§ Representation theory of $SU(2)$ [TODO ]

2x2
unitary matrices, so $AA^\dagger = I$.  Lie algebra is $su(2)$, which are of the form $A^\dagger = A$, and $Tr(A) = 0$.
 We write $M_v \equiv \begin{bmatrix} ix & y + iz \\ y + iz & ix \end{bmatrix}$.
 The group elements are matrices, so this is the standard representation, which goes from $SU(2)$ to $GL(2, \mathbb C)$. Turns out this is irreducible, 2D complex representation.
 We have a transformation which for a $g \in SU(2)$ creates a map which sends a matrix $M_v$ to $g M_v g^{1}$. so the representation is $g \mapsto \lambda M_v. g M_v g^{1}$, which has type signature $SU(2) \to GL(su(2))$. This is a 3D, real representation: the vectors $M_v$ have 3 degrees of freedom.
 We like complex representations, so we're going to build $SU(2) \to GL(su(2) \otimes \mathbb C)$.
 There is the trivial representation $\lambda g. (1)$.
 There is a zero dimensional representation $\lambda g. ()$ which maps $\star \in \mathbb C^0$ to $\star$. So it's the identity transformation on $\mathbb C^0$.
§ Theorem
For any integer $n$ there is an irrep $R_n: SU(2) \to GL(n, \mathbb C)$. Also, any irrep $R: SU(2) \to GL(v)$ is isomorphic to one of these.
§ New representations from old
 If we have $R: G \to GL(V)$ and $S: G \to GL(W)$, what are new representations?
 For one, we can build the direct sum $R \oplus S$. But this is useless, since we don't get irreps.
 We shall choose to take tensor product of representations.
 Symmetric power of $R: G \to GL(V)$ is $R^{\otimes n}: G \to GL(V^{\otimes n})$. This is not irreducible because it contains a subrep of symmetric tensors .
 Example, in $C^2 \otimes C^2$, we can consider $e1 \otimes e1$, $e2 \otimes e2$, and $e1 \otimes e2 + e2 \otimes e1$.
 Define $Av$ (for averaging) of $v_1 \otimes v_2 \dots v_n$ to be $1/n! \sum_{\sigma \in S_n} v_{\sigma(1)} \otimes v_{\sigma(2)} \dots v_{\sigma(n)}$. In other words, it symmetrizes an input tensor.
 Define $Sym^n (V) = Im(Av: V^{\otimes n} \to V^{\otimes n})$. We claim that $Sym^n(V)$ is a suprep of $V$. We do this by first showing that $Av$ is a morphism of representations, and then by showing that the image of a morphism is a subrepresentation.
§ Weight space decomposition
 $SU(2)$ contains a subgroup isomorphic to $U(1)$. Call this subgroup $T$, which is of the form $\begin{bmatrix} e^{i \theta} & 0 \\ 0 & e^{i \theta} \end{bmatrix}$.