# Trivial summand of a representation's symmetric power

The following comes from Exercise 13.17 of Fulton and Harris's book, Representation Theory: A First Course.

Let $V$ denote the standard representation of $\mathfrak{sl}_3\mathbb{C}$, with weights $L_1,L_2$, and $L_3$.

I would like to show that the only symmetric powers $\operatorname{Sym}^n(\operatorname{Sym}^2V)$ that contains trivial summands (e.g. $\mathbb{C}$ appears when decomposed into irreducible representations) are when $n=3k$ ($n$ is divisible by $3$), and that the trivial summand here is just the $k$-th power of the original trivial summand. So, for example, $$\operatorname{Sym}^3(\operatorname{Sym}^2V)=\operatorname{Sym}^6V\oplus \Gamma_{2,2}\oplus \mathbb{C},$$ where $\Gamma_{2,2}$ is the irreducible rep. with highest weight $2L_1-2L_3$. The goal is to say something about the appearance of $\mathbb{C}$'s for $n=3k$.

I have been hitting my head against the wall trying to do this with weight diagrams (finding $n$-wise sums of the weights of $\operatorname{Sym}^2V$ given by $\{L_i+L_j,2L_i\}$), and am having trouble finding a general pattern to the decomposition of $\operatorname{Sym}^n(\operatorname{Sym}^2V)$. I also know that the dimension of $\operatorname{Sym}^n(\operatorname{Sym}^2V)$ is ${n+5}\choose n$ and the dimension of each irreducible rep. with highest weight $aL_1-bL_3$ is $\dim(\Gamma_{a,b})=(a+1)(b+1)(\frac{a+b+2}{2})$, but I do not think this helps much.

If you look at this from an algebraic geometric standpoint, apparently the existence of this trivial summand in the decomposition above tells us that there exists a cubic hypersurface $X$ in $\mathbb{P}^5$ preserved under all automorphisms of $\mathbb{P}^5$ carrying the Veronese surface to itself. But my background here is lacking, and this is all Greek to me.

If anyone could possibly recommend a better way to approach this problem, I would greatly, greatly appreciate it.

• The weights of $Sym^2V$ are not in the root lattice. The root lattice is an index three subgroup in the weight lattice. Therefore the weights appearing in $Sym^n(Sym^2V)$ are in the root lattice if and only if $n$ is a multiple of three. The trivial rep has only one weight, zero, so for there to be trivial summands it is necessary that $3\mid n$. I don't see an easy way of finding the multiplicity of the trivial rep as a composition factor. At least not yet. – Jyrki Lahtonen Apr 6 '13 at 7:34
• Sorry, that was more like "stating the obvious". Surely you made the same observation while studying the weight diagrams. – Jyrki Lahtonen Apr 6 '13 at 7:40
• Yes, I do see that. But it is not entirely obvious to me that the weights being in the root lattice implies that the decomposition must contain a trivial summand, and for this it seems we need the multiplicity at zero. I have been trying to find the latter in a combinatorial fashion (finding all possible $n$-wise sums that sum to zero), but even so, wouldn't we still need the full decomposition to find out how many trivial summands are left after we take out all the irreducibles? – ff90 Apr 6 '13 at 17:11
• I agree. It isn't at all obvious there should be a trivial summand. In general the fact that all the weights are in the root lattice does not imply the existence of a trivial summand (a simple module of highest weight in the root lattice is a case in point). This is an interesting question actually. – Jyrki Lahtonen Apr 6 '13 at 17:17

Sym$^2(V)$ decomposes into one dimensional eigenspaces with respect to the given cartan subalgebra and has eigenvalues
$$E = \{2L_2,2L_1,2L_3,L_1+L_2,L_2+L_3,L_1+L_3\}$$
The eigenvalues of Sym$^n(V)$ are $E_n = \{a_1+\cdots+a_n\;|\;a_1,\cdots,a_n\in E\}$. The number of times $\mathbb{C}$ appears is the difference between multiplicity of eigenvalue 0 and eigenvalue $L_1-L_2$ in $E_n$. So this becomes a purely combinatorial question.