The Submodule Structure of the Symmetric Powers

of the Defining Representation of SL(n)

Leonid Krop

October 15, 1998

DePaul University, Chicago,Illinois 60614

Introduction

The purpose of this paper is to give a full and detailed account on the topic in the title. We intend to present the main results accumulated in this area since the appearance of ([C-C]). In what follows F is an infinite field of characteristic p > 0. SL(n, F), GL(n, F) and M(n, F) stand for the special linear group, general linear group and the monoid of n × n matrices over F, respectively. These notation will be simplified to SL(n), GL(n) and M(n) when F is clear from the context. For an F - vector space V S(V) denotes the symmetric algebra of V ([NO]). S(V) is graded by homogeneous degree, S(V) = ⨁_{d≥0} S^{d}(V),

where S^{d}(V) is the d-th symmetric power of V. S(V) as well as every S^{d}(V) depends only on dim_{F} V and the ground field F. When n = dim_{F} V we write S(n) and S^{d}(n) for them. When field F is to be emphasized the above notation will be modified to S(n, F) and S^{d}(n, F).

The functorial properties of S^{d}(•) enable us to lift the natural action of M(n, F) and its subgroups on V to an action on S(V) by graded algebra endomorphisms. Thus we obtain an infinite family {S^{d}(V)} of M(n, F)- modules.

If F were C, or more generally any field of characteristic zero, the S^{d}(n, C) would be simple SL(n, C)- modules, in fact the simplest instance of dual Schur-Weyl irreducible modules. Reverting to our F we notice that S^{d}(V) can be viewed as an ”F -form” of S^{d}(n, C) obtained by reduction modulo p of a maximal Z-lattice on the latter. Classification of the submodules of Weyl modules and their duals is central to the theory of modular representations of classical groups. The present paper provides a complete description of the lattice of SL(n, F)- submodules of every S^{d}(n, F), which seems to be the simplest among existing in the literature (cf. [C-C], [DE], [CL], [DO1], [SCH], [WI], [KO]). Some of its results are described in a survey article of Doty ([DO2]) devoted to various aspects of GL(n) action in S^{d}(n).

The main difference between this paper and previous publications on the same subject lies in the following. We show that the set of the composition factors of S^{d}(n) has the internal lattice structure. This lattice, call it P (n, d), is made of certain partitions of d. It forms the combinatorial underpinning of the SL(n)- representation theory in S^{d}(n). One manifestation of this is the fact that the position of every factor in a composition series of S^{d}(n) reflects its position in tne lattice P (n, d). That is, one factor occurs before another if and only if they are either incomparable or the earlier factor is greater than the later in the order of P (n, d). Thus this work describes not only the composition factors of S^{d}(n) but also the way they are ’stuck’ together.

This work grew out of our previous attempt ([KR1])at the classification of M(n)- submodules of S^{d}(n) under the additional condition n ≥ d. In the present paper we treat all cases of n and d uniformly via a study of the action of generators of the hyperalgebra of SL(n). This approach is common to almost all the papers referenced above.

The present exposition greatly benefited from the previous research on this subject listed above. We express our sincere thanks to the authors of these papers.

0. An Overview of the Results

The essence of our approach is this (cf. [KO], [KR1]). The lattice L(n, d) of the SL(n, F)- submodules of S^{d}(n) is finite and distributive. It follows immediately from the general theory of such lattices ([AI]) that every element of L(n, d) is a unique union of certain join-irreducible elements. In the module-theoretic setting, a join-irreducible is just a local module, i.e. one with a unique maximal submodule. The above cited property of L(n, d) translates into a unique representability of every U ∈ L(n, d) as an irredundant union

U = J_{1} + ⋯ + J_{s}  (01)

of local submodules J_{i}. The decomposition (01) recovers to a large extent the missing semisimplicity of SL(n, F)- modules S^{d}(n). It reduces the problem of classifying the elements of L(n, d) to same one for the set J (n, d) of join-irreducibles in L(n, d).

J (n, d) is partially ordered by the inclusion order (poset, for short) and the classification problem boils down to finding its structure. In this regard the key fact in an isomorphism between the poset J (n, d) and the poset P (n, d) of all non-negative integer vectors (x_{0}, x_{1}, …) subject to the conditions

∑_{i≥0} x_{i}p^{i} = d and 0 ≤ x_{i} ≤ n(p − 1)  (02)

and ordered by certain relation ”◁_{p}” (see 1a below). Due to this isomorphism every joinirreducible J is labeled by an λ ∈ P (n, d), J = J^{λ}. Then every composition factor of S^{d}(n) is of the form J^{λ} / r(J^{λ}), where r(J^{λ}) is the unique maximal submodule of J^{λ} - this also comes from the general theory of the lattice on hand. It remains to tie up the structure of composition factors to solutions of (02). This is done through an explicit simple construction. Let \overline{S}^{m}(n) be the unique top composition factor of S^{m}(n) for every m ≤ n(p−1). \overline{S}^{m}(n) are in fact the homogeneous components of the algebra \overline{S}(V) obtained from S(V) by factoring out the ideal generated by all v^{p}, v ∈ V. We recall that action of SL(n, F) on an F - space U can be twisted by composing it with the powers of the Frobenious map. The resulting SL(n, F)- modules are denoted U^{[i]}, i = 1, 2, . . .. Now one of our main results reads as follows: Let λ = (l_{0}, l_{1}, …) ∈ P (n, d). Then

J^{λ} / r(J^{λ}) ≃ ⨂_{i≥0} (\overline{S}^{l_{i}})^{[i]}  (03)

(03) can be viewed as concretization of a general theorem of R. Steinberg ([ST]) for S^{d}(n).

The paper is organized in 3 chapters. The results on P (n, d), J (n, d) and the isomorphism between them can be found in the first two of them. The reader also finds there a formula for the dimensions of \overline{S}^{m}(n) and the twisted tensor product theorem (03). The last chapter is an application of our technique to the special case of 2-dimensional groups. We work out a particularly simple realization of P (2, d) leading up to the computation of the decomposition numbers of the dual Weyl modules for GL(2, F). It should be mentioned though that these results are essentially a module - theoretic rendition of the relevant part of a deep paper of Winter ([WI]).

We add a few remarks that enlarge the scope of our results. First of all it is easy to show (cf. [KR2]) that an SL(n)- submodule of a homogeneous M(n)- module remains M(n)- invariant, i.e. the SL(n)- submodule lattice coincides with that for M(n). In the second place we recall two kinds of dual for a GL(n)- module U: the contragredient dual U^{∗} and the contravariant dual U^{o} (see [GR, ch.2]). Let Sym^{m}(V) be the space of all symmetric tensors in the m- fold tensor power V^{⊗m}. By ([GR,ch.5]) we have a GL(n)- isomorphism Sym^{m}(V) ≃ S^{m}(V)^{o}, while the usual contragredient duality leads up to S^{m}(V)^{∗} ≃ Sym^{m}(V^{∗}). On the other hand we have SL(n)- isomorphism V^{∗} ≃ ∧^{n−1}(V) for an n-dimensional V, where ∧^{m}(V) stands for the m-th exterior power. Combining these isomorphisms we conclude that the SL(n)- submodule lattice of Sym^{m}(V) is dual to L(n, d), while that of S^{m}(∧^{n−1}(V)) is isomorphic with L(n, d). Since both Sym^{m}(V) and S^{m}(∧^{n−1}(V)) afford a homogeneous polynomial representation of M(n) same is true for the latter.

One notational remark. N denotes the set of non-negative integers. For every n ∈ N, n ≥ 1, we put \underline{n} := \{1, 2, …, n\} and [0, n] := \{0, 1, …, n\}.

1. The Structure of P(n,d,p)

We open our investigation of L(n, d) with a study of certain poset, which turns out to be the combinatorial underpinning of SL(n) representation theory in S^{d}(n).

1a. The Poset P(n,d,p).

Let e be the greatest natural number such that p^{e} ≤ d. We define a set P (n, d, p) to be made up of all e + 1 − tuples κ = (k_{0}, k_{1}, …, k_{e}) of natural numbers including zero satisfying conditions

∑_{i≤e} k_{i}p^{i} = d and 0 ≤ k_{i} ≤ n(p − 1) for all i  (1)

We put a partial order ◁_{p} on P (n, d, p) by saying that κ\underline{◁}λ ⟷ ∑_{i≤m} k_{i}p^{i} ≤ ∑_{i≤m} l_{i}p^{i} for all

m ≤ e, where λ = (l_{0}, l_{1}, …). We usually abbreviate P (n, d, p) to just P, when n, d, and p are fixed, and simplify ◁_{p} by dropping p out.

