The finest countably generated free topological group so that $x^{m_{n}}_n\rightarrow 1$?

Question

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 $F\{x_1,x_2,...x_n\}$}?

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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).

Answer 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.