Some thoughts on random matrices

This post is to start writing down some thoughts on extending results in my paper with Yifeng Huang (to be referred as CH) about Haar random matrices over $\mathbb{Z}_{p}.$

Surjection moments. One thing we have not tried in our paper is to use the following strategy, popular in the literature. For example, given an odd prime $p,$ let $\mathcal{F}_{p}$ be the set of all isomorphism classes of finite abelian $p$-groups (or equivalently finite $\mathbb{Z}_{p}$-modules). Equip a structure of the probability space on $\mathcal{F}_{p},$ with the sample space consisting of all subsets of it, with a probability measure $$\mu : \mathcal{F}_{p} \rightarrow [0, 1].$$ The statement goes, if $$\mathbb{E}_{[H] \in \mathcal{F}_{p}}(|\mathrm{Sur}(H, G)|) = 1$$ for any finite abelian $p$-group $G,$ then $\mu$ is the Cohen-Lenstra measure, meaning we have $$\mu(G) := \mu(\{[G]\}) = \dfrac{1}{|\mathrm{Aut}(G)|}\prod_{i=1}^{\infty}(1 - p^{\infty}),$$ for any finite abelian $p$-group $G.$ (A simple proof can be found in the proof of Lemma 8.2 of this paper by Ellenberg, Venkatesh, and Westerland.) Here, we used the following notations:

• $\mathbb{E}$ meant the expected value;
• $\mathrm{Sur}(H, G)$ meant the set of surjective $\mathbb{Z}$-linear maps from $H$ to $G$;
• $\mathrm{Aut}(G)$ meant the set of $\mathbb{Z}$-linear automorphisms of $G.$

Surjection moments for Haar $p$-adic matrices. Now, consider the set $\mathrm{Mat}_{n}(\mathbb{Z}_{p})$ of $n \times n$ matrices over $\mathbb{Z}_{p},$ the ring of $p$-adic integers. The $p$-adic topology given by $\mathbb{Z}_{p}$ makes $\mathrm{Mat}_{n}(\mathbb{Z}_{p})$ a compact Hausdorff topological group with respect to its addition, so we can consider its Haar measure, which is a unique probability measure translation-invariant under addition. Can we compute $$\mathbb{E}_{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}(|\mathrm{Sur}(\mathrm{cok}(X), G)|),$$ for a given finite abelian $p$-group $G$?

We have $$\begin{align*}\mathbb{E}_{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}(|\mathrm{Sur}(\mathrm{cok}(X), G)|) &= \int_{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}|\mathrm{Sur}(\mathrm{cok}(X), G)| d\mu_{n}(X) \\ &= \int_{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}|\mathrm{Sur}(\mathbb{Z}_{p}^{n}/ \mathrm{im}(X), G)| d\mu_{n}(X) \\ &= \int_{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}\sum_{\bar{F} \in \mathrm{Sur}(\mathbb{Z}_{p}^{n}/ \mathrm{im}(X), G)}1 d\mu_{n}(X) \\ &= \int_{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}\sum_{F \in \mathrm{Sur}(\mathbb{Z}_{p}^{n}, G)}\boldsymbol{1}_{F(-) = 0}(X) d\mu_{n}(X) \\ &= \sum_{F \in \mathrm{Sur}(\mathbb{Z}_{p}^{n}, G)} \int_{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}\boldsymbol{1}_{F(-) = 0}(X) d\mu_{n}(X) \\ &= \sum_{F \in \mathrm{Sur}(\mathbb{Z}_{p}^{n}, G)} \mathrm{Prob}_{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}(FX = 0),\end{align*}$$ 

• where $\mu_{n}$ denotes the Haar measure (although it could be any probability measure);
• $\boldsymbol{1}_{F(-) = 0} : \mathrm{Mat}_{n}(\mathbb{Z}_{p}) \rightarrow \{0, 1\}$ is the indicator function such that $\boldsymbol{1}_{F(-) = 0}(X) = 0$ if and only if $FX = 0.$

Note that we have $FX = 0$ if and only if $\mathrm{im}(X) \subset \ker(F),$ which is the same as saying that all the columns of $X$ sits inside $\ker(F).$ When $\mu_{n}$ is the Haar measure, each column of $X$, which is an element of $\mathbb{Z}_{p}^{n}$ sits inside $\ker(F),$ with probability $$\frac{1}{|\mathbb{Z}_{p}^{n}/\ker(F)|} = \frac{1}{|G|}.$$ The event that a column $X$ sits inside $\ker(F)$ is independent to those that other columns sit inside $\ker(F)$ (according to the Haar measure), so we have

