## What is an infinite word?

In this post, we’ll explore the idea of non-commutative infinitary operations on groups, that is, multiplying together infinitely many elements in a group. This idea arises very naturally in “wild” or “infinitary” algebraic topology. In fact, lately, this has been at the forefront of my approach to the subject. Instead of sets with binary (or other finitary) operations, homotopy groups are viewed simply groups equipped with extra infinite product operations. It turns out that the well-definedness of infinite products as an operation is a crucial assumption of many important theorems.

Anyone who has taken a Calculus sequence or analysis course will be familiar with an infinite sum $\sum_{n=1}^{\infty}a_n=a_1+a_2+a_3+\dots$ which extends the commutative operation of addition on the reals/complex numbers. Even in that situation, you need to take a little bit of care to distinguish absolutely and conditionally convergent series. But here we’re considering the properties of infinite products in non-commutative groups; think free groups where the elements are words. In this situation, the difference between conditional and absolute convergence becomes a little trickier, relating to “infinite” commutativity.

First, let’s review finite words.

Finite Words: The free group $F(X)$ on a set $X$ can be constructed as the group of as reduced finite words in the “alphabet” $X$. In more detail, an element is a finite length word $w=x_{1}^{n_1}x_{2}^{n_2}...x_{m}^{n_m}$ where the “letters” $x_1,x_2,...,x_m$ are elements of $X$, the exponents $n_k$ are non-zero integers (negatives are necessary because a group must have inverses), and the word is reduced in the sense that consecutive letters can’t be the same otherwise we could combine or cancel them according to their exponents, i.e. $x_{j}\neq x_{j+1}$ for $j=1,\dots,m-1$. The exponent $x_{1}^{4}$ is meant to represent the 4-letter word $x_{1}x_{1}x_{1}x_{1}$. Hence the word $w$ above has word length $n_1+n_2+n_3+\dots +n_m$. Multiplication in $F(X)$ is given by concatenation (placing one word after another) and performing any necessary reduction at the place where the words are joined to get a reduced word. For example, if $w_1=x_{1}^{2}x_{2}^{-3}x_{1}x_{3}x_{4}^{-1}$ and $w_2=x_{4}x_{3}^{-2}x_{5}^{7}x_{1}^{-9}$, the product would be

$w_1w_2=x_{1}^{2}x_{2}^{-3}x_{1}x_{3}^{-1}x_{5}^{7}x_{1}^{-9}$

since first, the inverses $x_{4}$ and $x_{4}^{-1}$ would cancel and then $x_{3}$ and $x_{3}^{-2}$ are consecutive and need to be combined.

Here’s another way to look at it: Let $[n]=\{1,2,...,n\}$ be the finite linearly ordered set on $n$-elements and $[0]$ be the emptyset. Again, $X$ is a set of letters and $X^{-1}=\{x^{-1}\mid x\in X\}$ will denote a set of formal inverses. A word in $X$ is a function $w:[n]\to X\cup X^{-1}$ and we identify it with the $n$-letter word $w:=w(1)w(2)\dots w(n)$. If $n=0$, we call $w$ the empty word. A word is reduced if whenever $w(j)=x$, we have $w(j+1)\neq x^{-1}$. (i.e. a letter and it’s inverse never appear consecutively). In any word, one can delete consecutive inverse pairs to obtain a unique reduced representative, which is independent of the order of reduction. The product of the words $w_1:[n]\to X\cup X^{-1}$ and $w_2:[m]\to X\cup X^{-1}$ is now the reduced representative of $w_1w_2:[n+m]\to X\cup X^{-1}$.

Of course, this latter view is equivalent to the first one; however, it highlights an easily overlooked fact. The operation in free groups really comes from the ability to “add” linear orders together, that is, place them side by side and still have a linear order.

## Infinite products in groups

Infinite words and infinite products should extend their finite counterparts. However, we must be careful to jump to any conclusions about how these should work. Let’s start with a naive view that will lead us in the right direction. We want to to take a sequence $g_1,g_2,g_3,\dots$ of elements in a group $G$ and form a new infinite product element $g_{\infty}$ that behaves like the notation $g_1g_2g_3\dots$ suggests it should. In particular, we should also be able to form an infinite product element $g_{n}g_{n+1}g_{n+2}\dots\in G$ out of the terminal subsequence $g_{n},g_{n+1},g_{n+2}\dots$ so that

• $g_{1}^{-1}\left(g_1g_2g_3\dots\right)=g_2g_3g_4\cdots$
• $g_{2}^{-1}g_{1}^{-1}\left(g_1g_2g_3\dots\right)=g_3g_4g_5\dots$
• $g_{3}^{-1}g_{2}^{-1}g_{1}^{-1}\left(g_1g_2g_3\dots\right)=g_4g_5g_6\dots$
• and so on.

