On a $mathbb C$-linear map from $M(p-1,mathbb C)$ to $mathbb C^hat G$, where $p$ is an odd prime and...
$begingroup$
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
user521337user521337
1,1881416
$endgroup$
|
show 9 more comments
$begingroup$
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
user521337user521337
1,1881416
$endgroup$
$begingroup$
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
|
show 9 more comments
$begingroup$
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
user521337user521337
1,1881416
$endgroup$
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
user521337user521337
1,1881416
asked Dec 11 '18 at 21:52
user521337user521337
1,1881416
edited Dec 13 '18 at 8:40
user521337
asked Dec 11 '18 at 21:52
user521337user521337
1,1881416
asked Dec 11 '18 at 21:52
user521337user521337
1,1881416
asked Dec 11 '18 at 21:52
user521337user521337
1,1881416
1,1881416
$begingroup$
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
|
show 9 more comments
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
@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
$begingroup$
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
$begingroup$
@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
$begingroup$
It implies $A (omega_1,ldots,omega^n)^top = 0$ not $A = 0$
$endgroup$
– reuns
Dec 11 '18 at 22:49
|
show 9 more comments
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3035882%2fon-a-mathbb-c-linear-map-from-mp-1-mathbb-c-to-mathbb-c-hat-g-where%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3035882%2fon-a-mathbb-c-linear-map-from-mp-1-mathbb-c-to-mathbb-c-hat-g-where%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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