Character value need not determine similarity class of image under irreducible representation

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Statement

It is possible to have a finite group G and an irreducible representation \varphi of G over a splitting field in characteristic zero (and hence also over \mathbb{C}) such that, if \chi denotes the character of \varphi, we have elements g_1, g_2 \in G satisfying:

  • \chi(g_1) = \chi(g_2), and
  • \varphi(g_1) and \varphi(g_2) are not similar matrices, i.e., they are not conjugate in the general linear group to which the representation maps G. In particular, they may have different characteristic polynomials.

Related facts

Opposite facts

Proof

Example of the dihedral group

Further information: faithful irreducible representation of dihedral group:D8

Consider dihedral group:D8 (see also linear representation theory of dihedral group:D8) and its faithful irreducible two-dimensional representation.


The dihedral group of order eight has a two-dimensional irreducible representation, where the element a acts as a rotation (by an angle of \pi/2), and the element x acts as a reflection about the first axis. The matrices are:

a \mapsto \begin{pmatrix}0 & -1 \\ 1 & 0 \\\end{pmatrix}, \qquad x \mapsto \begin{pmatrix}1 & 0 \\ 0 & -1 \\\end{pmatrix}.

This particular choice of matrices give a representation as orthogonal matrices, and in fact, the representation is as signed permutation matrices (i.e., it takes values in the signed symmetric group of degree two). Thus, it is also a monomial representation.

Below is a description of the matrices based on the above choice as well as another formulation involving complex unitary matrices:

Element Matrix (orthogonal/monomial/signed permutation matrices) Matrix as complex unitary Characteristic polynomial Minimal polynomial Trace, character value Determinant
e \begin{pmatrix} 1 & 0 \\ 0 & 1 \\\end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 1 \\\end{pmatrix} (t - 1)^2 = t^2 - 2t + 1 t - 1 2 1
a \begin{pmatrix}0 & -1 \\ 1 & 0 \\\end{pmatrix} \begin{pmatrix}i & 0 \\ 0 & -i \\\end{pmatrix} t^2 + 1 t^2 + 1 0 1
a^2 \begin{pmatrix} -1 & 0 \\ 0 & -1 \\\end{pmatrix} \begin{pmatrix} -1 & 0 \\ 0 & -1 \\\end{pmatrix} (t + 1)^2 = t^2 + 2t + 1 t + 1 -2 1
a^3 \begin{pmatrix}0 & 1 \\ -1 & 0 \\\end{pmatrix} \begin{pmatrix}-i & 0 \\ 0 & i \\\end{pmatrix} t^2 + 1 t^2 + 1 0 1
x \begin{pmatrix}1 & 0 \\ 0 & -1 \\\end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \\\end{pmatrix} t^2 - 1 t^2 - 1 0 -1
ax \begin{pmatrix}0 & 1 \\ 1 & 0 \\\end{pmatrix} \begin{pmatrix} 0 & i \\ -i  & 0 \\\end{pmatrix} t^2 - 1 t^2 - 1 0 -1
a^2x \begin{pmatrix}-1 & 0 \\ 0 & 1 \\\end{pmatrix} \begin{pmatrix} 0 & -1 \\ -1 & 0 \\\end{pmatrix} t^2 - 1 t^2 - 1 0 -1
a^3x \begin{pmatrix}0 & -1 \\ -1 & 0 \\\end{pmatrix} \begin{pmatrix} 0 & -i \\ i & 0 \\\end{pmatrix} t^2 - 1 t^2 - 1 0 -1
Set of values used \{ 0,1,-1 \} \{ 0,1,-1,i,-i \} -- -- \{ 2,-2,0 \} \{ 1,-1 \}
Ring generated by values used (characteristic zero) \mathbb{Z} -- ring of integers \mathbb{Z}[i] -- ring of Gaussian integers -- -- \mathbb{Z} -- ring of integers \mathbb{Z} -- ring of integers
Field generated by values used (characteristic zero) \mathbb{Q} -- field of rational numbers \mathbb{Q}(i) = \mathbb{Q}[t]/(t^2 + 1) -- -- \mathbb{Q} -- field of rational numbers \mathbb{Q} -- field of rational numbers


We see from the table above that the elements a and x have the same character value, namely 0, but their images have different characteristic polynomials (the characteristic polynomial of a is t^2 + 1, that of x is t^2 - 1) and are hence the images are not similar matrices.