What are the representations of $U_2(\mathbb{C})$? I know that the (isomorphism types) of continuous representations of $SU_2(\mathbb{C})$ are given by the homogeneous parts of the natural change of variables action on $\mathbb{C}[X,Y]$. This clearly extends to an action of $U_2$.
Of course, any irreducible representation of $U_2(\mathbb{C})$ will give a representation of $SU_2$ by restriction, and so the restricted representation will be isomorphic to a direct sum of representations of $SU_2$. Will this representation of $SU_2$ remain irreducible?
We have the representations $det : U_2 \to C^*$, and $det^{-1}$, and the powers. Are these the only new representations? (Do it and its inverse generate the character ring along with the other representations of $SU_2$? Is the only information lost after restriction the twist of a power of the determinant?)
Mainly I'm after a good reference, unless the answer is easy. (The best thing would be a series of hints.)
 A: Following David Loeffler's Hint:


*

*We have the map $SU_2(C) \times U_1 \to U_2$ sending $(A,a) \to aA$. For any $T \in SU_2$ it is the image of $(T / \sqrt{det(T)} , \sqrt{det(T)})$, for some (local) choice of square root. So this map is surjective. It is also a continuous homomorphism. (Note that the necessity of choosing a square root is why we don't get a splitting.)

*The kernel is $\{(I,1), (-I, -1)\}$. This is because the only elements of $SU_2$ proportional to $I$ are $\pm I$.

*Thus the representations of $U_2(C)$ are thus exactly the representations of $SU_2 \times S^1$ which are trivial on the element $-1 = (-I,-1)$. 

*For such a representation, the images of the two groups in $U(V)$ are equal. Thus such a representation is irreducible as a representation of $SU_2 \times U_1$ if irreducible as a $U_2$ representation, and vica-versa.

*The irreducible representations of $U_2 \times U_1$ are $V \otimes_{\mathbb{C}} W$, where $(V, \pi)$ is an irreducible representation of $SU_2$ and $(W, \rho)$ is an irreducible representation of $U_1$. Here the action is given by $(A, \alpha) v \otimes w = (\pi(A) v) \otimes ( \rho(\alpha) w) = (\rho(\alpha) \pi(A) v) \otimes w$ (since $\rho$ is one dimensional) and extending linearly. (Or as the action on $Hom_{\mathbb{C}}(V^*,W)$.)

*In order for this representation to be trivial on $-1$, it must be the case that $\rho(-1)\pi(-I) = I$.

*The irreducible representations of $U_1$ are $\rho_n(z) = z^n$, for $n \in \mathbb{Z}$. The irreps of $SU_2$ are polynomial change of variables, indexed by degree in $\mathbb{N}_{\geq 0}$ - the action of $-I$ is by $X^k Y^{d - k} \to (-1)^d X^k Y^{d-k}$. Hence $\rho_n(-1) \pi_d(-I) = (-1)^n (-1)^d = (-1)^{d + n}$.

*It follows that the representations of $SU_2 \times S_1$ that descend to representations of $U_2$ are those of the form $\rho_n \otimes \pi_d$, with $d + n$ even.
