactly the number of elements in the ith row of T, that is λk, then a is a sequence of content λ. And second, we prove that a is 0 an Y-word. Let k ∈ {1, . . . , n} and consider the subsequence a = (a1,..., ak). Fixed i let ji and ji+1 be the number of times i and i + 1 appear in the subsequence respectively, that is ji is the number of elements of ith row of T smaller than k. If we have ji+1 elements in the i + 1th row of T smaller than k, then since T is standard, (T(i, l) ≤ T(i + 1, l), l = 1, . . . , λi+1), we have at least ji+1 elements in the ith row of T smaller than k too, so it is clear that ji ≥ ji+1. We define α : T → Y to be the map which sends T to a as above. Now let a = (a1,..., an) be a Y-word of content λ. For each i ∈ {1, . . . , n} we consider the subsequence of a with entries equal to i and we denote it by (ak ,..., ak ). Then 1 λi T(i, j) = kj define a row-standard tableau of shape λ with distinct entries in {1, . . . , n}. We want to prove that T is standard. Let (ak ,..., ak ) be the subsequence of a with entries 1 λi equal to i and (aj ,..., aj ) the subsequence of a with entries equal to i + 1. First we 1 λi+1 see that k1 ≤ j1. We know that k1 is the first subindex such that al = i and j1 is the first subindex such that al = i + 1. If j1 < k1, then for k = j1 the number of times i appears ( ) + in the sequence a1,..., aj1 is smaller than the number of i 1, which is a contradiction since a is an Y-word. With a symmetric argue we see that ki ≤ ji, i = 1, . . . , λi+1. We define β : Y → T to be the map which sends a to T as above. Clearly α and β are inverses.

Deﬁnition 2.5.10. Let a = (a1,..., an) a Y-word of content λ. If T ∈ T , then T˜ (i, j) = T(j, i) is a standard tableau of shape λ˜ with distinct entries from {1, . . . , n}. We deﬁne a˜ to be the Y-word α(βg(a)) of content λ˜ , we call a˜ the transpose of a.

Remark 2.5.11. a˜i is the number of ak such that ak = ai and k ≤ i. 32 Schur Functors and CoSchur Functors.

Definition 2.5.12. Let λ, µ be two partitions. We define A to be the set of pairs (U1, U2) where U1 is a standard tableau of shape λ and U2 is a standard tableau of shape µ. We define B to be the set of triples (ν, U, V) where ν is a partition containing λ, U is a standard tableau of shape ν and if the diagram of ν/λ is {(i, j) | i = 1, . . . , q, λi + 1 ≤ j ≤ νi, then V is a standard tableau of shape ν/λ of content µ˜ such that the sequence (V(q, λq + 1), V(q − 1, λq + 1),..., V(q, νq), V(q − 1, νq),..., V(1, λ1) is a Y-word.

Remark 2.5.13. Observe that the last sequence is the sequence obtained by reading each column of V from the bottom up, starting with the left-most column and moving to the right column by column.

For example, if λ = (3, 2, 1), µ = (3, 3, 2, 2, 2) and ν = (6, 4, 3, 3, 2), then V = 1 2 3 1 2 1 2 1 2 3 1 2 is a standard tableau of shape ν/λ, content (5, 5, 2) = µ˜ and with associated sequence a = (1, 1, 2, 2, 1, 3, 2, 1, 2, 1, 2, 3), which is a Y-word.

The main goal of this subsection is to construct an injection from A to B which will play an important role in the next subsection. Defining a map φ : A → B will take us a while, first we will obtain from a pair (U1, U2) of A a triple (ν, U, V) where ν and U satisfying B-conditions, the difficult step will be to verify V is of that such.

Let (U1, U2) ∈ A and let p1 ≥ p2 ≥ · · · ≥ pµ˜1 the entries of the first column of U2 (1) = ( → (· · · → ( → ( → )) ··· )) from the bottom up. We define U1 pµ˜1 p2 p1 U1 to be the tableau obtained from U1 by bumping in the column of U2 into U1. We repeat the (1) procedure with U1 and the second column of U2. Continuing in this manner we obtain a standard tableau U of some shape ν containing λ and, since we have bumped all elements of U2, |ν| = |λ| + |µ|. Each box of the diagram of λ/µ, {(i, j) | i = 1, . . . , q, λi + 1 ≤ j ≤ νi} was a a result of having bumped some entry U2(l, k) into U1, with the assignment V(i, j) = k we define a tableau V of shape ν/λ. The number of times i appears as an entry of V is clearly the number of elements in the i column of U2, therefore the content of V is µ˜. We say that (ν, U, V) is the triple associated to (U1, U2). We want to show that V is a standard tableau and the sequence associated to V is a word of Yamanouchi.

0 0 Lemma 2.5.14. Let U be a standard tableau of shape λ and let p1 ≥ · · · ≥ pr be elements 00 00 of S. Let U be the standard tableau pr → (· · · → (p2 → (p1 → U )) ··· ) of some shape λ00. Then, the diagram of λ00/λ contains at most one box in each row. Moreover, the box adjoined by bumping in pi+1 is below the box adjoined by bumping in pi, for i = 1, . . . , r − 1. 00 00 Proof. From the construction of U it is enough to consider p2 → (p1 → U ). If p1 does 0 0 not bump any entry in the first row of U , then p1 → U has p1 in the first row. Since 0 0 p2 ≤ p1, p2 must bump some element p2 in the first row of p1 → U , and hence the box 2.5 The Littlewood-Richardson rule for Schur functors. 33

0 adjoined by bumping p2 into p1 → U is below the box adjoined by bumping p1 verifying the lemma. 0 0 Otherwise, if p1 does bump some entry p1 in the first row of U , then p2 ≤ p1 must 0 0 0 0 0 also bump some entry p2 with p2 ≤ p1 in the first row of p1 → U . Let p1 = p1 and 0 0 0 0 p2 = p2, we repeat the argument with U and p1 → U without the first rows of U and 0 p1 → U . Continuing in the same manner we must stop when p1 does not bump any entry in the ith row of U0.

0 Lemma 2.5.15. Let U be a standard tableau of shape λ, let ls > ks ≥ · · · ≥ k1 and 0 0 let U be the tableau ls → (k1 → (· · · → (ks → U ) ··· )) of some shape λ . Then, 0 U = k1 → (· · · → (ks−1 → (ls → (ks → U ))) ··· ) Proof. It is sufficient to prove it for s = 2. We proceed by induction on the number of 0 0 0 rows of U . Suppose U has only one row. If l2 does not bump any element U , the result is 0 0 0 clear. Otherwise, if l2 bumps an element l2 of U U , then k2 must have bump and element 0 0 0 0 0 0 0 0 k2 and k1 an element k1, with k1 ≤ k2 < l2. We consider k1 < k2 < l2, the other case is similar. We have, 0 k2 → U = ... k2 ... , 0 k2 0 k1 → (k2 → U ) = ... k1 ... k2 ... 0 k1 0 k2 0 and l2 → (k1 → (k2 → U )) = ... k1 ... k2 ... l2 ... 0 0 k1 l2 0 k2 0 l2 → (k2 → U ) = ... k2 ... l2 ... 0 0 k2 l2 0 and l2 → (k1 → (k2 → U )) = ... k1 ... k2 ... l2 ... 0 0 k1 l2 0 k2 0 Assume that U has more than one row. If l2 does not bump any element in the 0 0 first row of U , the result is clear. Suppose that l2 does bump an element l2 in the first 0 0 0 0 0 row of U , as before k2 and k1 do bump elements l2 > k2 ≥ k1 in the first row of U . 0 0 Let U1 be the tableau U without the first row, clearly U is the tableau obtained from 0 0 0 0 l2 → (k1 → (k2 → U1) by adjoining the first row of U. The result follows by induction on 0 U1. 0 Lemma 2.5.16. Let U be a standard tableau of shape λ, let kt ≥ · · · ≥ k1, ls ≥ · · · ≥ l1, t ≥ s be elements of S such that ki < li, i = 1, . . . s and let U be the tableau 0 0 l1 → (··· ls → (··· k1 → (· · · → (kt → U ) ··· ))) ··· ) of some shape λ . Then, the box of 0 0 U adjoined to U by bumping in lα is strictly to the right of the box of U adjoined to U by bumping in kα, for α = 1, . . . , s. Proof. We can assume s = t, since it is the same to prove that the box of U0 adjoined to 0 ks+1 → (· · · → (kt → U ) ··· ) by bumping in lα is strictly to the right of the box of U 0 adjoined to ks+1 → (· · · → (kt → U ) ··· ) by bumping in kα, for α = 1, . . . , s. We proceed by induction on s. 34 Schur Functors and CoSchur Functors.

When s = 1, we have k1 > l1. By lemma 2.5.15, the box (i, λi + 1) adjoined by bumping in k1 is below that the box (j, λj + 1) adjoined by bumping in l1: So j > i ⇒ λi + 1 < λj + 1, and hence (j, λj + 1) is strictly to the right of (i, λi + 1). Now, suppose that is true for 0 0 k = s − 1. By lemma 2.5.15, if we let U1 = ls → (ks → U ), then U = l1 → (· · · → (ls−1 → 0 (k1 → (· · · → (ks−1 → U1) ··· ))) ··· ). The result follows by induction.

Proposition 2.5.17. Let (U1, U2) ∈ A and let (ν, U, V) be its triple associated. Then (ν, U, V) ∈ B. Proof. We only have to prove that V is a standard tableau whose sequence associated is a Y-word. Remember V is a tableau of shape ν/λ such that V(i, j) = k, where (i, j) is the box adjoined by bumping in the element U2(l, k). We start showing the standardness of V. Let (i, j), (i, j + 1) and (i + 1, j) be adjacent boxes of V. By Lemma 2.5.16, the tableau obtained by bumping in a standard column 2 p1 ≥ p2 ≥ · · · ≥ pµi of U never has two boxes in the same row, moreover the box adjoined by bumping in pi+1 is below to the box adjoined by bumping in pi. Therefore, if (i, j) is the box adjoined by bumping in the element U2(l, k), then (i, j + 1) is the box adjoined by 0 0 0 bumping in an element U2(l , k ) such that k > k, and hence V(i, j + 1) > V(i, j). The row- standardness of V is clear. Let (i + 1, j) be the box adjoined by bumping in the element U2(l0, k0) and let U0 be the tableau of some shape ν0 obtained by bumping in the ﬁrst k0 0 0 0 columns of U2 into U1. Since νi ≥ νi+1, necessarily the box (i, j) = (i, νi ) was adjoined ﬁrst, thus k ≤ k0. So V(i, j) = k ≤ k0 = V(i + 1, j) and the column-standardness of V follows. It remains to prove that the sequence associate to V is a Y-word. For convenience µ µ we name this sequence by a = (p1,..., p1 ,..., p 1 ,..., p 1 ). For each k = 1, . . . , |λ/ν| 1 µ˜1 1 µ˜q we consider the subsequence ak = (p1,..., p1 ,..., pi ,..., pi ) of a, we want to see that 1 µ˜1 1 k the content of ak is a partition. Let 1 ≤ j < k, the number of time j + 1 appears in ak is the number of boxes adjoined by bumping in the elements of the (j + 1)th column of U2 into U1. If j + 1 appears lj+1 times, then these boxes correspond to the element U2(j + 1, 1),..., U2(j + 1, l). By Lemma 2.5.16 we know that the boxes adjoined by bumping in U2(j + 1, 1),..., U2(j + 1, l) are strictly to the right of the boxes associated to U2(j, 1),..., U2(j, l). Moreover, if the box associated to U2(j + 1, h) appears in the sub- k 2 sequence a , then the box associated to U (j, h) must appear too. Clearly lj ≥ lj+1, and hence the content of ak is a partition.

Deﬁnition 2.5.18. We deﬁne a map Φ : A → B by mapping a pair (U1, U2) of A to its triple associated (ν, U, V) ∈ B.

Proposition 2.5.19. The map Φ : A → B is an injection. Proof. Let A¯ be the set of pairs (U1, U2) such that U1 ∈ Tabλ(S) and U2 ∈ Tabµ(S), we want to deﬁne a map Ψ : B → A¯ such that (Ψ ◦ Φ)(U1, U2) = (U1, U2) for all pairs (U1, U2) in A. Clearly, this implies the proposition. Let (ν, U, V) ∈ B. Remember V is a standard tableau of shape ν/λ over {1, . . . , µ1} and content µ˜ = (µ˜1,..., µ˜t). If µ1 is an entry in the k-th row of V, it is clear that the box associated to it is (k, µk). Since V is a standard tableau, if k is the last row in which µ1 appears, or equivalent, the ﬁrst time µ1 appears in the Y-word associated to V, this entry must be an extremal box of V, and thus of U. Therefore we can consider the standard 2.5 The Littlewood-Richardson rule for Schur functors. 35

tableau U(k, νk) ← U of some shape ν1. Let p1 be the element bumped out of U, since the second appearance of µ1 is an extremal box of U(k, νk) ← U too, we can repeat the 0 0 argument above. After µ˜t iterations we obtain a standard tableau U of some shape ν , which is a partition containing λ, and p1,..., pµ˜t elements bumpered out of U, we put 0 0 them into a column γt. Let V be the standard tableau of shape ν /λ obtained from V 0 by removing all boxes of V with the entry µ1. Clearly, the sequence associated to V is the sequence of V but deleting the elements µ1, and thus is an Y-word too. Repeat the 0 0 0 above argument with the triple (ν , U , V ) and µ2, and continue. Finally, we will obtain a standard tableau U1 of shape λ from U by bumping out the elements of V, and γt,..., γ1 columns. We define a map Ψ : B → A¯ sending (ν, U, V) to the pair (U1, U2), where U2 be the tableau of shape µ and columns γ1,..., γt. Let (U1, U2) ∈ A, we want to prove that Ψ(Φ(U1, U2))) = (U1, U2). We proceed by induction on the number of column of U2. Suppose that U2 is one column tableau, we 0 0 denote it by p1 ≥ · · · ≥ pr. Consider (ν, U, V) := Φ(U1, U2) and (U1, U2) := Ψ(ν, U, V). Remember U = pr → (· · · → (p1 → U1) ··· ) is a standard tableau of shape ν and V is a standard tableau of shape ν/λ with all entries 1. If r = 1, then U = p1 → U1 and V is a single box tableau with entry 1. Let U(i, λi + 1) be the box adjoined by bumping in p1, 0 0 since the bumped out element of U1 = U(i, λi + 1) ← U is p1, we must have U1 = U1, and 0 thus U2 = U2. Otherwise, if r > 0, the lowest extremal box of V must be the box adjoined by bumping in pr and it must be an extremal box of U too. Therefore, the first step of Ψ 0 give us back the tableau U = pr−1 → (· · · → (p1 → U1) ··· ) with pr the bumped out 0 0 0 0 element of U. By induction on r Ψ(ν , U , V ), with V to be the tableau V but removing pr, 00 00 is (U1, U2 ), where U2 is the tableau U2 but removing pr. Clearly Ψ(ν, U, V) = (U1, U2). 0 Assuming that U2 has more than one column, let U2 be the tableau U2 but removing 0 00 00 its last column p1 ≥ · · · ≥ pr. Let (ν , U , V ) be the triple obtained from the first step of Ψ applied to (ν, U, V), as we have seen before, U00 is the standard tableau obtained 00 by bumping out the boxes with entries µ1 in V and V is obtained from V by removing all boxes with entry µ1. From the definition of Ψ, clearly Ψ(Φ(U1, U2)) is obtained from 0 00 00 00 00 Ψ(ν , U , V ) = (U1, U2 ) by adjoining to U2 the bumped out column γt in the first step of 0 0 0 0 0 Ψ. By induction, Ψ(Φ(U1, U2)) = Ψ(ν , U , V ) = (U1, U2). So, in order to finish the proof 0 00 00 0 0 0 we have to show that (ν , U , V ) = (ν , U , V ) and the last column of U2 is the column 00 0 γt of bumped out elements. In this case, (U1, U2 ) = (U1, U2) and the result follows. It is 0 00 0 clear that V = V , since V is the tableau V but removing all boxes with entry µ1, which is the definition of V00. From definition of Φ, the tableau U0 is the tableau obtained from 0 U1 by bumping in since the last column of U2, and thus U = pr → (· · · → (p1 → U ) ··· ). 0 0 By the initial case, Ψ(Φ(U , (p1,..., pr)) = Ψ(ν, U, V1) = (U , (p1,..., pr)), where V1 is 0 obtained from V but deleting all boxes with entries 1, . . . , µ1 − 1. Because of that U is the 00 tableau obtained from U by bumping out the elements with entries µ1, that is U , and the bumping out column is the last column of U2.

