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Let $X$ be a topological affine space over a complete base field $\mathbb S := \mathbb C$, $\mathbb R$ or $\mathbb Q_p$. Let $X^*$ be the dual space of continuous affine functionals equipped with the weak-$*$ topology.

At this level of generality, what are the necessary properties that the weak-$*$ topology must satisfy? e.g., locally convex, Hausdorff, etc.

Here's a trivial property. Unlike the linear setting, the affine dual space contains a one-dimensional subspace of constant functionals $c(x) := c$.

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For $\mathbb R$ it is of course locally convex and Hausdorff. Indeed, $X^\ast$ is homeomorphic to a subset of the product space $\mathbb R^I$ for some index set $I$ (in fact we can take $I = X$). Similar for $\mathbb C$.

You will have to provide definitions of "convex" in other cases.

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    $\begingroup$ In fact it is locally convex for $\mathbb{C}$ and $\mathbb{Q}_p$ as well with the same proof. $\endgroup$
    – Johannes Hahn
    Commented Jul 30, 2013 at 23:37
  • $\begingroup$ Regarding the definition: In the world of valued fields the notion of an absolutely convex set still makes sense. $A\subseteq V$ is absolutely convex if $|\mu|+|\lambda|\leq 1 \implies \forall a,b\in A: \lambda a+ \mu b \in A$. Now the set $\lbrace\lambda\in K : |\lambda|\leq 1\rbrace$ is just the valuation ring $\mathcal{O}$ of the valued field $K$ and an absolutely convex subset is nothing else then an $\mathcal{O}$-submodule of $V$. Therefore we can just apply the definition "there is a neighborhood base of the origin consisting of absolutly convex subsets of V". $\endgroup$
    – Johannes Hahn
    Commented Jul 30, 2013 at 23:39

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