Let us consider the natural representation $\pi$ of the $D_4$ group on the vector space with orthonormal basis $\{e_1, e_2, e_3, e_4 \}$. Note that this gives us a $C^*$ algebra $A$ such that $A = \tilde{\pi} (\mathbb{C}(D_4))$, where $\tilde{\pi}: \mathbb{C}(D_4) \rightarrow B(H)$ is a *- homomorphism of $\mathbb{C}(D_4)$.
I have that $$ \mathrm{Rotation} = \rho= \begin{bmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{bmatrix} \text{.} $$
$$ \mathrm{Reflection}= \sigma = \begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix} \text{.} $$.
My question is: Let $A'$ be the commutant of $A$. What is $dim(A') $ in this case?
I know that if $x \in A'$, then we must have $x\rho=\rho x$ and $x \sigma = \sigma x$.
So this gives us some equations to satisfy. But I'm not sure how to proceed from here.
Thank you in advance!
By working out the matrix multiplications, a matrix $A$ commutes with both $\rho $ and $\sigma $ iff it of the form $$ A=\pmatrix{ a & b & c & b \cr b & a & b & c \cr c & b & a & b \cr b & c & b & a}. $$ So the dimension of the commutant is 3. The only 3-dimensonal C$^*$-algebra is ${\mathbb C}^3$, so $A'\simeq {\mathbb C}^3$. Moreover, all faithful 4-dimensional representations of ${\mathbb C}^3$ are, up to unitary equivalence, given by $$ (x,y,z) \mapsto A=\pmatrix{ x & 0 & 0 & 0 \cr 0 & y & 0 & 0 \cr 0 & 0 & z & 0 \cr 0 & 0 & 0 & *}, $$ where "$*$" can be taken to be $x$, $y$ or $z$.
In any case, this shows that $A'$ is unitarily equivalent to $$ \left\{\pmatrix{ x & 0 & 0 & 0 \cr 0 & y & 0 & 0 \cr 0 & 0 & z & 0 \cr 0 & 0 & 0 & z}: x, y, z\in {\mathbb C}\right\}. $$
The double commutant is therefore isomorphic to ${\mathbb C}\oplus {\mathbb C}\oplus M_2({\mathbb C})$, represented e.g. as block diagonal matrices.
This implies that the original representation of $\mathbb D_4$ splits as the direct sum of two one-dimensional, non-equivalent representations, plus one 2-dimensional irreducible representation.