A representation of $SU(2)$ is self dual Let $SU(2)$ be a set of $2 \times 2$ unitary matrices over $\mathbb{C}$ with determinant $1$. Let $H_j$ be a $2j+1$ dimensional vector space with basis $x^ay^b$ with $a+b=2j$.
A representation $U_j$ of $SU(2)$ on $H_j$ is defined by $(U_j(g)f)(v)=f(g^{-1}v)$, where $f \in H_j$ and $v\in \mathbb{C}^2$.
I would like to show that $U_j$ is equivalent to its dual $U_j^*$.
I would like to prove this directly using definitions. But so far I could not prove it. 
What I tried is as follows.
Let $F: H_j \to H_j^*$ be a isomorphisms defined by sending a basis to dual basis. Then I want to show that 
$$FU_j^*(g)=U_j(g)F$$
for all $g \in SU(2)$.
I evaluated both sides by an element of $V$ but could not show they are equal.
I don't know how to use the definition of the representation of $SU(2)$, especially I can not  use $v\in \mathbb{C}^2$ well.
Or should I use different isomorphism than $F$?
I am sorry I explained very poorly.
I hope someone can help me proving this.
 A: Firstly, any homogenous polynomial of degree $2j$ in $n$ variables in some field $K$  corresponds uniquely to a symmetric tensor of range $2j$ and components $t_{i_1,\ldots,t_{i_{2j}}}$ over $K^n$ given by
\begin{align}
f(x^1,\ldots,x^{n})
&=\sum_{i_1,\ldots,i_{2j}=1}^nt_{i_1,\ldots,i_{2j}}x^{i_1}\ldots x^{i_{2j}}
\\
&\equiv t_{f}\underbrace{\left(x\otimes\ldots\otimes x\right)}_{2j \text{ times}}\quad
\left(t\in\text{Sym}^{2j}(V),x=\sum_{i=1}^{n}x^{i}e_{i}\right)
\end{align}
where $V$ is an $n$-dimensional vector space over $K$. Expressing the dual basis as $\{e^i\}_{1\leq i\leq n}$, we then have an isomorphism
$$
t(f)=t_f = \sum_{i_1,\ldots,i_{2j}=1}^{n}t_{i_1,\ldots,i_{2j}}e^{i_{1}}\otimes\ldots\otimes e^{i_{2j}}
$$
Therefore we are considering representations in $\text{Sym}^{2j}(\mathbb{C}^2)$ where the components of the tensor transform contravariantly, according to
\begin{align}
(U_j(g)(f))(x^1,x^2)
&=\sum_{i_1,\ldots,i_{2j}=1}^2t_{i_1,\ldots,i_{2j}}\left(g^{-1}\right)^{i_1}_{i'_1}\ldots \left(g^{-1}\right)^{i_{2j}}_{i'_{2j}}x^{i'_1}\ldots x^{i'_{2j}}
\\
&=\sum_{i'_1,\ldots,i'_{2j}=1}^2(t')_{i'_1,\ldots,i'_{2j}}x^{i'_1}\ldots x^{i'_{2j}}
\\
&\equiv[U_{j}(g)t](x^{\otimes2j})
\end{align}
We have $\mathcal{H_{\frac{1}{2}}}\simeq \mathbb{C}^{2\ast}$ (this is a more natural point of view than $\mathbb{C}^{2}$, since the group acts by $g^{-1}$) and likewise $\text{Sym}^{2j}(\mathbb{C}^2)\subset\left(\mathbb{C}^{2\ast}\right)^{\otimes 2j}$, so it suffices to find an isomorphism
$$
T:\mathbb{C}^{2\ast}\rightarrow\mathbb{C}^{2}\quad\mid\quad U^{\ast}_{\frac{1}{2}}(g) T = TU_{\frac{1}{2}}(g)$$
