# Character of $n^{th}$ symmetric square $V_2$ of $SU(2)$

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...

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.

• 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? Apr 27, 2019 at 22:01
• 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". Apr 28, 2019 at 14:56