On a positivity conjecture in the character table of $S_n$

In previous work of this author it was conjectured that the sum of power sums $p_\lambda,$ for partitions $\lambda$ ranging over an interval $[(1^n), \mu]$ in reverse lexicographic order, is Schur-positive. Here we investigate this conjecture and establish its truth in the following special cases: for $\mu\in [(n-4,1^4), (n)]$ or $\mu\in [(1^n), (3,1^{n-3})], $ or $\mu=(3, 2^k, 1^r)$ when $k\geq 1$ and $0\leq r\leq 2.$ Many new Schur positivity questions are presented.


Introduction and Preliminaries
In this paper we consider Schur positivity questions related to the reverse lexicographic order on integer partitions. Recall that this total order is defined as follows [2, p. 6]. For partitions λ, µ of the same integer n, we say a partition λ is preceded by a partition µ in reverse lexicographic order if λ 1 > µ 1 or there is an index j ≥ 2 such that λ i = µ i for i < j and λ j > µ j . Thus for n = 4 we have the total order (1 4 ) < (2, 1 2 ) < (2 2 ) < (3, 1) < (4). In particular our convention is that the minimal and maximal elements in this total order are (1 n ) and (n) respectively. Our primary goal is to address the following conjecture: Conjecture 1. [6, Conjecture 1] Let L n denote the reverse lexicographic ordering on the set of partitions of n. Then the sum of power sum symmetric functions p λ , taken over any initial segment of the total order L n , i.e. any interval of the form [(1 n ), µ] for fixed µ, (and thus necessarily including the partition (1 n )), is Schur-positive.
In general, for arbitrary subsets T of partitions of n with (1 n ) ∈ T, the sums µ∈T p µ define (possibly virtual) representations of the symmetric group S n , of dimension n! There are many instances where Schur positivity fails; see the remarks following Example 5. Proposition 46 in Section 4 gives a lower bound for the number of failures.
Conjecture 1 has an equivalent formulation in terms of the character table of S n . If the columns of the table are indexed by the integer partitions of n corresponding to the conjugacy classes, in reverse lexicographic order, left to right, and the rows by the irreducible characters (hence also corresponding to partitions), then the conjecture states that, for each row, indexed by some fixed partition λ of n, the sum of the entries in the first k consecutive columns, beginning with the column indexed by (1 n ), is a nonnegative integer. If the kth column corresponds to the conjugacy class indexed by the partition µ, this row sum is the multiplicity of the Schur function s λ in the sum The questions treated in this paper are intimately connected to (and indeed motivated by) the representation theory of the symmetric group and the general linear group. For background on these topics we refer the reader to [2] and [5], in particular the Exercises in the latter reference.
Let ψ n denote the Frobenius characteristic of the conjugacy action on S n . The orbits of this action are the conjugacy classes. Let f n denote the Frobenius characteristic of the conjugacy action of S n on the class of n-cycles. Let h n , e n denote respectively the homogeneous and elementary symmetric functions of degree n, and let [ ] denote plethysm. By a general observation of Solomon [4] for finite groups (see also [5,Exercise 7.71], [6,Corollary 4.3]), we have the following facts. In view of Part (2) of the theorem below, Conjecture 1 may be seen as a generalisation of the Schur positivity of the sum of all power sums λ n p λ .
Theorem 2. The Frobenius characteristic ψ n of the conjugacy action of S n admits the following decompositions: (1) ψ n = λ n i h m i [f i ], where the partition λ has m i parts equal to i.
Definition 3. If µ is a partition of n, we write ψ µ for the sum of power sums λ∈[(1 n ),µ] p λ . More generally if T is any subset of partitions of n, define ψ T to be the sum µ∈T p µ .
Thus ψ (n) = ψ n , and the multiplicity of the Schur function s λ in ψ µ is the sum of the values of the irreducible character χ λ on the conjugacy classes in the interval [(1 n ), µ].
Our approach to Conjecture 1 proceeds in two directions. One can start at the bottom of the chain, with p n 1 (which contains all irreducibles), and add successive p λ 's going up the chain. The arguments in this case are subtle, and give an interesting decomposition of the corresponding representation. See Theorem 18. Alternatively, one can start at the top of the chain, with the known Schur-positive function ψ n , which is also known to contain all irreducibles (see Section 2), and examine what happens to the irreducibles upon subtracting successive p λ 's going down the chain, from ψ n . This is done in Theorem 23, and requires a careful analysis (Lemmas 19 to 21) of the Schur functions appearing in products of power sums. The technical difficulty here is in ensuring that the resulting expressions (Proposition 22) are reduced, i.e. each term corresponds to a unique Schur function. The argument now hinges on the following fact: no irreducible in the partial sum of power sums appears with multiplicity exceeding the lower bound, established in Lemma 13, for the multiplicity of each irreducible in ψ n .
