Number of irreducible representations of a finite group over a field of characteristic 0











up vote
27
down vote

favorite
9












Let $G$ be a finite group and $K$ a field with $mathbb{Q} subseteq K subseteq mathbb{C}$.



For $K=mathbb{C}$ the number of irreducible representations of $KG$ is equal to the number of conjugacy classes of $G$.



For $K=mathbb{R}$ the number of irreducible representations of $KG$ is equal to $frac{r+s}{2}$, where $r$ denotes the number of conjugacy classes of $G$ and $s$ the number of classes stable under inversion.



For $K=mathbb{Q}$ the number of irreducible representations of $KG$ is equal to the number of conjugacy classes of cyclic subgroups of $G$.



(You can find quick proofs of those results in the very recent book "A Journey Through Representation Theory: From Finite Groups to Quivers via Algebras" by Caroline Gruson and Vera Serganova, where a nice quick overview of representation theory of finite groups in characteristic 0 is given in chapters 1 and 2.)



Question:




Are there such nice closed forumlas for other fields $K$? For example quadratic, cubic or cyclotomic field extensions of $mathbb{Q}$.







co.combinatorics rt.representation-theory finite-groups







share|cite|improve this question




















  • 1




    To clarify your header, I'd suggest making it longer: "Number of irreducible representations of a finite group over a field of characteristic 0". It would also help to give explicit references for $mathbb{Q}$ and $mathbb{R}$. All of this depends on Richard Brauer's results over a splliting field. See also Chapter 9 of the classic book by I.M. Isaacs Character Theory of Finite Groups. (And as B. Steinberg's answer suggests, the results for a field of characteristic $p>0$ are also fairly explicit.)
    – Jim Humphreys
    Nov 4 at 14:30












  • @JimHumphreys Thanks. I followed your suggestions, although I do not know the original articles where this was proven.
    – Mare
    Nov 4 at 15:49















up vote
27
down vote

favorite
9












Let $G$ be a finite group and $K$ a field with $mathbb{Q} subseteq K subseteq mathbb{C}$.



For $K=mathbb{C}$ the number of irreducible representations of $KG$ is equal to the number of conjugacy classes of $G$.



For $K=mathbb{R}$ the number of irreducible representations of $KG$ is equal to $frac{r+s}{2}$, where $r$ denotes the number of conjugacy classes of $G$ and $s$ the number of classes stable under inversion.



For $K=mathbb{Q}$ the number of irreducible representations of $KG$ is equal to the number of conjugacy classes of cyclic subgroups of $G$.



(You can find quick proofs of those results in the very recent book "A Journey Through Representation Theory: From Finite Groups to Quivers via Algebras" by Caroline Gruson and Vera Serganova, where a nice quick overview of representation theory of finite groups in characteristic 0 is given in chapters 1 and 2.)



Question:




Are there such nice closed forumlas for other fields $K$? For example quadratic, cubic or cyclotomic field extensions of $mathbb{Q}$.







co.combinatorics rt.representation-theory finite-groups







share|cite|improve this question




















  • 1




    To clarify your header, I'd suggest making it longer: "Number of irreducible representations of a finite group over a field of characteristic 0". It would also help to give explicit references for $mathbb{Q}$ and $mathbb{R}$. All of this depends on Richard Brauer's results over a splliting field. See also Chapter 9 of the classic book by I.M. Isaacs Character Theory of Finite Groups. (And as B. Steinberg's answer suggests, the results for a field of characteristic $p>0$ are also fairly explicit.)
    – Jim Humphreys
    Nov 4 at 14:30












  • @JimHumphreys Thanks. I followed your suggestions, although I do not know the original articles where this was proven.
    – Mare
    Nov 4 at 15:49













up vote
27
down vote

favorite
9









up vote
27
down vote

favorite
9






9





Let $G$ be a finite group and $K$ a field with $mathbb{Q} subseteq K subseteq mathbb{C}$.



For $K=mathbb{C}$ the number of irreducible representations of $KG$ is equal to the number of conjugacy classes of $G$.



For $K=mathbb{R}$ the number of irreducible representations of $KG$ is equal to $frac{r+s}{2}$, where $r$ denotes the number of conjugacy classes of $G$ and $s$ the number of classes stable under inversion.



For $K=mathbb{Q}$ the number of irreducible representations of $KG$ is equal to the number of conjugacy classes of cyclic subgroups of $G$.



