John Jasper
Air Force Institute of Technology
(w/ Matthew Fickus, Joseph W. Iverson, Dustin G. Mixon)
The views expressed in this talk are those of the speaker and do not reflect the official policy
or position of the United States Air Force, Department of Defense, or the U.S. Government.
October 22, 2022
\(d\)
Definition. Let \[\Phi = \big[\varphi_{1}\ \ \varphi_{2}\ \ \cdots\ \ \varphi_{N}\big]\in \mathbb{F}^{d\times N},\]
where each column \(\varphi_{n}\) is unit norm
\[\|\varphi_{n} \|^{2}=1.\]
1) (Tightness) \(\exists\,A>0\) such that \((\Phi^{\ast}\Phi)^{2} = A\Phi^{\ast}\Phi\).
2) (Equiangular) \(\exists\,B>0\) such that \(|\frac{1}{B}\varphi_{m}^{\ast}\varphi_{n}^{}|=1\) for \(m\neq n\).
If both 1) and 2) hold, then \(\{\varphi_{n}\}_{n=1}^{N}\) is an ETF(\(d,N)\).
\[\Phi^{\ast}\Phi = \left[\begin{array}{cccc} 1 & \varphi_{1}^{\ast}\varphi_{2} & \cdots & \varphi_{1}^{\ast}\varphi_{N}\\[1ex] \varphi_{2}^{\ast}\varphi_{1} & 1 & \cdots & \varphi_{2}^{\ast}\varphi_{N}\\[1ex] \vdots & \vdots & \ddots & \vdots\\[1ex] \varphi_{N}^{\ast}\varphi_{1} & \varphi_{N}^{\ast}\varphi_{2} & \cdots & 1\end{array}\right]\]
\(1\)'s down the diagonal
1) \(\Phi^{\ast}\Phi \propto\) projection
2) \(|\varphi_{m}^{\ast}\varphi_{n}^{}|\) constant
Definition. Let \[\Phi = \big[\Phi_{1}\ \ \Phi_{2}\ \ \cdots\ \ \Phi_{N}\big]\in(\mathbb{F}^{d\times r})^{1\times N},\]
where the columns of each \(\Phi_{n}\) form ONB for a subspace (w/ dim\(=r\))
\[\Phi_{n}^{\ast}\Phi_{n} = \boldsymbol{I}.\]
1) (Tightness) \(\exists\,A>0\) such that \((\Phi^{\ast}\Phi)^{2} = A\Phi^{\ast}\Phi\).
2) (Equi-isoclinic) \(\exists\,B>0\) such that \(\frac{1}{B}\Phi_{m}^{\ast}\Phi_{n}\) is unitary for \(m\neq n\).
If both 1) and 2) hold, then \(\{\Phi_{n}\}_{n=1}^{N}\) is an EITFF(\(d,N,r)\).
\[\Phi^{\ast}\Phi = \left[\begin{array}{cccc} \boldsymbol{I} & \Phi_{1}^{\ast}\Phi_{2} & \cdots & \Phi_{1}^{\ast}\Phi_{N}\\[1ex] \Phi_{2}^{\ast}\Phi_{1} & \boldsymbol{I} & \cdots & \Phi_{2}^{\ast}\Phi_{N}\\[1ex] \vdots & \vdots & \ddots & \vdots\\[1ex] \Phi_{N}^{\ast}\Phi_{1} & \Phi_{N}^{\ast}\Phi_{2} & \cdots & \boldsymbol{I}\end{array}\right]\]
Identities down the diagonal
1) \(\Phi^{\ast}\Phi \propto\) projection
2) \(\Phi_{m}^{\ast}\Phi_{n}\propto\) unitaries
Take my favorite ETF, an ETF\((3,4)\):
\[\Phi = \frac{1}{\sqrt{3}}\left[\begin{array}{cccc} + & - & + & -\\ + & + & - & -\\ + & - & - & +\end{array}\right]\]
1) (ETFs) \(\Phi\) is an EITFF\((3,4,1)\)
2) (Tensors) \(\Phi\otimes\boldsymbol{I} = \dfrac{1}{\sqrt{3}}\left[\begin{array}{cc|cc|cc|cc} + & 0 & - & 0 & + & 0 & - & 0\\ 0 & + & 0 & - & 0 & + & 0 & -\\ + & 0 & + & 0 & - & 0 & - & 0\\ 0 & + & 0 & + & 0 & - & 0 & -\\ + & 0 & - & 0 & - & 0 & + & 0\\ 0 & + & 0 & - & 0 & - & 0 & + \end{array}\right]\)
\[\exists\,\operatorname{EITFF}(d,N,r)\quad\Rightarrow\quad \exists\,\operatorname{EITFF}(md,N,mr)\]
\[\exists\,\operatorname{ETF}(d,N)\quad\Leftrightarrow\quad\exists\,\operatorname{EITFF}(d,N,1)\]
We call these "tensor-sized"
is an EITFF\((6,4,2)\)
\[\Phi^{\ast}\Phi=\frac{1}{4}\left[\begin{array}{lllllllll}4&1&1&1&\omega&\omega^2&1&\omega&\omega^2\\1&4&1&\omega^2&1&\omega&\omega&\omega^2&1\\1&1&4&\omega&\omega^2&1&\omega^2&1&\omega\\1&\omega&\omega^2&4&1&1&1&\omega^2&\omega\\\omega^2&1&\omega&1&4&1&\omega&1&\omega^2\\\omega&\omega^2&1&1&1&4&\omega^2&\omega&1\\1&\omega^2&\omega&1&\omega^2&\omega&4&1&1\\\omega^2&\omega&1&\omega&1&\omega^2&1&4&1\\\omega&1&\omega^2&\omega^2&\omega&1&1&1&4\end{array}\right]\]
3) (\(\mathbb{C}\) to \(\mathbb{R}\) trick) Take \(\Phi\) to be an \(\operatorname{ETF}(6,9)\) with
(\(\omega=\exp(\frac{2\pi i}{3})\))
Replace: \(R\exp(\theta i)\mapsto R\left[\begin{array}{cc}\cos(\theta)&-\sin(\theta)\\\sin(\theta)&\cos(\theta)\end{array}\right]\)
We obtain the Gram matrix of a real EITFF\((12,9,2)\)
\[\frac{1}{4}\]
Tensor-sized, but no real ETF\((6,9)\) exists.
Goal: Find
Example. Let \(\mathcal{G} = \mathbb{Z}_{7} = \{0,1,2,3,4,5,6\}\),
\[\pi(g) = \begin{bmatrix} 1 & 0 & 0\\ 0 & e^{2\pi i g/7} & 0\\ 0 & 0 & e^{2 \pi i 3g/7}\end{bmatrix},\]
and \(\Phi_{0}= [1\ \ 1\ \ 1]^{\top}/\sqrt{3}\).
\(\Phi=\)
such that \(\{\pi(g)\Phi_{0}\}_{g\in\mathcal{G}}\) is an EITFF.
Example 2. Let \(\mathcal{G} = \mathbb{Z}_{5} = \{0,1,2,3,4\}\) and \(\omega=\exp(2\pi i/5)\).
\[\Psi = \big[\begin{array}{c|c|c|c|c} \Psi_{0} & \Psi_{1} & \Psi_{2} & \Psi_{3} & \Psi_{4}\end{array}\big] = \frac{1}{\sqrt{2}}\left[\begin{array}{c|c|c|c|c}\cdot & 1 & 1 & 1 & 1\\ \cdot & 1 & i & -i & -1\end{array}\right]\]
\[\Phi_{g}\doteq\pi(g)\Psi^{\ast}\doteq\left[\begin{array}{ccccc}\hspace{-5px}1 \boldsymbol{I}_{0}&&&&\\&\hspace{-5px}\omega^{g} \boldsymbol{I}_{1}&&&\\&&\hspace{-5px}\omega^{2g} \boldsymbol{I}_{1}&&\\&&&\hspace{-5px}\omega^{3g} \boldsymbol{I}_{1}&\\&&&&\hspace{-5px}\omega^{4g} \boldsymbol{I}_{1}\end{array}\right]\left[\begin{array}{cc}\cdot & \cdot\\\hline 1 & 1\\\hline 1 & -i\\\hline 1 & i\\\hline 1 & -1 \end{array}\right]\]
