On the Orthogonal Basis of the Symmetry Classes of Tensors Associated with the Dicyclic Group

M. R. Darafsheh, M. R. Pournaki

Abstract A necessary and sufficient condition for the existence of orthogonal basis of decomposable symmetrized tensors for the symmetry classes of tensors associated with the dicyclic group is given. In particular we apply these conditions to the generalized quaternion group, for which the dimensions of the symmetry classes of tensors are computed.

Keywords: Symmetry class of tensors, Orthogonal basis, Dicyclic group, 2-Adic valuation. 1995 Mathematics Subject Classification: Primary 20C30; Secondary 15A69.

1 Introduction

Let V be an m-unitary space. Let ⊗^{n}V be the n-th tensor power of V and write v_{1}⊗⋯⊗v_{n} for the decomposable tensor product of the indicated vectors. To each permutation σ in S_{n} there corresponds a unique linear operator P (σ): ⊗^{n}V → ⊗^{n}V determined by P (σ)(v_{1} ⊗ ⋯ ⊗ v_{n}) = v_{σ^{−1}(1)} ⊗ ⋯ ⊗ v_{σ^{−1}(n)}. Let G be a subgroup of S_{n} and let I(G) be the set of all the irreducible complex characters of G. It follows from the orthogonality relations for characters that

\{T (G, χ) : ⊗^{n}V → ⊗^{n}V | T (G, χ) = \frac{χ(1)}{|G|} ∑_{σ∈G} χ(σ)P (σ), χ ∈ I(G)\}

is a set of annihilating idempotents which sum to the identity. The image of ⊗^{n}V under the T (G, χ) is called the symmetry class of tensors associated with G and χ and it is denoted by V_{χ}^{n}(G). The image of v_{1} ⊗ ⋯ ⊗ v_{n} under T (G, χ) is denoted by v_{1} ∗ … ∗ v_{n} and it is called a decomposable tensor. It is well-known that

dim V_{χ}^{n}(G) = \frac{χ(1)}{|G|} ∑_{σ∈G} χ(σ)m^{c(σ)}  (1)

where c(σ) is the number of cycles, including cycles of length one, in the disjoint cycle decomposition of σ (see [5]).

The inner product on V induces an inner product on ⊗^{n}V whose restriction to V_{χ}^{n}(G) satisfies

⟨u_{1} ∗ … ∗ u_{n} | v_{1} ∗ … ∗ v_{n}⟩ = \frac{χ(1)}{|G|}d_{χ}^{G}(A)

where A = [a_{ij}]_{n×n} = [⟨u_{i}|v_{j}⟩]_{n×n} and d_{χ}^{G}(A) = ∑_{σ∈G} χ(σ)a_{1σ(1)} … a_{nσ(n)} is the generalized matrix function.

With respect to this inner product we have

⨂^{n}V = ⨁_{χ∈I(G)} V_{χ}^{n}(G)

which is an orthogonal direct sum.

Let Γ_{m}^{n} be the set of all sequences α = (α_{1}, …, α_{n}) with 1 ≤ α_{i} ≤ m, so α is a mapping from a set of n elements into a set of m elements. Then the group G acts on Γ_{m}^{n} by σ.α := (α_{σ^{−1}(1)}, …, α_{σ^{−1}(n)}) where σ ∈ G is a permutation on n letters and α ∈ Γ_{m}^{n} is a mapping from a set of n elements into a set of m elements. Therefore the action may be written as σ.α = ασ^{−1} which is a composition of two functions. Let O(α) = {σ.α| σ ∈ G} be the orbit with representative α, also let G_{α} be the stabilizer of α, i.e., G_{α} = {σ ∈ G| σ.α = α}. In this setting if α ∈ Γ_{m}^{n} and σ ∈ G, then we have G_{σ.α} = σG_{α}σ^{−1}.

Let Δ be a system of distinct representatives of the orbits of G acting on Γ_{m}^{n} and define

\overline{Δ} = \{α ∈ Δ| ∑_{σ∈G_{α}} χ(σ) ≠ 0\},

and let Ω be the union of those equivalence classes represented by elements of \overline{Δ}.