The proof of Theorem 18 hints at interesting properties of the representations ψ (2 k ) . In Section 3 we present conjectures suggested by that proof, and establish more Schur positivity results (the case µ = (3, 2 k , 1 r ) for 0 ≤ r ≤ 2, Proposition 36), as well as generalisations of Theorem 23 to the twisted conjugacy action as defined in [6]. Section 4 concludes the paper with an analysis of the number of subsets of partitions whose associated sum of power sums is not Schur-positive. Tables of the Schur expansion of ψ µ appear in Section 5.

Intervals in Reverse Lexicographic Order
The following fact about the representation ψ n was first proved by Avital Frumkin. See [5, Solution to Exercise 7.71] for more references.
If n = 2, the representation ψ n contains all irreducibles.
We will need the following stronger result of [7] characterising the conjugacy classes containing all irreducibles. Recall (see [7] for references to the literature) that a conjugacy class in a finite group G is called global if the orbit of the conjugacy action corresponding to that class contains all irreducibles of G.
Theorem 9. [7, Theorem 5.1] Let n = 4, 8. Then the conjugacy class indexed by a partition λ contains all irreducibles, i.e. it is a global class, if and only if λ has at least two parts, and all its parts are distinct and odd. If n = 8, the conjugacy class indexed by (7, 1) is global, while the class of the partition (5, 3) contains all irreducibles except those indexed by (4 2 ) and (2 4 ).
We also require some information on the irreducibles appearing in f n , the S n -action by conjugation on the class of n-cycles. Since this is a permutation representation with one orbit, the trivial representation appears exactly once. It is also easy to see that the sign representation appears only if n is odd. We will make use of the following definitive result of Joshua Swanson: Theorem 10. ( [9], [6, Lemma 4.2]) Let n ≥ 1. If n is odd, the representation f n contains all irreducibles except those indexed by (n − 1, 1) and (2, 1 n−2 ). If n is even, f n contains all irreducibles except (n − 1, 1) and (1 n ).
The result below was stated without proof in [6]; we sketch a proof here. (There is a misprint in the statement of Part (4) in [6,Proposition 4.21] which is corrected below.) Proposition 11. [6, Proposition 4.21] The multiplicity in ψ n of the irreducible indexed by the partition (1) (n) is p(n), the number of partitions of n.
(2) (1 n ) is the number of partitions of n into parts that are distinct and odd, which is also the number of self-conjugate partitions of n. This multiplicity is nonzero for n = 2.
(3) (n − 1, 1) is λ n (|{i : m i (λ) ≥ 1}| − 1) , which in turn equals the number of distinct parts in all the partitions of n, minus the number of partitions of n. In particular this multiplicity is at least the number of non-rectangular partitions of n, and hence at least (n − 1).
where the first sum runs over the set T 1 of all partitions of n λ with parts that are distinct and odd, and the set T 2 consists of partitions λ of n with all parts odd and distinct except for one part of multiplicity 2, while T 3 consists of partitions with all parts odd and distinct except for exactly one even part.
Proof. Part (1) is clear since p(n) is the number of conjugacy classes of n. As alluded to in the Introduction, Part (2) is a computation of the sum µ n (−1) n− (µ) , and follows from the standard generating function identity for integer partitions by number of parts.
Since the sign representation always occurs in f n if n is odd, it also occurs in f 1 f n−1 if n is even, i.e. in the conjugacy class (n − 1, 1) for n = 2. The second statement of Part (2) follows. For Parts (3) and (4), by Frobenius reciprocity, we compute the multiplicity of the trivial (respectively sign) representation in the restriction of ψ n to S n−1 , using the partial derivative with respect to p 1 (see e.g. [2]) and Theorem 2 (1). Here we also need the fact that f n always contains exactly one copy of the trivial representation, and contains exactly one copy of the sign representation when n is odd, and none if n is even. The restriction of ψ n to S n−1 has Frobenius characteristic Let λ n. For each i such that m i (λ) ≥ 1, the inner sum contributes 1 to the multiplicity of the trivial representation for each distinct part i of λ. Thus in Part (3), since s (n−1,1) = h n−1 h 1 − h n , the required multiplicity is p(n) less than the multiplicity of s (n−1) in the restriction of ψ n to S n−1 , and the result follows.
Part (4) follows by a similar analysis, using standard facts about how the sign representation restricts to wreath products. Again let λ n, and consider a fixed i such that m i ≥ 1. By restricting the sign to the appropriate subgroup, we see that the multiplicity in h m j [f j ] is nonzero (and thus equal to 1) if and only if j is odd and m j = 1. Similarly the multiplicity in h m i −1 [f i ] is nonzero (and thus equal to 1) if and only if either i is odd and m i − 1 = 1, or i is even and m i − 1 = 0.
This implies that every term to the multiplicity if λ has all parts odd and distinct, but if λ ∈ T 2 ∪ T 3 , only one of these terms, namely when i is either the odd part of multiplicity 2, or the unique even part, makes a nonzero contribution (equal to 1). For all other λ the multiplicity is zero. This completes the proof.
