# Representation Theory of Groups andD-Modules

Authors: Ibrahim Nonkané
Journal: International Journal of Mathematics and Mathematical Sciences
 (2021)

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## Abstract

In this paper, we study a decompositionD-module structure of the polynomial ring. Then, we illustrate a geometric interpretation of the Specht polynomials. Using Brauer’s characterization, we give a partial generalization of the fact that factors of the discriminant of a finite map π:specB⟶specA generate the irreducible factors of the direct image of B under the map π.


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## Body

## 1. Introduction

The main purpose of this paper is to generalize results on modules over the Weyl algebra which appeared in [1]. Results in [1] have been obtained in a geometric context. This paper is partially expository in nature. Section 2.2 has been presented at the 9th International Conference on Mathematical Modeling in Physical Sciences to describe the action of the rational quantum Calogero–Moser system on polynomials. For the sake of clarity, we reformulate it here in a more algebraic context.A prevailing idea in representation theory is that larger structures can be understood by breaking them up into their smallest pieces. Also the natural framework of algebraic geometry is one of the polynomials and, the development of modern algebra has given a particular status to polynomials. In this vein, we study polynomial rings as modules over a ring of invariant differential operators by elaborating its irreducible submodules. We know that the direct image of a simple module under a proper mapπ is semisimple by the decomposition theorem [2]. The simplest case is when the map π:X=specB⟶Y=specA is finite; in such case, it is easy to give an elementary and wholly algebraic proof, using essentially the generic correspondence with the differential Galois group, which equals the ordinary group G. The irreducible submodules of the direct image are in one-to-one correspondence with the irreducible representations of G (see [1]). In the case of the invariants of the symmetric group, B=ℂx1,…,xnA=ℂx1,…,xnSn=ℂy1,…,yn, an explicit basis of the A-module structure of B is given by Hn=xα|αi≤n−i,1≤i≤n. In what follows, we endowed B with a differential structure by using directly the action of the Weyl algebra associated to A after a localization. We use the representation theory of symmetric groups to exhibit the generators of its simple components.The approach in this paper is different from the one in [1].Secondly, we give a geometric interpretation of the ordinary Specht polynomials which are defined as combinatorial objects [3–5].Finally, using Brauer’s characterization of characters, we give a partial generalization to arbitrary finite maps of the fact that factors of the discriminant of the finite map generate the irreducible factors of the direct imageπ+OX.


### 1.1. Preliminaries: Specht Polynomials and Specht Modules

In this section, we recall some general facts about the actions of symmetric group on polynomial ring. The symmetric groupSn is the group of permutations of the set of variables x1,…,xn. Let g∈ℂx1,…,xn be a polynomial, and σ∈Sn; we define(1)σgx1,…,xn=gσx1,…,σxn.Peel gave the construction of irreducible submodules ofℂx1,…,xn in the following way [4].By a partition ofn, we mean a sequence λ=λ1,…,λr such that(2)λ1≥λ2≥…λr>0andλ1+λ2+⋯+λr=n.Letλ=λ1,…,λr be a partition; we arrange the variables x1,…,xn in an array, with r rows and λ1 columns, containing a variable in the first λi positions of the ith row; each variable occurs exactly once in the array. For example, one such array for the partition 4,2,1 of 7 is(3)and such an array is called a λ-tableau. There are n!λ-tableaux for each partition λ of n. We shall denote such tableaux by t. Suppose that the variables a1,…,al occur in jth column of λ-tableau t, with ai in the ith row. We form the difference product Δa1,…,al=∏i<kai−ak, if l>1, and if l=1,Δa1=1. Multiplying these difference products for all the columns of t, we obtain a polynomial which we denote by ft. For σ∈Sn, let σt be the tableau obtained from t by replacing xi in t by σxi. Then, σft=fσt. It follows that the set of all linear combinations of the n! polynomials ft, obtained from the λ-tableaux t, is a cyclic ℂSn-module generated by any ft. We denote this module by Sλ. A λ-tableau is said to be standard if the variables occur in increasing order (xi>xj if i>j) along each row from left to right and down each column. Peel proved the following in [4].Theorem 1.
Bλ = {ft:t is a standard λ-tableau} is a basis of Sλ.We callft the Specht polynomial corresponding to the λ-tableau t, we call Sλ the Specht module corresponding to the partition λ, and ft is a standard Specht polynomial if t is a standard tableau.Theorem 2.
Sλ for λ⊢n forms a complete list of irreducible Sn-module over the complex field.