2.5.2 The Littlewood-Richardson Rule.

This subsection is devoted to the proof of the Littlewood-Richardson rule:

Theorem 2.5.20. Let R be a ring of characteristic zero, let F be a free R-module and let λ 36 Schur Functors and CoSchur Functors. and µ two partitions. Then, there is a natural isomorphism

LλF ⊗ LµF ∑ u(λ, µ; ν)LνF ν where u(λ, µ; ν) is the number of standard tableau V ∈ Tabν/λ({1, . . . , |µ|}) of content µ˜ such that the sequence associated to V is a word of Yamanouchi.

Let S = {x1,..., xn} be an ordered basis of F. Since A is the set of all pairs (U1, U2) where U1 and U2 are standard basis in Tabλ(S) and Tabµ(S) respectively, from the stan- { ( ) ⊗ ( ) | ( ) dard basis theorem for Schur functors, (Theorem 2.1.21), dλ XU1 dµ XU2 U1, U2 ∈ A} is an R-basis of LλF ⊗ LµF. Now, if u(λ, µ; ν) is the number of standard tableau V ∈ Tabν/λ({1, . . . , |µ|}) of content µ˜ such that the sequence associated to V is a word of Ya- manouchi, then {dν/λ(X(ν,U,V)) | (ν, U, V) ∈ B} must be an R-basis of ∑ν u(λ, µ; ν)LνF. Since Proposition 2.5.19 tell us that the cardinality of A is less than the cardinality of B, to prove the Littlewood-Richardson rule we only have to construct an injection ∑νu(λ,µ;ν)LνF → LλF ⊗ LµF. We will proceed as follows. First we will define for each standard tableau V satisfying conditions in Theorem 2.5.20 a morphism ΦV : LνF → LλF ⊗ LµF which gives u(λ, µ; ν) independent copies of LνF inside LλF ⊗ LµF. That is ΦV (LνF) u(λ, µ; ν)LνF, and hence u(λ, µ; ν)LνF is a submodule of LλF ⊗ LµF, which give us the injection we are looking for. In order to define morphisms ΦV : LνF → LλF we need to introduce two more combinatorial definitions and one Lemma.

Deﬁnition 2.5.21. Let V be a standard tableau satisfying conditions in Theorem 2.5.20. We denote by a = (a1,..., am) the sequence associated to V and let a˜ = (a˜1,..., a˜m) the transpose of a. We deﬁne V˜ to be the tableau of shape ν/λ and content µ with associated sequence a˜.

As an example, V = 1 and V˜ = 3 . 1 2 2 1 1 1

Lemma 2.5.22. V˜ is decreasing in rows and strictly decreasing in columns. Proof. We start with the row condition. Let (k, l) and (k, l + 1) to adjacent boxes with entries ai < aj in V respectively. We want to see that a˜i ≥ a˜j. Remember a˜i is the number of ah such that ah = ai and h ≤ i, from this if we remove from V all boxes containing entries < ai we still have a standard tableau whose associated sequence is a Y-word and the picture a˜i a˜j not change in V˜ . Therefore we may assume that ai = 1 and l = λk + 1. We proceed by induction on the number of elements above the box (k, λk + 1). If there is nothing above the box of ai, then ai is the last appearance of 1 in a. So, a˜i and a˜j are the number of time ai and aj appear in the subsequence (a1,..., aj). Since a is a Y-word and ai < aj, it follows that a˜i > a˜j. Assuming that there are n boxes above ai, consider the boxes (k − 1, l) and (k − 1, l + 1) with entries ai+1 and aj+1 respectively. It is clear that a˜i = a˜i+1 + 1 and a˜j = a˜j+1 + 1. By induction, a˜i+1 ≥ a˜j+1 and hence, a˜i ≥ a˜j. In order to ﬁnish the proof we check the column condition. Let (k, l) and (k + 1, l) to adjacent boxes with entries ai+1 ≤ ai of V respectively. If ai+1 = ai, then clearly 2.5 The Littlewood-Richardson rule for Schur functors. 37

a˜i+1 = a˜i + 1 > a˜i. Suppose ai+1 < ai, a˜i is the number of ak = a˜i, k ≤ i, indeed a˜i is the number of time that ai appears in the subsequence (a1,..., ai). Since a is a Y-word and ai+1 ≤ ai, if ti+1 is the number of time ai+1 appears in the above subsequence, then ti+1 ≥ a˜i. We obtain the desire result since a˜i+1 = ti+1 + 1.

Deﬁnition 2.5.23. Let V be a tableau satisfying conditions in Theorem 2.5.20, let ∆ν/λ be the diagram of V. We denote µ = (µ1,..., µq) and a = (a1,..., am) the associated sequence of V. From deﬁnition of content, V˜ has entries in {1, . . . , µ˜1} and V has entries in {1, . . . , µ1}. Now, for each k ∈ {1, . . . , µ˜1} there exist µk boxes in ∆ν/λ such that V˜ (i, j) = k.

For a ﬁxed k, consider the sequence (i1, j1),..., (iµk , jµk ) of these boxes formed by listing them from bottom to top in each column, starting from the left-most column. In fact, k the entries of these boxes in this order is the subsequence a of a. Since a˜ = a, V(il, jl) k is the position of the entry in the box (il, jl) in the subsequence a . By Lemma 2.5.22 j1 < j2 < ··· < jµk , and hence V(il, jl) = l, l = 1, . . . , µk. Consequently, for each k ∈ {1, . . . , µ˜1} and l ∈ {1, . . . , µk} there exists a unique box (i, j) ∈ ∆ν/λ such that V˜ (i, j) = k and V(i, j) = l. We deﬁne a bijection σV : ∆ν/λ → ∆µ by σV (i, j) = (V˜ (i, j), V(i, j)).

Deﬁnition 2.5.24. Let V be a satisfying conditions in Theorem 2.5.20. We denote a = (a1,..., am) its associated sequence. We deﬁne a map φV to be the composition:

ı ∆ Id⊗ηV dλ⊗dµ LνF Kν˜ F −→ ΛνF −→ ΛλF ⊗ Λν/λF −−−→ ΛλF ⊗ ΛµF −−−→ LλF ⊗ LµF where the ﬁrst map is the isomorphism between LνF and Kν˜ F, ı is the natural inclusion → ⊗ of Kλ˜ F in ΛλF, ∆ is the tensor product of comultiplications Λνi F Λλi F Λνi−λi F, and 0 ı  m 0 ﬁnally ην is the composition Λν/λF −→ Tν/λF −→⊗µF −→ ΛµF, where ı is the tensor νi−λi νi−λi of the inclusions of Λ F in T F,  : ⊗ν/λF → ⊗µF sends (· · · ⊗ x(i,j) ⊗ · · · ) to (· · · ⊗ ⊗ · · · ) xσV (i,j) and m is the multiplication. Lemma 2.5.25. Let V be a satisfying conditions in Theorem 2.5.20. Then, the morphism Φν : LνF → LλF ⊗ LµF give u(λ, µ; ν) independent copies of LνF inside LλF ⊗ LµF. Proof. The proof of this lemma is found in [1], Lemma IV.2.5. It follows from arguments which imply knowledges about Representation Theory.

As we have discuss at the begging of this subsection, the Littlewood-Richardson rule follows from the above Lemma. We present here two consequence of Theorem 2.5.20.

Corollary 2.5.26. The map Φ : A → B is a bijection.

Corollary 2.5.27. Let R be a ring of characteristic zero, let F be a free R-module, let λ be a partition and p a positive integer. Then, there are natural isomorphisms: p (1) LλF ⊗ S F ∑ν LνF, where ν runs over all partitions containing λ such that |ν| = |λ| + p and νi ≤ λi + 1 for all i. p (2) LλF ⊗ Λ F ∑ν LνF, where ν runs over all partitions containing λ such that |ν| = |λ| + p and ν˜i ≤ λ˜ i + 1 for all i. p p Proof. Observe that S F = L(1,...,1)F, 1 + ··· + 1 = p and Λ F = L(p)F. In the ﬁrst case we have that if V is a standard tableau of shape ν/λ, content (p) with associated sequence a word of Yamanouchi. then, V must have in {1}. The standardness 38 Schur Functors and CoSchur Functors.

of V force that the diagram of V only has one box in each row. Then, νi is at most λi + 1. Clearly, |ν| = |λ| + p. In the second case, V has content (1, . . . , 1), 1 + ··· + 1 = p. Then, the tableau V˜ is of shape ν/λ, has content (p) and entries in {1}. Since V˜ is strictly decreasing in columns, for each column of V˜ there is only one box. Then, ν˜i ≤ λ˜ i + 1. Clearly, |ν| = |λ| + p.

We want to ﬁnish with some examples.

Examples 2.5.28. (1) We start computing the Pieri formulas when λ = (2, 1) and p = 2. First, the partitions ν containing (2, 1) such that |ν| = 5 are (2, 1, 1, 1), (2, 2, 1), (3, 1, 1), (3, 2) and (4, 1). The condition νi ≤ λi + 1 remove the partition (4, 1), and the condition ν˜i ≤ λ˜ i + 1 eliminate the partition (2, 1, 1, 1). We obtain the followings isomorphism.

2 L(2,1)F ⊗ S F L(2,1,1,1)F ⊗ L(2,2,1)F ⊕ L(3,1,1)F ⊕ L(3,2)F

2 L(2,1)F ⊗ Λ F L(2,2,1)F ⊕ L(3,1,1)F ⊕ L(3,2)F ⊕ L(4,1)F

(2) Let us see the direct decomposition of L(2,1)F ⊗ L(2,1)F. The partitions ν containing (2, 1) such that |ν| = 6 are (2, 1, 1, 1, 1), (2, 2, 1, 1), (2, 2, 2), (3, 1, 1, 1), (3, 2, 1), (3, 3), (4, 1, 1), (4, 2) and (5, 1). We must study the set of standard tableau V ∈ Tabν/λ{1, 2} of content (2, 1) which sequence associated is a Y-word for each ν in the above list.

ν = (2, 1, 1, 1, 1), V = 2 but the sequence associated to V is (2, 1, 1) which is not 1 1 a Y-word. u(λ, µ, (2, 1, 1, 1, 1)) = 0.

ν = (2, 2, 1, 1), V = 2 or 1 , (1, 1, 2) is a Y-word. u(λ, µ, (2, 2, 1, 1)) = 1. 1 1 1 2

ν = (2, 2, 2), V = 1 , (1, 2, 1) is a Y-word. u(λ, µ, (2, 2, 2)) = 1. 1 2

ν = (3, 1, 1, 1), V = 1 or 2 , (1, 1, 2) is a Y-word.

1 1 2 1 u(λ, µ, (3, 1, 1, 1)) = 1.

ν = (3, 2, 1), V = 1 or 1 or 2 , (1, 1, 2) and

1 2 1 2 1 1 (1, 2, 1) are Y-words. u(λ, µ, (3, 2, 1)) = 2.

ν = (3, 3), V = 1 , (1, 2, 1) is a Y-word. u(λ, µ, (3, 3)) = 1. 1 2 2.5 The Littlewood-Richardson rule for Schur functors. 39

ν = (4, 1, 1), V = 1 2 , (1, 1, 2) is a Y-word. u(λ, µ, (4, 1, 1)) = 1.

ν = (4, 2), V = 1 2 , (1, 1, 2) is a Y-word. u(λ, µ, (4, 2)) = 1. 1

ν = (5, 1), there is not standard tableau V with entries in {1, 2} and content (2, 1). u(λ, µ, (5, 1)) = 0.

We have, L(2,1)F ⊗ L(2,1)F L(2,2,1,1)F ⊕ L(2,2,2)F ⊕ L(3,1,1,1)F ⊕ 2L(3,2,1)F ⊕ L(3,3)F ⊕ L(4,1,1)F ⊕ L(4,2)F. 40 Schur Functors and CoSchur Functors. Chapter 3

A minimal free complex associated to the minors of a matrix.

Finding a minimal free resolution of the ideal associated to the minors of a matrix is a classical subject in Algebra and Algebraic Geometry. This problem was solved by Lascoux in [16]. In this chapter, we construct a minimal free complex associated to the minors of a matrix. Mainly, we follow [18] and [3], and the last section is based on [17], [4], [12], [16] and [13].

3.1 A minimal free complex.

Let R be a local commutative ring contained a ﬁeld K of characteristic zero and m its maximal ideal. Let n, m and t be positives integers such that n ≤ m and let F and G be free R-modules of ranks m + t − 1 and n + t − 1, with basis { f1,..., fm+t−1} and ∗ {g1,..., gn+t−1}, respectively. We denote G = Hom(G, R) the dual of G with basis ∗ ∗ {g1,..., gn+t−1} dual to the basis {g1,..., gn+t−1}. Let (ri,j) be a (n + t − 1) × (m + t − 1) ∗ matrix with entries in R associated to a map Φ : F → G . We denote by It to be the ideal of R generated by the t × t minors of the matrix (ri,j). Deﬁnition 3.1.1. Let M be a free R-module. We say that a complex

dn+1 dn d1 d0 M• : 0 −−→ Mn −→ Mn−1 → · · · → M1 −→ M0 −→ 0 of free R-modules is a minimal free complex of M if d1(M1) = M and for each i = 1, . . . , n, we have di(Mi) ⊂ mMi−1. The last condition is equivalent to the following: after choosing basis in Mi, for each morphism di the matrix associated to di has entries in m. We say that M• is a minimal free resolution of M if M• is a minimal free exact complex of M. Equivalently, if M• is a minimal free complex of M and for each i = 1, . . . , n + 1 the homology Hi(M•) = 0. Our aim is to construct a minimal free complex

dmn d1 C•(Φ, t) : 0 → Cmn(Φ, t) −−→ Cmn−1(Φ, t) → · · · → C1(Φ, t) −→ R → 0

41 42 A minimal free complex associated to the minors of a matrix.

of It.

