[Math] Symmetric tensor product of bosonic/fermionic Hilbert space

co.combinatoricslie-groupsmp.mathematical-physicsplethysmrt.representation-theory

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?

Best Answer

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.