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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?

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    $\begingroup$ Unless I am mistaken, it seems to me that the surreals in the order topology are not locally compact. Does this cause a problem? $\endgroup$ Sep 23, 2015 at 22:00
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    $\begingroup$ For Pontryagin it must be locally compact. If you do not mean the discrete topology, what do you mean by "suitably topologized"? $\endgroup$ Sep 23, 2015 at 22:00
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    $\begingroup$ There may be set/class issues, however, since I think perhaps every set-sized open cover of a bounded interval in the surreals has a finite subcover, but there are proper class open covers with no set-sized subcover. (But I have to think more about this to be sure.) $\endgroup$ Sep 23, 2015 at 22:05
  • $\begingroup$ As an amateur non-standard-analyst and occasional surreal-analyst ... or something ... I'd have the impression that the surreals are generally problematical by not being a set, in any case. Various incarnations of non-standard reals are sets, at least. The meaning of "finite subcover" maybe has to be non-standard-ized, or qualified by "non-standard open" or not... Probably @JoelDavidHamkins has good information at his fingertips about such. $\endgroup$ Sep 23, 2015 at 22:12
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    $\begingroup$ @JoelDavidHamkins: I used your comment as a basis for a new question mathoverflow.net/q/219167/454 $\endgroup$ Sep 24, 2015 at 13:43

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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).

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