Orthogonal Frobenius reciprocity.

Gabriele Nebe, Lehrstuhl B für Mathematik, RWTH Aachen, Templergraben 64, 52062 Aachen, Germany, gabi@math.rwth-aachen.de

Abstract: A quantitative version of Frobenius reciprocity for isometries is proved. It is applied to deduce recursion formulas for the rational class of some irreducible orthogonal modules of the symmetric groups.

1 Introduction

Let G be a finite group with subgroup H ⊂ G and K be a field of characteristic 0. If W is a right KH-module, then the induced module W^{G} defined as W^{G} = W ⊗_{KH} KG (cf. [CuR, Section 10]) is a KG-module. On the other hand, by restriction, any KGmodule V can also be viewed as a KH-module. Classical Frobenius reciprocity (cf. [CuR, Theorem (10.8)]) establishes a canonical isomorphism between the vector spaces of homomorphisms

(F) Hom_{KH}(W, V) ≅ Hom_{KG}(W^{G}, V), φ ↦ φ^{G}.

Now assume that W admits a non-degenerate symmetric bilinear form F_{W} : W × W → K that is H-invariant, i.e. F_{W} (v, w) = F_{W} (vh, wh) for all v, w ∈ W, h ∈ H. Call such a pair (W, F_{W}) an orthogonal KH-module. Then F_{W} defines a G-invariant form F_{W}^{G} on W^{G} such that (W^{G}, F_{W}^{G}) becomes an orthogonal KG-module.

If (V, F_{V}) is an orthogonal KG-module, then it is natural to ask what happens with the KH-isometries Isom_{KH}((W, F_{W}), (V, F_{V})) :=

{φ ∈ Hom_{KH}(W, V) | F_{W} (w, w′) = F_{V} (φ(w), φ(w′)) for all w, w′ ∈ W }

if one applies Frobenius reciprocity. Note that here by definition isometries are injective but not necessarily surjective. Since Frobenius reciprocity does not respect injectivity of the mappings, one has to dualise the right hand side of (F) to again get isometries: F_{V} induces a KG-isomorphism between V and its dual V^{∗} := Hom_{K}(V, K). Let F_{V}^{∗} be the form on V^{∗} such that this isomorphism is an isometry. Assume that V is a uniform KG-module, which means that F_{V} generates the space of all G-invariant symmetric bilinear forms on V. Then orthogonal Frobenius reciprocity gives a canonical bijection

Isom_{KH}((W, F_{W}), (V, F_{V})) ≅ Isom_{KG}((V^{∗}, F_{V}^{∗}), ((W^{G})^{∗}, (F_{W}^{G})^{∗}))

defined by

φ ↦ \frac{dim(V)}{dim(W^{G})}(φ^{G})^{∗}.

In Section 2, this orthogonal Frobenius reciprocity and two useful generalisations are proved. It is applied in Section 3 to determine the rational isometry class of some irreducible orthogonal ℚS_{n}-modules.

This paper is one section of my Habilitationsschrift [Neb]. There I give further applications of orthogonal Frobenius reciprocity to the determination of the rational isometry class of orthogonal G-modules.

2 Orthogonal Frobenius reciprocity.

Let K be a field of characteristic 0 and G be a finite group with subgroup H ⊆ G. Let G =∪^{.}_{i=1}^{s} Hg_{i} be a decomposition of G into H-cosets.

If W is a KH-module with K-basis (b_{1}, …, b_{m}) then (b_{1} ⊗ g_{1}, …, b_{m} ⊗ g_{1}, b_{1} ⊗ g_{2}, …, b_{m} ⊗ g_{s}) is a K-basis for the KG-module W^{G}. The action of g ∈ G on W^{G} is calculated combining the permutation of the cosets induced by g with the action of H on W: If g_{i}g = hg_{j} with h ∈ H then (w ⊗ g_{i})g = wh ⊗ g_{j} for all w ∈ W.

