# Exterior products of irreducible representations of sl_2(C)

It is well-known that $$\mathfrak{sl}_2(\mathbb{C})$$ admits exactly one irreducible representation $$V_n$$ of dimension $$n+1$$ for all $$n\geq 0$$. It is explicitly given by the action on homogeneous polynomials in two variables of degree $$n$$. We will concentrate on even $$n$$ here.

Also well-known is the Clebsch-Gordan decomposition: for $$m\leq n$$ we have $$V_{2n}\otimes V_{2m} \cong \bigoplus_{k=0}^{n-m} V_{2n+2m-2k}.$$ A similar formula holds for the anti-symmetric part of the tensor product, the exterior product: $$V_{2n}\wedge V_{2m}\cong \bigoplus_{n+m+k \text{ odd}} V_{2n+2m-2k}.$$

My question is: in terms of homogeneous polynomials, can one write down the projection map from $$V_{2n}\wedge V_{2m}$$ to one of the factors?

In the book of Fulton-Harris on representation theory, in Exercise 11.29 and 11.30, the claim is that the projection of $$V_{2n}\wedge V_{2n} = \Lambda^2(V_{2n})$$ onto $$V_{4n-2}$$ is given by $$F\wedge G \mapsto F\cdot dG-G\cdot dF$$. This can't be true since $$F\cdot dG-G\cdot dF$$ is a polynomial of degree $$4n-1$$, not $$4n-2$$. I think one has to divide by $$x dy-y dx$$. Is that true? Is there a similar formula for the projection on the other factors?

Hitchin in his paper "Lie groups and Teichmüller space", in Section 6, writes that the projection in the Clebsch-Gordan decomposition is given by "contracting $$k$$ times with the symplectic form". What does this mean?

• A simpler formula for the map of $V_{2n}\times V_{2n}\to V_{4n-2}$ is this: You just send $F\wedge G$ to $\mathrm{d}F\wedge\mathrm{d}G$ which is a $2$-form in the variables $x_1,x_2$ and then take the coefficient of the basis $2$-form $\mathrm{d}x_1\wedge\mathrm{d}x_2$. In other words, you send $F\wedge G$ to the Poisson bracket $\{F,G\}$. Feb 3 at 21:53

The explicit projection of $$V_m\otimes V_n$$ on $$V_{m+n-2k}$$ is given by the $$k$$-th transvectant. $$F\otimes G\mapsto (F,G)_k$$, which with classical mid 19-th century normalization is given by $$(F,G)_k=\frac{(m-k)!\ (n-k)!}{m!\ n!} \left.\left(\frac{\partial^2}{\partial x_1\partial y_2}-\frac{\partial^2}{\partial x_2\partial y_1}\right)^k F(x_1,x_2)G(y_1,y_2)\right|_{y:=x}$$ $$=\frac{(m-k)!\ (n-k)!}{m!\ n!}\sum_{j=0}^{k}\binom{k}{j}(-1)^j\ \partial_1^{k-j}\partial_2^{j}F\ \partial_1^{j}\partial_2^{k-j}G$$ where $$\partial_1$$ denotes $$\frac{\partial}{\partial x_1}$$ and $$\partial_2$$ denotes $$\frac{\partial}{\partial x_2}$$.
As for the construction mentioned by Hitchin, it's just how you construct the symmetric tensor (meaning just an array of numbers) for the transvectant using the symmetric tensors of $$F$$ and $$G$$. You just contract indices with the intermediation of $$k$$ epsilon Levi-Civita tensors which here are matrices $$\varepsilon=\begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix}$$ This is best explained in pictures. See Section 2 of my article