Wednesday, 19 November 2014

Joseph's conjectures on commuting probability, an ultrafinitary perspective

The commuting probability $\Pr(G)$ of a finite group $G$ is the proportion of pairs $(x,y)\in G^2$ which commute. In 1977 Keith Joseph made three conjectures about the set
$$\mathcal{P} = \{\Pr(G) : G \text{ a finite group}\},$$
namely the following.

Joseph's conjectures: 
J1. All limit points of $\mathcal{P}$ are rational.
J2. $\mathcal{P}$ is well ordered by >.
J3. $\{0\}\cup\mathcal{P}$ is closed.

Earlier this month I uploaded a preprint to the arxiv which proves the first two of these conjectures, and yesterday I gave a talk at the algebra seminar in Oxford about the proof. While preparing the talk I noticed that some aspects of the proof are simpler from an ultrafinitary perspective, basically because ultrafilters can be used to streamline epsilon management, and I gave one indication of this perspective during the talk. In this post I wish to lay out the ultrafinitary approach in greater detail.

Throughout this post we fix a nonprincipal ultrafilter $u\in\beta\mathbf{N}\setminus\mathbf{N}$, and we let $\mathbf{R}^*$ be the ultrapower $\mathbf{R}^\mathbf{N}/u$, where two elements of $\mathbf{R}^\mathbf{N}$ are considered equivalent iff they are equal in $u$-almost-every coordinate. The elements of $\mathbf{R}^*$ are called "nonstandard reals" or "hyperreals". There is a principle at work in nonstandard analysis, possibly called Łoś's theorem, which asserts, without going into the finer details, that all "first-order" things that you do with reals carry over in a natural way to the field of hyperreals, and everything more-or-less works just how you'd like it to. For instance, if $r$ and $s$ are hyperreals then $r<s$ naturally means that the inequality holds in $u$-almost-every coordinate, and similarly the field operations of $\mathbf{R}$ extend naturally, and with these definitions $\mathbf{R}^*$ becomes a totally ordered field. We will seldom spell out so explicitly how to naturally extend first-order properties in this way.

An ultrafinite group $G$ is an ultraproduct $\prod_u G_i = \prod_{i=1}^\infty G_i/u$ of finite groups. Its order $|G| = (|G_i|)/u$ is a nonstandard natural number, and its commuting probability $\Pr(G) = (\Pr(G_i))/u$ is a nonstandard real in the interval $[0,1]$. Joseph's first two conjectures can now be stated together in the following way.
Theorem: The commuting probability of every ultrafinite group $G$ has the form $q+\epsilon$, where $q$ is a standard rational and $\epsilon$ is a nonnegative infinitesimal.
Somewhat similarly, Joseph's third conjecture can be stated in an ultrafinitary way as follows: For every ultrafinite group $G$ there is a finite group $H$ such that $\text{st}\Pr(G)=\Pr(H)$. Here $\text{st}$ is the standard part operation, which maps a finite hyperreal to the nearest real. When phrased in this way it resembles a known result about compact groups. Every compact group $G$ has a unique normalised Haar measure, so we have a naturally defined notion of commuting probability $\Pr(G)$. However every compact group $G$ with $\Pr(G)>0$ has a finite-index center, so $\Pr(G)=\Pr(H)$, where $H$ is the finite group $G/Z(G)$. This is a theorem of Hofmann and Russo. Nevertheless, I find Joseph's third conjecture rather hard to believe.

For us the most important theorem about commuting probability is a theorem of P. Neumann, which states that if $\epsilon>0$ then every finite group $G$ such that $\Pr(G)\geq\epsilon$ has a normal subgroup $H$ such that $|G/H|$ and $|[H,H]|$ are both bounded in terms of $\epsilon$. To prove the above theorem we need the following "amplified" version:

Theorem (Neumann's theorem, amplified, ultrafinitary version): Every ultrafinite group $G$ has an internal normal subgroup $H$ such that $[H,H]$ is finite and such that almost every pair $(x,y)\in G^2$ such that $[x,y]\in[H,H]$ is contained in $H^2$.
Here if $G = \prod_u G_i$ we say that $S\subset G$ is internal if $S$ is itself an ultraproduct $\prod_u S_i$ of subsets $S_i\subset G_i$, and "almost every" needs little clarification because the set of pairs in question is an internal subset of $G^2$. (Otherwise we would need to introduce Loeb measure.)