## 1.1. Preliminaries: Specht Polynomials and Specht Modules

In this section, we recall some general facts about the actions of symmetric group on polynomial ring. The symmetric groupSn is the group of permutations of the set of variables x1,…,xn. Let g∈ℂx1,…,xn be a polynomial, and σ∈Sn; we define(1)σgx1,…,xn=gσx1,…,σxn.Peel gave the construction of irreducible submodules ofℂx1,…,xn in the following way [4].By a partition ofn, we mean a sequence λ=λ1,…,λr such that(2)λ1≥λ2≥…λr>0andλ1+λ2+⋯+λr=n.Letλ=λ1,…,λr be a partition; we arrange the variables x1,…,xn in an array, with r rows and λ1 columns, containing a variable in the first λi positions of the ith row; each variable occurs exactly once in the array. For example, one such array for the partition 4,2,1 of 7 is(3)and such an array is called a λ-tableau. There are n!λ-tableaux for each partition λ of n. We shall denote such tableaux by t. Suppose that the variables a1,…,al occur in jth column of λ-tableau t, with ai in the ith row. We form the difference product Δa1,…,al=∏i<kai−ak, if l>1, and if l=1,Δa1=1. Multiplying these difference products for all the columns of t, we obtain a polynomial which we denote by ft. For σ∈Sn, let σt be the tableau obtained from t by replacing xi in t by σxi. Then, σft=fσt. It follows that the set of all linear combinations of the n! polynomials ft, obtained from the λ-tableaux t, is a cyclic ℂSn-module generated by any ft. We denote this module by Sλ. A λ-tableau is said to be standard if the variables occur in increasing order (xi>xj if i>j) along each row from left to right and down each column. Peel proved the following in [4].Theorem 1.
Bλ = {ft:t is a standard λ-tableau} is a basis of Sλ.We callft the Specht polynomial corresponding to the λ-tableau t, we call Sλ the Specht module corresponding to the partition λ, and ft is a standard Specht polynomial if t is a standard tableau.Theorem 2.
Sλ for λ⊢n forms a complete list of irreducible Sn-module over the complex field.


## 2. Geometric Interpretation of the Specht Polynomials

In this section, we establish a decomposition theorem and give a geometric interpretation of the Specht polynomials.


### 2.1. Notation

LetDX=ℂx1,…,xn,∂/∂x1,…,∂/∂xn be the ring of differential operators associated to OX=ℂx1,…,xn, and let OY=ℂx1,…,xnSn=ℂy1,…,yn be the ring of invariant under the symmetric group Sn where(4)yj=∑i=1nxijj,forj=1,…,n.We denote byDY=ℂy1,…,yn,∂/∂y1,…,∂/∂yn the ring of differential operators associated to OY=ℂy1,…,yn. Since OX is a simple DX-module [6], the direct image π+OX of OX under the map π:ℂn⟶ℂn/Sn is semisimple [2]. We would like to study OX as a DY-module without the machinery of the direct image structure but by the direct actions of DY on OX. By localization, OX can be turned into a DY-module, as the following lemma states.Lemma 1.
LetA=xij−11≤i,j≤n, Δ=detA, and OX˜=OXΔ−1 and let DY˜=DYΔ−2 be the localization of DY at Δ2. Then, OX˜ is a DY˜-module.Proof.
Let us make clear the actions ofDY˜ on OX˜.
We haveyj=∑i=1nxij/j,j=1,…,n, and hence ∂/∂xi=∑j=1nxij−1∂/∂yj,i=1,…,n. We get the following equation:(5)∂∂xn⋮∂∂xn=A∂∂y1⋮∂∂yn.
SinceΔ≠0, it follows that(6)∂∂y1⋮∂∂yn=A−1∂∂xn⋮∂∂xn,and it is clear that OX˜ is a DY˜-module.
What are the simple componentsOX˜ as DY˜-module and their multiplicities?Example 1.
Forn=2, DY˜=ℂx1,x2,∂/∂y1,∂/∂y2,Δ−2 and OX˜=ℂx1,x2,Δ−1 where Δ=x1−x2 and(7)∂∂y1=x2x2−x1∂∂x1−x1x2−x1∂∂x2,∂∂y2=−1x2−x1∂∂x1+1x2−x1∂∂x2.
We have that(8)OX˜=M1⊕M2,where M1=OYΔ−2 and M2=D˜Yx1−x2 are D˜Y-simple modules.


