8
$\begingroup$

Consider two representation of the group $SU(n)$: $Sym^k(\mathbb{C}^n)$ and $\wedge^k\mathbb{C}^n$ ($k\leq n$) and take their symmetric tensor products: $Sym^2(Sym^k(\mathbb{C}^n))$, $Sym^2(\wedge^k\mathbb{C}^n)$. I have two questions concerning those representations:

Question 1:

It appears that $Sym^2(Sym^k(\mathbb{C}^n))$ and $Sym^2(\wedge^k\mathbb{C}^n)$ are multiplicity free (I checked this for low dimensional cases in Lie). Is that true in general? Is there an easy proof of this fact?

Question 2:

Which irreps of $SU(n)$ will appear in $Sym^2(Sym^k(\mathbb{C}^n))$ and $Sym^2(\wedge^k\mathbb{C}^n)$ respectively? In particular, which young diagrams of the symmetric group $S_{2k}$ will be present?

Improve this question
$\endgroup$
5
  • 3
    $\begingroup$ Isn't the polynomial representation theory of $\mathrm{SU}\left(n\right)$ (i. e., the part which considers only the Schur functors of the vector representation $\mathbb C^n$) the same as that of $\mathrm{GL}\left(n\right)$ ? For $\mathrm{GL}\left(n\right)$, I think both questions are very easy: any irreducible representation tensored with a $\mathrm{Sym}^k\left(\mathbb C^n\right)$ or with a $\wedge^k\left(\mathbb C^n\right)$ is multiplicity-free (by the Pieri and anti-Pieri rules), so in particular the tensor squares of $\mathrm{Sym}^k\left(\mathbb C^n\right)$ and ... $\endgroup$
    – darij grinberg
    Mar 2, 2012 at 1:06
  • $\begingroup$ ... $\wedge^k\left(\mathbb C^n\right)$ are multiplicity-free, and thus so are the symmetric squares. $\endgroup$
    – darij grinberg
    Mar 2, 2012 at 1:06
  • $\begingroup$ Oh, I have been talking about Question 1 only. For question 2, it is important to distinguish the representations occurring the symmetric square from those occurring in the wedge square. I don't know how to do this. $\endgroup$
    – darij grinberg
    Mar 2, 2012 at 1:08
  • $\begingroup$ I might have mixed up $\mathrm{SU}\left(n\right)$ and $\mathrm{U}\left(n\right)$, in which case you would have to replace $\mathrm{GL}\left(n\right)$ by $\mathrm{SL}\left(n\right)$ and get dirty with Young tableaux (two different Young tableaux can induce the same representation of $\mathrm{SL}\left(n\right)$, but only if they are "shifted" w.r.t. each other, and this should be manageable). $\endgroup$
    – darij grinberg
    Mar 2, 2012 at 1:14
  • $\begingroup$ Thank you Darji. Could you please provide some reference when the relation between the product of two Schur functions and decomposition onto corresponding irreps is discussed? $\endgroup$
    – Michał Oszmaniec
    Mar 19, 2012 at 14:16

1 Answer 1

Reset to default
6
$\begingroup$

For question two, you are asking about a composition of two Schur functors, i.e. a plethysm. More specifically you want to know $h_2 \circ h_k$ and $h_2 \circ e_k$. These can be found in Example 9 in the section on plethysm in Symmetric functions and Hall polynomials: $$ h_2 \circ h_k = \sum_{j \text{ even}} s_{(2k-j,j)}$$ and $$ h_2 \circ e_k = \sum_{j \text{ even}} s_{(k+j,k-j)^T}$$ where $(\cdot)^T$ denotes the transposed Young diagram, and the sum is taken over those even $j$ that make the subscript a valid Young diagram. These translate to universal identities between representations, i.e. an isomorphism of functors. They express how the composed functors $\mathrm{Sym}^2(\mathrm{Sym}^k(-))$ and $\mathrm{Sym}^2(\bigwedge^k(-))$ can be written as direct sums of Schur functors.

When you apply a Schur functor to the defining representation of $\mathrm{SU}(n)$, the result is nonzero if and only if the corresponding partition has at most $n$ parts. Think about how the functor $\bigwedge^k$ vanishes on any vector space of dimension less than $k$. This explains the observation you make in a comment below, that not all terms in the second sum will appear if $n$ is small.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks! This is exactly what I was suspecting. Yet I didn't know the reference to the proper terminology is used in this context. As for your answer I think you have to assume that in the second case $n+j\leq d$, where $d$ is the number that appears in definition of $SU(d)$ (If I understood you correctly $n$ in your answer is $k$ from my question). $\endgroup$
    – Michał Oszmaniec
    Mar 4, 2012 at 10:10
  • 1
    $\begingroup$ You're right, my $n$ was your $k$. I changed the notation. I also added some hopefully clarifying remarks. $\endgroup$
    – Dan Petersen
    Mar 4, 2012 at 11:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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