Proof: If $\text{st}\Pr(G)=0$ the theorem holds with $H=1$, so we may assume $\text{st}\Pr(G)>0$. By Neumann's theorem $G$ has an internal normal subgroup $K_0$ of finite index such that $[K_0,K_0]$ is finite. Since $G/K_0$ is finite there are only finitely many normal subgroups $K\leq G$ containing $K_0$ and each of them is internal, so we may find normal subgroups $K,L\leq G$ containing $K_0$ such that $[K,L]$ is finite, and which are maximal subject to these conditions.

Suppose that a positive proportion of pairs $(x,y)\in G^2$ outside of $K\times L$ satisfied $[x,y]\in[K,L]$. Then we could find $(x,y)\in G^2\setminus (K\times L)$, say with $x\notin K$, such that for a positive proportion of $(k,l)\in K\times L$ we have $[xk,yl]\in[K,L]$. After a little commutator algebra one can show then that for a positive proportion of $l\in L$ we have $[x,l]\in[K,L]$, or in other words that the centraliser
$$ N_0 = C_{L/[K,L]}(x) = C_{L/[K,L]}(\langle K,x\rangle)$$
of $x$ in $L/[K,L]$ has finite index. But this implies that the largest normal subgroup contained in $N_0$, namely
$$ N = C_{L/[K,L]}(K'),$$
where $K'$ is the normal subgroup of $G$ generated by $K$ and $x$, also has finite index. Since certainly $K\leq C_{K'/[K,L]}(L)$ a classical theorem of Baer implies that
$$[K'/[K,L], L/[K,L]] = [K',L]/[K,L]$$
is finite, and hence that $[K',L]$ is finite, but this contradicts the maximality of $K$ and $L$.

Hence almost every pair $(x,y)\in G^2$ such that $[x,y]\in[K,L]$ is contained in $K\times L$, and thus also in $L\times K$, so the theorem holds for $H=K\cap L$.$\square$

Now let $G$ be any ultrafinite group $G$ and let $H$ be as in the theorem. Then
$$\Pr(G) = \Pr(H) + \epsilon,$$
where $\epsilon$ is nonnegative and infinitesimal. Thus it suffices to show that $\Pr(H)$ is a standard rational whenever $[H,H]$ is finite. Note in this case that Hall's theorem implies that the second centre
$$Z_2(H) = \{h\in H : [h,H]\subset Z(H)\}$$
has finite index. One can complete the proof using a little duality theory of abelian groups, but the ultrafinite perspective adds little here so I refer the reader to my paper.

The other thing I noticed while preparing my talk is that the best lower bound I knew for the order type of $\mathcal{P}$, $\omega^2$, is easy to improve to $\omega^\omega$, just by remembering that $\mathcal{P}$ is a subsemigroup of $[0,1]$. In fact the order type of a well ordered subsemigroup of $(0,1]$ is heavily restricted: it's either $0$, $1$, or $\omega^{\omega^\alpha}$ for some ordinal $\alpha$. This observation reduces the possibilities for the order type of $\mathcal{P}$ to $\{\omega^\omega,\omega^{\omega^2}\}$. I have no idea which it is!

Thursday, 13 November 2014

The idempotent theorem

Let $G$ be a locally compact abelian group and let $M(G)$ be the Banach algebra of regular complex Borel measures on $G$. Given $\mu\in M(G)$ its Fourier transform
$$\hat{\mu}(\gamma) = \int \overline{\gamma}\,d\mu,$$
is a continuous function defined on the Pontryagin dual $\hat{G}$ of $G$. If the measure $\mu$ is "nice" in some way then we expect some amount of regularity from the function $\hat{\mu}$. For instance if $\mu$ is actually an element of the subspace $L^1(G)\subset M(G)$ of measures absolutely continuous with respect to the Haar measure of $G$ then the Riemann-Lebesgue lemma asserts $\hat{\mu}\in C_0(\hat{G})$.

