Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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!!

share|improve this question

3 Answers 3

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.

share|improve this answer

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.

share|improve this answer
3  
«at this point it's a matter of combinatorics»: great last words! :) –  Mariano Suárez-Alvarez Jan 24 '12 at 6:53
1  
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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.