THE EULERIAN REPRESENTATIONS OF Σ_{n} AS

RESTRICTIONS OF REPRESENTATIONS OF Σ_{n+1}

Sarah Whitehouse

Abstract

We show that each Eulerian representation of Σ_{n} is the restriction of a representation of Σ_{n+1}. We describe the new representations, giving character formulae, and identify the one which restricts to the first Eulerian representation as the tree representation.

Introduction

The Eulerian idempotents e_{n}^{(j)}, for j = 1, …, n, lying in the rational group algebra of the symmetric group ℚΣ_{n}, were defined by Gerstenhaber and Schack as follows [3]. An (i, n − i)-shuffle in Σ_{n} is a permutation π such that π(1) < π(2) < ⋯ < π(i) and π(i + 1) < π(i + 2) < ⋯ < π(n). Let s_{i,n−i} = ∑ (sgnπ)π ∈ ℚΣ_{n}, where the sum is over (i, n − i)-shuffles in Σ_{n}, and let s_{n} = ∑_{i=1}^{n−1} s_{i,n−i} ∈ ℚΣ_{n}. Now s_{n} has minimum polynomial ∏_{j=1}^{n}(x − μ_{j}), where μ_{j} = 2^{j} − 2. Then define

e_{n}^{(j)} = ∏_{i≠j} \frac{s_{n} − μ_{i}}{(μ_{j} − μ_{i})}.

The e_{n}^{(j)}’s for j = 1, …, n form a family of mutually orthogonal idempotents such that ∑_{j=1}^{n} e_{n}^{(j)} = 1 ([3]; Theorem 1.2).

These idempotents provide decompositions of Hochschild and cyclic homology of a commutative algebra over a ground ring which contains ℚ [3], [8]. We briefly recall the definitions since in particular we will need a property of Connes’ B map later. For A an associative algebra over k and M an A-bimodule, the Hochschild complex is C_{n}(A; M) = M ⊗ A^{⊗n}, with boundary b : C_{n}(A; M) → C_{n−1}(A; M) given by

b(m ⊗ a_{1} ⊗ ⋯ ⊗ a_{n}) = (ma_{1} ⊗ a_{2} ⊗ ⋯ ⊗ a_{n})

+ ∑_{i=1}^{n−1} (−1)^{i}(m ⊗ a_{1} ⊗ ⋯ ⊗ a_{i}a_{i+1} ⊗ ⋯ ⊗ a_{n})

+(−1)^{n}(a_{n}m ⊗ a_{1} ⊗ ⋯ ⊗ a_{n−1}).

Here ⊗ denotes ⊗_{k}. The Hochschild homology of A with coefficients in M, denoted HH_{n}(A; M), is the homology of this chain complex. The symmetric group Σ_{n} acts on the left on C_{n}(A; M) by

σ(m ⊗ a_{1} ⊗ ⋯ ⊗ a_{n}) = (m ⊗ a_{σ^{−1}(1)} ⊗ ⋯ ⊗ a_{σ^{−1}(n)}),

Typeset by AMS-TEX

and this extends linearly to an action of the group algebra kΣ_{n}. Then if A is commutative and the ground ring k contains ℚ, the Eulerian idempotents commute with the Hochschild boundary map b, be_{n}^{(j)} = e_{n−1}^{(j)}b, so that they yield a decomposition of Hochschild homology. The first part of this decomposition, given by the idempotents e_{n}^{(1)}, is Harrison homology [3].

Letting Ā = A/k, we may define the cyclic homology of A over k, denoted HC_{∗}(A), as the homology of the total complex corresponding to the normalised (b − B) bicomplex:

⋮  ⋮  ⋮
b↓  b↓  b↓
A ⊗ Ā^{⊗n} ⟵^{B} A ⊗ Ā^{⊗(n−1)} ⟵^{B} A ⊗ Ā^{⊗(n−2)} ⟵^{B} ⋯
b↓  b↓  b↓
⋮  ⋮  ⋮
b↓  b↓
A ⊗ Ā ⟵^{B}  A
b↓
A

where B : A ⊗ Ā^{⊗n} → A ⊗ Ā^{(n+1)} is defined by

B(a_{1}⊗a_{2}⊗⋯⊗a_{n+1}) = ∑_{j=1}^{n+1} (−1)^{n(j−1)}(1⊗a_{j}⊗a_{j+1}⊗⋯⊗a_{n+1}⊗a_{1}⊗a_{2}⊗⋯⊗a_{j−1}).

