Question

Let $S^{\lambda}$ be a Schur functor. Is there a known positive rule to compute the decomposition of $S^{\lambda}(\bigwedge^2 \mathbb{C}^n)$ into $GL_n(\mathbb{C})$ irreps?

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In response to Vladimir's request for clarification, the ideal answer would be a finite set whose cardinality is the multiplicity of $S^{\mu}(\mathbb{C}^n)$ in $S^{\lambda}(\bigwedge^2 \mathbb{C}^2)$. As an example, the paper Splitting the square of a Schur function into its symmetric and anti-symmetric parts gives such a rule for $\bigwedge^2(S^{\lambda}(\mathbb{C}^n))$.

Formulas involving evaluations of symmetric group characters, or involving alternating sums over stable rim hooks, are not good because they are not positive.

And, yes, it is easy to relate the answers for $\bigwedge^2 \mathbb{C}^n$ and $\mathrm{Sym}^2(\mathbb{C}^n)$, so feel free to answer with whichever is more convenient.

Answer 1

What kind of a formula will you find satisfactory? Formulas for the plethysm $s_\lambda\circ h_n$ where coefficients are expressed in terms of $S_n$-characters and generalized Kostka numbers are in Macdonald's book (see pp.138-140), so putting $n=2$ and applying the standard involution will give you some result for $e_2$ as well (which is your question, I presume)...

Answer 2

If I remember this correctly the cases $\mathrm{Sym}^k(\bigwedge^2 \mathbb{C}^n)$ and $\mathrm{Sym}^k(\mathrm{Sym}^2(\mathbb{C}^n))$ are known; and hence $\bigwedge^k(\bigwedge^2 \mathbb{C}^n)$ and $\bigwedge^k(\mathrm{Sym}^2(\mathbb{C}^n))$. I will look up the references tomorrow (if this is of interest).

Edit The result has now been stated. I learnt this from R.P.Stanley "Enumerative Combinatorics" Vol 2, Appendix 2. Specifically, A2.9 Example (page 449) which refers to (7.202) on page 503. This gives as the original reference (11.9;4) of the 1950 edition of:

Littlewood, Dudley E. "The theory of group characters and matrix representations of groups."

P.S. In the Notes at the end of 7.24 (bottom of page 404 in CUP 1999 edition) it discusses the origin and the etymology of "plethysm". It says:

Plethysm was introduced in
MR0010594 (6,41c) Littlewood, D. E. Invariant theory, tensors and group characters.
Philos. Trans. Roy. Soc. London. Ser. A. 239, (1944). 305--365

The term "plethysm" was suggested to Littlewood by M. L. Clark after the Greek word plethysmos $\pi\lambda\eta\theta\nu\sigma\mu o\zeta$ for "multiplication".

(the Greek is an approximation)

Answer 3

From Weyman's book "Cohomology of Vector bundles and Syzygies" Chapter 2 gives the following decompositions: $$\mathrm{Sym}^m \left(\bigwedge^2 E\right)=\bigoplus_{\lambda \in A_m}S^{\lambda}E$$ $$\bigwedge^m \left(\bigwedge^2E\right)=\bigoplus_{\lambda \in B_m}S^{\lambda}E$$ where $A_m$ is the set of all $\lambda$ with $|\lambda|=2m$ such that all parts $\lambda_i$ are even. $B_m$ is the set of all partitions $\lambda$ of $2m$ so that when you write it in hook notation $\lambda=(a_1,\dots,a_r|b_1,\dots,b_r)$ you have $a_i=b_i+1$ for all $i$. Also, maybe this article has some useful references.

Answer 4

Did you check the book by Procesi?

Answer 5

You may also use SAGE , (for example, the Sage online notebook )

Example:

The Riemann curvature tensor $R$ lives in the space $Sym^2(\Lambda^2 V)$ (after identifying $V$ with $V^{\vee}$)

Decomposing it in Sage:

$ s = SFASchur(QQ) $

(let s be the Schur functor)

$ s([2])(s([1,1])) $

(compute plethysm $ Sym^2 \Lambda^2 $)

s[1, 1, 1, 1] + s[2, 2]

-- i.e., $\Lambda^4 V + S_{[2,2]}$, as it should be

$ s([3])(s([1,1]))

s[1, 1, 1, 1, 1, 1] + s[2, 2, 1, 1] + s[3, 3]

-- though i understand that the explicit formula is better :)