Topology of a free group

Question

I wonder is there any general properties/ restrictions to the possible topologies of a free group (to make it a topological group ofc). More generally do such restrictions exist for any group written as presentation?

Answer 1

An interesting "restriction" is that if you want the topology to be locally compact Hausdorff or come from a complete metric then the topology will have to be the discrete topology. This follow from results in Continuity of homomorphisms by Dudley, where he proves "automatic continuity" results for certain types of groups.

A slight simplification is that if we have $G \to F$ a group homomorphism, where $F$ is free with the discrete topology and $G$ has a nice topology, like being locally compact Hausdorff, then the homomorphism is automatically continuous.

Using this we can suppose that $(F,\tau)$ is one of these nice topologies. Then $$Id:(F, \tau) \to F$$ is continuous by Dudley's automatic continuity results, which means that $\tau$ had to be the discrete topoology.

A consequence of this is that $F_\mathfrak{c}$, the free group on continuum many generators, can not be made into a Polish group. But $F_\mathfrak{c}$ obviously has quotients which can be made Polish, the any countable quotient or the group of real numbers.

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Note that $F_n$ can be given non-locally compact Hausdorff topolgies by embedding $F_n$ into nice groups like Lie groups and be dense in them(related to YCor's comment). These topolgies will "look like" $\mathbb Q$.

With that in mind there are finitely generated groups which have no Hausdorff group topology besides the discrete topology(hence quotients of free groups). The paper On topologizable and non-topologizable groups actually proves that there are lots satisfying even stronger properties.

Answer 2

Free groups are residually finite: for all $g\in F$ there exists a homomorphism $\phi_g: F\rightarrow H_g$ where $H_g$ is finite and $\phi_g(g)\neq1$. What the following says is that we can equip $F$ (and more generally, any residually finite group) with a natural topology, the "profinite topology", and construct a group, its "profinite completion" $\widehat{F}$, such that $F\hookrightarrow \widehat{F}$ is an monomorphism of topological groups (so the topology on $F$ is inherited from the topology on $\widehat{F}$).

The profinite topology of a group $G$ takes as a basis of open sets the finite index open subgroups of $G$. Then $G$ is residually finite if and only if its profinite topology is Hausdorff.

The profinite completion $\widehat{G}$ of a group $G$ is the inverse limit of the groups $G/N$, where $N$ runs through the normal subgroups in $G$ of finite index (these normal subgroups are partially ordered by inclusion, which translates into an inverse system of natural homomorphisms between the quotients: $$\{f_{i,j}: G/N_j \to G/N_i \mid i, j \in I, i \le j\}.$$ Formally, the profinite completion is as follows: $$ \begin{align*} \widehat{G}&:= \lim_{\leftarrow}{G/N_i}\\ &= \left\{ (gN_i)_{i \in I} \in \prod_{i \in I} G/N_i \:\:\middle|\:\: f_{i, j} (g_j) = g_i \text{ for all } j \ge i \right\}. \end{align*} $$ Bestow on $\widehat{G}$ a topology as follows: give each $G/N$ the discrete topology, the product the product topology, and the projective limit the restriction topology. Note that there is a natural homomorphism $\phi: G\rightarrow \widehat{G}$. Then $G$ is residually finite if and only if $\phi$ is an injection.

Suppose $G$ is residually finite (e.g. take $G$ to be free), and equip it with the profinite topology. Then $\phi$ is continuous. As it is an embedding, $G$ inherits the profinite topology from its profinite completion $\widehat{G}$.

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People care about profinite things. For example, Grothendieck asked, roughly, the following question: does $\widehat{G_1}\cong\widehat{G_2}$ imply $G_1\cong G_2$ (in the natural way)? The answer is no, as proven in Martin R. Bridson and Fritz J. Grunewald, Grothendieck’s problems concerning profinite completions and representations of groups, Ann. Math., 160 (2004), 359–373 (link).