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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 $\rho$.

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I would suggest to add "plethysm" tag. –  Sasha Oct 3 '12 at 18:11
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computing a few examples in SAGE suggests that the answer is definitely yes. –  Dan Petersen Oct 3 '12 at 18:21
    
@Sasha: Done. Thank you. –  semyon alesker Oct 3 '12 at 18:52
    
@Dan Petersen: This sounds encouraging, many thanks. Unfortunately I do not know SAGE. –  semyon alesker Oct 3 '12 at 18:53
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@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. –  semyon alesker Oct 4 '12 at 11:56

1 Answer 1

up vote 12 down vote accepted

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 $1$.

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This is nice. $\,$ –  Dan Petersen Oct 4 '12 at 6:55
    
Great!! Thanks a lot! –  semyon alesker Oct 4 '12 at 11:48
    
That's a great answer. –  Abdelmalek Abdesselam Oct 4 '12 at 13:11

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