The above definition of ” ◁ ” resembles that of the dominance order on the set of partitions of an integer ([BR]). We shall show that several salient features of the dominance order carry over to (P, ◁). Namely, our order is a lattice ordering and the corresponding lattice is distributive. Let N^{m} denotes the m-th cartesian power of N. N is always viewed as a lattice with respect to the natural ordering. The latter induces a lattice structure on N^{m}. The corresponding order relation is denoted by ≤. Clearly N^{m} being a product of distributive lattices is itself distributive. We define e + 1 functions σ_{i} : P → N, i = 0, 1, …, e, by σ_{i}(λ) = ∑_{j≤i} λ_{j}p^{i}, where λ = (λ_{0}, λ_{1}, …, λ_{e}). These functions give rise

to an embedding of posets φ : (P, ◁) → (N^{e+1}, ≤), φ(λ) = (σ_{0}(λ), …, σ_{i}(λ), …, d). The image of φ can be characterized by

Lemma 1. An α = (a_{0}, …, a_{e}) ∈ N^{e+1} belongs to imφ if and only if
a_{0} ≤ n(p − 1), a_{e} = d and ∀_{k≥1} : a_{k} − a_{k−1}/p^{k} ∈ [0, n(p − 1)]  (2)

Proof. Let α satisfies conditions (2). Define λ_{0} = a_{0} and λ_{k} := a_{k} − a_{k−1}/p^{k} and set

λ = (λ_{0}, …, λ_{e}). Then for every k ≥ 0 a_{k} = λ_{0}+λ_{1}p+⋯+λ_{k}p^{k}. Hence d = a_{e} = ∑_{i≤e} λ_{i}p^{i},

whence λ ∈ P. Clearly α = φ(λ), as desired. The converse part is immediate.

Corollary 2. For every α ∈ imφ and every k ∈ [0, e], d ≡ a_{k} mod p^{k+1}.

Proposition 3. The imφ is a sublattice of N^{e+1}. Consequently ”◁” induces a lattice structure on (P, ◁) with respect to which P is distributive.

Proof. We must show that if α, β ∈ φ(P), then so do α ∨ β and α ∧ β. There α ∨ β and α ∧ β are computed in N^{e+1}, i.e. if α = (a_{0}, a_{1}, …, d), β = (b_{0}, b_{1}, …, d) and α ∧ β = (c_{0}, c_{1}, …, d), α ∨ β = (f_{0}, f_{1}, …, d),then c_{k} = min(a_{k}, b_{k}) and f_{k} = max(a_{k}, b_{k}). Clearly c_{0}, c_{e} satisfy the relevant part of (2). Let k ≥ 1. One has to consider four cases: c_{k−1} = a_{k−1} or b_{k−1} and c_{k} = a_{k} or b_{k}. Suppose that c_{k−1} = a_{k−1}. If c_{k} = a_{k}, then (2) is trivially true for c_{k} − c_{k−1}. Assume that c_{k} = b_{k}. The latter means that a_{k−1} ≤ b_{k−1} and b_{k} ≤ a_{k}. Hence c_{k} − c_{k−1} = b_{k} − a_{k−1} ≤ a_{k} − a_{k−1}. On the other hand b_{k} − a_{k−1} = (b_{k} − b_{k−1}) + (b_{k−1} − a_{k−1}). Now by Corollary 2 b_{k−1} − a_{k−1} ≡ 0 mod p^{k}. Hence the above equality shows that b_{k} − a_{k−1} ≡ 0 mod p^{k}, while the preceding inequality yields b_{k} − a_{k−1}/p^{k} ≤ n(p−1). The argument for two other choices for c_{k−1} and c_{k} is entirely similar. An argument for the join α ∨ β is analogous to that given above for α ∧ β.

Corollary 4. (cf. [KO]) (P, ◁) has a unique minimal and maximal elements.

Proof. Every finite lattice has the property stated in the Corollary.

1b. The covering relation.

We aim here at a description of the covering relation in (P, ◁). This will be crucial to our investigation of (P, ◁) as well as in the applications to the lattice of SL_{n}(F)− submodules of S^{d}(n). We recall that λ is a cover of μ if λ ▷ μ and λ ▷ ν\underline{▷}μ implies ν = μ. We say that λ is a co-cover of μ if μ is a cover of λ.

Proposition 5. Let λ = (λ_{0}, λ_{1}, …, λ_{e}) and μ = (μ_{0}, μ_{1}, …, μ_{e}). λ is a cover of μ if and only if

(i)

∃k : 0 ≤ k < e such that  (3)
λ_{k} ≥ p, λ_{k+1} < n(p − 1) and μ = (⋯, λ_{k} − p, λ_{k+1} + 1, ⋯)

or (ii)

∃k : 0 < k ≤ e such that  (3′)
μ_{k} < (n − 1)(p − 1), μ_{k+1} > 0 and λ = (⋯, μ_{k} + p, μ_{k+1} − 1, ⋯)

where the dots replace the components common to λ and μ.

Proof. (ii) is a restatement of (i). To show (i) we define for each subscript k such that λ_{k} and λ_{k+1} satisfy (3) an element λ_{(k)} of P by λ_{(k)} = (⋯, λ_{k} − p, λ_{k+1} + 1, ⋯). Plainly σ_{t}(λ_{(k)}) = σ_{t}(λ) for all t ≠ k, while σ_{k}(λ_{(k)}) = σ_{k}(λ) − p^{k+1}. Hence λ_{(k)} ◁ λ and furthermore, if l is another subscript for which conditions (3) hold, then λ_{(l)} and λ_{(k)} are incomparable. It remains to show that whenever μ ◁ λ there exists a λ_{(k)} with μ\underline{◁}λ_{(k)}. To this end denote by k the first subscript such that σ_{k}(μ) < σ_{k}(λ). The choice of k yields μ_{k}p^{k} < λ_{k}p^{k} and moreover as σ_{k}(λ) − σ_{k}(μ) ≡ 0 mod p^{k} by Corollary 2 we deduce that λ_{k} − μ_{k} = rp for a natural integer r > 0. Thus λ_{k} ≥ p and we consider two possibilities : (a) λ_{k+1} < n(p − 1) and (b) λ_{k+1} = n(p − 1).

If (a) is the case we can define λ_{(k)}. As σ_{k}(μ) = σ_{k}(λ) − rp^{k+1} and σ_{k}(λ_{(k)}) = σ_{k}(λ) − p^{k+1}, we conclude that σ_{k}(μ) ≤ σ_{k}(λ(k)). Since σ_{t}(λ_{(k)}) = σ_{t}(λ) ≥ σ_{t}(μ) for every t ≠ k, relation μ\underline{◁}λ_{(k)} is true. Turning to (b) we designate m to be the least integer greater than k for which λ_{m} < n(p − 1). We observe first of all that such an

m does exist. For otherwise, since μ_{i} ≤ n(p − 1) for all i σ_{t}(μ) = σ_{k}(μ) + ∑_{i=k+1}^{t} μ_{i}p^{i}

would be strictly less than σ_{t}(λ) for all t > k contradicting to d = σ_{e}(μ) = σ_{e}(λ). Now we utilize λ_{(m−1)}. As we noticed earlier σ_{m−1}(λ_{(m−1)}) = σ_{m}(λ) − p^{m}. On the other hand σ_{(m−1)}(μ) is strictly less than σ_{(m−1)}(λ) because σ_{k}(μ) < σ_{k}(λ) by our assumption and μ_{i} ≤ λ_{i} = n(p − 1) for all i strictly between k and m by the choice of m. Hence σ_{m−1}(μ) = σ_{m−1}(λ) − rp^{m} for some r > 0 by Corollary 2 and therefore σ_{m−1}(μ) ≤ σ_{m−1}(λ_{m−1}) and the proof of (i) is complete.

We now can give description of the unique maximal and minimal elements of P

Corollary 6. Let d = ∑_{i=0}^{e} d_{i} p^{i}, 0 ≤ d_{i} ≤ p − 1, be the p−adic expansion of d. Then

α := (d_{0}, d_{1}, …, d_{e}) is a unique minimal element of P.

Proof. By the preceding proposition α is a minimal element of P. Uniqueness follows from Corollary 4.

As for the maximal element of P, say, ϵ := (ϵ_{0}, ϵ_{1}, …, ϵ_{e}) its description goes as follows. If d ≤ n(p − 1), then ϵ = (d, 0, …, 0). Otherwise we define the ϵ_{j} inductively. Namely, we let ϵ_{0} be a unique integer satisfying : (n − 1)(p − 1) ≤ ϵ_{0} ≤ n(p − 1) and ϵ_{0} ≡ d mod p. Assuming that ϵ_{j−1} is defined having properties (n − 1)(p − 1) ≤ ϵ_{j−1} ≤ n(p − 1) and ∑_{i≤j−1} ϵ_{i} p^{i} ≡ d mod p^{j}, we set ϵ_{j} to be a unique integer between (n − 1)(p − 1) and

n(p − 1) and ϵ_{j} ≡ (d − ∑_{i<j} ϵ_{i} p^{i})/p^{j} mod p, provided that the latter expression is greater

than n(p − 1) and ϵ_{j} = (d − ∑_{i<j} ϵ_{i} p^{i}/p^{j}, otherwise. If the second case occurs we set ϵ_{k} = 0 for all k > j.

Corollary 7. The element ϵ defined above is a unique maximal element of P.

Proof. From definition of ϵ_{j} we derive that ϵ_{j} p^{j} < d − ∑_{i<j} ϵ_{i} p^{i} and ϵ_{j} p^{j} ≡ d −

∑_{i<j} ϵ_{i} p^{i} mod p^{j+1}. Hence the inductive step can be carried out. Also as d − ∑_{i<j} ϵ_{i} p^{i}, i =

0, 1, … is a decreasing sequence, the procedure terminates after k + 1 steps for some k yielding ϵ = (ϵ_{0}, ϵ_{1}, …, ϵ_{k}, 0, …, 0) ∈ P. By Proposition 5(ii) ϵ does not admit of a cover and Corollary 4 completes the proof.