Now, for a commutative algebra A, over a ground ring k containing ℚ, the Eulerian idempotents are well-behaved with respect to B as well as b, Be_{n−1}^{(j−1)} = e_{n}^{(j)}B, so that they decompose cyclic homology ([8], 4.5.7).

The representation of Σ_{n} given by the right ideal e_{n}^{(j)}ℚΣ_{n}, which we denote by E_{n}^{(j)}, has been studied by Hanlon, who gives a character formula [5]. We show that this representation is the restriction of a representation of Σ_{n+1}, denoted F_{n+1}^{(j)}, given by a closely related idempotent f_{n+1}^{(j)} in ℚΣ_{n+1}. By first finding a simplified formula for the product e_{n}^{(j)}e_{n+1}^{(j)} in ℚΣ_{n+1}, we give a description of the representation F_{n+1}^{(j)} as a virtual representation:

F_{n+1}^{(j)} ⊕ ⨁_{i=1}^{j} E_{n+1}^{(i)} ≅ ⨁_{i=1}^{j} Ind_{Σ_{n}}^{Σ_{n+1}}E_{n}^{(i)}.

This leads to a character formula using Hanlon’s results. In the case j = 1, F_{n+1}^{(1)} is the tree representation [10].

THE EULERIAN REPRESENTATIONS OF Σ_{n} AS RESTRICTIONS 3

1. The idempotents f_{n+1}^{(j)}

We denote by λ_{n+1} the n + 1 cycle (1 2 … n + 1) in Σ_{n+1} and let Λ_{n+1} = \frac{1}{n+1} ∑_{i=0}^{n}(sgnλ_{n+1}^{i})λ_{n+1}^{i} ∈ ℚΣ_{n+1}. We will always regard Σ_{n} as contained in Σ_{n+1} as the subgroup of permutations fixing n + 1, and similarly ℚΣ_{n} ⊂ ℚΣ_{n+1}. 1.1 Proposition. Λ_{n+1}s_{n} = s_{n}Λ_{n+1}. Proof. A typical term on the right-hand side of this equation is πλ_{n+1}^{j}, appearing with sign, sgn(π).sgn(λ_{n+1}^{j}) = sgn(πλ_{n+1}^{j}), where π is some shuffle in Σ_{n}. Now we may write πλ_{n+1}^{j} = λ_{n+1}^{π(j)}π′, where π′ = λ_{n+1}^{−π(j)}πλ_{n+1}^{j} is in Σ_{n}. Let θ_{j} : Σ_{n} → Σ_{n} be defined by θ_{j}(π) = λ_{n+1}^{−π(j)}πλ_{n+1}^{j}. When j = n + 1 we simply have the identity map, and for j = n it was proved by Natsume and Schack that θ_{n} is a bijection which takes shuffles to shuffles ([9]; Lemma 9). Since it is easily checked that θ_{n−k} = (θ_{n})^{k+1} the same holds for each θ_{j}. So each term of the right-hand side, πλ_{n+1}^{j} with sign, appears in the left-hand side as λ_{n+1}^{π(j)}π′, with π′ a shuffle, and with sign sgn(λ_{n+1}^{π(j)}).sgn(π′) = sgn(λ_{n+1}^{π(j)}π′) = sgn(πλ_{n+1}^{j}). □

1.2 Corollary. Λ_{n+1}e_{n}^{(j)} = e_{n}^{(j)}Λ_{n+1} for j = 1, …, n.

Proof. Each e_{n}^{(j)} is a polynomial in s_{n}, so this is immediate from the above. □

Thus Λ_{n+1}e_{n}^{(j)} is an idempotent in ℚΣ_{n+1}.

1.3 Definitions. We denote by f_{n+1}^{(j)} the idempotent element Λ_{n+1}e_{n}^{(j)} in ℚΣ_{n+1}, for j = 1, …, n. We let E_{n}^{(j)} and F_{n}^{(j)} denote the ℚΣ_{n}-modules e_{n}^{(j)}ℚΣ_{n} and f_{n}^{(j)}ℚΣ_{n} respectively.

1.4 Proposition. The representation F_{n+1}^{(j)} of Σ_{n+1} when restricted to a representation of Σ_{n} is isomorphic to E_{n}^{(j)}.

