$V_j=$ polynomials functions on $\mathbb{C}^2$, that are homogeneous of degree $2j$. Could any one tell me how $U_{\frac{1}{2}}$ is equivalente to $\Gamma$?
What I know about a representation of $SU(2)$ by definition is, it must be a vector space $V$ together with a homomorphism $\phi:SU(2)\rightarrow GL(V)$.
Let $g\in SU(2)$ is just a linear map on $\mathbb{C}^2$, Define a Linear Transformation
$U_j(g):V_j\rightarrow V_j$ given by $[U_j(g)f](z)=f(g^{-1}z)$ where $f(z)=a_0z_1^{2j}+a_1z_1^{2j-1}z_2+\dots +a_{2j}z_2^{2j}$, $z=(z_1,z_2)\in\mathbb{C}^2$
And the second linear transformation:
$\Gamma: SU(2) \rightarrow \mathbb{C}^2$, given by $\Gamma(g)(z) = g z$, where $z=(z_1,z_2)\in\mathbb{C}^2$
These two represenentations are equivalente if exist na isomorphism $T: \mathbb{C}^2 \rightarrow \mathbb{C}^2$, so that $\Gamma(g)T = TU_\frac{1}{2}(g)$.
How to prove this equivalence ?
The two representations are dual to each other. However, for $SU(2)$ the representations are self-dual. A map like $(z_1,z_2)\mapsto(-z_2,z_1)$ should give the isomorphism if you write things explicitly in coordinates.