This week I was in my home town of Adelaide, Australia, for the 2009 annual meeting of the Australian Mathematical Society. This was a fairly large meeting (almost 500 participants). One of the highlights of such a large meeting is the ability to listen to plenary lectures in fields adjacent to one’s own, in which speakers can give high-level overviews of a subject without getting too bogged down in the technical details. From the talks here I learned a number of basic things which were well known to experts in that field, but which I had not fully appreciated, and so I wanted to share them here.

The first instance of this was from a plenary lecture by Danny Calegari entitled “faces of the stable commutator length (scl) ball”. One thing I learned from this talk is that in homotopy theory, there is a very close relationship between topological spaces (such as manifolds) on one hand, and groups (and generalisations of groups) on the other, so that homotopy-theoretic questions about the former can often be converted to purely algebraic questions about the latter, and vice versa; indeed, it seems that homotopy theorists almost think of topological spaces and groups as being essentially the same concept, despite looking very different at first glance. To get from a space ${X}$ to a group, one looks at homotopy groups ${\pi_n(X)}$ of that space, and in particular the fundamental group ${\pi_1(X)}$; conversely, to get from a group ${G}$ back to a topological space one can use the Eilenberg-Maclane spaces ${K(G,n)}$ associated to that group (and more generally, a Postnikov tower associated to a sequence of such groups, together with additional data). In Danny’s talk, he gave the following specific example: the problem of finding the least complicated embedded surface with prescribed (and homologically trivial) boundary in a space ${X}$, where “least complicated” is measured by genus (or more precisely, the negative component of Euler characteristic), is essentially equivalent to computing the commutator length of the element in the fundamental group ${\pi(X)}$ corresponding to that boundary (i.e. the least number of commutators one is required to multiply together to express the element); and the stable version of this problem (where one allows the surface to wrap around the boundary ${n}$ times for some large ${n}$, and one computes the asymptotic ratio between the Euler characteristic and ${n}$) is similarly equivalent to computing the stable commutator length of that group element. (Incidentally, there is a simple combinatorial open problem regarding commutator length in the free group, which I have placed on the polymath wiki.)

This theme was reinforced by another plenary lecture by Ezra Getzler entitled “${n}$-groups”, in which he showed how sequences of groups (such as the first ${n}$ homotopy groups ${\pi_1(X),\ldots,\pi_n(X)}$) can be enhanced into a more powerful structure known as an ${n}$-group, which is more complicated to define, requiring the machinery of simplicial complexes, sheaves, and nerves. Nevertheless, this gives a very topological and geometric interpretation of the concept of a group and its generalisations, which are of use in topological quantum field theory, among other things.

Mohammed Abuzaid gave a plenary lecture entitled “Functoriality in homological mirror symmetry”. One thing I learned from this talk was that the (partially conjectural) phenomenon of (homological) mirror symmetry is one of several types of duality, in which the behaviour of maps into one mathematical object ${X}$ (e.g. immersed or embedded curves, surfaces, etc.) are closely tied to the behaviour of maps out of a dual mathematical object ${\hat X}$ (e.g. functionals, vector fields, forms, sections, bundles, etc.). A familiar example of this is in linear algebra: by taking adjoints, a linear map into a vector space ${X}$ can be related to an adjoint linear map mapping out of the dual space ${X^*}$. Here, the behaviour of curves in a two-dimensional symplectic manifold (or more generally, Lagrangian submanifolds in a higher-dimensional symplectic manifold), is tied to the behaviour of holomorphic sections on bundles over a dual algebraic variety, where the precise definition of “behaviour” is category-theoretic, involving some rather complicated gadgets such as the Fukaya category of a symplectic manifold. As with many other applications of category theory, it is not just the individual pairings between an object and its dual which are of interest, but also the relationships between these pairings, as formalised by various functors between categories (and natural transformations between functors). (One approach to mirror symmetry was discussed by Shing-Tung Yau at a distinguished lecture at UCLA, as transcribed in this previous post.)

