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

I just want to consider the simplest case:

Let S=[0,0,0,0,1], how to derive the general formula for $Sym^k$ S?

My conjectured formula based on the results of LiE program for finite k values is:

$Sym^k(S)=\overset{\left[k/2\right]}{\underset{i=0}{\oplus}}[i,0,0,0,k-2i]$

But I have no clue how to prove it or even a derivation for the simplest case when k=2.

Any reference or tips would be greatly appreciated.

PS. Sorry for confusion. Now I changed the spinor representation "V" to "S".

share|improve this question
2  
I guess this is one of the two spin reps since $Sym^2(W)$ would contain the trivial rep. if $W$ is the $10$-dimensional rep. –  Eric Rowell Jun 5 '10 at 23:32
    
Could you clarify the notation? I take it that V=[0,0,0,0,1] is the 10 dimensional vector representation. Then I would expect [1,0,0,0,0] to be one of the two spin representations. Except you will not get a spinor in $Sym^k V$. The case k=2 is straightforward. You get [0,0,0,0,2] and the trivial representation which is [0,0,0,0,0]. –  Bruce Westbury Jun 5 '10 at 23:35
1  
(I wish I could edit comments!) Eric's guess is spot on: in Lie, with default Lie algebra of type D5, the representation with Dynkin labels [0,0,0,0,1] is one of the 16-dimensional complex half-spinor representations. The other, which is the complex conjugate representation, is [0,0,0,1,0]. –  José Figueroa-O'Farrill Jun 6 '10 at 0:00
    
Thanks. The case k=2 (and all spin representations) can be found in Adams posthumous book. I still find the formula in the question surprising. –  Bruce Westbury Jun 6 '10 at 0:12
add comment

3 Answers

up vote 7 down vote accepted

Based on comments from Eric Rowell and José Figueroa-O'Farrill, I assume that your $V$ is a half-spinor representation of $Spin_{10}$. The key issue is that your hypothetical decomposition of $S(V)$ is multiplicity-free, i.e. any module that appears in it has multiplicity one. There is a beautiful theory of multiplicity-free actions that allows you to do the following.

(1) Test that $V$ is a multiplicity-free $G$-module.

(2) Find the highest weights of the simple summands in $S(V)$.

For (1), the action is multiplicity-free if and only if the Borel subgroup $B$ has an open orbit on $V$. For (2), the highest weights form a free commutative semigroup whose generators are determined from the stabilizer in $B$ of a point in the open orbit.

An excellent survey is Roger Howe's Schur lectures.

All multiplicity-free linear actions (i.e. modules) of reductive algebraic groups have been classified. A convenient reference is

Howe and Umeda, The Capelli identity, the double commutant theorem and multiplicity-free-actions, Mathematische Annalen, 290, 565 - 620 (freely available through GDZ, but tricky to link to).

Item (x) of the list (11.0.1) on p.583 is your action $Spin_{10}\times GL_1.$ This module arises from the action of a Levi factor of a maximal parabolic of $E_6$ on its abelian nilradical (remove last node from the Dynkin diagram of $E_6$ to get $D_5$). The decomposition is described in 11.10 on p.602: there are two fundamental heighest weights, one in degree 1 corresponding to the Spin module itself and one in degree 2 corresponding to the standard 10-dimensional representation of $SO_{10}.$ This leads to the decomposition you have stated.

share|improve this answer
add comment

I think you can probably prove this by induction using Littelmann paths. Since the (16 dimensional) spin rep is minuscule, you can at least figure out the tensor powers by adding up weights of the form $(\pm 1/2,\ldots,\pm 1/2)$ where you take an even number of $-$ signs and throwing away any sums that are not in the dominant Weyl chamber.

Edit: So you can calculate the weights of the irreps. in $V^{\otimes 2}$ this way. Then use they Weyl dimension formula to pick the reps. whose dimension adds up to $136$. This will give you the $k=2$ case anyway.

share|improve this answer
    
This is a good way to decompose a tensor power. I don't see how to pick out the symmetric power. –  Bruce Westbury Jun 6 '10 at 0:17
    
@Bruce: one way is to cheat and use dimensions of the irreps to determine which ones add up to the dimension of $Sym^k(V)$. –  Eric Rowell Jun 6 '10 at 0:44
1  
There are philosophical reasons why methods such as Littelmann paths and crystal bases cannot be easily adapted to compute decomposition of symmetric powers. –  Victor Protsak Jun 6 '10 at 2:12
add comment

I'm not familiar with your notation, but is $V$ the standard representation of $\text{SO}_{10}\mathbb{C}$? If so, $\text{Sym}^k V$ is the representation of $\text{GL}_{10}\mathbb{C}$ corresponding to the partition $\lambda=(k,0,...,0)$. Then you can apply Littlewood's formula to see how this irreducible decomposes when you restrict from $\text{GL}_{10}\mathbb{C}$ to $\text{SO}_{10}\mathbb{C}$: the multiplicity of the irreducible representation $V_{[\mu]}$ corresponding to the partition $\mu$ will be

$\sum_\eta C_{\eta\mu}^\lambda$

where $C_{\bullet\bullet}^\bullet$ is the Littlewood-Richardson coefficient, and the sum is over all partitions $\eta=(\eta_1,\eta_2,\eta_3,\eta_4,\eta_5)$ with each $\eta_i$ even. (Fulton-Harris "Representation theory", Equation 25.37, p. 427) Hopefully you can check whether this agrees with your expected answer.

share|improve this answer
    
$V = S$ is a half-spinor representation. –  Steven Sam Jun 26 '13 at 6:51
add comment

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.