Character determines representation in characteristic zero

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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 \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_ia_{1i} = \langle \chi, \chi_i \rangle_G and m_1a_{2i} = \langle \chi, \chi_i \rangle_G for each 1 \le i \le 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 \le i \le s K has characteristic zero, so the manipulation makes sense Step (1)
3 a_{1i} = a_{2i} for each 1 \le i \le s K has characteristic zero Step (2) [SHOW MORE]
4 \rho_1 and \rho_2 are equivalent Step (3)