(2) Let n ≥ 5, and let n be odd. Then the product f n f 2 contains all irreducibles except the sign. (3) Let n ≥ 5. If n is odd, every irreducible except the sign appears in each of the conjugacy classes (n − 2, 2) and (n − 2, 1, 1). (4) Let n ≥ 6. If n is even, every irreducible except the sign appears in f n−3 f 2 f 1 . (5) Let n ≥ 6 be even. Then every irreducible appears in f n f 2 except for the sign and the one indexed by (2, 1 n ), which does however appear in f n−2 f 3 f 1 . In particular all the irreducibles except for the sign appear among the two conjugacy classes (n, 2) and (n − 2, 3, 1).
Proof. For Part (1): The result is clear for n = 1, 2 so assume n ≥ 4. If n is odd this is immediate from Theorem 9. If n is even, then by Theorem 10, f n contains all irreducibles except (1 n ) and (n − 1, 1). Now s (n−1,1) · f 1 = s (n+1) + s (n,1) + s (n−1,2) . But each of these summands appears in the product g n · f 1 for g n = s (n) , s (n) , s (n−2,2) respectively, and each g n appears in f n . The only irreducible that does not appear is (1 n+1 ).
So let n ≥ 5 be odd. By Theorem 10, f n contains all irreducibles except those indexed by (n − 1, 1) and (2, 1 n−2 ). We have The first two summands appear in g n · f 2 for g n = s (n) · f 2 . The last two summands appear in g n · f 2 for g n = s (n−1,2) . Finally s (n−1,1,1) appears in the product g n · f 2 for g n = s (n−2,1, 1) , and this appears in f n for n ≥ 5. Thus in all cases g n appears in f n .
Next consider the other missing irreducible, (2, 1 n−2 ). We have The first three appear in the product g n · f 2 for g n = s (3,1 n−3 ) , which is a constituent of f n , n ≥ 5. The last one appears in the product g n · f 2 for g n = s (2 2 ,1 n−4 ) , which again is a constituent of f n , n ≥ 5. This completes the argument. For Part (3), observe that the conjugacy classes indexed by (n, 2) and (n, 1, 1) both afford the same representation, namely f n · h 2 = f n · f 2 . The result now follows from Part (2).
To establish the claim, we now need only show that the irreducible (2, 1 n−2 ) appears in f n−2 f 3 f 1 . But f 3 = h 3 + e 3 , so f n−2 f 3 contains all the irreducibles in the product f n−2 e 3 . Since n − 2 is even, it contains s (2,1 n−4 ) and this finishes the argument.
Lemma 13. Let n ≥ 5. Let do n denote the number of partitions of n with at least two parts and with all parts odd and distinct. In the conjugacy representation ψ n , every irreducible except possibly the sign occurs with multiplicity at least 4 + do n , n odd; 3 + do n , n even. This number is at least 5 for odd n ≥ 7, and at least 4 for even n ≥ 6.
Proof. First let n be odd. By Lemma 12, the following conjugacy classes contain all irreducibles except the sign: (n − 1, 1), (n − 2, 2), (n − 2, 1 2 ). Also by Theorem 10, the conjugacy class (n) (or equivalently the symmetric function f n ) contains all irreducibles except for the one indexed by (n − 1, 1) and (1 n ). But the irreducible (n − 1, 1) appears at least n − 1 times in ψ n , by Proposition 11. Thus we have multiplicity at least 4 for each irreducible. Since none of the four conjugacy classes listed above is global by Theorem 8, we have a multiplicity of at least 4 plus the number of global classes. Now let n be even, n ≥ 8. (The case n = 6 can be checked by direct computation. See, e.g. [6, Table 1]. Then by Theorem 10, the conjugacy class (n) has all irreducibles except for (n − 1, 1) and (1 n ). Also by Lemma 12, f n−3 f 2 has all irreducibles except for (1 n ). Hence so does the conjugacy class (n − 3, 2, 1). Finally this is also true by Lemma 12 again, for the sum (f n−2 f 2 + f n−4 f 3 f 1 ). We have accounted for a multiplicity of at least 3 for every irreducible except the sign, in addition to the global classes.
We now show that the number of global classes is at least k 2 if n = 2k, 2k + 1. First let n = 2k ≥ 6 be even. In this case, applying Theorem 9, we have at least Tables of the decomposition into irreducibles for ψ n , n ≤ 10, are given in [6]. We point out a misprint in Table 1 of [6] for n = 7 : the fifth entry from the bottom, for the multiplicity of (3, 1 4 ) in ψ 7 , should be 13, not 7. From this data, the truth of the lemma follows for n ≤ 10. It is worth noting that the tables indicate far greater lower bounds than we have just established, for the multiplicity of the irreducible indexed by µ when µ = (1 n ), (2, 1 n−2 ). We begin our analysis by directing our attention to the bottom of the chain, to examine the representations ψ µ for µ > (1 n ). Our argument in this case is somewhat mysterious. One interesting aspect is the role played by the following calculation.
