A Universe of Sorts

§ 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 $\mathbb C^0$ .

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.

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 sub-representation.

$SU(2)$ contains a subgroup isomorphic to $U(1)$ . Call this subgroup $T$ , which is of the form $\begin{bmatrix} e^{i \theta} & 0 \\ 0 & e^{-i \theta} \end{bmatrix}$ .