If φ : W → V_{|H} is a KH-homomorphism, then φ^{G} : W^{G} → V defined by φ^{G}(∑_{i=1}^{s} w_{i} ⊗ g_{i}) = ∑_{i=1}^{s} φ(w_{i})g_{i} is a KG-homomorphism. The mapping φ ↦ φ^{G} is independent of the choice of the coset representatives g_{i} and defines a Kisomorphism Hom_{KH}(W, V_{|H}) → Hom_{KG}(W^{G}, V) with inverse Hom_{KG}(W^{G}, V) → Hom_{KH}(W, V_{|H}); φ ↦ φ_{|W}, the restriction to W, where W is identified with W ⊗ 1 ⊆ W^{G} (cf. [CuR, section 10]).

If (W, F_{W}) is an orthogonal KH-module, then F_{W}^{G} defined by

F_{W}^{G}(w ⊗ g_{i}, w′ ⊗ g_{j}) := δ_{ij}F_{W} (w, w′) (w, w′ ∈ W, 1 ≤ i, j ≤ s)

is a non-degenerate G-invariant symmetric bilinear form on W^{G}.

Any non-degenerate symmetric G-invariant bilinear form F_{V} on the KG-module V defines a KG-isomorphism F̃_{V} : v ↦ F_{V} (·, v) between V and the dual space V^{∗} = Hom_{K}(V, K). Note that V^{∗} is again a right KG-module, via (fg)(v) = f(vg^{−1}) for all v ∈ V, f ∈ V^{∗}, g ∈ G. Let F_{V}^{∗} be the form on V^{∗}, for which F̃_{V} is an isometry: For f ∈ V^{∗} let v_{f} := F̃_{V}^{−1}(f) ∈ V. Then

F_{V}^{∗}(f, h) := F_{V} (v_{f}, v_{h}) for all f, h ∈ V^{∗}.

Theorem 2.1 Let K be a field of characteristic 0, (W, F_{W}) an orthogonal KHmodule, and (V, F_{V}) a uniform orthogonal KG-module. If φ : (W, F_{W}) → (V, F_{V}) is a KH-isometry, then the transposed mapping

(φ^{G})^{∗} : (V^{∗}, \frac{dim(W^{G})}{dim(V)}F_{V}^{∗}) → ((W^{∗})^{G}, (F_{W}^{∗})^{G})

is a KG-isometry.

Remark 2.2 The constant \frac{dim(W^{G})}{dim(V)} can be easily remembered by taking V and W = ⟨w⟩ with F_{W} (w, w) = 1 to be the trivial modules. Then W^{G} and also (W^{G})^{∗} is the permutation module with orthonormal basis (w ⊗ g_{1}, …, w ⊗ g_{s}) and the trivial KGsubmodule of W^{G} is generated by v := ∑_{j=1}^{s} w⊗g_{j} with squared length (F_{W}^{G})^{∗}(v, v) = s.

To prove the theorem it is convenient to choose a basis of W and V and work with matrices. So F_{V}, F_{W} also denote the Gram matrices of F_{V} respectively F_{W} with respect to the chosen basis, g_{k} the matrix describing the action of g_{k} ∈ G, right multiplication with φ the corresponding mapping φ etc.

Lemma 2.3 Let (W, F_{W}) be an orthogonal KH-module and (V, F_{V}) be an orthogonal KG-module. Let φ ∈ Isom_{KH}((W, F_{W}), (V, F_{V})). Then the orthogonal projection P_{W} ∈ End_{KH}(V) onto φ(W) is given by right multiplication with

P_{W} := F_{V} φ^{tr}F_{W}^{−1}φ.

If V is a uniform KG-module, then

T r_{G/H}(P_{W}) := ∑_{j=1}^{s} g_{j}^{−1}P_{W} g_{j} = \frac{dim(W^{G})}{dim(V)}id_{V} .

