$SU(2)$ acting by conjugation, decomposition into irreducibles

Question

I am attempting past exam questions of the Cambridge Math Tripos. I know how to do the first few parts, which involves giving the irreducible representations of $U(1)$ and $SU(2)$. But I am not sure on how to do the last part.

"Let $SU(2)$ act on the space $M_3(\mathbb{C})$ of $3 \times 3$ matrices by conjugation, where $A \in SU(2)$ acts by $$A: M \mapsto A_1MA_1^{-1},$$in which $A_1$ denotes the $3 \times 3$ block diagonal matrix $\left( \begin{array}{ccc} A & 0\\ 0 & 1 \end{array} \right)$. Show that this gives a representation of $G$ and decompose it into irreducibles."

For starters, I was told by my supervisor this representation is $9$ dimensional. Any help would be appreciated.

Answer 1

We give a more elementary solution than the one already provided. Let $V$ denote the representation on $M_3(\mathbb{C})$. Every $A \in SU(2)$ is conjugate to a diagonal matrix $D = \text{diag}(e^{i\theta}, e^{-i\theta})$ so $\chi_V(A) = \chi_V(D)$. For $X \in M_3(\mathbb{C})$, written as $X = (x_1, x_2, \dots, x_9)$ by concatenating the rows of the matrix, we may compute that$$D \cdot X = (x_1, z^2 \cdot x_2, z \cdot x_3, z^{-1} \cdot x_4, x_5, z^{-1} \cdot x_6, z^{-1} \cdot x_7, z \cdot x_8, x_9)$$where $z = e^{i\theta}$. Hence$$\chi_V(D) = 3 + z^2 + z^{-2} + 2(z + z^{-1}) = 3 + 2\cos 2\theta + 4\cos\theta = (2\cos\theta + 1)^2 = \chi_{(V_\text{triv} \oplus V_1)^{\otimes 2}}.$$Now, using the Weyl integration formula, we can do some annoying computations and decomopse$$V \cong 2V_{\text{triv}} \oplus 2V_1 \oplus V_2.$$

Answer 2

Let $\rho :SU(2) \to End(M_3(\mathbb C^3))$ be the representation. For any $B \in M_3(\mathbb C)$ of the form

$$B = \left( \begin{array}{cc} 0 & b_1\\ b_2 & c \end{array} \right), $$

($b_1, b_2$ are vectors and $c$ a scalar), we have

$$\rho(A)B = \left( \begin{array}{cc} A & 0\\ 0& 1 \end{array} \right) \left( \begin{array}{cc} 0 & b_1\\ b_2 & c \end{array} \right)\left( \begin{array}{cc} A^{-1} & 0\\ 0 & 1 \end{array} \right)= \left( \begin{array}{cc} 0 & Ab_1\\ b_2 & c \end{array} \right)\left( \begin{array}{cc} A^{-1} & 0\\ 0 & 1 \end{array} \right)=\left( \begin{array}{cc} 0 & Ab_1\\ b_2A^{-1} & c \end{array} \right)$$

Thus we see that the action splits into three irreducible components:

$$W_1 = \left( \begin{array}{cc} 0 & b_1\\ 0 & 0 \end{array} \right),W_2 = \left( \begin{array}{cc} 0 & 0\\ b_2 & 0 \end{array} \right), W_3 = \left( \begin{array}{cc} 0 & 0\\ 0 & c \end{array} \right), $$

and $W_1$ is the standard one on $\mathbb C^2$, $W_2$ is the dual of $W_1$ and $W_3$ is the trivial one. Now for the remaining one:

$$W = \bigg\{B': B' = \left( \begin{array}{cc} M & 0\\ 0 & 0 \end{array} \right) \bigg\}, $$

we have

$$\rho(A) B' = \left( \begin{array}{cc} AMA^{-1} & 0\\ 0 & 0 \end{array} \right). $$

Now decompose $W$ into $W = W_4 \oplus \{cI : c\in \mathbb C\}$. where $W_4 = \{ B': tr M = 0\}$. Obviously $\rho$ acts trivially on $W_5 = \{cI : c\in \mathbb C\}$.

Lastly, note that $W_4 \cong sl(2, \mathbb C)$. And we know that $sl(2, \mathbb C)$ is the complexifiaction of $su(2)$. So $W_4$ is the usual adjoint representation of $SU(2)$ on $su(2)_\mathbb{C}$.

(To be explicit, let $X \in M_2(\mathbb C)$. Then

$$X = Y + \sqrt{-1} Z= \frac{X-X^*}{2} + \sqrt{-1} \frac{\sqrt{-1}(X^* + X)}{2}. $$

Note that both $Y, Z$ are in $su(2)$.)

To sum up, $\rho$ splits into two trivial representation ($W_3$ and $W_5$), one standard representation ($W_1$), one dual of it ($W_2$) and an adjoint representation $(W_4$).