# On a $\mathbb C$-linear map from $M(p-1,\mathbb C)$ to $\mathbb C^\hat G$, where $p$ is an odd prime and $G=\mathbb Z/(p) ^\times$

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 ?)

• 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 $A\ (\omega_1,\ldots,\omega^n)^\top$ – reuns Dec 11 '18 at 22:38
• @reuns: $f(A)(\chi)=\sum_{l,m} \chi(l)A_{l,m} \omega^m$ – user521337 Dec 11 '18 at 22:43
• 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 – reuns Dec 11 '18 at 22:46
• @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 $p=3$ ... – user521337 Dec 11 '18 at 22:48
• It implies $A\ (\omega_1,\ldots,\omega^n)^\top = 0$ not $A = 0$ – reuns Dec 11 '18 at 22:49