3.1.1 The group algebra and combinatorics.

Deﬁnition 3.1.2. The group algebra K(Sn) is formally the set of all ﬁnite linear combinations

∑ α(σ)σ, σ ∈ Sn, α(σ) ∈ K σ with a sum (∑σ α(σ)σ) + (∑σ β(σ)σ) := ∑σ(α(σ) + β(σ))σ and a product (∑σ α(σ)σ) · (∑ρ β(ρ)ρ) := ∑σ,ρ α(σ)β(ρ)σρ. The elements of Sn are called group elements, which are linearly independents and constitute a basis of K(Sn).

Deﬁnition 3.1.3. Let e ∈ K(Sn). We say that e is essentially idempotent if there is a nonzero element κ ∈ K such that e2 = κe. We say that e is idempotent if e2 = e.

Remark 3.1.4. If e is essentially idempotent with e2 = κe, then (κ−1e)(κ−1e) = κ−1e and hence κ−1e is idempotent.

Deﬁnition 3.1.5. Let λ = (λ1,..., λr) be a partition of weight n.A Young tableau is any tableau of shape λ with distinct entries on {1, . . . , n}.

Examples 3.1.6. For example, if λ = (4, 3, 2, 1) two possible Young tableau are T = 3 6 4 5 and R = 1 2 3 4 7 1 10 5 6 7 2 8 8 9 9 10

Remark 3.1.7. Let λ = (λ1,..., λr) be a partition of weight n. Clearly there are n! possible Young tableaux of shape λ. In fact, the n! possible ways of filling ∆λ with the elements {1, . . . , n}. We can define each Young tableau as follows. We define the Young tableau ( ) = i−1 + ∈ (S ) λid by λid i, j ∑k=1 λk j. Each α K n define a unique Young tableau λα by λα(i, j) = α(λid(i, j)), and hence any Young tableau is of the form λα for some α ∈ K(Sn).

Observe that the above tableau R is λid for λ = (4, 3, 2, 1). Each Young tableau λα deﬁnes two subgroups Pλα and Qλα of Sn. Pλα is the set of permutations that map each element to an element in the same row in λα. Qλα is the set of permutations that map each element to an element in the same column in λα.

Deﬁnition 3.1.8. Let λ be a partition of weight n and let λα be a Young tableau. We deﬁne sg(q) P = p ∈ K(Sn), Q = (−1) q ∈ K(Sn) and E = P Q . α ∑p∈Pλα α ∑q∈Qλα λα α α

Theorem 3.1.9. Let λ be a partition of weight n and let λα be a Young tableau. Then, Eλα is essentially idempotent. If E2 = κ E , we denote the idempotent element e(λ ) = λα λα λα α κ−1E . λα λα

Deﬁnition 3.1.10. Let λ be a partition of weight n and let λα and λβ be two Young tableaux. We deﬁne sα,β to be the permutation that takes λβ to λα. More precisely, λα(i, j) = sα,β(λβ(i, j)), for all (i, j) ∈ ∆λ. We denote λα = sα,βλβ. 3.1 A minimal free complex. 43

Theorem 3.1.11. Let λ be a partition of weight n and let λα and λβ be two Young tableaux. = −1 = −1 = −1 Then, Pλα sα,βPλβ sα,β, Qλα sα,βQλβ sα,β and Eλα sα,βEλβ sα,β.

Deﬁnition 3.1.12. Let λ be a partition of weight n and let λα and λβ be two Young tableaux. We say that λα λβ if not two elements in the same column of λα are in the same row of λβ.

Theorem 3.1.13. Let λ be a partition of weight n and let λα and λβ be two Young tableaux. = ∈ ∈ λα λβ if, and only if sα,β pq with p Pλβ and q Qλβ . We say that sαβ is a pq.

Proposition 3.1.14. Let λ be a partition of weight n and let λα be a Young tableau. For all sg(q) p ∈ Pλα and q ∈ Qλα , pP = Pp = P and qQ = Qq = (−1) Q.

Theorem 3.1.15. Let λ be a partition of weight n and let λα and λβ be two Young tableaux. If λα 3 λβ, then e(λα)e(λβ) = 0.

Remark 3.1.16. λα 3 λβ if, and only if there is two entries x and y in the same row of λβ and in the same column of λα. Let (xy) be the permutation which transposes x and y. ( ) ∈ ( ) ∈ ( ) ( ) ( ) ( ) Since xy Pλβ and xy Qλα , xy has the same coefﬁcient in e λβ and in xy e λα . Then, (xy)e(λβ) = e(λβ) and e(λα)(xy) = −e(λα) imply that e(λβ) is divisible on the right by id − (xy) and e(λα) is divisible on the left by id + (xy).

Theorem 3.1.17. Let λ be a partition of weight n and let λα and λβ be two standard Young tableaux. If λα λβ, then λα < λβ, where < is the usual order in tableaux induced lexicographically.

Theorem 3.1.18. Let λ be a partition of weight n and let λα and λβ be two Young tableaux. The elements sαβe(λβ) = e(λα)sαβ, where λα ranges over all the standard tableaux, form a basis for K(Sk)e(λβ).

3.1.2 The complex. We keep the notation introduced in section 3.1.1. In addition, we note (mn)= (n,..., n), such that n + ··· + n = nm, and we denote λ(mn) = λid, in the same manner we consider (m + t − 1, n) and λ((m+t−1)n). We proceed to construct the complex C•(Φ, t). First we define the modules Ck(Φ, t) by means of idempotents and we prove that they are a sum of tensor product of Schur functors. This implies that Ck(Φ, k) is a free R-module and we will describe a basis of Ck(Φ, k). In second place we determinate the morphisms of the complex, we often call them the boundary maps. And finally, we will prove that these elements define a minimal free complex of It.

Deﬁnition 3.1.19. Let λ be a partition. A diagonal box of ∆λ is a box of the form (i, i). The Durfee square of λ is the number of diagonal boxes in ∆λ. In other words, we say that λ has Durfee square d if λd ≥ d but λd+1 ≤ d. We arbitrarily assign (0) Durfee square 0. It is clear that λ and λ˜ have the same Durfee square.

Deﬁnition 3.1.20. Let k be an integer such that 1 ≤ k ≤ mn and let λ = (λ1,..., λm) be a partition of weight k such that λ ⊂ (mn). For t = 1, we deﬁne λF,1 to be the subtableau of 44 A minimal free complex associated to the minors of a matrix.

λ(mn) of shape λ, and we deﬁne λG,1 to be the tableau transpose of λF,1, i.e. the subtableau ˜ ˜ of λ(mn) of shape λ. Assume t > 1. To the square λ(mn)(i, j) we associate:

(i) The square λ((m+t−1)n)(i, j) if j > i.

(ii) The vertical string of t squares λ((m+t−1)n)(i, j),..., λ((m+t−1)n)(i + t − 1, j) if j = i.

(iii) The square λ((m+t−1)n)(i + t − 1, j) if j < i.

We deﬁne λF,t to be the tableau obtained from λF,1 by replacing each square of λF,1 by its associate square or string of t squares. If d is the Durfee square of λF,1, then the diagram of λF,t is obtained from the diagram of λF,1 by adjoining t − 1 boxes to the ﬁrst d columns. So, λF,t is a tableau of shape λ(F, t) = (λ1,..., λd, d,..., d, λd+1,..., λm), with + ··· + = − ˜ ( ) = ( ˜ + − ˜ + − ˜ ˜ ) d d t 1 and with transpose λ F, t λ1 t 1, . . . , λd t 1, λd+1,..., λλ1 . ˜ To the square λ(mn)(i, j) we associate:

(i) The square λ((m+t−1)n)(i, j) if j < i.

(ii) The vertical string of t squares λ((m+t−1)n)(i, j),..., λ((m+t−1)n)(i + t − 1, j) if i = j.

(iii) The square λ((m+t−1)n)(i + t − 1, j) if j > i.

We deﬁne the tableau λG,t in the same manner of λF,t with the above association. The ( ) = ( ) + ··· + = − shape of λG,t is λ G, t : eλ1,..., eλd, d,..., d, eλd+1,..., eλλ1 with d d t 1 and ˜ with transpose λ(G, t) = (λ1 + t − 1, . . . , λd + t − 1, λd+1,..., λm).

Remark 3.1.21. Note that if t > 1, in general λG,t is not the transpose of λF,t.

For example, consider m = 4, n = 3, t = 3, k = 5 and λ = (2, 2, 1). We have,

λ((m+t−1)n) = 1 2 3 λ(mn) = 1 2 3 λF,1 = 1 2 λG,1 = 1 4 7 4 5 6 4 5 6 4 5 2 5 7 8 9 7 8 9 7 10 11 12 10 11 12 13 14 15 16 17 18

λF,t = 1 2 λG,t = 1 10 13 4 5 4 5 7 8 7 8 10 11 2 11 13

Observe that λ((m+t−1)n) and λ(mn) have Durfee square 3, while the others tableaux have Durfee square 2. r Consider Sr. If M is a free R-module, Sr acts on T M as follows. If σ ∈ Sr, we deﬁne σ(m1 ⊗ · · · ⊗ mr) = mσ−1(1) ⊗ · · · ⊗ mσ−1(n), mi ∈ M for all i. So, for each σ, ρ ∈ Sr, (σρ)(m1 ⊗ · · · ⊗ mr) = σ(ρ((m1 ⊗ · · · ⊗ mr))). Since R contains K, the group algebra K(Sr) r acts on T M and each element of K(Sr) acts as an R-map. 3.1 A minimal free complex. 45

Keeping this in mind, we consider λF,t as in Definition 3.1.20 and Sk+d(t−1) to be the symmetric group on {λF(t)(1, 1),..., λF(t)(1, λ¯ 1),..., λF(t)(m + t − 1, λ¯ m+t−1)} with d the k+d(t−1) Durfee square of λF,1. The idempotent element e(λF,t) acts on T F as an R-map, k+d(t−1) (k+d(t−1) thus e(λF,t)(T F) is a R-module of T F. Apply the same argue with λG,t and k+d(t−1) k+d(t−1) G. For convenience we denote e(λF,t)(T F) and e(λG,t)(T G) by e(λF,t)F and e(λG,t)F, respectively. We define the modules Ck(Φ, t): L Definition 3.1.22. We define the R-module Ck(Φ, t) := |λ|=k e(λF(t))(F) ⊗ e(λG(t))(G) for each 1 ≤ k ≤ mn. For k = 0, let C0(Φ, t) R. L Proposition 3.1.23. For each 1 ≤ k ≤ mn, Ck(Φ, t) |λ|=k Lλ˜ (F,t)F ⊗ Lλ˜ (G,t)G. Hence, Ck(Φ, t) is a free R-module. Proof. We will show that e(λF,t)F Lλ˜ (F,t)F, the proof of e(λG,t)G Lλ˜ (G,t)G is analo- e( ) = 1 ( )( (− )sg(τ) ) P gous. Remember λF,t κ ∑σ∈Pλ σ ∑τ∈Qλ 1 τ where λF,t is the set of λF,t F,t F,t permutations that map each element to an element in the same row in λF,t and QλF,t is the set of permutations that map each element to an element in the same column in λF,t. Then, ∈ ( ) σ PλF,t has the form σ1,..., σm+t−1 where σi is a permutation on the i-th of λF,t and τ ∈ Q (τ τ ) τ j λ λF,t has the form 1,..., λ(F,t)1 where j is a permutation on the -column of F,t. For convenience we introduce the following notation to describe the factors of the module e(λF(t)). Since λF(t) is a tableau with k + d(t − 1) distinct entries in {1, . . . , (m + t − 1)n}, we can establish the one-one correspondence λF(t)(i, j) → (i, j) for each (i, j) ∈ ∆λ(F,t). Therefore, each permutation of the entries of λF,t induces a permutation of the boxes k+d(t−1) of ∆λ(F,t) and, clearly on ∆λ˜ (F,t). Then, we can consider the tensor product T F indexed by ∆ , more precisely, Tk+d(t−1)F = N F = T F, where F = F. λ(F,t) (i,j)∈∆λ(F,t) i,j λ(F,t) i,j k+d(t−1) Let f = f(1,1) ⊗ · · · ⊗ f(m+t−1,λ(F,t) ) ∈ T F, we have (∑τ∈Q τ)( f ) = m+t−1 λF,t sg(τ1) sg(τq) ∑τ∈Q (−1) ··· (−1) fτ (1,1) ⊗ · · · ⊗ fτ (1,λ(F,t) ) ⊗ · · · ⊗ fτ (m+t−1,1) ⊗ · · · ⊗ λF,t 1 λ(F,t)1 1 1 ( )( ) = ⊗ · · · ⊗ fτ (m+t−1,λ(F,t) + − ) and ∑σ∈P σ f ∑τ∈P fσ ((1,1)) fσ ((1,λ(F,t) )) λ(F,t)m+t−1 m t 1 λF,t λF(t) 1 1 1 ⊗ · · · ⊗ f ⊗ · · · ⊗ f T F → T σm+t−1((m+t−1,1)) σm+t−1((m+t−1,λ(F,t)m+t−1). Let Υ : λ(F,t) λ˜ (F,t) be the canonical isomorphism permuting factors, from the above Im(Υ ◦ (∑τ∈Q τ)) is λF,t the immersion of Λ ˜ ( )F in Tλ(F,t)F and Im(∑τ∈Q τ) is the immersion of Sλ(F,t)F λ F,t λF,t in Tλ(F,t)F. More precisely, (Υ ◦ (∑τ∈Q τ))( f ) corresponds bijectively to f(1,1) ∧ · · · ∧ λF,t f ∧ · · · ∧ f ⊗ · · · ⊗ f ( σ)( f ) (1,λ˜ (F,t) ) (λ(F,t)1,1 (λ(F,t) ,λ˜ (F,t) ) and ∑σ∈Pλ corresponds bijec- 1 1 λ(F,t)1 F,t f ····· f ⊗ · · · ⊗ f ····· f tively to (1,1) (1,λ(F,t)1) (m+t−1,1) (m+t−1,λ(F,t)m+t−1). −1 sg(τ1) sg(τq) Since the element e(λF,t)( f ) = κ ∑σ∈P ∑τ∈Q (−1) ··· (−1) fσ τ (1,1) λ(F,t) λ(F,t) λF(t) 1 1 ⊗ · · · ⊗ ⊗ · · · ⊗ ⊗ · · · ⊗ fσ τ (1,λ(F,t) ) fσ + − τ (m+t−1,1) fσ + − τ (m+t−1,λ(F,t) + − ), 1 λ(F,t)1 1 m t 1 1 m t 1 λ(F,t)m+t−1 m t 1 clearly e(λF,t)( f ) corresponds bijectively to d ( f ∧ · · · ∧ f ⊗ · · · ⊗ f ∧ · · · ∧ f ) λ˜ (F,t) (1,1) (1,λ˜ (F,t) ) (λ(F,t)1,1 (λ(F,t) ,λ˜ (F,t) ) . 1 1 λ(F,t)1

We can describe a basis of Ck(Φ, t) through a basis of F and G. Let T ∈ Tabλ(F,t){ f1,..., fm+t−1}, we deﬁne Tˆ to be tableau of shape λ˜ (F, t) transpose to T. From the above ˆ proposition, {e(λF,t)(XT) | T ∈ Tabλ(F,t){ f1,..., fm+t−1} and T is standard} form ˆ a basis of e(λF,t)F. Similarly, {e(λG,t)(YT) | S ∈ Tabλ(F,t){g1,..., gn+t−1} and S is standard} form a basis of e(λG,t)G. Finally, 46 A minimal free complex associated to the minors of a matrix.