There was a related theme in a talk by Dennis Gaitsgory entitled “The geometric Langlands program”. From my (very superficial) understanding of the Langlands program, the behaviour of specific maps into a reductive Lie group ${G}$, such as representations in ${G}$ of a fundamental group, étale fundamental group, class group, or Galois group of a global field, is conjecturally tied to specific maps out of a dual reductive Lie group ${\hat G}$, such as irreducible automorphic representations of ${\hat G}$, or of various structures (such as derived categories) attached to vector bundles on ${\hat G}$. There are apparently some tentatively conjectured links (due to Witten?) between Langlands duality and mirror symmetry, but they seem at present to be fairly distinct phenomena (one is topological and geometric, the other is more algebraic and arithmetic). For abelian groups, Langlands duality is closely connected to the much more classical Pontryagin duality in Fourier analysis. (There is an analogue of Fourier analysis for nonabelian groups, namely representation theory, but the link from this to the Langlands program is somewhat murky, at least to me.)

Related also to this was a plenary talk by Akshay Venkatesh, entitled “The Cohen-Lenstra heuristics over global fields”. Here, the question concerned the conjectural behaviour of class groups of quadratic fields, and in particular to explain the numerically observed phenomenon that about ${75.4\%}$ of all quadratic fields ${{\Bbb Q}[\sqrt{d}]}$ (with $d$ prime) enjoy unique factorisation (i.e. have trivial class group). (Class groups, as I learned in these two talks, are arithmetic analogues of the (abelianised) fundamental groups in topology, with Galois groups serving as the analogue of the full fundamental group.) One thing I learned here was that there was a canonical way to randomly generate a (profinite) abelian group, by taking the product of randomly generated finite abelian ${p}$-groups for each prime ${p}$. The way to canonically randomly generate a finite abelian ${p}$-group is to take large integers ${n, D}$, and look at the cokernel of a random homomorphism from ${({\mathbb Z}/p^n{\mathbb Z})^d}$ to ${({\mathbb Z}/p^n{\mathbb Z})^d}$. In the limit ${n,d \rightarrow \infty}$ (or by replacing ${{\mathbb Z}/p^n{\mathbb Z}}$ with the ${p}$-adics and just sending ${d \rightarrow \infty}$), this stabilises and generates any given ${p}$-group ${G}$ with probability

$\displaystyle \frac{1}{|\hbox{Aut}(G)|} \prod_{j=1}^\infty (1 - \frac{1}{p^j}), \ \ \ \ \ (1)$

where ${\hbox{Aut}(G)}$ is the group of automorphisms of ${G}$. In particular this leads to the strange identity

$\displaystyle \sum_G \frac{1}{|\hbox{Aut}(G)|} = \prod_{j=1}^\infty (1 - \frac{1}{p^j})^{-1} \ \ \ \ \ (2)$

where ${G}$ ranges over all ${p}$-groups; I do not know how to prove this identity other than via the above probability computation, the proof of which I give below the fold.

Based on the heuristic that the class group should behave “randomly” subject to some “obvious” constraints, it is expected that a randomly chosen real quadratic field ${{\Bbb Q}[\sqrt{d}]}$ has unique factorisation (i.e. the class group has trivial ${p}$-group component for every ${p}$) with probability

$\displaystyle \prod_{p \hbox{ odd}} \prod_{j=2}^\infty (1 - \frac{1}{p^j}) \approx 0.754,$

whereas a randomly chosen imaginary quadratic field ${{\Bbb Q}[\sqrt{-d}]}$ has unique factorisation with probability

$\displaystyle \prod_{p \hbox{ odd}} \prod_{j=1}^\infty (1 - \frac{1}{p^j}) = 0.$

The former claim is conjectural, whereas the latter claim follows from (for instance) Siegel’s theorem on the size of the class group, as discussed in this previous post. Ellenberg, Venkatesh, and Westerland have recently established some partial results towards the function field analogues of these heuristics.

— 1. ${p}$-groups —

Henceforth the prime ${p}$ will be fixed. We will abbreviate “finite abelian ${p}$-group” as “${p}$-group” for brevity. Thanks to the classification of finite abelian groups, the ${p}$-groups are all isomorphic to the products

$\displaystyle ({\mathbb Z}/p^{n_1}{\mathbb Z}) \times \ldots \times ({\mathbb Z}/p^{n_d}{\mathbb Z})$

of cyclic ${p}$-groups.

The cokernel of a random homomorphism from ${({\mathbb Z}/p^n{\mathbb Z})^d}$ to ${({\mathbb Z}/p^n{\mathbb Z})^d}$ can be written as the quotient of the ${p}$-group ${({\mathbb Z}/p^n{\mathbb Z})^d}$ by the subgroup generated by ${d}$ randomly chosen elements ${x_1,\ldots,x_d}$ from that ${p}$-group. One can view this quotient as a ${d}$-fold iterative process, in which one starts with the ${p}$-group ${({\mathbb Z}/p^n{\mathbb Z})^d}$, and then one iterates ${d}$ times the process of starting with a ${p}$-group ${G}$, and quotienting out by a randomly chosen element ${x}$ of that group ${G}$. From induction, one sees that at the ${j^{th}}$ stage of this process (${0 \leq j \leq d}$), one ends up with a ${p}$-group isomorphic to ${({\mathbb Z}/p^n{\mathbb Z})^{d-j} \times G_j}$ for some ${p}$-group ${G_j}$.

