Is there a preferred topology making the general linear group of a Frechet space into a topological group?

Question

Is there a preferred topology on the isomorphisms of a Frechet space giving them the structure of a topological group? I was playing with the metric and strong operator topology on the general linear group of Schwartz space $\mathcal{S}(\mathbb{R}^n)$, but wasn't able to show that inversion $T \mapsto T^{-1}$ is continuous. In Folland's harmonic analysis book, I saw that in a Banach algebra $A$, using properties of the geometric series, you can show that $GL(A)$ is a topological group, but I haven't seen an analogous tool in the locally convex case.

Answer 1

I will only address the topologies of uniform convergence on some family of sets. Clearly, there is the strongest topology of this kind -- topology of uniform convergence on bounded sets, and the weakest -- strong operator topology. I claim that all topologies that are stronger than the strong and weaker than the topology of uniform convergence on bounded sets do not turn $GL(E)$ into a topological group for all Frechet spaces $E$. More precisely, below I construct a Frechet space $E$ and a sequence of invertible continuous operators $u_n:E \to E$ such that $u_n \to \mathrm{id}_E$ uniformly on bounded sets, but $u_n^{-1}$ does not converge to $\mathrm{id}_E$ in strong operator topology.

Let $F = \prod_{n \in \mathbb N} \mathbb C$ be the Frechet space of all sequences with direct product topology (i.e. the topology is induced by the family of seminorms $p_k(a) = \max\{|a_1|,\dots, |a_k|\}$, $k \in \mathbb N$). Moreover, let $G \subset F$ denote the subspace of rapidly decreasing sequences, i.e. $a \in G$ iff $\|a\|_k = \sup\{n^k|a_n|: n \in \mathbb N\} < +\infty$ for all $k \in \mathbb N$. The space $G$ with respect to the family of norms $\{\|\cdot\|_k\}_{k \in \mathbb N}$ is a Frechet space. Finally, let $E = G \times F$. Clearly, the topology on $E$ is induced by the family of seminorms $q_k(a,b) = \|a\|_k + p_k(b)$, $k \in \mathbb N$. For an element $(a,b) \in G \times F$ we let $u_n(a,b) = (a',b')$, where $$ a' = (a_1,a_2,\dots,a_{n-1},a_{n+1},a_{n+2},\dots),\;\;b' = (b_1,\dots,b_{n-1},a_n,b_n,b_{n+1},\dots). $$ That is, $u_n$ takes the n-th term of the sequence $a$ (that rapisly decreases) and inserts it into the n-th place of $b$.

I claim that $u_n \to \mathrm{id}_E$ uniformly on bounded subsets of $E$. Let $A \subset G$, $B \subset F$, and let $k \in \mathbb N$. It suffices to prove that $q_k(u_n(a,b) - (a,b)) \to 0$ when $n \to \infty$ uniformly with respect to $a \in A$ and $b \in B$. Fix any $\varepsilon > 0$ and let $(a',b')$ denote $u_n(a,b)$ for brevity. It is easy to see that if $n > k$, then $p_k(b - b') = 0$. Moreover, $\|a' - a\|_k = \sup \{m^k |a_{m+1} - a_m|: m = n, n+1, \dots\}$. Let $C = \sup \{\|a\|_{k+1}: a \in A\} < +\infty$ ($A$ is bounded in $G$). Choose $N$ large enough, so $2C/N < \varepsilon$. It follows that $\|a' - a\|_k < \varepsilon$ for $n \ge N$. That is, if $n \ge \max\{N, k\}$, then $q_k(u_n(a,b) - (a,b)) < \varepsilon$ for all $(a,b) \in A \times B$.

Now consider $u_n^{-1}$. Clearly, this operator takes n-th terms from $b$ and inserts it into n-th place of $a$. It is easy to see that in general the sequence $u_n^{-1}(a,b)$ does not converge to $(a,b)$ (for instance, if $b$ is not a bounded sequence, then $u_n^{-1}(a,b)$ also is not bounded in $E$).