Question about Eigenvalues of group elements

Question

The quaternion algebra is given by
$\mathbb{H}$ = $\{a+bi+cj+dk \mid a, b, c, d \in \mathbb{R}, i^2 = j^2 = k^2 = -1, ij = k = -ji\} := \{z_1+z_2j \mid z_1, z_2 \in \mathbb{C}\}.$

I consider the 3-sphere as the set of unit quaternions (the quaternions of length 1) as follows: $\mathbb{S}^3 = \{a + bi + cj + dk \mid a^2 + b^2 + c^2 + d^2 = 1\} = \{z_1 + z_2j \mid |z_1|^2 + |z_2|^2 = 1\}$

The product in $\mathbb{H}$ induces a group structure on $\mathbb{S}^3$. For each pair $(p, q)$ of elements of $\mathbb{S}^3$, the function $\Phi_{p,q} : \mathbb{H} \rightarrow \mathbb{H}$ with $\Phi_{p,q}(h) = phq^{-1}$ leaves invariant the length of quaternions.

We can, therefore, define a homomorphism of groups: $\Phi : \mathbb{S}^3 \times \mathbb{S}^3 \rightarrow SO(4)$ such that $\Phi(p, q) = \Phi_{p,q}$.

I am looking at subgroup $C_4 \times D^*_{24}$ of $\mathbb{S}^3 \times \mathbb{S}^3$, where $C_4$ is the cyclic group of order 4 and $D^*_{12}$ is the binary dihedral group of order 12. I have looked at 3 different ways of representing this group and in each case I get resulting matrices in $SO(4)$ with different eigenvalues. Here are the 3 ways:

1. $C_4 = \{<g>$, where $g^4 = 1\}$, $D^*_{12} = \{<A, B>$, where $A^{6} = 1, B^4 = 1$ and $BAB^{-1} = A^{-1}\}$.

I represent $g$ as the matrix

\begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}

I represent $A$ as the matrix \begin{pmatrix} e^{i\pi/3} & 0 \\ 0 & e^{-i\pi/3} \end{pmatrix}.

I represent $B$ as the matrix

\begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}.

With this, $BA$ is the matrix \begin{pmatrix} 0 & e^{i\pi/6} \\ -e^{-i\pi/6} & 0 \end{pmatrix}.

Then $\Phi(g, BA)(h)$ is the $4 \times 4$ matrix

\begin{pmatrix} 0 & 0 & \sin \frac{\pi}{6} & \cos \frac{\pi}{6} \\\\ 0 & 0 & -\cos \frac{\pi}{6} & \sin \frac{\pi}{6} \\\\ \sin \frac{\pi}{6} & \cos \frac{\pi}{6} & 0 & 0 \\\\ -\cos \frac{\pi}{6} & \sin \frac{\pi}{6} & 0 & 0 \end{pmatrix}.

This has eigenvalues $\pm e^{\pm i\pi/3}$

2. I can also view $D^*_{12} = C_{6} \cup C_{6}j$

With $g$ and $A$ as before, I represent $j$ as the matrix \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}.

With this, $Aj$ is the matrix

\begin{pmatrix} \ 0 & e^{i\pi/3} \\ -e^{-i\pi/3} & 0 \end{pmatrix}.

Now, $\Phi(g, Aj)(h)$ is the $4 \times 4$ matrix

\begin{pmatrix} 0 & 0 & \sin \frac{\pi}{3} & \cos \frac{\pi}{3} \\\\ 0 & 0 & -\cos \frac{\pi}{3} & \sin \frac{\pi}{3} \\\\ \sin \frac{\pi}{3} & \cos \frac{\pi}{3} & 0 & 0 \\\\ -\cos \frac{\pi}{3} & \sin \frac{\pi}{3} & 0 & 0 \end{pmatrix}.

This has eigenvalues $\pm e^{\pm i\pi/6}$

3. I represent $g = i$, $A = e^{i\pi/3}$ and $j = j$.

   With this $\Phi(i, Aj)(h)$ = $i (a + bi + cj + dk) (-jcos(\pi/3) - ksin(\pi/3))$ = $c\sin(\pi/3)-d\cos(\pi/3) + i(c\cos(\pi/3) + d\sin(\pi/3)) + j(a\sin(\pi/3)+ b\cos(\pi/3)) + k(-a\cos(\pi/3) + b\sin(\pi/3))$

which gives the $4 \times 4$ matrix for $\Phi(i, Aj)$ to be

\begin{pmatrix} 0 & 0 & \sin \frac{\pi}{3} & -\cos \frac{\pi}{3} \\\\ 0 & 0 & \cos \frac{\pi}{3} & \sin \frac{\pi}{3} \\\\ \sin \frac{\pi}{3} & \cos \frac{\pi}{3} & 0 & 0 \\\\ -\cos \frac{\pi}{3} & \sin \frac{\pi}{3} & 0 & 0 \end{pmatrix}.

This has eigenvalues $\pm 1, \pm 1$

So I have 3 different ways to do this and in each case, I get a different matrix in SO(4) with different eigenvalues. I thought that if the three matrices represent the same element in SO(4), their eigenvalues should have been the same. What mistake am I making here?

Answer 1

It is not true that for a group $G$, and $g \in G$, that $g$ has the same eigenvalues under any representation $\rho$. You've found some examples already. Other examples are abound. In general, for a representation $\rho$, one can tensor it with any nontrivial one-dimensional representation and get different eigenvalues.

Depending on what you know about representation theory, you might remember that the trace of a representation, $\text{tr}(\rho(g))$ as a function on conjugacy classes of $G$ (or in your case, probably better is a $G$-adjoint-equivariant function on $G$), tells us a lot of information about the representation. If the eigenvalues of $g$ did not depend on a representation, then the trace (sum of eigenvalues) would only tell us the dimension of the representation, which is not so useful.