## Ordered algebraic geometry

Edit: Shortly after posting this, I found where the machinery I develop here was discussed in the literature. Real Algebraic Geometry by Bochnak, Coste, and Roy covers at least most of this material. I may eventually edit this to clean it up and adopt more standard notation, but don't hold your breath.

### Introduction

In algebraic geometry, an affine algebraic set is a subset of $\mathbb{C}^{n}$ which is the set of solutions to some finite set of polynomials. Since all ideals of $\mathbb{C}\left[x_{1},...,x_{n}\right]$ are finitely generated, this is equivalent to saying that an affine algebraic set is a subset of $\mathbb{C}^{n}$ which is the set of solutions to some arbitrary set of polynomials.

In semialgebraic geometry, a closed semialgebraic set is a subset of $\mathbb{R}^{n}$ of the form $\left\{ \bar{x}\in\mathbb{R}^{n}\mid f\left(\bar{x}\right)\geq0\,\forall f\in F\right\}$ for some finite set of polynomials $F\subseteq\mathbb{R}\left[x_{1},...,x_{n}\right]$. Unlike in the case of affine algebraic sets, if $F\subseteq\mathbb{R}\left[x_{1},...,x_{n}\right]$ is an arbitrary set of polynomials, $\left\{ \bar{x}\in\mathbb{R}^{n}\mid f\left(\bar{x}\right)\geq0\,\forall f\in F\right\}$ is not necessarily a closed semialgebraic set. As a result of this, the collection of closed semialgebraic sets are not the closed sets of a topology on $\mathbb{R}^{n}$. In the topology on $\mathbb{R}^{n}$ generated by closed semialgebraic sets being closed, the closed sets are the sets of the form $\left\{ \bar{x}\in\mathbb{R}^{n}\mid f\left(\bar{x}\right)\geq0\,\forall f\in F\right\}$ for arbitrary $F\subseteq\mathbb{R}\left[x_{1},...,x_{n}\right]$. Semialgebraic geometry usually restricts itself to the study of semialgebraic sets, but here I wish to consider all the closed sets of this topology. Notice that closed semialgebraic sets are also closed in the standard topology, so the standard topology is a refinement of this one. Notice also that the open ball $B_{r}\left(\bar{p}\right)$ of radius $r$ centered at $\bar{p}$ is the complement of the closed semialgebraic set $\left\{ \bar{x}\in\mathbb{R}^{n}\mid\left|\bar{x}-\bar{p}\right|^{2}-r^{2}\geq0\right\}$, and these open balls are a basis for the standard topology, so this topology is a refinement of the standard one. Thus, the topology I have defined is exactly the standard topology on $\mathbb{R}^{n}$.

In algebra, instead of referring to a set of polynomials, it is often nicer to talk about the ideal generated by that set instead. What is the analog of an ideal in ordered algebra? It's this thing:

Definition: If $A$ is a partially ordered commutative ring, a cone $C$ in $A$ is a subsemiring of $A$ which contains all positive elements, and such that $C\cap-C$ is an ideal of $A$. By "subsemiring", I mean a subset that contains $0$ and $1$, and is closed under addition and multiplication (but not necessarily negation). If $F\subseteq A$, the cone generated by $F$, denoted $\left\langle F\right\rangle$, is the smallest cone containing $F$. Given a cone $C$, the ideal $C\cap-C$ will be called the interior ideal of $C$, and denoted $C^{\circ}$.

$\mathbb{R}\left[x_{1},...,x_{n}\right]$ is partially ordered by $f\geq g\iff f\left(\bar{x}\right)\geq g\left(\bar{x}\right)\,\forall\bar{x}\in\mathbb{R}^{n}$. If $F\subseteq\mathbb{R}\left[x_{1},...,x_{n}\right]$ is a set of polynomials and $\bar{x}\in\mathbb{R}^{n}$, then $f\left(\bar{x}\right)\geq0\,\forall f\in F\iff f\left(\bar{x}\right)\geq0\,\forall f\in\left\langle F\right\rangle$. Thus I can consider closed sets to be defined by cones. We now have a Galois connection between cones of $\mathbb{R}\left[x_{1},...,x_{n}\right]$ and subsets of $\mathbb{R}^{n}$, given by, for a cone $C$, its positive-set is $P_{\mathbb{R}}\left(C\right):=\left\{ \bar{x}\in\mathbb{R}^{n}\mid f\left(\bar{x}\right)\geq0\,\forall f\in C\right\}$ (I'm calling it the "positive-set" even though it is where the polynomials are all non-negative, because "non-negative-set" is kind of a mouthful), and for $X\subseteq\mathbb{R}^{n}$, its cone is $C_{\mathbb{R}}\left(X\right):=\left\{ f\in\mathbb{R}\left[x_{1},...,x_{n}\right]\mid f\left(\bar{x}\right)\geq0\,\forall\bar{x}\in X\right\}$$P_{\mathbb{R}}\circ C_{\mathbb{R}}$ is closure in the standard topology on $\mathbb{R}^{n}$ (the analog in algebraic geometry is closure in the Zariski topology on $\mathbb{C}^{n}$). A closed set $X$ is semialgebraic if and only if it is the positive-set of a finitely-generated cone.

### Quotients by cones, and coordinate rings

An affine algebraic set $V$ is associated with its coordinate ring $\mathbb{C}\left[V\right]:=\mathbb{C}\left[x_{1},...,x_{n}\right]/I\left(V\right)$. We can do something analogous for closed subsets of $\mathbb{R}^{n}$.

Definition: If $A$ is a partially ordered commutative ring and $C\subseteq A$ is a cone, $A/C$ is the ring $A/C^{\circ}$, equipped with the partial order given by $f+C^{\circ}\geq g+C^{\circ}$ if and only if $f-g\in C$, for $f,g\in A$.

Definition: If $X\subseteq\mathbb{R}^{n}$ is closed, the coordinate ring of $X$ is $\mathbb{R}\left[X\right]:=\mathbb{R}\left[x_{1},...,x_{n}\right]/C\left(X\right)$. This is the ring of functions $X\rightarrow\mathbb{R}$ that are restrictions of polynomials, ordered by $f\geq g$ if and only if $f\left(\bar{x}\right)\geq g\left(\bar{x}\right)\,\forall\bar{x}\in X$. For arbitrary $X\subseteq\mathbb{R}^{n}$, the ring of regular functions on $X$, denoted $\mathcal{O}\left(X\right)$, consists of functions on $X$ that are locally ratios of polynomials, again ordered by $f\geq g$ if and only if $f\left(\bar{x}\right)\geq g\left(\bar{x}\right)\,\forall\bar{x}\in X$. Assigning its ring of regular functions to each open subset of $X$ endows $X$ with a sheaf of partially ordered commutative rings.

For closed $X\subseteq\mathbb{R}^{n}$, $\mathbb{R}\left[X\right]\subseteq\mathcal{O}\left(X\right)$, and this inclusion is generally proper, both because it is possible to divide by polynomials that do not have roots in $X$, and because $X$ may be disconnected, making it possible to have functions given by different polynomials on different connected components.