Theorem 17. [6, Theorem 4.23] If T = {λ n : λ i = 1, 2 for all i} then p n,T is Schur-positive. We have p 2m+1,T = p 1 p 2m,T and Then ψ µ is Schurpositive. Equivalently, the following are Schur-positive: , for k ≤ n/2; one has the recurrence (1): where T = {λ 2k : λ i = 1, 2 for all i}. But by Theorem 17 we know that p 2k,T is Schur-positive as a representation of S 2k . For Part (2): Writing m = n 2 , since in reverse lexicographic order, (3, 1 n−3 ) covers the partition with at most one part equal to 1 and all other parts equal to 2, we have, We will establish the stronger claim that is Schur-positive. From Theorem 17, it suffices to assume that n = 2m. In this case, with T being the set of partitions of 2m with parts 1,2, we have where the notation Odd(n) is used to signify 1 if n is odd, and 0 otherwise, and we have set From eqns. (3) and (4), we need to establish the Schur positivity of ) + e m 2 Odd(m + 1). In fact when m is odd, we will show that V 2m−2 − p 2m−2 1 itself is Schur-positive, whereas when m is even, we will need to multiply this by h 2 and examine the entire expression (5) in order to obtain Schur positivity.
Combining this with eqn. (6), we obtain where by convention m−1 t−1 is zero if t < 1. We will split the first sum in eqn. (7) (over even t) into three sums as follows: and for 2 ≤ t ≤ m − 2, Next consider the negated terms in eqn. (7). For these we write, for odd k, 1 ≤ k ≤ m−1, Our goal is to absorb every negated term N k into a Schur-positive term. We now describe a judicious grouping which will allow us to accomplish this.
Collect the terms in V 2m−2 − p 2m−2 1 as follows: where t is even.
But this is Schur-positive by Lemma 15. This absorbs the negative terms N k for k odd, k ≤ m − 3, into Schur-positive expressions. Looking at eqn. (7), there is only one more negated term to investigate.
Suppose m is odd, so that k = m − 2 is odd, and thus N m−2 is the last negated summand in eqn. (7), the only one not taken care of in eqn. (8). Group the terms as before, and noting that P ((m−3)+) was NOT used in the groupings of eqn. (8), we have and this is again Schur-positive by Lemma 15. (Admittedly this is something of a miracle.) Next suppose m is even, so that k = m − 1 is odd; then N m−1 is the last negated summand in eqn. (7), and the only one not absorbed in eqn. (8). Now we have (since P ((m−2)+) was not used in any of the groupings in eqn. (8)): While this is not itself Schur-positive, from eqn. (5) we can multiply by h 2 , and then we have (again somewhat fortuitously), since m + 1 is odd, and this is Schur-positive as before. By eqn. (5), this completes the argument, in which Lemma 15 clearly played a crucial role.
Next we examine the chain from the top down. Here the arguments are more direct, but also computationally technical. Our strategy for establishing Schur positivity will be to show that in the Schur expansion of the sum ν n:µ<ν≤(n) p ν , i.e. starting from the partition (n) and moving down the chain, the positive multiplicities of the irreducibles never exceed those in ψ n . For the cases we consider, we are able to show that, with the exception of the trivial representation, which clearly occurs as many times as the number of partitions in the interval [µ, (n)], this sum has multiplicities at most 4. This allows us to apply the lower bounds for the multiplicities in ψ n developed earlier in Lemma 13. The Schur function expansion of the product p n p m figures prominently in this analysis.
In the proofs that follow, for simplicity and clarity we write λ for the Schur function s λ indexed by λ. The context should make clear when (n − 2, 2) indicates the Schur function s (n−2,2) rather than the partition itself. Also for any statement S, δ S denotes the value 1 if and only if S is true, and is zero otherwise.
Recall [2,5] that ω is the involution defined on the ring of symmetric functions which sends h n to e n . Thus ω takes the Schur function indexed by a partition λ to the Schur function indexed by the conjugate partition λ t . (In the character table of S n , it corresponds to tensoring with the sign representation.) Lemma 19. Let n ≥ m and 4 ≥ m ≥ 1; if n = m assume m = 2. Then the Schur function expansion of the product p n p m has only the coefficients 0, ±1. More precisely, one has the following expansion into distinct irreducibles: The number of irreducibles appearing in the expansion is Proof. Note that these definitions imply that We begin with the well-known expansion of p n into Schur functions of hook shape (a special case of the Murnaghan-Nakayama rule): The Murnaghan-Nakayama rule says the Schur functions in the product p n p m are indexed by partitions obtained by attaching border strips (or rim hooks) of size m to each of the above hooks, with sign (−1) s where s is one less than the number of rows occupied by the border strip. (See [2] or [5].) We enumerate the disjoint possibilities in the figures below. Note that we have excluded the partition (1 m ) (respectively, (m)) from Figure 1a because it is counted in Figure 2b  The contribution to p n p m here is The contribution to p n p m is If m = 3 we have a conjugate pair of additional configurations: The contribution to p n p 3 is (−1) 2 (n − 1, 2, 2). The contribution to p n p 3 is (−1) n−2 · (−1)(3, 3, 1 n−3 ).
