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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 $\mathbb{C}[S_{a_1+\dotsb+a_k}]^{\mathbb{C}[S_{a_1}\times\dotsb\times S_{a_k}]}$?

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    $\begingroup$ @JohnMurray may be able to comment on this/ $\endgroup$ Jul 10 at 20:12

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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).

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    $\begingroup$ @ 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! $\endgroup$
    – Noah White
    Jul 18 at 4:57
  • $\begingroup$ 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. $\endgroup$ Jul 19 at 7:57
  • $\begingroup$ Thanks. Yes, this is precisely where I run out of ideas too. $\endgroup$
    – Noah White
    Jul 21 at 0:23

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