### 2.2. Simple Components and Their Multiplicities

In this section, we state our first main result. We use the representation theory of symmetric groups to yield results on modules over the ring of differential operators. It is well known that(9)OX=ℂSn⊗OYasOY−modules.Let us consider the multiplicative closed setS=Δkk∈ℕ⊂OX. It follows that(10)S−1OX=ℂSn⊗S−1OYasS−1OY−modules,where S−1OX and S−1OY are the localization of OX and OY at S, respectively. But S−1OX=O˜X and S−1OY=OY˜, whereby we get(11)O˜X=ℂSn⊗OY˜asSn−modules.Lemma 2.
There exists an injective map(12)ℂSn⟶HomℂO˜X,O˜X.Proof.
TheSn-module ℂSn acts on itself by multiplication, and this multiplication yields an injective map ℂSn⟶HomℂℂSn,ℂSn. Since OY˜ is invariant under this action of ℂSn, we get the expected injective map.Proposition 1.
There exists an injective map(13)ℂSn⟶HomDY˜O˜X,O˜X.Proof.
SinceDY˜=ℂy1,…,yn,∂y1,…,∂yn,Δ−2, we only need to show that every element of ℂSn commute with y1,…,yn,∂y1,…,∂yn.(i)
It is clear that every element ofℂSn commutes with yi,i=1,…,n.(ii)
Let us show that every element ofℂSn commutes with ∂yi,i=1,…,n. Let D be a derivation on the field O¯Y=ℂy1,…,yn (the field of fractions of OY); then, O¯Y,D is a differential field. Since O¯X=ℂx1,…,xn (the field of fractions of O˜X) is a Galois extension of O¯Y, by [7, Theorem 6.2.6] there exists a unique derivation on O¯X which extends D; then, O¯X,D is also a differential ring. In the same way, σ−1Dσ=D for every σ∈Sn. Therefore, σD=Dσ and σ commute with D.Corollary 1.
(14)ℂSn≅HomDY˜O˜X,O˜X.Proof.
see [[1], Corollary 2.6]. Before we state our first main result, let us recall some facts.
By Maschke’s theorem [[9], Chap XVIII], we know that ℂSn is a semisimple ring, and(15)ℂSn=⊕λ⊢nRλ,where the sum is taken over all the partitions of n and Rλ are simple rings. We have the following corresponding decomposition of the identity element of ℂSn:(16)1=∑λ⊢nrλ,where rλ is the identity element of Rλ, rλ2=1 and rλrμ=0 if λ≠μ, and the set rλλ⊢n is the set of central idempotents of ℂSn.
Letn be a positive integer, λ be a partition of n, Tabλ be the set of standard tableau of shape λ, and Tabn=∪λ⊢nTabλ. We have rλ=∑t∈Tabλet where et is the primitive idempotent associated to the standard tableau t∈Tabλ (see [10]).Theorem 3.
For every primitive idempotentei∈ℂSn,(1)
eiO˜X is a nontrivial D˜Y-submodule of O˜X(2)
TheD˜Y-module eiO˜X is simple(3)
There exist a partitionλ⊢n and a Specht polynomial pλ associated to a standard tableau of shape λ such that eiO˜X=D˜YpλProof.
(1)
Letei∈ℂSn be a primitive idempotent; by [10, Theorem 4.3], there exists a Specht polynomial pλ such that eipλ is a scalar multiple of pλ; then, 0≠pλ∈eiO˜X and eiO˜X≠0. Since ei commutes with every element of D˜Y and O˜X is a D˜Y-module, it follows that eiO˜X is a D˜Y-module.(2)
Assume that1=∑i=1sei where ei1≤i≤s is the set of primitive idempotents of ℂSn; then, O˜X=∑i=1seiO˜X. Let m∈eiO˜X∩ejO˜X with i≠j; then, m=eimi and m=ejmj, but eiej=0 then eim=eiejmj=0, and hence m=0. Therefore, O˜X=⊕i=1seiO˜X, and we get(17)HomD˜YO˜X,O˜X≅⊕i,j=1sHomD˜YeiO˜X,ejO˜X,and by Corollary1, we get(18)ℂSn≅⊕i,j=1sHomD˜YeiO˜X,ejO˜X.We also have, by [10, Proposition 3.29], that ℂSn≅⊕λ⊢nEndℂSλ where Sλ is the Specht module associated with the partition λ⊢n. But(19)ℂSn=⊕λ⊢nrλℂSnandrλℂSn≅Endℂℂfλ,wherefλ=dimSλ. We recall that each standard tableau ti is associated with an idempotent ei.Let us show thatℂSn≅⊕λ⊢n⊕ti,tj∈TabλHomD˜YeiO˜X,ejO˜X. Let x be an element of ℂSn and rλ be a central idempotent with λ⊢n. Then, x induces an D˜Y-homomorphism O˜X⟶O˜X,m↦x⋅m. Since rλ is in the center of ℂSn, x⋅rλO˜X=x⋅rλO˜X⊂rλO˜X, which means x∈⊕λ⊢nHomD˜YrλO˜X,rλO˜X. Then, HomD˜YeiO˜X,ejO˜X=0 if ti∈Tabλi,tj∈Tabλj and λi≠λj. We get(20)HomD˜YO˜X,O˜X≅⊕λ⊢n⊕ti,tj∈TabλHomD˜YeiO˜X,ejO˜X.The number of direct factors in the sum⊕ti,tj∈TabλHomD˜YeiO˜X,ejO˜X is fλ2.Let us show thatHomD˜YeiO˜X,ejO˜X≅ℂ if ti,tj∈Tabλ. Consider the following commutative diagram:
whereβλ:⊕μ⊢nHomD˜YrμO˜X,rμO˜X⟶HomD˜YrλO˜X,rλO˜X and αλ:⊕μ⊢nrμℂSn⟶rλℂSn are canonical projections and ϕ is the isomorphism of Corollary 1. It follows that ψ is an isomorphism, and hence rλℂSn≅HomD˜YrλO˜X,rλO˜X. Now we identify rλℂSn with the set Matfλℂ of square matrices of order fλ with coefficients in ℂ or with Endℂℂfλ.LetEij be the square matrix of order fλ with 1 at the position i,j and 0 elsewhere and Ei=Ei,i, and then we identify the idempotent ei∈rλℂSn with Ei in Matfλℂ. Let B=aij∈Matfλℂ; we get B=∑i,jai,jEi,j=∑i,jEiBEj; in fact EiBEj is the matrix with ai,j in the position i,j and 0 elsewhere; if R=Matfλ, we get that EiREj≅ℂ. This isomorphism ψ implies that ⊕ti,tj∈TabλEiREj≅⊕ti,tj∈TabλHomD˜YeiO˜X,ejO˜X; the restriction of ψ to EiREj yields a map EiREj⟶HomD˜YeiO˜X,ejO˜X and this map is surjective; moreover, we have EiREj≅HomD˜YeiO˜X,ejO˜X. Therefore, HomD˜YeiO˜X,eiO˜X≅ℂ. Let us assume that eiO˜X is not a simple D˜Y-module; then, eiO˜X may be written as eiO˜X=⊕j∈JNj where Nj are simple D˜Y-modules and J>1. It follows that dimℂHomD˜YeiO˜X,eiO˜X≥J but HomD˜YeiO˜X,eiO˜X≅ℂ, so we obtain that J=1, which necessarily implies that eiO˜X is a simple D˜Y-module.(3)
By proof (i), there exists a Specht polynomialpλ∈eiO˜X,λ⊢n such that eiO˜X=D˜Ypλ.Corollary 2.
With the above notations,eiO˜X≅D˜YejO˜X if ti and tj have the same size (if there is a partition λ⊢n such that ti,tj∈Tabλ.Proof.
TheD˜Y-modules eiO˜X are simple and HomD˜YeiO˜X,ejO˜X≅ℂ whenever there exists a partition λ⊢n such that ti,tj∈Tabλ. Since HomD˜YeiO˜X,ejO˜X≠0, we conclude by using the Schur lemma.Proposition 2.
For every young tableaut, let ft be the associated Specht polynomial; then, we have(1)
O˜X=⊕t∈TabnD˜Yft.(2)
O˜X≅D˜Y⊕λ⊢nfλD˜Yftλ,wheretλ∈Tabλ.Proof.
We have by the proof of Theorem3 that(21)O˜X=⊕eiO˜X,and eiO˜X are simple D˜Y-modules. Since to each ei corresponds a partition λ⊢n and a λ-tableau ti such that eiO˜X=D˜Yft, then O˜X=⊕t∈TabnD˜Yft. If t and t′ are two λ-tableaux, by Corollary 2, D˜Yft≅D˜YD˜Yft′. Therefore, O˜X≅⊕λ⊢nfλD˜Yftλ with tλ∈Tabλ.