The sum of hooks in Figures 2a-2b collapses as follows: if n − 1 ≥ m, we can split the first sum: If n = m, this latter expression coincides with ( * ) above, and is thus valid for all n ≥ m.
Note that in the double hook of Figure 3, we must have n − r ≥ m − s + 1 ≥ 1 and similarly r > s ≥ 0. This gives 0 ≤ s ≤ m − 1 and s < r < n − m + s. Hence Figure 3 contributes the sum of double hooks putting β(n, m) for the first sum above, we see that this can by rewritten as δ m≥2 β(n, m) + (−1) n−m ω(β(n, m) .
This completes the proof.
We now specialise this lemma to the values m ≤ 4. In what follows it will be convenient to write partitions of n as ( * , µ), where * will indicate a single part equal to n − |µ|, and µ is a partition whose largest part does not exceed the part indicated by * . For example, ( * , 2, 2, 1 t ) means the partition (n − 4 − t, 2, 2, 1 t ).
Lemma 20. One has the following Schur function expansions: (Note that the summations are nonzero if and only if n ≥ 6). For Part (7): We start with the expression for p n−3 p 2 and multiply by p 1 . One checks that this gives Note that the sums in (A1) and (A2) vanish identically unless n ≥ 7. The first and last summands in (A1) collapse to (−(n − 3, 3) + (−1) n−6 (3 2 , 1 n−6 ), and similarly the first and last summands in (A2) collapse to −(n − 4, 2 2 ) + (−1) n−6 (2 3 , 1 n−6 ). Also note that where we have written 1 t−1 for part 1 with multiplicity t − 1, there is no contribution unless t ≥ 1. So the third sum in (A1) and the second sum in (A2) cancel each other.
For (3), we compute the partial sum by using p 2 + p 2 1 = 2h 2 and thus it suffices to add the expansion of 2h 2 p n−2 from Lemma 20 to the preceding partial sum.
In general, in all cases we use the relevant computations of Lemma 20, the chief exception being the expression in (10), which requires the expansion of p n−4 p 2 2 computed in Lemma 21. The expression (11) of this proposition then follows cumulatively using Part (11) of Lemma 20.
It is important to note that the sums have been carefully rewritten so that there is no "collapsing": as an example, we give here an analysis of what happens in computing the partial sum (10). This sum is obtained by adding to the sum in (9) the expansion for p n−4 p 2 2 , which from Lemma 21 is The items of matching colour (or matching labels, in the absence of colour) can be combined as follows: Making these replacements finally yields the completely reduced expression (10). Likewise, the reduced expression (11) is obtained by adding to (10) the expansion of p n−4 p 2 p 2 1 . There is only one pair that recombines into one term here, namely We can now deduce the positivity of the functions ψ µ for µ ≥ (n − 4, 1 4 ). We use the partial sum computations in Proposition 22. Observe that, in each of those expansions, no Schur function appears with multiplicity greater than +4, except for the trivial representation, which appears with multiplicity equal to the number of partitions in the interval (µ, (n)]. For example, for ψ (n−4,3,1) we see from (8) that we need to subtract from ψ n a virtual representation in which no multiplicity in the sum exceeds +3, other than the multiplicity of the trivial representation which is now 8.
Similarly to obtain ψ (n−4,2 2 ) , from (9) it follows that we subtract from ψ n a representation in which no multiplicity in the sum exceeds +3, other than the multiplicity of the trivial representation which is now 9.
But Lemma 13 guarantees that ψ n has multiplicity at least 4 for each irreducible except the trivial module, and hence, examining the partial sums in Proposition 22, it is clear that the right-hand side is Schur-positive in all the cases enumerated. The fact that all the expressions are reduced (no further simplification occurs) is important in this argument. The multiplicity of the trivial representation is the partition number p(n), which is certainly at least the length of the interval (µ, (n)]. The theorem is proved.
Together Theorems 18 and 23 complete the proof of Theorem 4. It is difficult to see how to generalise this argument. Already for n = 8, 9, 10, computation with Maple shows that in the Schur function expansion of the sum λ≥(n−4,1 4 ) p λ , s (n−3,3) occurs with multiplicity −6; s (n−4,4) occurs with multiplicity −5 when n = 9, 10, and s (n−4,3,1) occurs with multiplicity −5 for n = 8. The lower bound that we were able to establish in Lemma 13 is therefore insufficient to guarantee Schur positivity of ψ µ by these arguments, in the case when µ is strictly below (n − 4, 1 4 ) in reverse lexicographic order.
From Theorem 18 and Proposition 11 it is also easy to derive the following information about the multiplicity of the sign representation. Clearly if µ and ν are consecutive partitions in reverse lexicographic order, this multiplicity differs by 1 in absolute value, since ψ µ − ψ ν = ±p ν . It is also clear that the multiplicity of the trivial representation decreases by one as we descend the chain from (n) to (1 n ).
