A question on invariant theory of $\mathrm{GL}_n(\mathbb{C})$

Question

$\DeclareMathOperator\GL{GL}\DeclareMathOperator\Sym{Sym}$Let $\rho$ denote the irreducible algebraic representation of $\GL_n(\mathbb{C})$ with the highest weight $(2,2,\underset{n-2}{\underbrace{0,\dots,0}})$.

Let $k\leq n/2$ be a non-negative integer. How to decompose into irreducible representations the representation $\Sym^k(\rho)$?

More specifically, I am interested whether $\Sym^k(\rho)$ contains the representation with the highest weight $(\underset{2k}{\underbrace{2,\dots,2}},\underset{n-2k}{\underbrace{0,\dots,0}})$, and if yes, whether the mutiplicity is equal to one.

A a side remark, the representation $\rho$ has a geometric interpretation important for me: it is the space of curvature tensors, namely the curvature tensor of any Riemannian metric on $\mathbb{R}^n$ lies in $\rho$.

Answer 1

The plethysm $\mathrm{Sym}^k \rho$ contains the irreducible representation with highest weight $(2,\ldots,2,0,\ldots,0)$ exactly once. It looks like a tricky problem to say much about its other irreducible constituents.

Let $\Delta^\lambda$ denote the Schur functor corresponding to the partition $\lambda$, and let $E$ be an $n$-dimensional complex vector space. Using symmetric polynomials (or other methods) one finds

$$\mathrm{Sym}^2 (\mathrm{Sym}^2 E) = \Delta^{(2,2)}E \oplus \mathrm{Sym}^4 E.$$

Therefore

$$ \mathrm{Sym}^k \mathrm{Sym}^2 \mathrm{Sym}^2 E \cong \sum_{r=0}^k \mathrm{Sym}^r (\Delta^{(2,2)}E) \otimes \mathrm{Sym}^{k-r} (\mathrm{Sym}^4 E) .$$

The irreducible representations contained in the $r$th summand are labelled by partitions with at most $2r+(k-r) = k+r$ parts. So to show that $\mathrm{Sym}^k(\Delta^{(2,2)}(E))$ contains $\Delta^{(2^{2k})}E$, it suffices to show that $\Delta^{(2^{2k})}E$ appears in $\mathrm{Sym}^k \mathrm{Sym}^2 \mathrm{Sym}^2 E$.

Let $U = \mathrm{Sym}^2 E$. There is a canonical surjection

$$ \mathrm{Sym}^k (\mathrm{Sym}^2 U ) \rightarrow \mathrm{Sym}^{2k} U. $$

given by mapping $(u_1u_1')\ldots (u_ku_k') \in \mathrm{Sym}^k (\mathrm{Sym}^2 U )$ to $u_1u_1'\ldots u_ku_k' \in \mathrm{Sym}^{2k} U$. Therefore $\mathrm{Sym}^k (\mathrm{Sym}^2 U )$ contains $ \mathrm{Sym}^{2k} U = \mathrm{Sym}^{2k} (\mathrm{Sym}^2 E)$. It is well known that

$$ \mathrm{Sym}^{2k} (\mathrm{Sym}^2 E) = \sum_{\lambda} \Delta^{2\lambda}(E) $$

where the sum is over all partitions $\lambda$ of $2k$ and $2(\lambda_1,\ldots,\lambda_m) = (2\lambda_1,\ldots, 2\lambda_m)$. Taking $\lambda = (1^{2k})$ we see that $\Delta^{(2^{2k})}E$ appears.

It remains to show that the multiplicity of $\Delta^{(2^{2k})}E$ in $\mathrm{Sym}^k (\Delta^{(2,2)}E)$ is $1$. We work over $\mathbb{C}$, so there is a chain of inclusions

$$ \mathrm{Sym}^k (\Delta^{(2,2)}(E)) \subseteq \mathrm{Sym}^k (\mathrm{Sym}^2 E \otimes \mathrm{Sym}^2 E) \subseteq (\mathrm{Sym}^2 E)^{\otimes 2k}.$$

By the Littlewood–Richardson rule (or the easier Young's rule), the multiplicity of $\Delta^{(2^k)}E$ in the right-hand side is $1$.

Answer 2

Here's another answer to the question about multiplicities which can be generalized to other partitions. First, for any partition $\lambda$, let $S_\lambda$ denote the irreducible representation (Schur functor) with highest weight $\lambda$. I'll explain how to get the following:

Theorem: Let $\lambda$ be a partition of even size and let $\mu$ be the result of scaling all column lengths of $\lambda$ by $k$. Then $Sym^k(S_\lambda({\bf C}^n))$ contains $S_\mu({\bf C}^n)$ with multiplicity 1 if $\mu$ has at most $n$ rows (otherwise $S_\mu({\bf C}^n)=0$).

For $\lambda = (2,2)$, scaling all column lengths by $k$ just means the partition $(2,2,2,\dots,2)$ ($2k$ instances of 2).

I'm going to assume basic fluency between symmetric functions and Schur functors.

Then $S_\lambda \otimes S_\mu$ always contains $S_{\lambda + \mu}$ with multiplicity 1: the tensor product of highest weight vectors gives the unique vector (up to scalar multiple) of this weight and it's also a highest weight vector (easy check). In particular, $S_{\lambda}^{\otimes k}$ contains $S_{k\lambda}$ with multiplicity 1. From the product description, the highest weight vector is symmetric so it also belongs to $Sym^k(S_\lambda)$.

Next, we use the involution $\omega$ on symmetric functions. One reference Section I.2 of Macdonald, Symmetric Functions and Hall Polynomials. This has the property that $\omega s_\lambda = s_{\lambda^T}$ where $\lambda^T$ is the transpose partition. It also behaves well with respect to plethysm (there are several kinds, here I refer to the one that corresponds to composition of Schur functors):

If $f$ and $g$ are homogeneous, then

$\omega (f \circ g) = \omega f \circ \omega g$ if $\deg g$ is odd, and

$\omega (f \circ g) = f \circ (\omega g)$ if $\deg g$ is even.

Reference: Macdonald, Example I.8.1(a).

In particular, this means that if $|\lambda|$ is even, then

$\omega (s_k \circ s_\lambda) = s_k \circ s_{\lambda^T}$.

Translating this back to representations:

The right side is $Sym^k(S_{\lambda^T})$ and hence contains $S_{k\lambda^T}$ with multiplicity 1. In particular, $\omega$ is an involution so $Sym^k(S_\lambda)$ contains $S_{(k\lambda^T)^T}$.

We can simplify further: $(k\lambda^T)^T$ is obtained from $\lambda$ by scaling all column lengths by $k$.