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?

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.


6 Answers 6


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 $\pi\lambda\eta\theta\upsilon\sigma\mu\acute o\varsigma$ for "multiplication".

  • $\begingroup$ Yes, this is of interest; please let me know what references you know of. $\endgroup$ May 4, 2010 at 22:06
  • $\begingroup$ This is a very special case, of course, because the decompositions are multiplicity free, and due to this fact, they are known very explicitly. I heartily recommend Roger Howe's Schur lectures. $\endgroup$ May 4, 2010 at 22:13

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.

  • $\begingroup$ Yes, these are also listed in Macdonald's book. $\endgroup$ May 5, 2010 at 13:13

Let $V = \mathbb{C}^n$ where $n$ is sufficiently large. This paper by Melanie de Boeck and Rowena Paget determines the constituents of $S^\lambda (\mathrm{Sym}^2 V)$ when $\lambda$ has either two rows, or two columns or is a hook partition of the form $(k-r,1^r)$. Since $S^\mu (V)$ appears in $S^\lambda (\mathrm{Sym}^2 V)$ if and only if $S^{\mu'}$ appears in $S^\lambda (\bigwedge^2V)$, these results apply to the question.

Explicit positive formulae are given for the multiplicities of irreducible consituents of $S^{(k-1,1)}(\mathrm{Sym}^2 V)$, $S^{(2,1^{k-2})}(\mathrm{Sym}^2V)$, $S^{(k-2,2)}(\mathrm{Sym}^2 V)$ and $S^{(k-2,1^2)}(\mathrm{Sym}^2 V)$. These results give a complete answer to the question in three new cases.

For example, Corollary 3.2 states that if $\mu$ is a partition of $2k$ then $S^\mu V$ appears in $S^{(k-1,1)}(\mathrm{Sym}^2 V)$ if and only if either $\mu$ has only even parts, or $\mu$ has exactly two odd parts of distinct sizes. In the latter case the multiplicity is $1$, in the former case the multiplicity is one less than the number of distinct part sizes of $\mu$.

Edit. Say that $S^\lambda(V)$ is a minimal constituent of a polynomial $\mathrm{GL}(V)$-module $W$ if $S^\lambda(V)$ appears in $W$ and $\lambda$ is minimal with this property. Define maximal constituent analogously. Let $m \in \mathbb{N}$. This paper by Rowena Paget and me characterizes, in terms of certain tuples of families of $m$-subsets of $\mathbb{N}$, all partitions $\mu$ such that $S^\mu$ is a minimal constituent of $S^\lambda(\mathrm{Sym}^m(V))$. There is an analogous characterization of the maximal constituents of $S^\lambda(\mathrm{Sym}^m(V))$ by replacing sets with multisets.

To give a very small example, the minimal constituent $S^{(4,3,1)}(V)$ of $S^{(1^4)}({\mathrm{Sym}^2(V)})$ corresponds to the family of $2$-sets $\bigl\{ \{1,2\}, \{1,3\}, \{2,3\}, \{1,4\} \bigr\}$ of multidegree $(4,3,1)' = (3,2,2,1)$.

These results give a practical sufficient condition on a partition $\nu$ for $S^\nu(V)$ to have multiplicity zero in $S^\lambda(\mathrm{Sym}^2(V))$, so are also relevant to the question.


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


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 :)


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)...

  • $\begingroup$ If I'm remembering correctly, the formulas in Macdonald contain the values of symmetric group characters evaluated on conjugacy classes, which probably doesn't meet David's criterion ("positive rule"). $\endgroup$
    – GS
    May 4, 2010 at 11:33
  • $\begingroup$ @Stephen: I surely seem to have overlooked "positive"... $\endgroup$ May 4, 2010 at 12:21
  • $\begingroup$ I've added some clarifications above. $\endgroup$ May 4, 2010 at 16:00

Did you check the book by Procesi?

  • 2
    $\begingroup$ It's better to give a complete reference. $\endgroup$ Jun 27, 2015 at 22:42

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