### Positivstellensätze

What is $C_{\mathbb{R}}\circ P_{\mathbb{R}}$? The Nullstellensatz says that its analog in algebraic geometry is the radical of an ideal. As such, we could say that the radical of a cone $C$, denoted $\text{Rad}_{\mathbb{R}}\left(C\right)$, is $C_{\mathbb{R}}\left(P_{\mathbb{R}}\left(C\right)\right)$, and that a cone $C$ is radical if $C=\text{Rad}_{\mathbb{R}}\left(C\right)$. In algebraic geometry, the Nullstellensatz shows that a notion of radical ideal defined without reference to algebraic sets in fact characterizes the ideals which are closed in the corresponding Galois connection. It would be nice to have a description of the radical of a cone that does not refer to the Galois connection. There is a semialgebraic analog of the Nullstellensatz, but it does not quite characterize radical cones.

Positivstellensatz 1: If $C\subseteq\mathbb{R}\left[x_{1},...,x_{n}\right]$ is a finitely-generated cone and $p\in\mathbb{R}\left[x_{1},...,x_{n}\right]$ is a polynomial, then $p\left(\bar{x}\right)>0\,\forall\bar{x}\in P_{\mathbb{R}}\left(C\right)$ if and only if $\exists f\in C$ such that $pf-1\in C$.

There are two ways in which this is unsatisfactory: first, it applies only to finitely-generated cones, and second, it tells us exactly which polynomials are strictly positive everywhere on a closed semialgebraic set, whereas we want to know which polynomials are non-negative everywhere on a set.

The second problem is easier to handle: a polynomial $p$ is non-negative everywhere on a set $S$ if and only if there is a decreasing sequence of polynomials $\left(p_{i}\mid i\in\mathbb{N}\right)$ converging to $p$ such that each $p_{i}$ is strictly positive everywhere on $S$. Thus, to find $\text{Rad}_{\mathbb{R}}\left(C\right)$, it is enough to first find all the polynomials that are strictly positive everywhere on $P_{\mathbb{R}}\left(C\right)$, and then take the closure under lower limits. Thus we have a characterization of radicals of finitely-generated cones.

Positivstellensatz 2: If $C\subseteq\mathbb{R}\left[x_{1},...,x_{n}\right]$ is a finitely-generated cone, $\text{Rad}_{\mathbb{R}}\left(C\right)$ is the closure of $\left\{ p\in\mathbb{R}\left[x_{1},...,x_{n}\right]\mid\exists f\in C\, pf-1\in C\right\}$, where the closure of a subset $X\subseteq\mathbb{R}\left[x_{1},...,x_{n}\right]$ is defined to be the set of all polynomials in $\mathbb{R}\left[x_{1},...,x_{n}\right]$ which are infima of chains contained in $X$.

This still doesn't even tell us what's going on for cones which are not finitely-generated. However, we can generalize the Positivstellensatz to some other cones.

Positivstellensatz 3: Let $C\subseteq\mathbb{R}\left[x_{1},...,x_{n}\right]$ be a cone containing a finitely-generated subcone $D\subseteq C$ such that $P_{\mathbb{R}}\left(D\right)$ is compact. If $p\in\mathbb{R}\left[x_{1},...,x_{n}\right]$ is a polynomial, then $p\left(\bar{x}\right)>0\,\forall\bar{x}\in P_{\mathbb{R}}\left(C\right)$ if and only if $\exists f\in C$ such that $pf-1\in C$. As before, it follows that $\text{Rad}_{\mathbb{R}}\left(C\right)$ is the closure of $\left\{ p\in\mathbb{R}\left[x_{1},...,x_{n}\right]\mid\exists f\in C\, pf-1\in C\right\}$.

proof: For a given $p\in\mathbb{R}\left[x_{1},...,x_{n}\right]$$\left\{ \bar{x}\in\mathbb{R}^{n}\mid p\left(\bar{x}\right)\leq0\right\} \cap P_{\mathbb{R}}\left(C\right)=\left\{ \bar{x}\in\mathbb{R}^{n}\mid p\left(\bar{x}\right)\leq0\right\} \cap\bigcap\left\{ P_{\mathbb{R}}\left(\left\langle f\right\rangle \right)\mid f\in C\right\}$, an intersection of closed sets contained in the compact set $P_{\mathbb{R}}\left(D\right)$, which is thus empty if and only if some finite subcollection of them has empty intersection within $P_{\mathbb{R}}\left(D\right)$. Thus if $p$ is strictly positive everywhere on $P_{\mathbb{R}}\left(C\right)$, then there is some finitely generated subcone $E\subseteq C$ such that $p$ is strictly positive everywhere on $P_{\mathbb{R}}\left(E\right)\cap P_{\mathbb{R}}\left(D\right)=P_{\mathbb{R}}\left(\left\langle E\cup D\right\rangle \right)$, and $\left\langle E\cup D\right\rangle$ is finitely-generated, so by Positivstellensatz 1, there is $f\in\left\langle E\cup D\right\rangle \subseteq C$ such that $pf-1\in\left\langle E\cup D\right\rangle \subseteq C$$\square$

For cones that are not finitely-generated and do not contain any finitely-generated subcones with compact positive-sets, the Positivstellensatz will usually fail. Thus, it seems likely that if there is a satisfactory general definition of radical for cones in arbitrary partially ordered commutative rings that agrees with this one in $\mathbb{R}\left[x_{1},...,x_{n}\right]$, then there is also an abstract notion of "having a compact positive-set" for such cones, even though they don't even have positive-sets associated with them.

### Beyond $\mathbb{R}^{n}$

An example of cone for which the Positivstellensatz fails is $C_{\infty}:=\left\{ f\in\mathbb{R}\left[x\right]\mid\exists x\in\mathbb{R}\,\forall y\geq x\, f\left(y\right)\geq0\right\}$, the cone of polynomials that are non-negative on sufficiently large inputs (equivalently, the cone of polynomials that are either $0$ or have positive leading coefficient). $P_{\mathbb{R}}\left(C\right)=\emptyset$, and $-1$ is strictly positive on $\emptyset$, but for $f\in C_{\infty}$$-f-1\notin C_{\infty}$.

However, it doesn't really look $C_{\infty}$ is trying to point to the empty set; instead, $C_{\infty}$ is trying to describe the set of all infinitely large reals, which only looks like the empty set because there are no infinitely large reals. Similar phenomena can occur even for cones that do contain finitely-generated subcones with compact positive-sets. For example, let $C_{\varepsilon}:=\left\{ f\in\mathbb{R}\left[x\right]\mid\exists x>0\,\forall y\in\left[0,x\right]\, f\left(y\right)\geq0\right\}$$P_{\mathbb{R}}\left(C_{\varepsilon}\right)=\left\{ 0\right\}$, but $C_{\varepsilon}$ is trying to point out the set containing $0$ and all positive infinitesimals. Since $\mathbb{R}$ has no infinitesimals, this looks like $\left\{ 0\right\}$.

