Pontryagin dual for non abelian $\widehat G := \operatorname{Hom}(G, T) $?

Question

For a locally compact abelian group $G$ , the Pontryagin dual is the group $\widehat G$ of continuous group homomorphisms from $G$ to the circle group $T$ . That is, :

$$\widehat G := \operatorname{Hom}(G, T). $$

The Pontryagin dual $\widehat G$ is usually endowed with the topology given by uniform convergence on compact sets (that is, the topology induced by the compact-open topology on the space of all continuous functions from $G$ to $T$).

For example,

$$\widehat {\mathbb Z} = T,\ \widehat {\mathbb R} = \mathbb R,\ \widehat T = \mathbb Z.$$

questions

1. do we have a relation that $$ \operatorname{Hom}(G_1, G_2)= \operatorname{Hom}(G_2, G_1)?$$ for any groups? I think the answer is no. If no, can you give the reasonings behind why and for counter examples? (for abelian groups examples? for non-abelian groups examples?)

2. How do we define $\widehat G$ for non-abelian $G?$ Is that $\widehat SU(2)=SO(3)$ and $\widehat SO(3)=SU(2)$?

Answer 1

For abelian groups $G_1$ and $G_2$, the map $$ \varphi \in \text{Hom}(G_1, G_2) \mapsto \hat \varphi \in \text{Hom}(\hat G_2, \hat G_1), $$ given by $$ \hat \varphi (\gamma ) = \gamma \circ \varphi , \quad \forall \gamma \in \hat G_2, $$ is a bijection. Now, for every finite abelian group $G$ one has that $\hat G$ is isomorphic to $G$. So, if $G_1$ and $G_2$ are finite abelian, one gets a bijection $$ \text{Hom}(G_1, G_2) \buildrel \widehat{} \over \longrightarrow \text{Hom}(\hat G_2, \hat G_1) = \text{Hom}(G_2, G_1). $$

Besides finite abelian groups, things fail miserably.

Without finiteness: Notice that $\text{Hom}(\mathbb T, \mathbb Z_2)$ has only one element, namely the trivial homomorphism, but in $\text{Hom}(\mathbb Z_2, \mathbb T)$ one may find the trivial homomorphism in addition to the map sending the generator to $-1$.

Without commutativity: Since the alternating group $A_n$ is simple for $n\geq 5$, one has that $\text{Hom}(A_n, \mathbb Z_2)$ only has the trivial homomorphism, while $\text{Hom}(\mathbb Z_2, A_n)$ has as many elements as order-two even permutations.

Regarding the dual of non-abelian groups, one usually defines $\hat G$ as the set of equivalence classes of irreducible unitary representations on Hilbert spaces. Here $\hat G$ does not have a group structure and the theory is simultaneously a lot more sophisticated and a lot less powerful than Pontryagin duality.

If $G$ is abelian, Schur's Lemma implies that every unitary irreducible representation $$ π:G→U(H) $$ is one-dimensional, that is, $H$ is a one-dimensional Hilbert space. In this case $U(H)=\mathbb T$, so $π$ is actually a homomorphism from $G$ to $\mathbb T$. In other words, the dual reduces to the Pontryagin dual.

However, for lots of non-abelian groups, e.g. $A_n$, there are no nontrivial one-dimensional representations, so $ \text{Hom}(G, \mathbb T) $ may be empty!