Proof. Consider the homomorphism of right ℚΣ_{n}-modules θ : E_{n}^{(j)} → F_{n+1}^{(j)} given by left multiplication by Λ_{n+1}. Now since Λ_{n+1} and e_{n}^{(j)} commute, and since we may write π ∈ Σ_{n+1} uniquely as λ_{n+1}^{i}π′ for some i and some π′ ∈ Σ_{n}, we have Λ_{n+1}(sgnλ_{n+1}^{i})e_{n}^{(j)}π′ = (sgnλ_{n+1}^{i})f_{n+1}^{(j)}π′ = f_{n+1}^{(j)}λ_{n+1}^{i}π′ = f_{n+1}^{(j)}π . Hence, the homomorphism of right ℚΣ_{n}-modules which is given by f_{n+1}^{(j)}π ↦ (sgnλ_{n+1}^{i})e_{n}^{(j)}π′ for π ∈ Σ_{n+1} as above, is an inverse for θ. So F_{n+1}^{(j)} and E_{n}^{(j)} are isomorphic as ℚΣ_{n}-modules as required. □ 1.5 Proposition.

⨁_{j=1}^{n} F_{n+1}^{(j)} ≅ Ind_{<λ_{n+1}>}^{Σ_{n+1}}(ϵ),

where ϵ denotes the sign representation of the cyclic subgroup < λ_{n+1} > of Σ_{n+1}. Proof.

∑_{j=1}^{n} f_{n+1}^{(j)} = Λ_{n+1} ∑_{j=1}^{n} e_{n}^{(j)} = Λ_{n+1},

since ∑_{j=1}^{n} e_{n}^{(j)} = 1, ([3]; Theorem 1.2). So the sum of the representations F_{n+1}^{(j)} is Λ_{n+1}ℚΣ_{n+1}. Since Λ_{n+1} is the standard idempotent for the sign representation of the cyclic subgroup of Σ_{n+1} generated by λ_{n+1}, Λ_{n+1}ℚΣ_{n+1} is the claimed induced representation. □

2. A relation between e_{n}^{(j)} and e_{n+1}^{(j)}

In this section we prove certain relations among the e_{n}^{(j)}’s and f_{n}^{(j)}’s, which will be needed in the following section to give descriptions of our representations. The main result is Proposition 2.5, giving a simplification of the product e_{n}^{(j)}e_{n+1}^{(j)}. We adopt the convention that f_{n}^{(k)} = e_{n}^{(k)} = 0 whenever k ≤ 0 or k > n.

2.1 Lemma. f_{n+1}^{(j−1)} = e_{n+1}^{(j)}Λ_{n+1}, for j = 1, …, n + 1.

Proof. We need to show that Λ_{n+1}e_{n}^{(j−1)} = e_{n+1}^{(j)}Λ_{n+1}. We start from Loday’s relation Be_{n}^{(j−1)} = e_{n+1}^{(j)}B ([8]; Theorem 4.6.6), where B is Connes’ boundary operator in the normalised setting. We may write B as (n+1)sΛ_{n+1}, where s is the standard extra degeneracy operator, s(a_{1}⊗a_{2}⊗⋯⊗a_{n+1}) = (1⊗a_{1}⊗a_{2}⊗⋯⊗a_{n+1}) ([8]; 1.1.12). So we have (n + 1)sΛ_{n+1}e_{n}^{(j−1)} = (n + 1)e_{n+1}^{(j)}sΛ_{n+1}. Since the action of e_{n+1}^{(j)} here is on the last n + 1 places, leaving the first unaffected, e_{n+1}^{(j)}sΛ_{n+1} = se_{n+1}^{(j)}Λ_{n+1}. So Λ_{n+1}e_{n}^{(j−1)} = e_{n+1}^{(j)}Λ_{n+1} as required. □

2.2 Corollary. F_{n+1}^{(j−1)} ⊂ E_{n+1}^{(j)}, for j = 1, …, n + 1. □

Note that, since F_{n+1}^{(j−1)} restricts to E_{n}^{(j−1)}, to understand the E_{n}^{(j)}’s it is sufficient to understand these submodules.

Since the group algebra ℚΣ_{n+1} is semi-simple, we may write

E_{n+1}^{(j)} = F_{n+1}^{(j−1)} ⊕ xℚΣ_{n+1}

for some x ∈ ℚΣ_{n+1}.

