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Number of irreducible representations of a finite group over a field of characteristic 0
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Let $G$ be a finite group and $K$ a field with $mathbbQ subseteq K subseteq mathbbC$.
For $K=mathbbC$ the number of irreducible representations of $KG$ is equal to the number of conjugacy classes of $G$.
For $K=mathbbR$ the number of irreducible representations of $KG$ is equal to $fracr+s2$, where $r$ denotes the number of conjugacy classes of $G$ and $s$ the number of classes stable under inversion.
For $K=mathbbQ$ 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 $mathbbQ$.
co.combinatorics rt.representation-theory finite-groups
asked 4 hours ago
Mare
3,26121030
add a comment |Â
up vote
15
down vote
favorite
Let $G$ be a finite group and $K$ a field with $mathbbQ subseteq K subseteq mathbbC$.
For $K=mathbbC$ the number of irreducible representations of $KG$ is equal to the number of conjugacy classes of $G$.
For $K=mathbbR$ the number of irreducible representations of $KG$ is equal to $fracr+s2$, where $r$ denotes the number of conjugacy classes of $G$ and $s$ the number of classes stable under inversion.
For $K=mathbbQ$ 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 $mathbbQ$.
co.combinatorics rt.representation-theory finite-groups
asked 4 hours ago
Mare
3,26121030
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 $mathbbQ$ and $mathbbR$. 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
1 hour ago
@JimHumphreys Thanks. I followed your suggestions, although I do not know the original articles where this was proven.
– Mare
6 mins ago
add a comment |Â
up vote
15
down vote
favorite
up vote
15
down vote
favorite
Let $G$ be a finite group and $K$ a field with $mathbbQ subseteq K subseteq mathbbC$.
For $K=mathbbC$ the number of irreducible representations of $KG$ is equal to the number of conjugacy classes of $G$.
For $K=mathbbR$ the number of irreducible representations of $KG$ is equal to $fracr+s2$, where $r$ denotes the number of conjugacy classes of $G$ and $s$ the number of classes stable under inversion.
For $K=mathbbQ$ 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 $mathbbQ$.
co.combinatorics rt.representation-theory finite-groups
asked 4 hours ago
Mare
3,26121030
Let $G$ be a finite group and $K$ a field with $mathbbQ subseteq K subseteq mathbbC$.
For $K=mathbbC$ the number of irreducible representations of $KG$ is equal to the number of conjugacy classes of $G$.
For $K=mathbbR$ the number of irreducible representations of $KG$ is equal to $fracr+s2$, where $r$ denotes the number of conjugacy classes of $G$ and $s$ the number of classes stable under inversion.
For $K=mathbbQ$ 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 $mathbbQ$.
co.combinatorics rt.representation-theory finite-groups
co.combinatorics rt.representation-theory finite-groups
asked 4 hours ago
Mare
3,26121030
asked 4 hours ago
Mare
3,26121030
edited 10 mins ago
asked 4 hours ago
Mare
3,26121030
asked 4 hours ago
Mare
3,26121030
asked 4 hours ago
Mare
3,26121030
3,26121030
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 $mathbbQ$ and $mathbbR$. 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
1 hour ago
@JimHumphreys Thanks. I followed your suggestions, although I do not know the original articles where this was proven.
– Mare
6 mins ago
add a comment |Â
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 $mathbbQ$ and $mathbbR$. 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
1 hour ago
@JimHumphreys Thanks. I followed your suggestions, although I do not know the original articles where this was proven.
– Mare
6 mins ago
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 $mathbbQ$ and $mathbbR$. 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
1 hour ago
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 $mathbbQ$ and $mathbbR$. 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
1 hour ago
@JimHumphreys Thanks. I followed your suggestions, although I do not know the original articles where this was proven.
– Mare
6 mins ago
@JimHumphreys Thanks. I followed your suggestions, although I do not know the original articles where this was proven.
– Mare
6 mins ago
add a comment |Â
1 Answer
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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.
answered 3 hours ago
Benjamin Steinberg
22.6k263123
2
Are there convenient references?
– Jim Humphreys
1 hour ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
15
down vote
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.
answered 3 hours ago
Benjamin Steinberg
22.6k263123
2
Are there convenient references?
– Jim Humphreys
1 hour ago
add a comment |Â
up vote
15
down vote
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.
answered 3 hours ago
Benjamin Steinberg
22.6k263123
2
Are there convenient references?
– Jim Humphreys
1 hour ago
add a comment |Â
up vote
15
down vote
up vote
15
down vote
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.
answered 3 hours ago
Benjamin Steinberg
22.6k263123
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.
answered 3 hours ago
Benjamin Steinberg
22.6k263123
answered 3 hours ago
Benjamin Steinberg
22.6k263123
answered 3 hours ago
Benjamin Steinberg
22.6k263123
answered 3 hours ago
Benjamin Steinberg
22.6k263123
22.6k263123
2
Are there convenient references?
– Jim Humphreys
1 hour ago
add a comment |Â
2
Are there convenient references?
– Jim Humphreys
1 hour ago
2
2
Are there convenient references?
– Jim Humphreys
1 hour ago
Are there convenient references?
– Jim Humphreys
1 hour ago
add a comment |Â
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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 $mathbbQ$ and $mathbbR$. 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
1 hour ago
@JimHumphreys Thanks. I followed your suggestions, although I do not know the original articles where this was proven.
– Mare
6 mins ago