On a $mathbb C$-linear map from $M(p-1,mathbb C)$ to $mathbb C^hat G$, where $p$ is an odd prime and...
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 $(f+g)(x):=f(x)+g(x),forall xin X$ ; $(f.g)(x)=f(x)g(x), forall xin X$, and $(k.f)(x):=kf(x),forall xin X$.
Let $n=p-1$, let $omega =e^{2pi i/p}$ 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 ?)
linear-algebra matrices number-theory characters gauss-sums
asked Dec 11 '18 at 21:52
user521337
9831315
|
show 9 more comments
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 $(f+g)(x):=f(x)+g(x),forall xin X$ ; $(f.g)(x)=f(x)g(x), forall xin X$, and $(k.f)(x):=kf(x),forall xin X$.
Let $n=p-1$, let $omega =e^{2pi i/p}$ 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 ?)
linear-algebra matrices number-theory characters gauss-sums
asked Dec 11 '18 at 21:52
user521337
9831315
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
|
show 9 more comments
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 $(f+g)(x):=f(x)+g(x),forall xin X$ ; $(f.g)(x)=f(x)g(x), forall xin X$, and $(k.f)(x):=kf(x),forall xin X$.
Let $n=p-1$, let $omega =e^{2pi i/p}$ 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 ?)
linear-algebra matrices number-theory characters gauss-sums
asked Dec 11 '18 at 21:52
user521337
9831315
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 $(f+g)(x):=f(x)+g(x),forall xin X$ ; $(f.g)(x)=f(x)g(x), forall xin X$, and $(k.f)(x):=kf(x),forall xin X$.
Let $n=p-1$, let $omega =e^{2pi i/p}$ 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 ?)
linear-algebra matrices number-theory characters gauss-sums
linear-algebra matrices number-theory characters gauss-sums
asked Dec 11 '18 at 21:52
user521337
9831315
asked Dec 11 '18 at 21:52
user521337
9831315
edited Dec 13 '18 at 8:40
asked Dec 11 '18 at 21:52
user521337
9831315
asked Dec 11 '18 at 21:52
user521337
9831315
asked Dec 11 '18 at 21:52
user521337
9831315
9831315
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
|
show 9 more comments
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
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
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
@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
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
@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
It implies $A (omega_1,ldots,omega^n)^top = 0$ not $A = 0$
– reuns
Dec 11 '18 at 22:49
|
show 9 more comments
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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