A question on invariant theory of $GL_n(\mathbb{C})$.

2012-10-03T17:31:23Z

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 by Mark Wildon for A question on invariant theory of $GL_n(\mathbb{C})$.

2012-10-04T00:42:22Z

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 $1$.

A question on invariant theory of $GL_n(\mathbb{C})$. - MathOverflow

A question on invariant theory of $GL_n(\mathbb{C})$.

Answer by Mark Wildon for A question on invariant theory of $GL_n(\mathbb{C})$.