2. Lattice L(n,d)

We let \underline{a} denote an element of N^{n}, i.e. \underline{a} = (a_{1}, …, a_{n}). We put |a| := ∑_{i∈\underline{n}} a_{i} and

refer to this number as the norm of \underline{a}. We denote by Λ(n, d) the set of all \underline{a} of norm d. Let us fix a basis e_{1}, …, e_{n}. The set of all monomials e\underline{a} := e_{1}^{a_{1}} ⋯ e_{n}^{a_{n}}, \underline{a} ∈ Λ(n, d), is denoted by M, its Boolean by B(M). Binomial coefficients \binom{m}{s} will play a significant rôle below. Under these we understand, as usual, the coefficients of t^{s} in the expansion of (1 + t)^{m}. Explicitly

\binom{m}{s} = \{ \begin{matrix}\frac{m!}{s! (m−s)!} & if m ≥ s ≥ 0 \\ (−1)^{s}\binom{s−m−1}{s} & if m < 0 ≤ s \\ 0, & in all other cases \end{matrix}

2a. The action of the hyperalgebra of SL(n).

We recall several well known facts about the algebra in the title and its action in S^{d}(n) (see [ST], [HU1], [J-K] for details). Let sl(n, F) be the special linear algebra over F. We observe that every linear transformation ϵ of V extends to a unique derivation ϵ_{D} of S(V). Restricted to sl(n, F) the correspondence ϵ ↦ ϵ_{D} affords a representation of sl(n, F) in S(V). The latter preserves homogeneous components of S(V) thereby giving rise to the representations in each S^{d}(n), d = 0, 1, . . .. In more detail define ϵ_{ij} to be the elementary matrix with 1 in the (ij)-position and zeros elsewhere. Let h_{i} := ϵ_{ii} − ϵ_{i+1 i+1}, i = 1, 2, …, n − 1. Then one readily checks the formulae

e\underline{a}ϵ_{ij} = a_{i} e_{1}^{a_{1}} ⋯ e_{i}^{a_{i}−1} ⋯ e_{j}^{a_{j}+1} ⋯ e_{n}^{a_{n}}

(4)

e\underline{a} h_{i} = (a_{i} − a_{i+1})e\underline{a}

We now recall that the hyperalgebra U_{F} (n) of SL(n, F) is generated as an algebra by the symbols ϵ_{ij}^{(s)}, i, j = 1, 2, …, n, s ∈ N. The ”diagonal” part of U_{F} (n) is the subalgebra generated by the symbols h_{i}^{(s)}, i = 1, 2, …, n − 1, s ∈ N. To compute the action of these generators on an e\underline{a} one invokes the standard argument with Z-forms of U(L) and S(V) (see the above references). Namely, in the Kostant lattice U_{Z}(L) of U(L) the generators ϵ_{ij}^{(s)} and h_{i}^{(s)} become ϵ_{ij}^{s}/s! and \binom{h_{i}}{s} := \frac{h_{i}(h_{i} − 1)···(h_{i} − s+1)}{s!}. Consequently formulae (4) tell us what is the action of these generators in S^{d}(n, Z). Passing on to U_{F} (n) and S^{d}(n) := S^{d}(n, Z) ⨂ F we arrive at the following

e\underline{a}ϵ_{ij}^{(s)} = \binom{a_{i}}{s}e_{1}^{a_{1}} ⋯ e_{i}^{a_{i}−s} ⋯ e_{j}^{a_{j}+s} ⋯ e_{n}^{a_{n}}

(5)

e\underline{a}h_{i}^{(s)} = \binom{a_{i} − a_{j}}{s}e\underline{a}

We pass on to a simple, but crucial proposition. Consider a map φ : L(n, d) ⟶ B(M) given by U → U ∧ M, U ∈ L(n, d). The proposition is -

Proposition 8. φ is a lattice embedding

Proof. The essence of the assertion is that for every submodule U a form u = ∑ f\underline{a}e\underline{a} lies in U if every f\underline{a}e\underline{a} lies in U.From the standpoint of SL(n)-representation theory every line F e\underline{a} affords a 1-dimensional representation of the standard torus D(n) of SL(n). The corresponding character χ\underline{a} : D(n) → F^{•} sends diag(t_{1}, …, t_{n}) to t_{1}^{a_{1}} ⋯ t_{n}^{a_{n}}. It is easy to see that distinct \underline{a} give rise to distinct characters. In other words F e\underline{a} is the weight subspace of S^{d}(n) of weight \underline{a}. Now the desired property follows at once from the linear independence of characters ([HU2,16.1]). Further the actions of SL(n) and U_{F} (n) coincide in the sense that their images in End_{F} (S^{d}(n) span the same subalgebra (cf [J-K,8.8.2]), thus completing the proof.

On the other hand if one wishes to stay within U_{F} (n), lemma 13 of ([ST]) furnishes the argument.

Clearly φ is a meet homomorphism. For a join U_{1} + U_{2} of two submodules notice that by the opening remark a monomial lies in U_{1} + U_{2} if and only if it lies in U_{1} or U_{2}. Hence (U_{1} + U_{2}) ∧ M = (U_{1} ∧ M) ∨ (U_{2} ∧ M).

Corollary 9. L(n, d) is a distributive lattice.

We put this corollary to work immediately by borrowing several basic facts from the general theory of finite distributive lattices. To state them we remind the reader that if L is such a lattice an element J ∈ L is called join-irreducible if it cannot be expressed as a join of its proper subelements. Clearly every join- irreducible J has a unique largest proper subelement which is denoted by r(J). Let J (L) be the poset of all join-irreducibles in L. The above mentioned facts are ([AI,Ch.II]):

(L1) L is uniquely determined by its poset J (L).

(L2) Every X ∈ L has a unique irredundant decomposition X = J_{1} ∨ … ∨ J_{m}, J_{i} ∈ J (L)

(L3) Y covers X if and only if Y = J ∨ X for a unique (for every X) J ∈ J (L) (L4) Let l(X) be the length of the maximal chain between X and O. Then l(1) = |

J (L) |.

Reverting to L = L(n, d) we put J = J (L(n, d) and translate (L1)- (L4) into the language of U_{F} (n)- submodules of S^{d}(n). Thus (L3) & (L4) mean that every composition factor of S^{d}(n) has the form J/r(J) for some J ∈ J and the length of a composition series in S^{d}(n) is | J |. As for (L1) - (L2) their effect is to reduce the problem of classifying all U_{F} (n)- submodules of S^{d}(n) (cf. [C-C], [DE], [DO1]) to that of all join-irreducible and the containment relation on them. To carry out this program we blend first (L4) with Proposition 8.

Proposition 10. The set {J/r(J) : J ∈ J } coincides with the set of all composition factors of S^{d}(n).

Proof. The proof of Proposition 8 shows that every weight subspace of S^{d}(n) is 1dimensional. Hence the multiplicity in S^{d}(n) of every irreducible SL(n)- module is ≤ 1. Further the irreducible modules J/r(J) provide | J | composition factors and by (L4) they account for all of them.

2b. The structure of J

The purpose of this section is to establish an isomorphism between J and P (n, d, p).

Proposition 11. A submodule of S^{d}(n) is join-irreducible if and only if it is U_{F} (n) − generated by a single monomial.

Proof. Assume that J is a join-irreducible submodule. Pick an e\underline{a} in J with the property that e\underline{a}U_{F} (n) is maximal among all submodules of J generated by a single monomial. Let Y be the set of monomials in J which do not belong to e\underline{a}U_{F} (n) ∧ M. Then J = e\underline{a}U_{F} (n) + Y U_{F} (n) and visibly Y U_{F} (n) ⊄ e\underline{a}U_{F} (n). Conversely assuming e\underline{a} ∈ Y U_{F} (n) we can derive from Proposition 8 that there exists a monomial e\underline{b} in Y such that e\underline{a} ∈ e\underline{b}U_{F} (n). Hence e\underline{b} belongs to e\underline{a}U_{F} (n) due to the maximality of the latter, a contradiction. Thus Y = ∅ and we are done. In the opposite direction the statement is obvious.

We can now establish that basic link between J and P (n, d, p) by associating to every e\underline{a}, \underline{a} ∈ Λ(n, d), an element of P (n, d, p). Namely, let \underline{a} = (a_{1}, …, a_{n}). Expand every a_{j} p−adically,viz.

a_{j} = ∑_{i≤e} a_{j}^{i} p^{i} , 0 ≤ a_{j}^{i} ≤ p − 1

and write the whole vector \underline{a} in the p − adic form

\underline{a} = ∑_{k=0}^{e} \underline{a}(k) p^{k}

where \underline{a}(k) = (a_{1}^{k}, …, a_{n}^{k}), k = 0, 1, …, e. Further we set t_{k} := | \underline{a}(k) | (:= ∑_{i∈\underline{n}} a_{i}^{k}). Now

we associate with the \underline{a} the (e + 1)-tuple τ(\underline{a}) := (t_{0}, …, t_{e}), called the type of \underline{a}. Clearly τ(\underline{a}) ∈ P (n, d, p) for every \underline{a} from Λ(n, d). Conversely, if κ = (k_{0}, …, k_{e}) is an element of P (n, d, p), then as k_{m} ≤ n(p − 1) there exists an n-tuple \underline{c}(m) of norm k_{m}. The monomial ∏ e^{\underline{c}(m)p^{m}} apparently has type κ.