Denote by , the inner product on the ring of symmetric functions for which the Schur functions form an orthonormal basis.
This establishes the induction step and hence claim (3). In view of (A) above, the positivity of the multiplicities of hooks in Hk n implies that Hk n is Schur-positive for all n. The last statement is clear.
The representations Hk n give rise to an interesting family of nonnegative integers. Let a n,r = s (n−r,1 r ) , Hk n , 0 ≤ r ≤ n − 1. Table 1 gives the first few values of this sequence.
Let Lie n denote the S n -module obtained by inducing a primitive nth root of unity from the cyclic subgroup generated by an n-cycle up to S n . It is a well-known fact that Lie n is also the representation of the symmetric group acting on the multilinear component of the free Lie algebra ([5, Ex. 7.88-89]). Write Lie for n≥1 ch Lie n . Recall from the Introduction that we denote by f n the conjugacy action on the n-cycles, and that f [g] denotes plethysm. The functions Hk n satisfy an interesting plethystic identity. In order to establish this, we need the following connection between Lie and the conjugacy action. In keeping with the notation of [8], we will write Conj n for f n in the remainder of this section. Proposition 27. Let W n be the representation with characteristic Hk n . Then W n satisfies the following properties: (1) The restriction W n+1 ↓ Sn from S n+1 to S n is isomorphic to the direct sum of W n and the induced module (W n ↓ S n−1 ) ↑ Sn . (2) Conj n is the degree n term in Proof. The following symmetric function identity is immediate from the definition of Hk n : Hk n .

The representations ψ 2 k and the twisted conjugacy action
The functions ψ 2 k of Theorem 18 appear to have interesting properties. We state separately the following consequence of the proof of Theorem 18: is Schur-positive, and hence so is Proof. The second statement follows by induction from the Schur positivity of expression (2), upon writing In fact the following stronger statement appears to be true.
is Schur-positive. This has been verified for k ≤ 16.
However, we do have the following: Lemma 31. The symmetric function ψ 2 m − h 2 ψ 2 m−1 is Schur-positive. More generally, for k ≤ m, the function ψ 2 m − h k 2 ψ 2 m−k is Schur-positive. Proof. Eqn. 3 of Theorem 18 gives The more general statement follows from the telescoping sum Proposition 32. Let k, r ≥ 1, and let m = r 2 . (So m = r+1 2 if r is odd, and m = r 2 if r is even.) We have Iterating this, the last two lines of this recurrence are , where in the last line a = 0 if r + 2k + 3 = 2m is even, i.e. if r is odd, and a = 1 if r + 2k + 3 = 2m + 1 is odd, i.e. if r is even.
This telescoping sum collapses to give

The proposition follows.
This leads us to make the following conjecture, which has been shown to be true in Lemma 31 for m = 1 : Conjecture 33. Let k, m ≥ 1. Then ψ 2 k+m − h 2 p 2m−2 1 ψ 2 k is Schur-positive, and hence so is ψ 2 k+m − h r 2 p 2m−2r 1 ψ 2 k , 1 ≤ r ≤ m. We have verified this for 1 ≤ k, m ≤ 5.
In view of Proposition 32, the truth of this conjecture would immediately imply Schur positivity of ψ (3,2 k ,1 r ) for all r, k ≥ 1.
Proof. We use Proposition 32. First let r = 1. Then we have and is thus Schur-positive by Lemma 35. But the last expression in brackets is also Schur-positive by Lemma 31.
If r = 0, Propostion 32 reduces to and again this is Schur-positive by Lemma 31. Finally if r = 2, Proposition 32 gives invoking Lemmas 31 and 35, this is Schur-positive as before.
This argument fails for r = 3. Proposition 32 then gives is no longer Schur-positive. In previous work of this author, a sign-twisted conjugacy action of S n was defined in terms of the exterior powers of the conjugacy action, and the following analogue of Theorem 2 was established (recall that f n is the characteristic of the conjugacy action on the class of n-cycles): Theorem 37. Note that ε n is self-conjugate, so in particular the multiplicities of the trivial and sign representations coincide (and are equal to the number of partitions of n with all parts odd). Based on character tables up to n = 10, we were led to make a conjecture in the spirit of Conjecture 1, which we have subsequently verified for n ≤ 28.
Let µ n be a partition with all parts odd. Define Conjecture 38. Let µ n be a partition with all parts odd. The symmetric function ε µ is Schur-positive.
The chains in reverse lexicographic order are now as follows: (1) If n is odd:  Proof. Note that ε (3 r ,1 n−3r ) = p n−3r 1 ε ν where ν = (3 r ). But ε ν is the sum of power sums for λ in the set T 3r consisting of all partitions with parts equal to 1 or 3. By [6,Theorem 4.23], this is Schur-positive.
An analogue of Theorem 8 holds here as well. It is also a consequence of Theorem 9, since the global classes defined there are also conjugacy classes appearing in ε n .
