In their celebrated paper "[A new approach to the representation theory of the symmetric group. II](https://doi.org/10.1007/s10958-005-0421-7)", 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.



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> Are there similar results about $\mathbb{C}[S\_{a\_1+\dotsb+a\_k}]^{\mathbb{C}[S\_{a\_1}\times\dotsb\times S\_{a\_k}]}$?
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What is the centralizer of a Young subgroup of $S_n$?
This answer is largely inspired by the wonderful paper of Samuel Creedon, [The Farahat-Higman Algebra of Centralizers of Symmetric Group Algebras](https://arxiv.org/abs/2206.02939), 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](https://en.wikipedia.org/wiki/Necklace_(combinatorics)) 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).


