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?

  • $\begingroup$ 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 $\{F,G\}$. $\endgroup$ Feb 3 at 21:53

1 Answer 1


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

and Section 4.2 of

See in particular equations (4.4) and (4.5) of the last reference (the published version, not the arXiv preprint), shown in the snippet below. enter image description here

  • $\begingroup$ @FrancoisZiegler: yes of course. Thank you for catching that. $\endgroup$ Feb 3 at 18:59
  • $\begingroup$ Thanks for your answer. The graphical calculus of classical invariant theory is amazing! $\endgroup$
    – AThomas
    Feb 4 at 8:25

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