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Boolean rings, ultrafilters, and Stone’s representation theorem

Recently, I have begun to appreciate the use of ultrafilters to clean up proofs in certain areas of mathematics. I’d like to talk a little about how this works, but first I’d like to give a hefty amount of motivation for the definition of an ultrafilter.

Terence Tao has already written a great introduction to ultrafilters with an eye towards nonstandard analysis. I’d like to introduce them from a different perspective. Some of the topics below are also covered in these posts by Todd Trimble.

A little more about zeta functions and statistical mechanics

In the previous post we described the following result characterizing the zeta distribution.

Theorem: Let $a_n = \mathbb{P}(X = n)$ be a probability distribution on $\mathbb{N}$. Suppose that the exponents in the prime factorization of $n$ are chosen independently and according to a geometric distribution, and further suppose that $a_n$ is monotonically decreasing. Then $a_n = \frac{1}{\zeta(s)} \left( \frac{1}{n^s} \right)$ for some real $s > 1$.

I have been thinking about the first condition, and I no longer like it. At least, I don’t like how I arrived at it. Here is a better way to conceptualize it: given that $n | X$, the probability distribution on $\frac{X}{n}$ should be the same as the original distribution on $X$. By Bayes’ theorem, this is equivalent to the condition that

$\displaystyle \frac{a_{mn}}{a_n + a_{2n} + a_{3n} + ...} = \frac{a_m}{a_1 + a_2 + ...}$

which in turn is equivalent to the condition that

$\displaystyle \frac{a_{mn}}{a_m} = \frac{a_n + a_{2n} + a_{3n} + ...}{a_1 + a_2 + a_3 + ...}$.

(I am adopting the natural assumption that $a_n > 0$ for all $n$. No sense in excluding a positive integer from any reasonable probability distribution on $\mathbb{N}$.) In other words, $\frac{a_{mn}}{a_m}$ is independent of $m$, from which it follows that $a_{mn} = c a_m a_n$ for some constant $c$. From here it already follows that $a_n$ is determined by $a_p$ for $p$ prime and that the exponents in the prime factorization are chosen geometrically. And now the condition that $a_n$ is monotonically decreasing gives the zeta distribution as before. So I think we should use the following characterization theorem instead.

Theorem: Let $a_n = \mathbb{P}(X = n)$ be a probability distribution on $\mathbb{N}$. Suppose that $a_{nm} = c a_n a_m$ for all $n, m \ge 1$ and some $c$, and further suppose that $a_n$ is monotonically decreasing. Then $a_n = \frac{1}{\zeta(s)} \left( \frac{1}{n^s} \right)$ for some real $s > 1$.

More generally, the following situation covers all the examples we have used so far. Let $M$ be a free commutative monoid on generators $p_1, p_2, ...$, and let $\phi : M \to \mathbb{R}$ be a homomorphism. Let $a_m = \mathbb{P}(X = m)$ be a probability distribution on $M$. Suppose that $a_{nm} = c a_n a_m$ for all $n, m \in M$ and some $c$, and further suppose that if $\phi(n) \ge \phi(m)$ then $a_n \le a_m$. Then $a_m = \frac{1}{\zeta_M(s)} e^{-\phi(m) s}$ for some $s$ such that the zeta function

$\displaystyle \zeta_M(s) = \sum_{m \in M} e^{-\phi(m) s}$

converges. Moreover, $\zeta_M(s)$ has the Euler product

$\displaystyle \zeta_M(s) = \prod_{i=1}^{\infty} \frac{1}{1 - e^{- \phi(p_i) s}}$.

Recall that in the statistical-mechanical interpretation, we are looking at a system whose states are finite collections of particles of types $p_1, p_2, ...$ and whose energies are given by $\phi(p_i)$; then the above is just the partition function. In the special case of the zeta function of a Dedekind abstract number ring, $M = M_R$ is the commutative monoid of nonzero ideals of $R$ under multiplication, which is free on the prime ideals by unique factorization, and $\phi(I) = \log N(I)$. In the special case of the dynamical zeta function of an invertible map $f : X \to X$, $M = M_X$ is the free commutative monoid on orbits of $f$ (equivalently, the invariant submonoid of the free commutative monoid on $X$), and $\phi(P) = \log |P|$, where $|P|$ is the number of points in $P$.

An interesting result that demonstrates, among other things, the ubiquity of $\pi$ in mathematics is that the probability that two random positive integers are relatively prime is $\frac{6}{\pi^2}$. A more revealing way to write this number is $\frac{1}{\zeta(2)}$, where
$\displaystyle \zeta(s) = \sum_{n \ge 1} \frac{1}{n^s}$
is the Riemann zeta function. A few weeks ago this result came up on math.SE in the following form: if you are standing at the origin in $\mathbb{R}^2$ and there is an infinitely thin tree placed at every integer lattice point, then $\frac{6}{\pi^2}$ is the proportion of the lattice points that you can see. In this post I’d like to explain why this “should” be true. This will give me a chance to blog about some material from another math.SE answer of mine which I’ve been meaning to get to, and along the way we’ll reach several other interesting destinations.