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 $(\mathbb{R},+)$.

Theorem 1: There exist exactly $2^{2^{\mathfrak{c}}}$ distinct group topologies on $(\mathbb{R},+)$.

Definition: Let $X$ be a normed vector space (over $\mathbb{R}$) and $\Phi$ be a linear subspace of the topological dual $X^*$ (ie. the space of continuous functionals). The weak topology $\sigma(X,\Phi)$ on $X$ is the initial topology associated to $\Phi$.

Lemma 1: A weak topology is a group topology.

Proof. Let $X$ be a normed vector space and $\sigma(X,\Phi)$ be a weak topology. To show that $\varphi : (x,y) \mapsto x+y$ is continuous, it is sufficient to notice that for all $\phi_1, \dots, \phi_n \in \Phi$ and $\epsilon>0$ the following set is open for $\sigma(X,\Phi)$:

$\displaystyle \varphi^{-1} \left( \bigcap\limits_{i=1}^n \phi_i^{-1} ((x_i- \epsilon_i,x_i+ \epsilon_i)) \right) \\ = \bigcap\limits_{i=1}^n \{(x,y) \in X \times X \mid x+y \in \phi_i^{-1}((x_i-\epsilon_i,x_i+ \epsilon_i)) \} \\ = \bigcap\limits_{i=1}^n \{(x,y) \in X \times X \mid x_i- \epsilon_i < \phi_i(x)+ \phi_i(y) < x_i+ \epsilon_i \} \\ = \bigcap\limits_{i=1}^n \bigcup\limits_{\delta \in (0,\epsilon_i)} \{ (x,y) \in X \times X \mid x_i- \delta < \phi_i(x) < x_i+ \delta, -\epsilon_i + \delta < \phi_i(y) < \epsilon_i - \delta \} \\ = \bigcap\limits_{i=1}^n \bigcup\limits_{\delta \in (0,\epsilon_i)} \phi_i^{-1} ((x_i- \delta,x_i+ \delta)) \times \phi_i^{-1} ((-\epsilon_i+ \delta,\epsilon_i- \delta))$

Then, $x \mapsto -x$ is clearly continuous, since $r-\epsilon < \phi(x) < r+ \epsilon$ is equivalent to $-r- \epsilon < \phi(-x) < -r + \epsilon$ for all $\phi \in \Phi$, $x \in X$, $r \in \mathbb{R}$ and $\epsilon >0$. $\square$

Lemma 2: Let $X$ be a normed vector space and $\sigma(X,\Phi)$ be a weak topology. With respect to $\sigma(X,\Phi)$, a functional $\varphi : X \to \mathbb{R}$ is continuous if and only if $\varphi \in \Phi$.

Proof. Let $\phi$ be a linear functional continuous for $\sigma(X,\Phi)$. Then there exist linear functionals $\phi_1, \dots, \phi_n \in \Phi$ and open neighborhoods of the origin $V_1, \dots, V_n \subset \mathbb{R}$ so that

$\displaystyle \phi^{-1}((-1,1)) \supset \bigcap\limits_{i=1}^n \phi_i^{-1}(V_i) \supset \bigcap\limits_{i=1}^n \mathrm{ker}(\phi_i)$.

In particular, $\phi$ is bounded on $\bigcap\limits_{i=1}^n \mathrm{ker}(\phi_i)$ hence $\bigcap\limits_{i=1}^n \mathrm{ker}(\phi_i) \subset \mathrm{ker}(\phi)$. Let

$f : \left\{ \begin{array}{ccc} X & \to & \mathbb{R}^n \\ x & \mapsto & (\phi_1(x), \dots,\phi_n(x)) \end{array} \right.$.

Because $\bigcap\limits_{i=1}^n \mathrm{ker}(\phi_i) \subset \mathrm{ker}(\phi)$, we may define $\varphi : f(X) \to \mathbb{R}$ by $\varphi(x)= \phi(y)$ when $f(y)=x$

Let $\tilde{\varphi} : \mathbb{R}^n \to \mathbb{R}$ be a linear extension of $\varphi$. In particular, $\phi = \tilde{\varphi} \circ f$.

