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## Ultrafilters in topology

Remark: To forestall empty set difficulties, whenever I talk about arbitrary sets in this post they will be non-empty.

We continue our exploration of ultrafilters from the previous post. Recall that a (proper) filter on a poset $P$ is a non-empty subset $F$ such that

1. For every $x, y \in F$, there is some $z \in F$ such that $z \le x, z \le y$.
2. For every $x \in F$, if $x \le y$ then $y \in F$.
3. $P$ is not the whole set $F$.

If $P$ has finite infima (meets), then the first condition, given the second, can be replaced with the condition that if $x, y \in F$ then $x \wedge y \in F$. (This holds in particular if $P$ is the poset structure on a Boolean ring, in which case $x \wedge y = xy$.) A filter is an ultrafilter if in addition it is maximal under inclusion among (proper) filters. For Boolean rings, an equivalent condition is that for every $x \in B$ either $x \in F$ or $1 - x \in F$, but not both. Recall that this condition tells us that ultrafilters are precisely complements of maximal ideals, and can be identified with morphisms in $\text{Hom}_{\text{BRing}}(B, \mathbb{F}_2)$. If $B = \text{Hom}(S, \mathbb{F}_2)$ for some set $S$, then we will sometimes call an ultrafilter on $B$ an ultrafilter on $S$ (for example, this is what people usually mean by “an ultrafilter on $\mathbb{N}$“).

Today we will meander towards an ultrafilter point of view on topology. This point of view provides, among other things, a short, elegant proof of Tychonoff’s theorem.

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## Zeta functions, statistical mechanics and Haar measure

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.

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## Graded representation theory

Let $G$ be a group and let

$\displaystyle V = \bigoplus_{n \ge 0} V_n$

be a graded representation of $G$, i.e. a functor from $G$ to the category of graded vector spaces with each piece finite-dimensional. Thus $G$ acts on each graded piece $V_i$ individually, each of which is an ordinary finite-dimensional representation. We want to define a character associated to a graded representation, but if a character is to have any hope of uniquely describing a representation it must contain information about the character on every finite-dimensional piece simultaneously. The natural definition here is the graded trace

$\displaystyle \chi_V(g) = \sum_{n \ge 0} \chi_{V_n}(g) t^n$.

In particular, the graded trace of the identity is the graded dimension or Hilbert series of $V$.

Classically a case of particular interest is when $V_n = \text{Sym}^n(W^{*})$ for some fixed representation $W$, since $V = \text{Sym}(W^{*})$ is the symmetric algebra (in particular, commutative ring) of polynomial functions on $W$ invariant under $G$. In the nicest cases (for example when $G$ is finite), $V$ is finitely generated, hence Noetherian, and $\text{Spec } V$ is a variety which describes the quotient $W/G$.

In a previous post we discussed instead the case where $V_n = (W^{*})^{\otimes n}$ for some fixed representation $W$, hence $V$ is the tensor algebra of functions on $W$. I thought it might be interesting to discuss some generalities about these graded representations, so that’s what we’ll be doing today.

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## Irreducible components

If it wasn’t clear, in this discussion all rings are assumed commutative.

Given a variety like $xy = 0$ we’d like to know if there’s a natural way to decompose it into its “components” $x = 0, y = 0$. These aren’t its connected components in the topological sense, but in any reasonable sense the two parts are unrelated except possibly where they intersect. It turns out that the Noetherian condition is a natural way to answer this question. In fact, we will see that the Noetherian condition allows us to write $\text{MaxSpec } R$ uniquely as a union of a finite number of “components” which have a natural property that is stronger than connectedness.

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## The Noetherian condition as compactness

Let’s think more about what an abstract theory of unique factorization of primes has to look like. One fundamental property it has to satisfy is that factorizations should be finite. Another way of saying this is that the process of writing elements as products of other elements (up to units) should end in a finite set of irreducible elements at some point. This condition is clearly not satisfied by sufficiently “large” commutative rings such as $\mathbb{C}[x, x^{ \frac{1}{2} }, x^{ \frac{1}{3} }, ... ]$, the ring of fractional polynomials.

Since we know we should think about ideals instead of numbers, let’s recast the problem in a different way: because we can write $x^{r} = x^{ \frac{r}{2} } x^{ \frac{r}{2} }$ for any $r$, the ascending chain of ideals $(x) \subset (x^{ \frac{1}{2} }) \subset (x^{ \frac{1}{4} }) \subset ...$ never terminates. In any reasonable theory of factorization writing $f = f_1 g_1$ and then comparing the ideals $(f) \subset (f_1)$, then repeating this process to obtain a chain of ideals $(f) \subset (f_1) \subset (f_2) \subset ...$, should eventually stabilize at a prime. This leads to the following definition.

Definition: A commutative ring $R$ is Noetherian if every ascending chain of ideals stabilizes.

Akhil’s posts at Delta Epsilons here and here describe the basic properties of Noetherian rings well, including the proof of the following.

Hilbert’s Basis Theorem: If $R$ is a Noetherian ring, so is $R[x]$.

Today we’ll discuss what the Noetherian condition means in terms of the topology of $\text{MaxSpec}$. The answer turns out to be quite nice.

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