(You can find quick proofs of those results in the very recent book "A Journey Through Representation Theory: From Finite Groups to Quivers via Algebras" by Caroline Gruson and Vera Serganova, where a nice quick overview of representation theory of finite groups in characteristic 0 is given in chapters 1 and 2.)



Question:




Are there such nice closed forumlas for other fields $K$? For example quadratic, cubic or cyclotomic field extensions of $mathbb{Q}$.







co.combinatorics rt.representation-theory finite-groups







share|cite|improve this question















Let $G$ be a finite group and $K$ a field with $mathbb{Q} subseteq K subseteq mathbb{C}$.



For $K=mathbb{C}$ the number of irreducible representations of $KG$ is equal to the number of conjugacy classes of $G$.



For $K=mathbb{R}$ the number of irreducible representations of $KG$ is equal to $frac{r+s}{2}$, where $r$ denotes the number of conjugacy classes of $G$ and $s$ the number of classes stable under inversion.



For $K=mathbb{Q}$ the number of irreducible representations of $KG$ is equal to the number of conjugacy classes of cyclic subgroups of $G$.



(You can find quick proofs of those results in the very recent book "A Journey Through Representation Theory: From Finite Groups to Quivers via Algebras" by Caroline Gruson and Vera Serganova, where a nice quick overview of representation theory of finite groups in characteristic 0 is given in chapters 1 and 2.)



Question:




Are there such nice closed forumlas for other fields $K$? For example quadratic, cubic or cyclotomic field extensions of $mathbb{Q}$.







co.combinatorics rt.representation-theory finite-groups




co.combinatorics rt.representation-theory finite-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 4 at 15:45

























asked Nov 4 at 11:43









Mare

3,39321131




3,39321131








  • 1




    To clarify your header, I'd suggest making it longer: "Number of irreducible representations of a finite group over a field of characteristic 0". It would also help to give explicit references for $mathbb{Q}$ and $mathbb{R}$. All of this depends on Richard Brauer's results over a splliting field. See also Chapter 9 of the classic book by I.M. Isaacs Character Theory of Finite Groups. (And as B. Steinberg's answer suggests, the results for a field of characteristic $p>0$ are also fairly explicit.)
    – Jim Humphreys
    Nov 4 at 14:30












  • @JimHumphreys Thanks. I followed your suggestions, although I do not know the original articles where this was proven.
    – Mare
    Nov 4 at 15:49














  • 1




    To clarify your header, I'd suggest making it longer: "Number of irreducible representations of a finite group over a field of characteristic 0". It would also help to give explicit references for $mathbb{Q}$ and $mathbb{R}$. All of this depends on Richard Brauer's results over a splliting field. See also Chapter 9 of the classic book by I.M. Isaacs Character Theory of Finite Groups. (And as B. Steinberg's answer suggests, the results for a field of characteristic $p>0$ are also fairly explicit.)
    – Jim Humphreys
    Nov 4 at 14:30












  • @JimHumphreys Thanks. I followed your suggestions, although I do not know the original articles where this was proven.
    – Mare
    Nov 4 at 15:49








1




1




To clarify your header, I'd suggest making it longer: "Number of irreducible representations of a finite group over a field of characteristic 0". It would also help to give explicit references for $mathbb{Q}$ and $mathbb{R}$. All of this depends on Richard Brauer's results over a splliting field. See also Chapter 9 of the classic book by I.M. Isaacs Character Theory of Finite Groups. (And as B. Steinberg's answer suggests, the results for a field of characteristic $p>0$ are also fairly explicit.)
– Jim Humphreys
Nov 4 at 14:30






To clarify your header, I'd suggest making it longer: "Number of irreducible representations of a finite group over a field of characteristic 0". It would also help to give explicit references for $mathbb{Q}$ and $mathbb{R}$. All of this depends on Richard Brauer's results over a splliting field. See also Chapter 9 of the classic book by I.M. Isaacs Character Theory of Finite Groups. (And as B. Steinberg's answer suggests, the results for a field of characteristic $p>0$ are also fairly explicit.)
– Jim Humphreys
Nov 4 at 14:30














@JimHumphreys Thanks. I followed your suggestions, although I do not know the original articles where this was proven.
– Mare
Nov 4 at 15:49




@JimHumphreys Thanks. I followed your suggestions, although I do not know the original articles where this was proven.
– Mare
Nov 4 at 15:49










1 Answer
1






active

oldest

votes

















up vote
26
down vote



accepted










There is a characterization due to Berman for any field. In your case let $n$ be the least common multiple of the orders of elements of $G$. Let $zeta$ be a primitive $n^{th}$ root of unity. Let $H=Gal(K(zeta)/K)$. We can identify $H$ with a subgroup of $(mathbb Z/nmathbb Z)^*$. Call $a,bin G$ $K$-conjugate if $b^j=gag^{-1}$ for some $gin G$ and $jin H$. This is an equivalence relation and the number of irreducible $KG$-modules is the number of $K$-conjugacy classes.