Let $\lambda_1, \dots, \lambda_n \in \mathbb{R}$ such that $\tilde{\varphi} : x \mapsto \sum\limits_{i=1}^n \lambda_i x_i$. Then for all $x \in X$,

$\phi(x)= \tilde{\varphi} \circ f(x)= \tilde{\varphi} (\phi_1(x) ,\dots \phi_n(x)) = \sum\limits_{i=1}^n \lambda_i \phi_i(x)$,

hence $\phi = \sum\limits_{i=1}^n \lambda_i \phi_i \in \Phi$. $\square$

Lemma 3: (Erdös-Kaplansky theorem) Let $V$ be an infinite-dimensional vector space. If $V^*$ denotes its dual space, $\dim(V^*)= 2^{\dim(V)}$.

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 $c_0$ be the set of eventually zero sequences endowed with the sup norm and let $B$ be a basis of $c_0^*$. For all $S \subset B$, let $\Phi_S= \mathrm{span}(S) \subset c_0^*$. According to lemma 2, for all $R,S \subset B$

$\sigma(X, \Phi_S)= \sigma(X,\Phi_R) \Leftrightarrow \Phi_S = \Phi_R \Leftrightarrow R=S$.

Therefore, according to lemmas 1 and 3, $\{ \sigma(X,\Phi_S) \mid S \subset B\}$ is a family of $2^{2^{\mathfrak{c}}}$ distinct group topologies on $(c_0,+)$.

Because $c_0$ and $\mathbb{R}$ are both a linear vector space over $\mathbb{Q}$ of dimension $\mathfrak{c}$, there exists an isomorphism $\phi$ between the abelian groups $(c_0,+)$ and $(\mathbb{R},+)$. Then we can use $\phi$ to transfer the $2^{2^{\mathfrak{c}}}$ distinct group topologies from $(c_0,+)$ to $(\mathbb{R},+)$.

So $(\mathbb{R},+)$ has at least $2^{2^{\mathfrak{c}}}$ distinct group topologies.  On the other hand, $(\mathbb{R},+)$ has at most $|\mathfrak{P}(\mathfrak{P}(\mathbb{R}))|=2^{2^{\mathfrak{c}}}$ distinct group topologies. $\square$

Then we may deduce the following corollaries:

Theorem 2: $(\mathbb{R},+)$ has exactly $2^{2^{\mathfrak{c}}}$ non-isomorphic group topologies.

Proof. $(\mathbb{R},+)$ has at most $| \mathbb{R}^{\mathbb{R}}|= 2^{\mathfrak{c}}$ automorphisms and has exactly $2^{2^{\mathfrak{c}}}$ group topologies. $\square$

Theorem 3: $(\mathbb{R},+)$ has exactly $2^{2^{\mathfrak{c}}}$ 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 $c_0$ in the proof of theorem 1 are not locally compact. $\square$

Despite theorems 1 and 3, $(\mathbb{R},+)$ admits compact group topologies:

Theorem 4: $(\mathbb{R},+)$ has infinitely many compact group topologies.

Definition: Let $G$ be a locally compact abelian group. The character group $\hat{G}$ of $G$ is the set of continuous morphisms from $G$ to the circle group $\mathbb{T}$ endowed with the compact-open topology.

Then $\hat{G}$ is also a locally compact abelian group and

Theorem: (Pontryagin duality) Let $G$ be a locally compact abelian group. Then $G$ and $\widehat{\widehat{G}}$ 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 $G$ be a discrete abelian group and $H \subset G$ be a proper subgroup. Then the annihilator $A(H)= \{ \chi \in \hat{G} \mid \forall h \in H , \chi(h)=1 \}$ is not trivial.

Proof. Because $H$ is a proper subgroup, $\widehat{G/H}$ is not trivial (otherwise, $G/H \simeq \widehat{\widehat{G/H}}$ would be trivial). Let $\hat{\chi} \in \widehat{G/H}$ be a non-trivial character and let $\chi = \hat{\chi} \circ \pi$ where $\pi : G \to G/H$ is the canonical projection. Then $\chi \in A(H) \backslash \{1\}$. $\square$

Lemma 5: Let $G$ be a discrete abelian group. Then $G$ is divisible (resp. torsion free) if and only if $\hat{G}$ is torsion free (resp. torsion free).

