Dihedral Group Frames with the Haar Property

Vignon Oussa; Brian Sheehan

arXiv:1705.00085·math.FA·May 2, 2017

Dihedral Group Frames with the Haar Property

Vignon Oussa, Brian Sheehan

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TL;DR

This paper proves that for odd integers n, the orbit of almost every vector under a specific dihedral group representation forms a basis for ^n, establishing the Haar property and providing explicit conditions.

Contribution

It demonstrates that for odd n, almost all vectors have orbits with the Haar property under the dihedral group representation, fully resolving a previously partial problem.

Findings

01

Almost every vector's orbit has the Haar property for odd n.

02

Explicit conditions are given for vectors with the Haar property.

03

The Haar property holds if and only if n is odd.

Abstract

We consider a unitary representation of the Dihedral group $D_{2 n}$ obtained by inducing the trivial character from the co-normal subgroup ${0} ⋊ Z_{2} .$ This representation is naturally realized as acting on the vector space $C^{n} .$ We prove that the orbit of almost every vector in $C^{n}$ with respect to the Lebesgue measure has the Haar property (every subset of cardinality $n$ of the orbit is a basis for $C^{n}$ ) if $n$ is an odd integer. Moreover, we provide explicit sufficient conditions for vectors in $C^{n}$ whose orbits have the Haar property. Finally, we derive that the orbit of almost every vector in $C^{n}$ under the action of the representation has the Haar property if and only if $n$ is odd. This completely settles a problem which was only partially answered in…

Equations119

A ∥ v ∥_{C^{n}}^{2} \leq k \in I \sum ⟨ v, v_{k} ⟩_{C^{n}} \leq B ∥ v ∥_{C^{n}}^{2}

A ∥ v ∥_{C^{n}}^{2} \leq k \in I \sum ⟨ v, v_{k} ⟩_{C^{n}} \leq B ∥ v ∥_{C^{n}}^{2}

T : C^{card (I)} \to C^{n}, T (c) = k \in I \sum c_{k} v_{k}

T : C^{card (I)} \to C^{n}, T (c) = k \in I \sum c_{k} v_{k}

T^{*} : C^{n} \to C^{card (I)}, T^{*} v = (⟨ v, v_{k} ⟩_{C^{n}})_{k \in I} .

T^{*} : C^{n} \to C^{card (I)}, T^{*} v = (⟨ v, v_{k} ⟩_{C^{n}})_{k \in I} .

S : C^{n} \to C^{n}, S v = k \in I \sum ⟨ v, v_{k} ⟩_{C^{n}} v_{k}

S : C^{n} \to C^{n}, S v = k \in I \sum ⟨ v, v_{k} ⟩_{C^{n}} v_{k}

v = k \in I \sum ⟨ v, v_{k} ⟩_{C^{n}} S^{- 1} v_{k} .

v = k \in I \sum ⟨ v, v_{k} ⟩_{C^{n}} S^{- 1} v_{k} .

v = k \in J \sum ⟨ v, v_{k} ⟩_{C^{n}} S_{J}^{- 1} v_{k} with S_{J} v = k \in J \sum ⟨ v, v_{k} ⟩_{C^{n}} v_{k} .

v = k \in J \sum ⟨ v, v_{k} ⟩_{C^{n}} S_{J}^{- 1} v_{k} with S_{J} v = k \in J \sum ⟨ v, v_{k} ⟩_{C^{n}} v_{k} .

T=\left[\begin{array}[c]{ccccc}0&0&\cdots&0&1\\ 1&0&\cdots&0&0\\ 0&1&\ddots&\vdots&\vdots\\ \vdots&\ddots&\ddots&0&0\\ 0&\cdots&0&1&0\end{array}\right]\text{ and }M=\left[\begin{array}[c]{ccccc}1&&&&\\ &\exp\left(\frac{2\pi i}{n}\right)&&&\\ &&\exp\left(\frac{4\pi i}{n}\right)&&\\ &&&\ddots&\\ &&&&\exp\left(\frac{2\pi i\left(n-1\right)}{n}\right)\end{array}\right].

T=\left[\begin{array}[c]{ccccc}0&0&\cdots&0&1\\ 1&0&\cdots&0&0\\ 0&1&\ddots&\vdots&\vdots\\ \vdots&\ddots&\ddots&0&0\\ 0&\cdots&0&1&0\end{array}\right]\text{ and }M=\left[\begin{array}[c]{ccccc}1&&&&\\ &\exp\left(\frac{2\pi i}{n}\right)&&&\\ &&\exp\left(\frac{4\pi i}{n}\right)&&\\ &&&\ddots&\\ &&&&\exp\left(\frac{2\pi i\left(n-1\right)}{n}\right)\end{array}\right].

{T^{ℓ} M^{κ} v : (ℓ, κ) \in Z_{n} \times Z_{n}}

{T^{ℓ} M^{κ} v : (ℓ, κ) \in Z_{n} \times Z_{n}}

T=\left[\begin{array}[c]{ccccc}0&0&\cdots&0&1\\ 1&0&\cdots&0&0\\ 0&1&\ddots&\vdots&\vdots\\ \vdots&\ddots&\ddots&0&0\\ 0&\cdots&0&1&0\end{array}\right],D=\left[\begin{array}[c]{ccccc}1&0&\cdots&0&0\\ 0&0&\cdots&0&1\\ \vdots&\vdots&&&0\\ 0&0&1&&\vdots\\ 0&1&0&\cdots&0\end{array}\right].

T=\left[\begin{array}[c]{ccccc}0&0&\cdots&0&1\\ 1&0&\cdots&0&0\\ 0&1&\ddots&\vdots&\vdots\\ \vdots&\ddots&\ddots&0&0\\ 0&\cdots&0&1&0\end{array}\right],D=\left[\begin{array}[c]{ccccc}1&0&\cdots&0&0\\ 0&0&\cdots&0&1\\ \vdots&\vdots&&&0\\ 0&0&1&&\vdots\\ 0&1&0&\cdots&0\end{array}\right].

\digamma_{n}=\left\{\begin{array}[c]{c}w\in\mathbb{C}^{n}:w=\left[\begin{array}[c]{c}\sum_{k=0}^{n-1}\lambda^{k}\\ \sum_{k=0}^{n-1}\lambda^{k}e^{-\frac{2\pi ki}{n}}\\ \vdots\\ \sum_{k=0}^{n-1}\lambda^{k}e^{-\frac{2\pi k\left(n-1\right)i}{n}}\end{array}\right]\\ \text{where }\lambda\text{ is a either transcendal}\\ \text{or an algebraic number over }\mathbb{\mathbb{\mathbb{Q}}}\left(\frac{2\pi i}{n}\right)\text{ whose degree is at least }n^{2}-n+1\end{array}\right\}.

