16
$\begingroup$

The symmetric group $S_n$ acts over $V=\mathbb{R}^n$ by permuting the canonical basis.

So it acts over $V^{\otimes p}$ with a diagonal action (acts the same over each element of the tensor product).

I'd like to find the decomposition into irreps of it. As I'm a physicist, I'm first interested in the simple cases $p=2$ (and if this is still difficult, $n=3,4$), even if something general would be nice!

I also need something constructive: I guess the proof will be constructive, but I'd prefer a reference where the basis vectors are easily tractable.

Thanks!

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
17
$\begingroup$

This question was recently completely solved in this paper.

As explained on page 15 of the paper, letting $v_1,...,v_n$ be a basis for $V$, the standard basis vectors $v_{i_1} \otimes \cdots \otimes v_{i_p}$ for $V^{\otimes p}$ naturally correspond to partitions of the set $\{1,...,p\}$ into at most $n$ blocks. The submodule corresponding to a particular partition into $t$ blocks is isomorphic to the permutation module $H^{(n-t,1^t)}=Ind_{S_{n-t}}^{S_n} 1$. Multiplicities of irreducibles in $H^{\lambda}$ are well-known to be Kostka numbers $K_{\lambda, \mu}$, so we get that the multiplicity of the irreducible $S_{\lambda}$ in $V^{\otimes p}$ is $$\sum_{t=0}^n S(p,t) K_{\lambda, (n-t,1^t)}$$

Where $S(p,t)$ denotes the Stirling number. The paper also gives a bijective proof of this equality using paths in the relevant Bratteli diagram.

The fact that you are working over $\mathbb{R}$ rather than $\mathbb{C}$ does not matter since all complex irreducible representations of $S_n$ can in fact be realized over $\mathbb{Z}$.

Improve this answer
$\endgroup$
10
  • $\begingroup$ Is this explicit, as in giving bases of the submodules? $\endgroup$
    – darij grinberg
    Commented Aug 22, 2017 at 17:52
  • $\begingroup$ @darijgrinberg Yes, there is an explicit basis for the submodules $H^{(n-t,1^t)}$ which one can identify with $S_{n-t}$ cosets and thereby with Young tabloids and then follow the usual story. $\endgroup$
    – Christian Gaetz
    Commented Aug 22, 2017 at 18:16
  • $\begingroup$ Ah, I see! I was misled by the mentions of Schur-Weyl duality and semisimplicity. (NB: Stirling, not Sterling.) $\endgroup$
    – darij grinberg
    Commented Aug 22, 2017 at 19:25
  • $\begingroup$ Ok. To be sure that I understand, I do it on the simplest case n=p=2: Basis $e_i\otimes e_j$ Partition $\{1,2\}$ and $\{1\},\{2\}$ The first partition corresponds to the "permutation module" $H_1=Span\{e_1\otimes e_1,e_2\otimes e_2\}$ The second partition corresponds to the "permutation module" $H_1=Span\{e_1\otimes e_2,e_2\otimes e_1\}$ $\endgroup$
    – MarcO
    Commented Aug 23, 2017 at 8:13
  • $\begingroup$ $H_1$ and $H_2$ can be decomposed further (with $e_1\otimes e_1\pm e_2\otimes e_2, ...$) but this is well known (Kosta number...). Thanks! $\endgroup$
    – MarcO
    Commented Aug 23, 2017 at 8:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .