# Properties of the weak-$*$ topology

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 $c(x) := c$.

• Why this question suddenly became "community wiki"? – TaQ Jul 30 '13 at 22:04
• @TaQ: the big-list tag states, "Such a question should typically be in Community Wiki (CW) mode; after asking, please, flag for moderators attention requesting the question to be made CW." – Tom LaGatta Jul 31 '13 at 4:23

## 1 Answer

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.

• In fact it is locally convex for $\mathbb{C}$ and $\mathbb{Q}_p$ as well with the same proof. – Johannes Hahn Jul 30 '13 at 23:37
• 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". – Johannes Hahn Jul 30 '13 at 23:39