$\omega$-modules
definition: $\omega$-module
Define an "$\omega$-module" as a tuple $(A, \varphi)$ such that
-
$A$ is an abelian group
-
$\varphi \colon A^\omega \rightarrow A$ is a linear map (where $A^\omega$ is the direct product, not the direct sum)
end of definition
We can regard $\varphi$ as an operation of countable arity on $A$.
(To be clear, I made up the term "$\omega$-module". But I don't know of a common name for this concept, so it'll have to do.)
commutative $\omega$-modules
definition: commutative $\omega$-module
Define a "commutative $\omega$-module" as an $\omega$-module $(A, \varphi)$ such that, for all families $(x_{ij} \in A)_{i, j \in \omega}$,
$\varphi(\varphi(x_{00}, x_{01}, x_{02}, \ldots), \varphi(x_{10}, x_{11}, x_{12}, \ldots), \varphi(x_{20}, x_{21}, x_{22}, \ldots), \ldots)$
= $\varphi(\varphi(x_{00}, x_{10}, x_{20}, \ldots), \varphi(x_{01}, x_{11}, x_{21}, \ldots), \varphi(x_{02}, x_{12}, x_{22}, \ldots), \ldots)$
Then we also say $\varphi$ is "commutative".
end of definition
Visually, this equational identity tells us that the two natural ways to evaluate $\varphi$ on a 2D table, actually agree:
$\begin{array}{c | c c c c c}
??? & \varphi(x_{00}, x_{10}, x_{20}, \ldots) & \varphi(x_{01}, x_{11}, x_{21}, \ldots) & \varphi(x_{02}, x_{12}, x_{22}, \ldots) & \cdots \\
\hline
\varphi(x_{00}, x_{01}, x_{02}, \ldots) & x_{00} & x_{01} & x_{02} & \cdots \\
\varphi(x_{10}, x_{11}, x_{12}, \ldots) & x_{10} & x_{11} & x_{12} & \cdots \\
\varphi(x_{20}, x_{21}, x_{22}, \ldots) & x_{20} & x_{21} & x_{22} & \cdots \\
\vdots & \vdots & \vdots & \vdots & \ddots \\
\end{array}$
We can either apply $\varphi$ to rows, then columns, or columns, then rows, and it doesn't matter, the result in the spot marked with $???$ will be the same.
do you actually have a motivation for this concept (you can safely skip this if you don't care)
Okay so, if we replace $\omega$ with some finite set $n = \{0, 1, 2, \ldots n-1\}$, then a commutative $n$-module is precisely a $\mathbb{Z}[f_0, f_1, \ldots, f_{n-1}]$-module.
In fact, we can reduct (i.e. this destroys a lot of information) a commutative $\omega$-module to a $\mathbb{Z}[f_0, f_1, \ldots]$-module by defining
$f_0 x := \varphi(x, 0, 0, 0, \ldots),$
$f_1 x := \varphi(0, x, 0, 0, \ldots),$
$f_2 x := \varphi(0, 0, x, 0, \ldots),$
so on and so forth.
The commutativity condition also guarantees similar things to the commutativity of a base ring.
For instance,
definition: $\omega$-module homomorphism
Define an "$\omega$-module homomorphism" $\ell \colon (A, \varphi_A) \rightarrow (B, \varphi_B)$ as a linear map $A \rightarrow B$ satisfying, for all families $(x_i \in A)_{i \in \omega}$,
$\ell(\varphi_A(x_0, x_1, x_2, \ldots)) = \varphi_B(\ell(x_0), \ell(x_1), \ell(x_2), \ldots)$
end of definition
Then the set of $\omega$-module homomorphisms between commutative $\omega$-modules should itself have an $\omega$-module structure.
I'm not sure if this is significant at all, as I have not actually utilized this property while thinking about commutative $\omega$-modules.
But, it's certainly where my motivation came from.