Notation. Let p_{n} = (1 n)(2 n−1)(3 n−2) ⋯ ∈ Σ_{n} and let op_{n} = (−1)^{n(n+1)/2}p_{n} = (−1)^{n}(sgnp_{n})p_{n} ∈ ℚΣ_{n}. Now consider the idempotents σ_{n}^{(j)} = \frac{1}{2}(1 + (−1)^{j}op_{n}) in ℚΣ_{n}. These two idempotents correspond to the sums of the even and odd Eulerian idempotents:

σ_{n}^{(j)} = ∑_{i≡j (mod 2)} e_{n}^{(i)}

by [3]. In particular, σ_{n}^{(j)} (and hence p_{n}) is a polynomial in s_{n}. Of course, since

the e_{n}^{(i)}’s are mutually orthogonal, e_{n}^{(i)}σ_{n}^{(j)} = σ_{n}^{(j)}e_{n}^{(i)} = \{ \begin{matrix}e_{n}^{(i)} & if i ≡ j (mod 2) \\ 0 & otherwise. \end{matrix}

Let s_{n+1}^{∗} = ∑(sgnπ)π, where the sum is over shuffles in Σ_{n+1} which do not fix n + 1. So we may write s_{n+1} = 1 + s_{n} + s_{n+1}^{∗}.

2.3 Lemma. (1 + s_{n+1}^{∗})σ_{n+1}^{(j)} = σ_{n}^{(j+1)}(1 + s_{n+1}^{∗}).

Proof. Equivalently, we show p_{n}(1 + s_{n+1}^{∗})p_{n+1} = (−1)^{n}(1 + s_{n+1}^{∗}). Firstly, on the left-hand side we have p_{n}p_{n+1} = λ_{n+1}^{−1}, and since this is a 1-shuffle with λ_{n+1}^{−1}(1) = n + 1, it also appears in the right-hand side. Any other term on the left-hand side has the form (sgnπ)p_{n}πp_{n+1} where π is some k-shuffle in Σ_{n+1} not fixing n + 1. Then π(k) = n + 1, and it is easy to see that p_{n}πp_{n+1} is an (n + 2 − k)-shuffle taking (n + 2 − k) to n + 1. (In the case k = 1, we must have π = λ_{n+1}^{−1}, and we get p_{n}πp_{n+1} = 1.) Since sgn(p_{n}πp_{n+1}) = (−1)^{n}sgn(π), the result follows. □

2.4 Lemma. e_{n}^{(j)}σ_{n+1}^{(j)}(s_{n+1} − μ_{j}) = 0.

Proof.

e_{n}^{(j)}σ_{n+1}^{(j)}(s_{n+1} − μ_{j}) = e_{n}^{(j)}(s_{n+1} − μ_{j})σ_{n+1}^{(j)}
= e_{n}^{(j)}(1 + s_{n} + s_{n+1}^{∗} − μ_{j})σ_{n+1}^{(j)}
= e_{n}^{(j)}(s_{n} − μ_{j})σ_{n+1}^{(j)} + e_{n}^{(j)}(1 + s_{n+1}^{∗})σ_{n+1}^{(j)}
= e_{n}^{(j)}(1 + s_{n+1}^{∗})σ_{n+1}^{(j)}
= e_{n}^{(j)}σ_{n}^{(j+1)}(1 + s_{n+1}^{∗}) by Lemma 2.3
= 0. □

2.5 Proposition. e_{n}^{(j)}e_{n+1}^{(j)} = e_{n}^{(j)}σ_{n+1}^{(j)}.

Proof. Recall that the minimum polynomial of s_{n+1} is ∏_{j=1}^{n+1}(x − μ_{j}), where μ_{j} = 2^{j} − 2. So right multiplication by s_{n+1} is an operator on the subspace of ℚΣ_{n+1} spanned by 1, s_{n+1}, …, s_{n+1}^{n}. It has n + 1 distinct eigenvalues μ_{1}, …, μ_{n+1} and it follows from the definition of the Eulerian idempotents that right multiplication by e_{n+1}^{(j)} is projection onto the eigenspace corresponding to eigenvalue μ_{j}. So, by 2.4, e_{n}^{(j)}σ_{n+1}^{(j)} is contained in the left ideal ℚΣ_{n+1}(e_{n+1}^{(j)}). Hence, e_{n}^{(j)}σ_{n+1}^{(j)} = e_{n}^{(j)}σ_{n+1}^{(j)}e_{n+1}^{(j)}. But σ_{n+1}^{(j)}e_{n+1}^{(j)} = e_{n+1}^{(j)}. □

In fact, using the same method, one can also show that e_{n}^{(j)}e_{n+1}^{(j)} = σ_{n}^{(j)}e_{n+1}^{(j)}.

3. F_{n+1}^{(j)} as a virtual representation

The main result of this section is Theorem 3.4, giving a description of the representation F_{n+1}^{(j)}.

3.1 Definition. We define certain elements of the group algebra ℚΣ_{n+1}:

x_{n+1}^{(j)} = \frac{2}{n + 1}((n − 1) + (−1)^{j}op_{n} + ∑_{i=3}^{n}(i − 2)(−1)^{ni}λ_{n+1}^{i}).

3.2 Lemma.

(1 + (−1)^{n+1}λ_{n+1}^{−1})x_{n+1}^{(j)} =

\frac{2}{n + 1}((n − 1) + (−1)^{j}op_{n} + (−1)^{n+j−1}λ_{n+1}^{−1}op_{n} − ∑_{i=2}^{n} (−1)^{ni}λ_{n+1}^{i}). □

3.3 Proposition. e_{n}^{(j)}e_{n+1}^{(j)}x_{n+1}^{(j)} = (1 − Λ_{n+1})e_{n}^{(j)}.

Proof. We will use e_{n}^{(j)} = (−1)^{j}e_{n}^{(j)}op_{n} and op_{n}op_{n+1} = (−1)^{n+1}λ_{n+1}^{−1}. Now

e_{n}^{(j)}e_{n+1}^{(j)}x_{n+1}^{(j)} = e_{n}^{(j)}σ_{n+1}^{(j)}x_{n+1}^{(j)} by Proposition 2.5

= \frac{1}{2}e_{n}^{(j)}(1 + (−1)^{j}op_{n+1})x_{n+1}^{(j)}

= \frac{1}{2}e_{n}^{(j)}(1 + op_{n}op_{n+1})x_{n+1}^{(j)}

= \frac{1}{2}e_{n}^{(j)}(1 + (−1)^{n+1}λ_{n+1}^{−1})x_{n+1}^{(j)}

So

e_{n}^{(j)}e_{n+1}^{(j)}x_{n+1}^{(j)} = \frac{e_{n}^{(j)}}{n + 1}((n − 1) + (−1)^{j}op_{n} + (−1)^{n+j−1}λ_{n+1}^{−1}op_{n} − ∑_{i=2}^{n} (−1)^{ni}λ_{n+1}^{i})

= \frac{e_{n}^{(j)}}{n + 1}(n + (−1)^{n+1}op_{n}λ_{n+1}^{−1}op_{n} − ∑_{i=2}^{n} (−1)^{ni}λ_{n+1}^{i})

= \frac{e_{n}^{(j)}}{n + 1}(n + (−1)^{n+1}λ_{n+1} − ∑_{i=2}^{n} (−1)^{ni}λ_{n+1}^{i})

= \frac{e_{n}^{(j)}}{n + 1}(n − ∑_{i=1}^{n} (−1)^{ni}λ_{n+1}^{i})

= e_{n}^{(j)}(1 − Λ_{n+1})

= (1 − Λ_{n+1})e_{n}^{(j)} by Corollary 1.2. □

Now we can prove the main result. We will use the fact that given an idempotent e in ℚΣ_{n}, giving a representation eℚΣ_{n} of Σ_{n}, then the induced representation of Σ_{n+1} is given by eℚΣ_{n+1}. We also need the result, due to Hanlon, that the dimension of the representation E_{n}^{(j)} is s(n, j), the number of permutations in Σ_{n} with exactly j cycles [5].

3.4 Theorem.

F_{n+1}^{(j)} ⊕ ⨁_{i=1}^{j} E_{n+1}^{(i)} ≅ ⨁_{i=1}^{j} Ind_{Σ_{n}}^{Σ_{n+1}}E_{n}^{(i)}.

Proof. The result will be proved by induction on j. First we consider the case j = 1. Here we need to show that F_{n+1}^{(1)} ⊕ E_{n+1}^{(1)} ≅ Ind_{Σ_{n}}^{Σ_{n+1}}E_{n}^{(1)}. That is,

F_{n+1}^{(1)} ⊕ E_{n+1}^{(1)} ≅ e_{n}^{(1)}ℚΣ_{n+1}.