ˆ Corollary 3.1.24. {e(λF,t)(XT) ⊗ e(λG,t)(YS) | T ∈ Tabλ(F,t){ f1,..., fm+t−1} and T is ˆ standard, S ∈ Tabλ(F,t){g1,..., gn+t−1} and S is standard, |λ| = k} form a basis of Ck(Φ, t).

Now, we proceed to construct the boundary maps dk : Ck(Φ, t) → Ck−1(Φ, t). As we have anticipated, the morphism dk are induced by Φ in a natural way. The canonical isomorphisms Hom(F, G∗) F∗ ⊗ G∗ Hom(F ⊗ G, R) allow us deﬁne

Deﬁnition 3.1.25. If TnF and TnG are indexed by sets in one-one correspondence I and ˜ n n N N I of cardinality n, we deﬁne a map Φ∗ : T F ⊗ T G → R by (( i∈I fi) ⊗ ( j∈I˜ gj)) → ( ) ∏i∈I Φ fi gi˜ . We say that Φ∗ is the natural contraction extending Φ. More general, for any corresponding subsets of I and I˜ of k elements in one-one correspondence we have a map n n n−k n−k Φ∗ : T F ⊗ T G → T F ⊗ T G contracting these k elements.

−1 Consider now the involution i : K(Sn) → K(Sn) induced by the map of Sn sending −1 −1 −1 ∗ σ → σ . That is, i (r = ∑ kss) = ∑ kss := r . The following proposition will be essential to proof that C•(Φ, t) is a complex.

n n Proposition 3.1.26. Let Φ∗ : T F ⊗ T G → R, where the two factors are indexed by sets I and I˜ in one-one correspondence i → i˜. Let x and y be elements of TnF and TnG respectively, and let s and t be elements of K(Sn), where Sn denotes the symmetric group on I. Each σ ∈ Sn induces a permutation σ˜ on I˜ such that σ(i) = j if, and only if ∗ σ˜ (i˜) = σ˜ (j˜). Then Φ∗((sx) ⊗ (ty˜ )) = Φ∗((t s)x ⊗ y). Proof. By linearity, it sufﬁces to prove it when σ and τ are elements of Sn. We consider ˜ −1 the permutation on I induced by τ . We denote σx = ⊗i∈I xσ−1(i) and τ˜y = ⊗j∈I0 yτ˜−1(j). Considering the induced one-one correspondence k0 = σ−1(i) → k˜ = τ˜−1(i˜), we have 0 −1 0 0 Φ∗(σx ⊗ τ˜y) = Φ y −1 . If k = τ˜ (i˜), then i˜ = τ˜(k ). This implies that k = l˜ ∏i∈I xσ−1(i) σ˜ (i˜) such that τ(l) = i and hence k = σ−1τ(l). The above expression becomes Φ y ∏l∈I xσ−1τ(l) l˜ −1 −1 which equals to Φ∗((τ σ)x ⊗ y) with the induced one-one correspondence σ τ(l) → l˜.

We consider Ck(Φ, t) as in Deﬁnition 3.1.22, where λ is a partition of weight k, λF,t and λG,t are the Young tableaux constructed in 3.1.20. Observe that λF,t and λG,t have the same entries. If a is an entry of λF,t we denote by a˜ the corresponding entry in λG,t. Clearly, a → a˜ deﬁnes a one-one correspondence ~, and thus we can consider the tensor (k+d(t−1) (k+d(t−1) products T F and T G indexed by the entries of the tableaux λF,t and λG,t in one-one correspondence ~.

For convenience we denote Cλ(t) the summand e(λF,t)F ⊗ e(λG,t)G of Ck(φ, t). Let λ and µ be partitions of weight k and k − 1, respectively. First we deﬁne a map dµλ : Cλ(t) → Cµ(t), we distinguish two cases: (i) If there is an entry of µF,1 that is not in λF,1, let dµλ = 0. (ii) Assume µF,1 ⊆ λF,1. Therefore µF,1 is obtained by removing one corner a of λF,1. We often denote µF,1 by λF,1 − a. If dµ is the Durfee square of µ, Cµ(t) is a submodule of Tk−1+dµ(t−1)F ⊗ Tk−1+dµ(t−1)G where the factors of Tk−1+dµ(t−1)F and Tk−1+dµ(t−1)G are indexed over the same sets as Tk+d(t−1)F and Tk+d(t−1)G except the entries associated to a 3.1 A minimal free complex. 47 and a˜ respectively, we call them the entries of a and a˜ respectively. In this situation we con- k+d(t−1) k+d(t−1) k−1+d (t−1) k−1+d (t−1) sider the map Φ∗ : T F ⊗ T G → T µ F ⊗ T µ G contracting the entries of a and a˜ with the one-one correspondence ~.

Deﬁnition 3.1.27. Let a be a corner of λF,1. If a is in the i-column of λF,1, we deﬁne s(λ, a) to be the number of squares in the i-column above a plus the number of squares in the λ(F,t)1 ˜ columns i + 1, . . . , λ(F, 1)1. Formally, s(λF,1, a) := (∑j=i λj) − 1.

s(λ ,a) Deﬁnition 3.1.28. We deﬁne the map dµλ : Cλ(t) → Cµ(t) to be (−1) F,1 (e(µF(t)) ⊗ e(µG(t))) ◦ φ∗, where φ∗ denotes the restriction of Φ∗ to Cλ(t).

With this ingredients we can finally define the boundary maps of the complex C•(Φ, t). L Definition 3.1.29. For each 1 ≤ k ≤ mn we define the morphism dk := |λ|=k dλ = L L |λ|=k |µ|=k−1 dµλ from Ck(Φ, t) to Ck−1(Φ, t).

Observe that φ∗ is deﬁned by a matrix with coefﬁcients in m, and hence [e(µF,t) ⊗ e(µG,t]φ∗ sends Cλ into mCµ. Therefore, dk(Ck(Φ, t)) ⊂ mCk−1(Φ, t)).

Proposition 3.1.30. d1(C1(Φ, t)) = It. Proof. There is only one partition λ = (1) satisfying |λ| = 1. In this case, λ(F, t) = (1, . . . , 1) = λ(G, t) and

λF,t = 1 = λG,t . . t ( ) = ( ) ⊗ ( ) { ( )( ⊗ · · · ⊗ ) ⊗ ( )( ⊗ Therefore, C1 Φ, t e λF,t F e λG,t G and e λF,t fi1 fit e λG,t gj1

· · · ⊗ gjt ) | 1 ≤ i1 < ··· < it ≤ m + t − 1, 1 ≤ j1 < ··· < jt < n + t − 1} form a basis of C1(Φ, t). Since ∑σ∈P σ = id, we have from Proposition 3.1.26 that φ∗(e(λF,t)( fi ⊗ · · · ⊗ λF,t 1 f ) ⊗ e(λ )(g ⊗ · · · ⊗ g )) = φ ( (− )sg(τ) f ⊗ · · · ⊗ f ⊗ g ⊗ · · · ⊗ g ) = it G,t j1 jt ∗ ∑σ∈St 1 σ(i1) σ(it) j1 jt sg(τ) sg(τ) ∑σ∈S (−1) Φ f gj ··· Φ f gjt = ∑σ∈S (−1) rσ(i )j ··· rσ(i )j , which is the t × t t σ(i1) 1 σ(it) t 1 1 t t minor of (ri,j) corresponding to the rows j1,..., jt and the columns i1,..., it of (ri,j).

It remains to prove that (C•(Φ, t), d•) is a complex. More speciﬁcally, we must show that M dνµdµλ = 0. |ν|=k−2,|µ|=k−1,|λ|=k−1

The partitions ν involved in this sum must be such that νF,1 = λF,1 − a − b = λF,1 − b − a, where a is a corner of λF,1 and b becomes a corner of λF,t − a or vice versa. There are two possible conﬁgurations of two entries a and b in λF,1 which can be removed. (1) a or b is not a corner of λF,1 and there is only one order to remove them. If a is corner of λF,1, then λF,1 is of the form: or b b a a 48 A minimal free complex associated to the minors of a matrix.

In this case, there is only one partition µ such that dνµ and dµλ are nonzero. (2) a and b are both corners of λF,1, and hence there are two possible ways of removing them. In this second case λF,1 is of the form: or a b b a

Then, there are two partitions µ such that dνµ and dµλ are non zero. We may assume that b is bellow a, the other case is analogous. Observe that s(λ − a, b) = s(λ, b) − 1 and s(λ − b, a) = s(λ, a). Therefore, (−1)s(λ−a,b)(−1)s(λ,a) = −(−1)s(λ−b,a)(−1)s(λ,b).

From those arguments to prove that C•(Φ, t) is a complex follows from showing that given two entries a and b which can be removed, with b a corner: a λ d d = (i) If is not a corner of F,1, then (λF,1−b−a)(λF,1−b) (λF,1−b)λF,1 0. (ii) If a is a corner of λF,1 above b, then

d d + d d = (λF,1−b−a)(λF,1−b) (λF,1−b)λF,1 (λF,1−b−a)(λF,1−a) (λF,1−a)λF,1 0.

1 2 Now, dνµdµλ module sign correspond to [e(νF,t) ⊗ e(νG,t)]φ∗[e(µF,t) ⊗ e(µG,t]φ∗. Since e(λF,t) ⊗ e(λG,t) as an idempotent is the identity to Cλ, we can write dνµdµλ as [e(νF,t) ⊗ 1 2 e(νG,t)]φ∗[e(µF,t) ⊗ e(µG,t]φ∗[e(λF,t) ⊗ e(λG,t)]. If µF,1 = λF,1 − b for some corner b of λF,1, 2 ˜ 2 then φ∗ contracts b and b and not acts on e(µF,t)F and e(µG,t)G. So, we can apply φ∗ 1 2 after the idempotent [e(µF,t) ⊗ e(µG,t]. The composition φ∗φ∗ deﬁnes a contraction on the union of the entries of a and b. Applying the above argument, dνµdµλ can be written as the composition

φ∗[e(νF,t) ⊗ e(νG,t)][e(µF,t) ⊗ e(µG,t)][e(λF,t) ⊗ e(λG,t)]. (∗)

For the remainder of this section we ﬁx k, λ a partition of weight k, b and a be squares which can be removed. We write λF,t − b = µF,t and λF,t − b − a = νF,t. For convenience we denote eˆ(λ) = [e(λF,t) ⊗ e(λG,t)], e(λ) = e(λF,t) and e(λ˜ ) = e(λG,t). The purpose of expressing dνµdµλ as (∗) is that identities involving eˆ(ν)eˆ(µ)eˆ(λ) become identities involving dνµdµλ just by adding φ∗ on the left. More precisely, conditions (i) and (ii) are equivalent to:

Lemma 3.1.31. Let b a corner of λF and let a be a square of λF directly above or to the left of b in such a way that it becomes a corner of λF − b. Then,

Φ∗eˆ(λ − a − b)eˆ(λ − b)eˆ(λ) = 0.

Lemma 3.1.32. Let a and b be corners of λF. Then,

Φ∗eˆ(λ − a − b)eˆ(λ − b)eˆ(λ) = Φ∗eˆ(λ − a − b)eˆ(λ − a)eˆ(λ).

Clearly, we can write eˆ(λ − a − b)eˆ(λ − b)eˆ(λ) = [e(λ − a − b)e(λ − b)e(λ)][e(λ˜ − a − b)eˆ(λ˜ − b)eˆ(λ˜ )]. We dedicate the following subsections to the proof of both Lemmas, which implies that C•(Φ, t) is a complex. First we consider the following. 3.1 A minimal free complex. 49

Remember Theorem 3.1.18 and notations in subsection 3.1.1, let Sk+d(t−1). In particular, noting sα,1 permutations from λF,t to λα. The following propositions are found in [18] and [3].

Proposition 3.1.33. We can write e(λ − b)e(λ) = ∑∗ kαsα,1e(λ), where ∗ denotes that the sum runs over the standard tableaux λα = sα,1λF,t. We denote it simply by ∑ kαsα,1e(λ).

Remark 3.1.34. There is not one way of write sα1 in the form sσ1 pq. Therefore, in general sg(q) (−1) kσ the coefficient of sα1 in e(λ − b)e(λ) is not kα. In fact, it is the sum of all the terms κ λF,t such that sα1 = sσ1 pq. = Definition 3.1.35. Let λα sα1λF,t be a standard tableaux. We define jα to be κλF,t times the coefficient of sα1 in e(λ − b)e(λ). = = ∈ Proposition 3.1.36. Let λα sα,1λF,t. We can write sα1 sσ1 pq where p PλF,t and ∈ q QλF,t if, and only if λα λσ.

Proposition 3.1.37. Consider e(λ − b)e(λ) = ∑ kαsα1e(λ). Suppose there are two elements in the same column in λα and in the same row in λF,t, and neither of these is an entry of b. Then the coefﬁcient of sα1 in e(λ − b)e(λ) is zero.

Proposition 3.1.38. Let a be an entry of λF,t which becomes a corner of λF,t − b. Consider e(λ − b)e(λ) = ∑ kαsα1e(λ). Suppose that

1. All entries of λF,t − a − b precede the entries of a and b. That is, the entries in a and b are the highest entries in λF,t.

2. The entry in a given row of λF,t − a − b precedes any element in any lower row. That is, given an entry (λF,t − a − b)(i, j), then (λF,t − a − b)(k, l) > (λF,t − a − b)(i, j), ∀k > i, ∀l.

Then, if sα1 moves any entry of λF,t − a − b, we have either

(i) kα = 0, or

(ii) e(λ − a − b)sα1e(λ) = 0.

In addition, if sα1 moves only elements of a and b, we have jα = kα.

3.1.3 The Proof of Lemma 3.1.31.

We consider Sk+d(t−1) on the entries of λF,t ordered ﬁrst row by row in λF,t − a − b and then the entries of a and b. In other words, λF,t is a tableau as in Proposition 3.1.38. The proof of both Lemmas are based on effectives calculations of the product of the idempotents e(λ − a − b), e(λ − a), e(λ − b) and e(λ). Our goal is to develop techniques which simplify this expressions. There are these possible conﬁgurations for b and a, depending on whether b and a are in the same row or the same column in λF,1, and depending on whether a or b corresponds to a single square or a string of t squares. The four possible arrangements in λF,t are 50 A minimal free complex associated to the minors of a matrix.

