Let $\mathbb{C}^n = V + \mathbb{C}$ be the defining representation of the symmetric group. Is there a nice formula for how $\Lambda^i V \otimes \Lambda^j V$ splits into irreps?


Congratulations, you have asked one of the few questions of this type for which there is a positive answer. Such a formula is the main result (Theorem 2.1) of Remmel's paper "A formula for the Kronecker products of Schur functions of hook shapes".

A few points of vocabulary to get you oriented: The irreducible representations of $S_n$ are indexed by the partitions of $n$. The partition $\bigwedge^k V$ corresponds to the partition $(n-k, 1,1,\ldots, 1)$ where there are $k$ $1$'s. These partitions are called "hooks". (Note the degenerate cases: $k=0$ is $(n)$ and gives the trivial representation; $k=n-1$ is $(1,1,\ldots,1)$ and gives the sign representation.) The tensor product of $S_n$ representations corresponds to the combinatorial operation known as Kronecker product. So the title of Remmel's paper tells you that it is answering your question, and googling on "Kronecker hook" will turn up more references.

Remmel defines a double hook to be a partition of the form $(q,p, 2, \ldots, 2, 1,\ldots, 1)$ where there are $k$ occurrences of $2$ and $\ell$ occurrences of $1$. His result is that the only irreps that appear in $\bigwedge^i V \otimes \bigwedge^j V$ are hooks and double hooks, that the multiplicities with which they appear are $\leq 2$, and he gives an exact formula for which ones appear with which multiplicities.

| cite | improve this answer | |
  • 4
    $\begingroup$ +1 for the first sentence. Alas. $\endgroup$ – Allen Knutson Apr 22 '11 at 2:45
  • $\begingroup$ Is there something similar for symmetric powers ? $\endgroup$ – Alexander Chervov Aug 10 '12 at 21:15
  • $\begingroup$ David, Thanks for your answer. A typo [The partition $\bigwedge^k V$ corresponds to the partition]--> [The representation $\bigwedge^k V$ corresponds to the partition] $\endgroup$ – Duchamp Gérard H. E. Apr 26 '13 at 1:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.