Let’s see how the group ${({\mathbb Z}/p^n{\mathbb Z})^{d-j} \times G_j}$ transforms to the next group ${({\mathbb Z}/p^n{\mathbb Z})^{d-j-1} \times G_{j+1}}$. We write a random element of ${({\mathbb Z}/p^n{\mathbb Z})^{d-j} \times G_j}$ as ${(x,y)}$, where ${x \in ({\mathbb Z}/p^n{\mathbb Z})^{d-j}}$ and ${y \in G_j}$. Observe that for any ${0 \leq i < n}$, ${x}$ is a multiple of ${p^i}$ (but not ${p^{i+1}}$) with probability ${(1-p^{-(d-j)}) p^{-i(d-j)}}$. (The remaining possibility is that ${x}$ is zero, but this event will have negligible probability in the limit ${n \rightarrow \infty}$.) If ${x}$ is indeed divisible by ${p^i}$ but not ${p^{i+1}}$, and ${i}$ is not too close to ${n}$, a little thought will then reveal that ${|G_{j+1}| = p^i |G_j|}$. Thus the size of the ${p}$-groups ${G_j}$ only grow as ${j}$ increases. (Things go wrong when ${i}$ gets close to ${n}$, e.g. ${p^i \geq p^n / |G_j|}$, but the total size of this event as ${j}$ ranges from ${0}$ to ${d}$ sums to be ${o(1)}$ as ${n \rightarrow \infty}$ (uniformly in ${d}$), by using the tightness bounds on ${|G_j|}$ mentioned below. Alternatively, one can avoid a lot of technicalities by taking the limit ${n \rightarrow \infty}$ before taking the limit ${d \rightarrow \infty}$ (instead of studying the double limit ${n,d \rightarrow \infty}$), or equivalently by replacing the cyclic group ${{\mathbb Z}/p^n {\mathbb Z}}$ with the ${p}$-adics ${{\mathbb Z}_p}$.)

The exponentially decreasing nature of the probability ${(1-p^{-(d-j)}) p^{-i(d-j)}}$ in ${i}$ (and in ${d-j}$) furthermore implies that the distribution of ${|G_j|}$ forms a tight sequence in ${n, j, d}$: for every ${\epsilon > 0}$, one has an ${R > 0}$ such that the probability that ${|G_j| \geq R}$ is less than ${\epsilon}$ for all choices of ${n,j,d}$. (This tightness is necessary to prove the equality in (2) rather than just an inequality (from Fatou’s lemma).) Indeed, the probability that ${|G_j| = p^m}$ converges as ${n,d \rightarrow \infty}$ to the ${t^m}$ coefficient in the generating function

$\displaystyle \prod_{k=1}^\infty \sum_{i=0}^\infty t^i (1-p^{-k}) p^{-ik} = \prod_{k=1}^\infty \frac{1-p^{-k}}{1 - t p^{-k}}. \ \ \ \ \ (3)$

In particular, this claim is true for the final cokernel ${G_d}$. Note that this (and the geometric series formula) already yields (1) in the case of the trivial group ${G = \{0\}}$ and the order ${p}$ group ${G = {\mathbb Z}/p{\mathbb Z}}$ (note that ${\hbox{Aut}(G)}$ has order ${1}$ and ${p}$ in these respective cases). But it is not enough to deal with higher groups. For instance, up to isomorphism there are two ${p}$-groups of order ${p^2}$, namely ${{\mathbb Z}/p^2{\mathbb Z}}$ and ${({\mathbb Z}/p{\mathbb Z})^2}$, whose automorphism group has order ${p^2-p}$ and ${(p^2-1)(p^2-p)}$ respectively. Summing up the corresponding two expressions (1) one can observe that this matches the ${t^2}$ coefficient of (3) (after some applications of the geometric series formula). Thus we see that (3) is consistent with the claim (1), but does not fully imply that claim.