Below we will identify ϵ_{ij}^{(s)} with the endomorphism induced by it in S^{d}(n). As a preliminary to the main theorem of this section we record a few familiar properties of ϵ_{ij}^{(s)}.

Lemma 12. (a) ϵ_{ij}^{(s)} · ϵ_{ij}^{(r)} = \binom{s+r}{s}ϵ_{ij}^{(s+r)}

(b) For every u, w ∈ S(V) (u · w)ϵ_{ij}^{(s)} = ∑_{k+l=s}(uϵ_{ij}^{(k)}) · (wϵ_{ij}^{(l)})

(c) For every m ∈ N  (u_{1} ⋯ u_{m})ϵ_{ij}^{(s)} = ∑_{k_{1}+···+k_{m}=s}(u_{1}ϵ_{ij}^{(k_{1})}) ⋯ (u_{m}ϵ_{ij}^{(k_{m})})

Proof. (a) reflects the fact that for every i ≠ j ϵ_{ij}^{(s)} span subalgebra of divided powers in U_{F} (n). Further (b) is an immediate consequence of (5) and (c) is derived from (b) by an induction on m.

We proceed to describe the action of the operators ϵ_{ij}^{(s)} on a basic monomial e\underline{a}. We remark that the p− adic decomposition \underline{a} = ∑_{m≤e} \underline{a}(m) p^{m} gives rise to a factorization

e\underline{a} = ∏_{m≤e}(e^{\underline{a}(m)})^{p^{m}}.

Lemma 13. (i) Let s be a natural number and s = ∑ s_{i} p^{i} be its p-adic expansion.

Then ϵ_{ij}^{(s)} = ∏_{k≤e} \frac{1}{s_{k}!}(ϵ_{ij}^{(p^{k})})^{s_{k}}

(ii) e\underline{a}ϵ_{ij}^{(s)} = ∏_{k≤e}(e^{\underline{a}(k)}ϵ_{ij}^{(s_{k})})^{p^{k}}

Proof. (i) By part (a) of the previous lemma ∏_{k≤e} ϵ_{ij}^{(s_{k}p^{k})} = ϵ_{ij}^{(s)}. Likewise ϵ_{ij}^{(s_{k}p^{k})} =

\frac{1}{s_{k}!}(ϵ_{ij}^{(p^{k})})^{s_{k}}.

(ii) In view of (i) it suffices to show that (∗) e\underline{a}ϵ_{ij}^{(p^{k})} = e^{\underline{a}(0)} ⋯ (e^{\underline{a}(k)}ϵ_{ij})^{p^{k}} ⋯ e^{\underline{a}(e)}.

But according to (c) of the previous lemma e\underline{a}ϵ_{ij}^{(p^{k})} = ∑_{m_{0}+···+m_{e}=p^{k}} ∏_{t=0}^{e} e^{\underline{a}(t)p^{t}}ϵ_{ij}^{(m_{t})}. Fur-

thermore we have e^{\underline{a}(t)p^{t}}ϵ_{ij}^{(m_{t})} = \binom{a_{i}^{t}p^{t}}{m_{t}}e^{\underline{b}(t)} for a suitable \underline{b}(t). The binomial coefficient in the last formula is non-zero if m_{t} = c_{t}p^{t} for some c_{t} ≤ a_{i}^{t}. Now the equality p^{k} = ∑ m_{t} forces m_{t} = 0 for all t ≠ k and (*) follows. Finally the equality e^{\underline{a}(k)p^{k}}ϵ_{ij}^{(p^{k})} = (e^{\underline{a}(k)}ϵ_{ij})^{p^{k}} can be easily checked.

The main utility of the last lemma lies in the fact that it enables us to remove p− powering from a divided power of an ϵ_{ij}, thereby reducing the action of ϵ_{ij}^{(s)} to that of ordinary power of ϵ_{ij}. We formalize this remark by introducing the set Λ_{p}(n, k) of all n− tuples of norm k with the coordinates less then p. We refer in general to every vector with the coordinates less than p as to a p− restricted vector. The set Λ_{p}(n, k) will be lexicographically ordered. The highest vector in this ordering is \underline{a}_{max} = (p−1, …, p−1, r), where k = s(p−1) + r with r < p−1. We notice that if \underline{b} is an arbitrary vector in Λ_{p}(n, k) distinct from \underline{a}_{max}, then b_{i} is less than the ith component of \underline{a}_{max} for some i ≤ s+1. Hence there is a j > i with b_{j} > 0 and in this case monomial \frac{1}{b_{j}} e\underline{b}ϵ_{ji} is lexicographically higher than e\underline{b}. Thus we can reach e^{\underline{a}^{max}} to within a scalar in a finite number of applications of ϵ_{ij}. Since our steps are reversible , this argument shows that for every \underline{a}, \underline{b} ∈ Λ_{p}(n, k) there exists a monomial ∏ ϵ_{ij}^{(s_{ij})} with s_{ij} < p that transforms e\underline{a} into e\underline{b} to within a scalar. More generally -

Lemma 14. Suppose that \underline{a}, \underline{b} ∈ Λ(n, d) have the same type. Then e\underline{a} can be transformed into e\underline{b} within a non-zero scalar by an application of some monomial from U_{F} (n).

Proof. By assumption | \underline{a}(m) | = | \underline{b}(m) | for every m. Since both those vectors lie in Λ_{p}(n, k) for k = | \underline{a}(m) |, our discussion preceding this lemma yields existence of a monomial ∏ ϵ_{ij}^{(s_{ij})} ,with s_{ij} < p, having the property e^{\underline{a}(m)} ∏ ϵ_{ij}^{(s_{ij})} = f ·e^{\underline{b}(m)}, 0 ≠ f ∈ F. Hence in view of Lemma 13(ii) e^{\underline{a}(m)p^{m}} ∏ ϵ_{ij}^{(s_{ij}p^{m})} = f · e^{\underline{b}(m)p^{m}}. Taking into account that the operators ϵ_{ij}^{(sp^{m})}, s < p, annihilate all factors of e\underline{a} except e^{\underline{a}(m)p^{m}} we can use this remark for every m from 0 to e and obtain a product of ϵ_{ij} that does the required job.

We now can proceed to the proof of the main-

Theorem 15. Let \underline{a} ∈ Λ(n, d), τ(\underline{a}) = κ and J = e\underline{a}U_{F} (n). A monomial e\underline{b} belongs to J if and only if τ(\underline{b}) \underline{◁} κ

Proof. For the ”if” part it will suffice to consider the case when θ is a co-cover of κ. In view of Proposition 5(i) if κ = (k_{0}, …, k_{e}), then θ equals to (…, k_{j} − p, k_{j+1} + 1, …) for a j such that k_{j} ≥ p and k_{j+1} < n(p − 1). By definition of κ k_{j} = | \underline{a}(j) | = ∑_{i∈\underline{n}} a_{i}^{j}.

Hence there exists an i and a c with c ≤ a_{i}^{j} satisfying a_{1}^{j} + ⋯ + a_{i−1}^{j} + c = p.Furthermore as k_{j+1} < n(p − 1) we can find a q ≤ n with a_{q}^{j+1} < p − 1. Now by Lemma 13(ii)

acting on e^{\underline{a}(j)p^{j}} by monomial M := ϵ_{1q}^{(a_{1}^{j}p^{j})} ⋯ ϵ_{iq}^{(cp^{j})} one obtains monomial e^{\underline{b}(j)p^{j}} · e_{q}^{p^{j+1}}, where \underline{b}(j) = (0, …, 0, a_{i}^{j} − c, …) with the dots replacing components common with \underline{a}(j). Moreover the other factors of e\underline{a} are not affected by the application of M- again by Lemma 13(ii). Therefore e\underline{a}M has type θ and evidently belongs to J. A reference to Lemma 14 completes the proof.

”Only if”. Because U_{F} (n) is generated by the ϵ_{ij}^{(s)} we must show the type of e\underline{a}ϵ_{ij}^{(s)} is less than κ for every i, j and s. Lemma 13 reduces this task to the case s = p^{r}. Now the action of ϵ_{ij}^{(p^{r})} on e\underline{a} affects only the factor e^{\underline{a}(r)p^{r}}, namely changing a_{i}^{r} to a_{i}^{r} − 1 and a_{j}^{r} to a_{j}^{r} + 1, provided a_{i}^{r} > 0 and zero, otherwise. If a_{j}^{r} < p − 1, then the type of e\underline{a}ϵ_{ij}^{(p^{r})} is κ. Otherwise, letting e\underline{b} := e\underline{a}ϵ_{ij}^{(p^{r})} and denoting by θ := (t_{0}, …, t_{e}) the type of \underline{b} we deduce from the above remarks that b_{q}^{r} = a_{q}^{r} for q ≠ i, j, b_{i}^{r} = a_{i}^{r} − 1 and b_{j}^{r} = 0. Consequently t_{r} :=| \underline{b}(r) | equals to ∑_{q≠i,j} a_{q}^{r} + a_{i}^{r} − 1 = k_{r} − p. Let m be the first integer greater than r

and satisfying a_{j}^{m} < p − 1. Such an m does exist, because were a_{j}^{m} = p − 1 for all m ≥ r, it would follow that a_{i}^{r}p^{r} + ∑ a_{j}^{m}p^{m} ≥ p^{e+1} > d, a contradiction. Now for every c between r + 1 and m − 1 \underline{b}(c) is obtained from \underline{a}(c) by changing a_{j}^{c} to zero, while b_{j}^{m} = a_{j}^{m} + 1 Therefore we conclude that t_{c} = k_{c} − (p − 1) for r + 1 ≤ c ≤ m − 1 and t_{m} = k_{m} + 1 The rest of components of \underline{b}(m) coincides with these of \underline{a}. It becomes evident that θ ◁ κ and the proof is complete.