$$\mathrm{Prob}_{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}(FX = 0) = \frac{1}{|G|^{n}} = \frac{1}{|\mathrm{Hom}(\mathbb{Z}_{p}^{n}, G)|}.$$ Thus, we have $$\begin{align*}\mathbb{E}_{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}(|\mathrm{Sur}(\mathrm{cok}(X), G)|) &= \sum_{F \in \mathrm{Sur}(\mathbb{Z}_{p}^{n}, G)} \mathrm{Prob}_{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}(FX = 0) \\ &= \frac{|\mathrm{Sur}(\mathbb{Z}_{p}^{n}, G)|}{|\mathrm{Hom}(\mathbb{Z}_{p}^{n}, G)|} \\ &= \frac{|\mathrm{Sur}_{\mathbb{F}_{p}}(\mathbb{F}_{p}^{n}, G/pG)|}{|\mathrm{Hom}_{\mathbb{F}_{p}}(\mathbb{F}_{p}^{n}, G/pG)|} \\ &= \frac{(p^{n}-1)(p^{n}-p) \cdots (p^{n}-p^{r_{p}(G)-1})}{p^{n r_{p}(G)}} \\ &= \left(1 - \frac{1}{p^{n-r_{p}(G)+1}}\right) \left(1 - \frac{1}{p^{n-r_{p}(G)+2}}\right) \cdots \left(1 - \frac{1}{p^{n}}\right) \\ &= \prod_{j=n-r_{p}(G)+1}^{n}(1 - p^{-j})\end{align*}$$ if $r_{p}(G) := \dim_{\mathbb{F}_{p}}(G/pG) \leq n$ and $0$ otherwise. Here, subscript $\mathbb{F}_{p}$ is to indicate linear maps over $\mathbb{F}_{p},$ and reduction over this field is possible due to Nakayama's lemma.

In particular, we have $$\lim_{n \rightarrow \infty} \mathbb{E}_{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}(|\mathrm{Sur}(\mathrm{cok}(X), G)|) = 1.$$

Question. Does the above computation guarantee that $$\lim_{n \rightarrow \infty}\mathrm{Prob}_{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}(\mathrm{cok}(X) \simeq G) = \frac{1}{|\mathrm{Aut}(G)|}\prod_{i=1}^{\infty}(1 - p^{-i}) \ ?$$

First, we need to know whether if $$\mathbb{E}_{H \in \mathcal{F}_{p}}(|\mathrm{Sur}(H, G)|) = 1$$ for any finite abelian $p$-group $G$ when $\mathcal{F}_{p}$ is equipped with the Cohen-Lenstra measure. This is known to be true, and one place to find a proof is Theorem 5.3 of this paper by Fulman and Kaplan.

Next, we need to know that if this is enough to answer the question. That is, we need a version of Lévy's theorem in this setting. This also turned out to be okay. (For example, see Theorem 3.1 of this paper by Wood--the proof of it seems to be in another paper of her mentioned in the reference.)

Friedman--Washington, Cheong--Huang, and further questions. Long ago, Friedman and Washington already proved that for $n \geq r_{p}(G),$ we have $$\mathrm{Prob}_{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}(\mathrm{cok}(X) \simeq G) = \frac{1}{|\mathrm{Aut}(G)|}\prod_{i=1}^{n}(1 - p^{-i}) \prod_{j=n-r_{p}(G)+1}^{n}(1 - p^{-j})$$ even without considering the surjection moments. Moreover, the probability is $0$ when $n < r_{p}(G),$ so from our previous computation, we have

$$\mathrm{Prob}_{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}(\mathrm{cok}(X) \simeq G) = \frac{\mathbb{E}(|\mathrm{Sur}(\mathrm{cok}(X), G)|) \prod_{i=1}^{n}(1 - p^{-i})}{|\mathrm{Aut}(G)| }.$$

Moreover, from another result of Friedman and Washington, it turns out that

$$\begin{align*}\lim_{n \rightarrow \infty}&\mathrm{Prob}_{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}(\mathrm{cok}(X) = 0, \mathrm{cok}(X - I_{n}) \simeq G) \\ &= \lim_{n \rightarrow \infty}\mathrm{Prob}_{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}(\mathrm{cok}(X) \simeq G, \mathrm{cok}(X - I_{n}) = 0) \\ &= \frac{1}{|\mathrm{Aut}(G)|}\left(\prod_{i=1}^{n}(1 - p^{-i})\right)^{2},\end{align*}$$ where $I_{n}$ is the $n \times n$ identity matrix.

Hence, the following conjecture from CH is reasonable:

Conjecture (Cheong and Huang). Given any finite abelian $p$-groups $G_{1}$ and $G_{2},$ we must have $$\lim_{n \rightarrow \infty}\mathrm{Prob}_{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}(\mathrm{cok}(X) \simeq G_{1}, \mathrm{cok}(X - I_{n}) \simeq G_{2}) = \frac{\left(\prod_{i=1}^{n}(1 - p^{-i})\right)^{2}}{|\mathrm{Aut}(G_{1})||\mathrm{Aut}(G_{2})|}.$$

Remark. We cannot find any possible use of surjection moments to attack this conjecture at the moment.

A special case of Conjecture. Conjecture implies that $$\begin{align*}&\sum_{m={1}}^{\infty}\lim_{n \rightarrow \infty}\underset{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}{\mathrm{Prob}}\begin{pmatrix}\mathrm{cok}(X) \simeq \mathbb{Z}/(p), \\ \mathrm{cok}(X - I_{n}) \simeq \mathbb{Z}/(p^{m}) \end{pmatrix} \\ &= \left(\frac{1}{(p-1)^{2}}\right)\left(\prod_{i=1}^{\infty}(1 - p^{-i})\right)^{2}(1 + p^{-1} + p^{-2} + \cdots) \\ &= \left(\frac{p}{(p-1)^{3}}\right)\left(\prod_{i=1}^{\infty}(1 - p^{-i})\right)^{2},\end{align*}$$ and we can compute this WITHOUT referring to Conjecture, which gives more evidence to the validity of the conjecture.

What can we do further? We believe that the trick we have in mind can compute many more sums like this, which may give more evidence to a more general conjecture given as Conjecture 2.3 in CH.

 

Update (6/6/2022). Nathan Kaplan and I proved the conjecture in our recent paper. Jungin Lee generalized this even further in his recent preprint.