## Monoids with infinite products

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Say a monoid $M$ has infinite products if, for any (possibly infinite) sequence $(m_1,m_2,\ldots)$ of elements of $M$, there exists an element $m_1m_2\cdots\in M$, satisfying some good properties. First, if the sequence is finite, it should coincide with the usual product on $M$. Second, concatenation of sequences results in multiplication of their products and is associative. Third, identities can be "thrown out," as can consecutive inverses ($m_{i+1}=m_i^{-1}$). (Other good properties to include?)

Another way to phrase this is: "$M$ is closed under small ordinal colimits." That is, if $M$ is considered as a one-object category, then for any small ordinal $[\kappa]$ and functor $m\colon[\kappa]\to M$, the colimit of $m$ exists in $M$.

Example: let ${\mathbb N}^+$ denote the monoid with underlying set ${\mathbb N}\cup${$\infty$} and whose operation on a sequence $m=(m_1,m_2,\ldots)$ is given by addition if $m$ has only finitely many non-zero elements, and by $\infty$ otherwise.

Now suppose that $M$ is any monoid and I want to replace it by a monoid that has infinite products. I'm hoping there are two ways to do this. One would be to add colimits freely, and the other would be to add a single "$\infty$" element that served as a catch-all (as in the example above).

Q: Do these both exist (functorially in $M$)? If so, can you describe them in elementary terms? For example, I'm worried about sequences like $1-1+1-1+\cdots$. So in a good answer I'd hope to see what happens with such infinite sums.

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So far the question does not make much sense (for me). What do you mean by "inverses"? Consider the monoid given by one defining relation $ab=1$ (the polycyclic monoid; note that $ba\not = 1$ in this monoid). What do you want to do with it? If your original monoid is commutative, do you expect the result to be commutative too?  – Mark Sapir Sep 8 2010 at 19:06
How about requiring that the product of (m_1, m_2, ...) is equal to the product of (1, m_1, m_2, ...)? – Qiaochu Yuan Sep 8 2010 at 19:50
I don't see how your "another way to phrase this" is the same thing...  how do the axioms in the first paragraph force the infinite products to be colimits? – Peter LeFanu Lumsdaine Sep 8 2010 at 20:14
Sorry, I've been absent -- you all have good points.  I'm not exactly sure what I want here; perhaps that is clear.  I'm looking for the right notion.  As for Mark's question -- you're right to ask about inverses: I was sloppy.  I think the question of whether the addition of infinite products to a commutative monoid leaves it commutative is a matter of choice, at least in the "add products freely" functor.  Under the "add a catch-all" it will remain commutative.   – David Spivak Sep 8 2010 at 21:01
As for Qiaochu Yuan's question -- yes, this is what I mean by "identities can be thrown out."  It also follows from the "concatenation of sequences" axiom.   As for Peter's question -- good point.  Just to make sure the idea is clear at least, the colimit of a finite sequence ($\kappa=n$ for some finite n) will include a canonical map from $0\in [n]$ and this map will be the product.  So in the finite case, the ideas coincide.  Colimits also take care of the other properties (think "final subcategory.")  But that's not to say that I can show that colimits are really what I want here. – David Spivak Sep 8 2010 at 21:06
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## 5 Answers

If I understand the question, the short answer is "yes, you can freely and functorially adjoin infinite products to monoids". The basic idea is that algebraic theories can accommodate arbitrary arities (bounded above by some cardinal), and one can discuss relative free-forgetful adjunctions between categories of algebras in great generality.

At first pass, let me just focus on the purely monoid-like aspects for now, because those are easier to visualize. Once the story for that is clear, one can work in inverses.

As a warm-up, let's recall that one way of defining an ordinary monoid $M$ is as an algebra of the terminal nonpermutative operad. In plainer English, this means we a single operation

$$\mu_n: M^n \to M$$

for each finite ordinal $n$, so that

$$\mu_{n_1 + \ldots + n_k} = \mu_k(\mu_{n_1} \times \ldots \times \mu_{n_k})$$

(generalized associativity equation).

Now let's generalize operads so as to allow operations of countably infinite arity. By "arities", I will really mean countable ordinals. Given an ordinal $k \lt \aleph_1$ and ordinals $n_j \lt \aleph_1$ for $j \lt k$, you can concatenate the $n_j$ to get a new ordinal $\sum_j n_j \lt \aleph_1$. Concatenation is associative in an evident sense. Now define an $\omega$-monoid to be a set $M$ equipped with operations

$$\mu_k: \hom(k, M) \to M,$$

one for each $k \lt \aleph_1$, such that $\mu_{\sum_j n_j} = \mu_k (\prod_j \mu_{n_j})$. (I am not completely certain we have to go all the way up to $\aleph_1$, but if not it will be some suitable initial segment. Let's just say $\aleph_1$ for now.) This condition can be interpreted in any category with countable products, such as $Set$.

The free $\omega$-monoid on a set $X$ will be $\sum_{k \lt \aleph_1} \hom(k, X)$. We get in this way a monad $T$ for a free-forgetful adjunction between $\omega$-monoids and sets.