To formalize this intuition, we can change the Galois connection. We could say that for a cone $C\subseteq\mathbb{R}\left[x_{1},...,x_{n}\right]$$P_{\text{*}\mathbb{R}}\left(C\right):=\left\{ \bar{x}\in\left(\text{*}\mathbb{R}\right)^{n}\mid f\left(\bar{x}\right)\geq0\,\forall f\in C\right\}$, where $\text{*}\mathbb{R}$ is the field of hyperreals. All you really need to know about $\text{*}\mathbb{R}$ is that it is a big ordered field extension of $\mathbb{R}$. $P_{\text{*}\mathbb{R}}\left(C_{\infty}\right)$ is the set of hyperreals that are bigger than any real number, and $P_{\text{*}\mathbb{R}}\left(C_{\varepsilon}\right)$ is the set of hyperreals that are non-negative and smaller than any positive real. The cone of a subset $X\subseteq\left(\text{*}\mathbb{R}\right)^{n}$, denoted $C_{\text{*}\mathbb{R}}$$\left(X\right)$ will be defined as before, still consisting only of polynomials with real coefficients. This defines a topology on $\left(\text{*}\mathbb{R}\right)^{n}$ by saying that the closed sets are the fixed points of $P_{\text{*}\mathbb{R}}\circ C_{\text{*}\mathbb{R}}$. This topology is not $T_{0}$ because, for example, there are many hyperreals that are larger than all reals, and they cannot be distinguished by polynomials with real coefficients. There is no use keeping track of the difference between points that are in the same closed sets. If you have a topology that is not $T_{0}$, you can make it $T_{0}$ by identifying any pair of points that have the same closure. If we do this to $\left(\text{*}\mathbb{R}\right)^{n}$ , we get what I'm calling ordered affine $n$-space over $\mathbb{R}$.

Definition: An $n$-type over $\mathbb{R}$ is a set $\Phi$ of inequalities, consisting of, for each polynomial $f\in\mathbb{R}\left[x_{1},..,x_{n}\right]$, one of the inequalities $f\left(\bar{x}\right)\geq0$ or $f\left(\bar{x}\right)<0$, such that there is some totally ordered field extension $\mathcal{R}\supseteq\mathbb{R}$ and $\bar{x}\in\mathcal{R}^{n}$ such that all inequalities in $\Phi$ are true about $\bar{x}$. $\Phi$ is called the type of $\bar{x}$. Ordered affine $n$-space over $\mathbb{R}$, denoted $\mathbb{OA}_{\mathbb{R}}^{n}$ is the set of $n$-types over $\mathbb{R}$.

Compactness Theorem: Let $\Phi$ be a set of inequalities consisting of, for each polynomial $f\in\mathbb{R}\left[x_{1},..,x_{n}\right]$, one of the inequalities $f\left(\bar{x}\right)\geq0$ or $f\left(\bar{x}\right)<0$. Then $\Phi$ is an $n$-type if and only if for any finite subset $\Delta\subseteq\Phi$, there is $\bar{x}\in\mathbb{R}$ such that all inequalities in $\Delta$ are true about $\bar{x}$.

proof: Follows from the compactness theorem of first-order logic and the fact that ordered field extensions of $\mathbb{R}$ embed into elementary extensions of $\mathbb{R}$. The theorem is not obvious if you do not know what those mean. $\square$

An $n$-type represents an $n$-tuple of elements of an ordered field extension of $\mathbb{R}$, up to the equivalence relation that identifies two such tuples that relate to $\mathbb{R}$ by polynomials in the same way. One way that a tuple of elements of an extension of $\mathbb{R}$ can relate to elements of $\mathbb{R}$ is to equal a tuple of elements of $\mathbb{R}$, so there is a natural inclusion $\mathbb{R}^{n}\subseteq\mathbb{OA}_{\mathbb{R}}^{n}$ that associates an $n$-tuple of reals with the set of polynomial inequalities that are true at that $n$-tuple.

A tuple of polynomials $\left(f_{1},...,f_{m}\right)\in\left(\mathbb{R}\left[x_{1},...,x_{n}\right]\right)^{m}$ describes a function $f:\mathbb{R}^{n}\rightarrow\mathbb{R}^{m}$, which extends naturally to a function $f:\mathbb{OA}_{\mathbb{R}}^{n}\rightarrow\mathbb{OA}_{\mathbb{R}}^{m}$ by $f\left(\Phi\right)$ is the type of $\left(f_{1}\left(\bar{x}\right),...,f_{m}\left(\bar{x}\right)\right)$, where $\bar{x}$ is an $n$-tuple of elements of type $\Phi$ in an extension of $\mathbb{R}$. In particular, a polynomial $f\in\mathbb{R}\left[x_{1},...,x_{n}\right]$ extends to a function $f:\mathbb{OA}_{\mathbb{R}}^{n}\rightarrow\mathbb{OA}_{\mathbb{R}}^{1}$, and $\mathbb{OA}_{\mathbb{R}}^{1}$ is totally ordered by $\Phi\geq\Psi$ if and only if $x\geq y$, where $x$ and $y$ are elements of type $\Phi$ and $\Psi$, respectively, in an extension of $\mathbb{R}$$f\left(\Phi\right)\geq0$ if and only if $\text{"}f\left(\bar{x}\right)\geq0\text{"}\in\Phi$, so we can talk about inequalities satisfied by types in place of talking about inequalities contained in types.

I will now change the Galois connection that we are talking about yet again (last time, I promise). It will now be a Galois connection between the set of cones in $\mathbb{R}\left[x_{1},...,x_{n}\right]$ and the set of subsets of $\mathbb{OA}_{\mathbb{R}}^{n}$. For a cone $C\subseteq\mathbb{R}\left[x_{1},...,x_{n}\right]$, $P\left(C\right):=\left\{ \Phi\in\mathbb{OA}_{\mathbb{R}}^{n}\mid f\left(\Phi\right)\geq0\,\forall f\in C\right\}$. For a set $X\subseteq\mathbb{OA}_{\mathbb{R}}^{n}$, $C\left(X\right):=\left\{ f\in\mathbb{R}\left[x_{1},...,x_{n}\right]\mid f\left(\Phi\right)\geq0\,\forall\Phi\in X\right\}$. Again, this defines a topology on $\mathbb{OA}_{\mathbb{R}}^{n}$ by saying that fixed points of $P\circ C$ are closed. $\mathbb{OA}_{\mathbb{R}}^{n}$ is $T_{0}$; in fact, it is the $T_{0}$ topological space obtained from $\left(\text{*}\mathbb{R}\right)^{n}$ by identifying points with the same closure as mentioned earlier. $\mathbb{OA}_{\mathbb{R}}^{n}$ is also compact, as can be seen from the compactness theorem. $\mathbb{OA}_{\mathbb{R}}^{n}$ is not $T_{1}$ (unless $n=0$). Note that model theorists have their own topology on $\mathbb{OA}_{\mathbb{R}}^{n}$, which is distinct from the one I use here, and is a refinement of it.

The new Galois connection is compatible with the old one via the inclusion $\mathbb{R}^{n}\subseteq\mathbb{OA}_{\mathbb{R}}^{n}$, in the sense that if $X\subseteq\mathbb{R}^{n}$, then $C_{\mathbb{R}}\left(X\right)=C\left(X\right)$ (where we identify $X$ with its image in $\mathbb{OA}_{\mathbb{R}}^{n}$), and for a cone $C\subseteq\mathbb{R}\left[x_{1},...,x_{n}\right]$$P_{\mathbb{R}}=P\left(C\right)\cap\mathbb{R}^{n}$.

Like our intermediate Galois connection $\left(P_{\text{*}\mathbb{R}},C_{\text{*}\mathbb{R}}\right)$, our final Galois connection $\left(P,C\right)$ succeeds in distinguishing $P\left(C_{\infty}\right)$ and $P\left(C_{\varepsilon}\right)$ from $\emptyset$ and $\left\{ 0\right\}$, respectively, in the desirable manner. $P\left(C_{\infty}\right)$ consists of the type of numbers larger than any real, and $P\left(C_{\varepsilon}\right)$ consists of the types of $0$ and of positive numbers smaller than any positive real.