Let {e_{1}, …, e_{m}} be an orthonormal basis of V. Denote by e_{α}^{∗} the tensor e_{α_{1}} ∗ … ∗ e_{α_{n}} where α = (α_{1}, …, α_{n}) ∈ Γ_{m}^{n}. We have

⟨e_{α}^{∗} | e_{β}^{∗}⟩ = \{ \begin{matrix}\frac{χ(1)}{|G|} ∑_{σ∈G_{β}} χ(στ^{−1}) & if & α = τ.β for some τ ∈ G, \\ 0 & if & O(α) ≠ O(β), \end{matrix}

in particular, taking the norm of e_{α}^{∗}, with respect to the induced inner product, one easily obtains the condition e_{α}^{∗} ≠ 0 if and only if α ∈ Ω.

For γ ∈ \overline{Δ}, V_{γ}^{∗} = ⟨e_{σ.γ}^{∗}| σ ∈ G⟩ is called the orbital subspace of V_{χ}^{n}(G). It follows that

V_{χ}^{n}(G) = ⨁_{γ∈\overline{Δ}} V_{γ}^{∗}  (2)

is an orthogonal direct sum. In [2] Freese proved that

dim V_{γ}^{∗} = \frac{χ(1)}{|G_{γ}|} ∑_{σ∈G_{γ}} χ(σ)  (3)

in particular, if χ is of degree one, then dim V_{γ}^{∗} = 1 for all γ ∈ \overline{Δ}.

If α = g.γ and β = g′.γ, then gg^{′−1}.β = α, so if we set τ = gg^{′−1} and use the above formula for ⟨e_{α}^{∗} | e_{β}^{∗}⟩, then we obtain

⟨e_{g.γ}^{∗} | e_{g′.γ}^{∗}⟩ = \frac{χ(1)}{|G|} ∑_{σ∈g′G_{γ}g^{−1}} χ(σ).  (4)

An orthogonal basis of the form {e_{α}^{∗}| α ∈ S}, where S is a subset of Γ_{m}^{n}, is called an orthogonal basis of decomposable symmetrized tensors for V_{χ}^{n}(G). By (2) V_{χ}^{n}(G) has an orthogonal basis of decomposable symmetrized tensors if and only if for all γ ∈ \overline{Δ}, the orbital subspace V_{γ}^{∗} has an orthogonal basis of decomposable symmetrized tensors. In particular, if χ is of degree one, since dim V_{γ}^{∗} = 1 for all γ ∈ \overline{Δ}, then V_{γ}^{∗} has an orthogonal basis of decomposable symmetrized tensors for all γ ∈ \overline{Δ} which implies that V_{χ}^{n}(G) has such a basis.

Several papers are devoted in investigation of the existence of an orthogonal basis of decomposable symmetrized tensors for V_{χ}^{n}(G), see for example [10]. In [3] a necessary and sufficient condition for the existence of orthogonal basis of decomposable symmetrized tensors for V_{χ}^{n}(G) is given, where G is a cyclic or a dihedral group. In [6] it was claimed that if {e_{1}, …, e_{m}} is an orthogonal basis of V, then there exists a subset S of Γ_{m}^{n} such that the set {e_{α}^{∗}| α ∈ S} forms an orthogonal basis of V_{χ}^{n}(G) if and only if χ(1) = 1. But later in [9] a counter-example to this claim was presented.

Now it is natural to consider a group G and an irreducible character χ ∈ I(G) and obtain necessary and sufficient conditions for the existence of an orthogonal basis for V_{χ}^{n}(G). With respect to this we consider the dicyclic group which will be explained below.

In this paper we find a necessary and sufficient condition for the existence of an orthogonal basis of decomposable symmetrized tensors of symmetry classes of tensors associated with the dicyclic group. Throughout this paper all characters are considered over the complex field ℂ. We adopt notations from [7] in this paper.

2 Dicyclic Group

The group T_{4n}, n ≥ 1, generated by the elements r, s such that r^{2n} = 1, r^{n} = s^{2}, s^{−1}rs = r^{−1} is called the dicyclic group of degree n, i.e., T_{4n} = ⟨r, s| r^{2n} = 1, r^{n} = s^{2}, s^{−1}rs = r^{−1}⟩ (see [4]). This group is of order 4n and in [1] page 7 it is denoted by ⟨2, 2, n⟩ and it is proved that T_{4n} = ⟨r, s| r^{n} = s^{2} = (rs)^{2}⟩. In any case we have T_{4n} = {r^{l}, r^{l}s| 0 ≤ l < 2n}.

By [4] T_{4n} has n + 3 conjugacy classes which are

\begin{matrix}\{1\}, \{r^{n}\}, \{r^{k}, r^{2n−k}\}, 1 ≤ k ≤ n − 1, \{r^{2k}s| 0 ≤ k ≤ n − 1\}, \\ \{r^{2k+1}s| 0 ≤ k ≤ n − 1\} \end{matrix}

and the character table of T_{4n} is indicated in Tables I and II.

Table I: The character table of T_{4n} where n is odd

|C_{T_{4n}} (g)|  4n  4n  2n  4  4
g  1  r^{n}  r^{k}(1 ≤ k ≤ n − 1)  s  rs
ψ_{1}  1  1  1  1  1
ψ_{2}  1  −1  (−1)^{k}  i  −i
ψ_{3}  1  1  1  −1  −1
ψ_{4}  1  −1  (−1)^{k}  −i  i
χ_{h}  2  2(−1)^{h}  2 cos(khπ/n)  0  0

(1 ≤ h ≤ n − 1)

Table II: The character table of T_{4n} where n is even

|C_{T_{4n}} (g)|  4n  4n  2n  4  4
g  1  r^{n}  r^{k}(1 ≤ k ≤ n − 1)  s  rs
ϕ_{1}  1  1  1  1  1
ϕ_{2}  1  1  (−1)^{k}  1  −1
ϕ_{3}  1  1  1  −1  −1
ϕ_{4}  1  1  (−1)^{k}  −1  1
χ_{h}  2  2(−1)^{h}  2 cos(khπ/n)  0  0

(1 ≤ h ≤ n − 1)

From the above tables we see that T_{4n} has four irreducible characters of degree 1 namely ψ_{1}, ψ_{2}, ψ_{3}, ψ_{4} if n is odd and ϕ_{1}, ϕ_{2}, ϕ_{3}, ϕ_{4} if n is even and n − 1 irreducible characters of degree 2 which are denoted by χ_{h}, 1 ≤ h ≤ n − 1.

By classical Cayley Theorem T_{4n} can be embed in S_{4n} and so we assume that T_{4n} is a subgroup of S_{4n}. In this case generators r, s of T_{4n} as permutations on 4n letters are given by

r = (1 2 3 … 2n)(2n + 1 2n + 2 2n + 3 … 4n),
s = (1 2n + 1 n + 1 3n + 1)(2 4n n + 2 3n)(3 4n − 1 n + 3 3n − 1) …
(n − 1 3n + 3 2n − 1 2n + 3)(n 3n + 2 2n 2n + 2).

In particular, the dicyclic group of degree 2^{n−1} is called the generalized quaternion group and denoted by Q_{2^{n+1}}, i.e., Q_{2^{n+1}} = T_{4(2^{n−1})} = T_{2^{n+1}}, and χ = χ_{h}, 1 ≤ h ≤ 2^{n−1}−1, are characters of degree 2 for Q_{2^{n+1}}. In the following theorems we find the dimensions of the symmetry classes of tensors associated with the dicyclic group T_{4n}.

Theorem 1 Let G = T_{4n}, n odd, and assume that V is an m-unitary space. Then considering G as a subgroup of the symmetric group on 4n letters we have the following, where (2n, k) denotes the greatest common divisor of 2n and k.

dim V_{ψ_{1}}^{4n}(G) = \frac{1}{4n} [m^{4n} + m^{2n} + 2 ∑_{k=1}^{n−1} m^{2(2n,k)} + 2nm^{n}] ,

dim V_{ψ_{2}}^{4n}(G) = \frac{1}{4n} [m^{4n} − m^{2n} + 2 ∑_{k=1}^{n−1}(−1)^{k}m^{2(2n,k)}] ,

dim V_{ψ_{3}}^{4n}(G) = \frac{1}{4n} [m^{4n} + m^{2n} + 2 ∑_{k=1}^{n−1} m^{2(2n,k)} − 2nm^{n}] ,

dim V_{ψ_{4}}^{4n}(G) = \frac{1}{4n} [m^{4n} − m^{2n} + 2 ∑_{k=1}^{n−1}(−1)^{k}m^{2(2n,k)}] ,

dim V_{χ_{h}}^{4n}(G) = \frac{1}{2n} [2m^{4n} + 2(−1)^{h}m^{2n} + 4 ∑_{k=1}^{n−1} cos(khπ/n)m^{2(2n,k)}] , 1 ≤ h ≤ n − 1.

