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 $\rho$.

  • 2
    $\begingroup$ I would suggest to add "plethysm" tag. $\endgroup$
    – Sasha
    Oct 3, 2012 at 18:11
  • 1
    $\begingroup$ computing a few examples in SAGE suggests that the answer is definitely yes. $\endgroup$ Oct 3, 2012 at 18:21
  • $\begingroup$ @Dan Petersen: This sounds encouraging, many thanks. Unfortunately I do not know SAGE. $\endgroup$
    – asv
    Oct 3, 2012 at 18:53
  • $\begingroup$ I would be interested to learn more about the connection with curvature tensors. Does anyone have a reference? $\endgroup$ Oct 4, 2012 at 9:24
  • 1
    $\begingroup$ @Mark Wildon: some discussion of the connection can be found in \S 10.3 of "Symmetry, Representations, and Invariants" by Goodman and Wallach. Another place is "Einstein manifolds" by Besse, Ch. 1, paragraphs G,H. $\endgroup$
    – asv
    Oct 4, 2012 at 11:56

1 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

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


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

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 $1$.

  • $\begingroup$ This is nice. $\,$ $\endgroup$ Oct 4, 2012 at 6:55
  • $\begingroup$ That's a great answer. $\endgroup$ Oct 4, 2012 at 13:11

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