2
$\begingroup$

Consider the finite symmetric group on $n$ letters $S_n$ and $\chi$ a representation on a finite dimensional complex vector space $V$. Further assume that you know the dimension of the invariant sub-vector space $I$ of $V$. In order to know how to decompose $\chi$ into irreducible representations of $S_n$, this already gives us some information. If $$\chi=\sum_{i=1}^{p(n)} a_i T_i$$ is the decomposition of $\chi$ into irreducibles with $T_1$ being the trivial representation, we have $$\dim V=\sum a_i \dim T_i$$ and $$a_1=\dim I.$$ Now consider the set of conjugation classes of $S_n$, that we denote by $\{C_i\}$. Pick an element $c_i$ in each conjugation class, and consider the group generated by it, that we denote by $Z_i$. By restricting $\chi$ to each $Z_i$, we obtain a $Z_i$ representation in $V$. Suppose you also know the dimension of the invariant sub-vector space $I_i$ for each $Z_i$. With this we can obtain a bunch of new equations, since $\dim I_i$ has to be equal to the coefficient of the trivial representation of $Z_i$ when we restrict $\chi$ to it.

So my question is the following: is this information enough to deduce all the $a_i$'s, that is, the decomposition of $\chi$ into irreducibles? In principle we are producing $p(n)+1$ equations, so it should be enough if those form a base of the vector space of dimension $p(n)$. I have done the calculations up to $n=6$, and it does work out. I am actually quite convinced this works out for all $n$, but haven't been able to come up with a proof..Any help would be greatly appreciated!

$\endgroup$
1
  • 4
    $\begingroup$ Yes, it is. More generally, by Artin's Induction Theorem, a representation of a finite group over $\mathbb{Q}$ is uniquely determined by the dimensions of fixed subspaces under all cyclic subgroups, and all representations of symmetric groups are realisable over $\mathbb{Q}$. $\endgroup$
    – Alex B.
    Commented Sep 17, 2019 at 17:24

0

You must log in to answer this question.

Browse other questions tagged .