### 2.3. Example

We consider the casen=3,4Forn=3, the Specht polynomials corresponding to standard tableaux are(22)ft1=1,ft2=x1−x2,ft3=x1−x3,ft4=x1−x2x1−x3x2−x3.Correspondingly, we have that(23)O˜X=O˜Y⊕D˜Yx1−x2⊕D˜Yx1−x3⊕D˜Yx1−x2x1−x3x2−x3.Forn=4, the Specht polynomials corresponding to standard tableaux are(24)ft1=1,ft2=x1−x2,ft3=x1−x3,ft4=x1−x4,ft5=x1−x2x3−x4,ft6=x1−x3x2−x4,ft7=x1−x2x1−x3x2−x3,ft8=x1−x2x1−x4x2−x4,ft9=x1−x3x3−x4x3−x4,ft10=x1−x2x1−x3x1−x4x2−x3x2−x4x3−x4.Correspondingly, we have that(25)O˜X=O˜Y⊕D˜Yx1−x2⊕D˜Yx1−x3⊕D˜Yx1−x4⊕D˜Yx1−x2x3−x4⊕D˜Yx1−x3x2−x4⊕D˜Yx1−x2x1−x3x2−x3⊕D˜Yx1−x2x1−x4x2−x4⊕D˜Yx1−x3x3−x4x3−x4⊕D˜Yx1−x2x1−x3x1−x4x2−x3x2−x4x3−x4.


### 2.4. Geometric Interpretation of Specht Polynomials

Specht polynomials were introduced as combinatoric objects [3, 4]. In fact, the Specht polynomials were first used by Wilhem Specht to generate rational representations of the symmetric group Sn [5].We give a geometric interpretation of those polynomials as follows.LetOX=ℤx1,…,xn and OY=OXSn. Let I⊂1,…,n, and we define(26)ΔI=∏i,j∈I,i<jxi−xj.LetJ=P1∪…∪Pk be a partition of 1,2,…,n as a set. To such partition, we associated the subgroup SJ=SP1×⋯×SPk of Sn and the ring OYJ=ℤx1,…,xnSJ. Let YJ=specOYJ,X=specOX and Y=specOY. The maps π′:X⟶YJ and π″:YJ⟶Y are the obvious ones, and we get the following commutative diagram.We clearly have the mapπIJ:YI⟶YJ, whenever I is a refinement of J. The Jacobian of π′ is(27)pJ=∏i=1kΔPi,a product of Vandermonde determinants on each of the set of variables with subscript in Pi. This is a Specht polynomial. Now J defines a numerical partition λ=λ1,…,λk such that λ1≤λ2≤⋯≤λk and λ1+λ2+⋯+λk=n where each λi=Pis for some is. This partition induces a Ferrers diagram where the first column has λ1 boxes, the second columns has λ2 boxes, and so on. Moreover, one gets an induced tableau by filling (in increasing order) the numbers in P1 in the first column, the numbers in P2 in the second column, and so on, where P1 is the subset of integers in the first columns, P2 is the subset of integers in the second, and so on. Conversely to every tableau corresponds a partition of {1, ⋯, n} given by letting the integers in the different columns form the partition Hence, to every tableau T, we can associate a Specht polynomial pT. These are geometrical objects, since they are Jacobians of certain polynomials maps.Proposition 3.
From the previous section, we get the following fact:(a)
pJ2∈OYJ.(b)
The submodulesOYJpJ2 and OYJpJ2⋅pJ are factors of rank one in the decomposition of OXpJ as OYJpJ2-module, i.e.,(28)OXpJ=OYJpJ2⊕OYJpJ2⋅pJ⊕….


