Sum over all elements in conjugacy class of $S_n$ has all integer eigenvalues in any representation?

Question

Sorry, for the question being obvious or well-known for some, just want to reconfirm not to mislead myself and colleagues. It seems the answer might follow from previous posts by N.Elkies and B.Steinberg, but again not sure in some details.

Question: Take any conjugacy class of $S_n$ and consider sum over ALL its elements - get an element in group algebra - is it true that eigenvalues of that element in ANY representation of $S_n$ are integers ?

Sketch of arguments.

Step 1. "Harder fact:" $S_n$ is rational group -- that means any character has rational values. (It might be deduced from rational realisations of irreps of $S_n$ (e.g. some Specht modules), or from the criteria like in N.Elkies post - element is rational if it is conjugated to any its power co-prime to its order). Any way that step I believe is well-known and true.

Here is point not so clear for me:

Step 2. One can express the sum over conjugacy class via the characters with RATIONAL coefficients - is that true ? Is it true only for $S_n$/(rational groups) ? Or it is true for general finite group ?

Step 3. "Easy" - from basic theory for ANY finite group - eigenvalues are algebraic integers , and since we deduced that they are rational - we get they are just integers (not just algebraic integers). That part again I believe is well-known and true.

Answer 1

Just repeating my comment: Any (ordinary) representation of a finite group $G$ is a direct sum of irreducibles, so it suffices to consider an irreducible representation. In an irreducible representation with character $\chi$ the element $z = \sum x^G$ acts (by Schur's lemma) as the scalar $\omega_\chi(z) := \mathrm{tr}(\rho(z)) / \chi(1) = |x^G| \chi(x) / \chi(1)$. It is a standard fact that $\omega_\chi(z)$ is an algebraic integer, so if $\chi(x)$ is a rational integer then so is $\omega_\chi(z)$. See for example in Isaacs "Character theory of finite groups", Theorem 3.7.

Here is essentially the proof in Isaacs. Consider the subring $S \le Z(\mathbf C G)$ generated by the class sums $z_1, \dots, z_k$. We have $z_i z_j = \sum_k a^k_{ij} z_k$ for some integers $a^k_{ij}$, namely $a^k_{ij} = |x_i^G \cap x_k (x_j^{-1})^G|$. It follows that $S$ is generated by $z_1, \dots, z_k$ as a $\mathbf Z$-submodule, so $S$ is actually integral over $\mathbf Z$. In particular each class sum $z_i$ is the root of a monic polynomial with integer coefficients. This is preserved under a representation $\rho : \mathbf C G \to \mathrm{End}(V)$, which implies that $\rho(z_i)$ has integral eigenvalues.

Answer 2

Here is a proof that is very specific to the case of symmetric groups and is probably rather involved for a general fact like this. (In fact, the primary reason for typing this answer is to record the statement and the proof of Proposition 1 below.)

Define the (Young-)Jucys-Murphy elements $\{X_k\}_{k=1}^{n}$ as $$ X_k=(1~k)+\ldots+(k-1~k)=\sum_{i=1}^{k-1}(i~k)\in\mathbb{Z}[S_n]. $$ The statement in question is a direct consequence of Propositions 1 and 2 below.

Proposition 1 (Jucys, Murphy). The center of the group ring $\mathbb{Z}[S_n]$ coincides with the subring of $\mathbb{Z}[S_n]$ generated by $f(X_1,\ldots,X_n)$, where $f$ runs over $\mathbb{Z}[x_1,\ldots,x_n]^{S_n}$ (symmetric polynomials in $n$ variables with integer coefficients).

Proof (sketch). We outline the main ideas, the details can be found in T. Ceccherini, F. Scarabotti, F. Tolli. Representation Theory of the Symmetric Groups. (The Okounkov-Vershik Approach, Character Formula, and Partition Algebras), Cambridge University Press (2010), see Section 4.4.2 and Theorem 4.4.5.