Now it is clear that

e_{n}^{(1)}ℚΣ_{n+1} = Λ_{n+1}e_{n}^{(1)}ℚΣ_{n+1}⊕(1−Λ_{n+1})e_{n}^{(1)}ℚΣ_{n+1} = F_{n+1}^{(1)}⊕(1−Λ_{n+1})e_{n}^{(1)}ℚΣ_{n+1}.

So we must show that (1−Λ_{n+1})e_{n}^{(1)}ℚΣ_{n+1} ≅ E_{n+1}^{(1)}, that is (1−Λ_{n+1})e_{n}^{(1)}ℚΣ_{n+1} ≅ e_{n+1}^{(1)}ℚΣ_{n+1}. Using the fact that E_{n}^{(1)} has dimension (n − 1)!, we see that both of these modules have dimension n!. We define

θ : e_{n+1}^{(1)}ℚΣ_{n+1} → (1 − Λ_{n+1})e_{n}^{(1)}ℚΣ_{n+1}

to be the homomorphism of right ℚΣ_{n+1}-modules given by left-multiplication by the element (1 − Λ_{n+1})e_{n}^{(1)}. Then

(1 − Λ_{n+1})e_{n}^{(1)} = e_{n}^{(1)}e_{n+1}^{(1)}x_{n+1}^{(1)} = (1 − Λ_{n+1})e_{n}^{(1)}e_{n+1}^{(1)}x_{n+1}^{(1)} = θ(e_{n+1}^{(1)}x_{n+1}^{(1)}),

by 3.3. Thus θ is surjective, and so an isomorphism, giving the result for j=1. Note that this identifies E_{n+1}^{(1)} with the submodule e_{n}^{(1)}(1 − Λ_{n+1})ℚΣ_{n+1} of e_{n}^{(1)}ℚΣ_{n+1}.

Now we assume the result for j−1 and consider j. Using the induction hypothesis it is sufficient to show F_{n+1}^{(j)} ⊕ E_{n+1}^{(j)} ≅ F_{n+1}^{(j−1)} ⊕ Ind_{Σ_{n}}^{Σ_{n+1}}E_{n}^{(j)}. That is

Λ_{n+1}e_{n}^{(j)}ℚΣ_{n+1} ⊕ E_{n+1}^{(j)} ≅ F_{n+1}^{(j−1)} ⊕ e_{n}^{(j)}ℚΣ_{n+1}.

Now we clearly have

e_{n}^{(j)}ℚΣ_{n+1} = Λ_{n+1}e_{n}^{(j)}ℚΣ_{n+1} ⊕ (1 − Λ_{n+1})e_{n}^{(j)}ℚΣ_{n+1},

so we must show that E_{n+1}^{(j)} ≅ F_{n+1}^{(j−1)} ⊕ (1 − Λ_{n+1})e_{n}^{(j)}ℚΣ_{n+1}. By Corollary 2.2, E_{n+1}^{(j)} = F_{n+1}^{(j−1)} ⊕ xℚΣ_{n+1}. Hence the above simplifies to showing that xℚΣ_{n+1} ≅ (1 − Λ_{n+1})e_{n}^{(j)}ℚΣ_{n+1}. We define θ : E_{n+1}^{(j)} → (1 − Λ_{n+1})e_{n}^{(j)}ℚΣ_{n+1} to be the homomorphism of right ℚΣ_{n+1}-modules given by left-multiplication by (1 − Λ_{n+1})e_{n}^{(j)}. By 1.2, (1 − Λ_{n+1})e_{n}^{(j)}Λ_{n+1}e_{n}^{(j−1)} = 0, so F_{n+1}^{(j−1)} ⊂ Kerθ. Hence θ induces a ℚΣ_{n+1}-module homomorphism: θ′ : xℚΣ_{n+1} → (1 − Λ_{n+1})e_{n}^{(j)}ℚΣ_{n+1}. Next we check the dimensions of these ℚΣ_{n+1}-modules. We have seen in 1.4 that F_{n+1}^{(j)} restricts to E_{n}^{(j)}, so has the same dimension, s(n, j). So xℚΣ_{n+1} has dimension s(n + 1, j) − s(n, j − 1), and (1 − Λ_{n+1})e_{n}^{(j)}ℚΣ_{n+1} has dimension (n + 1)s(n, j) − s(n, j) = ns(n, j). Since s(n + 1, j) = s(n, j − 1) + ns(n, j), (see [4], p261 equation 6.8), the two modules do have the same dimension. Hence, it is sufficient to show that θ′ is surjective to conclude that it is a ℚΣ_{n+1}-module isomorphism. But,

(1 − Λ_{n+1})e_{n}^{(j)} = e_{n}^{(j)}e_{n+1}^{(j)}x_{n+1}^{(j)} by Proposition 3.3
= (1 − Λ_{n+1})e_{n}^{(j)}e_{n+1}^{(j)}x_{n+1}^{(j)} since 1 − Λ_{n+1} is an idempotent
= θ′(e_{n+1}^{(j)}x_{n+1}^{(j)}).