## 2.1. Notation

LetDX=ℂx1,…,xn,∂/∂x1,…,∂/∂xn be the ring of differential operators associated to OX=ℂx1,…,xn, and let OY=ℂx1,…,xnSn=ℂy1,…,yn be the ring of invariant under the symmetric group Sn where(4)yj=∑i=1nxijj,forj=1,…,n.We denote byDY=ℂy1,…,yn,∂/∂y1,…,∂/∂yn the ring of differential operators associated to OY=ℂy1,…,yn. Since OX is a simple DX-module [6], the direct image π+OX of OX under the map π:ℂn⟶ℂn/Sn is semisimple [2]. We would like to study OX as a DY-module without the machinery of the direct image structure but by the direct actions of DY on OX. By localization, OX can be turned into a DY-module, as the following lemma states.Lemma 1.
LetA=xij−11≤i,j≤n, Δ=detA, and OX˜=OXΔ−1 and let DY˜=DYΔ−2 be the localization of DY at Δ2. Then, OX˜ is a DY˜-module.Proof.
Let us make clear the actions ofDY˜ on OX˜.
We haveyj=∑i=1nxij/j,j=1,…,n, and hence ∂/∂xi=∑j=1nxij−1∂/∂yj,i=1,…,n. We get the following equation:(5)∂∂xn⋮∂∂xn=A∂∂y1⋮∂∂yn.
SinceΔ≠0, it follows that(6)∂∂y1⋮∂∂yn=A−1∂∂xn⋮∂∂xn,and it is clear that OX˜ is a DY˜-module.
What are the simple componentsOX˜ as DY˜-module and their multiplicities?Example 1.
Forn=2, DY˜=ℂx1,x2,∂/∂y1,∂/∂y2,Δ−2 and OX˜=ℂx1,x2,Δ−1 where Δ=x1−x2 and(7)∂∂y1=x2x2−x1∂∂x1−x1x2−x1∂∂x2,∂∂y2=−1x2−x1∂∂x1+1x2−x1∂∂x2.
We have that(8)OX˜=M1⊕M2,where M1=OYΔ−2 and M2=D˜Yx1−x2 are D˜Y-simple modules.


