# What is the centralizer of a Young subgroup of $S_n$?

In their celebrated paper "A new approach to the representation theory of the symmetric group. II", Okounkov and Vershik prove that $$Z(n-1,1)$$, the centralizer of $$\mathbb{C}[S_{n-1}]$$ in $$\mathbb{C}[S_n]$$ is the (surprisingly commutative!) algebra generated by $$Z(\mathbb{C}[S_{n-1}])$$ and $$X_n$$, the Jucys–Murphy element $$(1,n)+\dotsb+(n-1,n)$$. In the same paper, they give a new proof of a result that generalizes this: the centralizer $$Z(l,k):=\mathbb{C}[S_{l+k}]^{\mathbb{C}[S_{l}]}$$ is generated by $$Z(\mathbb{C}[S_{l}])$$, the group $$S_k$$ permuting the elements $$l+1,\dotsc,l+k$$, and the JM elements $$X_{l+1},\dotsc,X_{l+k}$$.

Both of these are cases of centralizers of group algebras of Young subgroups, namely of $$S_{n-1}\times S_1$$ and $$S_{l}\times S_1\times \dotsb \times S_1$$ respectively.

Are there similar results about $$\mathbb{C}[S_{a_1+\dotsb+a_k}]^{\mathbb{C}[S_{a_1}\times\dotsb\times S_{a_k}]}$$?

• @JohnMurray may be able to comment on this/ Jul 10 at 20:12

This answer is largely inspired by the wonderful paper of Samuel Creedon, The Farahat-Higman Algebra of Centralizers of Symmetric Group Algebras, which studies in detail the case of $$\mathbb{C}S_n^{S_{n-m}}$$, where $$m$$ is considered fixed and $$n$$ varies. Let $$\lambda = (\lambda_1, \lambda_2, \ldots, \lambda_l)$$ be a partition of $$n$$, and write $$S_\lambda$$ for the Young subgroup $$S_{\lambda_1} \times S_{\lambda_2} \times \cdots \times S_{\lambda_l}$$ of $$S_n$$. It seems likely that one can extend some of the work in the paper mentioned above to the case where all parts of $$\lambda$$ may vary, though there might be less structure to that case.

First of all, let us describe what the centraliser looks like. Just as the centre of a group algebra, $$Z(\mathbb{C}G) = \mathbb{C}G^G$$, has a basis of conjugacy class sums, we have that $$\mathbb{C}S_n^{S_\lambda}$$ has a basis of sums of elements within an $$S_\lambda$$ orbit of $$S_n$$ (under the conjugation action).

To understand these $$S_\lambda$$-conjugacy classes, recall that if $$\sigma, \tau \in S_n$$, and $$(a_1, a_2, \ldots, a_k)$$ is a cycle of $$\sigma$$, then $$(\tau(a_1), \tau(a_2), \ldots, \tau(a_k))$$ is a cycle of $$\tau \sigma \tau^{-1}$$. If we were allowed to take $$\tau \in S_n$$, then we could replace the numbers appearing in the cycle with any other numbers (in the range $$1$$ to $$n$$), and only the size of the cycle would matter. However, since we can only conjugate by $$\tau \in S_{\lambda}$$, we have some restrictions on what each $$a_i$$ can be sent to. Let us say that a number $$k$$ ($$1 \leq k \leq n$$) is permuted by $$S_{\lambda_r}$$ if we have $$\lambda_1 + \cdots + \lambda_{r-1}+ 1 \leq k \leq \lambda_1 + \cdots + \lambda_r,$$ so that in $$S_{\lambda_1} \times S_{\lambda_2} \times \cdots \times S_{\lambda_l}$$, it is the factor $$S_{\lambda_r}$$ that acts non-trivially on $$k$$. The upshot of this is that the $$S_\lambda$$ conjugation action allows arbitrary renumberings of cycles that preserve which $$S_{\lambda_r}$$ permutes a given entry. So the appropriate data here is to remember which $$S_{\lambda_r}$$ permutes a given element in a cycle. We do this by replacing the number with a label (or "colour") to record this information. The simplest thing to do is to use the number $$r$$ corresponding to $$S_{\lambda_r}$$. Let's see an example.

