Mixing
Sum of characters of a non-trivial representation
I want to prove the following statement.
begin{align}
sum_{g in G}chi_{rho}(g)=begin{cases}
|G|quad hbox{if $rho$ is the trivial representation}\
0 hspace{6mm} hbox{if $rho$ is non-trivial}
end{cases}
end{align}
I understand the proof here that reindexes the sum:
The sum of character of finite groups
The problem I have with this is that they define character as a function $chi: G rightarrow mathbb{C}^*$, rather than $chi: G rightarrow mathbb{C}$. If we assume that $chi$ is non-trivial, then it must be that for some $g in G$, we have $chi(g)neq 1$. What if $chi(g)=0$? Clearly the linked argument no longer works.
proof-verification finite-groups representation-theory
asked Nov 20 '18 at 18:02
Bartolo Colon
497
|
show 4 more comments
I want to prove the following statement.
begin{align}
sum_{g in G}chi_{rho}(g)=begin{cases}
|G|quad hbox{if $rho$ is the trivial representation}\
0 hspace{6mm} hbox{if $rho$ is non-trivial}
end{cases}
end{align}
I understand the proof here that reindexes the sum:
The sum of character of finite groups
The problem I have with this is that they define character as a function $chi: G rightarrow mathbb{C}^*$, rather than $chi: G rightarrow mathbb{C}$. If we assume that $chi$ is non-trivial, then it must be that for some $g in G$, we have $chi(g)neq 1$. What if $chi(g)=0$? Clearly the linked argument no longer works.
proof-verification finite-groups representation-theory
asked Nov 20 '18 at 18:02
Bartolo Colon
497
We can't have it because $chi(g)chi(g^{-1}) = 1$
– Nicolas Hemelsoet
Nov 20 '18 at 18:09
I'm not sure that I follow. What is your "it" referring to?
– Bartolo Colon
Nov 20 '18 at 18:15
We can't have $chi(g) = 0$ because $chi(g)chi(g^{-1}) = 1$.
– Nicolas Hemelsoet
Nov 20 '18 at 18:16
1
But isn't this one of the Schur orthogonality relations?
– Tobias Kildetoft
Nov 20 '18 at 19:56
1
"Character of a group" means two different things: one of the meanings is the character of a linear representation, and the other one is the character of a one-dimensional linear representation. That's where the definition in terms of $mathbb{C}^{times}$ comes from.
– Qiaochu Yuan
Nov 21 '18 at 0:45
|
show 4 more comments
I want to prove the following statement.
begin{align}
sum_{g in G}chi_{rho}(g)=begin{cases}
|G|quad hbox{if $rho$ is the trivial representation}\
0 hspace{6mm} hbox{if $rho$ is non-trivial}
end{cases}
end{align}
I understand the proof here that reindexes the sum:
The sum of character of finite groups
The problem I have with this is that they define character as a function $chi: G rightarrow mathbb{C}^*$, rather than $chi: G rightarrow mathbb{C}$. If we assume that $chi$ is non-trivial, then it must be that for some $g in G$, we have $chi(g)neq 1$. What if $chi(g)=0$? Clearly the linked argument no longer works.
proof-verification finite-groups representation-theory
asked Nov 20 '18 at 18:02
Bartolo Colon
497
I want to prove the following statement.
begin{align}
sum_{g in G}chi_{rho}(g)=begin{cases}
|G|quad hbox{if $rho$ is the trivial representation}\
0 hspace{6mm} hbox{if $rho$ is non-trivial}
end{cases}
end{align}
I understand the proof here that reindexes the sum:
The sum of character of finite groups
The problem I have with this is that they define character as a function $chi: G rightarrow mathbb{C}^*$, rather than $chi: G rightarrow mathbb{C}$. If we assume that $chi$ is non-trivial, then it must be that for some $g in G$, we have $chi(g)neq 1$. What if $chi(g)=0$? Clearly the linked argument no longer works.
proof-verification finite-groups representation-theory
proof-verification finite-groups representation-theory
asked Nov 20 '18 at 18:02
Bartolo Colon
497
asked Nov 20 '18 at 18:02
Bartolo Colon
497
asked Nov 20 '18 at 18:02
Bartolo Colon
497
asked Nov 20 '18 at 18:02
Bartolo Colon
497
asked Nov 20 '18 at 18:02
Bartolo Colon
497
497
We can't have it because $chi(g)chi(g^{-1}) = 1$
– Nicolas Hemelsoet
Nov 20 '18 at 18:09
I'm not sure that I follow. What is your "it" referring to?
