Is there a generalization of the free group that includes infinitely long words?

Question

The free group over a set $S$ only includes finitely-long words made up of letters from $S$ and their inverses. It seems natural to me to also allow infinitely long words. While this would obviously be impossible to operationalize on a computer, in principle these infinitely-long group elements seem perfectly well-defined. Is there some fundamental logical problem with including infinitely long words? If not, has this concept ever been studied?

(I know that such a generalization would no longer count as being "generated" by $S$ - since by definition every group element needs to be reachable by a finite number of generator multiplications - so you'd need to find a new name.)

You may want to look at Perrin and Pin's Infinite Words: Automata, Semigroups, Logic and Games, Elsevier, 2004. — Fabio Somenzi, May 13 '18 at 22:40
Versions of this idea have been studied in relation with the fundamental group of the Hawaiian earring space. I don't know the details though. — Eric Wofsey, May 13 '18 at 22:50
Well, in a complete topological group, even a free pro-finite group, an infinite product (whether infinite from the right or from the left) is well defined, as long as the terms in the infinite tail approach the identity. — Lubin, May 13 '18 at 22:53
@tparker: Well, it depends what you mean by "infinitely long word". For instance, if you mean for the letters of your word to be indexed by natural numbers, that will no longer be true of the concatenation, because it has order type $\omega+\omega$. So you need to say something about what kind of ordered sets are allowed to index your "words". — Eric Wofsey, May 13 '18 at 22:58
I’m sorry, @tparker, how do you conatenate even a single generator on the right with an Infinite product extending to the right? — Lubin, May 13 '18 at 22:58
Well, there's a whole lot more to be said and I don't know how it all works. But, I'll write up what I know. — Eric Wofsey, May 13 '18 at 23:04
There's no reason to restrict to words that extend infinitely to the right. Consider the collection of all maps $f:I\to S$ where $I$ is a discrete totally ordered set up to isomorphism, and the composition of two maps $f:I\to S$, $g:J\to S$ is $f\sqcup g:I\sqcup J\to S$. This would be the "free monoid" with infinite words (but wouldn't be cancellative and thus doesn't sit inside any group as Eric points out with the swindle). — anon, May 14 '18 at 23:55

Eric Wofsey · Accepted Answer · 2018-05-14T16:15:05.333

This doesn't work if you just do it naively. The most naive definition of an "infinite word" would be an infinite string $s_1s_2s_3\dots$ where each $s_n$ is an element of $S$ or the formal inverse of an element of $S$. This fails horribly, since such strings are not closed under composition or inverses. For instance, the inverse of such an infinite string would be infinitely long on the left, instead of on the right. And if you concatenate two such infinite strings, you would get a string of the form $(s_1s_2s_3\dots)(t_1t_2t_3\dots)$ where the "letters" in the word are now arranged in a sequence indexed by the ordinal $\omega+\omega$, instead of just by the natural numbers. So, to make sense of this idea, you need to allow a more exotic variety of "infinite words" that can have many different infinite totally ordered sets as their index sets.

Another important obstacle is the Eilenberg swindle. Namely, let $s\in S$ (or more generally let $s$ be any word) and consider the infinite word $w=sss\dots$. Then any reasonable definition of a "group of infinite words" should have $sw=w$, which then implies $s=1$! So if you want your group to be nontrivial, you need to impose some restriction that disallows words of this type.

However, the news is not all bad! There does exist at least one interesting construction of a "free group with infinite words" (there are probably others too; I don't know the literature on this subject). You can find the details in Section 3 of the nice paper The Combinatorial Structure of the Hawaiian Earring Group by J. W. Cannon and G. R. Conner. Specifically, Cannon and Conner define a "transfinite word" on a set $S$ to be a map $f:I\to S\cup S^{-1}$ where $I$ is a totally ordered set, $S^{-1}$ is the set of formal inverses of elements of $S$, and each fiber of $f$ is finite. So, you can have a word indexed by any totally ordered set, as long as each element of $S$ only appears in it only finitely many times (this finiteness condition avoids Eilenberg swindles). Two words $f:I\to S\cup S^{-1}$ and $g:J\to S\cup S^{-1}$ are identified if there is an order-isomorphism $I\cong J$ which turns $f$ into $g$. They then define an equivalence relation on such words using a sort of cancellation, prove that each word is equivalent to a unique "reduced" word, and use this to define a "big free group" on $S$ consisting of transfinite words modulo equivalence (or equivalently, reduced transfinite words).

