What is the finest free topological group $H$ with generators ${x\_{1},x\_{2},...}$ so that $x^{m\_{n}}\_n\rightarrow 1$ for all sequences $m\_{1},m\_{2},...$?


Is $H \simeq K$, with $K$ the natural subspace (of eventually constant sequences) of the inverse limit of finitely generated discrete free groups $F\{x\_1,x\_2,...x\_n\}$}?




---


To make the question clear, let $G$ be the free group with generators ${x\_{1},x\_{2},...}$.


Let $L$ denote the partially ordered set of all topologies $\tau$ so that $(G,\tau)$ is a topological group and so that $\tau\_1 \leq \tau\_2 \iff \tau\_1 \subset \tau\_2$. Employing unions of topologies, it is straightforward to show $lub(A)$ exists for each $A \subset L$, and thus, (since $K$ satisfies all the conditions of the question except possibly being finest), $H$ exists.


The bonding maps $F\{x\_1,x\_2,...x\_n\} \rightarrow F\{x\_1,x\_2,...x\_{n-1}\}$ delete $x\_n$ from each word, and note $K \leq H$.


To see why the question is nontrivial, $H \neq K$ in the related category SeqGrp (in which all spaces are sequential, and group multiplication is merely required to be jointly continuous over convergent sequences).



The finest countably generated free topological group so that $x^{m_{n}}_n\rightarrow 1$?
I think I can show that $H \neq K$.


We will think about functions that can be constructed using the group operations and constant elements of the group, like $f(y\_1,y\_2,y\_3) = x\_4^2 y\_1 y\_2 x\_3^{-1} y\_3$.


Consider the following topology: A set $U$ is open if, for each $k$-variable function $f$ such that $f(1,\dots, 1) \in U$, there is a constant $N$ such that for all $(n\_1,\dots, n\_k)$ a tuple of natural numbers $\geq N$ and $(m\_1,\dots, m\_k) \in \mathbb Z$, $f( x\_{n\_1}^{m\_1},\dots, x\_{n\_k}^{m\_k} )\in U$. It is easy to see that this is a topology and that inversion and translation are continuous.


Next we check that composition is continuous. Let $U$ be an open neighborhood of the identity. We need to check that the inverse image of $U$ under the composition map contains the product of two open neighborhods of the identity. Choose a countable enumeration of all functions, where say $f\_i$ is the $i$th function. This gives a total ordering. For $U\_1$ to each function $f\_i$ such that $f\_i(1,\dots, 1) =1$ we assign the constant $N$ which is the min over all $f\_j$ such that $j\leq i$ and $f\_i(1,\dots,1)f\_j(1,\dots,1)=1$ of the constant assigned to $f\_if\_j$ by $U$. Then let $U\_1$ be the union of the image of every function with $x\_{n\_i}^{m\_i}$ plugged in for $n\_i \geq N(f)$. Similarly, let $U\_2$ be the same thing but with the composition in the opposite order. Then $U\_1$ and $U\_2$ are open, and $U\_1U\_2 \subset U$.


So this is a topological group. We can verify that $x\_n^{m\_n} \to 1$ using the definition of the open set $U$, and its topology is finer than $K$ - construct an open set where $N$ goes to $\infty$ very rapidly depending on the function. 


Indeed, this is $H$:


Let $U$ be an open neighborhood of the identity of $H$. For each sequence $m\_n$, there is an $N$ large enough such that $x\_n^{m\_n} \in U$ for $n \geq N$. So there exists a universal constant $N$ such that $x\_n^{m} \in U$ for all $n \geq N$ and all $m$. (If not, then build a sequence $m\_n$ from the counterexamples to the sequence of claims, for each $N$, that $N$ is a universal constant.)


Then for each open set and each function such that $f(1,\dots ,1 )$ is in the open set, choose a product of neighborhoods of the identity in the inverse image of the open set under the function, and take the max of the $N$ for each of them. So each open set of $H$ is open in our topology.


