Loop-augmented forests and a variant of the Foulkes' conjecture

A loop-augmented forest is a labeled rooted forest with loops on some of its roots. By exploiting an interplay between nilpotent partial functions and labeled rooted forests, we investigate the permutation action of the symmetric group on loop-augmented forests. Furthermore, we describe an extension of the Foulkes' conjecture and prove a special case. Among other important outcomes of our analysis are a complete description of the stabilizer subgroup of an idempotent in the semigroup of partial transformations and a generalization of the (Knuth-Sagan) hook length formula.


Introduction
This paper is a natural continuation of our earlier work [1], where we considered a permutation action of the symmetric group on the set of labeled rooted forests. The origin of this representation goes back to the semigroup of partial transformations on a finite set. We shall show in this paper that the circle of ideas here naturally leads to a new variant of a long standing conjecture in representation theory, namely the Foulkes' conjecture. Let us recall this famous conjecture. If m ≤ n, then the multiplicity of an irreducible representation V of S mn in the induced trivial representation Ind Smn Sm≀Sn 1 is less than or equal to the multiplicity of V in Ind Smn Sn≀Sm 1. We state this conjecture in a combinatorial way, which can be viewed as a precursor of our work.
Consider a set of vertices with mn elements which are placed along a line and grouped into blocks of m elements so that there are n blocks in total. We will distribute the labels 1, 2, . . . , mn on these vertices and call two resulting distributions equivalent if one configuration can be obtained from the other by permuting the labels within the same group and/or by permuting the blocks. For example, the three configurations in the first line of Figure 1, where n = 3 and m = 2, are in the same equivalence class. (For convenience we separated individual blocks in each configuration by putting a bar between them.) In the second line of Figure 1, we have two inequivalent configurations since the set of elements of the first block of the first configuration is not equal to any set of elements of any blocks in the second configuration. Let us denote by o ′ 1 the vector space spanned by all such equivalence classes. Next, we define another vector space, denoted by o 1 , by changing the roles of m and n. Clearly both of o 1 and o ′ 1 are S mn -modules. In this terminology, the Foulkes' conjecture states that the multiplicity of an irreducible S mn -module V in o ′ 1 is less than or equal to the multiplicity of V in o 1 .
Our conjecture is the following natural extension of Foulkes' conjecture. Let m and n be positive integers such that n > m ≥ 2. Consider m copies of the labeled rooted tree on n vertices where the root has exactly n − 1 children. See Figure 2, where n = 4 and m = 2. The symmetric group S mn acts on vertices without breaking the relations between labels imposed by the edges of the underlying trees. Hence we get a representation denoted by o ′ 2 . In a similar manner, next, we consider n copies of labeled rooted tree on m vertices where the root has exactly m − 1 children. This time the S mn representation is denoted by o 2 . Now, let V = V (λ) be an irreducible representation of S mn indexed by a partition λ which has more than two parts. Then the multiplicity of V in o 2 is greater than or equal to the multiplicity of V in o ′ 2 . One of the main results of our paper is the proof of this conjecture when m = 2.
As we mentioned before, we arrived at this conjecture by considering nilpotent partial transformations and rook theory. Our goal here is to further develop this topic, so we start with recalling the relevant concepts and results from [1]. This way, we hope that the reader will see a coherent picture of our work and the origins of our conjecture.
Let n denote a positive integer, [ The map ϕ is an injective semigroup homomorphism. By the rook monoid R n , we mean the monoid of injective partial transformations. We will often refer to the elements of R n as rook placements (on an n × n grid). Since P n is embedded into F ull n and since the latter monoid contains a zero transformation it makes sense to talk about nilpotent partial transformations in P n . In particular, C n and C n ∩R n are well defined although their definitions require an embedding of P n into F ull n . (In literature, many authors assume that 0 ∈ R n .) The unit groups of P n and R n are the same and are equal to S n . Moreover, the conjugation action of S n on itself extends to these semigroups as well as to the subset C n of nilpotent partial transformations. Therefore we have S n -representations on the vector spaces C P n , C R n and on C C n .
The combinatorial significance of C n , which is crucial to our purposes, stems from the fact that there is a bijection between C n and the set of labeled rooted forests on n vertices. Furthermore, the conjugation action of S n on functions in C n translates to the permutation action of S n on the labels. See [1].
Let τ be a rooted forest on n vertices. τ is called labeled if there exists a bijective map φ from [n] onto the vertex set of τ . Throughout this paper, when we talk about labeled rooted forests, we omit writing the corresponding labeling function despite the fact that the action of S n does not change the underlying forest but the labeling function only. In accordance with this convention, when we write S n · τ we actually mean the orbit (1) In [1], the S n -representation defined by the orbit (1) is called the odun of τ . It only depends on τ , so we write Stab Sn (τ ) to denote the stabilizer subgroup of the pair (τ, φ). As an S n -module, the vector space of functions on the right cosets, that is C[S n /Stab Sn (τ )] is isomorphic to the odun of τ , which is equivalent to the induced representation Ind Sn Stab Sn (τ ) 1. It turns out that the Frobenius character of this module is recursive in nature and can be computed by certain combinatorial rules. To avoid introducing too excessive notation, we illustrate these rules on an example. Let τ be the rooted forest depicted in Figure 3 and let τ i , i = 1, 2, 3 denote its connected components (from left to right in the figure). Let F τ and F τ i , i = 1, 2, 3 denote the corresponding Frobenius characters. Since τ 1 = τ 2 = τ 3 , we have Here, s 2 is the Schur function s (2) , the bracket stands for the plethysm of symmetric functions and the dot stands for ordinary multiplication. (We will give a brief introduction to plethysm in Section 2.2.) More generally, if a connected component τ ′ , which of course is a rooted tree appears k-times in a forest τ , then F τ has s k [F τ ′ ] as a factor. Now we proceed to explain the computation of the Frobenius character of a rooted tree. As an example we use F τ 1 of Figure 3. The combinatorial rule that we obtained in [1] is simple; it is the removal of the root from tree. The effect on the Frobenius character of this simple rule is as follows: Let τ ′ 1 denote the rooted forest that we obtain from τ 1 by removing the root. Then F τ 1 = s 1 · F τ ′ 1 . Thus, by the repeated application of this rule and the previous factorization rule, we obtain We compute F τ 2 by the same method and we arrive at the following satisfactory expression for the Frobenius character of τ : Let τ be a rooted forest and a be a vertex in τ . We let τ 0 a denote the rooted subforest whose set of vertices is {b ∈ τ : b < a} and we let τ a denote the rooted subtree whose set of vertices is {b ∈ τ : b ≤ a}. Then we let {a 1 , . . . , a r } of τ a such that the corresponding subtrees τ a i (i = 1, . . . , r) are mutually distinct. Finally, let m(a; a i ) = m τ (a; a i ) (i = 1, . . . , r) denote the number of copies of τ a i that appear in τ 0 a . In this notation, the dimension of the odun of τ turns out to be where n is the number of vertices of τ . See [1,Theorem 9.3]. This is a generalization of the Knuth-Sagan hook length formula for the rooted trees.
Another important advancement of [1] is the determination of the rooted trees whose odun afford the sign representation. We will give a detailed explanation of the arguments involved with this computation later in this paper. Such arguments were used in [1] to prove that the generating function for the number of rooted forests which afford the sign representation is n≥2 2 n−2 x n . Here, n is interpreted as the number of vertices.
We are now ready to introduce our new combinatorial object by means of which we will generalize the results of [1]. The algebraic significance of these objects will be developed later in this paper. A loop-augmented forest is a rooted forest with loops on some of its roots. See, for example, Figure 4, where we depict a loop-augmented forest on 22 vertices and four loops. It is a well known variation of the Cayley's theorem that the number of labeled forests on n vertices with k roots is equal to n−1 k−1 n n−k . See [3], Theorem D, pg 70. It follows that the number of loop-augmented forests on n vertices with k roots is 2 k n−1 k−1 n n−k . It follows from generating function manipulations that the number of loop-augmented forests on n vertices for n ≥ 2 is 2n n−3 .
Let τ be a rooted tree on n vertices. We group the maximal branches of τ according to the number of vertices on each maximal branch. As a convention, let us collect the maximal branches with more vertices on the left hand side of the graph and repeat this procedure inductively for every maximal branch and for their maximal branches, and so on. We do not worry about the ordering of the maximal branches in the same group. As an example, see Figure 5. Once the grouping is done, we label the vertices by [n] from left to right and from bottom to top, decreasing the label by 1 each time. Here, n is the number of vertices and the root of τ is labeled by n. Notwithstanding the ambiguity created by not ordering the maximal branches of the same size, the resulting incidence matrix from our labeling is a strictly upper triangular (nilpotent) matrix. We adopt this procedure to the rooted forests by first grouping (and depicting left to right) the rooted trees with respect to the number of vertices they have. The resulting incidence matrix of the forest is a direct sum of nilpotent matrices, hence it is nilpotent also. Notice that this procedure also explains why there is a bijection between labeled rooted forests and nilpotent partial functions. That is, given τ ∈ C n with respect to any labeling that we just described, we simply let f be the partial function defined by f (i) = j whenever i is a child of j in τ and there is an edge between i and j. Then f is nilpotent. In the light of this fact, from now on, we will will use the same notation τ for a nilpotent partial function and its corresponding labeled rooted forest. We now proceed to explain how to interpret loop-augmented forests in terms of partial functions. Let σ be a loop-augmented forest. Then some of the roots of σ have loops. There is still a partial function for σ, as defined in the previous paragraph for a rooted forest. The loops in this case correspond to the "fixed points" of the associated function. Indeed, with respect to labeling conventions of the previous paragraph, the incidence matrix of σ is upper triangular. However it can have both of 1's and 0's on its main diagonal. The resulting matrix can be viewed as the matrix representation of the partial function of σ. (Warning: Unlike our notation in this paragraph, in the sequel we will often use lower case Latin letters for denoting both of loop-augmented forests and partial functions.) We remark here that the correspondence between loop-augmented forests and (certain) partial functions fails if we consider adding loops on non-roots. In such a situation, the resulting objects do not always give partial functions, but, instead, give "partial relations." The subset of P n which corresponds to the loop-augmented forests will be formally defined in the next section. Nevertheless, we can briefly explain the underlying algebraic theme of our work. Recall that the permutation action of S n on the labels translates to the conjugation action on the incidence matrices. Clearly, the conjugates of a nilpotent (respectively unipotent) matrix are still nilpotent (respectively unipotent). Let U be an arbitrary upper triangular matrix. Then U is equal to a sum of the form D +N, where D is a diagonal matrix and N is a nilpotent matrix. If σ is a permutation matrix of the same size as U, then we conclude from the equalities σ · U = σUσ −1 = σDσ −1 + σNσ −1 that the conjugation action on loop-augmented forests is equivalent to the simultaneous conjugation action on labeled rooted forests and the representation of S n on diagonal matrices.
