constructing a unitary representation of $SU(2)$

Question

The group $SU(2)$ acts on $\mathbb{C}^2$. Hence, it has a representation on the space of all functions on $\mathbb{C}^2$. In particular, let $V_k$ be the space of homogeneous polynomials of degree $k$. We have

$$ V_k = span \{ z_1^ k , z_1^{k-1} z_2, \ldots, z_2^k \} . $$

If $A$ is any invertible $2\times 2 $ matrix and if $f\in V_k $ then clearly $\rho(A) f = f (A^{-1} \cdot )$ is again a homogeneous polynomial of degree $k$. We thus have a representation of $SU(2)$ on $V_k$.

It is then straightforward to construct the $(k+1)\times (k+1)$ representative matrix corresponding any $2\times 2 $ element in $SU(2)$. However, by no means is the representative matrix unitary.

The problem is that there is no natural or obvious inner product on the space $V_k$. Hence, we cannot say that $SU(2)$ acts on the space unitarily.

So, how to stipulate such an inner product structure and hence construct a unitary representation of $SU(2)$? Of course, the construction above will yield a representation equivalent to a unitary representation, but one favors a representation explicitly unitary.

Answer 1

Since $SU(2)$ is a compact Lie group, it has one and only one bi-invariant Haar measure $\mu$ such that $\mu\bigl(SU(2)\bigr)=1$. So, take any inner product $(\cdot,\cdot)$ on $V_k$ and then define$$\langle f_1,f_2\rangle=\int_{SU(2)}(g.f_1,g.f_2)\,\mathrm d\mu(g).$$Then $\langle\cdot,\cdot\rangle$ is an inner product and, with respect to this inner product, your action is unitary.

Answer 2

Let $F(z_1,z_2)$ be an element of $V_k$. There is a unique symmetric tensor $(F_{i_1,\ldots,i_k})_{1\le i_1,\ldots,i_k\le 2}$ such that for all $z_1,z_2\in\mathbb{C}$, $$ F(z_1,z_2)=\sum_{i_1,\ldots,i_k=1}^2 F_{i_1,\ldots,i_k}\ z_{i_1}\cdots z_{i_k}\ . $$ Now for two elements $F,G$ of $V_k$, you can define their inner product by $$ \langle F,G\rangle =\sum_{i_1,\ldots,i_k=1}^2 \overline{F_{i_1,\ldots,i_k}}\ G_{i_1,\ldots,i_k}\ . $$ This gives you the invariant inner product you need.