Just like for subsets of $\mathbb{R}^{n}$, a closed subset $X\subseteq\mathbb{OA}_{\mathbb{R}}^{n}$ has a coordinate ring $\mathbb{R}\left[X\right]:=\mathbb{R}\left[x_{1},...,x_{n}\right]/C\left(X\right)$, and an arbitrary $X\subseteq\mathbb{OA}_{\mathbb{R}}^{n}$ has a ring of regular functions $\mathcal{O}\left(X\right)$ consisting of functions on $X$ that are locally ratios of polynomials, ordered by $f\geq0$ if and only if $\forall\Phi\in X$, where $f=\frac{p}{q}$ is a representation of $f$ as a ratio of polynomials in a neighborhood of $\Phi$, either $p\left(\Phi\right)\geq0$ and $q\left(\Phi\right)>0$, or $p\left(\Phi\right)\leq0$ and $q\left(\Phi\right)<0$, and $f\geq g$ if and only if $f-g\geq0$. As before, $\mathbb{R}\left[X\right]\subseteq\mathcal{O}\left(X\right)$ for closed $X\subseteq\mathbb{OA}_{\mathbb{R}}^{n}$.

$\mathbb{OA}_{\mathbb{R}}^{n}$ is analogous to $\mathbb{A}_{\mathbb{C}}^{n}$ from algebraic geometry because if, in the above definitions, you replace "$\geq$" and "$<$" with "$=$" and "$\neq$", replace totally ordered field extensions with field extensions, and replace cones with ideals, then you recover a description of $\mathbb{A}_{\mathbb{C}}^{n}$, in the sense of $\text{Spec}\left(\mathbb{C}\left[x_{1},...,x_{n}\right]\right)$.

What about an analog of projective space? Since we're paying attention to order, we should look at spheres, not real projective space. The $n$-sphere over $\mathbb{R}$, denoted $\mathbb{S}_{\mathbb{R}}^{n}$, can be described as the locus of $\left|\bar{x}\right|^{2}=1$ in $\mathbb{OA}_{\mathbb{R}}^{n}$.

For any totally ordered field $k$, we can define $\mathbb{OA}_{k}^{n}$ similarly to $\mathbb{OA}_{\mathbb{R}}^{n}$, as the space of $n$-types over $k$, defined as above, replacing $\mathbb{R}$ with $k$ (although a model theorist would no longer call it the space of $n$-types over $k$). The compactness theorem is not true for arbitrary $k$, but its corollary that $\mathbb{OA}_{k}^{n}$ is compact still is true.

### Visualizing $\mathbb{OA}_{\mathbb{R}}^{n}$ and $\mathbb{S}_{\mathbb{R}}^{n}$

$\mathbb{S}_{\mathbb{R}}^{n}$ should be thought of as the $n$-sphere with infinitesimals in all directions around each point. Specifically, $\mathbb{S}_{\mathbb{R}}^{0}$ is just $\mathbb{S}^{0}$, a pair of points. The closed points of $\mathbb{S}_{\mathbb{R}}^{n+1}$ are the points of $\mathbb{S}^{n+1}$, and for each closed point $p$, there is an $n$-sphere of infinitesimals around $p$, meaning a copy of $\mathbb{S}_{\mathbb{R}}^{n}$, each point of which has $p$ in its closure.

$\mathbb{OA}_{\mathbb{R}}^{n}$ should be thought of as $n$-space with infinitesimals in all directions around each point, and infinities in all directions. Specifically, $\mathbb{OA}_{\mathbb{R}}^{n}$ contains $\mathbb{R}^{n}$, and for each point $p\in\mathbb{R}^{n}$, there is an $n-1$-sphere of infinitesimals around $p$, and there is also a copy of $\mathbb{S}_{\mathbb{R}}^{n-1}$ around the whole thing, the closed points of which are limits of rays in $\mathbb{R}^{n}$.

$\mathbb{OA}_{\mathbb{R}}^{n}$ and $\mathbb{S}_{\mathbb{R}}^{n}$ relate to each other the same way that $\mathbb{R}^{n}$ and $\mathbb{S}^{n}$ do. If you remove a closed point from $\mathbb{S}_{\mathbb{R}}^{n}$, you get $\mathbb{OA}_{\mathbb{R}}^{n}$, where the sphere of infinitesimals around the removed closed point becomes the sphere of infinities of $\mathbb{OA}_{\mathbb{R}}^{n}$.

More generally, if $k$ is a totally ordered field, let $k^{r}$ be its real closure. $\mathbb{OA}_{k}^{n}$ consists of the Cauchy completion of $\left(k^{r}\right)^{n}$ (as a metric space with distances valued in $k^{r}$), and for each point $p\in\left(k^{r}\right)^{n}$ (though not for points that are limits of Cauchy sequences that do not converge in $\left(k^{r}\right)^{n}$), an $n-1$-sphere $\mathbb{S}_{k}^{n-1}$ of infinitesimals around $p$, and an $n-1$-sphere $\mathbb{S}_{k}^{n-1}$ around the whole thing, where $\mathbb{S}_{k}^{n}$ is the locus of $\left|\bar{x}\right|^{2}=1$ in $\mathbb{OA}_{k}^{n}$. $\mathbb{OA}$ does not distinguish between fields with the same real closure.

### More Positivstellensätze

This Galois connection gives us a new notion of what it means for a cone to be radical, which is distinct from the old one and is better, so I will define $\text{Rad}\left(C\right)$ to be $C\left(P\left(C\right)\right)$. A cone $C$ will be called radical if $C=\text{Rad}\left(C\right)$. Again, it would be nice to be able to characterize radical cones without referring to the Galois connection. And this time, I can do it. Note that since $\mathbb{OA}_{\mathbb{R}}^{n}$ is compact, the proof of Positivstellensatz 3 shows that in our new context, the Positivstellensatz holds for all cones, since even the subcone generated by $\emptyset$ has a compact positive-set.

Positivstellensatz 4: If $C\subseteq\mathbb{R}\left[x_{1},...,x_{n}\right]$ is a cone and $p\in\mathbb{R}\left[x_{1},...,x_{n}\right]$ is a polynomial, then $p\left(\Phi\right)>0\,\forall\Phi\in P\left(C\right)$ if and only if $\exists f\in C$ such that $pf-1\in C$.

However, we can no longer add in lower limits of sequences of polynomials. For example, $-x+\varepsilon\in C_{\varepsilon}$ for all real $\varepsilon>0$, but $-x\notin C_{\varepsilon}$, even though $C_{\varepsilon}$ is radical. This happens because, where $\Sigma$ is the type of positive infinitesimals, $-\Sigma+\varepsilon>0$ for real $\varepsilon>0$, but $-\Sigma<0$. However, we can add in lower limits of sequences contained in finitely-generated subcones, and this is all we need to add, so this characterizes radical cones.