The outline of this paper is as follows. In Section 2, we review the most basic facts about the idempotents of the semigroups F ull n , P n , and R n . In addition we review the basic theory of plethysm for symmetric functions. In Section 3, we prove the first of our main results, Theorem 3.1, which states that the stabilizer subgroup in S n of a loop-augmented forest (viewed as a partial function on [n]) has a direct product decomposition of the form Z(σ) × Stab S n−k (τ ), where Z(σ) is the centralizer of a permutation σ in S k and Stab S n−k (τ ) is the stabilizer subgroup in S n−k of a labeled rooted forest τ . It follows from this result that the Frobenius character of the S n orbit of a loop-augmented forest f is of the form where τ 1 , . . . , τ r are the distinct connected components of τ where τ i occurs λ i times. We record this theorem as Theorem 4.1 in Section 4. Another consequence of the factorization of the stabilizer subgroup is a dimension formula for the S n representation defined by f (Corollary 3.1). If (1 m 1 2 m 2 . . . k m k ) denotes the conjugacy type of σ in S k , then the dimension of C[S n · f ] is given by where m(a; b) = m τ (a; b) is as defined in (2). The goal of Section 5 is to determine the loop-augmented forests that afford a sign representation. We prove in Theorem 5.1 that the number of loop-augmented forests on n vertices with k loops whose S n orbit admits a sign representation is 2 n−k−2 . In particular, the total number of loop-augmented forests on n vertices admitting a sign representation is 2 n−1 − 1. In Section 6, we prove the main algebraic result of our paper. Namely, if e ∈ P n is an arbitrary idempotent, then there exists a multiset of integers {β m 1 1 , . . . , β ms s } and m ∈ N such that We record this result in Corollary 6.2. Finally, in Section 7, we prove our variant of the Foulkes' conjecture in the special case where m = 2. (See Conjecture 7.1 for an equivalent statement in terms of symmetric functions and plethysm.)

Preliminaries
Structures of the semigroups P n and R n become more transparent by virtue of matrices; each element g of P n has a unique n × n matrix representative g = (g ij ) n i,j=1 defined by What is important for us is that the matrix product translates to the semigroup multiplication (composition). In particular, a partial transformation f ∈ P n is nilpotent if some power of the corresponding matrix f = (f i,j ) n i,j=1 is the zero matrix 0 n .

Idempotents.
An idempotent in a semigroup S is an element e ∈ S satisfying e 2 = e. The identity transformation id n : [n] → [n] of the unit group is an idempotent, and conversely for any idempotent e ∈ S := P n , there is a corresponding "local group" S e determined by e; it is defined as S e := eP n e. The identity of the local group S e is e. We denote the idempotent set of a semigroup S by E(S). By a diagonal idempotent in P n , we mean an idempotent of R n . The nomenclature follows from the fact that the matrix representation of e ∈ E(R n ) is a diagonal matrix. E(R n ) ⊂ R n has exactly 2 n elements (including the 0-matrix) and it forms a commutative submonoid of R n . It is not difficult to see that E(R n ) is isomorphic to the "face lattice" of the regular "n-simplex" in R n . For example, Let U denote the unit group of a semigroup S. For our purposes it is important to note that E(S) is closed under the conjugation action In particular we see that E(R n ) is an S n -set that is contained in E(P n ). For example, We list some easy facts about the set of idempotents of P n . For the proofs of some of these facts (and more) we recommend [4].
1. The number of idempotents in P n is given by n k=0 n k (k + 1) n−k . 2. The number of idempotents in F ull n is given by n k=1 n k k n−k . 3. The idempotents in P n do not form a subsemigroup of P n for n ≥ 2.
4. The idempotents in F ull n do not form a subsemigroup of F ull n for n ≥ 3.

The idempotents in
Let e be a diagonal idempotent in ∈ P n . Since all idempotents of rank k in R n are related to each other via conjugation, the S n orbit of e is uniquely determined by the number k = rank(e). We define which is closed under conjugation action of S n . Clearly, if k = 0, then N (n, k) is equal to C n , the set of nilpotent elements in P n .
Remark 2.1. By relaxing the requirement on diagonal property, we could have considered {f ∈ P n : f m is a rank k idempotent for some m ≥ 1}.
This set properly contains N (n, k). However, the matrix rank of idempotents is not a refined invariant with respect to conjugation action. To justify this claim, we consider e 1 := 1 0 0 0 and e 2 := 1 1 0 0 These idempotents are of the same rank but they are not conjugate. To capture differences between the conjugacy classes of e 1 and e 2 , we will introduce below the concept of "standard idempotent." Let x ∈ P n . The following set (not to be confused with N (n, k), where k is a number) will play an important role in the sequel: Lemma 2.1. If x = e be an idempotent, then N (n, e) is closed under conjugation. Furthermore, if e is a diagonal idempotent of rank k, then N (n, k) = N (n, e).
Remark 2.2. The reason for we introduce the set (4) is because N (n, k) is precisely the subset of P n that corresponds to the loop-augmented forests. Thus, our main goal is to understand the S n -module structure on N (n, k).

Symmetric functions and plethysm.