Proof. Note that if π is a cycle of length a and (k, a) = d, then π^{k} has d cycles of length a/d and therefore c(π^{k}) = d = (k, a). So c(1) = 4n, c(r^{n}) = 2n, c(r^{k}) = 2(2n, k) and c(s) = n, where c(π) denotes the number of cycles in the cycle structure of π including cycles of length one. Considering the cycle structures of r and s given above we obtain:

\begin{aligned}rs = & (1 2n + 2 n + 1 3n + 2)(2 2n + 1 n + 2 3n + 1)(3 4n n + 3 3n) … \\ & (n 3n + 3 2n 2n + 3), \end{aligned}

therefore c(rs) = n. Now using the character table of T_{4n} given in Table I the theorem holds by (1). □

Theorem 2 Let G = T_{4n}, n even, and assume that V is an m-unitary space. Then considering G as a subgroup of the symmetric group on 4n letters we have the following, where (2n, k) denotes the greatest common divisor of 2n and k.

dim V_{ϕ_{1}}^{4n}(G) = \frac{1}{4n} [m^{4n} + m^{2n} + 2 ∑_{k=1}^{n−1} m^{2(2n,k)} + 2nm^{n}] ,

dim V_{ϕ_{2}}^{4n}(G) = \frac{1}{4n} [m^{4n} + m^{2n} + 2 ∑_{k=1}^{n−1}(−1)^{k}m^{2(2n,k)}] ,

dim V_{ϕ_{3}}^{4n}(G) = \frac{1}{4n} [m^{4n} + m^{2n} + 2 ∑_{k=1}^{n−1} m^{2(2n,k)} − 2nm^{n}] ,

dim V_{ϕ_{4}}^{4n}(G) = \frac{1}{4n} [m^{4n} + m^{2n} + 2 ∑_{k=1}^{n−1}(−1)^{k}m^{2(2n,k)}] ,

dim V_{χ_{h}}^{4n}(G) = \frac{1}{2n} [2m^{4n} + 2(−1)^{h}m^{2n} + 4 ∑_{k=1}^{n−1} cos(khπ/n)m^{2(2n,k)}] , 1 ≤ h ≤ n − 1.

Proof. Similar to the proof of Theorem 1. □

The following lemma is useful in later considerations.

Lemma 1 Let H be a subgroup of T_{4n}. Then there is a natural number k, 0 ≤ k < 2n, such that H = ⟨r^{k}⟩ or ⟨r^{k}⟩ ≨ H and H ∩ ⟨r⟩ = ⟨r^{k}⟩. In the second case we have |H| ≥ 2|⟨r^{k}⟩|.

Proof. By definition of T_{4n} we see that elements of T_{4n} are of the forms r^{l} or r^{l}s where 0 ≤ l < 2n. If H is an arbitrary subgroup of T_{4n}, then H ∩ ⟨r⟩ is a cyclic subgroup of ⟨r⟩ and therefore there is a natural number k, 0 ≤ k < 2n, such that H ∩ ⟨r⟩ = ⟨r^{k}⟩. The rest by our lemma follows immediately. □

3 On the Existence of Orthogonal Basis for the Symmetry Classes of Tensors Associated with the Dicyclic Group

Let G = T_{4n}, n ≥ 1, and V be an m-unitary space, with orthonormal basis {e_{1}, …, e_{m}}. For n = 1, the dicyclic group T_{4} is cyclic, T_{4} ≅ ℤ_{4}, therefore all irreducible characters are of degree 1 and so V_{χ}^{4}(T_{4}) has an orthogonal basis of decomposable symmetrized tensors for all χ ∈ I(T_{4}). Therefore we assume that n ≥ 2. If m = 1, then dim ⊗^{4n}V = 1, so dim V_{χ}^{4n}(G) = 0 or 1, therefore we don’t have any problem about the existence of orthogonal basis of decomposable symmetrized tensors for V_{χ}^{4n}(G) for all χ ∈ I(G), therefore we assume that m ≥ 2.