1. b 2. 3. a 4. a b a a b b where we have considered the case where both a and b are single squares as a special case of either of these. First we study cases 1. and 2.

Proposition 3.1.39. Let b a corner of λF,1 and let a be a square directly to the left of b. Then, the product e(λ − a − b)e(λ − b)e(λ) is a sum of terms each of which is divisible on the left by id + (xy), for some distinct elements x and y in the entries of a and b. Proof. We let e(λ − b)e(λ) = ∑ kαsα1e(λ). By Proposition 3.1.38, in the above expression we only have to consider terms for which sα1 only moves the entries of a and b. Let sα1 be of that such, then there are two elements x and y which end up in a row of λα. x y

x y

Denoting (xy) the permutation which transposes x and y, hence (xy) is a p in Pα. Since e(λα) = (xy)e(λα), e(λα) is divisible on the left by id + (xy). Therefore, e(λ − a − b)e(λ − b)e(λ) is divisible on the left by id + (xy), because sα1e(λ) = e(λα)sα1 and e(λ − a − b) does not act over the entries of a and b.

Computation of cases 3. and 4. require a previous result.

Proposition 3.1.40. Let λ be any tableau, let c be a column of consecutive squares of λ all of which are at the right ends of their rows and such that the bottom square of c is a corner. Then, e(λ − c) is deﬁned and for any permutation q of the entries in c, we have qe(λ − c)e(λ) = (−1)sg(q)e(λ − c − d)e(λ). Similarly, if d is another corner square of λ, we have qe(λ − c − d)e(λ) = (−1)sg(q)e(λ − c − d)e(λ). Proof. Suppose c is in the nth column of λ. For i = 1, 2, . . . , n − 1, let qi be the permu- 0 tation on the ith column such that if q moves the entry in the box (k, n) to (k , n), then qi 0 moves the entry in the box (k, i) to (k , i). We deﬁne q¯ to be q¯ = q1q2 ··· qn−1 and consider qq¯, clearly qq¯ ∈ Qλ, q¯ ∈ Qλ−c and q¯ ∈ Qλ. Since qe(λ − c) = e(λ − c)q and qq¯ = qq¯ , we have qe(λ − c)e(λ) = (−1)sg(q)e(λ − c)e(λ). Let d be another corner squares of λ, since d must be below c and the left of c, q¯ ∈ Qλ−c−d and the results follows.

Proposition 3.1.41. Let b a corner of λF,1 and let a be a square directly above of b. Then e(λ − a − b)e(λ − b)e(λ) is divisible on the left by id − (xy) for all pairs x and y of entries in a and b. Proof. We let e(λ − b)e(λ) = ∑ kαsα1e(λ). By Proposition 3.1.38, we only need to consider factors sα1 moving elements in the entries of a and b. Since these entries are consecutive, they are in the same column in strictly increasing order. If λα is a standard tableau, sα1 must be the identity, and hence e(λ − b)e(λ) = kide(λ). Observe that for all pairs x and y of entries in a and b, the transposition (xy) is a q in a. From the above proposition e(λ − a − b)e(λ) = −qe(λ − a − b)e(λ). It follows that e(λ − a − b)e(λ) is divisible on the left by id − (xy) for all pairs x and y in the entries of a and b. 3.1 A minimal free complex. 51

Observe that b is a corner of λF,1 and a is directly on the left of b if, and only if b˜ is a ˜ corner of λG,1 and a˜ is directly above of b, and vice versa. Assume that 1. or 2. holds for λF,t, from Proposition 3.1.40 we can write e(λ − a − b)e(λ − b)e(λ) = ∑x,y(1 + (xy))Ax,y, where Ax,y are elements of K(Sk+d(t−1)) indexed over pairs x and y of entries of a and b. By Proposition 3.1.41, for each pair x˜ and y˜ of entries in a˜ and b˜ we can ﬁnd Bx,y such that e(λ˜ − a˜ − b˜)e(λ˜ − b˜)e(λ˜ ) = (1 − (x˜y˜)Bx,y. At this moment we are in disposition to prove Lemma 1. With the above expressions we have eˆ(λ − a − b)eˆ(λ − b)eˆ(λ) = e(λ − a − b)e(λ − ˜ ˆ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ b)e(λ)e(λ − a˜ − b)e(λ − b)e(λ) = ∑x,y(id + (xy))Ax,ye(λ − a˜ − b)e(λ − b)e(λ) = ∑x,y(id + (xy))Ax,y(id − (x˜y˜))Bx,y. Applying φ∗, it follows from Proposition 3.1.26 that φ∗(eˆ(λ − a − b)eˆ(λ − b)eˆ(λ)) = φ∗(∑x,y(1 + (xy))Ax,y(1 − (x˜y˜))Bx,y) = φ∗(∑x,y(1 + (xy))(1 − (xy))Ax,yBx,y)) = 0, where we have used that (id − (xy))∗ = id − (xy).

3.1.4 The proof of Lemma 3.1.32. We may assume that b is below a, the other case is similar. The proof of Lemma 3.1.32 is basically the same as Lemma 3.1.31, however the expressions involving the idempotents are signiﬁcantly more complicated as before. The three possible arrangements for a and b in λF,t are 1. 2. a 3. a a b b b We start estimating the coefﬁcient of id.

Proposition 3.1.42. Let e(λ − a)e(λ) = ∑ kαsα1e(λ). Then, kid = 1. Proof. Since id does not move any element of λ − b − a, by Proposition 3.1.38 we have 1 j = k . Remember j is k times the coefficient of id in e(λ − a)e(λ) = E − E . id id id λF,1 κλκλ−a λ a λ 1 0 0 In fact, j is times the coefficient of id in E − E . If Id = pqp q where p ∈ P − , q ∈ id κλ−a λ a λ λ a 0 0 0 Qλ−a, p ∈ Pλ and q ∈ Qλ, then since pq does not act over the entries of a, necessarily p and q0 cannot permute the entries of a either. So, Id = pqp0q0 if, and only if Id = pqρσ, where ρ ∈ Pλ−a and σ ∈ Pλ−a. This means that the coefficient of id in Eλ−aEλ equals to κλ−a the coefficient of id in E − E − = κ − E − , which is κ − . We obtain that j = = 1. λ a λ a λ a λ a λ a id κλ−a Proposition 3.1.43. e(λ − a − b)e(λ − b)e(λ) = e(λ − a − b)e(λ) Proof. Let e(λ − b)e(λ) = ∑ kαsα1e(λ), by Proposition 3.1.38 we only have to consider permutations sα1 which move only the entries in a and b, and we know that jα = kα. Let sα1 of that such and λα. Since λα is standard and the entries in a and b are in increasing order, if sα1 is not the identity, then at least one entry from a must go to a position in b 0 in λα. Let j and j be the columns of a and b respectively. If sα1 moves the entry of a in the box (i, j) to the box (i0, j0) in b, then the entries in the boxes (i, j)) and (i, j0) are in the same row in λF,t and in the same column of λα. Considering that b is below a and to the left of a, the entry in the box (i, j0) is not in b. Therefore, there are two elements in the 52 A minimal free complex associated to the minors of a matrix.

same column in λα and in the same row in λF,t, and neither of these is an entry of b. By Proposition 3.1.37 jα = 0 = kα and hence, the only term which contributes to the product ∑ kαsα1e is id. We have, e(λ − a − b)e(λ − b)e(λ) = e(λ − a − b)kide(λ) = e(λ − a − b)e(λ), since as we have just seen before kid = 1. However, the calculation of e(λ − a − b)e(λ − a)e(λ) is considerably more complicated. In this case one has e(λ − a)e(λ) = ∑ kαsα1e(λ) and again by Proposition 3.1.38 we only have to consider permutations sα1 which only move the entries of a and b. Nevertheless, we cannot apply the last proposition in that case. In fact, it is true that there are two elements in the same column in λα and in the same row in λF,t but, one of this elements is an entry of a and hence we cannot apply Proposition 3.1.37. At the beginning, there are nontrivial permutations sα1 with kα , 0 and we will have to ﬁnd these values. For convenience we denote the entries of a and b by 1, . . . , t + 1, where 1 denotes the entry in whichever a or b is a single square.

Proposition 3.1.44. Let s and s0 be permutations of 1, . . . , t + 1 such that s(1) = s0(1). We have:

0 (i) e(λ − a − b)se(λ) = (−1)s+s e(λ − a − b)s0e(λ).

0 (ii) If j is the coefﬁcient of s in e(λ − a)e(λ) and j0 of s0, then j = (−1)s+s j0.

Proof. (i) Let q = s−1s0, since s(1) = s0(1) we are in the hypothesis of Proposition 3.1.40. −1 0 Then, qe(λ − a − b)e(λ) = (−1)sg(q)e(λ) = (−1)sg(s )+sg(s )e(λ − a − b)e(λ) and the result follows. 0 sg(q) (ii) We have s = sq. Since q ∈ Qλ we know that e(λ)q = (−1) e(λ). Then, the coefﬁcient of s in e(λ − a)e(λ) is the coefﬁcient of s0 in (−1)sg(q)e(λ − a)e(λ).

Let e(λ − a)e(λ) = ∑ kαsα1e(λ). If sα,1 only moves 1, 2, . . . , t + 1, then sα1 is character- ized by the image of 1 and we know that jα = kα. Let sα1(1) = r , 1, from the above sg(s )−1 proposition considering (1r), e(λ − a − b)sα1e(λ) = (−1) α1 e(λ − a − b)(1r)e(λ) and sg(s )−1 the coefficient of sα1 equals to (−1) α1 times the coefficient of (1r) in e(λ − a)e(λ). sg(s )−1 We have e(λ − a − b)e(λ − a)e(λ) = ∑ kαe(λ − a − b)sα1e(λ) = ∑ kα(−1) α1 e(λ − a − t+1 b)(1r)e(λ) ∑ kre(λ − a − b)(1r)e(λ) = e(λ − a − b)(1 + ∑r=2 kr(1r))e(λ). 0 Proposition 3.1.45. If r and r are the entries in the t string of boxes, then kr = kr0 . Proof. We can write (1r0) = (rr0)(1r)(rr0). If r and r0 are in the nth column in the 0 positions (j, n) and (j , n), for i = 1, . . . , n − 1 we define qi to be the transposition on the ith 0 column which transposes the elements in the positions (j, i) and (j , i). Let q¯ = q1 ··· qn−1, we note that q¯2 = Id. Since (1r0) = (rr0)q¯q¯(1r)(rr0) = q¯(rr0)(1r)q¯(rr0) and q¯(rr0) commutes with e(λ − a) and e(λ), we have q¯(rr0)e(λ − a)e(λ)q¯(rr0) = e(λ − a)e(λ)[q¯(rr0)]2 = e(λ − a)e(λ). The coefficient of (1r0) in e(λ − a)e(λ) is the coefficient of (1r).

Proposition 3.1.46. Suppose a is a column of t squares. Let λ0 be the tableau obtained from λ by removing the last t − 1 entries of a. Then, j(12) is κλ0 times the coefficient of (12) in e(λ0 − a0)e(λ0). 0 0 1 1 Proof. Since λ − a = λ − a, = and the coefficient of (12) in Eλ−aEλ is the κλ−a κλ0−a0 coefficient in Eλ0−a0 Eλ. It is sufficient to see that the coefficient of (12) in Eλ−aEλ is the 3.1 A minimal free complex. 53

0 0 0 0 coefficient of (12) in Eλ0−a0 Eλ0 . If (12) = p q pq ∈ Eλ0−a0 Eλ, since p q and (12) do not act on the last entries of a, then pq does also and clearly pq ∈ Eλ0 . Hence, the coefficient of (12) in E(λ − a)E(λ) is the coefficient of (12) in E(λ0 − a0)E(λ0). From the above propositions if a is a column of t squares, we can write e(λ − a − t+1 b)e(λ − a)e(λ) = e(λ − a − b)(1 + k2 ∑r=2(1r))e(λ) and when b is a column of t squares, t+1 e(λ − a − b)e(λ − a)e(λ) = e(λ − a − b)(1 + kt+1 ∑r=2(1r))e(λ).

First we prove Lemma 3.1.32 in the case 3. when a and b are both single squares. Next, we will see that we can reduce the other cases to this. We need one more result which we can find in [18] Proposition 13, it is based on Young’s Semiformals representation of K(Sn). First a simple definition. Definition 3.1.47. The axial distance between two boxes x and y of a standard Young tableau T is the number of steps to get from x to y, where steps contribute with +1 when going down or to the left or −1 when going up or to the right. The axial distance does not depend on the path from x to y. Proposition 3.1.48. Let a and b be two corners of λ, with b below a. Then, the coefficient of (ab) in e(λ − a)e(λ) is 1 , where η is the axial distance from a to b in λ. Hence we (kλη ) 1 have k(ab) = j(ab) = η . Let η be the axial distance from a to b, we have e(λ − a − b)e(λ − a)e(λ) = e(λ − a − 1 b)[1 + η (ab)]e(λ) and e(λ − a − b)e(λ − b)e(λ) = e(λ − a − b)e(λ). In the same manner, e(λ˜ − a˜ − b˜)e(λ˜ − a˜)e(λ˜ ) = e(λ˜ − a˜ − b˜) and e(λ˜ − a˜ − b˜)e(λ˜ − b˜)e(λ˜ ) = e(λ˜ − a˜ − b˜)[1 + 1 ˜ ˜ η (a˜b)]e(λ). Since permutation (ab) commutes with e(λ − a − b) and (a˜b˜) with e(µ − a˜ − b˜), we 1 ˜ ˜ ˜ can write eˆ(λ − a − b)eˆ(λ − a)eˆ(λ) = [1 + η (ab)]e(λ − a − b)e(λ)e(λ − a˜ − b)e(λ) and 1 ˜ ˜ ˜ ˜ eˆ(λ − a − b)eˆ(λ − b)eˆ(λ) = e(λ − a − b)e(λ)[1 + η (a˜b)]e(λ − a˜ − b)e(λ). Applying φ∗, 1 ˜ ˜ ˜ φ∗(eˆ(λ − a − b)eˆ(λ − b)eˆ(λ)) = φ∗(e(λ − a − b)e(λ)[1 + η (a˜b)]e(µ − a˜ − b)e(λ) = φ∗([1 + 1 ˜ ˜ ˜ η (ab)]e(λ − a − b)e(λ)e(λ − a˜ − b)e(λ) = φ∗(eˆ(λ − a − b)eˆ(λ − a)eˆ(λ)). This completes the proof of Lemma 3.1.32 in case 3.