The idempotent theorem of Cohen, Helson, and Rudin describes the structure of measures $\mu$ whose Fourier transform $\hat{\mu}$ takes a discrete set of values, or equivalently, since $\|\hat{\mu}\|_\infty\leq\|\mu\|$, a finite set of values. To describe the theorem, note that we can define $P(\mu)$ for any polynomial $P$ by taking appropriate linear combinations of convolution powers of $\mu$, and moreover we have the relation $\widehat{P(\mu)} = P(\hat{\mu})$, where on the right hand side we apply $P$ pointwise. Thus if $\hat{\mu}$ takes only the values $a_1,\dots,a_n$ then by setting
$$P_i(x) = \prod_{j\neq i} (x-a_j)/(a_i-a_j)$$
we obtain a decomposition $\mu = a_1\mu_1 + \cdots + a_n\mu_n$ of $\mu$ into a linear combination of measures $\mu_i=P_i(\mu)$ whose Fourier transforms $\hat{\mu_i} = P_i(\hat{\mu})$ take only values $0$ and $1$. Such measures are called idempotent, because they are equivalently defined by $\mu\ast\mu=\mu$. By the argument just given it suffices to characterise idempotent measures: this explains the name of the theorem.

The most obvious example of an idempotent measure is the Haar measure $m_H$ of a compact subgroup $H\leq G$. Moreover we can multiply any idempotent measure $\mu$ by a character $\gamma\in\hat{G}$ to obtain a measure $\gamma\mu$ defined by
$$\int f \,d(\gamma\mu) = \int f\gamma\,d\mu.$$
This measure $\gamma\mu$ will again be idempotent, as
$$\int f\,d(\gamma \mu\ast\gamma \mu) = \int\int f(x+y)\gamma(x)\gamma(y)\,d\mu(x)d\mu(y) = \int\int f(x+y) \gamma(x+y)\,d\mu(x)d\mu(y) = \int f\gamma\,d\mu.$$
If we add or subtract two idempotent measures then though we may not have again an idempotent measure we certainly have a measure whose Fourier transform takes integer values. On reflection, it feels more natural in the setting of harmonic analysis to require that $\hat{\mu}$ takes values in a certain discrete subgroup than to require that it take values in $\{0,1\}$, so we relax our restriction so. The idempotent theorem states that we have already accounted for all those $\mu$ such that $\hat{\mu}$ is integer-valued.

Theorem (the idempotent theorem): For every $\mu\in M(G)$ such that $\hat{\mu}$ is integer-valued there is a finite collection of compact subgroups $G_1,\dots,G_k\leq G$, characters $\gamma_1,\dots,\gamma_k\in\hat{G}$, and integers $n_1,\dots,n_k\in\mathbf{Z}$ such that
$$\mu = n_1\gamma_1 m_{G_1} + \cdots + n_k\gamma_k m_{G_k}.$$
As a consequence we deduce a structure theorem for $\mu$ with $\hat{\mu}$ taking finitely many values, as we originally wanted: for every such $\mu$ there is a finite collection of compact subgroups $G_1,\dots,G_k\leq G$, characters $\gamma_1,\dots,\gamma_k\in\hat{G}$, and complex numbers $a_1,\dots,a_k\in\mathbf{C}$ such that
$$\mu = a_1\gamma_1 m_{G_1} + \cdots + a_k \gamma_k m_{G_k}.$$

The theorem was first proved in the case of $G=\mathbf{R}/\mathbf{Z}$ by Helson in 1953: in this case the theorem states simply that if $\hat{\mu}$ is integer-valued then it differs from some periodic function in finitely many places. In 1959 Rudin gave the theorem its present form and proved it for $(\mathbf{R}/\mathbf{Z})^d$. Finally in 1960 Cohen proved the general case, in the same paper in which he made the first substantial progress on the Littlewood problem. The proof was subsequently simplified a good deal, particularly by Amemiya and Ito in 1964. We reproduce their proof here.

First note that if $\hat{\mu}$ is integer-valued then $\mu$ is supported on a compact subgroup. Indeed by inner regularity there is a compact set $K$ such that $|\mu|(K^c)<0.1$, the set $U$ of all $\gamma\in\hat{G}$ such that $|1-\gamma|<0.1/\|\mu\|$ on $K$ is then open, and if $\gamma\in U$ then
$$\|\gamma\mu-\mu\| = \int_G |\gamma-1|\,d|\mu| \leq \int_K + \int_{K^c} < 0.1 + 0.1 < 1.$$
But if $\gamma\mu\neq\mu$ then
$$\|\gamma\mu-\mu\|\geq \|\widehat{\gamma\mu}-\hat{\mu}\|_\infty \geq 1,$$
so $\gamma\mu=\mu$ for all $\gamma\in U$. Thus $\Gamma=\{\gamma\in\hat{G}: \gamma\mu=\mu\}$ is an open subgroup of $\hat{G}$, so by Pontryagin duality its preannihilator $\Gamma^\perp = \{g\in G: \gamma(g)=1 \text{ for all }\gamma\in\Gamma\}$ is a compact subgroup of $G$. Clearly $\mu$ is supported on $\Gamma^\perp$. Thus from now on we assume $G$ is compact.