Corollary 16. The poset J is isomorphic with P (n, d, p).

Proof. We combine Proposition 11 with Theorem 15. By the former every J can be associated with a λ ∈ P (n, d, p) where λ = the type of a monomial generating J. By Theorem 15 this λ is well defined and the map J ↦ λ : J → P (n, d, p) is order preserving.

By this corollary every J can be labeled by a unique λ ∈ P (n, d, p). We call this λ the type of J and write J = J^{λ}.

2c. p - Truncated Symmetric Algebra.

It is apparent from the proof of Theorem 15 that the action of U_{F} (n) in S(V) can be reduced to a large extent to the one on the elements of p−restricted weights. These elements can be organized in to a graded algebra which we now describe.

Let I_{p} be the ideal of S(V) generated by v^{p} as v runs over V. We denote the algebra S(V)/I_{p} by \overline{S}_{p}(V) or \overline{S}_{p}(n) provided dimV = n and refer to it as the p− truncated symmetric algebra. Clearly I_{p} is a graded ideal of S(V) and so \overline{S}_{p}(V) inherits a grading from S(V). Namely, setting \overline{S}^{d}(V) := S^{d}(V)/S^{d}(V) ∧ I_{p} we obtain

\overline{S}_{p}(V) = ⊕\overline{S}_{p}^{d}(V)  (6)

It is clear that \overline{S}_{p}(V) is freely spanned by the images of e\underline{a} with p− restricted \underline{a} of norm d. Since | \underline{a} |≤ n(p − 1), the sum (6) is finite, viz.

\overline{S}_{p}(n) = ⨁_{d=0}^{n(p−1)} \overline{S}_{p}^{d}(n)

Let F_{p} be the prime subfield of F. The next proposition recaptures part of Lemma 14.

Proposition 17. For every d, 0 ≤ d ≤ n(p − 1) \overline{S}_{p}^{d}(n) is absolutely irreducible as either SL(n, F_{p})− or sl(n, F_{p})− module.

Proof . The generators ϵ_{ij}^{(s)} of U_{F} (n) annihilate \overline{S}_{p}(n) for all s ≥ p. Since ϵ_{ij}^{(s)} = \frac{1}{s!}ϵ_{ij}^{s} for s < p, the action of U_{F} (n) in \overline{S}_{p}(n) coincides with that of sl(n, F). The action of the latter in every F − space is, of course, the same as of its subalgebra sl(n, F_{p}). Since all elements of Λ_{p}(n, d) have the same type, Lemma 14 yields sl(n, F_{p})− irreducibility of \overline{S}_{p}^{d}(n). Because this argument goes through for every ground field K ⊃ F_{p} used in definition of S(V), the absolute irreducibility follows as well. Turning to SL(n, F_{p}) we notice first of all that for every transvection τ_{ij} = 1 + tϵ_{ij} and every form u ∈ \overline{S}_{p}^{d}(n) the action of transvection is connected to that of ϵ_{ij} by the formula (cf. [J-K, ch.8])

uτ_{ij}(t) = ∑_{s=0}^{p−1} t^{s}(uϵ_{ij}^{(s)})  (7)

Since SL(n, F_{p}) is generated by transvections its action in \overline{S}_{p}(n) coincides with that of sl(n, F_{p}).

We digress slightly from the main route of our discussion and comment on the dimensions of \overline{S}_{p}^{d}(n). Denote dim_{F} \overline{S}_{p}^{d}(n) by ν_{p,d}^{(n)}. We pick an indeterminate t over F and we

put f_{p}(t) := 1 + t + ⋯ + t^{p−1}. Furthermore we denote by g_{p}^{(n)}(t) := ∑_{d=0}^{n(p−1)} ν_{p,d}^{(n)}t^{d} the

generating function for ν_{p,d}^{(n)}.

Lemma 18. g_{p}^{(n)}(t) = (f_{p}(t))^{n}

Proof. Right from definition (f_{p}(t))^{n} = ∑_{0≤i_{j}<p} t^{i_{1}+···+i_{n}} = ∑_{d=0}^{n(p−1)} a_{d}t^{d}, where a_{d} =

#{(i_{1}, …, i_{n}) | ∑ i_{j} = d, i_{j} < p}. It becomes clear that a_{d} = ν_{p,d}^{(n)}.

By treating p as ∞ the above lemma includes the classical case p = 0. In this case f_{0}(t) = ∑_{i=0}^{∞} t^{i} or f_{0}(t) = (1 − t)^{−1} and g_{)}^{(n)} = (1 − t)^{−n}. From the standard expansion of

g_{0}^{(n)} we see that ν_{0,d} = dim S^{d}(n) = \binom{d+n−1}{n−1}. Therefore

Corollary 19. (i) If d < p, then ν_{p,d}^{(n)} = \binom{d+n−1}{n−1} (ii) For all p and n dim \overline{S}_{p}(n) = p^{n}

Proof. For (i) notice that if d < p, then ν_{p,d}^{(n)} = ν_{0,d}^{(n)}. For (ii) we have dim \overline{S}_{p}(n) =

∑_{d=0}^{n(p−1)} ν_{p,d}^{(n)} = g_{p}^{(n)}(1) = (f_{p}(1))^{n} = p^{n}.

Next we show that the sequence ν_{p,d}^{(n)} is symmetric about \frac{n(p−1)}{2}.

Lemma 20. For every p and n

ν_{p,d}^{(n)} = ν_{p,n(p−1)−d}^{(n)}

Proof. The assertion is equivalent to the identity t^{n(p−1)}g_{p}^{(n)}(\frac{1}{t}) = g_{p}^{(n)}(t) which can be readily verified.

We note that the above lemma is a by-product of the duality between \overline{S}_{p}^{d}(n) and \overline{S}_{p}^{n(p−10−d}(n) to be discussed later.

The only case we know of an easy expression for ν_{p,d}^{(n)} in terms of binomial coefficients for all n and d is when p = 2. Indeed \overline{S}_{2}^{d}(F^{n}) ≃ Λ^{d}(F^{n}) := the dth exterior power of F^{n}, whence ν_{2,d}^{(n)} = \binom{n}{d}.

Below we assume p to be an odd prime and it will be omitted from the notation. By Corollary 19 and Lemma 20, or directly one has

ν_{d}^{(2)} = \{ \begin{matrix}d + 1, & if d < p \\ 2(p − 1) − d + 1, & otherwise \end{matrix}

Proposition 21. The sequence ν_{d}^{(n)} is strictly increasing for d < \frac{n(p−1)}{2} and strictly decreasing for d ≥ \frac{n(p−1)}{2}.

Proof. The second part of the proposition follows from the first and Lemma 20. As for the first part, this is evident for n = 2 from the formula for ν_{d}^{(2)}. Further we induct on n. The definition of ν_{d}^{(n)} as the number of p− restricted solutions to the equation

x_{1} + ⋯ + x_{n} = d shows that (*) ν_{d}^{(n+1)} = ∑_{i=0}^{p−1} ν_{d−i}^{(n)}. If d < \frac{n(p−1)}{2}, then this formula and

the induction hypothesis readily yield ν_{d}^{(n+1)} < ν_{d+1}^{(n+1)}. Suppose that d = \frac{n(p−1)}{2} + k for a non-negative k strictly less than \frac{p−1}{2}. Then every summand of the sum (*) for ν_{d+1}^{(n+1)}, except for ν_{d+1}^{(n)}, matches up with the same summand in formula (*) for ν_{d}^{(n+1)}. To complete the proof it will suffice to show that ν_{d+1}^{(n)} > ν_{d−(p−1)}^{(n)}. But ν_{d+1}^{(n)} = ν_{\frac{n(p−1)}{2} −(k+1)}^{(n)} in view

of Lemma 20, so that this comes down to proving \frac{n(p−1)}{2} − (k + 1) > d − (p − 1) which in turn follows from k < \frac{p−1}{2}.

Our last remark on dim \overline{S}_{p}^{d}(n) is -

Theorem 22. For every n > 2, odd prime p and d ≤ n(p − 1) the ν_{d}^{(n)} is given by the formula

ν_{d}^{(n)} = ∑_{k=0}^{n−1}(−1)^{k}\binom{n}{k}\binom{d − kp + n − 1}{n − 1}

Proof. Let Z be the set of all solutions to x_{1}+⋯+x_{n} = d in the non-negative integers. Define \overline{Z} and Z_{i}, i = 1, 2, …, n as the subsets of all p- restricted solutions and solutions with x_{i} ≥ p, respectively. Clearly \overline{Z} ∧ Z_{i} = ∅ for all i and also Z = \overline{Z} ∨ Z_{1} ∨ … ∨ Z_{n}.