Suppose now a is a column of t squares, as we have seen e(λ − a − b)e(λ − a)e(λ) = e(λ − a − b)(1 + k2 ∑(1r))e(λ). From Proposition 3.1.46 k2 is κλ0 times the coefﬁcient of 0 0 0 (12) in e(λ − a)e(λ ), where λ is the tableau obtained from λ f ,t by removing the squares 3, . . . , t + 1. Considering λ0, a0 and b, we are in the hypothesis of Proposition 3.1.48, and 1 0 hence k2 is η where η is the axial distance from a to b which is, in fact, η0 + t − 1, where η0 is the axial distance from a to b in λF,1. Therefore, eˆ(λ − a − b)eˆ(λ − a)eˆ(λ) = 1 t+1 ˜ ˜ ˜ [1 + η ∑r=2(1r)]e(λ − a − b)e(λ)e(λ − a˜ − b)e(λ). ˜ In λG,t a˜ is below b, with a˜ a column of t squares and b a single square. As we have ˜ ˜ ˜ ˜ ˜ ]t+1 ˜ ˜ ˜ ˜ 0 seen before, e(λ − a˜ − b)e(λ − b)e(λ) = [1 + k2˜ ∑r˜=2(1r˜)]e(λ − a˜ − b)e(λ). If we consider λ to be the tableau obtained from λG,t by removing the last t − 1 squares from a˜, then k2˜ is κλ˜ 0 ˜ 0 0 ˜ 0 1 times the coefﬁcient of (12) in e(λ − a˜ )e(λ ), which is η˜ and where η˜ is the axial distance 0 from b˜ to a˜ , that is, the axial distance from b˜ to the top square of a˜. Clearly η˜ = η0 + t − 1 = 54 A minimal free complex associated to the minors of a matrix.

˜ η, since the axial distance distance from b to a˜ in λG,1 is the axial distance from a to b in 1 ˜ ˜ ˜ λF,1. Therefore, eˆ(λ − a − b)eˆ(λ − b)eˆ(λ) = e(λ − a − b)e(λ)[1 + η ∑(1r˜)]e(λ − a˜ − b)e(λ). 1 ∗ 1 Since [1 + η ∑(1r˜)] = [1 + η ∑(1r˜)], we have φ∗(eˆ(λ − a − b)eˆ(λ − b)eˆ(λ)) = φ∗(e(λ − a − 1 ˜ 1 ˜ ˜ ˜ b)e(λ)[1 + η ∑(1r˜)]e(µ − a˜ − b)e(µ)) = φ∗([1 + η ∑(1r)]e(λ − a − b)e(λ)e(λ − a˜ − b)e(λ) = φ∗(eˆ(λ − a − b)eˆ(λ − b)eˆ(λ)). When a is a single square and b is a column of t squares, the proof is exactly the same as in the previous case considering a simple modiﬁcation of Proposition 3.1.46.

Proposition 3.1.49. Suppose b is a column of t squares. Let λ0 be the tableau obtained from λ by removing the ﬁrst t − 1 entries of b. Then, j(1[t+1]) is κλ0 times the coefﬁcient of (1[t + 1]) in e(λ0 − b0)e(λ0).

3.2 Determinantal ideals and determinantal varieties.

Let K be an algebraically closed ﬁeld of characteristic zero, let Ps be the s-dimensional projective space over K, R = K[x0,..., xs] and m = (x0,..., xs) its homogeneous maximal ideal. Let m ≥ n be positives integers.

Deﬁnition 3.2.1. Let A be a commutative ring, let P ⊂ A be a prime ideal and let I ⊂ A be an ideal. The codimension of P is the maximum of the lengths of chains of prime ideals descending from P. We deﬁne the codimension or the height of I to be the minimum of the codimension of primes ideals containing I. We denote it by ht(I).

Deﬁnition 3.2.2. Let A be a commutative ring, let r1,..., rn ∈ A and let I ⊂ A be a proper ideal. We say that r1,..., rn is a regular sequence or A-sequence if (r1,..., rn) , A and for each i ∈ {1, . . . , n}, ri , Ann(A/(r1,..., ri−1)). We deﬁne the grade or depth of I to be the maximal length of a A-sequences contained in I. We denote it by depth(I).

In the particular case A = R, then if I ⊂ R is a proper ideal, ht(I) = depht(I).

Deﬁnition 3.2.3. Let (ri,j) be a m × n matrix with entries in R and let 1 ≤ t ≤ n. We say that (ri,j) is a t-homogeneous matrix if the k × k minors of (ri,j) are homogeneous polynomials for all k ≤ t. We say that (ri,j) is homogeneous matrix if all the minors of (ri,j) are homogeneous polynomials. If (ri,j) is t-homogeneous, then for each k ≤ t we deﬁne Ik to be the ideal generated by the k × k minors of (ri,j).

A classical result from Eagon-Northcott [7] gives us an estimation of the height of It.

Theorem 3.2.4. Let A be a commutative ring and let (ri,j) be a t-homogeneous matrix of size m × n with entries in A. Then, for any r ≤ t the ideal Ir has ht(Ir) ≤ (m − t + 1)(n − t + 1). Moreover, if A = R for any r ≤ t, depth(Ir) ≤ (m − t + 1)(n − t + 1).

Deﬁnition 3.2.5. Let (ri,j) be a t-homogeneous matrix of size m × n with entries in R and let k ≤ t. We say that Ik is determinantal if it has maximal height, that is, ht(Ik) = (m − k + 1)(n − k + 1).

Deﬁnition 3.2.6. Let J ⊂ R be a homogeneous ideal of R and let X ⊂ Ps be a projective variety. We say that J is a determinantal ideal if there is a t−homogeneous m × n matrix 3.2 Determinantal ideals and determinantal varieties. 55

(ri,j) with entries in R such that J = It. If J is determinantal, we say that J is a standard determinantal ideal if t = n and we say that J is a linear standard determinantal if J is standard determinantal and all entries of (ri,j) are linear forms. Finally, we say that X is a determinantal variety if its homogeneous ideal is determinantal.

Let us see a signiﬁcant example of determinantal ideals. Let m and n be positive integers and let R = K[Xi,j, 1 ≤ i ≤ m, 1 ≤ j ≤ n]. We call generic matrix to the m × n matrix of indeterminates (Xi,j), note that (Xi,j) is a homogeneous matrix of size m × n. We have,

Theorem 3.2.7. For each t ≤ min{m, n}, ht(It) = (m − t + 1)(n − t + 1). The above theorem is a particular case of a more general result found in [17] Theorem 2.5. Directly from Theorem 3.2.7, It is a determinantal ideal for all t ≤ min{m, n}. In fact, {It, t ≤ min{m, n} − 1} are the homogeneous ideal deﬁning a historically important class of determinantal varieties. Let M be the set of all m × n matrices over K and let Pmn−1 be the projective space associated to M. For each t ≤ min{m, n} − 1 we let Mt ⊂ M be the subset of matrices of rank t or less. Observe that Mt = {Z = [Zi,j, 1 ≤ i ≤ mn−1 m, 1 ≤ j ≤ n] ∈ P | rg((Xi,j)(Z)) ≤ t}, where (Xi,j)(Z) denotes the matrix (Zi,j). Thus, one easily can see that Mt is just the projective variety deﬁned by It+1. We call Mt generic determinantal variety. Moreover, as we can see in [12] Lecture 9, M1 is the Segre variety, M2 is the chordal variety of the Segre variety and more generally, Mt is the union of the secant (t − 1)-planes to the Segre variety for t > 2.

We end this section with another large family of determinantal ideals. Let R = K[x1,..., xs] and let (ri,j) be a matrix of linear forms in R.

Deﬁnition 3.2.8. A generalized row of (ri,j) is a nonzero linear combination of the rows of (ri,j).A generalized column of (ri,j) is a nonzero linear combination of the columns of (ri,j).A generalized entry of (ri,j) is a nonzero linear combination of the entries of some generalized row or a nonzero linear combination of some generalized column. We say that (ri,j) is 1-generic if every generalized entry of (ri,j) is nonzero. For example, the generic matrices are 1-generic. The homogeneous ideal deﬁned by the maximal minors of a 1-generic matrix is determinantal. More precisely,

Proposition 3.2.9. Let m ≥ n. If (ri,j) is a 1-generic matrix of size m × n in R, then ht(In) = m − n + 1. The proof of the above proposition is found in [8] Appendix 6B, Theorem 6.4. However the definition of 1-generic matrix seems sophisticated, there are well known examples of determinantal varieties defined by the maximal minor of a 1-generic matrix. One of this examples is the rational normal curve. As we can see in [12] Example 9.3 the rational normal curve X ⊂ Ps is the variety defined by the maximal minors of the 1-generic matrix x0 x1 ··· xs−1 x1 x2 ··· xs and hence X is a determinantal variety. 56 A minimal free complex associated to the minors of a matrix.

3.2.1 A minimal free resolution of determinantal ideals. Let K be an algebraically closed ﬁeld of characteristic zero, let Ps be the s-dimensional projective space over K, R = K[x0,..., xs] and m = (x0,..., xs) its homogeneous maximal ideal Let n, m and t be positives integers such that n ≤ m and let F and G be free R- modules of ranks m + t − 1 and n + t − 1 with basis { f1,..., fm+t−1} and {g1,..., gn+t−1} ∗ ∗ ∗ respectively. We denote G = Hom(G, R) the dual of G with basis {g1,..., gn+t−1} dual to the basis {g1,..., gn+t−1}. Let (ri,j) be a (n + t − 1) × (m + t − 1) homogeneous matrix ∗ with entries in R, we consider the ideal It. (ri,j) deﬁnes a R-map Φ : F → G sending fi to n+t−1 ∗ ∑j=1 ri,jgj , i = 1, . . . , m + t − 1. We note (mn) = (n,..., n), such that n + ··· + n = nm, and we denote λ(mn) = λid, in the same manner we consider (m + t − 1, n) and λ((m+t−1)n). We present an outline explanation of the work by Lascoux in [16]. We have based partially on notations used in [19] and [13] and our main purpose is connected it with the complex C•(Φ, t) and the Schur theory which we have developed. We start with a proposition which allows us to describe the modules of the resolution plainly.

Proposition 3.2.10. Let I = (i1,..., im+t−1) be a partition of weight k + d(t − 1) such that I ⊂ (m + t − 1, n) and d is the Durfee square of I. Then, 0 (i) If i˜d ≤ d + t − 1, we let I = 0. ˜ ≥ + − 0 = (˜ − + ˜ − + ˜ ˜ ) + (ii) If id d t 1, we let I i1 t 1, . . . , id t 1, d,..., d, id+1,..., iim+t−1 , d ··· + d = (t − 1)d. L Definition 3.2.11. ≤ k ≤ mn L = 0 L F ⊗ L G For each 1 we define k,t |I|−n(I)=k,I ,0 I˜ Ie0 , where 0 I = (i1,..., im) ⊂ (m + t − 1, n), n(I) = d(t − 1), d is the Durfee square of I and I is the partition describe in Proposition 3.2.10. Let L0,t = R. Let I be a partition with Durfee square d such that |I| ⊂ (m + t − 1, ), |I| = k + n(I) = 0 k + d(t − 1) and I , 0. Since i˜d ≥ d + t − 1, let λ˜ be the partition of terms λ˜ 1 = i˜1 − t + ˜ = ˜ − + ˜ = ˜ ˜ = ˜ 1, . . . , λd id t 1, λd+1 id+1,..., λim+t−1 iim+t−1 . Considering Definition 3.1.20 we see that λ is a partition of weight k + d(t − 1) and Durfee square d such that I = λ(F, t), ˜ 0 ˜ and hence I˜ = λ(F, t) and Ie = λ(G, t). Therefore, Lk,t = Ck(Φ, t). Definition 3.2.12. For any pairs of partitions (I, H) such that |I| − n(I) = k and |H| − n(H) = k − I0 H0 ψ L F ⊗ L G → L F ⊗ 1 with and nonzero we define a morphism I,H : I˜ Ie0 H˜ L G as follows, Hf0 (i) If H * I, let ψI,H = 0. (ii) Assume H ⊂ I, as we have seen I˜ is of the form λ˜ F,t = (λ˜ 1 + t − 1, . . . , λ˜ d + t − ˜ ˜ ) = ( ) ˜ 1, λd+1,..., λλ1 for a partition λ λ1,..., λm of weight k with Durfee square d. H is the shape of a tableau of the form λF,t − a where a is a corner of λF,1. Then, λF,t and λF,t − a coincide except in one square or a string of t squares in which case, I˜ and H˜ coincide 0 except in one term ij and hj such that ij − hj = 1 or ij − hj = t. The same is true for Ie and 0 0 Hf. Let q = im, we denote I˜ = (i1,..., il, i, il+1,..., iq), H˜ = (i1,..., il, h, il+1,..., iq), Ie = 0 0 0 0 0 (j1,..., jr, i , jr+1,..., js) and Hf = (j1,..., jr, h , jr+1,..., js). Denoting ρ = i − h = i − h , by the Pieri formulas (Corollary 2.5.27) we have injections

ρ ρ δ : L F → L F ⊗ Λ F and δ 0 : L G → L G ⊗ Λ G I,ρ I˜ H˜ I ,ρ Ie0 Hf0 3.2 Determinantal ideals and determinantal varieties. 57 which induce, together with the contraction ΛρF ⊗ ΛρG → R induced by Φ, a morphism ψ : L F ⊗ L G → L F ⊗ L G. I,H I˜ Ie0 H˜ Hf0

Deﬁnition 3.2.13. For each 1 ≤ k ≤ mn we deﬁne a morphism dk : Lk,t → Lk−1,t to be L ψI,H.

Proposition 3.2.14. d1(L1,t) = It. Proof. We are looking for partitions I˜ of Durfee square d such that |I˜| = 1 + d(t − 1) 2 and i˜1 ≥ d + t − 1. This last condition implies that |I˜| ≥ d(d + t − 1) = d + d(t − 1), then necessarily d = 1. Therefore, |I˜| = t and i˜1 ≥ t, and hence I˜ must be the line t t partition I˜ = (t). We obtain L1,t = Λ F ⊗ Λ G and d1 must be the natural contraction map t ⊗ t → ∧ · · · ∧ ⊗ ∧ · · · ∧ ( ) Λ F Λ G R sending fi1 fit gj1 gjt to the minors of ri,j corresponding to the i1,..., it rows and j1,..., jt columns, where 1 ≤ i1 < ··· < it ≤ m + t − 1 and 1 ≤ j1 < ··· < jt < n + t − 1.

Theorem 3.2.15. If ht(It) = mn, then L•,t is a minimal free resolution of It. We ﬁnish this chapter describing classical examples of minimal free resolutions of determinantal varieties. We start with the minimal resolution of the ideal generated by the maximal minors.

The Eagon-Northcott Complex.

We analyze the case n = 1, m ≥ 1 and t ≥ 1. Now (ri,j) is a matrix of size (m + t − 1, t) and It is the ideal of R deﬁned by the maximal minors of (ri,j). Assuming ht(It) = m, (L•,t, d•) is a minimal free resolution of It. First we describe the modules Lk,t. Since n = 1, each partition I ⊂ (m + t − 1, 1) has Durfee Square 1. Therefore, for each 1 ≤ k ≤ m there is only one partition I ⊆ (m + t − 1, 1) such that |I| = t − 1 + k, we have I = (1, . . . , 1), 1 + ··· + 1 = t − 1 + k and I˜ = (k + t − 1). Given that k + t − 1 ≥ t, from Proposition 3.2.10 I0 = (k, 1, . . . , 1), 1 + ··· + 1 = t − 1 and then Ie0 = (t, 1, . . . , 1), 1 + ··· + 1 = k − 1. We have, for 1 ≤ k ≤ m

k+t−1 Lk,t = L(k+t−1)F ⊗ L(t,1,...,1)G = Λ F ⊗ L(t,1,...,1)G.