Proof. For every group $H$, let $H_n'= \{ x \in H \mid nx=0 \}$ and $H_n''= \{x \in H \mid \exists H, x=ny\}$. First, we notice

$\begin{array}{cl} A(G_n'') & = \{ \chi \in \hat{G}\mid \forall y\in G, \chi(ny)=1 \} \\ & = \{ \chi \in \hat{G} \mid \forall y \in G, \chi^n(y)=1 \} \\ & = \{ \chi \in \hat{G} \mid \chi^n=1 \} = \hat{G}_n' \end{array}$

By Pontryagin duality, we also have $G_n' \simeq \hat{\hat{G}}_n'= A(\hat{G}_n'')$.

If $G$ is divisible, $G_n''=G$ hence $\{1\}=A(G)= A(G_n'')= \hat{G}_n'$. So $\hat{G}$ is torsion free.

If $G$ is torsion free ie. $G_n'=\{0\}$, then $A(\hat{G}_n'')= \{1\}$ hence $\hat{G}_n''=\hat{G}$ according to lemma 4. Therefore, $\hat{G}$ is divisible.

To conclude, we deduce that $G$ is torsion-free if and only if $\hat{G}$ is divisible by Pontryagin duality. $\square$

Proof of theorem 4. According to lemma 5, when $(\mathbb{Q},+)$ is endowed with the discrete topology, $\hat{\mathbb{Q}}$ is a divisible torsion-free abelian group, so $\hat{\mathbb{Q}}$ can be viewed as a vector space over $\mathbb{Q}$.

Notice that $| \hat{\mathbb{Q}}|= \mathfrak{c}$, because $\varphi_r : q \mapsto e^{irq}$ with $r \in (0,2\pi)$ defines a family of pairwise dijoint characters, so $\hat{\mathbb{Q}}$ is a vector space over $\mathbb{Q}$ of dimension $\mathfrak{c}$. Consequently $\hat{\mathbb{Q}}$ and $\mathbb{R}$ are isomorphic as abelian groups, so it is sufficient to show that $\hat{\mathbb{Q}}$ is compact to deduce a compact group topology on $\mathbb{R}$.

Because $\mathbb{Q}$ is discrete, the compact-open topology on $\hat{\mathbb{Q}}$ coincide with induced topology of $\mathbb{T}^{\mathbb{Q}}$ (endowed with the product topology) on $\hat{\mathbb{Q}}$. Since $\mathbb{T}^{\mathbb{Q}}$ is compact according to Tychonoff’s theorem, it is sufficient to show that $\hat{\mathbb{Q}}$ is closed in $\mathbb{T}^{\mathbb{Q}}$.

Let $\varphi \in \mathbb{T}^{\mathbb{Q}} \backslash \hat{\mathbb{Q}}$ that is $\varphi$ is not a homomorphism, so there exist $x,y \in \mathbb{Q}$ such that $\varphi(x-y) \neq \varphi(x) \cdot \varphi(y)^{-1}$. Let

$\Omega= \prod\limits_{q \in \mathbb{Q}} X_q$ where $X_q = \left\{ \begin{array}{cl} \{\varphi(z) \} & \text{if} \ q =z \in \{x,y,x-y\} \\ \mathbb{T} & \text{otherwise} \end{array} \right.$

Then $\Omega$ is an open neighborhood of $\varphi$ in $\mathbb{T}^{\mathbb{Q}}$ and does not meet $\hat{\mathbb{Q}}$. Hence $\hat{\mathbb{Q}}$ is closed in $\mathbb{T}^{\mathbb{Q}}$.

In the same way, we may induce a compact group topology on $\mathbb{R}$ using $\widehat{\mathbb{Q}^n}$. $\square$

In fact, $( \mathbb{R},+)$ turns out to have exactly $\aleph_0$ compact group topologies. For more information, see for example: