I guess this question only requires standard knowledge, but I'm a bit rusty with highest weight theory. I'm trying to catch up, but maybe I don't need the theory in full generality.


**Background**


Let $V$ be a Euclidean space of dimension $n$, and consider the representations of the group $G = O(V) \cong O(n)$. If I'm not wrong we can decompose
$$
\operatorname{Sym}^2 V = W \oplus Z,
$$
where $Z$ is the trivial one-dimensional representation and $W$ is an irreducible representation of $G$.


**More background**


This is how I can see the above decomposition. I don't know if it is of any use for the question itself, so feel free to skip it.


It is enough to decompose $\operatorname{Sym}^2 V^{\*}$, that is, degree $2$ homogeneous polynomials on $V$. Degree $n$ homogeneous polynomials are a representation of $G$ via
$$
A.f(x) = f(Ax),
$$
where $A$ is an orthogonal matrix and $f \in \operatorname{Sym}^n V^{\*}$.


Now for any $n$ we have a $G$-morphism $f : \operatorname{Sym}^n V^{\*} \to \operatorname{Sym}^{n + 2} V^{\*}$ which is given by multiplication by $x^2 = x\_1^2 + \cdots + x\_n^2$. This exhibits $\operatorname{Sym}^n V^{\*}$ as a subrepresentation of $\operatorname{Sym}^{n + 2} V^{\*}$.


The complement is easily found. The invariant scalar product on $\operatorname{Sym}^n V^{\*}$ is given by $(f, g) = f(D)g(x)$, where $D$ is the derivation operator. From this one finds that the adjoint of $f$ is the laplacian $\Delta$.


So one can decompose
$$
\operatorname{Sym}^{n + 2} V^{\*} = \operatorname{Sym}^{n} V^{\*} \oplus \mathcal{H}\_n,
$$
where $\mathcal{H}\_n$ is the space of harmonic homogeneous polynomials of degree $n$. If I recall well, $\mathcal{H}\_n$ is irreducible.


The decomposition which interests me is then
$$
\operatorname{Sym}^{2} V^{\*} = \mathcal{H}\_2 \oplus \mathcal{H}\_0.
$$


**Problem**


I'd like to understand how to decompose the tensor products of $W$; in particular



> 
> What is the decomposition of $W \otimes W$ and $W \otimes W \otimes W$ into irreducible representations of $G$?
> 
> 
> 



Decomposing a tensor product
For $SO(n)$ a calculation using LiE gives:  

(using partition notation so $W$ is [2])  

and assuming $n$ is not small


For $W\otimes W$, [4],[3,1],[2,2],[2],[1,1],[]  

(all with multiplicity one) 


and for $W\otimes W\otimes W$,  

1.[6] 2.[5,1] 3.[4,2] 1.[3,3] 1.[4,1,1] 2.[3,2,1] 1.[2,2,2] 3.[4] 6.[3,1] 2.[2,2] 3.[2,1,1] 6.[2] 3.[1,1] 1.[]


The same works for $Sp(n)$ by taking conjugate partitions. 


There is also a relationship with $SL(n)$.


This is taking your question at face value. If it is understanding you're after instead then the best approach is to use crystal graphs.


The notation I have used denotes a representation by a partition. I have put $m.$ in front to denote multiplicity is $m$. A partition is $[a\_1,a\_2,a\_3,...]$ where $a\_i\ge a\_j$ if $i$ less than $j$. To convert to a highest weight vector add the appropriate number of $0$s to the end.
Then take $[a\_1-a\_2,a\_2-a\_3,a\_3-a\_4,...]$. This gives a dominant integral weight. The fundamental weights are the partitions $[1,,,,1]$. If this has length $k$ this corresponds to the $k$-th exterior power of the vector representation (provided $2k-1$ less than $n$).


In particular the trivial representation is $[]$, the vector representation $V$ is $[1]$, the exterior square of $V$ is $[1,1]$, the symmetric square is $[2]+[]$.


For the $k$-th tensor power of $W$ you will see partitions of $2k-2p$ for $0\le p\le k$ only and it remains to determine the multiplicities (possibly $0$). For $SL(n)$ just take the partitions of $2k$ (with their multiplicities) and ignore the rest.