Let us to examine the boundary maps dk : Lk,t → Lk−1,t, for convenience we denote 1, . . . , 1 such that 1 + ··· + 1 = s by 1s. Let 2 ≤ k ≤ m, we have partitions I˜ = (k + t − 1), H˜ = (k + ∆ t − 2), Ie0 = (t, 1k−1) and Hf0 = (t, 1k−2). By the Pieri formula we have injections Λk+t−1F −→ k+t−2 Λ F ⊗ F and L(t,1k−1)G → L(t,1k−2)G ⊗ G, where ∆ is the appropriated component of the diagonal map. The second injection is the composition L(t,1k−1)G K(k,1t−1)G ,→ − → − ⊗ → − ⊗ ( ∧ ∧ · · · ∧ ⊗ ⊗ · · · ⊗ Λ(t,1k 1)G Λ(t,1k 2)G G L(t,1k 2)G G sending dI˜0 gi1 g2 gt gi2 gik = < ≤ · · · ≤ < k−1 ····· ····· ⊗ ⊗ · · · ⊗ ⊗ such that 1 i1 i2 ik−2 t to ∑u=1 gi1 gîu gik−1 g2 gt giu . ∧ · · · ∧ ≤ < ··· < ≤ And the first injection sends fj1 fjk+t−1 such that 1 j1 jk+t−1 m to k+t−1(− )s+1 ∧ · · · ∧ ˆ ∧ · · · ∧ ⊗ ˆ ∑s=1 1 fj1 fjs fk+t−1 fjs , where fis means that this element is not ( ( ∧ · · · ∧ ⊗ ∧ ∧ · · · ∧ in the exterior product. Hence, dk is defined by dk dI˜0 fj1 fjk+t−1 gi1 g2 ⊗ ⊗ · · · ⊗ ) = gt gi2 gik k−1 k+t−1 r (−1)s+1 f ∧ · · · ∧ fˆ ∧ · · · ∧ f ⊗ g ····· gˆ ····· g ⊗ g ⊗ · · · ⊗ g . ∑ ∑ u,ik j1 js k+t−1 i1 iu σ(ik−1) 2 t u=1 s=1 58 A minimal free complex associated to the minors of a matrix.

This is, in fact, the Eagon-Northcott complex (see [7]). Indeed, the Pieri formula gives k−1 k k us an isomorphism L(t,1k−1)G S G ⊗ Λ G. Since G has rank k, it follows that Λ G R k+t−1 k−1 and then, Lk,t Λ F ⊗ S G. Under this isomorphism we can describe more easily the boundary maps dk. In this case, dk is given by (( ∧ · · · ∧ ) ⊗ ( j1 ····· jt ) = dk fi1 fik+t−1 g1 gt

k+t−1 − ( (− )p+1r f ∧ · · · ∧ fˆ ∧ · · · ∧ f ) ⊗ gj1 ····· gjk 1 ····· gjt ∑ ∑ 1 k,ip i1 ip ik+t−1 1 k t k∗ p=1 where the sum runs over all values k such that jk > 0. This is the original form of the minimal resolution gives by Eagon-Northcott. The original construction of the resolution does not involve Schur functors and it is made over a commutative ring R of any characteristic.

Theorem 3.2.16. If ht(It) = m, then a minimal free resolution of R/It is given by

m+t−1 m−1 dm m+t−2 m−2 t d1 (L•,t, d•) : 0 → Λ F ⊗ S G −→ Λ F ⊗ S G →· · ·→ Λ F −→ R → R/It → 0.

The minimal free resolution of the rational normal curve is given by the Eagon-North- 7 cott complex. For example, consider K = [x0,..., x7] and the projective space P . The rational normal curve X of P7 is the variety is given by the ideal the maximal minors of the 2 × (6 + 2 − 1) matrix: x x ··· x 0 1 6 . x1 x2 ··· x7

7 2 k+1 k−1 So, F = R (−1) and G = R . We have, for 1 ≤ k ≤ 6, Lk,2 = Λ F ⊗ S G. Since k+1 7 k−1 2+k−2 rank(Λ F) = (k+1) and rank(S G) = ( k−1 ) = k, explicitly: 6 6 7 5 35 21 4 L6,2 = R(−7) ⊗ R R (−7), L5,2 = R (−6) ⊗ R R (−6), L4,2 = R (−5) ⊗ R 84 35 3 105 35 2 70 R (−5), L3,2 = R (−4) ⊗ R R (−4), L2,2 = R (−3) ⊗ R = R (−3), L1,2 = R21(−2) ⊗ R R21(−2). The resolution of the rational normal curve of P7 looks like: 0 → R6(−7) → R35(−6) → R84(−5) → R105(−4) → R70(−3) → R21(−2) → R → R/It → 0

The Gulliksen-Negard complex.

Now we will construct the minimal free resolution of determinantal ideals given by the submaximal minors of a square matrix, in particular the case m = n = 2 and t ≥ 1. Let (ri,j) be a matrix of size t × t. There are 4 free R-modules involved in the complex, to ﬁnd them we have to describe partitions I˜ = (i˜1, i˜2) of Durfee square d = 1 or d = 2 such that i˜d ≥ d + t − 2 and i˜1 + i˜2 = k + d(t − 2). We distinguish two cases according to the Durfee square:

(1) When d = 1, we obtain conditions i˜1 ≥ t − 1, i˜1 + i˜2 = k + t − 2 and i2 ≤ 1

(2) When d = 2, we obtain conditions i˜2 ≥ t and i˜1 + i˜2 = k + 2t − 4 and i2 ≥ 2. 3.2 Determinantal ideals and determinantal varieties. 59

For k = 2 and d = 2, since i˜1 = 2t − 2 − i˜2 ≤ t − 2 ≤ i1, there are no partitions I˜. If d = 1, since i˜2 = t − i1 and i1 ≥ t − 1, there are two partitions I˜ satisfying the above ˜ ˜ t conditions. We have I = (t − 1, 1) and I = (t, 0). In this case L2,t−1 = L(t−1,1)F ⊗ Λ G ⊕ t Λ F ⊗ L(t−1,1)G.

For k = 3 and d = 2, i˜1 = 2t − 1 − i˜2 and i˜2 ≥ t implies i˜1 ≤ t − 1 < i˜2, and hence (i˜1, i˜2) is not a partition. Otherwise, if d = 1 it results i˜1 ≥ t − 1 and i˜2 = t + 1 − i˜1 which 0 implies i2 = 1 and i2 = 2. Given that i2 ≤ 1, it follows that I˜ = (t, 1) and Ie = (t, 1). Therefore L3,t−1 = L(t,1)F ⊗ L(t,1)G.

For k = 4 and d = 1, i˜2 = t + 2 − i˜1 and i1 ≥ t − 1. Since i˜2 ≤ 1, we obtain the following possibilities (t + 2, 0) and (t + 1, 1), but i˜1 is the length of I ⊂ (t, 2). So, in this cases there are no partitions I˜. When d = 2, conditions says i˜1 = 2t − i˜2 and i˜2 ≥ t. There is only one ˜ ˜0 partition I = (t, t), and hence I = (t, t). Thus, L4,t−1 = L(t,t)F ⊗ L(t,t)G. The complex looks like

t t 0 → L(t,t)F ⊗ L(t,t)G → L(t,1)F ⊗ L(t,1)G → L(t−1,1)F ⊗ Λ G ⊕ Λ F ⊗ L(t−1,1)G →

→ Λt−1F ⊗ Λt−1G → R → 0.

t−1 t−1 It remains to describe the boundary maps d•. Let ϕ : Λ F ⊗ Λ G → R be the contraction map, d4 : L(t,t)F ⊗ L(t,t)G → L(t,1)F ⊗ L(t,1)G is the composition of the injections t−1 t−1 L(t,t)F → L(t,1)F ⊗ Λ F and L(t,t)G → L(t,1)G ⊗ Λ G followed by ϕ. One can easily i+j see that dk is deﬁned by sending d(t,t)( f1 ∧ · · · ∧ ft ⊗ f1 ∧ · · · ∧ ft) to ∑i,j(−1) ϕ( f1 ∧ · · · ∧ fˆi ∧ · · · ∧ ft ⊗ g1 ∧ · · · ∧ gˆj ∧ · · · ∧ gt) f1 ∧ · · · ∧ ft ⊗ fi ⊗ g1 ∧ · · · ∧ gt ⊗ gj. Since t t Λ F R Λ G which are generated by f1 ∧ · · · ∧ ft and g1 ∧ · · · ∧ gt respectively and ˆ ϕ( f1 ∧ · · · ∧ fi ∧ · · · ∧ ft ⊗ g1 ∧ · · · ∧ gˆj ∧ · · · ∧ gt) is the determinant of (ri,j) with the ith row and the jth column deleted, denoting them by 1 and deti,j respectively, d4(d(t,t)(1, 1)) = i+j ∑i,j(−1) deti,j fi ⊗ gj ⊗ 1.

d3 is the sum of the maps ψ(2,1t−1),(2,1t−2) and ψ(2,1t−1),(1t) as in Deﬁnition 3.2.12. The t j+1 ˆ injection L(t,1)F → L(t−1,1)F ⊗ F maps d(t,1)( f1 ∧ · · · ∧ ft ⊗ fi) to ∑j=1(−1) f1 ∧ · · · ∧ fj ∧ t j+1 ˆ · · · ∧ ft ⊗ fi ⊗ fj if i , 1 and to 2 ∑j=1(−1) f1 ∧ · · · ∧ fj ∧ · · · ∧ ft ⊗ f1 ⊗ fj if i = 1. While t the injection L(t,1)G → Λ G ⊗ G sends d(t,1)(g1 ∧ · · · ∧ gt ⊗ gs) to g1 ∧ · · · ∧ gt ⊗ gs if s , 1 and to 2g1 ∧ · · · ∧ gt ⊗ g1 if s = 1. Therefore, the respective images of ψ((t, 1), (t − 1, 1)) of this elements are in each case: t j+1 ˆ t j+1 ˆ ∑j=1 rj,s(−1) f1 ∧ · · · ∧ fj ∧ · · · ∧ ft ⊗ fi ⊗ g1 ∧ · · · ∧ gt = ∑j=1 rj,s(−1) f1 ∧ · · · ∧ fj ∧ · · · ∧ ft ⊗ fi ⊗ 1 if i, s , 1. t j+1 ˆ 2 ∑j=1(−1) rj,1 f1 ∧ · · · ∧ fj ∧ · · · ∧ ft ⊗ fi ⊗ 1 if i , 1 and s = 1. t j+1 ˆ 2 ∑j=1(−1) f1rj,s ∧ · · · ∧ fj ∧ · · · ∧ ft ⊗ f1 ⊗ f1 ⊗ 1 if i = 1 and s , 1. t j+1 ˆ 4 ∑j=1(−1) f1rj,1 ∧ · · · ∧ fj ∧ · · · ∧ ft ⊗ f1 ⊗ f1 ⊗ 1 if i = 1 = s. Symmetrically, the map ψ(t−1,1),(t,1) is the same as the above map changing f by g. Fi- nally, d2 is the sum of maps ψ(2,1t−2),(1t−1) and ψ(1t),(1t−1). Observe that these two maps coincide with the boundary maps of the Eagon- Northcott complex. The complex (L•,t−1, d•) is called the Gulliksen-Negard complex originally presented in [11]. 60 A minimal free complex associated to the minors of a matrix.

Two concrete examples computed with Macaulay2.

Let R = K[x0,..., x5] and consider the following 2 × 5 matrix: x0 + x1 x1 + x2 x2 + x3 x3 + x4 x4 + x5 M = 2 2 2 2 2 . x1 x2 x3 x4 x5

The ideal I2 generated by the maximal minors of M is prime and ht(I2) = 4. Therefore, I2 is determinantal and the homogeneous ideal of the variety X = {z ∈ P5 | rg(M(z)) = 1} is I2. Hence X is determinantal and a minimal free resolution of I2 is given by the Eagon- Northcott complex:

d d 0 → R(−6) ⊕ R(−7) ⊕ R(−8) ⊕ R(−9) −→4 R5(−5) ⊕ R5(−6) ⊕ R5(−7) −→3

10 10 d2 10 d1 R (−4) ⊕ R (−5) −→ R (−3) −→ R → R/I2 → 0.

In the next pages we give the matrix associated to the boundary maps dk for k = 1, . . . , 4. 3.2 Determinantal ideals and determinantal varieties. 61

10 The matrix associated to d1 : R (−3) → R is the transpose of the matrix:

 3 2 2 2  x1 + x1x2 − x0x2 − x1x2   x2x + x2x − x x2 − x x2  1 2 1 3 0 3 1 3    x3 + x2x − x x2 − x x2   2 2 3 1 3 2 3    x2x + x2x − x x2 − x x2  1 3 1 4 0 4 1 4    2 + 2 − 2 − 2 x2x3 x2x4 x1x4 x2x4    3 2 2 2   x3 + x3x4 − x2x4 − x3x4     2 2 2 2 x x4 + x x5 − x0x − x1x   1 1 5 5  2 2 2 2 x x4 + x x5 − x1x − x2x   2 2 5 5   x2x + x2x − x x2 − x x2  3 4 3 5 2 5 3 5 3 2 2 2 x4 + x4x5 − x3x5 − x4x5

10 10 10 d2 : R (−4) ⊕ R (−5) → R (−3)

 −x2 − x3 −x3 − x4 0 0 −x4 − x5 0 0 0  + + − − + − −  x1 x2 x3 x4 x3 x4 0 x4 x5 x4 x5 0 0   −x0 − x1 0 0 −x3 − x4 0 0 −x4 − x5 0   0 x1 − x3 x2 + x3 0 −x4 − x5 x4 + x5 0 −x4 − x5   0 −x0 − x 0 x2 + x3 0 0 x + x5 0  1 4  0 x + x −x − x −x − x 0 0 0 0  0 1 0 1 1 2  + − +  0 0 0 0 x1 x4 x2 x4 0 x3 x4   0 0 0 0 −x0 − x1 0 x2 − x4 0   0 0 0 0 x0 + x1 −x0 − x1 −x1 − x2 0 0 0 0 0 −x0 − x1 x0 + x1 x1 + x2 −x0 − x1

2 2 2  0 0 x3 x4 0 0 x5 0 0 0 0 0 −x2 x2 x2  0 0 2 0 4 0 0 5 0 0 0 0  2 2 2  0 0 x1 0 0 x4 0 0 x5 0 0 0  2 2 2  0 0 0 −x2 −x3 0 0 0 0 x5 0 0  2 2 2  −x − x5 0 0 x 0 −x 0 0 0 0 x 0  4 1 3 5  0 −x − x 0 0 x2 x2 0 0 0 0 0 x2  4 5 1 2 5  −x2 −x2 −x2  0 0 0 0 0 0 2 3 0 4 0 0  + 2 − 2 − 2  x3 x4 0 0 0 0 0 x1 0 x3 0 x4 0  2 2 2  0 x3 + x4 0 0 0 0 0 x1 x2 0 0 −x4  2 2 2 −x1 − x2 −x2 − x3 0 0 0 0 0 0 0 x1 x2 x3 62 A minimal free complex associated to the minors of a matrix.

