An Expectation of Cohen-Lenstra Measure

Question

The Cohen-Lenstra measure on the set of abelian p-groups assigns $\mathbb{P}(G) = \prod_{i \geq 1} \left( 1 - \frac{1}{p^i}\right) \cdot |\mathrm{Aut}(G)|^{-1} $. Apparently, this is equivalent to taking cokernels of random maps $f: (\mathbb{Z}_p)^N \to (\mathbb{Z}_p)^N$ and letting $N \to \infty$. These are the p-adics and there is a Haar measure on this linear space of maps. Alternatively, choose random maps between the finite groups: $f: (\mathbb{Z}/p^k\mathbb{Z})^N \to (\mathbb{Z}/p^k \mathbb{Z})^N$ and let $k, N \to \infty$.

Q: If G is a random abelian p-group according to the Cohen-Lenstra measure and A is a deterministic, why is the expected number of surjections $\phi: G \to A$ equal to 1? In fact, if G were deterministic I don't think this number could ever be 1 unless |G| = 1.

For references see Section 8 of Homological stability for Hurwitz spaces and the Cohen-Lenstra conjecture over function fields by Ellenberg, Venkatesh and Westerland or Terry Tao's blog entry At the AustMS conference.

Answer 1

Let me spell out the cokernel description of the Cohen-Lenstra distribution in more detail, as my answer will depend on it.

A map $(\mathbb{Z}_p)^N \to (\mathbb{Z}_p)^N$ is given by an $N \times N$ matrix of $p$-adic integers. Choose such a map by picking each of the digits of each integer uniformly at random from $\{ 0, 1, ..., p-1 \}$; this is the same as using the additive Haar measure on $\mathrm{Hom}((\mathbb{Z}_p)^N, (\mathbb{Z}_p)^N)$. With probability $1$, this map does not have determinant $0$, so its cokernel is a finite abelian $p$-group. Let $\mu_N$ be the probability measure on isomorphism classes of abelian $p$-groups which assigns each $p$-group the probability that it arises as this cokernel.

The Cohen-Lenstra distribution is the limit as $N \to \infty$ of $\mu_N$. As shown in several of the references you link to, it is given by the formula $$\lim_{N \to \infty} \mu_N(G) = \frac{1}{ |\mathrm{Aut}(G)|} \prod_{i=1}^{\infty} (1-1/p^i) .$$

For notational convenience, it will help to distinguish between the domain and range of a map in $\mathrm{Hom}((\mathbb{Z}_p)^N, (\mathbb{Z}_p)^N)$. I will call the former $U^N$ and the latter $V^N$.

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Now, to answer your question. Let $A$ be a fixed finite abelian $p$-group. Let $e_N(A)$ be the expected number of surjections from an abelian $p$-group $G$, picked according to measure $\mu_N$, to $A$. Ignoring issues about interchanging limits, we want to show that $\lim_{N \to \infty} e_N(A)=1$.

Lets start by considering $H_N(A) := \mathrm{Hom}(V^N, A)$. The set $H_N(A)$ has cardinality $|A|^N$, as any map is specified by giving the image of a basis for $V^N$. Inside this set, let $S_N(A)$ be the surjective maps and $C_N(A)$ the nonsurjective maps.

For any map $f \in S_N(A)$, let's consider the possibility that it extends to the cokernel of a random map $U^N \to V^N$. This will occur if and only if the $N$ generators of $U^N$ land in the kernel of $f$. Since $f$ is in $S_N(A)$, its kernel has index $|A|$. So the probability that $U^N$ is mapped into the kernel of $f$ is $1/|A|^N$.

We want to compute $$e_N(A) = |S_N(A)| \cdot (1/|A|^N) = 1 - |C_N(A)|/A^N.$$

If $A$ can be generated by $r$ elements, then $|C_N(A)|/A^N \leq (1-1/|A|^r)^{\lfloor N/r \rfloor}$, so the second term drops out as $N \to \infty$. (To see this bound, group the basis elements of $V^N$ into $N/r$ groups of size $r$; the probability that these $r$ basis elements are not sent to the $r$ generators of $A$ is $(1-1/|A|^r)$. This bound is probably much weaker than the true rate of convergence.)

Answer 2

To corroborate David's answer, I will compute $e_{n}(A)$, mainly because my guess in the above comment I made seems to be wrong, and I cannot erase it. (I think I did not see the possibility of choosing other $r$ cyclic generators.)

We have

$$e_{n}(A) = \frac{\#\text{Surj}(\mathbb{Z}_{p}^{n}, A)}{\#\text{Hom}(\mathbb{Z}_{p}^{n}, A)},$$

which is the proportion of $\mathbb{Z}_{p}$-linear maps $\mathbb{Z}_{p}^{n} \rightarrow A$ that are surjective. We may assume

$$A = \mathbb{Z}_{p}/(p^{\lambda_{1}}) \oplus \cdots \oplus \mathbb{Z}_{p}/(p^{\lambda_{l}}).$$

Now, note that any given map $\phi : \mathbb{Z}_{p}^{n} \rightarrow A$, its surjectivity can be checked by considering the surjectivity mod $p$, by Nakayama lemma. Moreover, the map

$$\text{Hom}_{\mathbb{Z}_{p}}(\mathbb{Z}_{p}^{n}, A) \twoheadrightarrow \text{Hom}_{\mathbb{F}_{p}}(\mathbb{F}_{p}^{n}, A/pA)$$

given by the mod $p$ reduction is a group homomorphism, so each of its fiber must have the same size. This implies that

$$e_{n}(A) = \frac{\#\text{Surj}_{\mathbb{F}_{p}}(\mathbb{F}_{p}^{n}, A/pA)}{\#\text{Hom}_{\mathbb{F}_{p}}(\mathbb{F}_{p}^{n}, A/pA)} = \frac{\#\text{Surj}_{\mathbb{F}_{p}}(\mathbb{F}_{p}^{n}, \mathbb{F}_{p}^{l})}{\#\text{Hom}_{\mathbb{F}_{p}}(\mathbb{F}_{p}^{n}, \mathbb{F}_{p}^{l})},$$

which is equal to $0$ if $n < l$ and to the proportion of $n \times l$ $\mathbb{F}_{p}$-matrices with rank $l$ if $n \geq l$. When $n \geq l$, we get

$$e_{n}(A) = \frac{(p^{n} - 1)(p^{n} - p)(p^{n} - p^{2}) \cdots (p^{n} - p^{l-1})}{p^{nl}} \\ = \left(1 - \frac{1}{p^{n-l+1}}\right)\left(1 - \frac{1}{p^{n-l+2}}\right) \cdots \left(1 - \frac{1}{p^{n}}\right).$$