For irreducible characters of T_{4n} of degree 1, ψ_{i}, ϕ_{i}, 1 ≤ i ≤ 4, V_{ψ_{i}}^{4n}(T_{4n}) and V_{ϕ_{i}}^{4n}(T_{4n}) have an orthogonal basis of decomposable symmetrized tensors and so we don’t deal with the ψ_{i}’s and ϕ_{i}’s.

Therefore we investigate the problem for irreducible characters of degree 2 of T_{4n}, i.e., χ_{h}, 1 ≤ h ≤ n − 1, which are given by

χ_{h}(r^{k}) = 2 cos \frac{khπ}{n} , χ_{h}(r^{k}s) = 0, 0 ≤ k < 2n.

Lemma 2 Suppose n ≥ 2, 1 ≤ h ≤ n − 1, 0 ≤ k < 2n. Let l = (2n/(2n, k)), where (2n, k) denotes the greatest common divisor of 2n and k. Then we have

∑_{t=1}^{l} cos \frac{tkhπ}{n} = \{ \begin{matrix}l & if & kh ≡^{2n} 0, \\ 0 & if & kh ≢^{2n} 0. \end{matrix}

Proof. It is straightforward. □

Lemma 3 Suppose G = T_{4n}, n ≥ 2, and χ = χ_{h}, 1 ≤ h ≤ n − 1. Let H be any subgroup of T_{4n}, i.e., H = ⟨r^{k}⟩ or ⟨r^{k}⟩ ≨ H and H ∩ ⟨r⟩ = ⟨r^{k}⟩, for some k, 0 ≤ k < 2n. If l = (2n/(2n, k)), where (2n, k) denotes the greatest common divisor of 2n and k, then we have

∑_{g∈H} χ(g) = \{ \begin{matrix}2l & if & kh ≡^{2n} 0, \\ 0 & if & kh ≢^{2n} 0. \end{matrix}

Proof. We have o(r^{k}) = (2n/(2n, k)) = l, so H = {r^{k}, r^{2k}, …, r^{lk}} or {r^{k}, r^{2k}, …, r^{lk}} ≨ H and H ∩ ⟨r⟩ = {r^{k}, r^{2k}, …, r^{lk}}. But χ vanishes outside ⟨r⟩, therefore by Lemma 2 we have

∑_{g∈H} χ(g) = ∑_{t=1}^{l} χ(r^{tk}) = 2 ∑_{t=1}^{l} cos \frac{tkhπ}{n} = \{ \begin{matrix}2l & if & kh ≡^{2n} 0, \\ 0 & if & kh ≢^{2n} 0. \end{matrix}

□

Lemma 4 Let G = T_{4n}, n ≥ 2, and χ = χ_{h}, 1 ≤ h ≤ n − 1. Then for γ ∈ \overline{Δ}, we have

G_{γ} = ⟨r^{k}⟩ or ⟨r^{k}⟩ ≨ G_{γ} and G_{γ} ∩ ⟨r⟩ = ⟨r^{k}⟩, for some k, 0 ≤ k < 2n, where kh ≡^{2n} 0. In

particular, if ⟨r^{k}⟩ ≨ G_{γ}, then we have |G_{γ}| ≥ 2|⟨r^{k}⟩|.

Proof. G_{γ} is a subgroup of G so by Lemma 1, G_{γ} = ⟨r^{k}⟩ or ⟨r^{k}⟩ ≨ G_{γ} and G_{γ} ∩ ⟨r⟩ = ⟨r^{k}⟩, for some k, 0 ≤ k < 2n. In particular if ⟨r^{k}⟩ ≨ G_{γ}, then |G_{γ}| ≥ 2|⟨r^{k}⟩|. But by Lemma 3 if kh ≢^{2n} 0, then ∑_{g∈G_{γ}} χ(g) = 0 and so γ ∉ \overline{Δ}. This contradiction show that

kh ≡^{2n} 0. □

Lemma 5 Let n ≥ 2 and 1 ≤ h ≤ n − 1. Then there exist t, t′, 0 ≤ t, t′ < 2n, such that cos ((t − t′)hπ/n) = 0 if and only if ν_{2}(h/n) < 0, where ν_{2} is the 2-adic valuation.

Proof. It is straightforward. For the definition of p-adic valuation we refer the reader to [8]. □

Lemma 6 Suppose G = T_{4n}, n ≥ 2, and let χ = χ_{h}, 1 ≤ h ≤ n − 1. Let γ ∈ \overline{Δ} and suppose that G_{γ} is of the form G_{γ} = ⟨r^{k}⟩, where 0 ≤ k < 2n, kh ≡^{2n} 0. If ν_{2}(h/n) < 0, where ν_{2} is the 2-adic valuation, then the orbital subspace V_{γ}^{∗} has an orthogonal basis of decomposable symmetrized tensors.

Proof. We have o(r^{k}) = (2n/(2n, k)) = l, so G_{γ} = {r^{k}, r^{2k}, …, r^{lk}} and therefore by (3) and Lemma 3

dim V_{γ}^{∗} = \frac{χ(1)}{|G_{γ}|} ∑_{g∈G_{γ}} χ(g) = \frac{2}{l}(2l) = 4.

Now for all g, g′ ∈ G we have

g′G_{γ}g^{−1} = \{ \begin{matrix}\{r^{k+b−a}, r^{2k+b−a}, …, r^{lk+b−a}\} & if & g = r^{a}, g′ = r^{b}, \\ \{r^{k+n+a+b}s, r^{2k+n+a+b}s, …, r^{lk+n+a+b}s\} & if & g = r^{a}s, g′ = r^{b}, \\ \{r^{−k+b−a}, r^{−2k+b−a}, …, r^{−lk+b−a}\} & if & g = r^{a}s, g′ = r^{b}s. \end{matrix}

For g = r^{a}, g′ = r^{b} by (4) we have

⟨e_{g.γ}^{∗}|e_{g′.γ}^{∗}⟩ = \frac{χ(1)}{|G|} ∑_{σ∈g′G_{γ}g^{−1}} χ(σ)
= \frac{2}{4n} ∑_{t=1}^{l} χ(r^{tk+b−a})
= \frac{1}{n} ∑_{t=1}^{l} cos \frac{(tk+b−a)hπ}{n}
= \frac{1}{n} ∑_{t=1}^{l} cos (\frac{tkhπ}{n} + \frac{(b−a)hπ}{n} )
= \frac{1}{n} ∑_{t=1}^{l} cos \frac{(b−a)hπ}{n}
= \frac{l}{n} cos \frac{(b−a)hπ}{n} .

If g = r^{a}s and g′ = r^{b} or g = r^{a}s and g′ = r^{b}s then by the same computation we obtain ⟨e_{g.γ}^{∗}|e_{g′.γ}^{∗}⟩ = 0 or ⟨e_{g.γ}^{∗}|e_{g′.γ}^{∗}⟩ = (l/n) cos ((b − a)hπ/n), respectively. Therefore

⟨e_{g.γ}^{∗}|e_{g′.γ}^{∗}⟩ = \{ \begin{matrix}\frac{l}{n} cos \frac{(b−a)hπ}{n} & if & g = r^{a}, g′ = r^{b}, \\ 0 & if & g = r^{a}s, g′ = r^{b}, \\ \frac{l}{n} cos \frac{(b−a)hπ}{n} & if & g = r^{a}s, g′ = r^{b}s. \end{matrix}

Since by the assumption ν_{2}(\frac{h}{n}) < 0, hence by Lemma 5 there exist t, t′, 0 ≤ t, t′ < 2n, such that cos ((t − t′)hπ/n) = 0. Let S = {r^{t}.γ, r^{t′}.γ, r^{t}s.γ, r^{t′}s.γ} ⊂ Γ_{m}^{4n}, then by the above computation, we have ⟨e_{α}^{∗}|e_{β}^{∗}⟩ = 0 for all α, β ∈ S, α ≠ β. But dim V_{γ}^{∗} = 4, so {e_{α}^{∗}| α ∈ S} is an orthogonal basis of decomposable symmetrized tensors for V_{γ}^{∗}. □

Lemma 7 Suppose G = T_{4n}, n ≥ 2, and χ = χ_{h}, 1 ≤ h ≤ n − 1. Let γ ∈ \overline{Δ} be such

that ⟨r^{k}⟩ ≨ G_{γ} and G_{γ} ∩ ⟨r⟩ = ⟨r^{k}⟩, where 0 ≤ k < 2n, kh ≡^{2n} 0. If ν_{2}(\frac{h}{n}) < 0, where ν_{2}

is 2-adic valuation, then the orbital subspace V_{γ}^{∗} has an orthogonal basis of decomposable symmetrized tensors.

