decompositions of a representation
$begingroup$
I am reading J.P Serre's book on Linear representations of finite groups. In chapter 2.6 it states:
Let $rho: G rightarrow GL(V)$ be a linear representation of $G$. We are going to define a direct sum decomposition of $V$ which is "coarser" than the decomposition into irreducible representations, but which has the advantage of being $unique$.
Now what does it mean by "coarser" in this context? I believe it means the former is less powerful than the latter, correct?
representation-theory
asked Jan 22 at 15:28
A.EA.E
1249
$endgroup$
add a comment |
$begingroup$
I am reading J.P Serre's book on Linear representations of finite groups. In chapter 2.6 it states:
Let $rho: G rightarrow GL(V)$ be a linear representation of $G$. We are going to define a direct sum decomposition of $V$ which is "coarser" than the decomposition into irreducible representations, but which has the advantage of being $unique$.
Now what does it mean by "coarser" in this context? I believe it means the former is less powerful than the latter, correct?
representation-theory
asked Jan 22 at 15:28
A.EA.E
1249
$endgroup$
add a comment |
$begingroup$
I am reading J.P Serre's book on Linear representations of finite groups. In chapter 2.6 it states:
Let $rho: G rightarrow GL(V)$ be a linear representation of $G$. We are going to define a direct sum decomposition of $V$ which is "coarser" than the decomposition into irreducible representations, but which has the advantage of being $unique$.
Now what does it mean by "coarser" in this context? I believe it means the former is less powerful than the latter, correct?
representation-theory
asked Jan 22 at 15:28
A.EA.E
1249
$endgroup$
I am reading J.P Serre's book on Linear representations of finite groups. In chapter 2.6 it states:
Let $rho: G rightarrow GL(V)$ be a linear representation of $G$. We are going to define a direct sum decomposition of $V$ which is "coarser" than the decomposition into irreducible representations, but which has the advantage of being $unique$.
Now what does it mean by "coarser" in this context? I believe it means the former is less powerful than the latter, correct?
representation-theory
representation-theory
asked Jan 22 at 15:28
A.EA.E
1249
asked Jan 22 at 15:28
A.EA.E
1249
asked Jan 22 at 15:28
A.EA.E
1249
asked Jan 22 at 15:28
A.EA.E
1249
asked Jan 22 at 15:28
A.EA.E
1249
1249
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It means that each piece of the former decomposition (that is, the decomposition into irreducible representations) will be part of some piece of the new decomposition.
answered Jan 22 at 15:32
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
$begingroup$
but this decomposition will not be unique, correct?
$endgroup$
– A.E
Jan 22 at 15:34
$begingroup$
I can't answer that question, since I don't know which is the specific decomposition defined by Serre.
$endgroup$
– José Carlos Santos
Jan 22 at 15:36
$begingroup$
ok. is it okay to say "less powerful" instead of "coarser"
$endgroup$
– A.E
Jan 22 at 15:38
$begingroup$
I'd think "less powerful" is a terrible term.
$endgroup$
– kimchi lover
Jan 22 at 15:56
$begingroup$
@kimchilover I agree.
$endgroup$
– José Carlos Santos
Jan 22 at 15:56
|
show 3 more comments
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%2f3083301%2fdecompositions-of-a-representation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It means that each piece of the former decomposition (that is, the decomposition into irreducible representations) will be part of some piece of the new decomposition.
answered Jan 22 at 15:32
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
$begingroup$
but this decomposition will not be unique, correct?
$endgroup$
– A.E
Jan 22 at 15:34
$begingroup$
I can't answer that question, since I don't know which is the specific decomposition defined by Serre.
$endgroup$
– José Carlos Santos
Jan 22 at 15:36
$begingroup$
ok. is it okay to say "less powerful" instead of "coarser"
$endgroup$
– A.E
Jan 22 at 15:38
$begingroup$
I'd think "less powerful" is a terrible term.
$endgroup$
– kimchi lover
Jan 22 at 15:56
$begingroup$
@kimchilover I agree.
$endgroup$
– José Carlos Santos
Jan 22 at 15:56
|
show 3 more comments
$begingroup$
It means that each piece of the former decomposition (that is, the decomposition into irreducible representations) will be part of some piece of the new decomposition.
answered Jan 22 at 15:32
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
$begingroup$
but this decomposition will not be unique, correct?
