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


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


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


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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1 Response to What is an infinite word?

  1. Pingback: Infinite Commutativity: Part I | Wild Topology

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