Proof. We have o(r^{k}) = (2n/(2n, k)) = l, so {r^{k}, r^{2k}, …, r^{lk}} ≨ G_{γ} and G_{γ} ∩ ⟨r⟩ = {r^{k}, r^{2k}, …, r^{lk}}. Note that, by Lemma 4, in this case we have |G_{γ}| ≥ 2l, and therefore by (3) and Lemma 3

dim V_{γ}^{∗} = \frac{χ(1)}{|G_{γ}|} ∑_{g∈G_{γ}} χ(g) ≤ \frac{2}{2l}(2l) = 2

so dim V_{γ}^{∗} = 1 or dim V_{γ}^{∗} = 2. If dim V_{γ}^{∗} = 1, then we don’t have any problem about the existence of orthogonal basis of decomposable symmetrized tensors for V_{γ}^{∗}, therefore we assume that dim V_{γ}^{∗} = 2. For g = r^{a}, g′ = r^{b} we have {r^{k+b−a}, r^{2k+b−a}, …, r^{lk+b−a}} ≨ g′G_{γ}g^{−1} and g′G_{γ}g^{−1} ∩ ⟨r⟩ = {r^{k+b−a}, r^{2k+b−a}, …, r^{lk+b−a}} therefore by (4) we have

⟨e_{g.γ}^{∗}|e_{g′.γ}^{∗}⟩ = \frac{χ(1)}{|G|} ∑_{σ∈g′G_{γ}g^{−1}} χ(σ)
= \frac{2}{4n} ∑_{t=1}^{l} χ(r^{tk+b−a})
= \frac{1}{n} ∑_{t=1}^{l} cos \frac{(tk+b−a)hπ}{n}
= \frac{1}{n} ∑_{t=1}^{l} cos (\frac{tkhπ}{n} + \frac{(b−a)hπ}{n} )
= \frac{1}{n} ∑_{t=1}^{l} cos \frac{(b−a)hπ}{n}
= \frac{l}{n} cos \frac{(b−a)hπ}{n} .

Since by assumption ν_{2}(\frac{h}{n}) < 0, hence by Lemma 5 there exist t, t′, 0 ≤ t, t′ < 2n, such that cos ((t − t′)hπ/n) = 0, therefore by the above computation ⟨e_{r^{t}.γ}^{∗}|e_{r^{t′}.γ}^{∗}⟩ = 0. For S = {r^{t}.γ, r^{t′}.γ} ⊂ Γ_{m}^{4n}, since dim V_{γ}^{∗} = 2, so {e_{α}^{∗}| α ∈ S} is an orthogonal basis of decomposable symmetrized tensors for V_{γ}^{∗}. □

Theorem 3 Let G = T_{4n}, n ≥ 2, and χ = χ_{h}, 1 ≤ h ≤ n − 1, dim V = m ≥ 2. Then V_{χ}^{4n}(G) has an orthogonal basis of decomposable symmetrized tensors if and only if ν_{2}(h/n) < 0, where ν_{2} is 2-adic valuation.

Proof. Assume V_{χ}^{4n}(G) has an orthogonal basis of decomposable symmetrized tensors, therefore by (2) for all γ ∈ \overline{Δ}, the orbital subspace V_{γ}^{∗} has an orthogonal basis of decomposable symmetrized tensors, in particular for γ = (1, 2, 2, …, 2). Note that in this case G_{γ} = \{1\}, and ∑_{g∈G_{γ}} χ(g) = 2 ≠ 0, so γ ∈ \overline{Δ}. For all g, g′ ∈ G, we have

g′G_{γ}g^{−1} = \{ \begin{matrix}\{r^{b−a}\} & if & g = r^{a}, g′ = r^{b}, \\ \{r^{n+a+b}s\} & if & g = r^{a}s, g′ = r^{b}, \\ \{r^{b−a}\} & if & g = r^{a}s, g′ = r^{b}s. \end{matrix}

Therefore by (4) we have

⟨e_{g.γ}^{∗}|e_{g′.γ}^{∗}⟩ = \{ \begin{matrix}\frac{1}{n} cos \frac{(b−a)hπ}{n} & if & g = r^{a}, g′ = r^{b}, \\ 0 & if & g = r^{a}s, g′ = r^{b}, \\ \frac{1}{n} cos \frac{(b−a)hπ}{n} & if & g = r^{a}s, g′ = r^{b}s. \end{matrix}

But by (3)

dim V_{γ}^{∗} = \frac{χ(1)}{|G_{γ}|} ∑_{g∈G_{γ}} χ(g) = \frac{2}{1}(2) = 4.