## 2.2. Simple Components and Their Multiplicities

In this section, we state our first main result. We use the representation theory of symmetric groups to yield results on modules over the ring of differential operators. It is well known that(9)OX=ℂSn⊗OYasOY−modules.Let us consider the multiplicative closed setS=Δkk∈ℕ⊂OX. It follows that(10)S−1OX=ℂSn⊗S−1OYasS−1OY−modules,where S−1OX and S−1OY are the localization of OX and OY at S, respectively. But S−1OX=O˜X and S−1OY=OY˜, whereby we get(11)O˜X=ℂSn⊗OY˜asSn−modules.Lemma 2.
There exists an injective map(12)ℂSn⟶HomℂO˜X,O˜X.Proof.
TheSn-module ℂSn acts on itself by multiplication, and this multiplication yields an injective map ℂSn⟶HomℂℂSn,ℂSn. Since OY˜ is invariant under this action of ℂSn, we get the expected injective map.Proposition 1.
There exists an injective map(13)ℂSn⟶HomDY˜O˜X,O˜X.Proof.
SinceDY˜=ℂy1,…,yn,∂y1,…,∂yn,Δ−2, we only need to show that every element of ℂSn commute with y1,…,yn,∂y1,…,∂yn.(i)
It is clear that every element ofℂSn commutes with yi,i=1,…,n.(ii)
Let us show that every element ofℂSn commutes with ∂yi,i=1,…,n. Let D be a derivation on the field O¯Y=ℂy1,…,yn (the field of fractions of OY); then, O¯Y,D is a differential field. Since O¯X=ℂx1,…,xn (the field of fractions of O˜X) is a Galois extension of O¯Y, by [7, Theorem 6.2.6] there exists a unique derivation on O¯X which extends D; then, O¯X,D is also a differential ring. In the same way, σ−1Dσ=D for every σ∈Sn. Therefore, σD=Dσ and σ commute with D.Corollary 1.
(14)ℂSn≅HomDY˜O˜X,O˜X.Proof.
see [[1], Corollary 2.6]. Before we state our first main result, let us recall some facts.
By Maschke’s theorem [[9], Chap XVIII], we know that ℂSn is a semisimple ring, and(15)ℂSn=⊕λ⊢nRλ,where the sum is taken over all the partitions of n and Rλ are simple rings. We have the following corresponding decomposition of the identity element of ℂSn:(16)1=∑λ⊢nrλ,where rλ is the identity element of Rλ, rλ2=1 and rλrμ=0 if λ≠μ, and the set rλλ⊢n is the set of central idempotents of ℂSn.
Letn be a positive integer, λ be a partition of n, Tabλ be the set of standard tableau of shape λ, and Tabn=∪λ⊢nTabλ. We have rλ=∑t∈Tabλet where et is the primitive idempotent associated to the standard tableau t∈Tabλ (see [10]).Theorem 3.
For every primitive idempotentei∈ℂSn,(1)
eiO˜X is a nontrivial D˜Y-submodule of O˜X(2)
TheD˜Y-module eiO˜X is simple(3)
There exist a partitionλ⊢n and a Specht polynomial pλ associated to a standard tableau of shape λ such that eiO˜X=D˜YpλProof.
(1)
Letei∈ℂSn be a primitive idempotent; by [10, Theorem 4.3], there exists a Specht polynomial pλ such that eipλ is a scalar multiple of pλ; then, 0≠pλ∈eiO˜X and eiO˜X≠0. Since ei commutes with every element of D˜Y and O˜X is a D˜Y-module, it follows that eiO˜X is a D˜Y-module.(2)
Assume that1=∑i=1sei where ei1≤i≤s is the set of primitive idempotents of ℂSn; then, O˜X=∑i=1seiO˜X. Let m∈eiO˜X∩ejO˜X with i≠j; then, m=eimi and m=ejmj, but eiej=0 then eim=eiejmj=0, and hence m=0. Therefore, O˜X=⊕i=1seiO˜X, and we get(17)HomD˜YO˜X,O˜X≅⊕i,j=1sHomD˜YeiO˜X,ejO˜X,and by Corollary1, we get(18)ℂSn≅⊕i,j=1sHomD˜YeiO˜X,ejO˜X.We also have, by [10, Proposition 3.29], that ℂSn≅⊕λ⊢nEndℂSλ where Sλ is the Specht module associated with the partition λ⊢n. But(19)ℂSn=⊕λ⊢nrλℂSnandrλℂSn≅Endℂℂfλ,wherefλ=dimSλ. We recall that each standard tableau ti is associated with an idempotent ei.Let us show thatℂSn≅⊕λ⊢n⊕ti,tj∈TabλHomD˜YeiO˜X,ejO˜X. Let x be an element of ℂSn and rλ be a central idempotent with λ⊢n. Then, x induces an D˜Y-homomorphism O˜X⟶O˜X,m↦x⋅m. Since rλ is in the center of ℂSn, x⋅rλO˜X=x⋅rλO˜X⊂rλO˜X, which means x∈⊕λ⊢nHomD˜YrλO˜X,rλO˜X. Then, HomD˜YeiO˜X,ejO˜X=0 if ti∈Tabλi,tj∈Tabλj and λi≠λj. We get(20)HomD˜YO˜X,O˜X≅⊕λ⊢n⊕ti,tj∈TabλHomD˜YeiO˜X,ejO˜X.The number of direct factors in the sum⊕ti,tj∈TabλHomD˜YeiO˜X,ejO˜X is fλ2.Let us show thatHomD˜YeiO˜X,ejO˜X≅ℂ if ti,tj∈Tabλ. Consider the following commutative diagram:
whereβλ:⊕μ⊢nHomD˜YrμO˜X,rμO˜X⟶HomD˜YrλO˜X,rλO˜X and αλ:⊕μ⊢nrμℂSn⟶rλℂSn are canonical projections and ϕ is the isomorphism of Corollary 1. It follows that ψ is an isomorphism, and hence rλℂSn≅HomD˜YrλO˜X,rλO˜X. Now we identify rλℂSn with the set Matfλℂ of square matrices of order fλ with coefficients in ℂ or with Endℂℂfλ.LetEij be the square matrix of order fλ with 1 at the position i,j and 0 elsewhere and Ei=Ei,i, and then we identify the idempotent ei∈rλℂSn with Ei in Matfλℂ. Let B=aij∈Matfλℂ; we get B=∑i,jai,jEi,j=∑i,jEiBEj; in fact EiBEj is the matrix with ai,j in the position i,j and 0 elsewhere; if R=Matfλ, we get that EiREj≅ℂ. This isomorphism ψ implies that ⊕ti,tj∈TabλEiREj≅⊕ti,tj∈TabλHomD˜YeiO˜X,ejO˜X; the restriction of ψ to EiREj yields a map EiREj⟶HomD˜YeiO˜X,ejO˜X and this map is surjective; moreover, we have EiREj≅HomD˜YeiO˜X,ejO˜X. Therefore, HomD˜YeiO˜X,eiO˜X≅ℂ. Let us assume that eiO˜X is not a simple D˜Y-module; then, eiO˜X may be written as eiO˜X=⊕j∈JNj where Nj are simple D˜Y-modules and J>1. It follows that dimℂHomD˜YeiO˜X,eiO˜X≥J but HomD˜YeiO˜X,eiO˜X≅ℂ, so we obtain that J=1, which necessarily implies that eiO˜X is a simple D˜Y-module.(3)
By proof (i), there exists a Specht polynomialpλ∈eiO˜X,λ⊢n such that eiO˜X=D˜Ypλ.Corollary 2.
With the above notations,eiO˜X≅D˜YejO˜X if ti and tj have the same size (if there is a partition λ⊢n such that ti,tj∈Tabλ.Proof.
TheD˜Y-modules eiO˜X are simple and HomD˜YeiO˜X,ejO˜X≅ℂ whenever there exists a partition λ⊢n such that ti,tj∈Tabλ. Since HomD˜YeiO˜X,ejO˜X≠0, we conclude by using the Schur lemma.Proposition 2.
For every young tableaut, let ft be the associated Specht polynomial; then, we have(1)
O˜X=⊕t∈TabnD˜Yft.(2)
O˜X≅D˜Y⊕λ⊢nfλD˜Yftλ,wheretλ∈Tabλ.Proof.
We have by the proof of Theorem3 that(21)O˜X=⊕eiO˜X,and eiO˜X are simple D˜Y-modules. Since to each ei corresponds a partition λ⊢n and a λ-tableau ti such that eiO˜X=D˜Yft, then O˜X=⊕t∈TabnD˜Yft. If t and t′ are two λ-tableaux, by Corollary 2, D˜Yft≅D˜YD˜Yft′. Therefore, O˜X≅⊕λ⊢nfλD˜Yftλ with tλ∈Tabλ.