– Bartolo Colon
Nov 20 '18 at 18:15
We can't have $chi(g) = 0$ because $chi(g)chi(g^{-1}) = 1$.
– Nicolas Hemelsoet
Nov 20 '18 at 18:16
1
But isn't this one of the Schur orthogonality relations?
– Tobias Kildetoft
Nov 20 '18 at 19:56
1
"Character of a group" means two different things: one of the meanings is the character of a linear representation, and the other one is the character of a one-dimensional linear representation. That's where the definition in terms of $mathbb{C}^{times}$ comes from.
– Qiaochu Yuan
Nov 21 '18 at 0:45
|
show 4 more comments
We can't have it because $chi(g)chi(g^{-1}) = 1$
– Nicolas Hemelsoet
Nov 20 '18 at 18:09
I'm not sure that I follow. What is your "it" referring to?
– Bartolo Colon
Nov 20 '18 at 18:15
We can't have $chi(g) = 0$ because $chi(g)chi(g^{-1}) = 1$.
– Nicolas Hemelsoet
Nov 20 '18 at 18:16
1
But isn't this one of the Schur orthogonality relations?
– Tobias Kildetoft
Nov 20 '18 at 19:56
1
"Character of a group" means two different things: one of the meanings is the character of a linear representation, and the other one is the character of a one-dimensional linear representation. That's where the definition in terms of $mathbb{C}^{times}$ comes from.
– Qiaochu Yuan
Nov 21 '18 at 0:45
We can't have it because $chi(g)chi(g^{-1}) = 1$
– Nicolas Hemelsoet
Nov 20 '18 at 18:09
We can't have it because $chi(g)chi(g^{-1}) = 1$
– Nicolas Hemelsoet
Nov 20 '18 at 18:09
I'm not sure that I follow. What is your "it" referring to?
– Bartolo Colon
Nov 20 '18 at 18:15
I'm not sure that I follow. What is your "it" referring to?
– Bartolo Colon
Nov 20 '18 at 18:15
We can't have $chi(g) = 0$ because $chi(g)chi(g^{-1}) = 1$.
– Nicolas Hemelsoet
Nov 20 '18 at 18:16
We can't have $chi(g) = 0$ because $chi(g)chi(g^{-1}) = 1$.
– Nicolas Hemelsoet
Nov 20 '18 at 18:16
1
1
But isn't this one of the Schur orthogonality relations?
– Tobias Kildetoft
Nov 20 '18 at 19:56
But isn't this one of the Schur orthogonality relations?
– Tobias Kildetoft
Nov 20 '18 at 19:56
1
1
"Character of a group" means two different things: one of the meanings is the character of a linear representation, and the other one is the character of a one-dimensional linear representation. That's where the definition in terms of $mathbb{C}^{times}$ comes from.
– Qiaochu Yuan
Nov 21 '18 at 0:45
"Character of a group" means two different things: one of the meanings is the character of a linear representation, and the other one is the character of a one-dimensional linear representation. That's where the definition in terms of $mathbb{C}^{times}$ comes from.
– Qiaochu Yuan
Nov 21 '18 at 0:45
|
show 4 more comments
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We can't have it because $chi(g)chi(g^{-1}) = 1$
– Nicolas Hemelsoet
Nov 20 '18 at 18:09
I'm not sure that I follow. What is your "it" referring to?
– Bartolo Colon
Nov 20 '18 at 18:15
We can't have $chi(g) = 0$ because $chi(g)chi(g^{-1}) = 1$.
– Nicolas Hemelsoet
Nov 20 '18 at 18:16
1
But isn't this one of the Schur orthogonality relations?
– Tobias Kildetoft
Nov 20 '18 at 19:56
1
"Character of a group" means two different things: one of the meanings is the character of a linear representation, and the other one is the character of a one-dimensional linear representation. That's where the definition in terms of $mathbb{C}^{times}$ comes from.
– Qiaochu Yuan
Nov 21 '18 at 0:45