Hence θ′ is surjective. □

Notation. Let Ψ_{n+1}^{j} denote the character of the representation F_{n+1}^{(j)} of Σ_{n+1} and let χ_{n}^{j} denote the character of the representation E_{n}^{(j)} of Σ_{n}.

3.5 Corollary.

Ψ_{n+1}^{j} = ∑_{i=1}^{j} Ind_{Σ_{n}}^{Σ_{n+1}}(χ_{n}^{i}) − ∑_{i=1}^{j} χ_{n+1}^{i}. □

We give the formula for the character Ψ_{n+1}^{1} explicitly.

3.6 Corollary. For g ∈ Σ_{n+1},

Ψ_{n+1}^{1}(g) = \{ \begin{matrix}(−1)^{n+s}(s − 1)!(r)^{s−1}μ(r) & if g has cycle type (r)^{s} with r > 1, or (r)^{s}(1) \\ 0 & if g has any other cycle type. \end{matrix}

Proof. We have shown that Ψ_{n+1}^{1} = Ind_{Σ_{n}}^{Σ_{n+1}}(χ_{n}^{1}) − χ_{n+1}^{1}. The result is a straightforward induced character calculation from Hanlon’s result: χ_{n}^{1} = ϵ · (Ind_{C_{n}}^{Σ_{n}}ρ_{n}), where ρ_{n} is a faithful linear character of the cyclic subgroup of Σ_{n+1} generated by an n-cycle and ϵ is the alternating character [5]. □

It is shown in [10] that this is exactly the character of the tree representation of Σ_{n+1}. This representation arises in Γ -homology, HΓ_{∗}, a homology theory for E_{∞}- ring spectra, introduced by Robinson in [11], which specialises to a new homology theory for commutative algebras. Let B be a flat commutative algebra over a commutative ground ring A and M a B-module. An application of our results allows us to prove that when the ground ring A contains ℚ, Γ -homology agrees with Harrison homology, HΓ_{p}(B/A; M) ≅ Harr_{p+1}(B/A; M). (In general the theories are different.) Since the definition of Γ -homology is rather long, here we only outline the idea of the proof. The result was announced in [14] and will appear elsewhere.

There is a first quadrant spectral sequence converging to Γ -homology,

ℰ_{p,q}^{1} = M ⊗ Tor_{q}^{AΣ_{p+1}}(V_{p+1}, B^{⊗p+1}) ⟹_{p} HΓ_{p+q}(B/A; M),

where V_{p+1} denotes the restriction to Σ_{p+1} of the tree representation of Σ_{p+2}. When A contains ℚ the spectral sequence collapses to the edge and, by 1.4, V_{p+1} is isomorphic to E_{p+1}^{(1)}. Then Γ -homology is the homology of a complex

M ⊗ E_{p+1}^{(1)} ⊗_{Σ_{p+1}} B^{⊗p+1}

with a differential which can be identified as induced by the Hochschild boundary b. This gives the first part of the decomposition of Hochschild homology, namely Harrison homology.

The representation F_{n+1}^{(1)} also occurs in the homology of partition lattices [12], the homology of configuration spaces [2], [7], and (up to sign) as the multilinear part of the free Lie algebra [1], [10].

4. Some Results on Decompositions

The following table lists the decompositions of the representations F_{n+1}^{(j)} of Σ_{n+1} for n = 1, …, 5 and j = 1, …, n.

4.1 Table: Decompositions of F_{n+1}^{(j)}.

The first column gives F_{n+1}^{(1)}, the tree representation of Σ_{n+1}. The sum along the n^{th} row of the table gives the representation Λ_{n+1}ℚΣ_{n+1}, the sign representation of < λ_{n+1} > induced to Σ_{n+1}.