This is because $U_{\frac{1}{2}}$ induces a (reducible) representation on the tensor product, given for $\mu=\bigotimes_{i=1}^{2j}\mu_{i}\in(\mathbb{C}^{2\ast})^{\otimes2j}$ by
$$
[U_j(g)(\mu)]=\bigotimes_{i=1}^{2j}U_{\frac{1}{2}}(g)\mu_{i}
$$
and extended linearly. It also induces a dual representation $U^{\ast}_j\in\text{End}\left[\left(\mathbb{C}^2\right)^{\otimes 2j}\right]$ in the usual manner. We can then define, for $\{\lambda_i,\mu_i\}_{1\leq i \leq 2j}\subset \mathbb{C^{2\ast}}$, the corresponding map $\tilde{T}:(\mathbb{C}^{2\ast})^{\otimes 2j}\rightarrow(\mathbb{C}^{2})^{\otimes 2j}$ given by
$$
(\mu_1\otimes\ldots\otimes \mu_{2j})[\tilde{T}(\lambda_1\otimes\ldots\otimes \lambda_{2j})]=\prod_{i=1}^{2j}\mu_i[T(\lambda_i)]
$$
and then restrict it to the symmetric tensors. We will have for a $t\in\text{Sym}^{2j}\left(\mathbb{C}^{2}\right)$ and any $\mu=\bigotimes_{i=1}^{2j}\mu_{i}$, that
\begin{align}
\mu\left[U^{\ast}_j(g)\tilde{T}(t)\right]=[U_j(g)(\mu)]\tilde{T}(t)&=\sum_{i_1,\ldots,i_{2j} =1}^2t_{i_1,\ldots,i_{2j}}[U_j(g)(\mu)]\tilde{T}(e^{i_1}\otimes\ldots\otimes e^{i_{2j}})
\\
&=\sum_{i_1,\ldots,i_{2j} =1}^2t_{i_1,\ldots,i_{2j}}\prod_{i=1}^{2j}\left[U_{\frac{1}{2}}(g)\mu_i\right][T(e^i)]
\\
&=\sum_{i_1,\ldots,i_{2j} =1}^2t_{i_1,\ldots,i_{2j}}\prod_{i=1}^{2j}\mu_i\left[U^{\ast}_{\frac{1}{2}}(g)T(e^i)\right]
\\
&=\sum_{i_1,\ldots,i_{2j} =1}^2t_{i_1,\ldots,i_{2j}}\prod_{i=1}^{2j}\mu_i\left[T\left(U_{\frac{1}{2}}(g)e^i\right)\right]
\\
&=\mu[\tilde{T}(U^\ast_j(g)t)]
\end{align}
and so $U^{\ast}_j(g)\tilde{T}(t)=\tilde{T}(U(g)t)$, which in turn means that $U_j^\ast(g)\tilde{T}f=\tilde{T}U_j(g)f$, where we are abusing notation by writing $\tilde{T}(f)\equiv\tilde{T}(t(f))$.

All that's left then is to show we can find a $$
T:\mathbb{C}^{2\ast}\rightarrow\mathbb{C}^{2}\quad\mid\quad U^{\ast}_{\frac{1}{2}}(g) T = TU_{\frac{1}{2}}(g)$$
for the action of $U^{\ast}_{\frac{1}{2}}(g)$ on $\mathbb{C}^2$ given by $(x,y)^{t}\mapsto g\cdot (x,y)^{t}$ for $g\in\text{SU}(2)$ with the usual matrix multiplication. Explicitly, we want
$$
\begin{pmatrix}
\alpha&-\overline{\beta}\\
\beta&\overline{\alpha}
\end{pmatrix}T(x,y)=T\left[(x,y)\begin{pmatrix}
\alpha&-\overline{\beta}\\
\beta&\overline{\alpha}
\end{pmatrix}^{-1}\right]
$$
This works if we set $(x,y)\mapsto(-y,x)^t$, as we effectively have (using $g^{-1}=g^{\dagger}$) that
$$
\begin{pmatrix}
\alpha&-\overline{\beta}\\
\beta&\overline{\alpha}
\end{pmatrix}\begin{pmatrix}-y\\x\end{pmatrix}=T\begin{pmatrix}
\overline{\alpha} x - \beta y\\
\overline{\beta}x + \alpha y\end{pmatrix}
$$