Positivstellensatz 5: If $C\subseteq\mathbb{R}\left[x_{1},...,x_{n}\right]$ is a cone, $\text{Rad}\left(C\right)$ is the union over all finitely-generated subcones $D\subseteq C$ of the closure of $\left\{ p\in\mathbb{R}\left[x_{1},...,x_{n}\right]\mid\exists f\in D\, pf-1\in D\right\}$ (again the closure of a subset $X\subseteq\mathbb{R}\left[x_{1},...,x_{n}\right]$ is defined to be the set of all polynomials in $\mathbb{R}\left[x_{1},...,x_{n}\right]$ which are infima of chains contained in $X$).

Proof: Suppose $D\subseteq C$ is a subcone generated by a finite set $\left\{ f_{1},...,f_{m}\right\}$, and $q$ is the infimum of a chain $\left\{ q_{\alpha}\right\} _{\alpha\in A}\subseteq\left\{ p\in\mathbb{R}\left[x_{1},...,x_{n}\right]\mid\exists f\in D\, pf-1\in D\right\}$. For any $\bar{x}\in\mathbb{R}^{n}$, if $f_{i}\left(\bar{x}\right)\geq0$ for each $i$, then $q_{\alpha}\left(\bar{x}\right)>0$ for each $\alpha$, and hence $q\left(\bar{x}\right)\geq0$. That is, the finite set of inequalities $\left\{ f_{i}\left(\bar{x}\right)\geq0\mid1\leq i\leq m\right\} \cup\left\{ q\left(\bar{x}\right)<0\right\}$ does not hold anywhere in $\mathbb{R}^{n}$. By the compactness theorem, there are no $n$-types satisfying all those inequalities. Given $\Phi\in P\left(C\right)$$f_{i}\left(\Phi\right)\geq0$, so $q\left(\Phi\right)\nless0$; that is, $q\left(\Phi\right)\geq0$.

Conversely, suppose $q\in\text{Rad}\left(C\right)$. Then by the compactness theorem, there are some $f_{1},...,f_{m}\in C$ such that $q\in\text{Rad}\left(\left\langle f_{1},...,f_{m}\right\rangle \right)$. Then $\forall\varepsilon>0$, $q+\varepsilon$ is strictly positive on $P\left(\left\langle f_{1},...,f_{m}\right\rangle \right)$, and hence by Positivstellensatz 4, $\exists f\in\left\langle f_{1},...,f_{m}\right\rangle$ such that $pf-1\in\left\langle f_{1},...,f_{m}\right\rangle$. That is, $\left\{ q+\varepsilon\mid\varepsilon>0\right\}$ is a chain contained in $\left\langle f_{1},...,f_{m}\right\rangle$, a finitely-generated subcone of $C$, whose infimum is $q$. $\square$

### Ordered commutative algebra

Even though they are technically not isomorphic, $\mathbb{C}^{n}$ and $\text{Spec}\left(\mathbb{C}\left[x_{1},...,x_{n}\right]\right)$ are closely related, and can often be used interchangeably. Of the two, $\text{Spec}\left(\mathbb{C}\left[x_{1},...,x_{n}\right]\right)$ is of a form that can be more easily generalized to more abstruse situations in algebraic geometry, which may indicate that it is the better thing to talk about, whereas $\mathbb{C}^{n}$ is merely the simpler thing that is easier to think about and just as good in practice in many contexts. In contrast, $\mathbb{R}^{n}$ and $\mathbb{OA}_{\mathbb{R}}^{n}$ are different in important ways. The situation in algebraic geometry provides further reason to pay more attention to $\mathbb{OA}_{\mathbb{R}}^{n}$ than to $\mathbb{R}^{n}$.

The next thing to look for would be an analog of the spectrum of a ring for a partially ordered commutative ring (I will henceforth abbreviate "partially ordered commutative ring" as "ordered ring" in order to cut down on the profusion of adjectives) in a way that makes use of the order, and gives us $\mathbb{OA}_{\mathbb{R}}^{n}$ when applied to $\mathbb{R}\left[x_{1},...,x_{n}\right]$. I will call it the order spectrum of an ordered ring $A$, denoted $\text{OrdSpec}\left(A\right)$. Then of course $\mathbb{OA}_{A}^{n}$ can be defined as $\text{OrdSpec}\left(A\left[x_{1},...,x_{n}\right]\right)$$\text{OrdSpec}\left(A\right)$ should be, of course, the set of prime cones. But what even is a prime cone?

Definition: A cone $\mathfrak{p}\subseteq A$ is prime if $A/\mathfrak{p}$ is a totally ordered integral domain.

Definition: $\text{OrdSpec}\left(A\right)$ is the set of prime cones in $A$, equipped with the topology whose closed sets are the sets of prime cones containing a given cone.

An $n$-type $\Phi\in\mathbb{OA}_{\mathbb{R}}^{n}$ can be seen as a cone, by identifying it with $\left\{ f\in\mathbb{R}\left[x_{1},...,x_{n}\right]\mid f\left(\Phi\right)\geq0\right\}$, aka $C\left(\left\{ \Phi\right\} \right)$. Under this identification, $\mathbb{OA}_{\mathbb{R}}^{n}=\text{OrdSpec}\left(\mathbb{R}\left[x_{1},...,x_{n}\right]\right)$, as desired. The prime cones in $\mathbb{R}\left[x_{1},...,x_{n}\right]$ are also the radical cones $C$ such that $P\left(C\right)$ is irreducible. Notice that irreducible subsets of $\mathbb{OA}_{\mathbb{R}}^{n}$ are much smaller than irreducible subsets of $\mathbb{A}_{\mathbb{C}}^{n}$; in particular, none of them contain more than one element of $\mathbb{R}^{n}$.

There is also a natural notion of maximal cone.

Definition: A cone $\mathfrak{m}\subseteq A$ is maximal if $\mathfrak{m}\neq A$ and there are no strictly intermediate cones between $\mathfrak{m}$ and $A$. Equivalently, if $\mathfrak{m}$ is prime and closed in $\text{OrdSpec}\left(A\right)$.

Maximal ideals of $\mathbb{C}\left[x_{1},...,x_{n}\right]$ correspond to elements of $\mathbb{C}^{n}$. And the cones of elements of $\mathbb{R}^{n}$ are maximal cones in $\mathbb{R}\left[x_{1},...,x_{n}\right]$, but unlike in the complex case, these are not all the maximal cones, since there are closed points in $\mathbb{OA}_{\mathbb{R}}^{n}$ outside of $\mathbb{R}^{n}$. For example, $C_{\infty}$ is a maximal cone, and the type of numbers greater than all reals is closed. To characterize the cones of elements of $\mathbb{R}^{n}$, we need something slightly different.

Definition: A cone $\mathfrak{m}\subseteq A$ is ideally maximal if $A/\mathfrak{m}$ is a totally ordered field. Equivalently, if $\mathfrak{m}$ is maximal and $\mathfrak{m}^{\circ}$ is a maximal ideal.

Elements of $\mathbb{R}^{n}$ correspond to ideally maximal cones of $\mathbb{R}\left[x_{1},...,x_{n}\right]$.

$\text{OrdSpec}$ also allows us to define the radical of a cone in an arbitrary partially ordered commutative ring.

Definition: For a cone $C\subseteq A$, $\text{Rad}\left(C\right)$ is the intersection of all prime cones containing $C$. $C$ is radical if $C=\text{Rad}\left(C\right)$.