5 5 5 10 10 d3 : R (−5) ⊕ R (−6) ⊕ R (−7) → R (−4) ⊕ R (−5)

 2 2 x3 + x4 x4 + x5 0 0 0 x4 x5  −x − x −x − x x + x −x2 −x2  2 3 4 5 4 5 0 0 3 5  − − − + 2 − 2 − 2  x1 x3 x4 x5 0 x4 x5 0 x2 x3 x5  2  −x0 − x1 0 0 0 x4 + x5 −x1 0  2 2  0 −x2 + x −x3 − x 0 0 0 −x + x  4 4 3 4  0 x + x 0 −x − x 0 0 x2 − x2 + x2  1 4 3 4 2 3 4  −x − x −x − x −x2  0 0 1 0 0 3 4 0 1   0 0 x1 + x4 x2 − x4 0 0 0   0 0 −x0 − x1 0 x2 − x4 0 0   0 0 x0 + x −x0 − x −x − x2 0 0  1 1 1  0 0 0 0 0 −x − x −x − x  3 4 4 5  + +  0 0 0 0 0 x2 x3 x4 x5   0 0 0 0 0 −x1 − x2 0   0 0 0 0 0 x0 + x1 0   0 0 0 0 0 0 x2 − x  4  0 0 0 0 0 0 −x − x  1 2  +  0 0 0 0 0 0 x0 x1   0 0 0 0 0 0 x1 + x2   0 0 0 0 0 0 −x0 − x1 0 0 0 0 0 0 0

 0 0 0 0 0 0 0 0 x2  5 0 0 0 0 0 0 0  − − 2 − − + 2 2  x2x4 x4 x2x5 x4x5 x5 x5 0 0 0 0 0 0  2  0 0 x 0 0 0 0 0  5  −x2 0 0 0 0 0 0 0  4  x x + x x + x x −x2 0 0 0 0 0 0  2 3 2 4 3 4 4  −x2  0 0 4 0 0 0 0 0  2 2 − 2  x4 x3 x4 0 0 0 0 0 0  2 2 2  −x1 0 x3 − x4 0 0 0 0 0  2 2  x0x2 + x x2 + x0x + x x −x −x 0 0 0 0 0  1 4 1 4 1 2  0 0 0 −x2 −x2 0 0 0  4 5  −x − x x2 −x2  4 5 0 0 3 0 5 0 0  − − − 2 − 2  0 x4 x5 0 x2 0 0 x5 0  2 2  0 0 −x4 − x5 x1 0 0 0 −x5  2 2  x3 + x 0 0 0 x x 0 0  4 3 4  0 x + x 0 0 −x2 0 x2 0  3 4 2 4  x + x x2 x2  0 0 3 4 0 1 0 0 4  − − − − − 2 − 2  x1 x2 x2 x3 0 0 0 x2 x3 0  2 2  x0 + x1 0 −x2 − x3 0 0 x1 0 −x3  2 2 0 x0 + x1 x1 + x2 0 0 0 x1 x2 3.2 Determinantal ideals and determinantal varieties. 63

5 5 5 d4 : R(−6) ⊕ R(−7) ⊕ R(−8) ⊕ R(−9) → R (−5) ⊕ R (−6) ⊕ R (−7)

 2  −x4 − x5 x5 0 0  x + x −x2   3 4 4 0 0   − + 2 − 2   x2 x4 x3 x4 0 0   2 2 2 2 2 3   x1 + x4 x3 − 2x4 −x2x3 − x3x4 + x2x4 + x4 0   2   −x0 − x x 0 0   1 1   0 x + x −x2 0   4 5 5   −x − x x2   0 3 4 4 0   − − 2 + 2   0 x2 x4 x3 x4 0 .  2   0 −x1 − x2 x2 0   2   0 x0 + x −x 0   1 1   0 0 −x − x x2   4 5 5   x + x −x2   0 0 3 4 4   − − 2   0 0 x2 x3 x3   2   0 0 x1 + x2 −x2  2 0 0 −x0 − x1 x1

Finally we want to describe a minimal free resolution of the determinantal ideal I3 generated by submaximal minors of the generic matrix:   x0 x1 x2 x3  x x5 x6 x7   4   x8 x9 x10 x11 x12 x13 x14 x15

Since I3 is prime, I3 is the homogeneous ideal of the determinantal variety M2. The Gulliksen-Negard complex gives a minimal free resolution of M2:

d4 16 d3 30 d2 16 d1 0 → R(−8) −→ R (−5) −→ R (−4) −→ R (−3) −→ R → R/I3 → 0

As before, we describe the boundary maps dk, k = 1, . . . , 4. 64 A minimal free complex associated to the minors of a matrix.

16 The matrix associated to d1 : R (−3) → R is the transpose of the matrix:

  x2x5x8 − x1x6x8 − x2x4x9 + x0x6x9 + x1x4x10 − x0x5x10  x x x − x x x − x x x + x x x + x x x − x x x   3 5 8 1 7 8 3 4 9 0 7 9 1 4 11 0 5 11     x3x6x8 − x2x7x8 − x3x4x10 + x0x7x10 + x2x4x11 − x0x6x11     x3x6x9 − x2x7x9 − x3x5x10 + x1x7x10 + x2x5x11 − x1x6x11     x2x5x12 − x1x6x12 − x2x4x13 + x0x6x13 + x1x4x14 − x0x5x14     x x x − x x x − x x x + x x x + x x x − x x x   3 5 12 1 7 12 3 4 13 0 7 13 1 4 15 0 5 15   x x x − x x x − x x x + x x x + x x x − x x x   3 6 12 2 7 12 3 4 14 0 7 14 2 4 15 0 6 15     x2x9x12 − x1x10x12 − x2x8x13 + x0x10x13 + x1x8x14 − x0x9x14     x3x9x12 − x1x11x12 − x3x8x13 + x0x11x13 + x1x8x15 − x0x9x15     x6x9x12 − x5x10x12 − x6x8x13 + x4x10x13 + x5x8x14 − x4x9x14     x x x − x x x − x x x + x x x + x x x − x x x   7 9 12 5 11 12 7 8 13 4 11 13 5 8 15 4 9 15   x x x − x x x − x x x + x x x + x x x − x x x   3 10 12 2 11 12 3 8 14 0 11 14 2 8 15 0 10 15     x7x10x12 − x6x11x12 − x7x8x14 + x4x11x14 + x6x8x15 − x4x10x15     x3x6x13 − x2x7x13 − x3x5x14 + x1x7x14 + x2x5x15 − x1x6x15     x3x10x13 − x2x11x13 − x3x9x14 + x1x11x14 + x2x9x15 − x1x10x15  x7x10x13 − x6x11x13 − x7x9x14 + x5x11x14 + x6x9x15 − x5x10x15

30 16 d2 : R (−4) → R (−3)

 −x x −x 0 0 −x 0 0 0 x 0 −x 0 0 0 3 7 11 12 13 14  x −x x 0 0 0 −x 0 0 0 x 0 0 0 −x  2 6 10 12 13 14   −x1 x5 −x9 0 0 0 0 −x12 0 0 0 0 0 x13 0   x0 −x4 x8 0 0 0 0 0 0 0 0 0 −x12 0 0   0 0 0 −x3 x7 x8 0 0 0 −x9 0 x10 x11 0 0   0 0 0 x −x 0 x 0 0 0 −x 0 −x 0 x  2 6 8 9 10 10  0 0 0 −x x 0 0 x 0 0 0 0 x −x 0  1 5 8 9 9   0 0 0 0 0 −x4 0 0 −x3 x5 0 −x6 0 −x7 0   0 0 0 0 0 0 −x4 0 x2 0 x5 0 0 x6 −x6   0 0 0 0 0 x0 0 0 0 −x1 0 x2 0 0 x3   0 0 0 0 0 0 x 0 0 0 −x 0 0 0 0  0 1  0 0 0 0 0 0 0 −x −x 0 0 0 0 0 0  4 1   0 0 0 0 0 0 0 x0 0 0 0 0 0 −x1 x1   0 0 0 x0 −x4 0 0 0 0 0 0 0 0 0 0   0 0 0 0 0 0 0 0 x0 0 0 0 −x4 x4 0

000000000000 x0 0 −x0 3.2 Determinantal ideals and determinantal varieties. 65

 −x15 0 0 −x15 0 0 0 0 −x15 0 0 0 0 0 0

0 −x 0 x 0 0 0 0 x 0 0 0 0 0 0  15 14 14  − −  0 0 0 0 x14 x15 0 0 x13 0 0 0 0 0 0   0 0 0 0 0 0 0 0 0 −x13 x14 −x15 0 0 0   x11 0 0 x11 0 0 0 0 0 0 0 0 −x15 0 0   0 x 0 −x 0 0 0 0 0 0 0 0 x 0 0  11 10 14  −x x −x  7 0 0 0 0 0 11 0 0 0 0 0 0 15 0   0 −x7 0 0 0 0 −x10 0 0 0 0 0 0 x14 0   0 0 −x7 0 0 0 0 x11 0 0 0 0 0 0 −x15   x x x 0 0 0 0 −x 0 0 0 0 0 0 x  2 3 6 10 14  0 0 0 x x −x x 0 0 0 0 0 0 −x 0  5 6 7 9 13   −x1 0 −x5 −x1 −x2 x3 0 x9 0 0 0 0 0 0 −x13   0 0 0 0 0 0 0 0 x8 x9 −x10 x11 x12 0 0   0 0 0 −x 0 0 −x 0 −x −x x −x 0 x 0  4 8 4 5 6 7 12 x0 0 x4 x0 0 0 0 −x8 x0 x1 −x2 x3 0 0 x12

16 30 d3 : R (−5) → R (−4)

  x12 0 −x13 x14 0 x15 0 0 0 0 0 0 0 0 0 0  x −x x −x   0 12 0 0 13 0 15 0 14 0 0 0 0 0 0 0     0 0 0 0 0 0 0 x12 0 −x14 x15 x13 0 0 0 0     −x8 0 x9 −x10 0 −x11 0 0 0 0 0 0 0 0 0 0     0 −x 0 0 x 0 −x 0 x 0 0 0 0 0 0 0   8 9 11 10   −x x 0 0 0 0 0 −x 0 0 0 0 x 0 0 0   3 7 11 15     x2 −x6 0 0 0 0 0 x10 0 0 0 0 −x14 0 0 0     −x1 x5 0 0 0 0 0 −x9 0 0 0 0 x13 0 0 0     x 0 −x x 0 x 0 0 0 0 0 0 0 0 0 0   4 5 6 7   0 0 −x 0 x 0 0 0 0 0 0 x 0 −x 0 0   3 7 11 15     0 0 x2 0 −x6 0 0 0 0 0 0 −x10 0 x14 0 0     0 0 0 −x3 0 0 0 0 −x7 x11 0 0 0 0 −x15 0     x −x 0 0 0 0 0 x 0 0 0 0 0 x −x x   0 4 8 13 14 15   x −x −x 0 x 0 0 0 0 x −x 0 0 0 −x x   0 4 1 5 10 11 14 15   − −   x0 0 x1 x2 0 0 x7 0 0 0 x11 0 0 0 0 x15     −x0 0 x1 0 0 −x3 0 0 x6 −x10 0 0 0 0 x14 0     0 0 0 0 0 x2 −x6 0 0 0 x 0 0 0 0 −x   10 14   0 x 0 0 −x 0 x 0 −x 0 0 0 0 0 0 0   0 1 3 2   − − −   0 x4 x1 0 0 0 x7 0 x6 0 0 x9 0 x13 0 0     0 0 0 x1 0 0 0 0 x5 −x9 0 0 0 0 x13 0     0 0 0 0 0 −x1 x5 0 0 0 −x9 0 0 0 0 x13     0 0 0 0 0 0 0 −x 0 x −x −x 0 0 0 0   4 6 7 5   −   0 0 0 0 0 0 0 x0 0 x2 x3 x1 0 0 0 0     x0 −x4 0 0 0 0 0 0 0 x10 −x11 −x9 −x12 0 0 0     0 0 −x0 0 x4 0 0 0 0 0 0 x8 0 −x12 0 0     0 0 0 −x 0 0 0 0 −x x 0 0 0 0 −x 0   0 4 8 12   x −x x −x   0 0 0 0 0 0 4 0 0 0 8 0 0 0 0 12     000000000000 x8 x9 −x10 x11     000000000000 −x4 −x5 x6 −x7  000000000000 x0 x1 −x2 x3 66 A minimal free complex associated to the minors of a matrix.

16 d4 : R(−8) → R (−5)

  −x7x10x13 + x6x11x13 + x7x9x14 − x5x11x14 − x6x9x15 + x5x10x15  −x x x + x x x + x x x − x x x − x x x + x x x   3 10 13 2 11 13 3 9 14 1 11 14 2 9 15 1 10 15   −x x x + x x x + x x x − x x x − x x x + x x x   7 10 12 6 11 12 7 8 14 4 11 14 6 8 15 4 10 15     −x7x9x12 + x5x11x12 + x7x8x13 − x4x11x13 − x5x8x15 + x4x9x15     −x3x10x12 + x2x11x12 + x3x8x14 − x0x11x14 − x2x8x15 + x0x10x15     x x x − x x x − x x x + x x x + x x x − x x x   6 9 12 5 10 12 6 8 13 4 10 13 5 8 14 4 9 14   x x x − x x x − x x x + x x x + x x x − x x x   2 9 12 1 10 12 2 8 13 0 10 13 1 8 14 0 9 14   − + + − − +   x3x6x13 x2x7x13 x3x5x14 x1x7x14 x2x5x15 x1x6x15     x3x9x12 − x1x11x12 − x3x8x13 + x0x11x13 + x1x8x15 − x0x9x15     x3x5x − x x7x − x3x x + x0x7x + x x x − x0x5x   12 1 12 4 13 13 1 4 15 15   x x x − x x x − x x x + x x x + x x x − x x x   2 5 12 1 6 12 2 4 13 0 6 13 1 4 14 0 5 14   − − + + −   x3x6x12 x2x7x12 x3x4x14 x0x7x14 x2x4x15 x0x6x15     −x3x6x9 + x2x7x9 + x3x5x10 − x1x7x10 − x2x5x11 + x1x6x11     x3x6x8 − x2x7x8 − x3x4x10 + x0x7x10 + x2x4x11 − x0x6x11     x3x5x8 − x1x7x8 − x3x4x9 + x0x7x9 + x1x4x11 − x0x5x11  x2x5x8 − x1x6x8 − x2x4x9 + x0x6x9 + x1x4x10 − x0x5x10 Bibliography

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