In characteristic $p$ you do the analogous thing but only look at this equivalence relation on $p$-regular elements.






share|cite|improve this answer

















  • 3




    Are there convenient references?
    – Jim Humphreys
    Nov 4 at 14:38






  • 1




    The characteristic 0 case is in Curtis and Reiner. Reiner has a paper maybe in proceedings of the ams doing the modular case with Brauer characters.
    – Benjamin Steinberg
    Nov 4 at 17:40










  • Im traveling now so don't have the books handy.
    – Benjamin Steinberg
    Nov 4 at 17:43










  • In particular, this says that every irrep can be defined over $mathbb{Q}(zeta)$, which is an important consequence (maybe ~equivalent to?) Brauer's induction theorem.
    – Kevin Casto
    Nov 4 at 17:58










  • I believe the induction theorem is used in the proof.
    – Benjamin Steinberg
    Nov 4 at 18:13











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: "504"
};
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',
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
});


}
});














 

draft saved


draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f314532%2fnumber-of-irreducible-representations-of-a-finite-group-over-a-field-of-characte%23new-answer', 'question_page');
}
);

Post as a guest


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
26
down vote



accepted










There is a characterization due to Berman for any field. In your case let $n$ be the least common multiple of the orders of elements of $G$. Let $zeta$ be a primitive $n^{th}$ root of unity. Let $H=Gal(K(zeta)/K)$. We can identify $H$ with a subgroup of $(mathbb Z/nmathbb Z)^*$. Call $a,bin G$ $K$-conjugate if $b^j=gag^{-1}$ for some $gin G$ and $jin H$. This is an equivalence relation and the number of irreducible $KG$-modules is the number of $K$-conjugacy classes.



In characteristic $p$ you do the analogous thing but only look at this equivalence relation on $p$-regular elements.






share|cite|improve this answer

















  • 3




    Are there convenient references?
    – Jim Humphreys
    Nov 4 at 14:38






  • 1




    The characteristic 0 case is in Curtis and Reiner. Reiner has a paper maybe in proceedings of the ams doing the modular case with Brauer characters.
    – Benjamin Steinberg
    Nov 4 at 17:40










  • Im traveling now so don't have the books handy.
    – Benjamin Steinberg
    Nov 4 at 17:43










  • In particular, this says that every irrep can be defined over $mathbb{Q}(zeta)$, which is an important consequence (maybe ~equivalent to?) Brauer's induction theorem.
    – Kevin Casto
    Nov 4 at 17:58










  • I believe the induction theorem is used in the proof.
    – Benjamin Steinberg
    Nov 4 at 18:13















up vote
26
down vote



accepted










There is a characterization due to Berman for any field. In your case let $n$ be the least common multiple of the orders of elements of $G$. Let $zeta$ be a primitive $n^{th}$ root of unity. Let $H=Gal(K(zeta)/K)$. We can identify $H$ with a subgroup of $(mathbb Z/nmathbb Z)^*$. Call $a,bin G$ $K$-conjugate if $b^j=gag^{-1}$ for some $gin G$ and $jin H$. This is an equivalence relation and the number of irreducible $KG$-modules is the number of $K$-conjugacy classes.



In characteristic $p$ you do the analogous thing but only look at this equivalence relation on $p$-regular elements.






share|cite|improve this answer

















  • 3




    Are there convenient references?
    – Jim Humphreys
    Nov 4 at 14:38






  • 1




    The characteristic 0 case is in Curtis and Reiner. Reiner has a paper maybe in proceedings of the ams doing the modular case with Brauer characters.
    – Benjamin Steinberg
    Nov 4 at 17:40










  • Im traveling now so don't have the books handy.
    – Benjamin Steinberg
    Nov 4 at 17:43










  • In particular, this says that every irrep can be defined over $mathbb{Q}(zeta)$, which is an important consequence (maybe ~equivalent to?) Brauer's induction theorem.
    – Kevin Casto
    Nov 4 at 17:58










  • I believe the induction theorem is used in the proof.
    – Benjamin Steinberg
    Nov 4 at 18:13













up vote
26
down vote



accepted







up vote
26
down vote



accepted






There is a characterization due to Berman for any field. In your case let $n$ be the least common multiple of the orders of elements of $G$. Let $zeta$ be a primitive $n^{th}$ root of unity. Let $H=Gal(K(zeta)/K)$. We can identify $H$ with a subgroup of $(mathbb Z/nmathbb Z)^*$. Call $a,bin G$ $K$-conjugate if $b^j=gag^{-1}$ for some $gin G$ and $jin H$. This is an equivalence relation and the number of irreducible $KG$-modules is the number of $K$-conjugacy classes.