\digamma_{n}=\left\{\begin{array}[c]{c}w\in\mathbb{C}^{n}:w=\left[\begin{array}[c]{c}\sum_{k=0}^{n-1}\lambda^{k}\\ \sum_{k=0}^{n-1}\lambda^{k}e^{-\frac{2\pi ki}{n}}\\ \vdots\\ \sum_{k=0}^{n-1}\lambda^{k}e^{-\frac{2\pi k\left(n-1\right)i}{n}}\end{array}\right]\\ \text{where }\lambda\text{ is a either transcendal}\\ \text{or an algebraic number over }\mathbb{\mathbb{\mathbb{Q}}}\left(\frac{2\pi i}{n}\right)\text{ whose degree is at least }n^{2}-n+1\end{array}\right\}.

v\mapsto O\left(v\right)=\left[\begin{array}[c]{ccc}\left(g_{1}v\right)_{0}&\cdots&\left(g_{1}v\right)_{n-1}\\ \vdots&\ddots&\vdots\\ \left(g_{m}v\right)_{0}&\cdots&\left(g_{m}v\right)_{n-1}\end{array}\right].

v\mapsto O\left(v\right)=\left[\begin{array}[c]{ccc}\left(g_{1}v\right)_{0}&\cdots&\left(g_{1}v\right)_{n-1}\\ \vdots&\ddots&\vdots\\ \left(g_{m}v\right)_{0}&\cdots&\left(g_{m}v\right)_{n-1}\end{array}\right].

p (v) = lcm ({det (X (v)) : X (v) \in P_{n}}) .

p (v) = lcm ({det (X (v)) : X (v) \in P_{n}}) .

⟨ ϕ, ψ ⟩ = x \in Z_{n} \sum ϕ (x) \overline{ψ (x)} for ϕ, ψ \in l^{2} (Z_{n}) .

⟨ ϕ, ψ ⟩ = x \in Z_{n} \sum ϕ (x) \overline{ψ (x)} for ϕ, ψ \in l^{2} (Z_{n}) .

[F ϕ] (ξ) = \frac{1}{n ^{1/2}} k \in Z_{n} \sum ϕ (k) exp (\frac{2 π ik ξ}{n}), for ϕ \in l^{2} (Z_{n}) .

[F ϕ] (ξ) = \frac{1}{n ^{1/2}} k \in Z_{n} \sum ϕ (k) exp (\frac{2 π ik ξ}{n}), for ϕ \in l^{2} (Z_{n}) .

[F^{- 1} φ] (k) = \frac{1}{n ^{1/2}} ξ \in Z_{n} \sum φ (ξ) exp (- \frac{2 π ik ξ}{n}) .

[F^{- 1} φ] (k) = \frac{1}{n ^{1/2}} ξ \in Z_{n} \sum φ (ξ) exp (- \frac{2 π ik ξ}{n}) .

(T v) (j) = v ((j - 1) mod n) and (D v) (j) = v ((n - j) mod n) .

(T v) (j) = v ((j - 1) mod n) and (D v) (j) = v ((n - j) mod n) .

s \in T (n, p) \sum (- 1)^{\sum_{k = 1}^{p} (s_{k} + 1) + \sum_{k = 1}^{p} (t_{k + 1})} det (X (t, s)) det (X (t, s)^{c}) .

s \in T (n, p) \sum (- 1)^{\sum_{k = 1}^{p} (s_{k} + 1) + \sum_{k = 1}^{p} (t_{k + 1})} det (X (t, s)) det (X (t, s)^{c}) .

V=\left[\begin{array}[c]{ccccc}1&a_{1}&a_{1}^{2}&\cdots&a_{1}^{n-1}\\ 1&a_{2}&a_{2}^{2}&\cdots&a_{2}^{n-1}\\ \vdots&\vdots&\vdots&\cdots&\vdots\\ 1&a_{n}&a_{n}^{2}&\cdots&a_{n}^{n-1}\end{array}\right]

V=\left[\begin{array}[c]{ccccc}1&a_{1}&a_{1}^{2}&\cdots&a_{1}^{n-1}\\ 1&a_{2}&a_{2}^{2}&\cdots&a_{2}^{n-1}\\ \vdots&\vdots&\vdots&\cdots&\vdots\\ 1&a_{n}&a_{n}^{2}&\cdots&a_{n}^{n-1}\end{array}\right]

det (V) = 1 \leq j < k \leq n \prod (a_{k} - a_{j})

det (V) = 1 \leq j < k \leq n \prod (a_{k} - a_{j})

ξ = e^{\frac{2 π i}{n}} .

ξ = e^{\frac{2 π i}{n}} .

A\left(z\right)=\left[\begin{array}[c]{ccc}\left(\gamma_{1}z\right)_{0}&\cdots&\left(\gamma_{1}z\right)_{n-1}\\ \vdots&\ddots&\vdots\\ \left(\gamma_{2n}z\right)_{0}&\cdots&\left(\gamma_{2n}z\right)_{n-1}\end{array}\right].

A\left(z\right)=\left[\begin{array}[c]{ccc}\left(\gamma_{1}z\right)_{0}&\cdots&\left(\gamma_{1}z\right)_{n-1}\\ \vdots&\ddots&\vdots\\ \left(\gamma_{2n}z\right)_{0}&\cdots&\left(\gamma_{2n}z\right)_{n-1}\end{array}\right].

S\left(z\right)=\left[\begin{array}[c]{cccc}z_{0}&z_{1}&\cdots&z_{n-1}\\ z_{0}&\xi z_{1}&\cdots&\xi^{\left(n-1\right)}z_{n-1}\\ \vdots&\ddots&\ddots&\vdots\\ z_{0}&\xi^{\left(n-1\right)}z_{1}&\cdots&\xi^{\left(n-1\right)\left(n-1\right)}z_{n-1}\end{array}\right]\text{ }

S\left(z\right)=\left[\begin{array}[c]{cccc}z_{0}&z_{1}&\cdots&z_{n-1}\\ z_{0}&\xi z_{1}&\cdots&\xi^{\left(n-1\right)}z_{n-1}\\ \vdots&\ddots&\ddots&\vdots\\ z_{0}&\xi^{\left(n-1\right)}z_{1}&\cdots&\xi^{\left(n-1\right)\left(n-1\right)}z_{n-1}\end{array}\right]\text{ }

S\left(z\right)=\left[\begin{array}[c]{cccc}z_{0}&z_{\iota\left(1\right)}&\cdots&z_{i\left(n-1\right)}\\ z_{0}&\xi z_{\iota\left(1\right)}&\cdots&\xi^{\left(n-1\right)}z_{i\left(n-1\right)}\\ \vdots&\ddots&\ddots&\vdots\\ z_{0}&\xi^{\left(n-1\right)}z_{\iota\left(1\right)}&\cdots&\xi^{\left(n-1\right)\left(n-1\right)}z_{i\left(n-1\right)}\end{array}\right]

S\left(z\right)=\left[\begin{array}[c]{cccc}z_{0}&z_{\iota\left(1\right)}&\cdots&z_{i\left(n-1\right)}\\ z_{0}&\xi z_{\iota\left(1\right)}&\cdots&\xi^{\left(n-1\right)}z_{i\left(n-1\right)}\\ \vdots&\ddots&\ddots&\vdots\\ z_{0}&\xi^{\left(n-1\right)}z_{\iota\left(1\right)}&\cdots&\xi^{\left(n-1\right)\left(n-1\right)}z_{i\left(n-1\right)}\end{array}\right]

\left[\begin{array}[c]{cccccccc}z_{0}&\xi^{1l_{0}}z_{1}&\cdots&\xi^{\left(m-1\right)l_{0}}z_{m-1}&\xi^{ml_{0}}z_{m}&\xi^{\left(m+1\right)l_{0}}z_{m+1}&\cdots&\xi^{\left(n-1\right)l_{0}}z_{n-1}\\ z_{0}&\xi^{1l_{1}}z_{1}&\cdots&\xi^{\left(m-1\right)l_{1}}z_{m-1}&\xi^{ml_{1}}z_{m}&\xi^{\left(m+1\right)l_{1}}z_{m+1}&\cdots&\xi^{\left(n-1\right)l_{1}}z_{n-1}\\ \vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots\\ z_{0}&\xi^{1l_{m-1}}z_{1}&\cdots&\xi^{\left(m-1\right)l_{m-1}}z_{m-1}&\xi^{ml_{m-1}}z_{m}&\xi^{\left(m+1\right)l_{m-1}}z_{m+1}&\cdots&\xi^{\left(n-1\right)l_{m-1}}z_{n-1}\\ z_{0}&\xi^{1j_{0}}z_{\iota\left(1\right)}&\cdots&\xi^{\left(m-1\right)j_{0}}z_{\iota\left(m-1\right)}&\xi^{mj_{0}}z_{\iota\left(m\right)}&\xi^{\left(m+1\right)j_{0}}z_{\iota\left(m+1\right)}&\cdots&\xi^{\left(n-1\right)j_{0}}z_{i\left(n-1\right)}\\ z_{0}&\xi^{1j_{1}}z_{\iota\left(1\right)}&\cdots&\xi^{\left(m-1\right)j_{1}}z_{\iota\left(m-1\right)}&\xi^{mj_{1}}z_{\iota\left(m\right)}&\xi^{\left(m+1\right)j_{1}}z_{\iota\left(m+1\right)}&\cdots&\xi^{\left(n-1\right)j_{1}}z_{i\left(n-1\right)}\\ \vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots\\ z_{0}&\xi^{1j_{s-1}}z_{\iota\left(1\right)}&\cdots&\xi^{\left(m-1\right)j_{s-1}}z_{\iota\left(m-1\right)}&\xi^{mj_{s-1}}z_{\iota\left(m\right)}&\xi^{\left(m+1\right)j_{s-1}}z_{\iota\left(m+1\right)}&\cdots&\xi^{\left(n-1\right)j_{s-1}}z_{i\left(n-1\right)}\end{array}\right]

\left[\begin{array}[c]{cccccccc}z_{0}&\xi^{1l_{0}}z_{1}&\cdots&\xi^{\left(m-1\right)l_{0}}z_{m-1}&\xi^{ml_{0}}z_{m}&\xi^{\left(m+1\right)l_{0}}z_{m+1}&\cdots&\xi^{\left(n-1\right)l_{0}}z_{n-1}\\ z_{0}&\xi^{1l_{1}}z_{1}&\cdots&\xi^{\left(m-1\right)l_{1}}z_{m-1}&\xi^{ml_{1}}z_{m}&\xi^{\left(m+1\right)l_{1}}z_{m+1}&\cdots&\xi^{\left(n-1\right)l_{1}}z_{n-1}\\ \vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots\\ z_{0}&\xi^{1l_{m-1}}z_{1}&\cdots&\xi^{\left(m-1\right)l_{m-1}}z_{m-1}&\xi^{ml_{m-1}}z_{m}&\xi^{\left(m+1\right)l_{m-1}}z_{m+1}&\cdots&\xi^{\left(n-1\right)l_{m-1}}z_{n-1}\\ z_{0}&\xi^{1j_{0}}z_{\iota\left(1\right)}&\cdots&\xi^{\left(m-1\right)j_{0}}z_{\iota\left(m-1\right)}&\xi^{mj_{0}}z_{\iota\left(m\right)}&\xi^{\left(m+1\right)j_{0}}z_{\iota\left(m+1\right)}&\cdots&\xi^{\left(n-1\right)j_{0}}z_{i\left(n-1\right)}\\ z_{0}&\xi^{1j_{1}}z_{\iota\left(1\right)}&\cdots&\xi^{\left(m-1\right)j_{1}}z_{\iota\left(m-1\right)}&\xi^{mj_{1}}z_{\iota\left(m\right)}&\xi^{\left(m+1\right)j_{1}}z_{\iota\left(m+1\right)}&\cdots&\xi^{\left(n-1\right)j_{1}}z_{i\left(n-1\right)}\\ \vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots\\ z_{0}&\xi^{1j_{s-1}}z_{\iota\left(1\right)}&\cdots&\xi^{\left(m-1\right)j_{s-1}}z_{\iota\left(m-1\right)}&\xi^{mj_{s-1}}z_{\iota\left(m\right)}&\xi^{\left(m+1\right)j_{s-1}}z_{\iota\left(m+1\right)}&\cdots&\xi^{\left(n-1\right)j_{s-1}}z_{i\left(n-1\right)}\end{array}\right]

q (t) = k = 1 \prod n - 1 t^{k} .

q (t) = k = 1 \prod n - 1 t^{k} .

\mathbb{C}\backepsilon t\mapsto\mathfrak{c}\left(t\right)=\left[\begin{array}[c]{ccccc}1&t&t^{2}&\cdots&t^{n-1}\end{array}\right]^{T}\in\mathbb{C}^{n}.

\mathbb{C}\backepsilon t\mapsto\mathfrak{c}\left(t\right)=\left[\begin{array}[c]{ccccc}1&t&t^{2}&\cdots&t^{n-1}\end{array}\right]^{T}\in\mathbb{C}^{n}.

\det\left(S\left(1,t,t^{2},\cdots,t^{n-1}\right)\right)=\mathbf{q}\left(t\right)\left|\begin{array}[c]{cccc}1&1&\cdots&1\\ 1&\xi&\cdots&\xi^{\left(n-1\right)}\\ 1&\xi^{2}&\cdots&\xi^{2\left(n-1\right)}\\ \vdots&\ddots&\ddots&\vdots\\ 1&\xi^{\left(n-1\right)}&\cdots&\xi^{\left(n-1\right)\left(n-1\right)}\end{array}\right|=\left({\prod\limits_{0\leq j<k\leq n-1}}\left(\xi^{k}-\xi^{j}\right)\right)\times{\prod\limits_{k=1}^{n-1}}t^{k}.

\det\left(S\left(1,t,t^{2},\cdots,t^{n-1}\right)\right)=\mathbf{q}\left(t\right)\left|\begin{array}[c]{cccc}1&1&\cdots&1\\ 1&\xi&\cdots&\xi^{\left(n-1\right)}\\ 1&\xi^{2}&\cdots&\xi^{2\left(n-1\right)}\\ \vdots&\ddots&\ddots&\vdots\\ 1&\xi^{\left(n-1\right)}&\cdots&\xi^{\left(n-1\right)\left(n-1\right)}\end{array}\right|=\left({\prod\limits_{0\leq j<k\leq n-1}}\left(\xi^{k}-\xi^{j}\right)\right)\times{\prod\limits_{k=1}^{n-1}}t^{k}.

det (S (1, t, t^{2}, \dots, t^{n - 1})) = 0 \leq j < k \leq n - 1 \prod (ξ^{k} - ξ^{j}) \times q (t) .

det (S (1, t, t^{2}, \dots, t^{n - 1})) = 0 \leq j < k \leq n - 1 \prod (ξ^{k} - ξ^{j}) \times q (t) .

\left[\begin{array}[c]{ccc}\xi^{0l_{0}}t&\cdots&\xi^{\left(m-1\right)l_{0}}t^{m-1}\\ \xi^{0l_{1}}t&\cdots&\xi^{\left(m-1\right)l_{1}}t^{m-1}\\ \vdots&\ddots&\vdots\\ \xi^{0l_{m-1}}t&\cdots&\xi^{\left(m-1\right)l_{m-1}}t^{m-1}\end{array}\right],\left[\begin{array}[c]{cccc}\xi^{ml_{0}}t^{m}&\xi^{\left(m+1\right)l_{0}}t^{m+1}&\cdots&\xi^{\left(n-1\right)l_{0}}t^{n-1}\\ \xi^{ml_{1}}t^{m}&\xi^{\left(m+1\right)l_{1}}t^{m+1}&\cdots&\xi^{\left(n-1\right)l_{1}}t^{n-1}\\ \vdots&\vdots&\ddots&\vdots\\ \xi^{ml_{m-1}}t^{m}&\xi^{\left(m+1\right)l_{m-1}}t^{m+1}&\cdots&\xi^{\left(n-1\right)l_{m-1}}t^{n-1}\end{array}\right]

\left[\begin{array}[c]{ccc}\xi^{0l_{0}}t&\cdots&\xi^{\left(m-1\right)l_{0}}t^{m-1}\\ \xi^{0l_{1}}t&\cdots&\xi^{\left(m-1\right)l_{1}}t^{m-1}\\ \vdots&\ddots&\vdots\\ \xi^{0l_{m-1}}t&\cdots&\xi^{\left(m-1\right)l_{m-1}}t^{m-1}\end{array}\right],\left[\begin{array}[c]{cccc}\xi^{ml_{0}}t^{m}&\xi^{\left(m+1\right)l_{0}}t^{m+1}&\cdots&\xi^{\left(n-1\right)l_{0}}t^{n-1}\\ \xi^{ml_{1}}t^{m}&\xi^{\left(m+1\right)l_{1}}t^{m+1}&\cdots&\xi^{\left(n-1\right)l_{1}}t^{n-1}\\ \vdots&\vdots&\ddots&\vdots\\ \xi^{ml_{m-1}}t^{m}&\xi^{\left(m+1\right)l_{m-1}}t^{m+1}&\cdots&\xi^{\left(n-1\right)l_{m-1}}t^{n-1}\end{array}\right]

\left[\begin{array}[c]{ccc}\xi^{0\times j_{0}}t^{n-1}&\cdots&\xi^{\left(m-1\right)j_{0}}t^{n-\left(m-1\right)}\\ \xi^{0\times j_{1}}t^{n-1}&\cdots&\xi^{\left(m-1\right)j_{1}}t^{n-\left(m-1\right)}\\ \vdots&\ddots&\vdots\\ \xi^{0\times j_{s-1}}t^{n-1}&\cdots&\xi^{\left(m-1\right)j_{s-1}}t^{n-\left(m-1\right)}\end{array}\right]\text{ and }\left[\begin{array}[c]{cccc}\xi^{mj_{0}}t^{n-m}&\xi^{\left(m+1\right)j_{0}}t^{n-m-1}&\cdots&\xi^{\left(n-1\right)j_{0}}t\\ \xi^{mj_{1}}t^{n-m}&\xi^{\left(m+1\right)j_{1}}t^{n-m-1}&\cdots&\xi^{\left(n-1\right)j_{1}}t\\ \vdots&\vdots&\ddots&\vdots\\ \xi^{mj_{s-1}}t^{n-m}&\xi^{m+1j_{s-1}}t^{n-m-1}&\cdots&\xi^{\left(n-1\right)j_{s-1}}t\end{array}\right]

\left[\begin{array}[c]{ccc}\xi^{0\times j_{0}}t^{n-1}&\cdots&\xi^{\left(m-1\right)j_{0}}t^{n-\left(m-1\right)}\\ \xi^{0\times j_{1}}t^{n-1}&\cdots&\xi^{\left(m-1\right)j_{1}}t^{n-\left(m-1\right)}\\ \vdots&\ddots&\vdots\\ \xi^{0\times j_{s-1}}t^{n-1}&\cdots&\xi^{\left(m-1\right)j_{s-1}}t^{n-\left(m-1\right)}\end{array}\right]\text{ and }\left[\begin{array}[c]{cccc}\xi^{mj_{0}}t^{n-m}&\xi^{\left(m+1\right)j_{0}}t^{n-m-1}&\cdots&\xi^{\left(n-1\right)j_{0}}t\\ \xi^{mj_{1}}t^{n-m}&\xi^{\left(m+1\right)j_{1}}t^{n-m-1}&\cdots&\xi^{\left(n-1\right)j_{1}}t\\ \vdots&\vdots&\ddots&\vdots\\ \xi^{mj_{s-1}}t^{n-m}&\xi^{m+1j_{s-1}}t^{n-m-1}&\cdots&\xi^{\left(n-1\right)j_{s-1}}t\end{array}\right]

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Full text

Dihedral Group Frames with the Haar Property

Vignon Oussa, Brian Sheehan

Dept. of Mathematics

Bridgewater State University

Bridgewater, MA 02324 U.S.A.

[email protected]

Abstract.