Fix a measure $\mu\in M(G)$ and let $A=\{\gamma\mu: \gamma\in\hat{G}\}$.
Lemma 1: If $\nu$ is a weak* limit point of $A$ then $\|\nu\|<\|\mu\|$.
Proof: Fix $\epsilon>0$ and suppose we could find $f\in C(G)$ such that $\|f\|_\infty\leq 1$ and $\int f\,d\nu > (1-\epsilon)\|\mu\|$. Let $\gamma\mu$ be close enough to $\nu$ that $\Re\int f\gamma\,d\mu > (1-\epsilon)\|\mu\|$. Write $\mu = \theta|\mu|$ and $f\gamma\theta = g + ih$. Then if $Z$ is the complex number $Z = \int (g+i|h|)\,d|\mu|$, then $|Z|\leq\|\mu\|$ and
$$\Re Z = \int g \,d|\mu| = \Re\int f\gamma\,d\mu > (1-\epsilon)\|\mu\|,$$
so we must have
$$\Im Z = \int |h|\,d|\mu| \leq (1-(1-\epsilon)^2)^{1/2}\|\mu\| \leq 2\epsilon^{1/2}\|\mu\|.$$
Thus also
$$\int |1 - f\gamma\theta| \,d|\mu| \leq \int |1 - g|\,d|\mu| + \int |h|\,d|\mu| \leq 3\epsilon^{1/2}\|\mu\|.$$
But if this holds for both $\gamma_1\mu$ and $\gamma_2\mu$, say with $\gamma_1\mu\neq\gamma_2\mu$, then we have
$$ 1\leq \|\gamma_1\mu-\gamma_2\mu\| \leq \int |\gamma_1 - f\gamma_1\gamma_2\theta|\,d|\mu| + \int |\gamma_2 - f\gamma_1\gamma_2\theta|\,d|\mu| \leq 6\epsilon^{1/2}\|\mu\|,$$
so $\epsilon \geq 1/(36\|\mu\|^2)$, so
$$\|\nu\| \leq \|\mu\| - \frac{1}{36\|\mu\|}.\square$$
Lemma 2: If $\nu$ is a weak* limit point of $A$ then $\nu$ is singular with respect to the Haar measure $m_G$ of $G$.
Proof: By the Radon-Nikodym theorem we have a decomposition $\mu = f m_G + \mu_s$ for some $f\in L^1(G)$ and some $\mu_s$ singular with respect to $m_G$. By the Riemann-Lebesgue lemma then $\nu$ is a limit point of $\{\gamma\mu_s:\gamma\in\hat{G}\}$. Thus for any open set $U$ and $f\in C(G)$ such that $\|f\|_\infty\leq 1$ and $f=0$ outside of $U$ we have
$$\left|\int f\,d\nu\right| \leq \sup_\gamma \left|\int f\gamma \,d\mu_s\right| \leq |\mu_s|(U),$$
so $|\nu|(U)\leq |\mu_s|(U)$. This inequality extends to Borel sets in the usual way, so $\nu$ is singular.$\square$

The theorem follows relatively painlessly from the two lemmas. Fix $\mu\in M(G)$ with $\hat{\mu}$ integer-valued and let $A = \{\gamma\mu: \int\gamma\,d\mu\neq 0\}$. Then $\overline{A}$ is weak* compact, so because $\|\cdot\|$ is lower semicontinuous in the weak* topology there is some $\nu\in\overline{A}$ of minimal norm. Since $\int d\nu$ is an integer different from $0$ we must have $\nu\neq 0$. Thus by Lemma 1 the set $\{\gamma\nu: \int\gamma\,d\nu\neq 0\}$ is finite. But this implies that
$$\nu = (n_1 \gamma_1 + \cdots + n_k \gamma_k) m_H\qquad(\ast)$$
for some $n_1,\dots,n_k\in\mathbf{Z}$, $\gamma_1,\dots,\gamma_k\in\hat{G}$, and $H=\{\gamma:\gamma\nu=\nu\}^\perp$ the support group of $\nu$. In particular $\nu$ is absolutely continuous with respect to $m_H$, so because $\nu|_H$ is in the weak* closure of $\{\gamma\mu|_H:\gamma\in\hat{G}\}$ we deduce from Lemma 2 that $\nu|_H = \gamma\mu|_H$ for some $\gamma$.  Thus $\mu|_H$ is a nonzero measure of the form $(\ast)$ and we have an obvious mutually singular decomposition
$$\mu = \mu|_H + (\mu-\mu|_H).$$
Since $\|\mu-\mu|_H\| = \|\mu\| - \|\mu|_H\|\leq\|\mu\|-1$ the theorem follows by induction.