In characteristic $p$ you do the analogous thing but only look at this equivalence relation on $p$-regular elements.






share|cite|improve this answer












There is a characterization due to Berman for any field. In your case let $n$ be the least common multiple of the orders of elements of $G$. Let $zeta$ be a primitive $n^{th}$ root of unity. Let $H=Gal(K(zeta)/K)$. We can identify $H$ with a subgroup of $(mathbb Z/nmathbb Z)^*$. Call $a,bin G$ $K$-conjugate if $b^j=gag^{-1}$ for some $gin G$ and $jin H$. This is an equivalence relation and the number of irreducible $KG$-modules is the number of $K$-conjugacy classes.



In characteristic $p$ you do the analogous thing but only look at this equivalence relation on $p$-regular elements.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 4 at 12:36









Benjamin Steinberg

22.8k265124




22.8k265124








  • 3




    Are there convenient references?
    – Jim Humphreys
    Nov 4 at 14:38






  • 1




    The characteristic 0 case is in Curtis and Reiner. Reiner has a paper maybe in proceedings of the ams doing the modular case with Brauer characters.
    – Benjamin Steinberg
    Nov 4 at 17:40










  • Im traveling now so don't have the books handy.
    – Benjamin Steinberg
    Nov 4 at 17:43










  • In particular, this says that every irrep can be defined over $mathbb{Q}(zeta)$, which is an important consequence (maybe ~equivalent to?) Brauer's induction theorem.
    – Kevin Casto
    Nov 4 at 17:58










  • I believe the induction theorem is used in the proof.
    – Benjamin Steinberg
    Nov 4 at 18:13














  • 3




    Are there convenient references?
    – Jim Humphreys
    Nov 4 at 14:38






  • 1




    The characteristic 0 case is in Curtis and Reiner. Reiner has a paper maybe in proceedings of the ams doing the modular case with Brauer characters.
    – Benjamin Steinberg
    Nov 4 at 17:40










  • Im traveling now so don't have the books handy.
    – Benjamin Steinberg
    Nov 4 at 17:43










  • In particular, this says that every irrep can be defined over $mathbb{Q}(zeta)$, which is an important consequence (maybe ~equivalent to?) Brauer's induction theorem.
    – Kevin Casto
    Nov 4 at 17:58










  • I believe the induction theorem is used in the proof.
    – Benjamin Steinberg
    Nov 4 at 18:13








3




3




Are there convenient references?
– Jim Humphreys
Nov 4 at 14:38




Are there convenient references?
– Jim Humphreys
Nov 4 at 14:38




1




1




The characteristic 0 case is in Curtis and Reiner. Reiner has a paper maybe in proceedings of the ams doing the modular case with Brauer characters.
– Benjamin Steinberg
Nov 4 at 17:40




The characteristic 0 case is in Curtis and Reiner. Reiner has a paper maybe in proceedings of the ams doing the modular case with Brauer characters.
– Benjamin Steinberg
Nov 4 at 17:40












Im traveling now so don't have the books handy.
– Benjamin Steinberg
Nov 4 at 17:43




Im traveling now so don't have the books handy.
– Benjamin Steinberg
Nov 4 at 17:43












In particular, this says that every irrep can be defined over $mathbb{Q}(zeta)$, which is an important consequence (maybe ~equivalent to?) Brauer's induction theorem.
– Kevin Casto
Nov 4 at 17:58




In particular, this says that every irrep can be defined over $mathbb{Q}(zeta)$, which is an important consequence (maybe ~equivalent to?) Brauer's induction theorem.
– Kevin Casto
Nov 4 at 17:58












I believe the induction theorem is used in the proof.
– Benjamin Steinberg
Nov 4 at 18:13




I believe the induction theorem is used in the proof.
– Benjamin Steinberg
Nov 4 at 18:13


















 

draft saved


draft discarded



















































 


draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f314532%2fnumber-of-irreducible-representations-of-a-finite-group-over-a-field-of-characte%23new-answer', 'question_page');
}
);

Post as a guest






































































-