\underline{Proof.} A straightforward calculation shows that P_{W} is the orthogonal projection onto φ(W). Therefore P_{W} F_{V} is the Gram matrix of a symmetric H-invariant bilinear form on V. By construction T r_{G/H}(P_{W}) ∈ End_{KG}(V) is a KG-endomorphism of V. Since g_{j}F_{V} = F_{V} (g_{j}^{tr})^{−1}, the trace

T r_{G/H}(P_{W})F_{V} = ∑_{j=1}^{s} g_{j}^{−1}P_{W} F_{V} (g_{j}^{−1})^{tr}

is the Gram matrix of a G-invariant symmetric bilinear form on V. Since V is a uniform KG-module, this implies that T r_{G/H}(P_{W}) is a scalar matrix. The trace of the matrix P_{W} is

tr(P_{W}) = tr(F_{V} φ^{tr}F_{W}^{−1}φ) = tr(φF_{V} φ^{tr}F_{W}^{−1}) = dim(W).

Hence the trace of T r_{G/H}(P_{W}) is s·dim(W) = dim(W^{G}) and T r_{G/H}(P_{W}) = \frac{dim(W^{G})}{dim(V)}id_{V}.

□

Proof of Theorem 2.1. Let (w_{a}|1 ≤ a ≤ m) be a K-basis of W and v_{a} := φ(w_{a}) (1 ≤ a ≤ m). Then the set {v_{a}g_{i} | 1 ≤ i ≤ s, 1 ≤ a ≤ m} generates the vector space V over K, because V is an irreducible KG-module. Therefore V^{∗} is generated by the functions

f_{a,i} = F̃_{V} (v_{a}g_{i}) (1 ≤ a ≤ m, 1 ≤ i ≤ s).

Since F_{W}^{G} is non-degenerate there are unique w_{j}^{(a,i)} ∈ W (1 ≤ j ≤ s) such that

(φ^{G})^{∗}(f_{a,i}) = F̃_{W}^{G}(∑_{j=1}^{s} w_{j}^{(a,i)} ⊗ g_{j}) for all 1 ≤ a ≤ m, 1 ≤ i ≤ s.

For 1 ≤ a, b ≤ m, 1 ≤ i, k ≤ s one has

(φ^{G})^{∗}(f_{a,i})(w_{b} ⊗ g_{k}) = f_{a,i}(φ^{G}(w_{b} ⊗ g_{k})) = f_{a,i}(v_{b}g_{k}) = F_{V} (v_{b}g_{k}, v_{a}g_{i}).

On the other hand

(φ^{G})^{∗}(f_{a,i})(w_{b} ⊗ g_{k}) = F_{W}^{G}(w_{b} ⊗ g_{k}, ∑_{j=1}^{s} w_{j}^{(a,i)} ⊗ g_{j}) = F_{W} (w_{b}, w_{k}^{(a,i)}).

Choosing K-bases and working with matrices one therefore gets

F_{W} (w_{k}^{(a,i)})^{tr} = φg_{k}F_{V} g_{i}^{tr}v_{a}^{tr} for all 1 ≤ i, k ≤ s, 1 ≤ a ≤ m.

Hence the scalar product of (φ^{G})^{∗}(f_{b,k}) and (φ^{G})^{∗}(f_{a,i}) with respect to (F_{W}^{∗})^{G} is

∑_{j=1}^{s} w_{j}^{(b,k)}F_{W} (w_{j}^{(a,i)})^{tr} = ∑_{j=1}^{s} v_{b}g_{k}(F_{V} g_{j}^{tr}φ^{tr}F_{W}^{−1}φg_{j}F_{V})g_{i}^{tr}v_{a}^{tr}.

