2
$\begingroup$

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 $c(x) := c$.

$\endgroup$
  • $\begingroup$ Why this question suddenly became "community wiki"? $\endgroup$ – TaQ Jul 30 '13 at 22:04
  • $\begingroup$ @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." $\endgroup$ – Tom LaGatta Jul 31 '13 at 4:23
1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ In fact it is locally convex for $\mathbb{C}$ and $\mathbb{Q}_p$ as well with the same proof. $\endgroup$ – Johannes Hahn Jul 30 '13 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 Jul 30 '13 at 23:39

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.