In other words, we need to have a sequence $t_1,t_2,t_3,\dots\in G$ of tails where $t_n$ represents $g_ng_{n+1}g_{n+2}\cdots$. Specifically, $t_1=g_{\infty}$ is the desired infinite product itself. The above equations simply mean that the tails must satisfy the equations $t_n=g_nt_{n+1}$ for all $n\in\mathbb{N}$. Hence, to find an infinite product value for $g_1,g_2,g_3,\dots$ we should look for a tail sequence $\{t_n\}$ and check these equations.

But this leaves us in an awkward situation. Given any sequence $g_1,g_2,g_3,\dots$ in any group and any given $g\in G$, I can set $t_1=g$ and start inductively solving $t_{n+1}=g_{n}^{-1}t_n$ to find a tail sequence $t_1,t_2,t_3,\dots$ that realizes the arbitrary element $g$ as the “infinite product.” We conclude that algebraic structure alone is not sufficient to make infinite products in groups become well-defined. However, the equations do tell us that a tail sequence should uniquely determine an infinite product and conversely that the infinite product value should uniquely determine the entire tail sequence.

Warning: Here is a another way to understand why we should not expect to be able to create a well-defined infinite product element $g_1g_2g_3\dots\in G$ out of every sequence $g_1,g_2,g_3,\dots$ in $G$. Suppose $g\neq e$ is a non-identity element and $g_n=g$ for all $n\in\mathbb{N}$. If $g_{\infty}=ggg\dots$, then the tail $t_2$ is also the infinite product of the sequence $g,g,g,\dots$. Hence

$g_{\infty} = ggg\dots = g (ggg\dots ) = g(g_{\infty})$

and cancelling $g_{\infty}$ on the right gives $e=g$; a contradiction.

Introducing a notion of convergence to the identity: Recalling how classical infinite sums and products in $\mathbb{R}$ are defined, we quickly realize that such infinitary operations depend heavily on the topological structure of $\mathbb{R}$. Hence, we must add a notion of convergence in our group $G$. Using infinite sums as a guide, we should be able to form an infinite product of a sequence $g_1,g_2,g_3,\dots$, which is “shrinking” in some sense. So let’s suppose $G$ is a group with identity element $e$. Moreover, $G$ is equipped with a descending filtration of subgroups $G_1\supseteq G_2\supseteq G_3\supseteq \cdots$.

This extra structure will be the data that tells us when sequences are “converging” to $e$. A difference choice of filtration might yield difference notions of convergence and hence a difference infinite product operation.

Definition: A sequence $g_1,g_2,g_3,\dots$ in $G$ is shrinking if for every $m\geq 1$, there exists a $N\geq 1$ such that $g_n\in G_m$ for all $n\geq N$.

Definition: Given shrinking sequences $g_1,g_2,g_3,\dots$ and $t_1,t_2,t_3,\dots$ such that $t_{n}=g_{n}t_{n+1}$ for all $n\in\mathbb{N}$, we call $t_1,t_2,t_3,\dots$ a tail sequence for $g_1,g_2,g_3,\dots$ and $g_{\infty}=t_1$ an infinite product of the sequence $g_1,g_2,g_3,\dots$.

Definition: We say that infinite products are well-defined in the filtered group $(G,\{G_m\})$ if the infinite product value of a sequence $g_1,g_2,g_3,\dots$ is unique.

Theorem: Infinite products in the filtered group $(G,\{G_n\})$ are well-defined if and only if $\bigcap_{m\in\mathbb{N}}G_m=\{e\}$.