The fact that any expression of the form $f(X_1,\ldots,X_n)$ with $f\in\mathbb{Z}[x_1,\ldots,x_n]^{S_n}$ follows from the identities $$ s_iX_js_i^{-1}=X_j~\text{for}~j\neq i,i+1, \\ s_i(X_{i}+X_{i+1})s_i^{-1}=X_{i}+X_{i+1},\quad s_i(X_{i}X_{i+1})s_i^{-1}=X_{i}X_{i+1}, $$ where $s_i=(i~i+1)$. (Recall that $s_iX_{i+1}s_i^{-1}=X_i+s_i$.)

The harder part is to show that any element of the center of $\mathbb{Z}[S_n]$ can be represented as $f(X_1,\ldots,X_n)$ for $f\in\mathbb{Z}[x_1,\ldots,x_n]^{S_n}$.

The center of $\mathbb{Z}[S_n]$ is additively generated by the elements $$ C_{\mu}=\sum_{\sigma\in S_n,~\mathrm{ct}(\sigma)=\mu}\sigma, $$ where $\mu$ runs over partitions of $n$, and $\mathrm{ct}(\sigma)$ stands for the cycle type of permutation $\sigma$. In other words, for $\mu=(\mu_1,\ldots,\mu_k)\vdash n$ with $\mu_1\ge\ldots\ge\mu_k\ge 1$ the element $C_{\mu}$ is the sum of all permutations that are products of $k$ disjoint cycles of lengths $\mu_1,\ldots,\mu_k$, respectively.

Let $m_{\lambda}(x_1,\ldots,x_n)$ be the monomial symmetric polynomial in $n$ variables $x_1\ldots,x_n$. The main observation is the following

Claim. For any $\mu=(\mu_1,\ldots,\mu_k)\vdash n$ with $\mu_1\ge \ldots\ge\mu_k\ge 1$ we have the identity: $$ m_{\mu_1-1,\ldots,\mu_k-1}(X_1,\ldots,X_n)=C_{\mu}+\sum_{\nu}k_{\nu}C_{\nu},\quad k_{\nu}\in\mathbb{Z}, $$ where the summation runs over partitions $\nu\vdash n$ which have more parts equal to $1$ than $\mu$.

This proposition is proven via the analysis of permutations arising after the expansion of the left-hand side. The proof is now concluded by proving by induction on the number of parts of $\mu$ equal to $1$ that $C_{\mu}=f(X_1,\ldots,X_n)$ for some $f\in\mathbb{Z}[x_1,\ldots,x_n]^{S_n}$. $\square$

Remark. If one does not care about integer coefficients, only about $\mathbb{Q}$ or $\mathbb{C}$, then one can induct on the number of fixed points, use $X_1^{r-1}+\ldots+X_{n}^{r-1}$ to generate $r$-cycles and use the expansion $C_{\mu}=\frac{1}{\operatorname{Stab}(\mu)}C_{(\mu_1)}\ldots C_{(\mu_k)}+\ldots$ to generate other conjugacy classes.

Proposition 2. The elements $\{X_k\}_{k=1}^{n}\subset\mathbb{Z}[S_n]$ can be simultaneously diagonalised in any finite-dimensional representation of $S_n$. Moreover, all eigenvalues of the corresponding operators are integers.

A nice and rather elementary proof of the proposition above is given by Igor Makhlin in this answer.

Remark. In fact, as follows from the Okounkov-Vershik approach to the representation theory of $S_n$, in the irreducible representation of $S_n$ corresponding to a partition $\lambda\vdash n$ one can choose a basis $\{v_T\}$ parameterised by the standard tableaux $T\in\operatorname{Tab}(\lambda)$ of shape $\lambda$. Then, $\{X_k\}_{k=1}^{n}$ act diagonally on this basis and the eigenvalue of $X_k$ equals the content $c_k(T)=\mathrm{col}_k-\mathrm{row}_k$ of the box of $T$ containing $k$.