TITLE: Centralizer of a cyclic subgroup within the group algebra $\mathbb{C} S_N$ of the symmetric group
QUESTION [6 upvotes]: Let us take the group algebra $\mathbb{C} S_N$ and the subgroup $H=Z_N$ generated by the element
$\sigma=(123\dots N)$, which is a cyclic shift. What is the structure of the centralizer of $H$ within $\mathbb{C} S_N$? If we
are looking at the Lie algebra $\mathcal{L}(\mathbb{C} S_N)$ obtained from the usual 
commutator of two elements, then what is the centralizer of the commutative sub-algebra $\mathbb{C} H$?
My motivation for this comes from physics, where the special ordering $1,2,3,\dots N$ has a
concrete physical meaning, for example actual atoms are placed next to each other. In this case those
group algebra elements that commute with $\sigma$ are called ``translationally invariant''
permutation operators, under periodic boundary conditions.
Ultimately I would like to know what is the maximal number of mutually commuting operators within
$\mathbb{C} S_N$ that commute with $\mathbb{C} H$, and how to find explicit bases for them. I wanted to approach this through the structure of
the Lie algebra $\mathcal{L}(\mathbb{C} S_N)$, decomposition into simple Lie algebras, etc. I understand
that the irreducible subspaces in $\mathbb{C} S_N$ correspond to the Young symmetrizers. But it is not
so clear what to do from here, how to add the action of the concrete element $\sigma$ into this.

REPLY [2 votes]: $\newcommand{\IC}{\mathbb{C}}$
Let $V_\lambda$ be the irreducible $S_N$-module (Specht module). The representations $\rho_\lambda: \IC S_N \to \operatorname{End}_\IC(V_\lambda)$ give us an isomorphism of algebras $\IC[S_N] \to \prod_{\lambda \vdash N} \operatorname{End}_\IC(V_\lambda)$.
Since we work with $\IC$, the endomorphisms of finite order are always diagonalisable. In particular we can decompose $V_\lambda = \bigoplus_{k=0}^{N-1} V_{\lambda,k}$ such that $\sigma$ acts as multiplication by $\exp(\frac{2\pi i}{N}k)$ on $V_{\lambda,k}$. The centraliser of $\sigma$ is therefore exactly equal to $\prod_{\lambda\vdash N, 0\leq k<N} \operatorname{End}_\IC(V_{\lambda,k})$.
This reduces the problem to finding the commutative subalgebras of $\IC^{d\times d}$ of maximal dimension for all dimensions $d$. If I'm not mistaken, the subalgebra of diagonal matrices is of maximal dimension among the commutative subalgebras of $\IC^{d\times d}$, i.e. the maximal dimension is just $d$ itself.
Therefore the maximal dimension of a commutative subalgebra inside the $C_{\IC S_N}(\sigma)$ is $\sum_{\lambda,k} \dim(V_{\lambda,k}) = \sum_\lambda \dim(V_\lambda)$. The dimensions $\dim(V_\lambda)$ can be calculated by the hook length formula.
In principle this approach also tells you how to find a set of basis elements: Find an eigen-basis for $\sigma$ inside each $V_\lambda$. The diagonal matrices w.r.t. to this basis will give you a basis of a commutative subalgebra of maximal dimension.