Suppose that $$\lambda = (2,2)$$ and $$n=4$$. Then some examples of $$S_\lambda$$-conjugacy classes are: $$(1)(2)(3)(4)$$ (the identity), whose label becomes (1)(1)(2)(2) (since the first two 1-cycles are permuted by $$S_{\lambda_1}$$, and the second two 1-cycles are permuted by $$S_{\lambda_2}$$) $$(13)(2)(4), (23)(1)(4), (14)(2)(3), (24)(1)(3)$$ The label for this class is (12)(1)(2), since the two-cycle contains one element permuted by $$S_{\lambda_1}$$ and one permuted by $$S_{\lambda_2}$$. (Note that the 2-cycles $$(12)(3)(4)$$ and $$(34)(1)(2)$$ are in classes by themselves, and in particular not with the remaining 2-cycles above. Their labels are $$(1,1)(2)(2)$$ and $$(2,2)(1)(1)$$, respectively.)

So now it is not difficult to check that the $$S_\lambda$$-conjugacy classes in $$S_n$$ are in bijection with the following combinatorial objects (which are the labels we have constructed:

multisets of necklaces using the "colours" $$1,2,\ldots,l$$, where the colour $$r$$ appears $$\lambda_r$$ times among all necklaces in the multiset.

If $$M$$ is such a multiset of necklaces, let us write $$X_M$$ for the sum of elements of the corresponding $$S_\lambda$$-conjugacy class. We can check that the centraliser is generated by those $$X_M$$ indexed by multisets of necklaces where exactly one necklace has size larger than 1. This is a filtration argument. Let us put $$X_M$$ in filtration degree equal to $$n$$-$$(\mbox{number of necklaces of size 1})$$. This way, the $$S_\lambda$$-conjugacy class of a permutation $$\sigma$$ is in filtration degree equal to the number of elements of $$1,2,\ldots,n$$ that are moved by $$\sigma$$ (not fixed points).

It is not difficult to check that (1) this is indeed a filtration, (2) the associated graded multiplication is computed by multiplying the two elements and assuming that they move disjoint elements in $$1,2,\ldots,n$$.

Having said all this, things are neater if instead of writing every element of $$M$$, we only write the necklaces of size greater than one. This is enough information to reconstruct the original multiset of necklaces because the sizes $$\lambda_r$$ tell us how many necklaces of size 1 and colour $$r$$ have neen omitted. So let us adopt this simplified notation. Then $$X_M$$ is in filtration degree equal to the sum of sizes of the necklaces in $$M$$.

In the (non-identity) conjugacy-class example above, where in our new notation $$M = (12)$$, squaring this element in the associated graded would give us $$2X_N$$, where $$N = (12)(12)$$.

From this associated graded rule, it follows up to some rational scalar, any $$X_M$$ is the product of $$X_N$$ where $$N$$ ranges across the constituent necklaces of $$M$$ (counted with multiplicity), plus lower order terms. In particular $$X_N$$ with $$N$$ having one necklace generates the centraliser.

So we've found a description of the centraliser, and a generating set. Let us see how it generalises the case $$\lambda = (n-m, 1^m)$$. I would like to take the case $$m=1$$ for granted, so that the centraliser of $$S_{n-1}$$ in $$\mathbb{C}S_n$$ is generated by conjugacy-class sums of $$S_{n-1}$$ and the JM element $$L_n = (1,n) + (2,n) + \ldots + (n-1,n)$$. As an indication of what is to come, note that for the partition $$(n-1,1)$$, we have $$L_n = X_{(12)}$$ (it would be $$X_{(12)(1)(1)\cdots(1)}$$ if we weren't removing necklaces of size 1).

Claim: when $$\lambda = (n-m,1^m)$$ the centraliser of $$S_{\lambda}$$ in $$\mathbb{C}S_n$$ is generated by $$\mathbb{Z}(\mathbb{C}S_{n-m})$$ and those $$X_M$$ where $$M$$ has one necklace of size 2.

I will sketch the proof (since this post is already quite verbose):

• The base case $$m=1$$ was mentioned above.

• For general $$m$$, pick out any one part of size 1 in the partition, and think about the sub-partition $$(n-m,1)$$ of $$\lambda$$. This embeds inside $$S_n$$ as $$H = \mathrm{Sym}(1,2,\ldots,n-m) \times \mathrm{Sym}(k)$$, where $$k$$ depends on which part of size 1 we chose. We refer to this smaller centraliser problem as a subcase.