Conjecture: $\text{Rad}\left(C\right)$ is the union over all finitely-generated subcones $C\subseteq D$ of the closure of $\left\{ p\in A\mid\exists f\in D\, pf-1\in D\right\}$ (as before, the closure of a subset $X\subseteq A$ is defined to be the set of all elements of $A$ which are infima of chains contained in $X$).

### Order schemes

Definition: An ordered ringed space is a topological space equipped with a sheaf of ordered rings. An ordered ring is local if it has a unique ideally maximal cone, and a locally ordered ringed space is an ordered ringed space whose stalks are local.

$\text{OrdSpec}\left(A\right)$ can be equipped with a sheaf of ordered rings $\mathcal{O}_{A}$, making it a locally ordered ringed space.

Definition: For a prime cone $\mathfrak{p}\subseteq A$, the localization of $A$ at $\mathfrak{p}$, denoted $A_{\mathfrak{p}}$, is the ring $A_{\mathfrak{p}^{\circ}}$ equipped with an ordering that makes it a local ordered ring. This will be the stalk at $\mathfrak{p}$ of $\mathcal{O}_{A}$. A fraction $\frac{a}{b}\in A_{\mathfrak{p}}$ ($b\notin\mathfrak{p}^{\circ}$) is also an element of $A_{\mathfrak{q}}$ for any prime cone $\mathfrak{q}\subseteq A$ whose interior ideal does not contain $b$. This is an open neighborhood of $\mathfrak{p}$ (its complement is the set of prime cones containing $\left\langle b,-b\right\rangle$). There is a natural map $A_{\mathfrak{p}}\rightarrow\text{Frac}\left(A/\mathfrak{p}\right)$ given by $\frac{a}{b}\mapsto\frac{a+\mathfrak{p}^{\circ}}{b+\mathfrak{p}^{\circ}}$, and the total order on $A/\mathfrak{p}$ extends uniquely to a total order on the fraction field, so for $a,b\in A_{\mathfrak{p}}$, we can say that $a\geq b$ at $\mathfrak{p}$ if this is true of their images in $\text{Frac}\left(A/\mathfrak{p}\right)$. We can then say that $a\geq b$ near $\mathfrak{p}$ if $a\geq b$ at every point in some neighborhood of $\mathfrak{p}$, which defines the ordering on $A_{\mathfrak{p}}$.

Definition: For open $U\subseteq\text{OrdSpec}\left(A\right)$, $\mathcal{O}_{A}\left(U\right)$ consists of elements of $\prod_{\mathfrak{p}\in U}A_{\mathfrak{p}}$ that are locally ratios of elements of $A$. $\mathcal{O}_{A}\left(U\right)$ is ordered by $a\geq b$ if and only if $\forall\mathfrak{p}\in\text{OrdSpec}\left(A\right)$ $a\geq b$ near $\mathfrak{p}$ (equivalently, if $\forall\mathfrak{p}\in\text{OrdSpec}\left(A\right)$ $a\geq b$ at $\mathfrak{p}$).

$A\subseteq\mathcal{O}_{A}\left(\text{OrdSpec}\left(A\right)\right)$, and this inclusion can be proper. Conjecture: $\text{OrdSpec}\left(\mathcal{O}_{A}\left(U\right)\right)\cong U$ as locally ordered ringed spaces for open $U\subseteq\text{OrdSpec}\left(A\right)$. This conjecture says that it makes sense to talk about whether or not a locally ordered ringed space looks locally like an order spectrum near a given point. Thus, if this conjecture is false, it would make the following definition look highly suspect.

Definition: An order scheme is a topological space $X$ equipped with a sheaf of ordered commutative rings $\mathcal{O}_{X}$ such that for some open cover of $X$, the restrictions of $\mathcal{O}_{X}$ to the open sets in the cover are all isomorphic to order spectra of ordered commutative rings.

I don't have any uses in mind for order schemes, but then again, I don't know what ordinary schemes are for either and they are apparently useful, and order schemes seem like a natural analog of them.

## Nonabelian modules

This is a rough overview of my thoughts on a thing I've been thinking about, and as such is incomplete and may contain errors. Proofs have been omitted when writing them out would be at all tedious.

Edit: It has been pointed out to me that near-ring modules have already been defined, and the objects I describe in this post are just near-ring modules where the near-ring happens to be a ring.

### Introduction

As you all know (those of you who have the background for this post, anyway), an $R$-module is an abelian group $M$ (written additively) together with a multiplication map $R\times M\rightarrow M$ such that for all $\alpha,\beta\in R$ and $x,y\in M$, $\alpha\cdot\left(x+y\right)=\alpha\cdot x+\alpha\cdot y$$\left(\alpha+\beta\right)\cdot x=\alpha\cdot x+\beta\cdot x$, $\left(\alpha\beta\right)\cdot x=\alpha\cdot\left(\beta\cdot x\right)$, and $1\cdot x=x$.

What if we don't want to restrict attention to abelian groups? One could attempt to define a nonabelian module using the same axioms, but without the restriction that the group be abelian. As it is customary to write groups multiplicatively if they are not assumed to be abelian, we will do that, and the map $R\times M\rightarrow M$ will be written as exponentiation (since exponents are written on the right, I'll follow the definition of right-modules, rather than left-modules). The axioms become: for all $\alpha,\beta\in R$ and $x,y\in M$, $\left(xy\right)^{\alpha}=x^{\alpha}y^{\alpha}$$x^{\alpha+\beta}=x^{\alpha}x^{\beta}$, $x^{\left(\alpha\beta\right)}=\left(x^{\alpha}\right)^{\beta}$, and $x^{1}=x$.

What has changed? Absolutely nothing, as it turns out. The first axiom says again that $M$ is abelian, because $yx=x^{-1}\left(xy\right)^{2}y^{-1}=x^{-1}\left(x^{2}y^{2}\right)y^{-1}=xy$. We'll have to get rid of that axiom. Our new definition, which it seems to me captures the essence of a module except for abelianness:

A nonabelian $R$-module is a group $M$ (written multiplicatively) together with a scalar exponentiation map $R\times M\rightarrow M$ such that for all $\alpha,\beta\in R$ and $x\in M$, $x^{1}=x$$x^{\alpha+\beta}=x^{\alpha}x^{\beta}$, and $x^{\left(\alpha\beta\right)}=\left(x^{\alpha}\right)^{\beta}$.

These imply that $x^{0}=1$, $1^{\alpha}=1$, and $x^{-1}$ is the inverse of $x$, because $x\cdot x^{0}=x^{1}x^{0}=x^{1+0}=x^{1}=x$$1^{\alpha}=\left(1^{0}\right)^{\alpha}=1^{\left(0\alpha\right)}=1^{0}=1$, and $x\cdot x^{-1}=x^{1-1}=1$.

Just like a $\mathbb{Z}$-module is just an abelian group, a nonabelian $\mathbb{Z}$-module is just a group. Just like a $\mathbb{Z}/n\mathbb{Z}$-module is an abelian group whose exponent divides $n$, a nonabelian $\mathbb{Z}/n\mathbb{Z}$-module is a group whose exponent divides $n$.