By Lemma 2.3

∑_{j=1}^{s} F_{V} g_{j}^{tr}φ^{tr}F_{W}^{−1}φg_{j}F_{V} = ∑_{j=1}^{s} g_{j}^{−1}P_{W} g_{j}F_{V} = \frac{dim(W^{G})}{dim(V)}F_{V}

and therefore

(F_{W}^{G})^{∗}((φ^{G})^{∗}(f_{b,k}), (φ^{G})^{∗}(f_{a,i})) = \frac{dim(W^{G})}{dim(V)}F_{V}^{∗}(f_{b,k}, f_{a,i})

for all 1 ≤ i, k ≤ s, 1 ≤ a, b ≤ m, which proves the theorem. □

If W^{G} contains simple KG-modules with multiplicity > 1 then Theorem 2.1 does not give a complete decomposition of the orthogonal KG-module (W^{G}, F_{W}^{G}).

Theorem 2.4 Let (W, F_{W}) be an orthogonal KH-module and let (V, F_{V}) be an absolutely irreducible orthogonal KG-module. Assume, there is C = C^{tr} ∈ GL_{n}(K) such that (φ_{1}, …, φ_{n}) : (W^{n}, C ⊗ F_{W}) → (V, F_{V}) is a KH-isometry. Then

((φ_{1}^{G})^{∗}, …, (φ_{n}^{G})^{∗}) : ((V^{∗})^{n}, \frac{dim(W^{G})}{dim(V)}C ⊗ F_{V}^{∗}) → ((W^{∗})^{G}, (F_{W}^{∗})^{G})

is a KG-isometry.

\underline{Proof.} With the notation from the proof of Theorem 2.1 let v_{a}^{(x)} := φ_{x}(w_{a}) and f_{a,i}^{(x)} = F̃_{V} (v_{a}^{(x)}g_{i}) for 1 ≤ a ≤ m, 1 ≤ x ≤ n, 1 ≤ i ≤ s. As above one calculates

(1) (F_{W}^{∗})^{G}((φ_{y}^{G})^{∗}(f_{b,k}^{(y)}), (φ_{x}^{G})^{∗}(f_{a,i}^{(x)})) = ∑_{j=1}^{s}(v_{b}^{(y)}g_{k}F_{V} g_{j}^{tr}φ_{y}^{tr}F_{W}^{−1})(φ_{x}g_{j}F_{V} g_{i}^{tr}(v_{a}^{(x)})^{tr}).

Now F_{V} φ_{y}^{tr}F_{W}^{−1}φ_{x} ∈ End_{KH}(V) has trace c_{x,y}dim(W), where c_{x,y} is the x, y-entry of C, since φ_{x}F_{V} φ_{y}^{tr} = c_{x,y}F_{W}. Therefore

∑_{j=1}^{s} F_{V} g_{j}^{tr}φ_{y}^{tr}F_{W}^{−1}φ_{x}g_{j} ∈ End_{KG}(V)

is an endomorphism of trace c_{x,y}dim(W^{G}). Since V is an absolutely irreducible KGmodule, this endomorphism is scalar, hence the right hand side of (1) is

c_{x,y} \frac{dim(W^{G})}{dim(V)}v_{b}^{(y)}g_{k}F_{V} g_{i}^{tr}(v_{a}^{(x)})^{tr} = c_{x,y} \frac{dim(W^{G})}{dim(V)}F_{V}^{∗}(f_{b,j}^{(y)}, f_{a,i}^{(x)}). □

In Theorem 2.1 the KG-module V was assumed to be uniform. If one drops this assumption, one has to know more about the KH-isometry φ : W → V to identify the invariant form F_{V}.

Let (W, F_{W}) be an orthogonal KH-module and (V, F_{V}) be an irreducible orthogonal KG-module. Let C be the space of symmetric KG-endomorphisms of V,

C := {φ ∈ End_{KG}(V) | φF̃_{V} = F̃_{V} φ^{∗}}

and D the space of symmetric KH-endomorphisms of W. Assume that there is a K-linear mapping α : C → D satisfying tr(c)/dim_{K}(V) = tr(α(c))/dim_{K}(W) for all c ∈ C.

