We have that for a locally compact abelian group,

$$A \mapsto \operatorname{Hom}(A, \mathbb{R}/\mathbb{Z})$$

is a biduality map when endowing $\operatorname{Hom}$ with the compact-open topology.

My question is why $\mathbb{R}/\mathbb{Z}$? This sounds like a rather counter-intuitive result. I can see one why one would give $\operatorname{Hom}$ the compact-open topology (there are many categorical and topological reasons to do this) but it looks rather ad-hoc that we use the circle group.

How did this come up? Are there other bidualities using different groups?

  • $\begingroup$ mathoverflow.net/questions/294019/… $\endgroup$ Commented Feb 15, 2021 at 22:02
  • $\begingroup$ It is in fact explained in this post. Read the first comment by Kcd. His $U(1)$ is your $\Bbb R/\Bbb Z$. And he cited Hewitt and Ross vol. 1 p. 424 as well as Pontryagin's book Topological Groups Example 72. I didn't check any of the citations. $\endgroup$
    – WhatsUp
    Commented Feb 15, 2021 at 22:27

1 Answer 1


The fundamental notion behind the theory of groups is the concept of symmetries and the best place to visualize symmetries is Euclidean space and, by extension, Hilbert's space.

By a symmetry on a Hilbert space $H$ one means any map $$ U:H\to H $$ that preserves distance, and sends the origin to itself. In fact any such map is necessarily a unitary operator.

The group $\mathscr U(H)$, formed by all unitary operators on Hilbert's space is, according to this, the archetype of symmetry!

Given any group $G$, it therefore makes a lot of sense to try to model $G$ via $\mathscr U(H)$, and this is usually done by considering group homomorphisms $$ u:G\to \mathscr U(H), $$ often called group representations.

If $u_1$ and $u_2$ are representations of $G$, one may define a direct sum representation $$ u=u_1\oplus u_2 \tag 1 $$ on $H\oplus H$, but everything there is to know about $u$ is already present in either $u_1$ or $u_2$.

On the other hand, should we be interested in studying a given representation $u$, we might ask whether or not it has an expression such as (1), in which case it would be better to study $u_1$ and $u_2$ separately.

This means that one should concentrate in the study of irreducible group representations, namely those which cannot be split as in (1).

For abelian groups, Schur's Lemma says that all irreducible unitary representations are necessarily one-dimensional, meaning that the underlying Hilbert space is $\mathbb C$, and since $$ \mathscr U(\mathbb C) = \mathbb T, $$ we are inexorably led to studying homomorphisms from $G$ to $\mathbb T$!

  • $\begingroup$ This is a very vivid, nice explanation. $\endgroup$
    – Bumblebee
    Commented Jun 15, 2021 at 1:39

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