(If you are not satisfied, here is a miscellaneous observation that does not fit into the rest of this post:
the commutativity condition sort of resembles fubini's theorem, in the sense that an $\omega$-module can be thought of as an infinite weighted sum.)
a nice result
Before I can state a nice result I proved, I need to give a few more definitions.
definition: $A^C$
Given a subset $C \subseteq \omega$, define $A^C := \{(x_i) \in A^\omega | \forall i \notin C, x_i = 0\}$
end of definition
In English, $A^C$ is the submodule of $A^\omega$ consisting of elements whose support is contained in $C$.
This is sometimes called "extension by 0".
Because $A^C \subseteq A^\omega$, we can restrict $\varphi$ to $A^C$.
For convenience, I will introduce the following notation:
definition: $\varphi_C$
Given a subset $C \subseteq \omega$ and an $\omega$-module $(A, \varphi)$, define $\varphi_C := \varphi |_{A^C}$
end of definition
theorem: commutative $\omega$-modules are "eventually non-surjective"
Given a commutative $\omega$-module $(A, \varphi)$, if $\varphi_C$ is surjective for all cofinte $C$, then $A = 0$.
end of theorem statement
the proof of this theorem
We may prove the contrapositive, i.e., there must exist a witness $x = (x_{ij})$ to $\varphi$ failing to be commutative.
We can construct it "one row and column at a time":
To start, let $y = (y_i)$ be a family such that $\varphi(y) \neq 0$.
Our goal will be to have $\varphi(x) = 0$ when $\varphi$ is applied by columns and $\varphi(x) = y$ when $\varphi$ is applied by rows.
Firstly, let $(x_{0j})$ be a family such that $\varphi(x_{0j}) = y_0$.
We can visualize what we're doing via this table:
$\begin{array}{c | c c c c c}
??? & ? & ? & ? & \cdots \\
\hline
\mathbf{y_0} & \mathbf{x_{00}} & \mathbf{x_{01}} & \mathbf{x_{02}} & \boldsymbol{\cdots} \\
? & ? & ? & ? & \cdots \\
? & ? & ? & ? & \cdots \\
\vdots & \vdots & \vdots & \vdots & \ddots \\
\end{array}$
The $?$ represent values we haven't filled in yet, and the bolded variables are what we just filled in.
Now, let $(x_{i0})$ be a family, that agrees with our previous family on $x_{00}$, such that $\varphi(x_{i0}) = 0$.
We can do this because $\varphi_{\{1, 2, 3, \ldots\}}$ is surjective, so we can always "adjust" our value to compensate for $\varphi(x_{00}, 0, 0, \ldots)$.
Here is the table visualization:
$\begin{array}{c | c c c c c}
??? & \mathbf{0} & ? & ? & \cdots \\
\hline
y_0 & x_{00} & x_{01} & x_{02} & \cdots \\
? & \mathbf{x_{10}} & ? & ? & \cdots \\
? & \mathbf{x_{20}} & ? & ? & \cdots \\
\vdots & \pmb{\vdots} & \vdots & \vdots & \ddots \\
\end{array}$
Continue on in this way, each time using that $\varphi_C$ is surjective for cofinite $C$.
Here are the further visualizations:
$\begin{array}{c | c c c c c}
??? & 0 & ? & ? & \cdots \\
\hline
y_0 & x_{00} & x_{01} & x_{02} & \cdots \\
\mathbf{y_1} & x_{10} & \mathbf{x_{11}} & \mathbf{x_{12}} & \boldsymbol{\cdots} \\
? & x_{20} & ? & ? & \cdots \\
\vdots & \vdots & \vdots & \vdots & \ddots \\
\end{array}$
$\begin{array}{c | c c c c c}
??? & 0 & \mathbf{0} & ? & \cdots \\
\hline
y_0 & x_{00} & x_{01} & x_{02} & \cdots \\
y_1 & x_{10} & x_{11} & x_{12} & \cdots \\
? & x_{20} & \mathbf{x_{21}} & ? & \cdots \\
\vdots & \vdots & \pmb{\vdots} & \vdots & \ddots \\
\end{array}$
etc., so on and so forth.
Once this is all done, we have $x$ as desired!
okay, what's the catch?
Oh yeah, er... this inequality is very sharp.
The dumbest examples of commutative $\omega$-modules are constructed as follows:
let $(A, \varphi)$ be any $\omega$-module, i.e. it can be non-commutative.
Let
$$0 \rightarrow A \xrightarrow{\iota} B \xrightarrow{\rho} A \rightarrow 0$$
be a short exact sequence of abelian groups.
We have the composition
$$\varphi' \colon B^\omega \xrightarrow{\rho^\omega} A^\omega \xrightarrow{\varphi} A \xrightarrow{\iota} B$$
which has the same image as $\varphi$; we also have $\varphi'(\varphi'(x)) = 0$ for all $x$, so $\varphi'$ is commutative.
(Note that this construction can never produce an example where $A = B \neq 0$.)
both dumb and non-dumb concrete examples
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Let $A = \mathbb{Z}/p$ for some prime $p > 0$.
$\omega$-module structures on $A$ in general can be very wild because of field theory nonsense.
In particular, we can think of $A^\omega$ as a vector space over $\mathbb{F}_p$, so it has a basis which uniquely determines maps out of it.
This basis is very non-canonical and requires the axiom of choice to construct.
However, because commutative $\omega$-modules must be eventually non-surjective, we have that $\varphi$ must actually have finite support!
Interestingly, one can show these are precisely the "constructable maps" in the sense that you don't need the axiom of choice to prove they exist.
Alternatively, they're precisely continuous maps, putting the product topology on $A^\omega$.
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However, the fun does not last.
If we let $A = \mathbb{Z}/4$, using the "dumb" construction earlier, we can get very non-continuous but commutative $\omega$-modules on $A$.
This also works on $(\mathbb{Z}/2)^2$, etc.
I wish I knew how to fix this issue, but I don't have any natural ideas yet.
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Interestingly, any $\omega$-module structure on $\mathbb{Z}$ is continuous.
This is group theory folklore, if you're curious about the details, search up the "Baer-Specker group".
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Here's possibly the most interesting example: suppose that
$$A \cong \varprojlim A_0 \leftarrow A_1 \leftarrow \ldots.$$
Then, by the universal property of limits, maps $\varphi \colon A^\omega \rightarrow A$ are determined by a family of compatible maps $A^\omega \rightarrow A_i$ for all $i \in \omega$.
Additionally, we always have natural projection maps $A^\omega \rightarrow A$ and $A \rightarrow A_i$, and we can always add them together.
The end result is a huge zoo of examples of commutative $\omega$-modules.
A good example is when $A = \mathbb{Z}_p$, the $p$-adics, where we define $\varphi(x_0, x_1, x_2, \ldots) := x_0 + p x_1 + p^2 x_2 + \ldots$.
This recovers the "weighted sum" intuition mentioned briefly in the motivation section.
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Of course, who doesn't want to look at the free object of a category?
Because our operation here has countable arity, our syntax trees can actually have depth indexed by $\omega_1,$ the first uncountable ordinal.
That being said, although I haven't checked the fine details, I'm pretty sure one can still abstractly define the free $\omega$-module in terms of operations in the signature of abelian groups with $\varphi$ added.
(Of course, we would be using the definition of an $\omega$-module homomorphism provided earlier, so that a "free" $\omega$-module would be a well-defined concept.)
We can also freely mod out by the commutativity condition to get the free commutative $\omega$-module from the free non-commutative one.
I believe the free commutative $\omega$-module on one generator can be viewed as a subring of
$$\varprojlim \mathbb{Z}[f_0] \leftarrow \mathbb{Z}[f_0, f_1] \leftarrow \ldots$$
with each projection sending $f_n$ to $0$; we have $\varphi(x) = f_0 x_0 + f_1 x_1 + \ldots$, and we take the smallest subring closed under $\varphi$.
However, I haven't checked all the details.
conclusion
I needed to put all my thoughts in one place, as usual.
By now, this post has gone through several revisions, primarily to aid clarity.
I am still actively looking for feedback, so feel free to reach out to me if you have any.