Proposition 2.5 With the notations above let φ : (W, α(c)F_{W}) → (V, cF_{V}) be a KHisometry for all 0 ≠ c ∈ C^{+}. Then

(φ^{G})^{∗} : (V^{∗}, \frac{dim(W^{G})}{dim(V)}F_{V}^{∗}) → ((W^{∗})^{G}, (F_{W}^{∗})^{G})

is a KG-isometry.

\underline{Proof.} The proof of the proposition is analogous to the one of Theorem 2.1. It only remains to show that the statement of Lemma 2.3 holds with the assumption of the proposition. But the assumptions on φ guarantee that for all c ∈ C

tr(cF_{V} g_{j}^{tr}φ^{tr}F_{W}^{−1}φg_{j}) = tr(φcF_{V} φ^{tr}F_{W}^{−1}) = tr(α(c)) = tr(c)\frac{dim(W)}{dim(V)}.

Now C is the eigenspace of the mapping End_{KG}(V) → End_{KG}(V), φ ↦ F̃_{V} φ^{∗}F̃_{V}^{−1}, which is orthogonal with respect to the trace bilinear form. Therefore the restriction of the trace bilinear form of the separable algebra End_{KG}(V) to C is non-degenerate, and therefore

∑_{j=1}^{s} F_{V} g_{j}^{tr}φ^{tr}F_{W}^{−1}φg_{j} = \frac{dim(W^{G})}{dim(V)}id_{V} . □

3 The Specht modules S^{(n−k,k)}.

The representation theory of the symmetric group S_{n} is very well understood (cf. [Jam], [JaK]). The irreducible representations, so called Specht modules S^{λ}, of S_{n} over a field of characteristic 0 are in bijection with the partitions λ of n. They have the remarkable property that S^{λ} occurs with multiplicity one as a submodule of a permutation module M^{λ}, such that all the other constituents of M^{λ} belong to partitions that are smaller than λ for a suitable ordering. So the Specht modules are good candidates to apply orthogonal Frobenius reciprocity (Theorem 2.1).

A particular easy construction for S^{λ} can be given, if the partition λ of n has only two parts. So let k, l, n ∈ ℕ with 1 ≤ k ≤ l ≤ \frac{n}{2} and let S_{l} × S_{n−l} denote the Young subgroup of the symmetric group S_{n}, which is the set stabiliser of the subset {1, …, l} of {1, …, n}.

Let M^{(n−k,k)} be the S_{n}-permutation module having the k-element subsets of {1, …, n} as an orthonormal ℚ-basis. Denote the corresponding S_{n}-invariant symmetric bilinear form by I\binom{n}{k}. Then

dim_{ℚ}(M^{(n−k,k)}) = \binom{n}{k} and M^{(n−k,k)} = 1_{S_{k}×S_{n−k}}^{S_{n}}.

For a fixed subset T ⊂ {1, …, n} let σ_{T} : M^{(n−k,k)} → ℚ be the ℚ-linear mapping defined by

σ_{T} (S) := \{ \begin{matrix}0 & T ⊈ S \\ 1 & T ⊆ S \end{matrix}

for all k-element subsets S ⊂ {1, …, n}. Then the Specht module S^{(n−k,k)} ⊆ M^{(n−k,k)} is

S^{(n−k,k)} = ⋂_{T ⊆{1,...,n},|T |<k} Ker(σ_{T}).

(cf. [Jam, Corollary (17.18)]).

S^{(n−k,k)} is an absolutely irreducible S_{n}-submodule of M^{(n−k,k)}. Therefore the S_{n}- invariant symmetric bilinear forms on S^{(n−k,k)} are rational multiples of the restriction F_{k} of I\binom{n}{k}.

Young’s rule [Jam, (14.4)] says that the ℚS_{n}-module M^{(n−l,l)} is the direct sum of all S^{(n−k,k)} with k ≤ l. Hence by classical Frobenius reciprocity the fixed space of S_{l} × S_{n−l} on S^{(n−k,k)} is one-dimensional, say spanned by some v ≠ 0. To apply orthogonal Frobenius reciprocity it suffices to calculate the length of v:

Theorem 3.1 Let 1 ≤ k ≤ l ≤ \frac{n}{2}. Then there is v ∈ S^{(n−k,k)} with vg = v for all g ∈ S_{l} × S_{n−l} satisfying

F_{k}(v, v) = a(l, k) := \binom{n + 1 − k}{k}\binom{n − l}{k}\binom{l}{k}^{−1}.

To prove this theorem we need 2 lemmata on binomial coefficients.

Lemma 3.2

∑_{j=0}^{k} \binom{l − k + j}{j}\binom{n − l − j}{k − j} = \binom{n + 1 − k}{k}.

\underline{Proof.} First assume l = k. Then the left hand side ist

∑_{j=0}^{k} \binom{n − k − j}{k − j} = ∑_{j=0}^{k} \binom{n − 2k + j}{j} = \binom{n − 2k + k + 1}{k} = \binom{n + 1 − k}{k}.

To show the statement in the general case let d(l, l + 1) denote the difference of the left hand sides for l and l + 1. Then

d(l, l + 1) = ∑_{j=0}^{k} \binom{l − k + j}{j}\binom{n − l − j}{k − j} − \binom{l − k + j + 1}{j}\binom{n − l − j − 1}{k − j} =

∑_{j=0}^{k} \binom{n − l − j}{k − j}(\binom{l − k + j}{j}−\binom{l − k + j + 1}{j})+∑_{j=0}^{k} \binom{l − k + j + 1}{j}\binom{n − l − j − 1}{k − j − 1}

since \binom{n−l−j−1}{k−j} = \binom{n−l−j}{k−j}−\binom{n−l−j−1}{k−j−1}. The difference in brackets is \binom{l−k+j}{j}−\binom{l−k+j+1}{j} = −\binom{l−k+j}{j−1}.

If one substitutes the summation index i = j + 1 in the second sum, one finds

d(l, l + 1) = − ∑_{j=0}^{k} \binom{n − l − j}{k − j}\binom{l − k + j}{j − 1} + ∑_{i=1}^{k+1} \binom{l − k + i}{i − 1}\binom{n − l − i}{k − i} = 0.

Hence the left hand side is independent of l and the lemma follows. □

Lemma 3.3 Let k ≤ l ≤ \frac{n}{2}. For 0 ≤ i ≤ k define

a_{i} := (−1)^{i} ∏_{j=0}^{i−1} \frac{n − l − k + j + 1}{l − j}.

Then

∑_{i=x}^{x+b} a_{i}\binom{n − l − k + x + b}{x + b − i}\binom{l − x}{i − x} = 0 for all 0 ≤ x ≤ x + b ≤ k.

\underline{Proof.} Let A := ∏_{j=1}^{x+b}(n − l − k + j) be the product of the numerator of the first binomial coefficient with the numerator of a_{i} and let B := ∏_{j=0}^{x−1}(l − j)^{−1} be quotient of the numerator of the second binomial coefficient with the denominator of a_{i}. Then the sum in the lemma simplifies to

BA ∑_{i=x}^{x+b}(−1)^{i} \frac{1}{(i − x)!(b + x − i)!} = \frac{BA}{b!}(−1)^{x} ∑_{j=0}^{b}(−1)^{j} \frac{b!}{j!(b − j)!} = 0. □

Proof of Theorem 3.1: The orbits of S_{l} × S_{n−l} = Stab_{S_{n}}({1, …, l}) on the k-element subsets T of {1, …, n} are parametrised by |T ∩ {1, …, l}|. For 0 ≤ j ≤ k let v_{j} ∈ M^{(n−k,k)} be the sum over all k-element subsets of {1, …, n} that intersect {1, …, l} in j elements. Then (v_{0}, …, v_{k}) is a basis of the fixed space of S_{l} × S_{n−l} on M^{(n−k,k)}. For the standard scalar product one finds

I\binom{n}{k}(v_{i}, v_{j}) = δ_{ij}\binom{n − l}{k − j}\binom{l}{j}.

