I've been trying to show that the tensor product $V\otimes V$ of a finite dimensional $\mathfrak{so}(2n)$ representation of dimension at least 3 decomposes as a direct sum of at least 3 irreducible representations. I've been trying to use the Steinberg formula as in Humphreys section 24 but I feel like it's too complicated to do so for arbitrary $n$ or I'm just missing something. Is this the way to go or is there a simpler proof maybe using highest weight vectors?
Decomposing tensor products of $\mathfrak{so}(2n)$-representations
lie-algebrasrepresentation-of-algebrasrepresentation-theorysemisimple-lie-algebras
Related Solutions
The statement "Remove a copy of this irreducible representation from $\mathbb{C}^3\otimes V_{1,1}$" just means remove the weights occurring in the irreducible representation from those appearing in your representation, and see what's left. In your case, the irreducible representation with highest weight (1,1) is 8-dimensional (and is actually isomorphic to the adjoint representation), and its weights are listed in Section 6.5 of Hall: they are (1,1), (-1,2), (2,-1), (0,0) (twice), (1,-2), (-2,1), and (-1,-1). Removing them from your list of weights leaves only (0,0), which is the weight of the 1-dimensional trivial representation (with highest weight (0,0)). So your tensor product representation decomposes into the 8-dimensional adjoint representation plus a 1-dimensional trivial representation. This is sometimes written as $3\otimes\overline{3} = 8\oplus 1$.
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.
Best Answer
Combining some of these comments into an answer. It is always true that the tensor square breaks up into (at least) an alternating part and a symmetric part. This is because the action of the representation commutes with the action of the symmetric group $S_2$ swapping the order of the tensor product (for higher tensor powers this becomes more complicated and that's what young tableaux are for). Thus this is true for any Lie group action on a tensor square. These may then decompose even further of course.
If the representation is self-dual there is a $\mathfrak{g}$-invariant bilinear form (it is not too hard to see that this forces it to be symmetric or symplectic in fact). We can think of this form as an element of $V\otimes V$ and $\mathfrak{g}$-invariance means that its span is a trivial representation. As we've noted this should live in one of our symmetric/alternating representations.
Now for $n$ even, all the representations are self-dual. For $n$ odd, this is only true if the multiplicities of the last two fundamental weights in the highest weight are equal.
A quick check shows that it doesn't hold for $n$ odd: $\mathfrak{so}(6)=\mathfrak{sl}(4)$ and one of the half spin representations here is isomorphic to the usual $4$-dimensional $\mathfrak{sl}(4)$ representation (the other to its dual). The tensor product of this with itself decomposes into only $2$ irreducible representations: the symmetric and alternating parts.