In representation theory, plethysm refers to a general idea on a functorial operation, which is defined from a representation to another one. More concretely, it is a description of the decompositions, with multiplicities, of representations derived from a given representation V such as V ⊗ V, k (V ), k ( l (V )), Sym k (V ), Sym l (Sym k (V )), V * .. In practice one often considers functors attached to the weight of an irreducible representation such as Schur functors. Let S µ resp. S λ denote the Schur functors indexed by the partitions µ and λ, respectively.. Then S λ • S µ := S λ (S µ (C n )) is called the plethysm of S λ with S µ . There is a corresponding operation on the ring of symmetric functions. The k-th power-sum symmetric function, denoted by p k , is the sum of k-th powers of the variables. For a partition λ = (λ 1 ≤ · · · ≤ λ l ), we define p λ to be equal to i p λ i . (We will maintain the convention of writing parts of a partition in the increasing order throughout our paper.) The k-th complete symmetric function, h k is defined to be the sum of all monomials The Schur function associated with λ, denoted by s λ , is the symmetric function defined by the determinant det(h λ i +j−i ) l i,j=1 . The power sum and Schur symmetric functions will play special roles in our computations via the Frobenius character map F : class functions on S n → symmetric functions where σ ⊂ S n is a conjugacy class of type λ and δ σ is the indicator function It turns out that if χ λ is the irreducible character of S n indexed by the partition λ, then F(χ λ ) = s λ . In the sequel, we will not distinguish between representations of S n the corresponding characters. In particular, we will often write the Frobenius character of an orbit to mean the image under F of the character of the representation of S n that is defined by the action on the orbit. For two irreducible characters χ µ and χ λ indexed by partitions λ and µ, the Frobenius characteristic image of the plethysm χ λ [χ µ ] is the plethystic substitution s λ [s µ ] of the corresponding Schur functions. Roughly speaking, the plethysm of the Schur function s λ with s µ is the symmetric function obtained from s λ by substituting the monomials of s µ for the variables of s λ . In the notation of [6]; the plethysm of symmetric functions is the unique map [·] : Λ × Λ → Λ satisfying the following three axioms: Although the problem of computing the plethysm of two (arbitrary) symmetric functions is very difficult, there are some useful formulas for Schur functions: and Here, g and h are arbitrary symmetric functions, c λ µ,ν is a scalar, and γ λ µ,ν is 1 n! χ λ , χ µ χ ν , where the pairing stands for the standard Hall inner product on characters.
Finally, let us mention the conjugation property of plethysm:

Nilpotents and diagonal idempotents
Let f be an element of N (n, k), m be a positive integer such that f m is a diagonal idempotent in P n . Without loss of generality we assume that f m is of the form Our first result is about the actual "shape" of this matrix representation of f .
1. If f is a partial transformation as in the previous paragraph, then there exist a permutation σ ∈ S k and a nilpotent partial transformation τ from C n−k such that Proof. First, we are going to show that f has the block-diagonal form as in (13). We will prove this by contradiction, so we assume that there exists j ∈ {k, . . . , n} such that f (j) = i < k. On one hand, since f m is as in (12), we have f m (i) = i. On the other hand, by our assumption . Therefore, f m (i) = 0, but this is a contradiction. Therefore, f is as in (13) for some σ ∈ P k and τ ∈ C n−k .
Since the m-th power of σ 0 0 τ is as in (12), the m-th power of σ is the k × k identity matrix and the m-th power of τ is 0 n−k . In other words, σ is a permutation matrix, τ is a nilpotent matrix. The proof is complete.
a permutation and τ ∈ C n−k is a nilpotent partial transformation. In this case, the stabilizer subgroup in S n of f has the following decomposition: Stab Proof. Let π = A B C D ∈ S n be a permutation matrix given in block form so that A is a Assume that πf π −1 = f . We claim that both B and C are 0 matrices. To see this, we use πf = f π to get four matrix equations: We are going to show only that B = 0 since for C the idea of the proof is the same. First of all, if N = 0, then it immediately follows from the above equations that B = 0, so we assume that N is a nonzero nilpotent matrix. Let r be the smallest integer such that BN r = 0. If r = 1, then since σ is invertible, it follows from equation (15) that B = 0, so we continue with the assumption that r > 1. In this case, by multiplying equation (15) with N r−1 on the right we obtain σBN r−1 = BN r = 0. By the same token, since σ is invertible, we see that BN r−1 = 0. But this contradicts with our assumption on the minimality of r. Therefore, r = 1, hence B = 0. Since both of B and C are 0, the proof follows from equations (14) and (17). By Proposition 3.1 and Remark 2.2, we see that a partial transformation f ∈ P n represents a loop-augmented forest, then its matrix form is as in (13). By this notation, as an application of Theorem 3.1, we will now record a representation theoretic generalization of the famous Knuth-Sagan hook length formula.
where m(a; b) = m τ (a; b) is as defined in (2).
Proof. By Theorem 3.1 we know that the S n representation defined by f is isomorphic to C[S n /Z(σ) × Stab S n−k (τ )], hence its dimension is given by n!/(|Z(σ)||Stab S n−k (τ )]|). We express this number in a different way: Note that |Z(σ)| is equal to k!/|C(σ)|, where C(σ) is the conjugacy class of σ in S k . The cardinality of C(σ) is equal to k!/ j j m j (m j !). Thus, the rest of the proof follows from (2) and (18).

Characters
Let f be a partial transformation in N (n, k) and let H(f ) denote the stabilizer subgroup Stab Sn (f ). As a consequence of Theorem 3.1 we see that the vector space o(f ) is S nequivariantly isomorphic to the vector space C[S n /H(f )] and that H(f ) ≃ Z(σ)×Stab S n−k (τ ) for some permutation σ ∈ S k and a nilpotent partial transformation τ ∈ C n−k . The S n module structure on C[S n /H(f )] is given by the left translation action of S n on the cosets of H(f ). In other words, o(f ) is isomorphic to the induced trivial representation from H(f ) to S n . Since H(f ) decomposes as a direct product, the character of o(f ) is equal to the product of characters of the induced trivial representations on the corresponding ambient symmetric subgroups. We paraphrase the (isomorphism of) representations in induction notation: Notation : Let H ⊂ G be a subgroup. We denote by χ H,G the character of the induced trivial representation Ind G H 1 H . If τ is a loop-augmented forest on m vertices, then we denote the character of the induced trivial representation from Stab Sm (τ ) to S m by χ o(τ ) . Lemma 4.1. We preserve our notation from the previous paragraph and let f = σ 0 0 τ denote a partial function representing a loop-augmented forest. In this case the character of o(f ) has the following factorization: Proof. The proof is a straightforward consequence of (19).