For some values of j, it is possible to describe the decomposition of the representation F_{n+1}^{(j)} into irreducible components. Let ω^{λ} be the irreducible character of the symmetric group Σ_{n+1} corresponding to a partition λ of n + 1. Let a(T) denote the sum of ascents of a standard tableau T, that is the sum of those i such that i + 1 appears to the right of i in T. Then we denote by λ(j, n) the number of standard tableaux T of shape λ such that a(T) ≡ j (mod n).

4.2 Proposition.

(1) (j = 1) The multiplicity of ω^{λ} in Ψ_{n+1}^{1} is λ(1, n) − λ(1, n + 1)

(2) (j = n − 2) Ψ_{n+1}^{n−2} = ω^{2^{2}1^{n−3}} ⊕ ω^{321^{n−4}} ⊕ ω^{3^{2}1^{n−5}} ⊕ ω^{51^{n−4}}

(3) (j = n − 1) Ψ_{n+1}^{n−1} = ω^{31^{n−2}}

(4) (j = n) Ψ_{n+1}^{n} = ω^{1^{n+1}}

(5) The multiplicity of ω^{λ} in the sum of characters ∑_{j=1}^{n} Ψ_{n+1}^{j} is λ(0, n + 1).

Proof.

(1) It is easily seen that T is a standard tableau for Σ_{n+1} such that a(T) ≡ 1 (mod n) if and only if it is obtained from a standard tableau T′ for Σ_{n}, with a(T′) ≡ 1 (mod n), by attaching n + 1 to the end of some row or column. Now in χ_{n}^{1}, ω^{λ} has multiplicity the number of standard tableaux T′ of shape λ for Σ_{n} such that a(T′) ≡ 1 (mod n) by a result of Kraskiewicz and Weyman [6]. So in Ind_{Σ_{n}}^{Σ_{n+1}}(χ_{n}^{1}), ω^{λ} has multiplicity λ(1, n). Since Ψ_{n+1}^{1} = Ind_{Σ_{n}}^{Σ_{n+1}}(χ_{n}^{1}) − χ_{n+1}^{1}, the result follows.

(2,3,4) These results follow directly from those of [5] for χ_{n}^{n−2}, χ_{n}^{n−1} and χ_{n}^{n}. That is, for j = n − 2, n − 1, n, the decomposition of Ψ_{n+1}^{j} given above is the only one which will restrict back to give the correct decomposition of χ_{n}^{i}. (Of course, in the case j = n, we have f_{n+1}^{(n)} = Λ_{n+1}e_{n}^{(n)} = Λ_{n+1}ε_{n} = ε_{n+1}, where ε_{n} = \frac{1}{n!} ∑_{π∈Σ_{n}}(sgnπ)π, and we see directly that we have the sign representation.)

(5) We have seen that this sum of characters is just the sign character of < λ_{n+1} > induced to Σ_{n+1}. The formula for the decomposition can be deduced from the work of Stembridge [13] . □

We also give the relationship between our characters and the trivial character.

4.3 Proposition. The trivial character ω^{n+1} appears only in Ψ_{n+1}^{n/2} if n is even and does not appear in any Ψ_{n+1}^{i} if n is odd.

Proof. Let e_{n+1} = \frac{1}{(n+1)!} ∑_{π∈Σ_{n+1}} π. It is easily checked that

Λ_{n+1}e_{n+1} = \{ \begin{matrix}e_{n+1} & if n is even \\ 0 & if n is odd. \end{matrix}

Hence, the trivial representation does not appear in Λ_{n+1}ℚΣ_{n+1} when n is odd. When n is even it appears once, and this must be in Λ_{n+1}e_{n}^{(n/2)}ℚΣ_{n+1}, since Hanlon shows that the trivial representation of Σ_{n} always appears in e_{n}^{([(n+1)/2])}ℚΣ_{n}. □

4.4 Corollary. The character ω^{n1} does not appear in any Ψ_{n+1}^{i} if n is even and appears only in Ψ_{n+1}^{(n+1)/2} if n is odd.

Proof. The irreducible character ω^{n1} of Σ_{n+1} is the only one apart from ω^{n+1} which gives a copy of the trivial character of Σ_{n} on restriction. Hence the result follows from the above. □

Acknowledgement

I acknowledge the financial support of the Science and Engineering Research Council for some of the period when this work was carried out.

References

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[11] C.A. Robinson, “Γ-homology 1,” in preparation.

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Mathematics Institute, University of Warwick, Coventry, England CV4 7AL E-mail address: sarah@uk.ac.warwick.maths