When $S$ is countably infinite, this group is isomorphic to the fundamental group of the Hawaiian earring. Sadly, this construction is only really of interest when $S$ is infinite, since when $S$ is finite all words must be finite and you just get the ordinary free group.

You can find all the details in the paper I linked. Multiplication is just concatenation (given $f:I\to S\cup S^{-1}$ and $g:J\to S\cup S^{-1}$, their product is the induced map $I\sqcup J\to S\cup S^{-1}$ where $I\sqcup J$ is ordered by putting $I$ before $J$). The group is not free if $S$ is infinite (see Theorem 2.5(4) of the paper, where the "Hawaiian earring group" is the fundamental group of the Hawaiian earring, which is isomorphic to the "big free group" on a countably infinite set). — Eric Wofsey, May 14 '18 at 20:49

Qiaochu Yuan · Answer 2 · 2018-05-14T21:05:11.190

8

You can consider the profinite completions $\widehat{F_n}$ of the free groups. These allow some infinite words, to the left or right or both, but not all. For example, $aba^{2!} b^{2!} a^{3!} b^{3!} \dots$ is an element of $\widehat{F_2}$, but $ababab \dots$ isn't.

edited May 14 '18 at 21:05

answered May 13 '18 at 23:19

Qiaochu Yuan

468,795

5

What gives the product of $aba^{2!} b^{2!} a^{3!} b^{3!} \dots$ with itself? – Alphonse May 14 '18 at 06:48
1

It doesn’t have a simpler description other than that. – Qiaochu Yuan May 14 '18 at 17:24
1

Then I don't see why this element belongs to $\hat{F_2}$ while $ababa...$ does not (because I don't understand how the question you wrote in brackets can help). – Alphonse May 14 '18 at 20:05
Yeah, you're right, the thing in parentheses isn't very helpful. – Qiaochu Yuan May 14 '18 at 21:05
On a somewhat-related note - is there some description/explanation of the normal subgroups of $F_n$? Where can I find such a description? – Itai May 15 '18 at 06:32
@Itai: they correspond to pairs consisting of a group and a tuple of $n$ elements generating it. Geometrically they correspond to normal covering spaces of the wedge of $n$ circles. Fortunately you don't really need an answer to this question to understand the profinite completion. – Qiaochu Yuan May 15 '18 at 07:15
Wait, are you saying any (finitely-generated) subgroup of a free group is normal? (as, given $k$ elements $g_1, ..., g_k$ you have $<g_1, ..., g_k> \leq F_n$, so any choice of $k$ elements gives such a pair). For $F_2 = <a,b>$, is $ = {a^k | k \in \mathbb{Z}}$ a normal subgroup? I am equally confused about your second comment - since the profinite completion of a group is the inverse limit of quotients by normal subgroups of finite index, how am I to understand the profinite completion of a group without understanding its normal subgroups? – Itai May 15 '18 at 07:56
@Itai: 1) no, I'm not saying that. I'm saying that a normal subgroup $N$ can be classified by classifying the corresponding quotient $F_n/N$, and that these quotients correspond exactly to groups together with a choice of $n$ elements generating them. 2) Inverse limits are highly redundant; for example, you can omit finitely many elements of the inverse limit and still get the same thing. Instead of thinking explicitly about all the finite quotients you can try to understand what convergence in the profinite topology looks like. The elements of $\widehat{F_n}$ are limits of sequences... – Qiaochu Yuan May 15 '18 at 20:23
...of elements of $F_n$, and what determines whether these sequences converge is whether they're eventually constant when interpreted in any particular finite group. Hence the sequence $ab, aba^{2!} b^{2!}, aba^{2!} b^{2!} a^{3!} b^{3!} \dots$ converges because in any particular finite group $G$ and for any $a, b \in G$ this sequence is constant once the exponents get as high as $|G|!$ at worst. This is much easier than understanding every finite group generated by $2$ elements. – Qiaochu Yuan May 15 '18 at 20:24

score 0 · Answer 3 · answered May 16 '18 at 23:00

0

The Hawaiian earring group involves infinite (but not all) group words. It's a proper subgroup of the profinite completion.

answered May 16 '18 at 23:00

YCor

18,715

Is there a generalization of the free group that includes infinitely long words?

3 Answers3