Character determines representation in characteristic zero

Statement

Suppose $G$ is a finite group and $K$ is a field of characteristic zero. Then, the character of any finite-dimensional representation of $G$ over $K$ completely determines the representation, i.e., no two inequivalent finite-dimensional representations can have the same character.

Note that $K$ does not need to be a splitting field.

Related facts

Opposite facts

• Character does not determine representation in any prime characteristic: The problem is that we can construct representations whose character is identically zero simply by adding $p$ copies of an irreducible representation to itself.

Applications

• Equivalent linear representations of finite group over field are equivalent over subfield in characteristic zero

Facts used

1. Character orthogonality theorem
2. Maschke's averaging lemma
3. Orthogonal projection formula

Proof

Given: A group $G$, two linear representations $\rho _{1},\rho _{2}$ of $G$ with the same character $\chi$ over a field $K$ of characteristic zero.

To prove: $\rho _{1}$ and $\rho _{2}$ are equivalent as linear representations.

Proof: By Fact (2), both $\rho _{1}$ and $\rho _{2}$ are completely reducible, and are expressible as sums of irreducible representations. Suppose $\varphi _{1},\varphi _{2},\dots ,\varphi _{s}$ is a collection of distinct irreducible representations obtained as the union of all the representations occurring in a decomposition of $\rho _{1}$ into irreducible representations and a decomposition of $\rho _{2}$ into irreducible representations. In other words, there are nonnegative integers $a_{11},a_{12},\dots ,a_{1s},a_{21},a_{22},\dots ,a_{2s}$ such that:

$\rho _{1}\cong a_{11}\varphi _{1}\oplus a_{12}\varphi _{2}\oplus \dots \oplus a_{1s}\varphi _{s}$

and

$\rho _{2}\cong a_{21}\varphi _{1}\oplus a_{22}\varphi _{2}\oplus \dots \oplus a_{2s}\varphi _{s}$

Let $\chi _{i}$ denote the character of $\varphi _{i}$ and denote by $m_{i}$ the value $\langle \chi _{i},\chi _{i}\rangle _{G}$ (note: this would be 1 if $K$ were a splitting field, and in general it is the sum of squares of multiplicities of irreducible constituents over a splitting field).

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$m_{i}a_{1i}=\langle \chi ,\chi _{i}\rangle _{G}$ and $m_{1}a_{2i}=\langle \chi ,\chi _{i}\rangle _{G}$ for each $1\leq i\leq s$	Fact (3)			Direct application of fact
2	$a_{1i}={\frac {1}{m}}\langle \chi ,\chi _{i}\rangle _{G}$ and $a_{2i}={\frac {1}{m}}\langle \chi ,\chi _{i}\rangle _{G}$ for each $1\leq i\leq s$		$K$ has characteristic zero, so the manipulation makes sense	Step (1)
3	$a_{1i}=a_{2i}$ for each $1\leq i\leq s$		$K$ has characteristic zero	Step (2)	[SHOW MORE] Note that if $K$ had characteristic $p$, we could only conclude equality modulo $p$, and not equality as nonnegative integers.
4	$\rho _{1}$ and $\rho _{2}$ are equivalent			Step (3)