Let a_{i} be as in Lemma 3.3, T be a (k − b)-element subset of {1, …, n} and x := |T ∩ {1, …, l}|. Then

σ_{T} (∑_{i=0}^{k} a_{i}v_{i}) = ∑_{i=x}^{x+b} a_{i}\binom{n − l − k + x + b}{x + b − i}\binom{l − x}{i − x} = 0

by Lemma 3.3. Therefore

∑_{i=0}^{k} a_{i}v_{i} ∈ S^{(n−k,k)}.

Now

a_{i} = (−1)^{i} ∏_{j=0}^{i−1} \frac{n − l − k + j + 1}{l − j} = (−1)^{i} \frac{(n − l − (k − i))!}{(n − l − k)!} \frac{(l − i)!}{l!}.

Substituting j = k − i, the length of ∑_{i=0}^{k} a_{i}v_{i} becomes

∑_{j=0}^{k} a_{k−j}^{2}\binom{n − l}{j}\binom{l}{k − j} =

∑_{j=0}^{k} \frac{(n − l)!l!((l − k + j)!)^{2}((n − l − j)!)^{2}}{j!(n − l − j)!(k − j)!(l − k + j)!(l!)^{2}((n − l − k)!)^{2}}

= \frac{(n − l)!}{l!(n − l − k)!} ∑_{j=0}^{k} \frac{(l − k + j)!(n − l − j)!}{j!(k − j)!(n − l − k)!} =

\frac{(n − l)!(l − k)!}{l!(n − l − k)!} ∑_{j=0}^{k} \binom{n − l − j}{k − j}\binom{l − k + j}{j}.

By Lemma 3.2 this equals a(l, k). □

Orthogonal Frobenius reciprocity (Theorem 2.1) now allows to deduce from Theorem 3.1 the following recursion formula for the rational isometry class of F_{k}.

Corollary 3.4 For 0 ≤ l, k ≤ \frac{n}{2} and 0 ≠ a ∈ ℚ let [aF_{k}] respectively [I\binom{n}{l}] denote the class of aF_{k} respectively I\binom{n}{l} in the Witt group W (ℚ). Then

[I\binom{n}{l}] = ∑_{k=0}^{l}[\frac{\binom{n}{l}}{\binom{n}{k} − \binom{n}{k−1}}a(l, k)F_{k}] = ∑_{k=0}^{l}[\binom{n − 2k}{l − k}F_{k}]

where a(l, k) is as in Theorem 3.1.

Remark 3.5 In principle this method can also be used to obtain the rational isometry classes of the other irreducible orthogonal S_{n}-modules using Theorem 2.4 (see [Neb, Section 2.3]). However the combinatorics to determine explicit bases for the fixed space of the corresponding Young subgroups on these modules gets much more involved so one cannot hope to get formulas for arbitrary n.

References

[CuR] C. W. Curtis, I. Reiner, Methods of Representation Theory Vol I, Wiley Classics (1990)

[Jam] G. James, The Representation Theory of the Symmetric Groups. Springer LNM 682 (1978)

[JaK] G. James, A. Kerber, The representation theory of the symmetric group. Encyclopedia of Mathematics and Its Applications, Vol. 16. Reading, Massachusetts, etc.: Addison-Wesley Publishing Company, Advanced Book Program. XXVIII (1981).

[Neb] G. Nebe, Orthogonale Darstellungen endlicher Gruppen und Gruppenringe. Habilitationsschrift, Aachener Beiträge zur Mathematik, Verlag Mainz Aachen (1999)