Therefore | Z | = | \overline{Z} | + | Z_{1} ∨ … ∨ Z_{n} | while | Z_{1} ∨ … ∨ Z_{n} | = ∑_{k=1}^{n}(−1)^{k−1}S_{k}, where

S_{k} = ∑_{1≤i_{1}<...<i_{k}≤n} | Z_{i_{1}} ∧ ⋯ ∧ Z_{i_{k}} |. Next we note that every set Z_{i_{1}} ∧ ⋯ ∧ Z_{i_{k}} stands

in a 1 - 1 correspondence with the set of all solutions to t_{1} + ⋯ + t_{n} = d − kp. Therefore S_{k} = \binom{n}{k}\binom{d−kp+n−1}{n−1} and taking into account that Z_{1} ∧ ⋯ ∧ Z_{n} = 0 we arrive at the above formula.

We proceed to a discussion of the dual to \overline{S}_{p}(n). In what follows we fix p and n and omit them from the notation. We begin by identifying the 1- dimensional representation afforded by \overline{S}^{n(p−1)}. We recall (e.g. [KR2]) that every polynomial 1-dimensional representation of M_{n}(F) is given by γ ↦ (det γ)^{a}, γ ∈ M_{n}(F) for a non-negative integer a. We denote by D^{a} the corresponding M_{n}(F)− module. In case of GL_{n}(F) we extend definition of D^{a} to include the negative a. Let ∂ := (e_{1} ⋯ e_{n})^{p−1} be a unique F − generator of \overline{S}^{n(p−1)}. Clearly ∂γ = (det γ)^{p−1}∂ for every γ = diag(u, …, u) and this implies identification \overline{S}^{n(p−1)} ≃ D^{p−1}. Further for every d between 0 and n(p−1) we let d^{∗} := n(p−1)−d. Moreover for an \underline{a} ∈ Λ_{p}(n, d) we define \underline{a}^{∗} as the n−tuple (p − 1 − a_{1}, …, p − 1 − a_{n}). Apparently \underline{a} ∈ Λ_{p}(n, d) if and only if \underline{a}^{∗} ∈ Λ_{p}(n, d^{∗}). Therefore every basic monomial in \overline{S}^{d^{∗}} has the form e^{\underline{a}^{∗}} for a unique \underline{a} ∈ Λ_{p}(n, d). We now define a fundamental pairing \overline{S}^{d} × \overline{S}^{d^{∗}} ⟶ F by sending a pair (v, u) to ⟨v, u⟩ := vu/∂. We adopt the usual convention of writing V^{∗} for the contragredient dual of a G-module V, i.e. V^{∗} = Hom_{F} (V, F) with the action of G determined through (ug)(u) = u(vg^{−1}), u ∈ V^{∗}, v ∈ V and g ∈ G.

Proposition 23. In the foregoing notation

(i) \overline{S}^{d^{∗}} is the linear dual of \overline{S}^{d}
(ii) As a GL(n)-module \overline{S}^{d^{∗}} ≃ (\overline{S}^{d})^{∗} ⊗ D^{p−1}

Proof. (i) It’s obvious that ⟨e\underline{a}, e^{\underline{a}^{∗}}⟩ = 1 for every \underline{a}. If \underline{b} is distinct from \underline{a}, then b_{k} > a_{k} for at least one k. But then b_{k} + (p − 1) − a_{k} ≥ p, whence e\underline{b} · e^{\underline{a}^{∗}} = 0. Thus the sets \{e\underline{a}\} and \{e^{\underline{a}^{∗}}\} form dual bases.

(ii) First of all we notice that for every g ∈ G one has ⟨vg, ug⟩ = (det g)^{p−1}⟨v, u⟩. For assuming vu = α · ∂ for some α ∈ F we derive (vu)g = α · ∂g = (det g)^{p−1}(vu) which gives the desired result. Next consider a GL(n)- module W := \overline{S}^{d^{∗}} ⊗ D^{−(p−1)}. Because every element of W is of the form u ⊗ ∂^{−(p−1)} with u ∈ \overline{S}^{d^{∗}} we can define a pairing [,] : \overline{S}^{d} × W ⟶ F by [v, u ⊗ ∂^{−(p−1)}] = ⟨v, u⟩. A routine computation shows that this pairing is GL(n)-invariant, so that W ≃ (\overline{S}^{d})^{∗}. Tensoring both sides of this isomorphism with D^{p−1} yields (ii).

Since D^{p−1} becomes trivial upon restriction to either SL(n) or GL_{n}(F_{p}) we arrive at

Corollary 24. (\overline{S}^{d})^{∗} ≃ \overline{S}^{d^{∗}} as SL(n)- or GL_{n}(F_{p})- module.

A little bit more can be said about \overline{S}^{d}(2). Namely

Proposition 25. (i) For every d = p − 1 + a with 0 ≤ a ≤ p − 1 \overline{S}^{d} ≃ \overline{S}^{p−1−a} ⊗ D^{a} (ii) For every d ≤ p − 1 (\overline{S}^{d})^{∗} ≃ \overline{S}^{d} ⊗ D^{−d}

Proof. (i) We use a well known fact that polynomial irreducible GL(n)-modules are distinguished by their highest weight vectors. A direct inspection shows that both modules in (i) have (p − 1, a) as their highest weight vector, as needed.

(ii) By the previous proposition (\overline{S}^{d})^{∗} ≃ \overline{S}^{d^{∗}} ⊗ D^{−(p−1)}. Now d^{∗} ≥ p − 1 so that an application of (i) yields (ii).

2d. The Twisted Tensor Product Theorem.

We know due to Proposition 10 and Corollary 16 that every composition factor of S^{d}(n) has the form J^{μ}/r(J^{μ}) for a μ ∈ P (n, d). It remains to determine the structure of those quotients. To this end we recall the notion of Frobenious twist of a module. Namely, every GL(n)- module U gives rise to infinitely many new GL(n)- modules, denoted by U^{[i]}, and obtained by composing the given representation GL(n) ⟶ GL(U) with the embedding F r^{i} : F → F, t ↦ t^{p^{i}} of the ground field F. We now state the main result

Theorem 26. A composition factor of S^{d}(n) labeled by μ = (m_{o}, m_{1}, …, m_{e}) ∈ P (n, d) has the form

C^{μ} := J^{μ}/r(J^{μ}) ≃ \overline{S}^{m_{0}} ⊗ (\overline{S}^{m_{1}})^{[1]} ⊗ … (\overline{S}^{m_{e}})^{[e]}

Proof. We begin with a remark on twisted copies of a GL(n)- submodule U of S^{d}(n). For a g = (g_{ij}) in GL(n) we set g^{[k]} := (F r^{k}(g_{ij})) = (g_{ij}^{p^{k}}). Then for every monomial e\underline{a} one checks (e\underline{a})^{p^{k}}g = e\underline{a}g^{[k]}. The last formula together with the fact that U admits of a basis of monomials makes it evident that U^{[k]} can be identified with the span of all u^{p^{k}}, u ∈ U, and the latter lies in S^{mp^{k}}(n). We proceed with a few remarks on S^{m} for m ≤ n(p − 1) and its twisted copies. Firstly, for every m in this range there are monomials of type (m, 0, …, 0) and S^{m} is U_{F} (n)- generated by every one of them. For by Corollary 7 (m, 0, …, 0) is a unique maximal element in P (n, m) and the assertion follows from Theorem 15. In addition Proposition 11 shows that S^{m} is join-irreducible with radical r(S^{m}) spanned by the monomials of the type lesser than (m, 0, …, 0) and with S^{m}/r(S^{m}) as its top composition factor. On the other hand we know that \overline{S}^{m} is irreducible, so we must have \overline{S}^{m} ≃ S^{m}/r(S^{m}). Likewise , taking into account that (S^{m})^{[i]} can be realized as ⟨u^{p^{i}} | u ∈ S^{m}⟩ we conclude that the latter is U_{F} (n)- generated by every monomial of type (0, …, m, …, 0), where m occupies the ith place, and hence it is join-irreducible whose radical is the span of all monomials of the type less than (0, …, m, …, 0). Further the composition factor (S^{m})^{[i]}/r(S^{m})^{[i]} is naturally isomorphic with (S^{m}/r(S^{m}))^{[i]}, that is (\overline{S}^{m})^{[i]}. Consider a natural map S^{m_{0}} ⊗ (S^{m_{1}})^{[1]} ⊗ ⋯ (S^{m_{e}})^{[e]} ⟶ S^{m_{0}} · (S^{m_{1}})^{p} ⋯ (S^{m_{e}})^{p^{e}} ↪ S^{d} which replaces the tensor sign by multiplication in S(n). Pick a generating monomial in every (S^{m_{k}})^{[k]} and let P be their product. One can readily check using Lemma 13(ii) that P generates S^{m_{0}} ⋯ (S^{m_{e}})^{p^{e}}. On the other hand P has type μ and therefore generates J^{μ} by the very definition of the latter. Factoring through the canonical epimorphism J^{μ} ⟶ J^{μ}/r(J^{m}u) we arrive at an epimorphism φ : ⨂_{k=0}^{e}(S^{m_{k}})^{[k]} ⟶ J^{μ}/r(J^{μ}). As we