### Exponentiation-like families of operations

Perhaps a bit more revealing is what nonabelian modules over free rings look like, since then the generators are completely generic ring elements. Where $A$ is the generating set, a $\mathbb{Z}\left\langle A\right\rangle$-module is an abelian group together with endomorphisms $\left\{ x\mapsto\alpha x\mid\alpha\in A\right\}$, which tells us that modules are about endomorphisms of an abelian group indexed by the elements of a ring. Nonabelian modules are certainly not about endomorphisms. After all, in a nonabelian group, the map $x\mapsto x^{2}$ is not an endomorphism. I will call the things that nonabelian modules are about "exponentiation-like families of operations'', and give four equivalent definitions, in roughly increasing order of concreteness and decreasing order of elegance. Definition 2 uses basic model theory, so skip it if that scares you. Definition 3 is the "for dummies'' version of definition 2.

Definition 0: Let $G$ be a group, and let $A$ be a family of functions from $G$ to $G$ (not necessarily endomorphisms). If $G$ can be made into a nonabelian $\mathbb{Z}\left\langle A\right\rangle$-module such that $x^{\alpha}=\alpha\left(x\right)$ for $x\in G$ and $\alpha\in A$, then $A$ is called an exponentiation-like family of operations on $G$. If so, the nonabelian $\mathbb{Z}\left\langle A\right\rangle$-module structure on $G$ with that property is unique, so define $x^{p}$ to be its value according to that structure, for $p\in\mathbb{Z}\left\langle A\right\rangle$ and $x\in G$.

Definition 1: $A$ is an exponentiation-like family of operations on $G$ if for all $x\in G$, the smallest subgroup containing $x$ which is closed under actions by elements of $A$ (which I will call $\overline{\left\{ x\right\} }$) is abelian, and the elements of $A$ restrict to endomorphisms of it. Using the universal property of $\mathbb{Z}\left\langle A\right\rangle$, this induces a homomorphism $\mathbb{Z}\left\langle A\right\rangle \rightarrow\text{End}\left(\overline{\left\{ x\right\} }\right)^{\text{op}}$. Let $x^{p}$ denote the action of $p$ on $x$ under that map, for $p\in\mathbb{Z}\left\langle A\right\rangle$. By $\text{End}\left(\overline{\left\{ x\right\} }\right)^{\text{op}}$, I mean the endomorphism ring of $\overline{\left\{ x\right\} }$ with composition running in the opposite direction (i.e., the multiplication operation given by $\left(f,g\right)\mapsto g\circ f$). This is because of the convention that nonabelian modules are written as nonabelian right-modules by default.

Definition 2: Let consider the language $\mathcal{L}_{Rings}\sqcup A$, where $\mathcal{L}_{Rings}:=\left\{ 0,1,+,-,\cdot\right\}$ is the language of rings, and each element of $A$ is used as a constant symbol. Closed terms in $\mathcal{L}_{Rings}\sqcup A$ act as functions from $G$ to $G$, with the action of $t$ written as $x\mapsto x^{t}$, defined inductively as: $x^{0}:=1$, $x^{1}:=x$, $x^{\alpha}:=\alpha\left(x\right)$ for $\alpha\in X$, $x^{t+s}:=x^{t}x^{s}$, $x^{-t}:=\left(x^{t}\right)^{-1}$, and $x^{ts}:=\left(x^{t}\right)^{s}$ for closed $\mathcal{L}_{Rings}\sqcup A$-terms $t$ and $s$. $A$ is called an exponentiation-like family of operations on $G$ if $x^{t}=x^{s}$ whenever $T_{Rings}\models t=s$, where $T_{Rings}$ is the theory of rings. If $A$ is an exponentiation-like family of operations on $G$ and $p\in\mathbb{Z}\left\langle A\right\rangle$ is a noncommutative polynomial with variables in $A$, then for $x\in G$$x^{p}$ is defined to be $x^{t}$ where $t$ is any term representing $p$.

Definition 3: Pick a total order on the free monoid on $A$ (e.g. by ordering $A$ and then using the lexicographic order). The order you use won't matter. Given $x\in G$ and $w:=\alpha_{1}...\alpha_{n}$ in the free monoid on $A$, let $x^{w}=\alpha_{n}\left(...\alpha_{1}\left(x\right)\right)$. Where $p\in\mathbb{Z}\left\langle A\right\rangle$ is a noncommutative polynomial, $p=c_{1}w_{1}+...+c_{n}w_{n}$ for some $c_{1},...,c_{n}\in\mathbb{Z}$ and decreasing sequence $w_{1},...,w_{n}$ of noncommutative monomials (elements of the free monoid on $A$). Let $x^{p}=\left(x^{c_{1}}\right)^{w_{1}}...\left(x^{c_{n}}\right)^{w_{n}}$$A$ is called an exponentiation-like family of operations on $G$ if for every $x\in G$ and $p,q\in\mathbb{Z}\left\langle A\right\rangle$$x^{pq}=\left(x^{p}\right)^{q}$ and $x^{p+q}=x^{p}x^{q}$.

These four definitions of exponentiation-like family are equivalent, and for exponentiation-like families, their definitions of exponentiation by a noncommutative polynomial are equivalent.

Facts: $\emptyset$ is an exponentiation-like family of operations on $G$. If $A$ is an exponentiation-like family of operations on $G$ and $B\subseteq A$, then so is $B$. If $G$ is abelian, then $\text{End}\left(G\right)$ is exponentiation-like. Given a nonabelian $R$-module structure on $G$, the actions of the elements of $R$ on $G$ form an exponentiation-like family. In particular, if $A$ is an exponentiation-like family of operations on $G$, then so is $\mathbb{Z}\left\langle A\right\rangle$, with the actions being defined as above.

[The following paragraph has been edited since this comment.]

For an abelian group $A$, the endomorphisms of $A$ form a ring $\text{End}\left(A\right)$, and an $R$-module structure on $A$ is simply a homomorphism $R\rightarrow\text{End}\left(A\right)$. Can we say a similar thing about exponentiation-like families of operations of $G$? Let $\text{Exp}\left(G\right)$ be the set of all functions $G\rightarrow G$ (as sets). Given $\alpha,\beta\in\text{Exp}\left(G\right)$, let multiplication be given by composition: $x^{\left(\alpha\beta\right)}=\left(x^{\alpha}\right)^{\beta}$, addition be given by $x^{\alpha+\beta}=x^{\alpha}x^{\beta}$, negation be given by $x^{-\alpha}=\left(x^{\alpha}\right)^{-1}$, and $0$ and $1$ be given by $x^{0}=1$ and $x^{1}=x$. This makes $\text{Exp}\left(G\right)$ into a near-ring. A nonabelian $R$-module structure on $G$ is a homomorphism $R\rightarrow\text{Exp}\left(G\right)$, and a set of operations on $G$ is an exponentiation-like family of operations on $G$ if and only if it is contained in a ring which is contained in $\text{Exp}\left(G\right)$.

### Some aimless rambling

What are some interesting examples of nonabelian modules that are not abelian? (That might sound redundant, but "nonabelian module'' means that the requirement of abelianness has been removed, not that a requirement of nonabelianness has been imposed. Perhaps I should come up with better terminology. To make matters worse, since the requirement that got removed is actually stronger than abelianness, there are nonabelian modules that are abelian and not modules. For instance, consider the nonabelian $\mathbb{Z}\left[\alpha\right]$-module whose underlying set is the Klein four group (generated by two elements $a,b$) such that $a^{\alpha}=a$, $b^{\alpha}=b$, and $\left(ab\right)^{\alpha}=1$.)