There is a general bit of nonsense that for any morphism of monads $\phi: S \to T$ on $Set$, there is a forgetful functor $Alg_T \to Alg_S$, and this forgetful functor has a left adjoint. This follows from an adjoint functor theorem, although if I'm not mistaken, in this particular scenario a more direct construction is available: if $S$ is the monad for ordinary monoids and $T$ is as above, the evident inclusion $S \to T$ induces a forgetful functor

$$\omega-Mon \to Mon$$

which has a left adjoint $L$ described by the type of coequalizer familiar from tensor products:

$$L(M) = coeq((\mu_T \circ \phi) M, T\theta: TSM \stackrel{\to}{\to} T M)$$

where $\mu_T: TT \to T$ is the monad multiplication and $\theta: S M \to M$ is the structure of $M$ as $S$-algebra. This left adjoint $L$ would correspond to what I think you were asking for with "freely adjoined colimits", and the left adjoint means we indeed have a functorial construction.

If you want to work inverses in, you can do that too. Long story short: for any set of formal operation symbols of arbitrary arity, subject to any set of well-formed equations you jolly well please, you can form a monad whose algebras are precisely the models of for the corresponding algebraic theory. So: together with operations of countable arity as above, subject to generalized associativity equations, you can certainly toss in an unary inversion operation as well. I leave it to you to decide what, in addition to associativity, are the sensible equations to impose on inversion $i$, but it seems to me you might want to impose only

$$\mu_2(id \times i) = id = \mu_2(i \times id)$$

and stop there. (Operations involving infinitely many instances of inversion are still permissible, but the equations would enforce only finitely many cancellations at a time.) You get in this way a monad for "$\omega$-groups", and again the forgetful functor from $\omega$-groups to groups admits a left adjoint, constructed in a way analogous to the above.

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Thanks Todd.  Is the forgetful functor $\omega-Mon\to Mon$ not also a left adjoint?   If you work out your construction for the case of the free monoid on one generator ${\mathbb N}$, do you get this ${\mathbb N}^+$ thing I mention above, or something else?  I was more hoping to have a "unique point at infinity," to sew it all up there, than to have so many different styles of infinite product under this completion process.  Any chance of that?     – David Spivak Sep 9 2010 at 12:52
David, I've nothing intelligent yet to say about the $\mathbb{N}^+$ thing, but I don't think that forgetful functor has a right adjoint. It would have to preserve coproducts. The coproduct of two copies of the free $\omega$-monoid on the 1-element set, $F(1)$, is the free $\omega$-monoid on the 2-element set, $F(2)$. You'd need the canonical map $F(1) + F(1) \to F(2)$ on the underlying monoids, where $+$ is monoid coproduct, to be invertible. So, each countable word in two letters would need to be a finite concatenation of countable words, each in one or the other letter. Which is false. :-( – Todd Trimble Sep 9 2010 at 21:33
While I'm here: on second thought I don't think $\omega$-groups are all that sensible, because of the swindle mentioned by both you and Theo. [The "logic" of an algebraic theory involving operations $p$ of infinite arity, like that of $\omega$-groups, means we must have $p(a_1, \ldots) = p(b_1, \ldots)$ whenever $a_i = b_i$. So the swindle is derivable in the theory.] In other words, while you can consider or construct the theory or monad of $\omega$-groups, it would unfortunately collapse to something trivial. Silly of me to have suggested otherwise. :-(  – Todd Trimble Sep 9 2010 at 22:02

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Semigroups with an $\omega$-product play an important role in automata accepting infinite words and the second order monadic theory of the natural numbers with the successor operator. See the book Infinite Words: Automata, Semigroups, Logic and Games by Perrin and Pin.

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Monoids with infinite products were considered by Kharlampovich, Myasnikov and Serbin. They use it in relation with the Tarski problems about free groups, and groups acting on $\Lambda$-trees. See, for example, http://uk.arxiv.org/PS_cache/arxiv/pdf/0911/0911.0209v1.pdf . Probably more details can be found in Serbin's thesis (McGill).

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Depending on exactly what you want, if you take some monoid and force it to allow infinite products, you might end up with something rather trivial. In particular, the Mazur Swindle puts rather severe limits on the structure of monoids with infinite products and strongly-behaving inverses. Note that the Swindle requires, among other things, an infinite notion of associativity, which your proposal may or may not have.

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Mazur's swindle depends strongly on the fact that you can embed the infinite sums in space.  That allows you to do infinite cancellation, which is definitely not assumed here. – S. Carnahan♦ Sep 9 2010 at 1:05
My point was: if you do allow infinite cancellations, then you're hosed.  The only options are either disallow infinite cancellations, or disallow inverses all together. – Theo Johnson-Freyd Sep 9 2010 at 14:21

I've also thought about monoids with infinite products in the context of this question. I just want to remark that every group, which has infinite products, which are invariant under permutations (this is a natural generalization of commutativity) is trivial:

$a^\mathbb{N} = a a^{\mathbb{N} - \{0\}} = a a^{\mathbb{N}} \Rightarrow a=1$.

Of course, this does not happen when you just impose invariance under permutations with finite support, but I don't think this is natural.

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