By the above computation if there are 4 decomposable symmetrized tensors for which any distinct pair are mutually orthogonal, then there should exist t, t′, 0 ≤ t, t′ < 2n, such that cos ((t − t′)hπ/n) = 0. Therefore by Lemma 5 we obtain ν_{2}(h/n) < 0.

Conversely assume ν_{2}(h/n) < 0, then by Lemmas 4, 6 and 7, for all γ ∈ \overline{Δ}, V_{γ}^{∗} has the orthogonal basis of decomposable symmetrized tensors, and therefore by (2) so does V_{χ}^{4n}(G). □

Corollary 1 Let G = T_{4n}, n ≥ 2 is odd, and χ = χ_{h}, 1 ≤ h ≤ n − 1, dim V = m ≥ 2. Then V_{χ}^{4n}(G) does not have an orthogonal basis of decomposable symmetrized tensors.

Proof. Since n is odd, therefore ν_{2}(h/n) ≥ 0, and by Theorem 3 the corollary holds. □

Theorem 4 Let G = Q_{2^{n+1}}, n ≥ 2, the generalized quaternion group, and χ = χ_{h}, 1 ≤ h ≤ 2^{n−1} − 1, dim V = m ≥ 2. Then V_{χ}^{2^{n+1}}(G) has an orthogonal basis of decomposable symmetrized tensors.

Proof. Note that G = Q_{2^{n+1}} = T_{4(2^{n−1})}, and since 1 ≤ h ≤ 2^{n−1} − 1 therefore ν_{2}(h/2^{n−1}) < 0 and by Theorem 3, this result follows. □

Acknowledgment: Both authors would like to thank the referee for his interest in the subject and making useful suggestions.

References

[1] H. S. M. Coxeter, W. O. J. Moser, “Generators and Relations for Discrete Groups”, Third edition, Springer-Verlag, Berlin, Heidelberg, New York, 1972.

[2] R. Freese, Inequalities for Generalized Matrix Functions Based on Arbitrary Characters, Linear Algebra Appl. 7 (1973), 337-345.

[3] R. R. Holmes, T. Y. Tam, Symmetry Classes of Tensors Associated with Certain Groups, Linear and Multilinear Algebra 32 (1992), 21-31.

[4] G. James, M. Liebeck, “Representations and Characters of Groups”, Cambridge University Press, 1993.

[5] M. Marcus, “Finite Dimensional Multilinear Algebra”, Part 1, Marcel Dekker, New York, 1973.

[6] M. Marcus, J. Chollet, Construction of Orthonormal Bases in Higher Symmetry Classes of Tensors, Linear and Multilinear Algebra 19 (1986), 133-140.

[7] R. Merris, “Multilinear Algebra”, Gordon and Breach Science Publishers, 1997.

[8] J. P. Serre, “A Course in Arithmetic”, First edition, Springer-Verlag, Berlin, Heidelberg, New York, 1973.

[9] B. Y. Wang, M. P. Gong, A Higher Symmetry Classes of Tensors with an Orthogonal Basis of Decomposable Symmetrized Tensors, Linear and Multilinear Algebra 30 (1991), 61-64.

[10] B. Y. Wang, M. P. Gong, The Subspace and Orthonormal Bases of Symmetry Classes of Tensors, Linear and Multilinear Algebra 30 (1991), 195-204.

The Authors’ Addresses

M. R. Darafsheh, Department of Mathematics and Computer Science, University of Tehran, and Institute for Studies in Theoretical Physics and Mathematics, Tehran, Iran.

E-mail address: darafshe@vax.ipm.ac.ir

M. R. Pournaki, Department of Mathematics and Computer Science, University of Tehran, and Institute for Studies in Theoretical Physics and Mathematics, Tehran, Iran.

E-mail address: pournaki@vax.ipm.ac.ir