know J^{μ}/r(J^{μ}) is spanned by the (image of) monomials of type μ. Each one of them has the form ∏_{k=0}^{e} e^{\underline{a}(k)p^{k}} with the p- restricted \underline{a}(k) satisfying | \underline{a}(k) | = m_{k} for every k,

which shows that dim J^{μ}/r(J^{μ}) equals ∏_{k=0}^{e} dim \overline{S}^{m_{k}}. It remains to estimate Kerφ. To this

end we recall that our description of r(S^{m_{k}})^{[k]} earlier in the proof shows that if e\underline{b} lies in r(S^{m_{k}})^{[k]}, then | \underline{b}(k) |< m_{k}. Therefore for every product e^{\underline{b}^{0}} ⊗ … e^{\underline{b}^{e}}, e^{\underline{b}^{k}} ∈ (S^{m_{k}})^{[k]} taken from S^{m_{0}} ⊗ ⋯ ⊗ r(S^{m_{k}})^{[k]} ⊕ ⋯ (S^{m_{e}})^{[e]} there exists an i ≤ k with | \underline{b}^{i}(i) |< m_{i} and | \underline{b}^{j} | = m_{j} for all j < i. The image of this product under φ apparently falls in r(J^{μ}) which forces the above tensor product in Kerφ. Hence φ factors through the canonical

map ⨂_{k=0}^{e}(S^{m_{k}})^{[k]} ⟶ ⨂_{k=0}^{e}(S^{m_{k}})^{[k]}/r(S^{m_{k}})^{[k]} ≃ ⨂_{k=0}^{e}(\overline{S}^{m_{k}})^{[k]}. By the reason of dimension

J^{μ}/r(J^{μ}) ≃ ⨂_{k=0}^{e}(\overline{S}^{m_{k}})^{[k]}, as desired.

Two additional remarks are in order. First of all we want to connect the previous theorem with an important result in the representation theory of GL(n), the Steinberg tensor product theorem([ST, Thm.4.1]). According to this theorem a GL(n)- irreducible module L(\underline{a}) labeled by the highest weight vector \underline{a} decomposes into the tensor product of L(\underline{a}(k))^{[k]}, where \underline{a} = ∑ \underline{a}(k)p^{k}. In light of this basic result Theorem 26 identifies which twisted tensor product of GL(n)- irreducibles occur in S(n). Namely, it says that in order for this to happen every L(\underline{a}(k)) must be isomorphic to \overline{S}^{|\underline{a}(k)|}. In the second place we note that a generalization of the last theorem to universal enveloping algebras of a class of non-abelian Lie algebras can be found in ([KR3]).

It is easy to find the highest weight vector of C^{μ}. Let \underline{a}_{max}(m) denote the highest weight in \overline{S}^{m} regarded as GL(n)- module. As we know (see p.10) \underline{a}_{max}(m) = (p−1, …, p− 1, r, 0, …, 0), where m = s(p − 1) + r with 0 ≤ r < p − 1 and likewise \underline{a}_{max}(m)p^{k} is the highest weight in (\overline{S}^{m})^{[k]}. Therefore -

Corollary 27. The highest weight of C^{μ} is ∑_{k=0}^{e} \underline{a}_{max}(m_{k})p^{k}.

We remark that this corollary readily implies that C^{μ} ≃ C^{ν} if and only if μ = ν.

3. The 2 - dimensional Case. Decomposition Numbers for SL(2) and GL(2).

In this section we concentrate on the n = 2- case. We shall work out all the polynomial irreducible modules for GL(2, F) for any infinite field F of characteristic p > 0. Like in the previous sections the centerpiece of our study will be the lattice P = P (2, d). We will describe the structure of an element of this lattice to a very fine detail. That will allow us to construct an imbedding of P (2, d) in B(N_{e−1}) which turns out to be generically an isomorphism, i.e. whenever d is in the generic position ([DE, Def1]). Furthermore since the elements of P label the composition factors of S^{d}(2) we derive an explicit characterization of the decomposition numbers for SL(2) and GL(2).

3a. The Lattice P(2,d).

We begin with the p-adic expansion of d, d = ∑_{i=0}^{e} d_{i}p^{i}.

Definition 1. A natural number d is called generic relative to p if 0 < d_{i} < p − 1 holds for all i, 1 ≤ i ≤ e − 1 and d_{0} < p − 1, d_{e} > 0.

We refer to an element of P (2, d) as a 2-type for d. The next lemma limits the choices for values of the coordinates of a 2-type.

Lemma 28. For every 2-type β = (b_{0}, …, b_{e}) for d b_{j} = d_{j}, d_{j}−1, d_{j}+p−1 or d_{j}+p.

Proof. Our assertion obviously holds for the minimal element α = (d_{0}, d_{1}, …, d_{e}) of P. In addition α satisfies the following property : (*) For every γ = (c_{0}, …, c_{e}) ∈ P and every i ≥ 1, if c_{i} equals to d_{i} − 1 or d_{i} + p − 1, then c_{i−1} ≥ p − 1. This enables us to start induction on the order ’◁’. Assume that for a given β, every γ ◁ β satisfies the lemma along with property (*). By Proposition 5(ii) β is obtained from a co-cover γ by changing just two neighboring components of it. Hence there is a k ≥ 1 such that if γ = (c_{0}, …, c_{e}), then b_{k−1} = c_{k−1} + p and b_{k} = c_{k} − 1. It remains to notice that the restriction b_{k} ≤ 2(p − 1) placed by definition on the coordinates of every 2-type implies that c_{k} can not be either d_{k} − 1 or d_{k} + p − 1 and likewise c_{k−1} must be d_{k−1} or d_{k−1} − 1. This confirms both the lemma and property (*) for β.

We proceed to a description of a 2-type. Essentially its structure is as follows. When read from left to right every β ∈ P is made up of the strings of the form (d_{i} + p, d_{i+1} + p − 1, …, d_{j} + p − 1, d_{j+1} − 1) separated by components common with α. We now associate with β a subset Z(β) of N_{e} := {0, 1, …, e} consisting of all those i for which b_{i} is a d_{i} + p or d_{i} + p − 1. We want to express Z(β) as a union of maximal, distinct intervals [i_{k}, j_{k}],viz. Z(β) = ⋃_{k=1}^{r}[i_{k}, j_{k}], meaning that i_{k} and j_{k} are subject to i_{k} ≤ j_{k} and j_{k} + 1 < i_{k+1} for

all k. As usual [i, j] denotes the set of all k, i ≤ k ≤ j and (i, j] := [i, j] \ {i}.

Proposition 29. In the foregoing notation

(i) The set Z(β) determines β by the rule : b_{r} = d_{r} if and only if r ∉ Z(β) ∪ {j_{1} +

1, …, j_{r} + 1}, b_{i_{k}} = d_{i_{k}} + p for all k, b_{s} = d_{s} + (p − 1) for all s in ⋃_{k=1}^{r}(i_{k}, j_{k}] and b_{j_{k}+1} =

d_{j_{k}+1} − 1 for all k. The association β → Z(β) defines a map ζ : P (2, d) → B(N_{e−1}). A subset X ⊂ N_{e−1} belongs to the image of ζ if and only if when expressed as the union of maximal intervals, viz. X = ⋃_{k=1}^{r}[i_{k}, j_{k}] the i_{k} and j_{k} satisfy conditions d_{i_{k}} < p − 1 and

d_{j_{k}+1} > 0.

(ii) ζ is an embedding of lattices. It is a lattice isomorphism if and only if d is generic. Proof. (i) Notice that for α the assertions are vacuously true, hence the induction on

’◁’ can be started. Thus we assume that for all γ, γ ◁ β the first part of (i) is true. We can assume that β results from γ = (c_{0}, …, c_{e}) by changing two components of γ, c_{m} and c_{m+1}, say, to c_{m} + p and c_{m+1} − 1. From the property (*) of the last lemma it follows that c_{m+1} = d_{m+1} or c_{m+1} = d_{m+1} + p. Assuming Z(γ) = ⋃_{k=1}^{r}[i_{k}, j_{k}] we see that

m + 1 ∉ Z(γ) ∪ {j_{1} + 1, …, j_{r} + 1} and hence Z(β) obtains from Z(γ) either by adding a new one point interval or extending an interval of Z(γ) by adding a point just to the right of it. In the second case m + 1 is one of the i_{k} and the passage from Z(γ) to Z(β) consists of adding a point just to the left of an interval of Z(γ) or merging some two intervals of the latter. In either of these cases the relationship between γ and Z(γ) is sustained under the inductive step. As for the second assertion it is equivalent to saying that Z(γ) ⊂ N_{e−1} for every γ. Indeed, were e in Z(γ) for some γ, say, e = j_{r}, it would follow from the proven part of the proposition that d_{j_{r}+1} = d_{e+1} > 0,which is absurd.

