2
$\begingroup$

As I was going through various representation-theory posts in the site, I stumbled upon this one: Characters of the symmetric group corresponding to partitions into two parts. Now, that question hasn't been answered but there is a link in the comments to the following paper by Mark Wildon, Labelling the character tables of symmetric and alternating groups (you can find the published version here).

In that paper, Mark shows that for $n\neq 4,6$ given an unlabelled character table of $S_n$, there is a unique way to label its rows and columns naturally (that is, in such a way that $\chi_{\lambda\nu}=\chi^{\lambda}(\nu)$). From what I understand, this says that the character table isn't that symmetric.

Now, this is definitely a beautiful result, but my question regards his comment after Proposition 2.3. The proposition gives examples of conjugacy classes whose multisets of character values are the same (so you cannot distinguish between them). After that he asks whether such examples exist for the rows of the character table too. In particular, he says this never happens for $n\leq 30$.

Can two different characters of $S_n$ have the same multiset of values? Does anything change if we consider reducible characters as well? What happens in arbitrary (nonabelian) groups $G$, are there easy counterexamples there?

$\endgroup$
  • $\begingroup$ For $S_4$ at least it is easy to check that the class functions $\chi_1,\chi_2$ determined by $\chi_1(1)=24=\chi_2(1)$, $\chi_1((12))=\chi_2((1234))=4$, the rest $=0$, are both characters. This is essentially because there are two conjugacy classes of same size (here six each). I suppose a similar pair can be constructed whenever there are two equal size conjugacy classes. I haven't looked at the list too long, so don't know if this ever happens when $n\neq4,6$. All the books have the result that 2-cycles form the smallest non-trivial conjugacy class, when $n>6$, but what about others? $\endgroup$ – Jyrki Lahtonen Apr 2 '15 at 19:11
  • 1
    $\begingroup$ For the arbitrary group case there are plenty of examples when you have outer automorphisms. For a simple example, consider a cyclic group of prime order. Any two nontrivial characters differ by an outer automorphism (corresponding to multiplication mod $p$) and therefore have the same multiset of character values. Now for $S_n$ we never have outer automorphisms when $n \ne 6$ so this construction certainly won't work. $\endgroup$ – Nate Apr 2 '15 at 19:22
  • 1
    $\begingroup$ @JyrkiLahtonen I am not sure how you'd show that the resulting class functions are actually characters... For that matter it is not uncommon for conjugacy classes to have the same size: groupprops.subwiki.org/wiki/… It is however true that you can deduce quite a lot about the group if you know the sizes (and maybe their multiplicites) of the conjugacy classes: arxiv.org/abs/1002.3960 $\endgroup$ – Theo Apr 2 '15 at 19:50
1
$\begingroup$

Fleshing out the comment a bit.

Assume that the group $S_n$ has two non-trivial conjugacy classes of the same size, say $|[\sigma]|=|[\tau]|=C$ for some non-conjugate permutations $\sigma,\tau\in S_n$ and some positive integer $C\mid n!$. Define two class function $\psi_1$ and $\psi_2$ by declaring that $\psi_1(1)=Kn!=\psi_2(1)$, $\psi_1(\tau)=n!/C=\psi_2(\sigma)$ and setting the remaining values of both to zero.

Let $\chi_1,\chi_2,\ldots,\chi_\ell$ be the irreducible characters of $S_n$. It is known that $\chi_i(\pi)$ is a rational integer for all $\pi\in S_n, i=1,2,\ldots,\ell$.

Let's calculate the inner products: $$ (\chi_i\mid \psi_1)=\frac1{n!}\left[\chi_i(1)\psi_1(1)+C\cdot\chi_i(\tau)\psi_1(\tau)\right]=K\cdot\chi_i(1)+\chi_i(\tau), $$ and $$ (\chi_i\mid \psi_2)=\frac1{n!}\left[\chi_i(1)\psi_2(1)+C\cdot\chi_i(\tau)\psi_2(\sigma)\right]=K\cdot\chi_i(1)+\chi_i(\sigma). $$ These inner products are thus automatically rational integers. Whenever $K$ is large enough, all these integers are non-negative (and at least one is positive). For such a choice of $K$ both class functions are thus characters.

Incrementing $K$ by one amounts to summing the regular representation to the virtual character.

Of course, both $\psi_1,\psi_2$ are both highly reducible. The point of this note was to make it clear that the question becomes quite easy, if we allow reducible characters.

$\endgroup$
  • $\begingroup$ Thanks @Jyrki, I get it now, I like how the fact that the characters are integer-valued forces so much... Am I correct to think that all integer-valued class functions are actually virtual characters of the symmetric group? $\endgroup$ – Theo Apr 2 '15 at 21:05

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.