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If $V_n$ is $n+1$-dimensional repn of $SU(2)$ how to compute decomposition into irreducibles of $S^nV_2$ efficiently?

Doing some rather tedious computation by picking a basis in $V_2$ and considering action of diagonal representatives of classes of $SU(2)$ then computing coefficients in the character, I arrived at the result:

$$ S^nV_2 = V_{2n} \oplus V_{2n-4} \oplus V_{2n-6} \oplus \dots \oplus V_0$$

Can this be obtained using properties of the symmetric product and some clever observation? The final formula looks rather nice and it'd be nice if it had some intuitive sense...

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Unfortunately, there is no nice property of the symmetric product to be used here. Your computation is the $m=2$ case of the plethysm of symmetric powers $$ S^n(S^m(V_1))\ . $$ Understanding plethysm in general is a very difficult open problem. That's for the bad news. The good news is for $SL_2$ the decomposition is known since about 1856 and it is called the Cayley-Sylvester formula. It gives the multiplicity of the irreducible module $V_k$ as $$ \left\{m,n,\frac{mn-k}{2}\right\}-\left\{m,n,\frac{mn-k}{2}-1\right\} $$ where $\{a,b,w\}$ denotes the number of integer partitions of $w$ which fit inside an $a$ by $b$ rectangle. The easiest proof is what you did, namely a character computation which essentially produces the $q$-binomials or Gaussian polynomials.

Note that in your particular case, this classically corresponds to counting covariants of degree $n$ of the generic quadratic binary form $Q$. These are linear combinations of $Q^i \Delta^j$ where $\Delta=b^2-4ac$ is the discriminant of the quadratic.

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  • $\begingroup$ I'm a little confused here, you're talking about $SL_2$, I was taught about $SU(2)$. Searching for this Cayley-Sylvester formula it would seem that these are connected in some simple way. Am I missing something obvious here? $\endgroup$
    – Radost
    Apr 27, 2019 at 22:01
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    $\begingroup$ When looking at finite dimensional representations from an algebra standpoint, the theory is the same for $SL_2(\mathbb{C})$ and $SU_2(\mathbb{C})$. The irrep if dimension $n+1$ is the symmetric power $S^n(\mathbb{C}^2)$. It is a unitary representation for the compact group $SU_2(\mathbb{C})$. On the other hand it is not a unitary representation for $SL_2(\mathbb{C})$ which is noncompact and only has infinite dimensional unitary representations. The connection between the two groups is that $SU_2(\mathbb{C})$ is Zariski dense in $SL_2(\mathbb{C})$. Look up "unitary trick". $\endgroup$ Apr 28, 2019 at 14:56

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