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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.


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 $$ \operatorname{Sym}^{2} V^{*} = \mathcal{H}_2 \oplus \mathcal{H}_0. $$


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 $G$?

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For small $n = \dim V$, you can compute these decompositions using the Weyl Character Formula. For large $n$, I think this becomes computationally slow. Note that you always have the decomposition into the various characters of the symmetric group, which acts by permuting the copies of $W$. But for $G = O(n)$, these are usually not irreducible as $G$-reps. For example, I believe that there is a copy of $W$ in $W \otimes W$. –  Theo Johnson-Freyd Mar 23 '10 at 19:20
@Theo: Can you point out the proper reference for using Weyl Character Formula to decompose this tensor product? Thank you! –  Shizhuo Zhang Apr 20 '10 at 0:44

1 Answer 1

up vote 3 down vote accepted

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.

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As I said in the question, my highest weight theory is a bit rusty. Can you explain the notation you are using? –  Andrea Ferretti Mar 24 '10 at 0:34
I believe the notation is that of Young tableaux but written in terms of partitions, so that [4] stands for the Young tableau with four horizontal boxes. In other words, it's $\mathrm{Sym}^4V$ with all traces removed. –  José Figueroa-O'Farrill Mar 24 '10 at 1:41
It should be possible to explain the multiplicities as well using a version of Pieri's rule. –  Bruce Westbury Mar 24 '10 at 6:33

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