Let $\omega_1$ and $\omega_1$ be fundatmentl weights and $V(\lambda)$ be the unique irreducible representations of highest weight $\lambda$ of $\mathfrak{sl_3}$.
I want to decompose, for $m>n$,
$V(m\omega_1) \otimes V(n\omega_2)$.
I have been given the hint to consider symmetric powers of 'the obvious modules'. I know that $Sym^a(V(\omega_i))=V(a\omega_i)$ but I am not sure how to employ that here as I am not sure how symmetric powers interact with the tensor product.
Best Answer
Question: "I want to decompose, for $m>n$, $V(m\omega_1) \otimes V(n\omega_2)$. I have been given the hint to consider symmetric powers of 'the obvious modules'. I know that $Sym^a(V(\omega_i))=V(a\omega_i)$ but I am not sure how to employ that here as I am not sure how symmetric powers interact with the tensor product."
Comment: Let $\Gamma_{(a,b)} \subseteq Sym^a(V) \otimes Sym^b(\wedge^2 V)$ be the irreducible module on $SL_3$ corresponding to $\lambda:=(a,b)$ for $a,b \geq 0$. It follows $\Gamma_{(0,1)}:=\wedge^2 V$ where $V$ is the "standard representation" of $SL_3$.
You may find this in the Fulton/Harris book:
There is a canonical contraction map
$$i_{a,b}: Sym^a(V) \otimes Sym^b(\wedge^2 V) \rightarrow Sym^{a-1}(V) \otimes Sym^{b-1}(\wedge^2 V)$$
with kernel $\Gamma_{(a,b)}$. Hence there is a direct sum decomposition
$$Sym^a(V) \otimes Sym^b(\wedge^2 V) \cong \Gamma_{(a,b)} \oplus Sym^{a-1}(V) \otimes Sym^{b-1}(\wedge^2 V).$$
Example: In the case of the map from the comments: There is a canonical surjective map
$$i_{2,1}:Sym^2(V) \otimes V^* \rightarrow V$$
defined by
$$i_{2,1}(uv \otimes f):=f(u)v+f(v)u.$$
The kernel $\Gamma_{(2,1)}$ has highest weight vector $e_1^2 \otimes (e_1 \wedge e_2)= e_1^2 \otimes x_3$.
In the book they prove the following general formula:
$$dim(\Gamma_{(a,b)}) = \frac{(a+b+2)(a+1)(b+1)}{2}$$
and this formula implies there is an exact sequence of $SL_3$-modules
$$(*)0 \rightarrow \Gamma_{(a,b)} \rightarrow Sym^a(V) \otimes Sym^b(\wedge^2 V) \rightarrow Sym^{a-1}(V) \otimes Sym^{b-1}(\wedge^2 V) \rightarrow 0 .$$
When you count the dimensions in $(*)$ you get the following:
$$\binom{a+2}{2}\binom{b+2}{2}-\binom{a+1}{2}\binom{b+1}{2}=$$
$$\frac{(a+1)(b+1)}{4}((a+2)(b+2)-ab)= $$
$$\frac{(a+b+2)(a+1)(b+1)}{2} = dim(\Gamma_{(a,b})$$
hence the sequence $(*)$ above is exact (this is exercise 15.19 in the mentioned book).
Question: "How does this apply directly to solve my question? – RickSmith"
Answer: Since $SL_3$ is semi simple it follows any short exact sequence of representations split hence there is by $(*)$ an isomorphism
$$Sym^a(V) \otimes Sym^b(\wedge^2 V) \cong \Gamma_{(a,b)} \oplus Sym^{a-1}(V) \otimes Sym^{b-1}(\wedge^2 V).$$
Your formula follows using an induction.