\[=\frac{1}{2}\left[\begin{array}{cc|cc|cc|cc|cc}1&1&\mu^{4}&\mu^{4}&\mu^{8}&\mu^{8}&\mu^{12}&\mu^{12}&\mu^{16}&\mu^{16}\\1&\mu^{15}&\mu^{8}&\mu^{3}&\mu^{16}&\mu^{11}&\mu^{4}&\mu^{19}&\mu^{12}&\mu^{7}\\1&\mu^{5}&\mu^{12}&\mu^{17}&\mu^{4}&\mu^{9}&\mu^{16}&\mu^{1}&\mu^{8}&\mu^{13}\\1&\mu^{10}&\mu^{16}&\mu^{6}&\mu^{12}&\mu^{2}&\mu^{8}&\mu^{18}&\mu^{4}&\mu^{14}\end{array}\right]\]
\[\Phi_{g}\doteq\pi(g)\Psi^{\ast}\doteq\left[\begin{array}{cccc}&\hspace{-5px}\omega^{g} \boldsymbol{I}_{1}&&&\\&&\hspace{-5px}\omega^{2g} \boldsymbol{I}_{1}&&\\&&&\hspace{-5px}\omega^{3g} \boldsymbol{I}_{1}&\\&&&&\hspace{-5px}\omega^{4g} \boldsymbol{I}_{1}\end{array}\right]\left[\begin{array}{cc} 1 & 1\\ 1 & -i\\ 1 & i\\ 1 & -1 \end{array}\right]\]
\[\Phi = \big[\ \Phi_{0}\ \ \vert\ \ \Phi_{1}\ \ \vert\ \ \Phi_{2}\ \ \vert\ \ \Phi_{3}\ \ \vert\ \ \Phi_{4}\ \big]\]
\(\big(\mu = \exp(2\pi i/20)\big)\)
For each \(\chi\in\hat{\mathcal{G}}\) the matrix \(\Psi_{\chi}\) is an \(r\times D_{\chi}\) matrix. (\(D_{\chi}\) could be zero!)
Start with:
Characters: \(\hat{\mathcal{G}} = \{\chi_{0},\chi_{1},\ldots,\chi_{N-1}\}\)
\[\pi(g) := \bigoplus_{\chi\in\hat{\mathcal{G}}}\chi(g)\boldsymbol{I}_{D_{\chi}} = \left[\begin{array}{ccccc}\hspace{-2px}\chi_{0}(g) \boldsymbol{I}_{D_{\chi_{0}}}&&&&\\&\hspace{-10px}\chi_{1}(g) \boldsymbol{I}_{D_{\chi_{1}}}&&&\\&&\hspace{-10px}\chi_{2}(g) \boldsymbol{I}_{D_{\chi_{2}}}&&\\&&&\hspace{-15px}\ &\\&&&&\hspace{-10px}\chi_{N-1}(g) \boldsymbol{I}_{D_{\chi_{N-1}}}\hspace{-2px}\end{array}\right]\]
Harmonic TFF generated by \(\Psi\):
\[\Phi \doteq\left[\begin{array}{c|c|c|c|c} \pi(g_{0})\Psi^{\ast} & \pi(g_{1})\Psi^{\ast} & \pi(g_{2})\Psi^{\ast} & \cdots & \pi(g_{N-1})\Psi^{\ast}\end{array}\right]\]
\(\ddots\)
Harmonic TFF generated by \(\Psi\):
\[\Phi \doteq \left[\begin{array}{c|c|c|c|c} \pi(g_{0})\Psi^{\ast} & \pi(g_{1})\Psi^{\ast} & \pi(g_{2})\Psi^{\ast} & \cdots & \pi(g_{N-1})\Psi^{\ast}\end{array}\right]\]
Set \(\Phi_{g} \doteq \pi(g)\Psi^{\ast}\).
\[\Phi_{g}^{\ast}\Phi_{h}^{} \doteq (\pi(g)\Psi^{\ast})^{\ast}(\pi(h)\Psi^{\ast}) = \Psi\pi(g^{-1})\pi(h)\Psi^{\ast} \doteq \Phi_{g_{0}}^{\ast}\Phi_{h^{-1}g}^{}\]
Cross-Grams:
The Gram matrix is block circulant.
Also,
\[\Phi_{g_{0}}^{\ast}\Phi_{g} \doteq \sum_{\chi\in\hat{\mathcal{G}}}\chi(g)\Psi_{\chi}\Psi_{\chi}^{\ast} = \sum_{\chi\in\hat{\mathcal{G}}}\chi(g)\mathbf{P}_{\chi}=:\mathbf{M}_{g}\]
\(\mathbf{P}_{\chi}\) is the projection onto \(\operatorname{ran}\Psi_{\chi}\)
\(\mathbf{M}_{g}\) is the entrywise Fourier transform of the \(\mathbf{P}_{\chi}\)'s.
Theorem (Fickus,Iverson,J,Mixon '21) Suppose
Then,
Example 2 (redux). Let \(\mathcal{G} = \mathbb{Z}_{5} = \{0,1,2,3,4\}\) and \(\omega=\exp(2\pi i/5)\).
\[\Psi = \big[\begin{array}{c|c|c|c|c} \Psi_{0} & \Psi_{1} & \Psi_{2} & \Psi_{3} & \Psi_{4}\end{array}\big] = \frac{1}{\sqrt{2}}\left[\begin{array}{c|c|c|c|c}\cdot & 1 & 1 & 1 & 1\\ \cdot & 1 & i & -i & -1\end{array}\right]\]
\[=\frac{1}{2}\left[\begin{array}{cc}-1 & \omega^{-k}-i\omega^{-2k}+i\omega^{-3k}-\omega^{-4k}\\\omega^{-k}+i\omega^{-2k}-i\omega^{-3k}-\omega^{-4k} & -1\end{array}\right]\]
\[\mathbf{P}_{0} = \begin{bmatrix}0&0\\0&0\end{bmatrix}\]
\[\mathbf{P}_{1} = \frac{1}{2}\begin{bmatrix}1&1\\1&1\end{bmatrix}\]
\[\mathbf{P}_{2} = \frac{1}{2}\begin{bmatrix}1&-i\\i&1\end{bmatrix}\]
\[\mathbf{P}_{3} = \frac{1}{2}\begin{bmatrix}1&i\\-i&1\end{bmatrix}\]
\[\mathbf{P}_{4} = \frac{1}{2}\begin{bmatrix}1&-1\\-1&1\end{bmatrix}\]
\[\mathbf{M}_{k} = \sum_{j=0}^{4}\omega^{-jk}\mathbf{P}_{j}\]
Indeed, \(\mathbf{M}_{k}^{\ast}\mathbf{M}_{k}^{} = \frac{3}{2}\boldsymbol{I}_{2}\) for each nonzero \(k\in\mathbb{Z}_{5}\).
Therefore \(\Phi\) is a EITFF\((4,5,2)\) (five two-dimensional subspaces of \(\mathbb{C}^{4}\))
Example 1 (redux). Let \(\mathcal{G} = \mathbb{Z}_{7} = \{0,1,2,3,4,5,6\}\) and \(\omega=\exp(2\pi i/7)\)
\[\Psi = \big[\begin{array}{c|c|c|c|c|c|c} \Psi_{0} & \Psi_{1} & \Psi_{2} & \Psi_{3} & \Psi_{4} & \Psi_{5} & \Psi_{6}\end{array}\big] = \big[\begin{array}{c|c|c|c|c|c|c} 1 & 1 & \cdot & 1 & \cdot & \cdot & \cdot\end{array}\big]\]
\[\mathbf{P}_{0} = \begin{bmatrix}1\end{bmatrix},\ \mathbf{P}_{1} = \begin{bmatrix}1\end{bmatrix},\ \mathbf{P}_{2} = \begin{bmatrix}0\end{bmatrix},\ \mathbf{P}_{3} = \begin{bmatrix}1\end{bmatrix},\ \mathbf{P}_{4} = \begin{bmatrix}0\end{bmatrix},\ \mathbf{P}_{5} = \begin{bmatrix}0\end{bmatrix},\ \mathbf{P}_{6} = \begin{bmatrix}0\end{bmatrix}\]
\[\mathbf{M}_{k} = \omega^{0k}+\omega^{-k}+\omega^{-3k}\]
\(\mathbf{M}_{k}\) has the same modulus for all \(k\neq 0\)
The Fourier transform of \([1\ \ 1\ \ 0\ \ 1\ \ 0\ \ 0\ \ 0]^{\top}\) is "spike+flat"
\(\{0,1,3\}\) is a difference set in \(\mathbb{Z}_{7}\).
\(\Updownarrow\)
\(\Updownarrow\)
Theorem. (Fickus, Iverson, J, Mixon '21) Write \(\mathbb{F}_{q}^{\times} = \langle \alpha\rangle = \{\alpha^{j}\}_{j=0}^{q-2}\)
Are there more harmonic EITFFs?
Can we combine these?