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Suppose $V=\mathbb{C}^2$ and $G=SL(V)=SL_2(\mathbb{C})$. We define $C_n = H_{\mathbb{C},n}(V,\mathbb{C}) \cong S^n(V^*)$, the n-th symmetric power of the dual of $V$, i.e. the homogeneous polynomials of degree $n$ in 2 variables $X,Y$. Then for $f,g$ functions of $X,Y$ we define the r-th transvectant of $f,g$ to be

$\tau_r (f,g) = \sum \limits_{j=0}^r \frac{(-1)^j}{j!(r-j)!} \frac{\partial^r f}{\partial X^{r-j}\partial Y^j} \frac{\partial^r g}{\partial X^{j}\partial Y^{r-j}} $.

I want to show that any $\mathbb{C}G$-module map $C_p \otimes C_q \to C_{p+q-2r}$ is a multiple of $\tau_r$. My worksheet has the ''hint'' "use Clebsch-Gordan"; I believe this refers to the formula

$D_{p,o}(V) \otimes D_{q,0}(V) = \bigoplus \limits_{r=0}^{min(p,q)}D_{p+q-r,r}(V)$,

where $D_{\lambda_1,\ldots,\lambda_m}(V) = h_{\lambda_1,\ldots,\lambda_m}V^{\otimes n}$ for $\lambda_1\geq\lambda_2\geq...\geq 0$ a partition of $n$ and $h_{\lambda}$ the Young symmetrizer for this partition (where $m=dim(V)$, so in our particular example $m=2$). If $\lambda_m < 0$ we define $D_{\lambda_1,\ldots,\lambda_m}(V) = D_{\lambda_1-\lambda_m,\ldots,\lambda_{m-1}-\lambda_m,0}(V) \otimes det^{\lambda_m}$, with $det$ the determinant module $\cong \Lambda^mV$. Finally note that $D_{p,0}(V) = S^p(V) = C_p^*$.

Sorry for all the information, I'm not sure how much of this is standard and standard notation. So, that's basically all the information I have, and all I can see to do is the following:

$C_p \otimes C_q \cong (D_{p,o}(V) \otimes D_{q,0}(V))^* \cong \left(\bigoplus \limits_{r=0}^{min(p,q)}D_{p+q-r,r}(V)\right)^* \cong \bigoplus \limits_{r=0}^{min(p,q)}D_{p+q-r,r}(V)^* \cong \bigoplus \limits_{r=0}^{min(p,q)}(D_{p+q-2r,0}(V) \otimes det^{r})^* = \bigoplus \limits_{r=0}^{min(p,q)}C_{p+q-2r}\otimes (det^r)^*$. Thus any map to $C_{p+q-2r}$, if I've done that right, is a map $\bigoplus \limits_{r=0}^{min(p,q)}C_{p+q-2r}\otimes det^{-r} \to C_{p+q-2r}$.

Even if I have done the right thing, it isn't clear to me how on earth we then deduce that any such map must be a multiple of $\tau_r$; is this some sort of application of Schur's lemma? Your thoughts would be appreciated. If anything is unclear from my question (as I said, I forget what is standard and what is not), then please let me know and I will clarify.

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up vote 2 down vote accepted

One consequence of the Clebsch-Gordan theorem for $SL_2(\mathbb C)$ is that in the decomposition of $C_p\otimes C_q$ as a direct sum of $G$-modules each $C_w$ appears at most once. It follows from this and Shur's lemma that $$\dim_{\mathbb C}\hom_G(C_p\otimes C_q, C_w)\leq1$$ for all $p$, $q$, $w\geq0$. In particular, since you know that $\tau_r$ is a non-zero $G$-module map $C_p\otimes C_q\to C_{p+q-2r}$, it follows that the space $\hom_G(C_p\otimes C_q, C_{p+q-2r})$ is non-zero and spanned by $\tau_r$.

This is precisely what you wanted to show.

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Thanks Mariano! Looks like I was almost there. – Spyam Apr 30 '12 at 6:38

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