## 2.3. Example

We consider the casen=3,4Forn=3, the Specht polynomials corresponding to standard tableaux are(22)ft1=1,ft2=x1−x2,ft3=x1−x3,ft4=x1−x2x1−x3x2−x3.Correspondingly, we have that(23)O˜X=O˜Y⊕D˜Yx1−x2⊕D˜Yx1−x3⊕D˜Yx1−x2x1−x3x2−x3.Forn=4, the Specht polynomials corresponding to standard tableaux are(24)ft1=1,ft2=x1−x2,ft3=x1−x3,ft4=x1−x4,ft5=x1−x2x3−x4,ft6=x1−x3x2−x4,ft7=x1−x2x1−x3x2−x3,ft8=x1−x2x1−x4x2−x4,ft9=x1−x3x3−x4x3−x4,ft10=x1−x2x1−x3x1−x4x2−x3x2−x4x3−x4.Correspondingly, we have that(25)O˜X=O˜Y⊕D˜Yx1−x2⊕D˜Yx1−x3⊕D˜Yx1−x4⊕D˜Yx1−x2x3−x4⊕D˜Yx1−x3x2−x4⊕D˜Yx1−x2x1−x3x2−x3⊕D˜Yx1−x2x1−x4x2−x4⊕D˜Yx1−x3x3−x4x3−x4⊕D˜Yx1−x2x1−x3x1−x4x2−x3x2−x4x3−x4.


## 2.4. Geometric Interpretation of Specht Polynomials

Specht polynomials were introduced as combinatoric objects [3, 4]. In fact, the Specht polynomials were first used by Wilhem Specht to generate rational representations of the symmetric group Sn [5].We give a geometric interpretation of those polynomials as follows.LetOX=ℤx1,…,xn and OY=OXSn. Let I⊂1,…,n, and we define(26)ΔI=∏i,j∈I,i<jxi−xj.LetJ=P1∪…∪Pk be a partition of 1,2,…,n as a set. To such partition, we associated the subgroup SJ=SP1×⋯×SPk of Sn and the ring OYJ=ℤx1,…,xnSJ. Let YJ=specOYJ,X=specOX and Y=specOY. The maps π′:X⟶YJ and π″:YJ⟶Y are the obvious ones, and we get the following commutative diagram.We clearly have the mapπIJ:YI⟶YJ, whenever I is a refinement of J. The Jacobian of π′ is(27)pJ=∏i=1kΔPi,a product of Vandermonde determinants on each of the set of variables with subscript in Pi. This is a Specht polynomial. Now J defines a numerical partition λ=λ1,…,λk such that λ1≤λ2≤⋯≤λk and λ1+λ2+⋯+λk=n where each λi=Pis for some is. This partition induces a Ferrers diagram where the first column has λ1 boxes, the second columns has λ2 boxes, and so on. Moreover, one gets an induced tableau by filling (in increasing order) the numbers in P1 in the first column, the numbers in P2 in the second column, and so on, where P1 is the subset of integers in the first columns, P2 is the subset of integers in the second, and so on. Conversely to every tableau corresponds a partition of {1, ⋯, n} given by letting the integers in the different columns form the partition Hence, to every tableau T, we can associate a Specht polynomial pT. These are geometrical objects, since they are Jacobians of certain polynomials maps.Proposition 3.
From the previous section, we get the following fact:(a)
pJ2∈OYJ.(b)
The submodulesOYJpJ2 and OYJpJ2⋅pJ are factors of rank one in the decomposition of OXpJ as OYJpJ2-module, i.e.,(28)OXpJ=OYJpJ2⊕OYJpJ2⋅pJ⊕….


