0

In the subject of topological groups. The discussion starts with defining the topological group $G$, then a dual object, $G^{*}$ is defined. This is the following definition that I am working with which I think is a standard one:

Let $G$ be an abelian topological group. Define the dual group $G^{*}$ to be the set of all continuous homomorphism from $G$ into the circle group.

Now, equip this set with a binary operation

$$(\phi + \xi)(x)=\phi(g)+\xi(g)$$ for all $g\in G$ and $\phi,\xi\in G^{*}$ which in turn makes this set to be an abelian group. One particular objective is to make this group to be a topological group. The idea of compact-open topology, which is intuitive for these set-ups provides a framework we needed: Let $K$ be a compact subset of $G$ and $U$ an open set of the circle group $T$. Then the set $P(K,U)\subseteq G^*$ is defined to be

$$\{ \psi \ : \ \psi\in G^* \ \text{and} \ \psi(K)\subseteq U \}.$$

My question in this stage is, how to prove that if we define sets of the form $P(K,U)$ to be open sets of $G^*$, then $G^*$ is a Hausdorff topological group?

J. W. Tanner
  • 63,683
  • 4
  • 43
  • 88
Julius Monk
  • 5
  • 1
    What have you been able to prove and what is blocking you? Have you proved Hausdorffness? continuity of the product? Of the inverse? – Olivier Bégassat Oct 22 '21 at 16:02
  • As a general topology remark: the compact-open topology will only be nice if the domain (here $G$) has enough compact sets (you could have that only finite subsets are compact and the topology would reduce to the product topology on $(S^1)^G$ etc.). Local compactness is a common assumption. Here you have no such condition on $G$ so beware. – Henno Brandsma Oct 23 '21 at 08:08
  • This question from the sidebar (always check it!) might interest you.. This one too perhaps. The local compactness does seem to be needed a lot. – Henno Brandsma Oct 23 '21 at 08:09

0 Answers0