We now turn to the last assertion of (i). Suppose that X ⊂ N_{e} is written as a union of its maximal intervals [i_{k}, j_{k}] satisfying d_{i_{k}} < p − 1 and d_{j_{k}+1} > 0 for all k. We define a β = β(X) through b_{i_{k}} = d_{i_{k}} + p, b_{j_{k}+1} = d_{j_{k}+1} − 1, b_{s} = d_{s} + (p − 1) for every

s ∈ ⋃_{k=1}^{r}(i_{k}, j_{k}] and b_{t} = d_{t} for all other t. It is immediate to check that β is a 2-type for

d and moreover ζ(β) = X. The opposite direction is trivial.

(ii) First of all we notice that ζ is an order-preserving embedding, i.e. γ\underline{◁}β ⟺ ζ(γ) ⊆ ζ(β). This fact is an easy consequence of (i). Hence the first claim in (ii) is equivalent to checking that for every 2-types γ and β ζ(γ) ∨ ζ(β) as well as ζ(γ) ∧ ζ(β) belong to imζ. This is readily obtainable from our criteria for a set to lie in imζ. As for the last assertion it will suffice to show, because ζ is a lattice monomorphism, that every one element subset of N_{e−1} is in the imζ if and only if d is generic. In one direction, viz. if d is generic, this is quite obvious. Conversely, assume that d_{i} is 0 or p − 1 for some i between 1 and e − 1. If d_{i} = 0, then i − 1 ∉ imζ and d_{i} = p − 1 implies i ∉ imζ. Likewise {0} ∉ imζ, if d_{0} = p − 1 and {e − 1} ∉ imζ, if d_{e} = 0.

We end this section with a property peculiar to n = 2. As we know thanks to Proposition 17 if d < p S^{d}(n) is SL(n, F)- irreducible for every n, p and the ground field F. In the opposite direction we have

Proposition 30. For d ≥ p S^{d}(n) is SL(n)- irreducible if and only if n = 2 and d = ap^{e} − 1 with an e ≥ 0 and 1 ≤ a ≤ p.

Proof. S^{d}(n) is irreducible if and only if P (n, d) consists of α = (d_{0}, …, d_{e}) only. If n > 2, then d_{i} + p ≤ 3(p − 1) for every prime p and every d_{i}. Now by Proposition 5(ii) atom α has exactly m covers, where m is the number of non-zero coordinates of α. Since d ≥ p, there is at least one i ≥ 1 with d_{i} > 0, and so S^{d}(n) is reducible.

Let now n = 2. α does not admit of a cover if and only if d_{0} = d_{1} = … = d_{e−1} = p − 1, which leads up to d having the asserted form.

3b. Decomposition Numbers.

We begin with a brief review of some basic facts of the modular representation theory of GL(2). We adhere on the whole to the notation and convention of ([GR]). We recall that to every partition λ of d into at most n parts there is associated GL(n)- module D_{λ} ([GR,4.1]). For n = 2 and λ = (λ_{1}, λ_{2}) it is fairly immediate that D_{λ} is isomorphic to S^{λ_{1}−λ_{2}} ⊗ D^{λ_{2}}, where D denotes, as before, the determinant representation of GL(2). D_{λ} has irreducible socle F_{λ} which in the case on hand take on the form Soc(S^{λ_{1}−λ_{2}}) ⊗ D^{λ_{2}}.

We define the decomposition number m_{λ,μ}(GL(2)) to be the composition multiplicity of F_{μ} in D_{λ}. Restricting the above defined modules to SL(2) we arrive at two families of SL(2)- modules, viz. S^{a} and F_{a} = Soc(S^{a}), a = 0, 1, …, and we denote m_{a,b}(SL(2)) the composition multiplicity of F_{b} in S^{a}. We pass on to the main result of this section. It gives a criteria for m_{λ,μ} to be equal 1. We notice that as S^{d} has 1-dimensional weight subspaces, the only other possible value of it is 0, so that our result completely describes these numbers. The above mentioned criteria run in two directions. One can either fix a λ and express the μ satisfying m_{λ,μ} = 1 in terms of combinatorial objects associated with λ or go the other way round and express λ in terms of a fixed μ. We now introduce some further notation and make slight adjustments to some old ones. To begin with we put e(d) = e_{p}(d) for the last integer such that p^{e} ≤ d. We write ζ_{d} for the imbedding of P (d) in B(N_{e(d)−1}) defined in the last section. Further we define two numbers n_{Z,d}

and m_{Y,d} as follows. Let Z = ⋃_{k=1}^{r}[i_{k}, j_{k}] (see Prop. 29) be a set from imζ_{d}. Then we

set n_{Z,d} := ∑_{i∈Z} d_{i}p^{i} + ∑_{k=1}^{r} p^{i_{k}}. As for the second number let us first append the p- adic

decomposition of d by an infinite tail of zeros. Next we define Δ(d) to be the infinite set of all i ∈ N for which d_{i} ≠ p − 1. Further for each finite subset Y of Δ(d) we set m_{Y,d} := ∑_{i∈Y} (p − 1 − d_{i})p^{i}. Passing on to a partition μ = (μ_{1}, μ_{2}) we name a subset Y of

Δ(μ_{1} − μ_{2}) admissible if m_{Y,μ_{1}−μ_{2}} ≤ μ_{2}.

Proposition 31. In the foregoing notation

(i) For a given λ m_{λ,μ} = 1 if and only if there exists a Z ∈ imζ_{λ_{1}−λ_{2}} such that μ_{1} = λ_{1} − n_{Z,λ_{1}−λ_{2}} and μ_{2} = λ_{2} + n_{Z,λ_{1}−λ_{2}}

(ii) For a fixed μ m_{λ,μ} = 1 if and only if λ_{1} = μ_{1} + m_{Y,μ_{1}−μ_{2}} and λ_{2} = μ_{2} − m_{Y,μ_{1}−μ_{2}} for an admissible Y ⊂ Δ(μ_{1} − μ_{2}).

Proof. Every composition factor of D_{λ} has the form C^{β} ⊗D^{λ_{2}} for a β = (b_{0}, b_{1}, …) ∈ P (λ_{1}−λ_{2}), thanks to the isomorphism D_{λ} ≃ S^{λ_{1}−λ_{2}} ⊗D^{λ_{2}}. In order to identify this factor all we have to do is to compute its highest weight. Let us write h.w.(M) for the unique highest weight of an irreducible module M. Taking into account ([GR]) that h.w.(F_{λ}) = λ we conclude that C^{β} ⊗ D^{λ_{2}} ≃ F_{μ} if and only if μ = h.w.C^{β} ⊗ D^{λ_{2}}. Let Z = ζ_{λ_{1}−λ_{2}}(β) and for every j ∈ Z express b_{j} in the form b_{j} = p − 1 + r_{j}, 0 ≤ r_{j} ≤ p − 1. By Corollary 27 h.w.(C^{β}) = ( ∑_{i∉Z} b_{i}p^{i} + ∑_{j∈Z}(p − 1)p^{j}, ∑_{j∈Z} r_{j}p^{j}) and , of course, h.w.(D^{λ_{2}}) = (λ_{2}, λ_{2}).

It follows that μ = h.w.(C^{β} ⊗ D^{λ_{2}}) if and only if μ_{1} = λ_{1} − ∑_{j∈Z}(b_{j} − p + 1)p^{j} and

μ_{2} = λ_{2} + ∑_{j∈Z}(b_{j}−p+1)p^{j}. It remains to notice that due to Proposition 29(i) ∑_{j∈Z}(b_{j}−p+1)

is exactly our n_{Z,λ_{1}−λ_{2}}.

Turning to the second part of this proposition we observe that the above formulae for μ_{i} show μ_{1} − μ_{2} = λ_{1} − λ_{2} − 2n_{Z,λ_{1}−λ_{2}}. Using the fact that λ_{1} − λ_{2} = ∑ b_{k}p^{k} we derive equality μ_{1} −μ_{2} = ∑_{i∉Z} b_{i}p^{i} + ∑_{j∈Z}(2(p−1)−b_{j})p^{j}. Therefore assuming μ_{1} −μ_{2} = ∑ m_{k}p^{k}

we conclude that m_{j} = 2(p − 1) − b_{j} for j ∈ Z and m_{i} = b_{i}, otherwise. That is to say, the above β is obtained from (m_{0}, m_{1}, …) by replacing some of m_{j} ≠ p − 1 by 2(p − 1) − m_{j}. In the opposite direction, let Y be an admissible set in Δ(μ_{1} − μ_{2}). Define β by changing every m_{j}, j ∈ Y, to 2(p − 1) − m_{j}. Put λ_{1} = μ_{1} + m_{Y,μ_{1}−μ_{2}} and λ_{2} = μ_{2} − m_{Y,μ_{1}−μ_{2}}. One can check readily that β ∈ P (λ_{1} −λ_{2}), so that C^{β} ⊗D^{λ_{2}} is the factor of D_{λ} isomorphic to F_{μ}.

The case of SL(2) is essentially a repetition of the previous one. One small change occurs in definition of the admissible set. Since the restriction of D to SL(2) is trivial, every subset of Δ(d) is now admissible.

Proposition 31. In the foregoing notation

(i) For an a ∈ N m_{a,b} = 1 if and only if b = a − 2n_{Z,a} for some Z ∈ imζ_{a}.

(ii) For a b ∈ N m_{a,b} = 1 if and only if a = b + 2m_{Y,b} for some finite set Y ⊂ Δ(b).

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