Pontryagin dual of the surreal numbers? Has any work been done on the Pontryagin dual of the surreal numbers (suitably topologized)? I have not been able to find anything and am not sure if this is still unknown.
Alternatively, has this been worked out for the various hyperreal fields, or real-closed fields in general?
 A: For any infinite cardinal $\kappa$, let $S_\kappa$ be the surreal numbers of rank $<\kappa$, considered as a group under addition and topologized with the order topology (if you want to consider all the surreal numbers, suppose $\kappa$ is inaccessible).  Suppose now that $\kappa$ has uncountable cofinality and $f:S_\kappa\to U(1)$ is a continuous homomorphism.  Since $\kappa$ has uncountable cofinality, $S_\kappa$ is countably saturated as an ordered set, and it follows that a countable intersection of open sets in $S_\kappa$ is still open.  In particular, this implies that $\ker(f)$ is an open subgroup of $S_\kappa$.
That is, every continuous homomorphism $f:S_\kappa\to U(1)$ factors through a discrete group.  The open subgroups $K_\alpha=\bigcup_{n\in \mathbb{N}}(-n\omega^{-\alpha},n\omega^{-\alpha})$ for ordinals $\alpha<\kappa$ are a neighborhood base at $0$, so the group $G$ of continuous homomorphisms $f:S_\kappa\to U(1)$ can be considered as the direct limit of the Pontryagin duals of the discrete groups $S_\kappa/K_\alpha$.  It is easy to see that for each $\alpha<\kappa$, $K_\alpha/\bigcup_{\beta<\alpha} K_\beta$ is a $\mathbb{Q}$-vector space of dimension $\kappa$, as is $S_\kappa/K_0$.  Choosing bases for all these vector spaces, we obtain the following description of the group $G$.  Consider $\mathbb{Q}$ as a discrete group, let $B$ be its Pontryagin dual, and let $A=B^\kappa$.  Then $G$ is isomorphic to the group of functions $\kappa\to A$ which are eventually $0$.  Note that in this description, the group of all (possibly discontinuous) homomorphisms $S_\kappa\to U(1)$ can be identified with the full product $A^\kappa$.
I don't know whether there is any particularly natural topology to put on $G$, but it is not hard to check that the compact-open topology is just the product topology on $G$ as a subgroup of $A^\kappa$ (to show this, first show that any compact subset of $S_\kappa$ is finite).