$endgroup$
– A.E
Jan 22 at 15:34
$begingroup$
I can't answer that question, since I don't know which is the specific decomposition defined by Serre.
$endgroup$
– José Carlos Santos
Jan 22 at 15:36
$begingroup$
ok. is it okay to say "less powerful" instead of "coarser"
$endgroup$
– A.E
Jan 22 at 15:38
$begingroup$
I'd think "less powerful" is a terrible term.
$endgroup$
– kimchi lover
Jan 22 at 15:56
$begingroup$
@kimchilover I agree.
$endgroup$
– José Carlos Santos
Jan 22 at 15:56
|
show 3 more comments
$begingroup$
It means that each piece of the former decomposition (that is, the decomposition into irreducible representations) will be part of some piece of the new decomposition.
answered Jan 22 at 15:32
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
It means that each piece of the former decomposition (that is, the decomposition into irreducible representations) will be part of some piece of the new decomposition.
answered Jan 22 at 15:32
José Carlos SantosJosé Carlos Santos
164k22131234
answered Jan 22 at 15:32
José Carlos SantosJosé Carlos Santos
164k22131234
answered Jan 22 at 15:32
José Carlos SantosJosé Carlos Santos
164k22131234
answered Jan 22 at 15:32
José Carlos SantosJosé Carlos Santos
164k22131234
164k22131234
$begingroup$
but this decomposition will not be unique, correct?
$endgroup$
– A.E
Jan 22 at 15:34
$begingroup$
I can't answer that question, since I don't know which is the specific decomposition defined by Serre.
$endgroup$
– José Carlos Santos
Jan 22 at 15:36
$begingroup$
ok. is it okay to say "less powerful" instead of "coarser"
$endgroup$
– A.E
Jan 22 at 15:38
$begingroup$
I'd think "less powerful" is a terrible term.
$endgroup$
– kimchi lover
Jan 22 at 15:56
$begingroup$
@kimchilover I agree.
$endgroup$
– José Carlos Santos
Jan 22 at 15:56
|
show 3 more comments
$begingroup$
but this decomposition will not be unique, correct?
$endgroup$
– A.E
Jan 22 at 15:34
$begingroup$
I can't answer that question, since I don't know which is the specific decomposition defined by Serre.
$endgroup$
– José Carlos Santos
Jan 22 at 15:36
$begingroup$
ok. is it okay to say "less powerful" instead of "coarser"
$endgroup$
– A.E
Jan 22 at 15:38
$begingroup$
I'd think "less powerful" is a terrible term.
$endgroup$
– kimchi lover
Jan 22 at 15:56
$begingroup$
@kimchilover I agree.
$endgroup$
– José Carlos Santos
Jan 22 at 15:56
$begingroup$
but this decomposition will not be unique, correct?
$endgroup$
– A.E
Jan 22 at 15:34
$begingroup$
but this decomposition will not be unique, correct?
$endgroup$
– A.E
Jan 22 at 15:34
$begingroup$
I can't answer that question, since I don't know which is the specific decomposition defined by Serre.
$endgroup$
– José Carlos Santos
Jan 22 at 15:36
$begingroup$
I can't answer that question, since I don't know which is the specific decomposition defined by Serre.
$endgroup$
– José Carlos Santos
Jan 22 at 15:36
$begingroup$
ok. is it okay to say "less powerful" instead of "coarser"
$endgroup$
– A.E
Jan 22 at 15:38
$begingroup$
ok. is it okay to say "less powerful" instead of "coarser"
$endgroup$
– A.E
Jan 22 at 15:38
$begingroup$
I'd think "less powerful" is a terrible term.
$endgroup$
– kimchi lover
Jan 22 at 15:56
$begingroup$
I'd think "less powerful" is a terrible term.
$endgroup$
– kimchi lover
Jan 22 at 15:56
$begingroup$
@kimchilover I agree.
$endgroup$
– José Carlos Santos
Jan 22 at 15:56
$begingroup$
@kimchilover I agree.
$endgroup$
– José Carlos Santos
Jan 22 at 15:56
|
show 3 more comments
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%2f3083301%2fdecompositions-of-a-representation%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
