## 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 abelian subgroup, and with a little more work one can actually find a finite group $H$ with $\Pr(G)=\Pr(H)$. 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 Peter 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
$$\Pr(G) = \frac1{|G/H|^2} \Pr(H) + \epsilon,$$
where $\epsilon$ is nonnegative and infinitesimal. Thus it suffices to show that $\Pr(H)$ has the form (standard rational) + (nonnegative infinitesimal) 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.