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Let $p$ be an odd prime and $G=(\mathbb Z/(p))^\times=\{1,2,...,p-1\}$ i.e. $G$ is a cyclic group of order $p-1$. Let $\hat G:=\{\chi:G \to \mathbb C^\times : \chi $ is a group homomorphism $\}$. For any set $X$, let $\mathbb C^X$ denote the set of all functions from $X$ to $\mathbb C$, and note that this can be given a usual $\mathbb C$-algebra structure as $(f+g)(x):=f(x)+g(x),\forall x\in X$ ; $(f.g)(x)=f(x)g(x), \forall x\in X$, and $(k.f)(x):=kf(x),\forall x\in X$.

Let $n=p-1$, let $\omega =e^{2\pi i/p}$ and define a function

$f:M(n,\mathbb C) \to \mathbb C^\hat G$ as $f(A)(\chi)=\begin{pmatrix} \chi(1) & ... & \chi(p-1) \end{pmatrix} A \begin{pmatrix} \omega \\ \omega^2 \\ .\\.\\. \\ \omega^n \end{pmatrix} , \forall A \in M(n,\mathbb C), \forall \chi \in \hat G$.

It easily follows that $f$ is a $\mathbb C$-linear function.

Moreover, $f(A)=0 \implies A \begin{pmatrix} \omega \\ \omega^2 \\ .\\.\\. \\ \omega^n \end{pmatrix}=0$. From this, it follows that since the minimal polynomial of $\omega $ over $\mathbb Q$ has degree $p-1=n$, so $A \in M(n, \mathbb Q)$ and $A \begin{pmatrix} \omega \\ \omega^2 \\ .\\.\\. \\ \omega^n \end{pmatrix}=0 \implies A=O$, thus $A \in M(n, \mathbb Q)$ and $f(A)=0 \implies A=O$.

Now my questions are the following :

(1) For every $A,B \in M(n, \mathbb Q)$, does there exist $C \in M(n, \mathbb Q)$ such that $f(A).f(B)=f(C)$ ? (Notice that such a $C$, if exists, must be unique)

(2) How to show that there exists Hermitian matrices $A_1,...,A_n$ of rank $1$ such that $f(I)=f(A_1)+...+f(A_n)$ and $f(A_j)f(A_k)=0, \forall j \ne k$ ? (may be this has something to do with Orthogonality of characters ?)

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  • $\begingroup$ So you meant $f(A)(\chi) =\chi(l) \sum_{m=1}^n \sum_{l=1}^n A_{l,m} \omega^m$. Then $A \mapsto f(A)(.)$ is not injective as it depends only on $A\ (\omega_1,\ldots,\omega^n)^\top$ $\endgroup$ – reuns Dec 11 '18 at 22:38
  • $\begingroup$ @reuns: $f(A)(\chi)=\sum_{l,m} \chi(l)A_{l,m} \omega^m$ $\endgroup$ – user521337 Dec 11 '18 at 22:43
  • $\begingroup$ Sure that was a typo. So what do you mean ? Letting the $p$-th root of unity vary it becomes injective as a function on pairs of multiplicative and additive character $\endgroup$ – reuns Dec 11 '18 at 22:46
  • $\begingroup$ @reuns: I mean to say that $f(A)(\chi)=0, \forall \chi \in \hat G \implies A=O$ ... at least that's what I believe and verified by brute force for $p=3$ ... $\endgroup$ – user521337 Dec 11 '18 at 22:48
  • $\begingroup$ It implies $A\ (\omega_1,\ldots,\omega^n)^\top = 0$ not $A = 0$ $\endgroup$ – reuns Dec 11 '18 at 22:49

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