# Why unitary characters for the dual group in Pontryagin duality if $G$ is not compact?

In harmonic analysis, for any locally compact abelian group, one constructs the dual group as the group of homomorphisms into the unit circle with the compact open topology. In other words, unitary linear characters of the group. We know from representation theory that we should expect irreducible reps of an abelian group to be 1 dimensional, so it makes sense to use linear characters. We also know that if the group is compact, we can always take the (finite dim) rep to be unitary, hence for compact groups, our characters should be unitary.

So what about when $G$ is not compact? Are we losing something by skipping nonunitary characters (homomorphisms into $\mathbb{C}^\times$)? Obviously not, since Pontryagin duality works, but is there a nice way to understand that? If there is no invariant inner product, how do we even define unitarity? More generally, what's so special about $\text{U}(1)$ that it should be the "universal dualizer"?

• The choice $U(1)$ is crucial for Pontryagin duality. For any abelian topological groups $G$ and $H$, the continuous homomorphisms ${\rm Hom}(G,H)$ with pointwise multiplication and the compact-open topology form a topological group (see Hewitt and Ross vol. 1 p. 374). Suppose for a locally compact abelian group $A$ that $G \cong {\rm Hom}({\rm Hom}(G,A),A))$ for all locally compact abelian groups $G$. Then $A$ must be isomorphic to $U(1)$! In fact this follows from the isom. just for $G = U(1)$, and it could be a wacky top. gp. isom. See H&R p. 424 or Pontryagin's book Top. Gps, Example 72. – KCd Mar 25 '12 at 23:39
• By "wacky" isomorphism I mean the hypothesis is that $U(1) \cong {\rm Hom}({\rm Hom}(U(1),A),A))$ by some top. group isomorphism, not necessarily one induced by the natural mapping from $U(1)$ to ${\rm Hom}({\rm Hom}(U(1),A),A))$. – KCd Mar 25 '12 at 23:40
• My previous comments indicated what is so special about $U(1)$ as a "universal dualizer", but the more general continuous homomorphisms $G \rightarrow {\mathbf C}^\times$ are also quite useful, depending on the context. They just won't lead to a duality theory. – KCd Mar 25 '12 at 23:42