Theorem 40. [6, Theorem 4.9, Proposition 4.22] The representation ε n contains all irreducibles. The multiplicity of the trivial representation (and hence also the sign) is the number of partitions of n into odd parts. In particular this multiplicity is at least n 2 ≥ 3 for n ≥ 5. The last statement in the theorem is simply a consequence of the observation that if n is odd, the partitions (n − 2r, 1 2r ), 0 ≤ r ≤ n−1 2 all have odd parts, while if n is even, the partitions (n − 1 − 2r, 1 2r ), 0 ≤ r ≤ n−2 2 all have odd parts. Proposition 41. Let n be odd. Then ε (n−2,1 2 ) , ε (n−4,3,1) and ε (n−4,1 3 ) are all Schurpositive.
Recall that ω denotes the involution on the ring of symmetric functions which sends h n to e n . Another result of [6] states that Theorem 43. [6,Theorem 4.11] The sum λ n n− (λ) even p λ equals 1 2 (ψ n + ω(ψ n )) and is Schur-positive.
Similarly we have, for any partition µ of n, 1 2 (ψ µ + ω(ψ µ )) = (1 n )≤λ≤µ n− (λ) even p λ . This leads us to make the following conjecture, which has been verified for n ≤ 20 : Conjecture 44. Let µ n. The sum Clearly Conjecture 1 implies Conjecture 44. Maple computations with the character table of S n show that the sum λ∈T p λ is NOT Schur-positive for arbitrary subsets T containing (1 n ) and consisting of all partitions λ with n − (λ) even. The first counterexample occurs only for n = 14, and there are then at least 2 11 such subsets for which Schur positivity fails.
Note that if we require that n− (λ) be odd, but also include the regular representation in the sum, the preceding conjecture is false: Question 45. In [6] and [8], S n -modules are constructed whose characteristics are multiplicity-free sums of power sums, thereby settling the Schur positivity question in these cases. Is there a representation-theoretic context for the sums ψ µ ?

Arbitrary subsets of conjugacy classes
In this section we examine the following more general question: Let f (n) be the number of subsets of {p λ : λ n} containing p n 1 , and having the property that the sum of their elements is NOT Schur-positive. What can be said about f (n)? Richard Stanley computed the values of f (n) for n ≤ 7 after seeing a preprint of [6]. Table 2 extends these values up to n = 10.
Recall from Section 1 that ψ T denotes the Schur function µ∈T p µ . The analysis of the multiplicity of the sign representation in Example 5 suggests a way to obtain a lower bound for the numbers f (n). Indeed, let A(n) = {µ n : n − (µ) is even}, and let B(n) = {µ n : n − (µ) is odd}. Let α(n), β(n) respectively be the cardinalities of A(n), B(n). Clearly α(n) + β(n) = p(n). As in [6,Proposition 4.21] (see also eqn. (1)), is the number of self-conjugate partitions of n, and hence α(n) ≥ β(n). By manipulating generating functions it can be seen that α(n) is also the number of partitions of n with an even number of even parts, and an arbitrary number of odd parts. The sequence appears in [3, A046682].
Proposition 46. Let T be a subset of the set of partitions of n not containing the partition (1 n ). The Schur function indexed by (1 n ) appears with negative multiplicity in the Schur expansion of ψ T ∪{(1 n )} if and only if |T ∩ B(n)| ≥ 2 + |T ∩ A(n)|. Hence the number of such subsets gives the following lower bound for f (n): In particular f (n) is positive for all n ≥ 4.
Proof. This is immediate from the preceding discussion and the fact that each p µ contributes (−1) n− (µ) to the multiplicity of (1 n ) in ψ T . A simple count then tells us that this multiplicity is negative for exactly as many subsets T as given by the following sum: This is precisely the sum of the coefficients of the powers of x j , j ≥ 2, in the Laurent series expansion of Since α(n) + β(n) = p(n), this in turn is the sum of the coefficients of the terms x j in (1 + x) p(n)−1 , for j ≥ α(n) + 1, i.e: Now replace j with p(n) − 1 − i. The last claim follows because α(n) is the number of partitions with an even number of even parts and thus α(n) ≤ p(n) − 2. (If n ≥ 4, exclude the partitions (2, 1 n−2 ) and (n) if n is even, (n − 1, 1) if n is odd.) Table 2 includes data up to n = 10, and the resulting lower bound b(n) on the number f (n) of non-Schur-positive functions ψ T , omitting the trivial values f (n) = 0 for n ≤ 3.  Proof. Write χ µ for the irreducible character indexed by the partition µ. Recall that the value of χ (n−1,1) (λ) is one less than the number m 1 (λ) of parts of λ which are equal to 1, and is therefore never less than −1. Hence we have for the (p(n) − p(n − 1)) partitions λ with m 1 (λ) = 0, 0 for the(p(n − 1) − p(n − 2)) partitions λ with m 1 (λ) = 1, ≥ 1 for the p(n − 2) partitions λ with m 1 (λ) ≥ 2.
