Finite nilpotent orbits: GL(n,q)-conjugacy classes and a partial order on partitions

Question

I have a question regarding a partial order $<$ on the set ${\rm Part}(n)$ of partitions of $n$. Given $\lambda=(\lambda_1,\lambda_2,\ldots)\in{\rm Part}(n)$ with $\sum_{i\geq1} \lambda_i=n$ and $\lambda_1\geq\lambda_2\geq\cdots\geq0$, let $J_\lambda$ denote the $n\times n$ block diagonal matrix $\bigoplus_{i\geq1}J_{\lambda_i}$. For example, $J_{(3,2,1)}=\left(\begin{smallmatrix}0&1&0&&&\\0&0&1&&&\\0&0&0&&&\\&&&0&1&\\&&&0&0&\\&&&&&0\end{smallmatrix}\right)$. Consider the ${\rm GL}(n,F)$-conjugacy classes of the set ${\rm M}(n,F)$ of all $n\times n$ matrices over a field $F$. A nilpotent matrix $X\in{\rm M}(n,F)$ lies in a conjugacy classes $\mathcal{O}_\lambda:=J_\lambda^{{\rm GL}(n,F)}$ for a unique $\lambda\in{\rm Part}(n)$. (Nilpotent means $X^n=0$.)

If $F=\mathbb{F}_q$ is a finite field, then set $n_\lambda:=|J_\lambda^{{\rm GL}(n,q)}|$. A formula for $n_\lambda$ is given in Fulman, Cycle indices for finite classical groups. It turns out that $n_\lambda=n_\lambda(q)$ is a polynomial in $q$ with integer coefficients. Define a partial order $<$ on ${\rm Part}(n)$ as follows: $\lambda<\mu$ if and only if $n_\lambda(q)$ divides $n_\mu(q)$. I call this the divisibility partial order.

When $F$ is the complex field $\mathbb{C}$, define $\lambda\triangleleft\mu$ if $\overline{\mathcal{O}_\lambda}\subset\overline{\mathcal{O}_\mu}$ where $\overline{\mathcal{O}_\lambda}$ denotes the Zariski closure of $\mathcal{O}_\lambda$. It is shown in Collingwood and McGovern, Nilpotent orbits of semisimple Lie algebras, pp 93--95, that $\triangleleft$ is the dominance partial order on ${\rm Part}(n)$. That is, $\lambda\triangleleft\mu$ if and only if $\sum_{i=1}^{k-1}\lambda_i=\sum_{i=1}^{k-1}\mu_i$ and $\lambda_k<\mu_k$ for some $k\geq1$.

If $n\leq5$, then the partial orders $<$ and $\triangleleft$ are identical and are total orders. However, when $n=6$ the partition $(3,2,1)$ of 6 has three partitions divisibility larger, and has five partitions dominance larger.

Does anyone have any insight into divisibility partial order? or know of its appearance in the literature? (I have not found a reference to $<$ in Roger Carter's book Finite groups of Lie type: conjugacy classes and complex characters, but $\triangleleft$ appears in 5.5 and 5.11.) For specific $\lambda$, I can (theoretically) factor $n_\lambda(q)$ and so can determined whether $\lambda<\mu$ for specific $\lambda$ and $\mu$, but I have few global results.

Answer 1

I think the right way is to factor $n_\lambda(q)$ in general. In particular, it is obviously a quotient of the order of $GL_n(q)$. The formula for the order of $GL_n(q)$ does not have very many prime factors: just $q$ and the first $n$ cyclotomic polynomials.

One could consider an alternate question, the order of the centralizer of $J$ in $GL(n,\mathbb F_q)$. This gives the reverse of the partial order, since the divisibility relation is reversed. The centralizer is the automorphism group of the corresponding $\mathbb F_q[x]/x^n$-module, $M$. The subgroup that fixes $M/x$ is a $q$-group, since it is unipotent. Its quotient is a product of copies of $GL(k,\mathbb F_q)$: one for each type of block, with $k$ equal to the number of that block that appears.

The number of times that the $k$th elementary cyclotomic polynomial appears in the size of the centralizer is just the sum over all sizes $n$ of the floor of $a_n/k$, where $a_n$ is the number of blocks of size $n$.

This function behaves very erratically, so we can conclude that the partial order behaves erratically as well.