hermitian inner product that is invariant on a Lie Group $C[x_1, x_2]_n$ is the space of homogeneous polynomial of degree n and is the irreducible representation of $\mathfrak{gl}_2(\mathbb{C})$ with $e_{ij}$ acting as $x_i\frac{\partial}{\partial x_j}$. Find a Hermitian inner product on $C[x_1, x_2]_n$  that is $SU(2)$-invariant.
I feel like I have to use integration for the inner product but I cannot seem to make it invariant under $SU(2)$ action. what should be the correct form for the inner product?
In addition, how am I supposed to define this $SU(2)$ action on a polynomial? My initial thought was using the basis of $SU(2)
\begin{pmatrix}
    i & 0\\
   0 & -i
  \end{pmatrix}, \begin{pmatrix}
    0 & 1\\
   -1 & 0
  \end{pmatrix}, \begin{pmatrix}
    0 & i\\
   i & 0
  \end{pmatrix}, \begin{pmatrix}
    1&0 \\
   0 & 1
  \end{pmatrix}. Should they transformed to, for example, $ix_1\frac{\partial}{\partial x_1}-ix_2\frac{\partial}{\partial x_2}$ for the first matrix?
 A: Here's a very general strategy (that is kind of overkill). Let $G$ be a compact Hausdorff topological group and let $\rho:G \to \mathrm{GL}_n(\Bbb C)$ be continuous representation. Then one can construct a $G$-invariant Hermitian inner product on $\Bbb C^n$ as follows: take the standard Hermitian inner product $\langle-,-\rangle$ (or really, any Hermitian inner product on $\Bbb C^n$) and then define a new inner product via
$$(v,w):=\int_{G}\langle gv,gw\rangle\mathrm{d}\mu$$ where $\mu$ is a Haar measure on $G$. Because $G$ is compact, $\mu(G) < \infty$ and thus the integral converges. One may readily check that it is a Hermitian inner product and also $G$-invariant.
For your specific situation, here's another approach (this also explains the action of $SU(2)$ on the vecor space $\Bbb C[x_1,x_2]_n$).
Note that $SU(2)$ comes with a "standard" representation $SU(2) \to \mathrm{GL}_2(\Bbb C)$ and essentially by definition, the standard Hermitian inner product on $\Bbb C^2$ is $SU(2)$-invariant. The usual representation on the space $\Bbb C[x_1,x_2]_n$ is the $n$-th symmetric power of this standard representation if we identify $x_1$ and $x_2$ with the standard basis $(e_1=(1,0),e_2=(0,1))$ for $\Bbb C^2$ and replace polynomial multiplication with (symmetrical) tensor products. So for example $x_1^2x_2+2x_2^3$ would correspond to the vector $e_1 \otimes e_2 \otimes e_2 + 2 e_2 \otimes e_2 \otimes e_2 \in \mathrm{Sym}^3(\Bbb C^2)$. The action we get on polynomials may be most easily described if we think of the arguments of a two-variable polynomial as a vector $\begin{pmatrix}x_1 \\ x_2\end{pmatrix}$, then we have an action for $A \in SU(2),f \in \Bbb C[x_1,x_2]_n$ $$(A \cdot f) \begin{pmatrix}x_1 \\ x_2\end{pmatrix}:=f\left(A^T\begin{pmatrix}x_1 \\ x_2\end{pmatrix}\right) \in \Bbb C[x_1,x_2]_n$$
Now we are lead to consider the following situation: Let $G$ be a group and let $V$ be a representation of $G$. Suppose we have a $G$-invariant Hermitian inner product, then can we use this to define a $G$-invariant Hermitian inner product on $\mathrm{Sym}^n(V)$? The answer is yes. $\mathrm{Sym}^n(V)$ is generated by elementary (symmetrical) tensors $v_1 \otimes v_2 \otimes \dots \otimes v_n$, so we can define $$\langle v_1 \otimes \dots \otimes v_n,w_1 \otimes \dots \otimes w_n \rangle:=\prod_{i=1}^n \langle v_i,w_i\rangle$$
One can check that this is well-defined, Hermitian, an inner product and $G$-invariant.