Tuesday, 1 April 2014

Erdős-Turán statistical group theory

What is Erdős-Turán "statistical group theory"? Erdős and Turán published a series of seven papers with this title, from 1965 to 1972, in which they proved many beautiful statistical or counting results about permutations. For example, if $|\sigma|$ denotes the order of a permutation $\sigma\in S_n$, they showed that when $\sigma$ is chosen uniformly at random $\log|\sigma|$ is approximately normally distributed, with mean $(\log n)^2/2$ and variance $(\log n)^3/3$. Another typical result, though much simpler, is that if $A$ is any subset of $\{1,\dots,n\}$, then the probability that a random permutation contains no cycle with length in $A$ is at most $2\left(\sum_{a\in A} 1/a\right)^{-1}$.

Although most of their results are of the above approximate nature, they prove at least one beautiful exact counting result, and I thought I might relate it here.

Theorem: If $q$ is a prime power then the proportion of $\sigma\in S_n$ with order not divisible by $q$ is exactly
\[\left(1-\frac{1}{q}\right)\left(1-\frac{1}{2q}\right)\cdots\left(1-\frac{1}{\lfloor n/q\rfloor q}\right).\]

Proof: The number of $\sigma\in S_n$ with $m_1$ cycles of length $v_1$, $m_2$ cycles of length $v_2$, $\dots$, and $m_k$ cycles of length $v_k$, where $m_1 v_1 + \cdots m_k v_k = n$, is
\[\frac{n!}{m_1!\cdots m_k! v_1^{m_1}\cdots v_k^{m_k}}:\]
indeed, one can partition $\{1,\dots,n\}$ into $m_i$ sets of size $v_i$ for each $i$ in
\[\frac{n!}{m_1!\cdots m_k! v_1!^{m_1}\cdots v_k!^{m_k}}\]
ways, and then one can arrange each of the $m_i$ sets of size $v_i$ for each $i$ into cycles in
\[(v_1 - 1)!^{m_1} \cdots (v_k - 1)!^{m_k}\]
ways. Moreover, the order of every such $\sigma$ is $\text{lcm}(v_1,\dots,v_k)$, so the order of $\sigma$ is divisible by $q$ if and only if some $v_i$ is divisible by $q$. (This is the where we use the hypothesis that $q$ is a prime power.) Thus the proportion of $\sigma\in S_n$ with order not divisible by $q$ is
\[\sum\frac{1}{m_1!\cdots m_k! v_1^{m_1} \cdots v_k^{m_k}},\]
where the sum runs over all $k\geq 0$ and $2k$-tuples $(m_1,\dots,m_k,v_1,\dots,v_k)$ of positive integers such that $m_1 v_1 + \cdots m_k v_k = n$ and such that no $v_i$ is divisible by $q$. But this is just the coefficient of $X^n$ in
\[\prod_{v:q\nmid v} \sum_{m\geq 0} \frac{X^{mv}}{m! v^m}\\=\prod_{v:q\nmid v} \exp\left(\frac{X^v}{v}\right)\\= \exp\left(\sum_{v:q\nmid v} \frac{X^v}{v}\right)\\= \exp\left(\sum_v \frac{X^v}{v} - \sum_v \frac{X^{qv}}{qv}\right)\\= \exp\left(-\log(1-X) + \log(1-X^q)/q\right)\\ = \frac{(1-X^q)^{1/q}}{1-X}\\= (1 + X + \cdots + X^{q-1}) (1-X^q)^{-\frac{q-1}{q}}\\= (1+X+\cdots+X^{q-1}) \left(1 + \left(1-\frac{1}{q}\right) X^q + \left(1-\frac{1}{q}\right)\left(1-\frac{1}{2q}\right)X^{2q} + \cdots\right),\]
where the last equality follows from the binomial formula. This completes the proof.

Such an explicit formula should make us feel foolish for having used generating functions. Is there a direct combinatorial proof?

By the way, for those who are not already aware of this invaluable resource, almost all of Erdős's papers have been made freely available by the Rényi institute here: