Character determines representation in characteristic zero

From Groupprops

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

Applications

Facts used

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

Proof

Given: A group G, two linear representations ρ1,ρ2 of G with the same character χ over a field K of characteristic zero.

To prove: ρ1 and ρ2 are equivalent as linear representations.

Proof: By Fact (2), both ρ1 and ρ2 are completely reducible, and are expressible as sums of irreducible representations. Suppose φ1,φ2,,φs is a collection of distinct irreducible representations obtained as the union of all the representations occurring in a decomposition of ρ1 into irreducible representations and a decomposition of ρ2 into irreducible representations. In other words, there are nonnegative integers a11,a12,,a1s,a21,a22,,a2s such that:

ρ1a11φ1a12φ2a1sφs

and

ρ2a21φ1a22φ2a2sφs

Let χi denote the character of φi and denote by mi the value χi,χiG (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 mia1i=χ,χiG and m1a2i=χ,χiG for each 1is Fact (3) Direct application of fact
2 a1i=1mχ,χiG and a2i=1mχ,χiG for each 1is K has characteristic zero, so the manipulation makes sense Step (1)
3 a1i=a2i for each 1is K has characteristic zero [SHOW MORE]
4 ρ1 and ρ2 are equivalent Step (3)