• By the base case, we can get any element in $$\mathbb{C}\mathrm{Sym}(1,2,\ldots,n-m,k)^H$$ from $$Z(\mathbb{C}S_{n-m})$$ and a "modified" JM element, which will equal $$(1,k) + (2,k) + \cdots + (n-m,k)$$. If we add $$(n-m+1,k) + (n-m+2,k) + \cdots + (k-1,k)$$ we will recover the usual JM element $$L_k$$. Each of the transpositions $$(p,k)$$ we added is of the form $$X_M$$ where $$M$$ consists of the 2-element necklace $$(p-(n-m-1), k-(n-m-1))$$. So up to introducing some terms in our generating set, we recover the usual JM element.

• The base case tells us that using $$Z(\mathbb{C}S_{n-m})$$ and the appropriate JM element we can generate the whole centraliser in the subcase we are working. This includes $$X_M$$ where $$M$$ has one necklace $$(A,1,1,\ldots,1)$$, where $$A=k-(n-m-1)$$ is the label of the part of size 1 that we are working with in this subcase.

• We can "stitch together" $$X_M$$ where $$M$$ are of the fom $$(A,1,1,\ldots,1)$$ for varying $$A$$ as follows. Consider $$X_{(A,B)} X_{(B,1,1,\ldots,1)} X_{(A,1,1,\ldots,1)}$$. This turns out to be $$X_{(A,1,1,\ldots,1,B,1,1,\ldots,1)}$$ plus lower-order terms (lower-order terms arise because the entries of cycles labelled by 1 can coincide in the two terms, but the leading order term in the product is where they are all distinct). Iterating this we can construct $$X_M$$ with an arbitrary necklace $$M$$.

• So we can construct arbitrary $$X_M$$, and in particular the whole centraliser using the following generators: $$Z(\mathbb{C}S_{n-m})$$, the modified JM elements, and $$X_{(A,B)}$$ where $$A,B > 1$$. In turn these generators can be expressed in terms of the usual JM elements and $$X_{(A,B)}$$.

Thus we have deduced the two (equivalent) statements:

• The centraliser of $$S_{\lambda}$$ in $$\mathbb{C}S_n$$ is generated by $$Z(\mathbb{C}S_{n-m})$$ and the usual JM elements and $$S_{m}$$ (which is generated by transpositions).
• The centraliser of $$S_{\lambda}$$ in $$\mathbb{C}S_n$$ is generated by $$Z(\mathbb{C}S_{n-m})$$ and $$X_{M}$$ (where we may take $$M$$ to have a single necklace of size 2).

It would be nice to have similar statements (that we can restrict ourselves to a smaller set of generators) when $$\lambda$$ has more than one part of size larger than one. I don't currently know of such a statement, but maybe they could be obtained from a good understanding of the case where $$\lambda = (n-d,d)$$ has two parts ($$d=1$$ was the base case above).

• @ Christopher Ryba, this is a really nice answer! The elements $X_M$ (for $M$ a single necklace) also appear in limits of Bethe algebras. Do you know anything about their eigenvalues? If $X_M$ is in degree $2$, the eigenvalues are given by a generalisation of contents for tableaux, and in particular are integers. If $X_M$ is cubic or larger, then the eigenvalues are often algebraic integers, but I don't have a combinatorial description of them. I'm not sure if it is reasonable to expect a description, but it would be nice! Jul 18 at 4:57
• Thanks for the kind words. All I know is the following. If $\lambda=(a,b)$ then $X_{(1,2)}$ equals (sum of 2-cycles in $S_{a+b}$) - (sum of 2-cycles in $S_a$) - (sum of 2-cycles in $S_b$). To understand the eigenvalues on an irrep of $S_{a+b}$, we restrict to $S_a \times S_b$ and use the Frobenius formula: sum of 2-cycles acts by sum of contents. This breaks down for larger necklaces because 3-cycles come in four kinds: (1,1,1), (1,1,2), (1,2,2), (2,2,2). The first and last can be dealt with as above, but I don't know how to disentangle $X_{(1,1,2)}+X_{(1,2,2)}$ into its summands. Jul 19 at 7:57
• Thanks. Yes, this is precisely where I run out of ideas too. Jul 21 at 0:23