Decomposing tensor products of $\mathfrak{so}(2n)$-representations 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?
 A: 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.
