How to prove the rep of SU(2) on homogeneous polynomials in 2 variables is irreducible? The group $\mathrm{SU}(2)$ has a tautologous representation on the space $\mathbb{C}^2$ and thus a representation on the $d$th symmetric power $S^d (\mathbb{C}^2)$.  What's the easiest way to prove that this representation is irreducible?
I want to explain this in a course I'm teaching.  I want to prove this "from scratch", not using powerful tools like Schur-Weyl duality or the general classification of representations of compact simple Lie groups or simple Lie algebras.   I have an elementary method but it involves some mildly annoying calculations.  Ideally there would be some slick, quick, elementary and insightful proof.
If it helps, I'm happy to think of $S^d(\mathbb{C}^2)$ as the space of homogeneous degree-d polynomial functions on $\mathbb{C}^2$ and describe the representation in this way.  I'm happy to consider it either as a representation of $\mathrm{SU}(2)$, $\mathrm{SL}(2,\mathbb{C})$, or the Lie algebra $\mathfrak{su}(2)$ or $\mathfrak{sl}(2,\mathbb{C})$: it's pretty painless to prove irreducibility of the representation of any one of these groups or Lie algebras on $S^d(\mathbb{C}^2)$ implies irreducibility of all the others.
I'd be equally happy to prove indecomposability, since I've proved an indecomposable representation of a compact Lie group is irreducible.
 A: The action of $\mathrm{SL}(2,\mathbb{C})$ on $S^d(\mathbb{C}^2)$ preserves the image of the Veronese embedding, which is $\{ (\alpha x + \beta y)^d \mid \alpha,\beta \in \mathbb{C} \}$, and acts transitively on the nonzero elements. This set spans $S^d(\mathbb{C}^2)$.
The decomposition of a symmetric tensor into elements of this form ("rank 1" tensors) is the Waring decomposition. You can prove that this decomposition holds by using the Vandermonde determinant. Just choose any $d+1$ pairwise distinct elements of $\mathbb{C}$, say $\{\alpha_i\}$, and then $\{ (\alpha_i x + y)^d \}$ spans $S^d(\mathbb{C}^2)$. There are a few proofs of the Vandermonde determinant formula on the linked Wikipedia page, hopefully at least one of them is quick and elementary enough for your purposes.
A: Depending on the background of your audience the following may be simple enough.
Let $x,y$ be the indeterminates, so $S^d(\Bbb{C}^2)$ is the linear span of $x^jy^{d-j},0\le j\le d$.
Assume contrariwise that there exists a non-trivial submodule $V\subset S^d(\Bbb{C}^2)$.
Claim 1. $V$ contains a monomial. That is, a weight vector.
Proof. Let $P=\sum_{j=0}^dc_j x^{d-j}y^j$ be the element of $V$ with the lowest possible number $m$ of non-zero terms. Assume contrariwise that $m>1$. Then there
exist indices $\ell\neq k$ such that $c_\ell\neq0\neq c_k$.
Action by $g(\theta)=\operatorname{diag}(e^{i\theta},e^{-i\theta})$ will map $x\mapsto e^{i\theta}x$ and $y\mapsto e^{-i\theta}y$. Therefore
$$
g(\theta)\cdot P=\sum_{j=0}^dc_je^{i(d-2j)\theta}x^jy^{d-j}.
$$
By selecting $\theta$ to be small but positive we can ensure that $e^{i(d-2\ell)\theta}$
and $e^{i(d-2k)\theta}$ are distinct. It follows that
$$
g(\theta)\cdot P-e^{i(d-2\ell)\theta}P
$$
is an element of $V$ that contains no monomials other than those of $P$, is missing the $x^{d-\ell} y^{\ell}$-term, and still had the $x^{d-k}y^k$-term. This contradicts the choice of $P$, and proves the claim.
Claim 2. $V$ must be all of $S^d(\Bbb{C}^2)$
Proof. If you have explained the ladder operators (here acting as $x\partial_y$ and
$y\partial_x$) this is easy starting from the monomial created in the first Claim. If you have $SL_2(\Bbb{C})$, then an option is to
use elements of a root subgroup (together with a Vandermonde matrix) to show that all the monomials must be there as well. You probably know the drill.