In particular, what do free nonabelian modules look like? The free nonabelian $\mathbb{Z}$-modules are, of course, free groups. The free nonabelian $\mathbb{Z}/n\mathbb{Z}$-modules have been studied in combinatorial group theory; they're called Burnside groups. (Fun but tangential fact: not all Burnside groups are finite (the Burnside problem), but despite this, the category of finite nonabelian $\mathbb{Z}/n\mathbb{Z}$-modules has free objects on any finite generating set, called Restricted Burnside groups.)

The free nonabelian $\mathbb{Z}\left[\alpha\right]$-modules are monstrosities. They can be constructed in the usual way of constructing free objects in a variety of algebraic structures, but that construction seems not to be very enlightening about their structure. So I'll give a somewhat more direct construction of the free nonabelian $\mathbb{Z}\left[\alpha\right]$-module on $d$ generators, which may also not be that enlightening, and which is only suspected to be correct. Define an increasing sequence of groups $G_{n}$, and functions $\alpha_{n}:G_{n}\rightarrow G_{n+1}$, as follows: $G_{0}$ is the free group on $d$ generators. Given $G_{n}$, and given a subgroup $X\leq G_{n}$, let the top-degree portion of $X$ be $\alpha_{n-1}^{k}\left(X\right)$ for the largest $k$ such that this is nontrivial. Let $H_{n}$ be the free product of the top-degree portions of maximal abelian subgroups of $G_{n}$. Let $G_{n+1}$ be the free product of $G_{n}$ with $H_{n}$ modulo commutativity of the maximal abelian subgroups of $G_{n}$ with the images of their top-degree portions in $H_{n}$. Given a maximal abelian subgroup $X\leq G_{n}$, let $\alpha_{n}\restriction_{X}$ be the homomorphism extending $\alpha_{n-1}\restriction_{X\cap G_{n-1}}$ which sends the top-degree portion identically onto its image in $H_{n}$. Since every non-identity element of $G_{n}$ is in a unique maximal abelian subgroup, this defines $\alpha_{n}$. $G:=\bigcup_{n}G_{n}$ with $\alpha:=\bigcup_{n}\alpha_{n}$ is the free nonabelian $\mathbb{Z}\left[\alpha\right]$-module on $d$ generators. If $A$ is a set, the free nonabelian $\mathbb{Z}\left\langle A\right\rangle$-modules can be constructed similarly, with $\left|A\right|$ copies of $H_{n}$ at each step. Are these constructions even correct? Are there nicer ones?

A nonabelian $\mathbb{Z}\left[\frac{1}{2}\right]$-module would be a group with a formal square root operation. As an example, any group of odd exponent $n$ can be made into a $\mathbb{Z}\left[\frac{1}{2}\right]$-module in a canonical way by letting $x^{\frac{1}{2}}=x^{\frac{n+1}{2}}$. More generally, any group of finite exponent $n$ can be made into a $\mathbb{Z}\left[\left\{ p^{-1}|p\nmid n\right\} \right]$-module in a similar fashion. Are there any more nice examples of nonabelian modules over localizations of $\mathbb{Z}$?

In particular, a nonabelian $\mathbb{Q}$-module would be a group with formal $n$th root operations for all $n$. What are some nonabelian examples of these? Note that nonabelian $\mathbb{Q}$-modules cannot have any torsion, for suppose $x^{n}=1$ for some $n\neq0$. Then $x=\left(x^{n}\right)^{\frac{1}{n}}=1^{\frac{1}{n}}=1$. More generally, nonabelian modules cannot have any $n$-torsion (meaning $x^{n}=1\implies x=1$) for any $n$ which is invertible in the scalar ring.

The free nonabelian $\mathbb{Z}\left[\frac{1}{m}\right]$-modules can be constructed similarly to the construction of free nonabelian $\mathbb{Z}\left[\alpha\right]$-modules above, except that when constructing $G_{n+1}$ from $G_{n}$ and $H_{n}$, we also mod out by elements of $G_{n}$ being equal to the $m$th powers of their images in $H_{n}$. Using the fact that $\mathbb{Q}\cong\mathbb{Z}\left\langle \left\{ p^{-1}|\text{primes }p\right\} \right\rangle$, this lets us modify the construction of free nonabelian $\mathbb{Z}\left\langle A\right\rangle$-modules to give us a construction of free nonabelian $\mathbb{Q}$-modules. Again, is there a nicer way to do it?

### Topological nonabelian modules

It is also interesting to consider topological nonabelian modules over topological rings; that is, nonabelian modules endowed with a topology such that the group operation and scalar exponentiation are continuous. A module over a topological ring has a canonical finest topology on it, and the same remains true for nonabelian modules. For finite-dimensional real vector spaces, this is the only topology. Does the same remain true for finitely-generated nonabelian $\mathbb{R}$-modules? Finite-dimensional real vector spaces are complete, and topological nonabelian modules are, in particular, topological groups, and can thus be made into uniform spaces, so the notion of completeness still makes sense, but I think some finitely-generated nonabelian $\mathbb{R}$-modules are not complete.

A topological nonabelian $\mathbb{R}$-module is a sort of Lie group-like object. One might try constructing a Lie algebra for a complete nonabelian $\mathbb{R}$-module $M$ by letting the underlying set be $M$, and defining $x+y=\lim_{\varepsilon\rightarrow0}\left(x^{\varepsilon}y^{\varepsilon}\right)^{\left(\varepsilon^{-1}\right)}$ and $\left[x,y\right]=\lim_{\varepsilon\rightarrow0}\left(x^{\varepsilon}y^{\varepsilon}x^{-\varepsilon}y^{-\varepsilon}\right)^{\left(\varepsilon^{-2}\right)}$. One might try putting a differential structure on $M$ such that this is the Lie algebra of left-invariant derivations. Does this or something like it work?

A Lie group is a nonabelian $\mathbb{R}$-module if and only if its exponential map is a bijection between it and its Lie algebra. In this case, scalar exponentiation is closely related to the exponential map by a compelling formula: $x^{\alpha}=\exp\left(\alpha\exp^{-1}\left(x\right)\right)$. As an example, the continuous Heisenberg group is a nonabelian $\mathbb{R}$-module which is not abelian. This observation actually suggests a nice class of examples of nonabelian modules without a topology: given a commutative ring $R$, the Heisenberg group over $R$ is a nonabelian $R$-module.

The Heisenberg group of dimension $2n+1$ over a commutative ring $R$ has underlying set $R^{n}\times R^{n}\times R^{1}$, with the group operation given by $\left(\boldsymbol{a}_{1},\boldsymbol{b}_{1},c_{1}\right)*\left(\boldsymbol{a}_{2},\boldsymbol{b}_{2},c_{2}\right):=$$\left(\boldsymbol{a}_{1}+\boldsymbol{a}_{2},\boldsymbol{b}_{1}+\boldsymbol{b}_{2},c_{1}+c_{2}+\boldsymbol{a}_{1}\cdot\boldsymbol{b}_{2}-\boldsymbol{a}_{2}\cdot\boldsymbol{b}_{1}\right)$. The continuous Heisenberg group means the Heisenberg group over $\mathbb{R}$. Scalar exponentiation on a Heisenberg group is just given by scalar multiplication: $\left(\boldsymbol{a},\boldsymbol{b},c\right)^{\alpha}:=\left(\alpha\boldsymbol{a},\alpha\boldsymbol{b},\alpha c\right)$.