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.

  • $\begingroup$ 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 $h \in H$? $\endgroup$ – LSpice Dec 19 '19 at 17:05
  • $\begingroup$ Yes. I just mentioned the Lie algebras, because it is easier for me to think about them, I am more used to that. $\endgroup$ – 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 $\prod_{\lambda\vdash N, 0\leq k<N} \operatorname{End}_\IC(V_{\lambda,k})$.

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.

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    $\begingroup$ 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? $\endgroup$ – Mark Wildon Dec 18 '19 at 21:06
  • $\begingroup$ 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 $d_{\lambda,k}$. $\endgroup$ – Johannes Hahn Dec 18 '19 at 21:10
  • $\begingroup$ 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 $k$. $\endgroup$ – Balázs Pozsgay Dec 19 '19 at 10:58
  • $\begingroup$ 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. $\endgroup$ – Balázs Pozsgay Dec 19 '19 at 11:05
  • $\begingroup$ 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 $\mathbb{C}^{d\times d}$. $\endgroup$ – 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.

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    $\begingroup$ 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 $N \ge 4$. $\endgroup$ – Mark Wildon Dec 18 '19 at 19:09
  • $\begingroup$ I agree with Mark, this seems to be too large. $\endgroup$ – Balázs Pozsgay Dec 18 '19 at 19:39
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    $\begingroup$ 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 $\mathrm{End}_\mathbb{C}(S^{(2,1)})$. $\endgroup$ – Mark Wildon Dec 19 '19 at 15:29
  • $\begingroup$ Yes, thanks, I had realised that too. I'll try to write more carefully what it really should be. $\endgroup$ – Geoff Robinson Dec 19 '19 at 16:28

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