We consider a unitary representation of the Dihedral group $D_{2n}=\mathbb{Z}_{n}\rtimes\mathbb{Z}_{2}$ obtained by inducing the trivial character from the co-normal subgroup $\left\{0\right\}\rtimes\mathbb{Z}_{2}.$ This representation is naturally realized as acting on the vector space $\mathbb{C}^{n}.$ We prove that the orbit of almost every vector in $\mathbb{C}^{n}$ with respect to the Lebesgue measure has the Haar property (every subset of cardinality $n$ of the orbit is a basis for $\mathbb{C}^{n}$ ) if $n$ is an odd integer. Moreover, we provide explicit sufficient conditions for vectors in $\mathbb{C}^{n}$ whose orbits have the Haar property. Finally, we derive that the orbit of almost every vector in $\mathbb{C}^{n}$ under the action of the representation has the Haar property if and only if $n$ is odd. This completely settles a problem which was only partially answered in [7].

Key words and phrases:

Linearly independent frames, Haar property

2000 Mathematics Subject Classification:

15A15,42C15

1. Introduction

Let $\mathfrak{F}$ be a set of $m\geq n$ vectors in an $n$ -dimensional vector space $\mathbb{C}^{n}$ over $\mathbb{C}.$ We say that $\mathfrak{F}$ has the Haar property or is a full spark frame if any subset of $\mathfrak{F}$ of cardinality $n$ is a basis for $\mathbb{C}^{n}$ . We recall that a set $\mathfrak{F}$ of vectors contained in $\mathbb{C}^{n}$ is called a frame for $\mathbb{C}^{n}$ [3] if $\mathfrak{F}$ is a spanning set for the vector space $\mathbb{C}^{n}.$ Moreover, it is well-known that a countable family of vectors $\mathfrak{F}=\left(v_{k}\right)_{k\in I}$ is a frame for $\mathbb{C}^{n}$ if and only if there exist positive constants $A,B>0$ such that

[TABLE]

for every vector $v\in\mathbb{C}^{n}.$ Let

[TABLE]

be the synthesis operator corresponding to $\mathfrak{F}$ . The adjoint of $T$ also known as the analysis operator is defined as follows

[TABLE]

The composition of the synthesis operator with its adjoint

[TABLE]

is called the frame operator. It is a well-known fact [3] that if $\mathfrak{F}=\left(v_{k}\right)_{k\in I}$ is a frame then $S$ is a self-adjoint invertible operator and every vector $v\in\mathbb{C}^{n}$ can be represented as

[TABLE]

In other words, any vector $v$ can be reconstructed from the sequence $\left(\left\langle v,v_{k}\right\rangle_{\mathbb{C}^{n}}\right)_{k\in I}.$ In fact if $K$ is a proper subset of $I$ such that $\left(v_{k}\right)_{k\in K}$ is also a frame then any vector $v$ can be reconstructed from the sequence $\left(\left\langle v,v_{k}\right\rangle_{\mathbb{C}^{n}}\right)_{k\in K}$ . Thus, full spark frames are very flexible basis-like tools in the sense that if $\left(v_{k}\right)_{k\in I}$ is a full spark frame, then for any subset $J$ of $I$ of cardinality $n,$ every vector $v\in\mathbb{C}^{n}$ can be uniquely represented as

[TABLE]

We shall now present some examples of full spark frames induced from the action of group representations. Let $\mathbb{H}_{n}<GL\left(n,\mathbb{\mathbb{C}}\right)$ be the Heisenberg group which is a finite group generated by

[TABLE]

It is shown in [6, 8] that for almost every vector $v\in\mathbb{C}^{n}$ (with respect to Lebesgue measure) the collection

[TABLE]

is a full spark frame. Next, let $n$ be a natural number greater than two. Let $D_{2n}$ be the Dihedral group generated by $T,D\in GL\left(n,\mathbb{\mathbb{C}}\right)$ where

[TABLE]

It is proved in [7] that if $n$ is even then it is not possible to find a full spark frame of the type $D_{2n}v.$ Moreover, if $n$ is prime then there exists a Zariski open subset $\Omega$ of $\mathbb{C}^{n}$ such that $D_{2n}v$ is a full spark frame for every vector $v$ in $\Omega.$ While this result establishes the existence of vectors $v$ such that $D_{2n}v$ has the Haar property when $n$ is prime, the work in [7] is not constructive. Moreover, the case where $n$ is odd but not prime was left open in [7]. The main objective of the present paper is to fill this gap by settling the following. Let $n$ be an odd number which is not necessarily odd.

•

Is there a vector $v\in\mathbb{\mathbb{C}}^{n}$ such that $D_{2n}v$ has the Haar property?

•

Can we provide an explicit construction of a vector $v\in\mathbb{\mathbb{C}}^{n}$ such that $D_{2n}v$ has the Haar property?

1.1. Main results

Here is a summary of our results. For a fixed natural number $n>2$ , we define $\digamma_{n}\subset\mathbb{C}^{n}$ such that

[TABLE]

Theorem 1.

If $n$ is odd and $w\in\digamma_{n}$ then $D_{2n}w$ is a full spark frame.

Suppose now that $G$ is a finite group of matrices of order $m$ acting on $\mathbb{C}^{n}.$ Moreover, let us assume that the order of $G$ is greater than or equal to $n.$ It is not hard to verify that if there exists one vector $w$ in $\mathbb{C}^{n}$ such that the orbit of $w$ under the action of $G$ has the Haar property then for almost every vector $v$ (with respect to Lebesgue measure) in $\mathbb{C}^{n},$ the orbit of $v$ under the action of $G$ has the Haar property. Indeed, assume that there exists a vector $w$ in $\mathbb{C}^{n}$ such that $Gw$ has the Haar property. Consider the matrix-valued function

[TABLE]

By assumption, there exists an element $O\left(w\right)$ in the range of the function above such that any sub-matrix of order $n$ of $O\left(w\right)$ has a non-zero determinant. Let $P_{n}$ be the set of all sub-matrices of $O\left(v\right)$ of order $n.$ Next, let $p\left(v\right)$ be the least common multiple of all polynomials obtained by computing the determinant of $X\left(v\right)$ where $X\left(v\right)\in P_{n}.$ That is,

[TABLE]

Clearly $w$ is not a zero for the polynomial $p.$ Next, let $W$ be an open set around $p\left(w\right)\neq 0$ such that $W$ does not contain zero. Since $p$ is continuous, the inverse image of the open set $W$ under the map $p$ is an open subset of $\mathbb{C}^{n}.$ So, there exists an open subset of $\mathbb{C}^{n}$ containing $w$ which is disjoint from the zero set of $p.$ This implies that $p$ is not the zero polynomial. Since the zero set of $p$ is a co-null set with respect to the Lebesgue measure on $\mathbb{C}^{n}$ , it immediately follows that for almost every vector $v$ in $\mathbb{C}^{n},$ the set $Gv$ has the Haar property. From the preceding discussion, the following is immediate.

Corollary 2.

Let $n$ be an odd natural number greater than two. There exists a Zariski open subset $\Omega\subset\mathbb{C}^{n}$ such that if $v\in\Omega$ then $D_{2n}v$ is a full spark frame.