The set of cosets of the subgroup Z(σ) in S k is in natural bijection with the set of conjugates of σ ∈ S k . In other words, viewed as an S n -set, S k /Z(σ) is identified with the conjugacy class of σ in S k . It is well known that the conjugacy classes (of S k ) are parametrized by the partitions (of k), and a representation on a conjugacy class is irreducible. By the Frobenius characteristic map the irreducible representation on a conjugacy class that is indexed by a partition ν of k is mapped to the Schur function s ν . Clearly, this argument takes care of the character of the induced representation Ind S k Z(σ) 1 in (19). For the other tensor factor, we refer to our earlier work: [1, Corollary 6.5] states that if λ = (λ 1 , . . . , λ r ) is a partition of n − k and τ is a rooted forest consisting of λ 1 copies of a rooted tree τ 1 , λ 2 copies of a rooted tree τ 2 and so on, then the character of o(τ ) is given by We summarize these observations as follows: Theorem 4.1. Let f be a partial function representing a loop-augmented forest on n vertices.
Then f is similar to a block diagonal matrix of the form σ 0 0 τ for some σ ∈ S k and τ ∈ N (n − k, 0). Furthermore, if ν, which is a partition of k, is the conjugacy type of σ in S k and if the underlying rooted forest of the nilpotent partial transformation τ has λ 1 copies of the rooted tree τ 1 , λ 2 copies of the rooted forest τ 2 and so on, then the character of o(f ) is given by Proof. The proof follows from the discussion in the previous paragraph and Lemma 4.1.

Sign representation
In [1, Section 7], the first author found a characterization of labeled rooted forests that afford the sign representation.
Definition 5.1. A subtree a of a rooted tree is called a terminal branch (TB for short) if any vertex of a has at most 1 successor. A maximal terminal branch (or, MTB for short) is a terminal branch that is not a subtree of any terminal branch other than itself. The length (or height) of a TB is the number of vertices it has. We call a rooted tree blossoming if all of its MTB's are of even length, or no two odd length MTB's of the same length are connected to the same parent. We call a non-blossoming tree dry. Finally, a rooted forest is called blossoming if the rooted tree that is obtained by adding a new root to the forest is a blossoming tree; otherwise, the forest is called dry.
Note that if a forest is blossoming, then all of its connected components are blossoming. Also, if a single connected component is dry, then the whole forest is dry. In Figure 6 we have the list of all blossoming forests up to 5 vertices. 2. If τ is a dry forest, then s λ [F τ ], s (1 ml ) = 0.
The following consequence of the above lemma is not very difficult to see. We will now combine these results with our master character formula from Theorem 4.1. Let f be a loop-augmented forest on n vertices as in Theorem 4.1, that is, the partial transformation corresponding to f is S n -conjugate to a block diagonal matrix of the form σ 0 0 τ for some σ ∈ S k and τ ∈ N (n − k, 0). Let ν denote the conjugacy type of σ in S k .
Then it follows from Theorem 4.1 and Lemma 5.1 that is the odun of a blossoming forest; 0 otherwise.
Theorem 5.1. The number of loop-augmented forests on n vertices with k loops whose odun affords the sign representation is 2 n−k−2 . The total number of loop-augmented forests on n vertices admitting the sign representation is 2 n−1 − 1.
Proof. The proof of the first claim follows from Corollary 5.1 and the formula (21). To prove the second claim, we first note that k can be any number from 2 to n. Since 1+2+· · ·+2 n−2 = 2 n−1 − 1, the proof is finished.

Non-diagonal idempotents
Let e denote the i-th constant idempotent c (i) . Observe that e stays unchanged by multiplication by permutations on the right and therefore the conjugation action on e is given by the left multiplication action; this is equivalent to the "standard representation" of S n on C n . In particular, the stabilizer subgroup of e is isomorphic to S 1 × S n−1 ≃ S n−1 .
Definition 6.1. Let e be an idempotent with k 1's, for 0 < k ≤ n, on its i-th row and 0's elsewhere. Then there exists σ ∈ S n such that the nonzero entries of σeσ −1 are in the first row and in the first k columns. In this case, we call σeσ −1 the standard form of e.
Note that a constant idempotent c (r) is already in its standard form. Note also that the standard form of an idempotent is unique. We will show in an algorithmic way that it is always possible to determine the permutation σ that leads to the standard form of e. We proceed to explain our algorithmic construction of the permutation σ for the idempotent e as in Definition 6.1. If i = 1, then skip the next three sentences. As e is an idempotent, the i-th entry in the i-th row of e has to be 1. Apply σ 1 := (1, i) to bring the nonzero entry at the (i, i)-th position to (1, 1)-th position. In fact, the application of σ 1 to e interchanges the i-th row with the first row and interchanges the i-th column with the first column. We denote the resulting idempotent by e 1 . Assuming i = 1, we look for the first occurrence of 1 in the first row (other than the one in the first position). If this 1 is at the j-th position with j > 2, then apply the transposition σ 2 := (2, j) to e 1 . σ 2 interchanges the second and the j-th rows and it interchanges the second and the j-th columns in e 1 . Thus, the 2-nd entry of the first row of e 2 := σ 2 · e 1 is now 1. We repeat this process, by choosing transpositions, σ 3 , σ 4 , . . . until we move all of the 1's in the first row to the left of the row. In summary, by conjugating e by a product σ s · · · σ 1 of not all disjoint transpositions we bring e to its standard form, the idempotent having 1's in the first k entries of its first row. Lemma 6.1. Let e be an idempotent from P n and let j 1 ≤ · · · ≤ j r denote the indices of those rows of e which have nonzero entries in them. If α i (i = 1, . . . , r) denotes the number of 1's in the j i -th row, then there exists a permutation σ ∈ S n such that σeσ −1 is a block diagonal matrix of the form where 0 m is the square 0 matrix of dimension m := n − α i .
Proof. The proof uses induction by first applying the standardization procedure outlined in the above paragraphs to the row j 1 . Instead of repeating what is written above, we demonstrate this procedure by an example.  The proof of our next observation is now straightforward; it follows from the similar arguments as above.
Corollary 6.1. Let e be an idempotent from P n . Then there exists a permutation σ ∈ S n such that σeσ −1 is a block diagonal matrix of the form where 0 m is the square 0 matrix of dimension m := n − β i and β 1 ≥ β 2 · · · ≥ β r .
We now generalize our previous definition of the "standard form of an idempotent". For an arbitrary idempotent e ∈ P n , we call its conjugate σeσ −1 = c (β 1 ) ⊕ · · · ⊕ c (βr) ⊕ 0 m as in (22) the standard form of e and denote it by st(e). An idempotent e is called standard if it satisfies the equation x = st(x). Our next task is to compute the stabilizer of a standard idempotent. Theorem 6.1. Let β 1 ≥ β 2 ≥ · · · ≥ β r be a sequence of positive integers and let e ∈ P n be a standard idempotent in which c (β 1 ) appears m 1 times, c (β 2 ) appears m 2 times, and so on. If s i=1 m i β i = n, then the stabilizer subgroup in S n of e is Stab Sn (e) = (S m 1 ≀ S β 1 −1 ) × (S m 2 ≀ S β 2 −1 ) × · · · × (S ms ≀ S βs−1 ).
To show that Stab S m i β i (c ii ) ≃ S m i ≀ S β i −1 (i = 1, 2), we assume without loss of generality that i = 1 and we decompose σ 11 into m 2 1 square blocks as in Note that each τ ij is a rook matrix. Since c 11 has copies of c (β 1 ) 's along its main diagonal and zero blocks elsewhere, it follows from the commutation relation σ 11 c 11 = c 11 σ 11 that for all i, j = 1, . . . , m 1 . It is not difficult to verify that unless τ ij is a permutation matrix with 1 at its (1, 1)-th entry, in order for relation (24) to hold, τ ij must be a zero matrix. In other words, in each row (of blocks) in (23) there exists exactly 1 block which is a permutation matrix of dimension β 1 with 1 at its (1, 1)-th entry. It is not difficult to verify that a permutation matrix τ ∈ S m 1 β 1 with m 2 1 square blocks just as described satisfies the commutation relation τ c 11 = c 11 τ , also. But such a permutation τ is uniquely determined by a permutation τ ′ ∈ S m 1 describing the positions of nonzero blocks of τ and in addition m 2 1 permutation matrices τ ′ ij ∈ S β 1 −1 which forms that lower-right sub-matrices of the nonzero blocks. This is precisely our definition of S m 1 ≀ S β 1 −1 . Corollary 6.2. If e ∈ P n is an idempotent, then there exists a multiset of integers {β m 1 1 , . . . , β ms s } and m ∈ N such that Proof. We know from Corollary 6.1 that by conjugating with an appropriate permutation σ ∈ S n we get σeσ −1 = e ⊕ 0 m , where e is a standard idempotent in P k for some 1 ≤ k ≤ n and m = n − k. The rest of the proof follows from Theorem 6.1.

The conjecture
Throughout this section, when the parts of a partition λ = (λ 1 , . . . , λ k ) satisfy λ 1 = · · · = λ k , we will write λ = λ k 1 . Let τ be a rooted forest with the Frobenius character F τ . Recall from Introduction that adding a root to τ has the effect of multiplying F τ with h 1 . Recall also that, if τ is a rooted tree, then the Frobenius character of the rooted forest that is comprised of m copies of τ is equal to h m [F τ ].
Example 7.1. Since h 1 is equal to the power-sum symmetric function p 1 , the Frobenius Example 7.2. The Frobenius character of the rooted tree that is obtained from n − 1 disjoint nodes (for n ≥ 2) by attaching them to a root is h 1 h n−1 . The rest of our paper is devoted to a proof of our conjectured extension of the Foulkes' conjecture in the special case where m = 2. In the light of Examples 7.1-7.3, our general conjecture can be stated in the language of symmetric functions as follows.
Conjecture 7.1. If n > m ≥ 2, then for any λ which has more than two parts, We will prove this conjecture in the case where m = 2. In particular, we shall show that for all n > 2, and λ = (2, 2n − 2), Before we can carry out our computations, we need to recall Littlewood's formulas for the plethysms h n [h 2 ] and h n [e 2 ], the Murnaghan-Nakayama rule to compute the product of power symmetric function p n times a Schur function s µ , and Chen's rule [2] which allows us to compute the plethysm p k [h n ].
We say that a partition λ = (λ 1 , . . . , λ k ) is even if λ i is even for all i. Then Littlewood [5] proved the following two results.
Next we introduce the notion of rim hook tableaux, special rim hook tabloids, and transposed special rim hook tabloids.
Given a Ferrers diagram F (λ) of a partition, a rim hook h of λ is a consecutive sequence of cells along the northeast boundary of λ such that any two consecutive cells of h share an edge and the removal of the cells of h from λ results in a Ferrers diagram of another partition. We let r(h) denote the number of rows of h and c(h) denote the number of columns of h. We say that h is special if h has at least one cell in the first column of F (λ) and h is transposed special if h has at least one cell in the first row of F (λ). We define the sign of h, sgn(h), to be sgn(h) = (−1) r(h)−1 . Figure 7: A rim hook of (2, 3, 4).
The Murnaghan-Nakayama rule gives us a rule for the Schur function expansion of a power symmetric function p n times a Schur function s µ . That is, p n s µ = µ⊆λ λ/µ is a rim hook of λ of size n sgn(λ/µ)s λ .
Chen's rule [2] states that p k [h n ], s λ = sgn(T ) if T is a special rim hook tabloid of shape λ and type k n ; 0 otherwise.
