Is there an infinite topological meadow with non-trivial topology?

Question

For reference meadows are a generalization of fields that were designed to be compatible with the requirements of universal algebra. Specifically a meadow is a commutative ring equiped with an involution $x\mapsto x^{-1}$ which obeys

$$x\cdot x\cdot x^{-1}=x$$

for all $x$. (A field can be turned into a meadow by letting $x^{-1}$ be the multiplicative inverse when $x\neq0$ and $0^{-1}=0$.) A topological meadow would be a commutative topological ring with such a $x\mapsto x^{-1}$ such that $x\mapsto x^{-1}$ is a continuous function.

It was established in a previous question that none of the standard fields can be thought of as topological meadows with their standard topologies. (I'm also fairly certain the $p$-adics aren't topological meadows under their standard topologies either.)

So my question is: Does there exist an infinite topological meadow with non-trivial topology? Non-trivial is somewhat vague, but I think essentially what I'm looking for is the space being Hausdorff and connected (if you take the topology of a topological field and make 0 an isolated point I believe you get a topological meadow, but it's disconnected), although failing these I would like to know 'how close' you can get. And if there is an infinite Hausdorff, connected topological meadow I would like to know how much more geoemtric structure you can have. Is there a topological meadow that is a connected (smooth) manifold (that isn't the zero meadow)?

Answer 1

The argument given in this answer on MO shows that if $A$ is any set equipped with a collection of finitary operations which satisfy some equational axioms such that $2\leq|A|\leq\aleph_0$, then there is an infinite-dimensional contractible CW-complex $X$ which can be equipped with corresponding finitary operations which are continuous and which satisfy the same axioms (explicitly, $X$ is the classifying space of the contractible groupoid with $A$ as the set of objects; I believe that actually $X$ is always homeomorphic to $\mathbb{R}^\infty$, but I don't see an easy way to prove this). In particular, every nontrivial countable meadow gives rise to a nontrivial connected topological meadow.

As another way to get examples of topological meadows, any Boolean ring (i.e., a ring in which $x^2=x$ for all $x$) has a meadow structure for which $x^{-1}=x$ for all $x$. Since the identity map is always continuous, this means any topological Boolean ring can be made into a topological meadow. It is easy to find topological Boolean rings with Hausdorff but non-discrete topologies. For instance, consider an infinite product of copies of $\mathbb{F}_2$, with the product topology. Finding a connected topological Boolean ring seems harder; the only way I know how is using the construction in the previous paragraph.

However, there are no nontrivial connected topological meadows that are manifolds. To show this, note first that any connected topological ring which is a manifold must have $\mathbb{R}^n$ as its additive group for some $n$ (every abelian Lie group is of the form $\mathbb{R}^n\times (S^1)^m$, and it is not hard to show that this cannot admit a ring structure if $m>0$). In particular, unless $n=0$, it must contain $\mathbb{Q}$ as a topological subring, with its usual topology. But if $M$ is a meadow and $u\in M$ has a multiplicative inverse, then $u^{-1}$ must be its multiplicative inverse. Since the inverse map (together with $0\mapsto 0$) is not continuous on $\mathbb{Q}$, this means our ring cannot admit a topological meadow structure unless $n=0$.