# Simple question in the representation of SL(2,C)

Let $V$ the standard two dimensional representation of SL(2,C). The Fulton's book in representation theory say in pag 156 that $Sym^3(Sym^2V)=Sym^6(V) \oplus Sym^2(V)$. In the excercises 11.23, the books asks to prove the decomposition
$$Sym^3(Sym^3V) = Sym^9(V) \oplus Sym^5(V) \oplus Sym^3V$$

In my work, I found $Sym^5(Sym^3V)$,and $Sym^k(Sym^3V)$, So I was looking for a similar decomposition. I am not familiar enought with the theory, and to study the subject will take a little to far from my current work. So I decided to ask here (sorry if the question is too simple).

Thanks for any help!!

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You're looking at plethysm of $SL_2(\mathbb{C})$-modules. According to a paper of Manivel (An extension of the Cayley-Sylvester formula, 2008) the answer is given by the Cayley-Sylvester formula. In your case it states that the multiplicity of $Sym^e(V)$ in $Sym^n(Sym^3(V))$ is $$Par(n,3;(3n-e)/2) - Par(n,3;(3n-e)/2 - 1),$$ where $Par(n,k,m)$ is the number of partitions in an $n$-by-$k$ box of size $m$. For example, if $n=3$ and $e=5$ then $Par(3,3;2)=2$ and $Par(3,3;1) = 1$, which agrees with what you have above.

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Highest weight theory is ideally suited to answer just this sort of question. Here's how to figure out your problem.

1. First recall that $Sym^k(V)$ is the irreducible representation of highest weight $k$. So, it has weight spaces with weights $-k,-k+2,\ldots,k-2,k$ occurring with multiplicity one. In particular, $Sym^3(V)$ has weights $-3,-1,1,3$.

2. Then the weights occurring in $Sym^k(Sym^3(V))$ correspond to all possible ways of adding $k$ of the weights $-3,-1,1,3$ together. For example, $3k$ will be a weight occurring, and in fact it will be the highest weight, corresponding to the fact that $Sym^{3k}(V)$ will always be a direct summand of $Sym^k(Sym^3(V))$.

3. Having found all the weights, you know need to know the multiplicity with which they occur. For example, the reason that $Sym^3(Sym^2(V)) = Sym^6(V) \oplus Sym^2(V)$ is that the weights $-6,-4,\ldots,6$ all occur, but the weights $-2,0,2$ all occur with multiplicity two.

I am not sufficiently motivated to write out a formula for $Sym^k(Sym^3(V))$ right now, but at this point it's a matter of combinatorics.

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«at this point it's a matter of combinatorics»: great last words! :) – Mariano Suárez-Alvarez Jan 24 '12 at 6:53
If insufficiently motivated, you can always ask the computer (young.sp2mi.univ-poitiers.fr/cgi-bin/form-prep/marc/…) to do the work (available from www-math.univ-poitiers.fr/~maavl/LiE/form.html). Of course you'll need to look at the source code to understand how the answer was obtained. – Marc van Leeuwen Jan 24 '12 at 14:53

This is a bit of expansion on the answer of Mike Skirvin; in particular it gives one way of explicitly calculating the combinatorics involved. My previous answer, although correct in its result, is horribly roundabout and overly computational, a mathematical Rube Goldberg machine if you will; so after waking up this morning I realized there is a much easier approach using a recursion on Symmetric powers.

Define:

$L = U^3 + U + U^{-1} + U^{-3}$

$M = U^4 + U^2 + 2 + U^{-2} + U^{-4}$

Now define a sequence $S_i$ by:

$S_{-2} = 0$

$S_{-1} = 0$

$S_0 = 1$

$S_1 = L$

$S_k = L\cdot S_{k-1} - M\cdot S_{k-2} + L\cdot S_{k-3} - S_{k-4}$ for $k\geq 2$.

Note that the exponents of $U$ in $S_1$ are exactly the weights mentioned by Mike in his comment (1). It turns out the same is true for all the $S_k$: the exponents of $U$ in $S_k$ are exactly the set of weights of $Sym^k(Sym^3(V))$ and the coefficient of $U^\ell$ is exactly the multiplicity of the weight $\ell$ in $Sym^k(Sym^3(V))$.

From this, you can pick out the subrepresentations by looking at where coefficients change; since the weights of any $Sym^\ell(V)$ occur with multiplicity 1, the only time the coefficients change is when a new summand occurs.

For example, working out $Sym^3(Sym^3(V))$ one gets the following expression:

$U^9 + U^7 + 2U^5 + 3U^3 + 3U + 3U^{-1} + 3U^{-3} + 2U^{-5} + U^{-7} + U^{-9}$

For the module corresponding to the leading coefficient, subtract 1 from each exponent giving a copy of $Sym^9(V)$ and leaving:

$U^5 + 2U^3 + 2U + 2U^{-1} + 2U^{-3} + U^{-5}$

Repeat this process to pull out a copy of $Sym^5(V)$ and finally a copy of $Sym^3(V)$; there are no more terms left, so this is the complete decomposition of $Sym^3(Sym^3(V))$. In general, the expression for $Sym^k(Sym^3(V))$ in $U$ so obtained is of the form:

$a_0U^{3k} + a_2U^{3k-2} + a_4U^{3k-4} + ... + a_4U^{-3k+4} + a_2U^{-3k+2} + a_0U^{-3k}$

Then $a_0 = 1$ by Mike's comment (2) and the multiplicity of $Sym^\ell(V)$ for $\ell\geq 0$ in the decomposition is just $(a_\ell - a_{\ell+2})$ and the multiplicity of $Sym^{3k}(V)$ is 1 since $a_{-2} = 0$.

As for the recursion, it ultimately expresses symmetric powers in terms of lower symmetric powers and exterior powers; this can be proven using multiplication of Young diagrams and inclusion-exclusion although I don't have a good reference at hand.

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