Branching Rules of Symmetric group to cyclic group formula

Question

The cyclic group $\mathbb{Z}_n = \langle a \rangle$ has sort of a "natural embedding" into the symmetric group $S_n$ via $$ a \mapsto (1,2, \cdots, n). $$ Given an irreducible representation $\pi$ of $S_n$ we can consider the restricted representation. This representation is reducible and can be decomposed as a direct sum of irreps of $\mathbb{Z}_n$:

$$ \pi = \bigoplus_\lambda (m_{\lambda,\pi}) \lambda. $$ where $m_{\lambda,\pi} = \langle \chi_\pi, \chi_\lambda \rangle$ is the multiplicity of (the $\mathbb{Z}_n$ irrep) $\lambda$ in $\pi$.

Question: Is there a closed formula for the multiplicity $m_{\lambda,\pi}$ in this specific case?

For example, consider $n=3$. Denote the irreps of $S_3$ by $\pi_{ [3] }$, $\pi_{[2,1]}$, and $\pi_{ [1,1,1]}$ (corresponding to the respective Young diagram) and denote the irreps of $\mathbb{Z}_3$ by $\lambda_0$, $\lambda_1$, $\lambda_2$ (where if $\zeta_3$ is a 3rd root of unity then $\lambda_j(\zeta_3) = \zeta_3^j$). The multiplicities in this case are given in the following table.

$$ \begin{array}{|c|c|c|c|} \hline m_{\lambda,\pi} & \lambda_0 & \lambda_1 & \lambda_2 \newline \hline \pi_{ [3] } & 1 & 0 & 0 \newline \hline \pi_{ [2,1] } & 0 & 1 & 1 \newline \hline \pi_{ [1,1,1] } & 1 & 0 & 0 \newline \hline \end{array} $$

Answer 1

Your question fits comfortably into the somewhat more general framework of (complex) reflection groups. I'll start by giving the general setup and then say what happens in the case you are interested in (as usually happens for the symmetric group, the general theory may be rephrased using some nice combinatorics). For now, in case you are interested only in the answer for the symmetric group, here it is:

The multi-set of integers $k$ such that $e^{2 \pi i k/n}$ is an eigenvalue of the $n$-cycle $(12\cdots n)$ on $S^\lambda$ is equal to the multi-set of major indices of standard Young tableaux of shape $\lambda$, read modulo $n$.

See the end of this answer for the terminology, if any of it is unfamiliar (and note that describing the eigenvalues of the $n$-cycle is the same as describing the branching rule you ask about). The proof of this statement is outlined below; it starts with what I believe is the natural generality for this circle of ideas---complex reflection groups.

So let $V$ be a finite dimensional $\mathbf{C}$-vector space. A reflection group acting on $V$ is a finite subgroup $W \leq \mathrm{GL}(V)$ that is generated by the set $R=\{r \in W \ | \ \mathrm{codim}_V(\mathrm{fix}(r))=1 \}$ of reflections that it contains. As an example, the symmetric group $S_n$ acting on $\mathbf{C}^n$ by permuting the coordinates is a reflection group, with set of reflections equal to the set of transpositions.

Let $W$ be a reflection group acting on $V$. By a theorem of Chevalley, the quotient map $\pi:V \to V/W$ is flat and so the (scheme-theoretic) fiber over $0$ is of degree $|W|$ and in fact carries the regular representation of $W$. Algebraically, this means that the $\mathbf{C}$-algebra $$\mathbf{C}[\pi^{-1}(\pi(0))]=\mathbf{C}[V]/ \langle f \in \mathbf{C}[V]^W \ | \ f(0)=0 \rangle$$ is, as a $\mathbf{C} W$-module, isomorphic to the regular representation of $W$. It is (unfortunately) traditional to call this ring of functions on the zero fiber of $\pi$ the co-invariant ring of $W$, but as you can see a more appropriate name would be zero fiber ring. With this object in hand, we define the multi-set of exponents of an irreducible representation $E$ of $W$ to be the multi-set of polynomial degrees in which $E$ occurs in the zero-fiber ring. This multi-set therefore contains $\mathrm{dim}_\mathbf{C}(M)$ non-negative integers.

The last ingredient we need is the notion of regular element. First, a vector $v \in V$ is regular if it is not fixed by any non-trivial element of $W$, or equivalently (by a theorem of Steinberg) if it is not fixed by any reflection in $W$. An element $w \in W$ is regular if it possesses a regular eigenvector. For example, the cycle $(12\cdots n)$ in $S_n$ is a regular element since its eigenvector $(1,\zeta,\zeta^2,\dots,\zeta^{n-1})$ is regular, where $\zeta$ is some (any) fixed primitive $n$th root of $1$.

There is a beautiful theorem relating regular elements to the exponents of irreducible representations of $W$. Here it is:

Theorem [Springer, Regular elements of finite reflection groups, Prop. 4.5]: Let $W$ be a reflection group in $V$, let $c \in W$ and $v \in V$ be regular with $cv=\zeta v$ for some eigenvalue $\zeta$ of order $h$, and let $E$ be an irreducible representation of $W$. Then the multi-set of exponents of $E$ modulo $h$ is equal to the multi-set of integers $k$ such that $\zeta^{-k}$ is an eigenvalue of $c$ acting on $E$.

In the case in which $W$ is a real reflection group, an important class of regular elements is given by the Coxeter elements, in which case $h$ is the Coxeter number of $W$. When $W=S_n$ the $n$-cycles are the Coxeter elements and $h=n$.

In case $W$ is the symmetric group, there is a combinatorial description of the exponents of a representation $E$ (and hence, of the eigenvalues of regular elements acting on irreducible representations). Namely, the irreducible representations of $\mathbf{C} S_n$ are indexed by integer partitions $\lambda$ of $n$, in such a way that the dimension of $S^\lambda=E$ is equal to the number of standard Young tableaux of shape $\lambda$ (in this answer I use the English convention for visualizing Young diagrams and standard Young tableaux). There is a refinement of this fact that describes the exponents: the multi-set of exponents of $S^\lambda$ is equal to the multi-set of major indices of the standard Young tableaux of shape $\lambda$. Here the major index of a standard Young tableau $T$ is the sum of the descents $j$ of $T$, where an integer $1 \leq j \leq n$ is a descent of $T$ if $j+1$ appears in a row strictly below $j$.