Consider the conjugacy classes indexed by the p(n) − p(n − 1) partitions with no part equal to 1, and the partition (1 n ). The row sum indexed by (n − 1, 1) in the character table of S n will then be n − 1 − (p(n) − p(n − 1)). The first claim follows by observing that p(n) − p(n − 1) first exceeds χ (n−1,1) (1 n ) = n − 1 when n = 10, and the fact that the values p(n) − p(n − 1) increase.
Of course we could also append to the set T above any of the 2 p(n−1)−p(n−2) subsets of conjugacy classes with exactly one fixed point (since these do not contribute to the multiplicity of (n − 1, 1)), to obtain even more non-Schur-positive instances of ψ T ; for the number of subsets with negative multiplicity for (n − 1, 1) this gives a lower bound of (12) 2 p(n−1)−p(n−2) p(n)−p(n−1)−n j=0 p(n) − p(n − 1) n + j .
Combining these two quantities, we have that the number of conjugacy classes for which the value of χ (2,1 n−2 ) is negative is |C 1 | + |C 2 |, where C 1 = {µ n : n − (µ) is even and µ has no singleton parts} and C 2 = {µ n : n − (µ) is odd, µ has at least two singleton parts}.
Note that the lower bound of Proposition 46 surpasses the two lower bounds obtained above. In order to test the Schur positivity of ψ T , we need to examine the multiplicity of the irreducible indexed by each λ n in ψ T . This is given by a T (λ) = f λ + µ∈T :µ =(1 n ) χ λ (µ), where f λ = χ λ ((1 n )) is the number of standard Young tableaux of shape λ. Now χ λ ((1 n )) is larger than any other value of the character χ λ . Hence one way in which we can see how to make these values negative is to find µ such that f λ + χ λ (µ) is small relative to the number p(n) of conjugacy classes. For instance: Proposition 48. Let λ n, and let τ = (2, 1 n−2 ) be the (conjugacy class of ) a single transposition. Then χ λ ((1 n )) + χ λ (τ ) equals (1) 2 if λ = (2, 1 n−2 ), Proof. The first part has already been observed in the proof of Proposition 47. For the rest, we use the formula . When λ dominates λ we must have b(λ ) < b(λ) and thus χ λ (τ ) is negative.
An examination of the character tables of S n leads to the following observations. The use of character tables eliminates the need for Stembridge's SF package for Maple, by means of which the values f (n) were originally calculated, up to n = 8.
• For n = 6, of the 184 subsets that fail to be Schur-positive, exactly 176 fail to be Schur-positive because of the irreducible (1 6 ), another 4 fail because the irreducible (2, 1 4 ) appears with negative coefficient, and the remaining 4 fail because of the irreducible (3 2 ). From the character table of S 6 , it is easy to identify these 8 subsets. (In each of these cases no other irreducibles occur with negative coefficient.) • For n = 7, the number of subsets failing Schur positivity because of (a negative coefficient for) (1 7 ) is 3473, and 384 were identified as failing (in part) because of the irreducible (2, 1 5 ). The count for subsets in which both irreducibles appear with negative coefficient is 183, and this confirms f (7) = 3674. From the character table of S 7 , it is easy to verify that the number of subsets T resulting in a negative coefficient for (2, 1 5 ) in ψ T is exactly 384, and also that no other irreducibles occur with negative coefficient in any subset. = 0.22913. The character table shows that in addition to (1 9 ) and (2, 1 7 ), only the irreducibles (2 2 , 1 5 ) and (3, 1 6 ) will appear with negative coefficient in some subsets. The value of f (9) was calculated by exploiting this fact, and took 6.8 hours in Maple. However, the C code ran in only 36 seconds.
Tables 3a and 3b below contain, for each n, the values of the function g(n), defined to be the number of partitions µ of n such that, for some subset T containing (1 n ), the irreducible indexed by µ appears with negative multiplicity in ψ T . Table 3a n 4 5 6 7 8 9 10 11 12 p(n) 5 7 11 15 22 30 42 56 77 g(n) 1 1 3 2 2 4 5 6 8 Based on our computations, we make the following conjecture: Conjecture 49. For n ≥ 6, the numbers f (n) 2 p(n)−1 are bounded below by 1 16 , above by 1 2 , and are strictly increasing.
This would imply an affirmative answer to a question raised by Richard Stanley: Conjecture 50. The numbers f (n) 2 p(n)−1 approach a limit strictly between 0 and 1. We close this section with a list of the cases of Schur positivity known to us, for subsets of conjugacy classes.
The symmetric function ψ T is Schur-positive for the following subsets T of the set of partitions of n. In nearly all cases one can describe an S n -module whose Frobenius characteristic is given by ψ T . (The analogous statement holds trivially for arbitrary finite groups, since for any irreducible χ, the value of the character χ(g) is a sum of χ(1) roots of unity, and hence its absolute value cannot exceed the degree χ(1). )