Context: A complex Hadamard matrix is a matrix with orthogonal rows, whose entries are all on the complex unit circle. We say that $H$ is in the $p$th Butson class if $[H_{ij}]^{p} = 1$ for all $i, j$.
In the case where $p = 2$, we call $H$ a real Hadamard matrix. For any $H$ in the $p$th Butson class (with $p$ a prime), we can attach a graph in a nice way. Let $\zeta_p$ be a $p$th root of unity; the adjacency matrix of this graph can be expressed in terms of the matrices $H_{1}, H_{\zeta}, H_{\zeta^2}, ..., H_{\zeta^{p - 1}}$, where $H_{\zeta^{k}}$ is a matrix with entries in $\{0, 1\}$ s.t. $[H_{\zeta^{k}}]_{ij} = 1 \iff H_{ij} = \zeta^k$. This is an indicator matrix, looking for $\zeta^k$ in $H$. Now, for $p = 2$, the real Hadamard case, it turns out to be really useful that we can write $H_{\pm 1}$ in terms of $J$, the all ones matrix; in particular, $H_{1} = \frac{1}{2}(J + H)$ and $H_{-1} = \frac{1}{2}(J - H)$. This does a lot of good for us computationally.
Question: For $p > 2$ (and maybe even just $p = 3$), is there a way to write $H_{\zeta^k} = f(J, H)$ where $f \in \mathbb{C}[x, y]$ is a noncommuting polynomial in two variables? What if we also allow the conjugation/transpose/conjugate transpose of $H$? If not, is there any "nice" way to write these $H_{\zeta^k}$ in terms of $J$, $H$, and any other matrices? I will accept an answer for just $p = 3$.
My thoughts: For $p = 2$, note that we exploited that $1$ vanishes at $x + 1$ and $-1$ vanishes at $x - 1$; I thought there might be a way to exploit that all our cube roots of one vanish at $1 + x + x^2$. However, taking $J + \omega H + \omega^2 H$, or $J + \omega J + H$, or any variant thereof seems to only kill one of the three roots. For the classic example of a Hadamard matrix in the 3rd Butson class, there's the matrix
$$\begin{pmatrix}1 & 1 & 1 \\ 1 & \omega & \omega^2 \\ 1 & \omega^2 & \omega \end{pmatrix}$$
When you square it, you get the matrix $3\begin{pmatrix}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}$, which gets us ones where the $\omega^2$ lives, but there's also a $1$ in the top right. If we could kill that, we could use this to selectively kill $\omega^2$. I'm, however, unconvinced that you can kill just the top right, and I'm unconvinced that this generalizes nicely.
I've tagged this "Fourier Analysis" since the smallest example of such a matrix (and lots more matrices) arise from Fourier coupling. I've also tagged "reference request", since Hadamard matrices are well studied, and any paper around this