To get the full asymptotic (1) we try a slightly different tack. Fix a ${p}$-group ${G}$, and consider the event that the cokernel of a random map ${T: ({\mathbb Z}/p^n{\mathbb Z})^d \rightarrow ({\mathbb Z}/p^n{\mathbb Z})^d}$ is isomorphic to ${G}$. We assume ${n}$ so large that all elements in ${G}$ have order at most ${p^n}$. If this is the case, then there must be a surjective homomorphism ${\phi: ({\mathbb Z}/p^n {\mathbb Z})^d \rightarrow G}$ such that the range of ${T}$ is equal to the kernel of ${\phi}$. The number of homomorphisms from ${({\mathbb Z}/p^n{\mathbb Z})^d}$ to ${G}$ is ${|G|^d}$ (one has to pick ${d}$ generators in ${G}$). If ${d}$ is large, it is easy to see that most of these homomorphisms are surjective (the proportion of such homomorphisms is ${1-o(1)}$ as ${d \rightarrow \infty}$). On the other hand, there is some multiplicity; the range of ${T}$ can emerge as the kernel of ${\phi}$ in ${|\hbox{Aut}(G)|}$ different ways (since any two surjective homomorphisms ${\phi, \phi': ({\mathbb Z}/p^n {\mathbb Z})^d \rightarrow G}$ with the same kernel arise from an automorphism of ${G}$). So to prove (1), it suffices to show that for any surjective homomorphism ${\phi: ({\mathbb Z}/p^n{\mathbb Z})^d \rightarrow G}$, the probability that the range of ${T}$ equals the kernel of ${\phi}$ is

$\displaystyle (1+o(1)) |G|^{-d} \prod_{j=1}^\infty (1 - \frac{1}{p^j}).$

The range of ${T}$ is the same thing as the subgroup of ${({\mathbb Z}/p^n{\mathbb Z})^d}$ generated by ${d}$ random elements ${x_1,\ldots,x_d}$ of that group. The kernel of ${\phi}$ has index ${|G|}$ inside ${({\mathbb Z}/p^n{\mathbb Z})^d}$, so the probability that all of those random elements lie in the kernel of ${\phi}$ is ${|G|^{-d}}$. So it suffices to prove the following claim: if ${\phi}$ is a fixed surjective homomorphism from ${({\mathbb Z}/p^n{\mathbb Z})^d}$ to ${G}$, and ${x_1,\ldots,x_d}$ are chosen randomly from the kernel of ${\phi}$, then ${x_1,\ldots,x_d}$ will generate that kernel with probability

$\displaystyle (1+o(1)) \prod_{j=1}^\infty (1 - \frac{1}{p^j}). \ \ \ \ \ (4)$

But from the classification of ${p}$-groups, the kernel of ${\phi}$ (which has bounded index inside ${({\mathbb Z}/p^n{\mathbb Z})^d}$) is isomorphic to

$\displaystyle ({\mathbb Z}/p^{n-O(1)}{\mathbb Z}) \times \ldots \times ({\mathbb Z}/p^{n-O(1)}{\mathbb Z}) \ \ \ \ \ (5)$

where ${O(1)}$ means “bounded uniformly in ${n}$“, and there are ${d}$ factors here. As in the previous argument, one can now imagine starting with the group (5), and then iterating ${d}$ times the operation of quotienting out by the group generated by a randomly chosen element; our task is to compute the probability that one ends up with the trivial group by applying this process.

As before, at the ${j^{th}}$ stage of the iteration, one ends up with a group of the form

$\displaystyle ({\mathbb Z}/p^{n-O(1)}{\mathbb Z}) \times \ldots \times ({\mathbb Z}/p^{n-O(1)}{\mathbb Z}) \times G_j \ \ \ \ \ (6)$

where there are ${d-j}$ factors of ${({\mathbb Z}/p^{n-O(1)}{\mathbb Z})}$. The group ${G_j}$ is increasing in size, so the only way in which one ends up with the trivial group is if all the ${G_j}$ are trivial. But if ${G_j}$ is trivial, the only way that ${G_{j+1}}$ is trivial is if the randomly chosen element from (6) has a ${({\mathbb Z}/p^{n-O(1)}{\mathbb Z}) \times \ldots \times ({\mathbb Z}/p^{n-O(1)}{\mathbb Z})}$ component which is invertible (i.e. not a multiple of ${p}$), which occurs with probability ${1 - p^{-(d-j)}}$ (assuming ${n}$ is large enough). Multiplying all these probabilities together gives (4).