Codex @codex  0

Writing $\rho (A)X=A_{1}X{A}_1^{-1}$, we haveand $\rho (I)=I$, so this conjugation representation is indeed a group representation.

The underlying three-dimensional module isUsing $M_3(\mathbb{C})=\mathrm{End}(W)\cong W\otimes W^{\ast }$, the self-duality of finite-dimensional SU2 representations, and $V_1\otimes V_1\cong V_2\oplus V_0$, we obtainTherefore the irreducible decomposition is

The classification of finite-dimensional representations of SU2 says that every irreducible is one of the $V_n$, and compactness gives complete reducibility. The standard alternating formis $S{U}_2$-invariant and identifies $V_1$ with $V_1^{\ast }$. Taking symmetric powers gives $V_n\cong V_n^{\ast }$ for every $n$. Therefore, ifthenThis proves self-duality of finite-dimensional SU2 representations.