## 3. A Generalization

Consider the mapπ:X=specOX⟶Y=specOY. We proved in [1, Theorem 2.10] that the irreducible D-module factors of the direct image π+OX are generated by the Specht polynomials which are divisors of the Jacobian of π.We will now consider a general finite mapπ:X⟶Y. A consequence of that situation is that the simple submodules of π+OX are generated by divisors of the Jacobian of π. A natural question is in what generality this is true. We will prove a similar though weaker result in general. To describe this generalization, let us recall some facts established in [1].Letk be an algebraically closed field of characteristic 0. Denote for a k-algebra B, by TB/K the k-linear derivations of B.There is a general correspondence between representations of the differential Galois group of aDX-module M, defined using a Picard–Vessiot extension and the category of modules generated by M (for the case of one variable see [8]).LetL and K be two fields, say that a TK/k-module M is L-trivial if L⊗KM≅Ln as TL/k-modules. Denote by ModLTK/k the full subcategory of finitely generated TK/k-modules that are L-trivial.IfG is a finite group, let ModkG be the category of finite-dimensional representations of kG. Let now k⟶K⟶L be a tower of fields such that K=LG.Proposition 4 (see [1]).
The functor(29)∇:ModkG⟶ModTK/k,V↦L⊗kVGis fully faithful and defines an equivalence of categories(30)ModkG⟶ModLTK/k.The quasi-inverse of∇ is the functor(31)Loc:ModLTK/k⟶ModkG,LocM=L⊗KMϕTK/k.The above equivalence(32)ModTK⟶LocModG⟶∇ModTK,can be extended to a Galois correspondence. Fix K and L and consider intermediate fields L⊃E⊃K. Given two such fields E1⊃E2, we have the categories ModLTE1/k and ModLTE2/k. The map π:specE1⟶specE2 induces an isomorphism E2⊗E1TE2/k≅TE1/k, in particular a canonical lifting DE2⟶DE1. Corresponding to this ring homomorphism, we have the usual pair of adjoint functors. First the inverse image:(33)π+:ModLTE2/k⟶ModLTE1/k,is given by(34)M⟶E1⊗E2M.It is immediate byL⊗E1E1⊗E2M≅L⊗E2M that the image of the inverse image lies in ModLTE1/k. Secondly, we have the direct image functor π+, between the same categories, given by restricting the action on M to TE2/k using the canonical lifting TE2/k⟶TE1/k. The direct image is right adjoint to the inverse image.Thus, the direct image landing inModLTE2/k is clear, e.g., in the following way. By the proposition in the preceding section, it suffices to prove this for E1, since it then follows for any direct factor. Now the category ModLTL/k is closed under submodules and quotients, and L⊗E2E1 is a submodule of L⊗E2L, which is a quotient of L⊗KL∈ModLTL/k. So L⊗E2E1∈ModLTE2/k. By the proposition, there are equivalences of categories ModkHi⟶ModTEi/k, where Ei=LHi, and we now want to express the direct and inverse images of D-modules in terms of the corresponding group representation categories and functors.Proposition 5 .
(see [1]).(i)
Defineπ+:ModkH2⟶ModkH1 as the restriction associated to the injection kH1⊂kH2. Then,(35)π+∇V=∇π+V.(ii)
Defineπ+:ModkH1⟶ModkH2 as the coinduction associated to the injection kH1⊂kH2, defined in the following way:(36)π+V=HomkH1kH2,V.Then,(37)π+∇V=∇π+V.The study of the decomposition factors ofπ+OX can be reduced to the behavior of the direct image over the complement to the branch locus or even over the generic point. Let j:U⊂X and i:V⊂Y be the inclusions.Proposition 6 (see [1]).
Letπ:X⟶Y be a finite map. Then,(i)
π+OX is semisimple as a DY-module.(ii)
Ifπ+OX=⊕Mk,k∈I is a decomposition into simple (non-zero) DY-modules, then π+OU=⊕i∗Mk,k∈I, is a decomposition of π+OU into simple (nonzero) DV-modules.For simplicity, we assume that we are in a generic situation, working with fields. Consider a factorization(38)π=π2°π1:X⟶Z⟶Y.Generically, this corresponds to extensions of fieldsK⊂E⊂L, and we have isomorphisms TL/k≅L⊗KTK/k. Assume that K⊂L is Galois with group G. The field E corresponds to a subgroup H, and π1+OX to 1↑1H≅kH. Maps λ:H⟶k∗ may be factored through the quotient H/H′ with the commutator subgroup H′=H,H and give rise to one-dimensional representations. Corresponding functions xλ∈L exist using the functor ∇, characterized by the property that hxλ=λhxλ. Hence, π1+OX contains as direct factor the DE-submodule ∑λExλ, where the sum runs over all homomorphisms λ:H⟶k∗. This is actually the sum of all one-dimensional DE-submodules of π1OX. Each gives rise to submodules π2+Exλ. As is seen in the example below, these functions may sometimes be thought of as powers of discriminants (corresponding to the extension E⊂LH′). In the case of the symmetric group, the Specht polynomials were such functions for the sign representation Sλ⟶k∗.An immediate consequence of Brauer’s characterization of characters [11, Theorem 20], saying that any character of a finite group is a linear sum of monomial characters with integer coefficients, is then the generic case of the following theorem.Theorem 4.
Letπ:X⟶Y be an affine finite map. The Grothendieck group generated by the DY submodules of π+OX, is also generated by π2+R, where R is a rank 1 DZ submodule of π1+OX and Z is an intermediate factorization (38).Proof.
The corresponding generic result follows from Brauer’s theorem and the correspondence of irreducible modules with group representations. Then, the result follows by Proposition6.


### 3.1. Example: Cyclic Affine Covering

Consider the affine finite mapf:A1⟶A1 given by the map on rings ℂy⟶ℂx that identifies y=xn. Then, we claim that, as (left) DY-modules,(39)f∗ℂx≅ℂy⊕ℂy,y−1x⊕…⊕ℂy,y−1xn−1⊂ℂx,y−1.We use the description of the direct image in [1, Lemma 2.3]; it is thus a question of finding the DY-module generated by ℂx⊂ℂx,y−1.The extension of function fieldsK=ℂy⊂ℂx is a Galois extension with Galois group, the cyclic group of order n. Abelian groups only have one-dimensional irreducible representations. Hence, by the theory in [1], we know that f∗ℂx splits as a sum of rank 1 simple DY-modules.But we may of course also easily see this using the action ofDY. The relation between DX and DY is described by noting that(40)∂x=nxn−1∂y.This implies that all the factors in (39) above are DY-modules:(41)∂yy−kx=−k+1ny−k−1x.Hence,DYx=ℂy,y−1x, and this module, as rank 1 and torsionfree ℂy-module, is irreducible.


## 3.1. Example: Cyclic Affine Covering

Consider the affine finite mapf:A1⟶A1 given by the map on rings ℂy⟶ℂx that identifies y=xn. Then, we claim that, as (left) DY-modules,(39)f∗ℂx≅ℂy⊕ℂy,y−1x⊕…⊕ℂy,y−1xn−1⊂ℂx,y−1.We use the description of the direct image in [1, Lemma 2.3]; it is thus a question of finding the DY-module generated by ℂx⊂ℂx,y−1.The extension of function fieldsK=ℂy⊂ℂx is a Galois extension with Galois group, the cyclic group of order n. Abelian groups only have one-dimensional irreducible representations. Hence, by the theory in [1], we know that f∗ℂx splits as a sum of rank 1 simple DY-modules.But we may of course also easily see this using the action ofDY. The relation between DX and DY is described by noting that(40)∂x=nxn−1∂y.This implies that all the factors in (39) above are DY-modules:(41)∂yy−kx=−k+1ny−k−1x.Hence,DYx=ℂy,y−1x, and this module, as rank 1 and torsionfree ℂy-module, is irreducible.



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