# Centralizer of a cyclic subgroup within the group algebra $\mathbb{C} S_N$ of the symmetric group

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.

• Aren't the Lie centraliser and the group/algebra centraliser the same, since the first demands that $h X - X h = 0$ and the latter that $h X = X h$ for all $h \in H$? – LSpice Dec 19 '19 at 17:05
• Yes. I just mentioned the Lie algebras, because it is easier for me to think about them, I am more used to that. – Balázs Pozsgay Dec 19 '19 at 21:01

$$\newcommand{\IC}{\mathbb{C}}$$ 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.

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.

• The complete matrix algebra of $d \times d$ matrices over $\mathbb{C}$ ($\mathbb{C}^{d \times d}$ in the notation above) has (unital) commutative subalgebras of dimension about $d^2/4$: take the nilpotent block matrices with non-zero entries only in the top-right corner, and extend by the identity matrix. Your claim is correct for semisimple algebras. Maybe this is enough? – Mark Wildon Dec 18 '19 at 21:06
• Well, this certainly complicated things. is that the maximum? In this case one would need to actually know the dimensions $\dim(V_{\lambda,k})$ to get the precise value. I'm sure this is doable given how much is known about the representations of the symmetric group, but off the top of my head, I don't know how to calculate $d_{\lambda,k}$. – Johannes Hahn Dec 18 '19 at 21:10
• Thanks for the answer. But again I am confused. The endomorphism you are mentioning is conjugation with $\sigma$, right? But then I specifically need the eigenstates with $k=0$, and not all $k$. – Balázs Pozsgay Dec 19 '19 at 10:58
• And a small comment about what I want later: this is for physics, these will be some operators in Quantum Mechanics. So they have to be Hermitian. In this language I define Hermitian conjugate for the basis of the algebra simply as the inverse of the permutation. And then it is extended to the whole group algebra. So when I understand the matrix algebra decomposition, I will basically want the Cartan subalgebra of each block. Sorry I was not precise about this. I am not interested of other possibilities of abelian subalgebras. – Balázs Pozsgay Dec 19 '19 at 11:05
• Well if you want hermitian operators that commute with $\sigma$, then my answer stands, because all hermitian operators are diagonalisable and commuting operators are simultaneously diagonalisable so that the algebra of diagonal matrices is in fact the unique-up-to-conjugation maximal commutative sub-*-algebra of $\mathbb{C}^{d\times d}$. – Johannes Hahn Dec 19 '19 at 13:26

Thanks to Mark Wildon and the OP for pointing out that my answer was incorrect- I was considering the centralizer of $$\sigma$$ in the wrong algebra. However, it does seem to me that the structure of $$A = C_{\mathbb{C}S_{N}}(\sigma)$$ depends on the prime factorization of $$N.$$

In general it is well-known that if $$\lambda$$ is a partition of $$N$$ and $$\chi_{\lambda}$$ is the associated complex irreducible character of $$S_{N}$$, then $$\chi_{\lambda}(\sigma) \in \{0,1,-1 \}$$.

When $$N= p$$ is prime, this implies that $${\rm Res}^{S_{N}}_{\langle \sigma \rangle}(\chi_{\lambda})$$ has the form $$t\rho + \chi_{\lambda}(\sigma)1,$$ where $$t= \frac{\chi_{\lambda}(1) - \chi_{\lambda}(\sigma)}{p}$$ is a non-negative integer and $$\rho$$ is the regular character of $$\sigma$$. But this is never the case for all $$\chi_{\lambda}$$ when $$N$$ is not prime.

Now $$\mathbb{C}S_{N}$$ is isomorphic to $$\bigoplus_{\lambda} M_{\chi_{\lambda}(1)}( \mathbb{C})$$ as $$\lambda$$ runs through partitions of $$N$$. Now $$\sigma$$ acts as a matrix of trace $$0$$ or $$\pm 1$$ inside $$M_{\chi_{\lambda}(1)}( \mathbb{C}).$$

In the former case, the fixed subalgebra of $$\sigma$$ on the matrix algebra $$M_{\chi_{\lambda}(1)}( \mathbb{C})$$ has dimension $$\frac{\chi_{\lambda}(1)^{2}}{p}.$$

In the latter cases, we may compute the dimension of the fixed fixed subalgabra of $$\sigma$$ in the relevant matrix algebra.

If $${\rm Res}^{S_{N}}_{\langle \sigma \rangle }(\chi_{\lambda})= t_{\lambda} \rho \pm 1,$$ then the fixed subalgebra of $$\sigma$$ in the matrix algebra has dimension $$(p-1)t_{\lambda}^{2} + (t_{\lambda} \pm 1)^{2} = pt_{\lambda}^{2} \pm 2t_{\lambda} +1$$.

This means that when $$N = p$$ is prime, the dimension of the centralizer algebra of $$\sigma$$ in $$\mathbb{C}S_{N}$$ is totally detrmined by the values of the $$\chi_{\lambda}(1)$$ and $$\chi_{\lambda}(\sigma)$$, but his is not the case when $$N$$ is not prime.

• I'm confused: $\dim \mathbb{C}S_N = |S_N| = N!$ whereas the dimension of any full matrix algebra of $(N-1)! \times (N-1)!$ matrices is at least $(N-1)!^2$, which is more whenever $N \ge 4$. – Mark Wildon Dec 18 '19 at 19:09
• I agree with Mark, this seems to be too large. – Balázs Pozsgay Dec 18 '19 at 19:39
• I can't follow the penultimate paragraph. For $\mathbb{C}S_3$, the fixed subalgebra of $\langle (1,2,3) \rangle$ (acting by conjugation) has linear basis $\mathrm{id}$, $(1,2,3)$, $(1,3,2)$ and $(1,2) + (2,3) + (3,1)$. So its dimension is $4$, not $(3-1)!+1 = 3$. The factors in each matrix algebra are $1$ and $1$-dimensional (for the trivial and sign representations) and a $2$-dimensional semisimple commutative subalgebra of $\mathrm{End}_\mathbb{C}(S^{(2,1)})$. – Mark Wildon Dec 19 '19 at 15:29
• Yes, thanks, I had realised that too. I'll try to write more carefully what it really should be. – Geoff Robinson Dec 19 '19 at 16:28