What is the topology on the set of field orders

Question

Inspired by this question I was wondering whether there is a natural topology on the set of all orders on a field (that extend a given order on a subfield)?

For example for the function field $\mathbb{Q}(X)$ there are the following examples of total orders:

1. The convention $X>q$ for all $q\in \mathbb{Q}$;
2. the convention $0<X<1/n$ for all $n\in \mathbb{N}$; the pullback of
3. the standard-order on the reals along an embedding $\mathbb{Q(X)}\rightarrow \mathbb{R}$ sending $X$ to any transcendental element;
   4. the pullback of the total order on the surreal numbers along an embedding.

I suspect that different embeddings in (4) still can give the same total order, but maybe all orders arise that way.

Apart from the question how to classify these field orderings as a set, it seems interesting to ask how to topologize them.

Answer 1

The topology you are looking for is called the Harrison topology. If we denote the set of ordering of a field $F$ with $\mathrm{Sper}\,F$ (more on that in a moment), this is the subspace topology given by the embedding $$\mathrm{sign}:\mathrm{Sper}\,F\to \prod_{x\in F^\times}\{+1,-1\}$$ sending an order to the collection of signs of the elements of $F^\times$. This identifies $\mathrm{Sper}\,F$ as a closed subset of that product, and hence as totally disconnected compact Hausdorff space. A basis of clopen sets is given by $U_x:=\{\le\mid x> 0\}$ for every $x\in F^\times$.

This is a special case of the construction of the real spectrum of a ring $A$. That is the collection of the ``partial orders'' of $A$, which are essentially couples $(p,\le)$ where $p\in\mathrm{Spec}\,A$ is a prime ideal and $\le$ is an order on the residue field $k(p)$. We can topologize it as the closed subspace of $$\mathrm{sign}:\mathrm{Sper}\,A\to \prod_{x\in A}\{+1,0,-1\}$$ where $\{+1,0,-1\}$ has the coarsest topology such that $\{0\}$ is closed and $\{+1\}$ and $\{-1\}$ are open.

To give a concrete example, the points of $\mathrm{Sper}\,\mathbb{Q}(x)$ are of one of the following forms

• $a^+,a^-$ where $a$ is a real algebraic number. The order is given by $f\ge 0$ iff $f$ is non-negative in $(a,a+\epsilon)$, resp. $(a-\epsilon,a)$ for some $\epsilon>0$
• $r$ where $r$ is a trascendental real number. Here $f\ge0$ iff $f(r)\ge0$
• $+\infty,-\infty$. Here the order is given by $f\ge0$ iff $f$ is non-negative on $(M,\infty$), resp. $(-\infty,-M)$ for some $M>0$. Equivalently, this picks the sign of the leading coefficient (resp. $(-1)^{\deg f}$ times the sign of the leading coefficient).

These are all distinct, as one can easily show.

The topology of $\mathrm{Sper}\,\mathbb{Q}(x)$ has a basis of clopen subsets of the form $(a,b)$ where $a,b$ are either real algebraic numbers or $\pm\infty$. This is the set of orders $\ge$ such that $f\ge0$ if $f$ is non-negative on $(a,b)$. Note that $a^+\in (a,b)$, but $a^-\not\in(a,b)$.

There is a more comprehensive treatment of this, with more examples, in chapter 7 of

Bochnak, Jacek; Coste, Michel; Roy, Marie-Françoise, Real algebraic geometry. Transl. from the French., Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge. 36. Berlin: Springer. ix, 430 p. (1998). ZBL0912.14023.