Finally, a direct application of the results in [7] and Theorem 1 together with Corollary 2 gives the following.

Corollary 3.

The $D_{2n}$ -orbit of almost every vector (with respect to the Lebesgue measure) in $\mathbb{C}^{n}$ is a full spark frame if and only if $n$ is odd.

The paper is organized as follows. In the second section, we fix notation and we review a variety of relevant concepts. The third section contains intermediate results leading to the proof of the main result which is given in the last section.

2. Preliminaries

Let us start by fixing notation. Given a matrix $M,$ the transpose of $M$ is denoted $M^{T}.$ The determinant of a matrix $M$ is denoted by $\mathrm{det}(M)$ or $|M|.$ The $k$ -th row of $M$ is denoted $\operatorname{row}_{k}\left(M\right)$ and similarly, the $k$ -th column of the matrix $M$ is denoted $\operatorname{col}_{k}\left(M\right).$ Let $G$ be a group acting on a set $S.$ We denote this action multiplicatively. For any fixed element $s\in S,$ the $G$ -orbit of $s$ is described as $Gs=\left\{gs:g\in G\right\}.$

2.1. Fourier Analysis on $\mathbb{Z}_{n}$

Let $\mathbb{Z}_{n}=\left\{0,1,\cdots,n-1\right\}.$ Let $l^{2}\left(\mathbb{Z}_{n}\right)=\left\{f:\mathbb{Z}_{n}\rightarrow\mathbb{C}\right\}$ be the vector space of all complex-valued functions on $\mathbb{Z}_{n}$ endowed with the dot product

[TABLE]

The norm of a given vector $\phi$ in $l^{2}\left(\mathbb{Z}_{n}\right)$ is computed as follows $\left\|\phi\right\|_{l^{2}\left(\mathbb{Z}_{n}\right)}=\left\langle\phi,\phi\right\rangle^{1/2}.$ We recall that the discrete Fourier transform is a map $\mathcal{F}:l^{2}\left(\mathbb{Z}_{n}\right)\rightarrow l^{2}\left(\mathbb{Z}_{n}\right)$ defined by

[TABLE]

The following facts are also well-known [10].

•

The discrete Fourier transform is a bijective linear operator.

•

The Fourier inverse of a vector $\varphi$ is given by

[TABLE]

•

The Fourier transform is a unitary operator. More precisely, given $\phi,\psi\in l^{2}\left(\mathbb{Z}_{n}\right),$ we have $\left\langle\phi,\psi\right\rangle=\left\langle\mathcal{F}\phi,\mathcal{F}\psi\right\rangle.$

The followig is proved in [4].

Lemma 4.

Let $\mathbf{F}$ be the matrix representation of the Fourier transform. If $n$ is prime then every minor of $\mathbf{F}$ is nonzero.

Identifying $l^{2}\left(\mathbb{Z}_{n}\right)$ with $\mathbb{C}^{n}$ via the map $v\mapsto\left[\begin{array}[c]{ccc}v\left(0\right)&\cdots&v\left(n-1\right)\end{array}\right]^{T},$ we may write

[TABLE]

It is not hard to verify that for any $\xi\in\mathbb{Z}_{n},$ we have $\left[\mathcal{F}Bv\right]\left(\xi\right)=\left(\mathcal{F}v\right)\left(\left(n-\xi\right)\operatorname{mod}n\right)$ and $\left[\mathcal{F}Av\right]\left(\xi\right)=e^{\frac{2\pi i}{n}\xi}\left(\mathcal{F}v\right)\left(\xi\right).$ Next, let $\mathbf{F}$ be the matrix representation of the Fourier transform with respect to the canonical basis of $\mathbb{C}^{n}.$ The matrix $T$ is diagonalizable and its Jordan canonical form is obtained by conjugating $T$ by the discrete Fourier matrix. More precisely if $M=\mathbf{F}T\mathbf{F}^{-1}$ then the $k^{th}$ diagonal entry of $M$ is given by $M_{kk}=e^{\frac{2\left(k-1\right)\pi i}{n}}.$ Finally, it is worth noting that $D$ commutes with the Fourier matrix $\mathbf{F.}$

2.2. Determinants

([5], Section $3.7$ ) Let $X$ be a square matrix of order $n.$ A minor of $X$ is the determinant of any square sub-matrix $Y$ of $X.$ Let $\left|Y\right|=\det Y$ be an $m$ -rowed minor of $X.$ The determinant of the sub-matrix obtained by deleting from $X$ the rows and columns represented in $Y$ is called the complement of $\left|Y\right|$ and is denoted $\left|Y^{c}\right|=\det\left(Y^{c}\right).$ Let $\left|Y\right|$ be the $m$ -rowed minor of $X$ in which rows $i_{1},\cdots,i_{m}$ and columns $j_{1},\cdots,j_{m}$ are represented. Then the algebraic complement, or cofactor of $\left|Y\right|$ is given by $\left(-1\right)^{\sum_{k=1}^{m}i_{k}+\sum_{k=1}^{m}j_{k}}\left|Y^{c}\right|.$ According to Laplace’s Expansion Theorem (see $3.7.3,$ [5]) a formula for the determinant of $X$ can be obtained as follows. Let $T\left(n,p\right)$ be the set of all ordered sets of cardinality $p$ of the type $s=\left\{0\leq s_{1}<s_{2}<\cdots<s_{p}\leq n-1\right\}.$ Given any two elements $s$ and $t$ in $T\left(n,p\right),$ we define $X\left(s,t\right)$ to be the sub-matrix of $X$ of order $p$ such that $X\left(s,t\right)_{i,j}=X_{s_{i}+1,t_{j}+1}.$ According to Laplace’s Expansion Theorem, once we fix an ordered set $t$ in $T\left(n,p\right),$ the determinant of $X$ can be computed via the following expansion

[TABLE]

The following well-known result will also be useful to us.

Lemma 5.

(Page $127,$ [5]) Let

[TABLE]

be a Vandermonde matrix. Then

[TABLE]

3. Intermediate Results

In this section, we will prove several results which lead to the proof of Theorem 1. Put

[TABLE]

Let $\iota\left(k\right)$ is be the inverse of $k$ modulo $n.$ That is $\iota\left(k\right)=\left(n-k\right)\operatorname{mod}n.$ Let $M$ be the linear operator obtained by conjugating $T$ by the Fourier transform. Let $\Lambda$ be the group of linear operators generated by $M$ and $D.$ Fixing an ordering for the elements of $\Lambda$ , we may write $\Lambda=\left\{\gamma_{1},\cdots,\gamma_{2n}\right\}.$ We identify the Hilbert space $l^{2}\left(\mathbb{Z}_{n}\right)$ in the usual way with the Cartesian product of $n$ copies of the complex plane. For any $z\in\mathbb{C}^{n},$ we define the map $z\mapsto A\left(z\right)$ where $A\left(z\right)$ is the $2n$ by $n$ matrix-valued function whose $k^{th}$ row corresponds to the vector $\gamma_{k}z$ which is an element of the $\Lambda$ -orbit of $z.$ From our definition, $A$ is regarded as a matrix-valued function taking the vector $z$ to the matrix

[TABLE]

Let $S\left(z\right)$ be a fixed submatrix of $A\left(z\right)$ of order $n.$ Fixing an ordering for the rows of $S\left(z\right)$ , we may assume that the following holds true.