Proof. If $e\neq g\in \bigcap_{m\in\mathbb{N}}G_m$, then the constant sequences $e,e,e,\dots$ and $g,g,g,\dots$ are both shrinking tail sequences for the sequence $e,e,e,\dots$. These two tail sequences define infinite product value of the sequence $e,e,e,\dots$ as both $e$ and $g$ respectively. Hence infinite product values are not uniquely defined. For the converse, suppose infinite product values are not unique. Then we have shrinking sequence $g_1,g_2,g_3,\dots$ that allows for two shrinking tail sequences $t_1,t_2,t_3,\dots$ and $s_1,s_2,s_3,\dots$ such that $s_1\neq t_1$. Define $g=s_{1}^{-1}t_1$ and notice $g\neq e$. From our tail sequence equations, we have $s_{n}s_{n+1}^{-1}=g_n=t_{n}t_{n+1}^{-1}$ for all $n$ and so $g_n=s_{n}^{-1}t_{n}=s_{n+1}^{-1}t_{n+1}$ for all $n$. In particular, $g=s_{n}^{-1}t_{n}$ for all $n\in\mathbb{N}$. Fix $m\in \mathbb{N}$. Since $\{s_n\}$ and $\{t_n\}$ are shrinking sequences, these sequences are eventually in $G_m$. Since $G_m$ is a subgroup, $\{s_{n}^{-1}t_{n}\}$ is eventually in $G_m$. Since $m$ was arbitrary, we see that $g\in G_m$ for all $m\in\mathbb{N}$. Therefore, $e\neq g\in \bigcap_{m\in\mathbb{N}}G_m$. $\square$

So given a descending filtration which has trivial intersection, one can meaningfully define infinite products. If one uses a filtration with $G_m=\{e\}$ for sufficiently large $m$, then the infinite product structure will be trivial, i.e. only sequences terminating in $e,e,e,\dots$ will have infinite products.

For full generality: To do this in more generality so that it applies to $(\mathbb{R},+)$ one should use a descending filtration of subsets $\{G_n\}$ containing $e$ such that for given $n$, we have $G_mG_m\subseteq G_n$ for sufficiently large $m$. For complete generality, one should fix a subgroup $S\subseteq G^{\omega}$ of shrinking sequences.

Another warning: We know that even from the nicest situation possible (infinite sums in $\mathbb{R}$) that just because a sequence $g_1,g_2,g_3,\dots$ is shrinking (converging to $0$) does not mean that it yields a “convergent” infinite sum, e.g. the harmonic series $\sum\frac{1}{n}$. Hence the necessity of tails really is apparent even in basic analysis.

## Connection to Infinitary Homotopy Groups

The motivation here is topological in nature: given a first countable space $X$, a point $x_0\in X$, and a sequence of maps $\alpha_k:([0,1]^n,\partial [0,1]^n)\to (X,x_0)$ converging to the constant loop at $x_0$, we can define the infinite concatenation $\alpha_{\infty}:([0,1]^n,\partial [0,1]^n)\to (X,x_0)$ to be the map which is (after a suitable scaling) $\alpha_k$ on $\left[\frac{k-1}{k},\frac{k}{k+1}\right]\times [0,1]^{n-1}$ and $\alpha_{\infty}(\{1\}\times [0,1]^{n-1})=x_0$. Then we could consider the homotopy class $[\alpha_{\infty}]$ to be an infinite product of the sequence $[\alpha_k]$ in the n-th homotopy group $\pi_n(X,x_0)$. What’s really happening here is that we have a well-defined infinitary operation on the space of maps $([0,1]^n,\partial [0,1]^n)\to (X,x_0)$ and we’re hoping it descends to a well-defined operation on homotopy classes.

If we want to use a filtration, we should assume $X$ is first countable at $x_0$. If you don’t, you may not have any natural infinite products. For example, if $Y=\omega_1+1$ is the first compact uncountable ordinal, then the reduced suspension $\Sigma Y$ doesn’t allow for sequences like $\alpha_k$; the fundamental group is isomorphic to the free group $F(\omega_1)$ where there are no meaningful infinite products. Let $U_1\supseteq U_2\supseteq U_3\supseteq...$ be a countable neighborhood base at $x_0$ and let $G_m$ be the image of the homomorphism $\pi_n(U_m,x_0)\to\pi_n(X,x_0)$ induced by inclusion $U_m\to X$. Now $G_1\supseteq G_2\supseteq G_3\supseteq...$ is a descending filtration of $G=\pi_n(X,x_0)$. According to the theorem above, infinite products in $\pi_n(X,x_0)$ defined using infinite concatenations of maps at $x_0$ and those defined using the filtration agree and are unique if there aren’t any non-trivial classes in $\pi_n(X,x_0)$ that have arbitrarily small representatives. You have such problematic classes in the fundamental group of the Harmonic Archipelago and it’s higher dimensional analogues. In those groups, you have a notion of infinite product coming from a filtration but they are not uniquely defined, i.e. you may have $[\alpha_k]=[\beta_k]$ for all $k\in\mathbb{N}$ but the infinite products $\prod_{k=1}^{\infty}[\alpha_k]:=[\alpha_{\infty}]$ and $\prod_{k=1}^{\infty}[\beta_k]:=[\beta_{\infty}]$ are not equal.

## Beyond $\omega$-type products

So far, we have only considered infinite products $g_1g_2g_3\dots$ ordered by $\omega=\{1,2,3,\dots\}$. However, if $G$ is non-commutative, we are forced to expand the kinds of products we must be able to form.

Suppose our filtration satisfies $\bigcap_{n\in\mathbb{N}}G_n=\{e\}$ and gives us a well-defined notion of “infinite product” in a group $G$. Then we have elements $g_1g_2g_3\dots$ that behave like the notation suggests it should. However, a group has inverses and inverting a product $g_1g_2\cdots g_n$ requires reversing the order to $g_{n}^{-1}\cdots g_{2}^{-1}g_{1}^{-1}$. So if $t_1,t_2,t_3,...$ is a tail sequence for $g_1,g_2,g_3,\dots$, then $g_{\infty}^{-1}=t_{1}^{-1}$ satisfies $t_{1}^{-1}=t_{n+1}^{-1}g_{n}^{-1}\cdots g_{2}^{-1}g_{1}^{-1}$ for all $n\in\mathbb{N}$. This means the inverse $g_{\infty}^{-1}$ of the infinite product $g_{\infty}=g_1g_2g_3\cdots$ behaves exactly as an element written $...g_{3}^{-1}g_{2}^{-1}g_{1}^{-1}$ should behave. Hence we have infinite products $g_1g_2g_3\cdots$ on the right of order type $\omega=\{1,2,3,\dots\}$ and infinite products $\cdots h_3h_2h_1$ on the left with the order type of the negative integers $\{\dots,-3,-2,-1\}$. Multiplying infinite products $\cdots h_3h_2h_1$ and $g_1g_2g_3\cdots$ gives a product $\cdots h_3h_2h_1g_1g_2g_3\cdots$ with the order type of the integers $\mathbb{Z}$. From here, you can probably see how things begin to take off into linear-order-land. There’s nothing stopping us from creating infinite products of infinite products, and infinite products of infinite products of infinite products, and so on. We could have products that looks like

$(g_{1,1}g_{1,2}g_{1,3}...)(g_{2,1}g_{2,2}g_{2,3}...)(g_{3,1}g_{3,2}g_{3,3}...)(g_{4,1}g_{4,2}g_{4,3}...)(g_{5,1}g_{5,2}g_{5,3}...)...$

of order type $\omega\cdot\omega$, i.e. $\mathbb{N}\times\mathbb{N}$ in the dictionary ordering. Hence we think of the above product as being represented by function $w:\mathbb{N}\times\mathbb{N}\to G$, $w(n,m)=g_{n,m}\in G$. We also need to make sure the products respect the filtration and one way to ensure this is to demand that for each $k\in\mathbb{N}$, only finitely many values $g_{n,m}$ can lie in $G_{k}\backslash G_{k+1}$.

We could even form a product like

$...(...x_{a-1,b-1}x_{a-1,b}x_{a-1,b+1}...)(...x_{a,b-1}x_{a,b}x_{a,b+1}...)(...x_{a+1,b-1}x_{a+1,b}x_{a+1,b+1}...)...$

for $x_{n,m}\in G$, $n,m\in\mathbb{Z}$. This latter product has the order type of $\mathbb{Z}\times \mathbb{Z}$ in the dictionary ordering and is represented by a function $v:\mathbb{Z}\times\mathbb{Z}\to G$, $v(n,m)=x_{n,m}$. Again, to respect the filtration, we should insist that for each $k$, only finitely many $x_{n,m}$ lie in $G_{k}\backslash G_{k+1}$.

Ultimately, if we use a countable version of Hausdorff’s Theorem, given any countable scattered linear order $L$ (one not containing a copy of the dense order $\mathbb{Q}$), we can create products of order type $L$. Warning: this definition of “scattered” is related to, but not equivalent to, the notion of “scattered space” in topology. Can we know how a product in $G$ of order type $L$ will reduce in $G$, i.e. what its value will be? Sure, but you’ll need to know all of the relations in $G$ and keep track of values after each successive infinite product. Regardless of relations, the main take-away is that it is worth considering iterated infinite products and words as filtration-respecting functions $w:L\to G$ from a countable linear order into $G$.

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