# Symmetric tensor product of bosonic/fermionic Hilbert space

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?

• 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 ... – darij grinberg Mar 2 '12 at 1:06
• ... $\wedge^k\left(\mathbb C^n\right)$ are multiplicity-free, and thus so are the symmetric squares. – darij grinberg Mar 2 '12 at 1:06
• 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. – darij grinberg Mar 2 '12 at 1:08
• 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). – darij grinberg Mar 2 '12 at 1:14
• 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? – Michał Oszmaniec Mar 19 '12 at 14:16

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.
• 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). – Michał Oszmaniec Mar 4 '12 at 10:10
• You're right, my $n$ was your $k$. I changed the notation. I also added some hopefully clarifying remarks. – Dan Petersen Mar 4 '12 at 11:45