**Case $\left(a\right)$ : **

Either

[TABLE]

or

[TABLE]

**Case $\left(b\right)$ : **

or $S\left(z\right)$ is equal to

[TABLE]

for some integers $l_{0},\cdots,l_{m-1},j_{0},j_{1},\cdots,j_{s-1}$ such that $m+s=n,$ if $k\neq k^{\prime}$ then $l_{k}\neq l_{k^{\prime}}$ and if $k\neq k^{\prime}$ then $j_{k}\neq j_{k^{\prime}}.$

Let $\mathbf{q}$ be the non-trivial polynomial defined by

[TABLE]

3.1. Case $\left(a\right)$

Let us consider the complex curve

[TABLE]

The image of this curve under the function $S$ is simply computed by replacing each variable $z_{k}$ by $t^{k}.$ If we assume that $S\left(z\right)$ is equal to the first matrix described in Case $\left(a\right)$ then

[TABLE]

In a similar fashion, it is easy to verify that if $S\left(z\right)$ is equal to the second matrix described in Case $\left(a\right)$ then

[TABLE]

Thus, the determinants of the matrices described in Case (a) are non-zero polynomials.

3.2. Case $\left(b\right)$

Define matrices $A_{1}\left(t\right),A_{2}\left(t\right)$ as

[TABLE]

respectively. Additionally, we define $A_{3}\left(t\right),$ and $A_{4}\left(t\right)$ as

[TABLE]

respectively. Note that $A_{1}\left(t\right)$ is a square matrix of order $m,$ $A_{4}\left(t\right)$ is a square matrix of order $n-m=s,$ $A_{2}\left(t\right)$ and $A_{3}\left(t\right)$ are matrices of order $m\times s$ and $s\times m$ respectively. If we assume that $S\left(z\right)$ is as described in Case $\left(b\right)$ then $S\left(\mathfrak{c}\left(t\right)\right)$ is given in block form as follows

[TABLE]

Lemma 6.

The following holds true.

(1)

$\det A_{1}\left(t\right)={\prod\limits_{0\leq k<j\leq m-1}}\left(\xi^{l_{j}}-\xi^{l_{k}}\right)\left(t\cdots t^{m-1}\right).$ ** 2. (2)

$\det A_{4}\left(t\right)=\left(\xi^{mj_{0}}\xi^{mj_{1}}\cdots\xi^{mj_{s-1}}\right){\prod\limits_{0\leq k<j\leq n-1-m}}\left(\xi^{j_{\ell}}-\xi^{j_{k}}\right)\left(t^{n-m}t^{n-m-1}\cdots t^{2}t\right).$ **

Proof.

The multi-linearity of the determinant function together with Formula (2.2) give the desired result. ∎

Let $\mathbf{p}_{m}=\left\{p_{1}<p_{2}<\cdots<p_{m-1}<p_{m}\right\}$ be an ordered subset of $\mathbb{Z}_{n}$ of cardinality $m$ and set

[TABLE]

Let $S\left(\mathfrak{c}\left(t\right)\right)$ be the matrix given in (3.8). To avoid cluster of notation, from now on, we shall write $S\left(\mathfrak{c}\left(t\right)\right)=\mathbf{S}_{t}\mathbf{.}$ Put

[TABLE]

For $n,m\in\mathbb{N}$ such that $n>2$ and $m\leq n,$ define

[TABLE]

Lemma 7.

The product of the determinants of $\mathbf{S}_{t}\left(\mathbf{i}_{m},\mathbf{i}_{m}\right)$ and $\mathbf{S}_{t}\left(\mathbf{i}_{m},\mathbf{i}_{m}\right)^{c}$ is equal to the non-trivial polynomial $Ct^{\kappa_{\left(n,m\right)}}.$

Proof.

We check that the determinant of the matrix $\mathbf{S}_{t}\left(\mathbf{i}_{m},\mathbf{i}_{m}\right)$ is a monomial in the variable $t$ which is given by

[TABLE]

Moreover, the determinant of $\left[\mathbf{S}_{t}\left(\mathbf{i}_{m},\mathbf{i}_{m}\right)\right]^{c}$ is given by

[TABLE]

In summary, taking the product of (3.11) and (3.12) we obtain $Ct^{\kappa_{\left(n,m\right)}}.$ Moreover, since $\xi^{j_{\ell}}-\xi^{j_{k}}\neq 0$ and $\xi^{l_{j}}-\xi^{l_{k}}\neq 0$ for $0\leq k<j\leq n-1-m$ and $0\leq k<j\leq m-1$ , it must be case that the monomial $Ct^{\kappa_{\left(n,m\right)}}$ is a non-trivial polynomial. ∎

Observe that $\mathbf{p}_{m}=\left(\mathbf{i}_{m}\cap\mathbf{p}_{m}\right)\cup\left(\mathbf{p}_{m}-\mathbf{i}_{m}\right)$ and $\mathbf{p}_{m}-\mathbf{i}_{m}$ and $\mathbf{i}_{m}-\mathbf{p}_{m}$ are subsets of $\mathbb{Z}_{n}$ of the same cardinality. Furthermore if $\mathbf{p}_{m}$ and $\mathbf{i}_{m}$ are not equal as sets, we shall write $\mathbf{i}_{m}=\left(\mathbf{i}_{m}\cap\mathbf{p}_{m}\right)\cup\left(\mathbf{i}_{m}-\mathbf{p}_{m}\right)$ and

[TABLE]

such that each $\ell_{k}\left(\mathbf{p}_{m}\right)$ is a positive number belonging to the set $\mathbb{Z}_{n}$ , and $u\left(\mathbf{p}_{m}\right)=\left|\mathbf{i}_{m}-\mathbf{p}_{m}\right|>0.$ We recall that $\mathbf{S}_{t}\left(\mathbf{i}_{m},\mathbf{p}_{m}\right)$ is the matrix obtained by selecting all entries of the matrix $\mathbf{S}_{t}$ of the type $\left[\mathbf{S}_{t}\right]_{k,k^{\prime}}$ where $k-1$ is in $\mathbf{i}_{m}$ and $k^{\prime}-1$ is an element of $\mathbf{p}_{m}.$

Lemma 8.

If $n$ is odd then

[TABLE]

Proof.

According to Lemma 7, if $\mathbf{i}_{m}=\mathbf{p}_{m}$ then

[TABLE]

For the other way around, let us suppose that $\mathbf{i}_{m}\neq\mathbf{p}_{m}.$ Then $u\left(\mathbf{p}_{m}\right)=\mathrm{card}\left(\mathbf{i}_{m}-\mathbf{p}_{m}\right)>0.$ First, we observe that, there exists integer $\delta$ and a complex number $q$ such that

[TABLE]

Secondly, the monomial $t^{\delta}$ can be computed by multiplying

[TABLE]

by a suitable rational function of the variable $t$ . Indeed, since the set $\mathbf{p}_{m}$ is obtained by taking the union of $\mathbf{i}_{m}\cap\mathbf{p}_{m}$ and $\mathbf{p}_{m}-$$\mathbf{i}_{m}$ , the above-mentioned monomial $t^{\delta}$ is the product of $\left(t\cdots t^{m-1}\right)\left(t^{n-m}t^{n-\left(m+1\right)}\cdots t^{2}t\right)$ and $r\left(t\right)$ where

[TABLE]

Next, it is clear that $r\left(t\right)=t^{\sigma}$ for some integer $\sigma,$ and $\sigma$ is computed as follows. If zero is an element of $\mathbf{i}_{m}\mathbf{-\mathbf{p}_{m}}$ then

[TABLE]

Next,

[TABLE]

Since $n$ is odd, the integer $\sigma$ cannot be equal to zero. On the other hand, if zero is not an element of $\mathbf{i-p}$ then

[TABLE]

In this case, since each $\ell_{k}\left(\mathbf{p}_{m}\right)$ is positive, $r\left(t\right)$ must be a non-trivial homogeneous polynomial in the variable $t.$ It follows that

[TABLE]

and $\delta$ is a not equal to zero. Thus,

[TABLE]

∎

Lemma 9.

The following holds true.

(1)

Every submatrix of order $n$ of $A\left(\mathfrak{c}\left(t\right)\right)$ is a non-vanishing element of the ring $\mathbb{Q}\left(\xi\right)\left[t\right].$ 2. (2)

Every submatrix of order $n$ of $A\left(\mathfrak{c}\left(t\right)\right)$ is a non-vanishing element of the ring $\mathbb{Q}\left(\xi\right)\left[t\right]$ of degree less than or equal to $n\left(n-1\right).$

Proof.

For the first part, let $S\left(\mathfrak{c}\left(t\right)\right)$ be a submatrix of $A\left(\mathfrak{c}\left(t\right)\right)$ of order $n$ . Assume that $S\left(\mathfrak{c}\left(t\right)\right)$ is given as shown in (3.8). Let

[TABLE]

We fix $t\in T\left(n,m\right).$ Put $\mathbf{S}_{t}=S\left(\mathfrak{c}\left(t\right)\right).$ According to Laplace’s Expansion Theorem,

[TABLE]

Next,

[TABLE]

where

[TABLE]

From our previous discussion, the monomial $C\times t^{\frac{2m^{2}-2mn-2m+n^{2}+n}{2}}$ appears once in the expansion above. Moreover, since $C$ is never equal to zero, it follows that the determinant of $S\left(\mathfrak{c}\left(t\right)\right)$ is a non-vanishing polynomial in the variable $t.$ For the second part, it is clear that the degree of the polynomial $\det\left(A_{1}\left(t\right)A_{4}\left(t\right)\right)$ satisfies the following condition

[TABLE]

Next, since the determinant of any sub-matrix of order $n$ of $A\left(\mathfrak{c}\left(t\right)\right)$ must be a sum of polynomials of degree less or equal to $n\left(n-1\right)$ then the second part of the lemma holds. ∎

4. Proof of Theorem 1

First, observe that $n\left(n-1\right)+1=n^{2}-n+1$ . Fix $\lambda\in\digamma_{n}$ where $\digamma_{n}$ is as described in (1.2). Since the determinant of every submatrix of $A\left(\mathfrak{c}\left(t\right)\right)$ of order $n$ is a polynomial in $\mathbb{Q}\left(\xi\right)\left[t\right]$ of degree bounded above by $n\left(n-1\right),$ it is clear that the determinant of every submatrix of $A\left(\mathfrak{c}\left(\lambda\right)\right)$ of order $n$ cannot be equal to zero. As such, it immediately follows that $\Lambda\left[1,\lambda,\cdots,\lambda^{n-1}\right]^{T}$ is a set of cardinality $2n$ in $\mathbb{C}^{n}$ which is a full spark frame. Consequently, since $\Gamma=\mathbf{F}^{-1}\Lambda\mathbf{F}\Leftrightarrow\Gamma\mathbf{F}^{-1}=\mathbf{F}^{-1}\Lambda$ it follows that

[TABLE]

is a set of cardinality $2n$ in $\mathbb{C}^{n}$ enjoying the Haar property.

Bibliography10

The reference list from the paper itself. Each links out to its DOI / PubMed record.

1[1] A. C, Aitken Determinants and Matrices, Oliver and Boyd, Ltd., Edinburgh (1939).
2[2] B. Alexeev, J. Cahill, D. G. Mixon, Full spark frames, preprint (2011)
3[3] P. Casazza, J. Kovacevi, Equal-Norm Tight Frames with Erasures, Adv. Comput. Math. 18(2-4), 387–430 (2003).
4[4] R.J. Evans, I.M. Isaacs, Generalized Vandermonde Determinants and Roots of Unity of Prime Order, Proc. Amer. Math. Soc. 58, 51–54 (1976).
5[5] E. Howard Elementary matrix theory, Reprint of the 1966 edition. Dover Books on Advanced Mathematics. Dover Publications, Inc., New York, 1980.
6[6] R. D. Malikiosis, A note on Gabor frames in finite dimensions, Appl. Comput. Harmon. Anal. 38(2), 318330 (2015).
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TL;DR

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Dihedral Group Frames with the Haar Property

Abstract.

Key words and phrases:

2000 Mathematics Subject Classification:

1. Introduction

1.1. Main results

Theorem 1**.**

Corollary 2**.**

Corollary 3**.**

2. Preliminaries

2.1. Fourier Analysis on Zn\mathbb{Z}_{n}Zn​

Lemma 4**.**

2.2. Determinants

Lemma 5**.**

3. Intermediate Results

3.1. Case (a)\left(a\right)(a)

3.2. Case (b)\left(b\right)(b)

Lemma 6**.**

Proof.

Lemma 7**.**

Proof.

Lemma 8**.**

Proof.

Lemma 9**.**

Proof.

4. Proof of Theorem 1

Theorem 1.

Corollary 2.

Corollary 3.

2.1. Fourier Analysis on $\mathbb{Z}_{n}$

Lemma 4.

Lemma 5.

3.1. Case $\left(a\right)$

3.2. Case $\left(b\right)$

Lemma 6.

Lemma 7.

Lemma 8.

Lemma 9.