Thus to compute p k [s n ] = s n [p k ], we need only generate all the t-special rim hook tabloids T of type k n and replace T by sgn(T )s sh(T ) where sh(T ) denotes the shape of T . For example, Figure 10 pictures all the t-special rim hook tabloids of type 3 2 where instead of drawing a Ferrers diagram, we have indicated the cells of the Ferrers diagram by dots. Thus (6) . The computation of p 2 [h n ] is particularly straightforward via Chen's rule. That is, a t-special rim hook tabloid of type 2 n consists of bunch of vertical rim hooks followed by a bunch of horizontal rim hooks as pictured in Figure 11.
We want to compare h 2 [h 1 h n−1 ] to h n [h 1 h 1 ]. Obviously, the two expressions are equal when n = 2. Our first goal is to prove the following theorem which gives an explicit formula for the Schur function expansion of h 2 [h 1 h n−1 ].
Notation: For any logical statement A, we set Proof. Note that by (30) and the Pieri rules we have It is easy to see from the Pieri rules and the Murnaghan-Nakayama rules that the only shapes λ of Schur functions s λ that can arise in the Schur function expansions of products of the from h 2 1 s (a,2n−2−a) and p 2 s (a,2n−2−a) are (1 2 , a, b), (2, a, b), (1, a, b), (a, b), or (2n). We will consider each type of shape in turn. It follows from the Pieri rules that a product h 2 1 s (a,2n−2−a) is the sum over all s λ where we first add single cell on the outside of (a, 2n − 2 − a) to get a shape µ, which we will indicate by a cell containing the number 1, and then adding a single cell on the outside of the µ, which we will indicate by cell containing 2.
By the Murnaghan-Nakayama rule, a product of the form p 2 s (a,2n−2−a) is the sum over all ±s λ that arises by putting a rim hook of length 2 along the outside of (a, 2n − 2 − a), which we will indicate by a rim hook colored red, where the sign is plus if the rim hook is horizontal and −1 if the rim hook is vertical.
In this case, there is only one way to get a term of the form s (1 2 ,a,2n−2−a) , where 1 ≤ a ≤ n−1, from the products h 2 1 h 2 n−1 and p 2 [h n−1 ]p 2 which are pictured in Figure 12. That is, it is easy to see that s (1 2 ,a,2n−2−a) can only arise in one way from h 2 1 h 2 n−1 which comes the product h 2 1 s (a,2n−2−a) by placing the cells with 1 and 2 as pictured at the top of Figure 12. Similarly, s (1 2 ,a,2n−2−a) can only arise in one way from which comes the product p 2 (−1) a s (a,2n−2−a) by placing the rim hook as pictured in red at the bottom of Figure 12. Hence we get a contribution of (−1) a+1 s (1 2 ,a,2n−2−a) from p 2 p 2 [h n−1 ]. These two cancel out if 1 ≤ a ≤ n − 1 and a is even and they give a contribution of 1 2 2s (1 2 ,a,2n−2−a) = s (1 2 ,a,2n−2−a) if a is odd.
In this case, there is only one way to get a term of the form s (1 2 ,a,2n−2−a) , where 1 ≤ a ≤ n−1, from the products h 2 1 h 2 n−1 and p 2 [h n−1 ]p 2 which are pictured in Figure 13. That is, it is easy to see that s (1 2 ,a,2n−2−a) can only arise in one way from h 2 1 h 2 n−1 which comes the product h 2 1 s (a,2n−2−a) by placing the cells with 1 and 2 as pictured at the top of Figure 13. Similarly, s (1 2 ,a,2n−2−a) can only arise in one way from which comes the product p 2 (−1) a s (a,2n−2−a) by placing the rim hook as pictured in red at the bottom of Figure 13. Hence we get a contribution of (−1) a s (1 2 ,a,2n−2−a) from p 2 p 2 [h n−1 ]. This two cancel out if 1 ≤ a ≤ n − 1 and a is odd and a contribution of 1 2 2s (1 2 ,a,2n−2−a) = s (1 2 ,a,2n−2−a) if a is even. Thus the Schur functions in Case 2 contribute 2≤a≤n−1,a even In this case, we must consider two subcases.   Then it is easy to see that we can not obtain a Schur function of shape (1, a, 2n − 1 − a) by adding a rim hook of size two on the outside of two row shape. Thus s (1,a,2n−1−a) does not appear in the expansion of p 2 p 2 [h n−1 ]. However it can appear in the expansion of h 2 1 h n−1 in 4 different ways, namely, twice from the product product h 2 1 s (a−1,2n−a−1) and twice from the product h 2 1 s (a,2n−a−2) . These are pictured in Figure 15. Thus in Subcase 3.2, we get an additional contribution of 2 2≤a≤n−1 s (1,a,2n−1−a) . In this case, we have three subcases. In this case we have three ways to obtain s (1,2n−1) from h 2 1 h 2 n−1 , namely, two ways from the product h 2 1 s (2n−1) and one way from the product h 2 1 s (1,2n−3) . These three possibilities are pictured at the top of Figure 16. There is only one way to obtain s (1,2n−1) from p 2 p 2 [h n−1 ], namely, from the product p 2 s (1,2n−3) which comes with a sign of −1. Thus Subcase 4.1 contributes a factor of s (1,2n−1) . Subcase 4.2. a = n.
In this case we have two ways to obtain s (n,n) from h 2 1 h 2 n−1 , namely, one way from the product h 2 1 s (n−2,n) and one way from the product h 2 1 s (n−1,n−1) . These possibilities are pictured at the top of Figure 17. There are two ways to obtain s (n,n) from p 2 p 2 [h n−1 ], namely, once from the product p 2 s (n−2,n) which comes with a sign of (−1) n−2 and once from the product p 2 s (n−1,n−1) which comes with a sign of (−1) n . These possibilities are pictured at the bottom of Figure 17. Thus Subcase 4.2 contributes a factor of 2s (n,n) if n is even and 0 if n is odd.  In this case we have four ways to obtain s (a,2n−a) from h 2 1 h 2 n−1 , namely, once from the product h 2 1 s (a−2,2n−a) , once from the product h 2 1 s (a,2n−a−2) , and twice from the product h 2 1 s (a−1,2n−a−1) . These possibilities are pictured at the top of Figure 18. There are two ways to obtain s (a,2n−1) from p 2 p 2 [h n−1 ], namely, once from the product p 2 s (a−2,2n−a) which comes with a sign of (−1) a−2 and once from the product p 2 s (a,2n−a−2) which comes with a sign of (−1) a . These possibilities are pictured at the bottom of Figure 18 In this case, we have one way to obtain s (2n) from h 2 1 h 2 n−1 , namely, it can arise in one way from the product h 2 1 s (2n−2) . There also is only one way to obtain s (2n) from p 2 p 2 [h n−1 ], namely, from the product p 2 s (2n−2) which comes with a sign of 1. Thus Subcase 5 contributes a factor of s (2n) to h 2 [h 1 h n−1 ].

Note that
λ even,λ⊢2k Our goal is to show that for all partitions of 2n except λ = (2, 2n − 2), It follows from our previous theorem that the only s λ such that h 2 [h 1 h n−1 ], s λ = 0 is when λ is either contained in (2, 2n, 2n) or (1 2 , 2n, 2n). Now the only such s λ that appear in λ even,λ⊢2k s λ are of the form 1. s (2,a,2k−a−2) where a is even and 2 ≤ a ≤ n − 1, 2. s (a,2k−a) where a is even and 2 ≤ a ≤ n, or The only such s λ that appear in λ ′ even,λ⊢2(n−k) s λ are of the form Our goal now is to go through the six cases of the proof of Theorem 7.1 to show that in all but one case, h n [h 1 h 1 ], s λ ≥ h 2 [h 1 h n−1 ], s λ . However, we will need to apply the skew Schur function expansion rule of Remmel and Whitney [7]. Given a skew shape, λ/µ written in French notation, the reverse lexicographic filling of λ/µ, RLF (λ/µ), is the filling of the Ferrers diagram with the integers 1, . . . , |λ/µ|, which is obtained by filling the cells in order from right to left in each row, starting at the bottom row and proceeding upwards. For example, the RLF ((2, 4, 4, 4)/(1, 2)) is pictured in Figure 19. A standard tableau T of shape ν is said to λ/µ-compatible if and only if T satisfies the following two conditions.

R1
. If x and x + 1 lie in the same row in RLF (λ/µ), then x + 1 appear strictly to the right and weakly below x in T .
R2. If y is immediately above x in RLF (λ/µ), then y must appear strictly above and weakly to the left of x in T .

R2
x y x y Figure 20: The two rules for λ/µ-compatible standard tableaux.
A pictorial representation of these two conditions are depicted in Figure 20. We let λ/µ-ST denote the set of all λ/µ-compatible standard tableaux.
Note that expanding a product of Schur functions is a special case of expanding a skew Schur function. That is, it easy to see that s λ s µ is equal to the skew Schur function s λ * µ where λ * µ is the skew shape that arises by placing λ above µ so that the bottom row of λ starts just to the left of top row of µ. For example, the if λ = (2, 2) and µ = (1, 2), then λ * µ is pictured at the top-right of Figure 22. The other thing to observe is that if µ = (µ 1 , . . . , µ k ) where 0 < µ 1 ≤ · · · ≤ µ k , then in the expansion of s λ * µ , the numbers in the part of the reverse lexicographic filling of µ are completely determined. That is, 1, . . . , µ k must appear in row 1, reading from left to right, the numbers µ k + 1, . . . , µ k + µ k−1 must appear in row 2, reading from left to right, the numbers µ k + µ k−1 + 1, . . . , µ k + µ k−1 + µ k−2 must appear in row three, etc.. This means that in constructing our tree, we will eventually reach a single filling of shape µ and then we will continue to build our tree using the numbers of the reverse lexicographic filling that corresponding to λ. One can see an example of this phenomenon in Figure 21 in processing the numbers 1, 2, and 3 corresponding to the shape (1,2). This means that we might as will ignore the numbers in µ and simply build our tree starting with a the Ferrers diagram of µ and using the numbers 1, 2, . . . , |λ| to fill the part corresponding to λ in reverse lexicographic filling of λ * µ. An example of this process is pictured in Figure 22 for the Schur function expansion of s (2,2) s (1,2) . In this case, we see that s (2,2) s (1,2) = s (1,2,2,2) + s (1,1,2,3) + s (2,2,3) + s (1,3,3) + s (1,2,4) + s (3,4) .
RLF (λ/µ) In this case, we can give an explicit formula for h n [h 1 h 1 ], s (a,2n−a) . The key observation to make is that the only Schur function s λ which appears in the expansion of h n−k [e 2 ] where λ is a 2-part partition is s (n−k,n−k) . In addition, it is easy to see from our Skew Schur function expansion rule that the only s λ which appears in the expansion of a product s (n−k,n−k) s (a,b) where λ is a 2-part partition is s (n−k+a,n−k+b) . That is, one can see that when multiplying a Schur function s (a,b) by any Schur function s (c,c) , the only way to get a a shape with two parts is to put all the elements in first row of the reverse lexicographic filling corresponding to (c, c) in the first row and all the elements in the second row of the reverse lexicographic filling corresponding to (c, c) in the second row. This process is pictured in Figure 24.
Reverse lexicographic filling The only other Schur function corresponding to a two row shape that we have not considered is s (n,n) . In this case,