In this note, we show how functional analysis can be used to compute the number of group topologies on the real line and how representation theory of locally compact abelian groups leads to the existence of a compact group topology on .
Theorem 1: There exist exactly distinct group topologies on .
Definition: Let be a normed vector space (over ) and be a linear subspace of the topological dual (ie. the space of continuous functionals). The weak topology on is the initial topology associated to .
Lemma 1: A weak topology is a group topology.
Proof. Let be a normed vector space and be a weak topology. To show that is continuous, it is sufficient to notice that for all and the following set is open for :
Then, is clearly continuous, since is equivalent to for all , , and .
Lemma 2: Let be a normed vector space and be a weak topology. With respect to , a functional is continuous if and only if .
Proof. Let be a linear functional continuous for . Then there exist linear functionals and open neighborhoods of the origin so that
In particular, is bounded on hence . Let
Because , we may define by when
Let be a linear extension of . In particular, .
Let such that . Then for all ,
Lemma 3: (Erdös-Kaplansky theorem) Let be an infinite-dimensional vector space. If denotes its dual space, .
We do not prove Erdös-Kaplansky theorem here and refer to Bourbaki’s book, Algebra. Chapters 1 to 3 (exercice 3 page 400).
Proof of theorem 1. Let be the set of eventually zero sequences endowed with the sup norm and let be a basis of . For all , let . According to lemma 2, for all
Therefore, according to lemmas 1 and 3, is a family of distinct group topologies on .
Because and are both a linear vector space over of dimension , there exists an isomorphism between the abelian groups and . Then we can use to transfer the distinct group topologies from to .
So has at least distinct group topologies. On the other hand, has at most distinct group topologies.
Then we may deduce the following corollaries:
Theorem 2: has exactly non-isomorphic group topologies.
Proof. has at most automorphisms and has exactly group topologies.
Theorem 3: has exactly non-locally-compact group topologies.
Proof. It is known that every locally compact topological vector space is finite-dimensional (the reader might see Tao’s article on his blog or Bourbaki’s book, Topological vector spaces). Therefore, the weak topologies defined on in the proof of theorem 1 are not locally compact.
Despite theorems 1 and 3, admits compact group topologies:
Theorem 4: has infinitely many compact group topologies.
Then is also a locally compact abelian group and
Theorem: (Pontryagin duality) Let be a locally compact abelian group. Then and are isomorphic.
Pontryagin duality theorem is a difficult result so we do not prove it here; a proof for discrete and compact abelian groups (the case we deal with here) can be found here. We deduce:
Lemma 4: Let be a discrete abelian group and be a proper subgroup. Then the annihilator is not trivial.
Proof. Because is a proper subgroup, is not trivial (otherwise, would be trivial). Let be a non-trivial character and let where is the canonical projection. Then .
Lemma 5: Let be a discrete abelian group. Then is divisible (resp. torsion free) if and only if is torsion free (resp. torsion free).
Proof. For every group , let and . First, we notice
By Pontryagin duality, we also have .
If is divisible, hence . So is torsion free.
If is torsion free ie. , then hence according to lemma 4. Therefore, is divisible.
To conclude, we deduce that is torsion-free if and only if is divisible by Pontryagin duality.
Proof of theorem 4. According to lemma 5, when is endowed with the discrete topology, is a divisible torsion-free abelian group, so can be viewed as a vector space over .
Notice that , because with defines a family of pairwise dijoint characters, so is a vector space over of dimension . Consequently and are isomorphic as abelian groups, so it is sufficient to show that is compact to deduce a compact group topology on .
Because is discrete, the compact-open topology on coincide with induced topology of (endowed with the product topology) on . Since is compact according to Tychonoff’s theorem, it is sufficient to show that is closed in .
Let that is is not a homomorphism, so there exist such that . Let
Then is an open neighborhood of in and does not meet . Hence is closed in .
In the same way, we may induce a compact group topology on using .
In fact, turns out to have exactly compact group topologies. For more information, see for example: