Size-degree-weighted characters are algebraic integers

Statement

For an algebraically closed field of characteristic zero

Suppose $k$ is an algebraically closed field of characteristic zero, and $G$ is a finite group. Let $\rho$ be an Irreducible linear representation (?) of $G$ over $k$, and $\chi$ be the character corresponding to $\rho$. Let $c$ be a conjugacy class in $G$ and $g\in c$ be an element. Then:

${\frac {|c|\chi (g)}{\chi (1)}}$

is an algebraic integer.

Note that both aspects: irreducibility of the representation and the fact that we are working over an algebraically closed field of characteristic zero, are required. Getting rid of either condition makes the statement false.

For a splitting field

We can weaken the condition on $k$ somewhat: instead of requiring it to be an algebraically closed field of characteristic zero, it suffices to require $k$ to be a Splitting field (?) for $G$. In particular, any sufficiently large field for $G$ will do.

Related facts

• Characters are cyclotomic integers: This statement holds in much greater generality. In particular, it holds over any field and it does not require the linear representation to be irreducible.
• Characters are algebraic integers

Applications and further results

• Degree of irreducible representation divides group order
• Zero-or-scalar lemma
• Conjugacy class of prime power order implies not simple
• Order has only two prime factors implies solvable

Breakdown for a field that is not algebraically closed

Further information: cyclic group:Z3

Let $G$ be the cyclic group of order three and $\mathbb {R}$ be the field. $G$ has an irreducible two-dimensional linear representation over $\mathbb {R}$ given by rotation by multiples of $2\pi /3$. For a non-identity element $g$ of $G$, $\chi (g)=-1$ for the corresponding character, while $\chi (1)=2$. Thus, the expression works out to $-1/2$, which is not an algebraic integer.

Breakdown for a representation that is not irreducible

The same example as the above (the one for breakdown over a field that is not algebraically closed) works. Specifically, the irreducible representation over $\mathbb {R}$ can be viewed as a reducible representation over $\mathbb {C}$.

Proof

The proof is based on the idea of the convolution algebra on conjugacy classes.

Description of the convolution algebra on conjugacy classes

Let $C(G,\mathbb {Z} )$ be a $\mathbb {Z}$-subalgebra of the group ring $\mathbb {Z} (G)$ defined as follows: as a group, it is the free Abelian group on all indicator class functions for conjugacy classes. In other words, for each conjugacy class, we have a free generator that corresponds to the sum of elements of that conjugacy class.

The structure constant for multiplication of elements of $C(G,\mathbb {Z} )$ is defined as follows: given conjugacy classes $c_{1},c_{2},c_{3}$, the coefficient of the $c_{3}$-indicator function in the product of the $c_{1}$-indicator function and the $c_{2}$-indicator function is the number of ways of writing $g_{1}g_{2}=a$ where $g_{i}\in c_{i}$, and $a$ a fixed element of $c_{3}$.

Note that all the structure constants are integers.

A homomorphism from this convolution algebra to the matrix ring

The representation $\rho$ gives rise to a homomorphism from $C(G,\mathbb {Z} )$ to the matrix ring $M_{n}(k)$. The indicator function for a conjugacy class $c$ goes to the matrix given by:

$\sum _{g\in c}\rho (g)$.

This sum commutes with $\rho (h)$ for all $h$, and thus, by Schur's lemma, the sum is a scalar matrix. The trace of the sum is $|c|\chi (g)$, so the sum must be a scalar matrix with scalar entry:

${\frac {|c|\chi (g)}{\chi (1)}}$.

Thus, the set of scalar matrices with entries described as above additively generate a group that is a ring under multiplication. The structure constants for this ring are the same as the structure constants for the convolution algebra. A result from algebraic number theory now tells us that this forces the entire ring to be a ring of algebraic integers, and in particular, the generating elements are algebraic integers.