Questions on the group $\mathrm{GL}(H)$

Question

$\DeclareMathOperator\GL{GL}\DeclareMathOperator\U{U}$Let $H$ be an infinite dimensional complex Hilbert space. Consider the group $\GL(H)$ of bounded invertible operators on $H$.

Question 1. I've heard that $\GL(H)$ is not a topological group regardless we take (the subspace topology of) the strong-$*$, strong, weak, ultra-strong-$*$, ultra-strong or ultra-weak topologies. It quickly becomes a tedious exercise to see how the continuity of multiplication or inverse fails in each situation, can anyone point out a quick proof or some references of these facts?

Question 2. If $G$ is a subgroup of $\GL(H)$ that is bounded in norm, is it then a topological group with respect to any of the above topologies? Note that if $G$ is conjugate to a subgroup of the group $\U(H)$ of unitary operators, the answer is affirmative with respect to all of the above mentioned topologies, since they all coincide on $\U(H)$, and multiplication on either side by a fixed $a \in \mathcal{B}(H)$ is always continuous with respect to all of these topologies.

Question 3. Is it true that the subgroup $G$ in question 2 is always a conjugate of a subgroup of $\U(H)$?

I know questions 2 & 3 are related to Kadison's famous problem of whether every strongly continuous uniformly bounded representations of a locally compact group is equivalent to a unitary one, to which various counter-examples have been found. Though I am not familiar with these works, they still make me suspect that the answer to questions 2 and 3 are both likely to be negative, and it seems to me that Q3 might be easier.

Answer 1

I think it is possible to give a "nearly unified" approach to Q1 and Q2 (Q3 being answered in the comments), and also to show why a completely unified approach is not possible. I cannot answer one part of Q2 however.

General comments: As observed in the OP, translation (on left or right) by a fixed operator is always continuous for any of the topologies. So it suffices to consider continuity at the identity.

Question 1

We construct nets $(T_i), (S_i)$ which converge $\sigma$-strong-$^\ast$ to $1$ yet $(T_iS_i)$ does not converge strongly to $1$. This deals with all the "strong" topologies.

Let $X$ be the collection of all families $(\xi_n)$ with $\sum_n \|\xi_n\|^2<\infty$, let $X_{<\infty}$ be the finite subsets of $X$, and let $I=X_{<\infty}\times\mathbb N$ with the obvious ordering. Fix a unit vector $\xi_0$. For $i=(F,m)\in I$ choose a unit vector $\eta_i$ with $(\eta_i|\xi_0)=0$ and with $\sum_n |(\eta_i|\xi_n)|^2 < m^{-3}$ for each $(\xi_n)\in F$. Define $T_i:\xi\mapsto m (\eta_i|\xi) \eta_i$ and $S_i:\xi\mapsto m^{-1} (\xi_0|\xi) \eta_i$, so $\|S_i\|=m^{-1}$ and hence $1+S_i\rightarrow 1$ in norm. As $S_i$ has small norm, $1+S_i$ is invertible, and as the spectrum of $T_i$ is $\{0,m\}$, $1+T_i$ is invertible. For any $(\xi_n)\in X$, $$ \sum_n \|T_i(\xi_n)\|^2 = \sum_n m^2 |(\eta_i|\xi_n)|^2 < m^{-1}, $$ so $1+T_i\rightarrow 1$ $\sigma$-strong$^\ast$ (as $T_i$ is self-adjoint). However, as $$ T_iS_i(\xi) = m^{-1} (\xi_0|\xi) T_i(\eta_i) = (\xi_0|\xi)\eta_i, $$ we see that the net $(T_iS_i(\xi_0))$ does not converge in norm, and so $(1+T_i)(1+S_i)$ does not converge strongly to $1$.

We cannot extend this to the "weak" topologies, because:

Theorem: Let $(T_i), (S_i)$ be nets converging strong$^*$ to $1$. Then $(T_iS_i)$ converges $\sigma$-weakly to $1$.

Proof: It suffices to show weak convergence, the $\sigma$-weak case following by replacing $H$ by $H\otimes\ell^2$. Thus, let $\xi,\eta\in H$ can consider that $T_i^*\xi\rightarrow\xi$ and $S_i\eta\rightarrow\eta$ in norm, because of strong$^*$ convergence. Hence $$ (\xi|T_iS_i\eta) = (T_i^*\xi|S_i\eta) \rightarrow (\xi|\eta), $$ as required.

Thus we seek a new counter-example for the "weak" topologies. Let $(e_n)_{n\geq 0}$ be an orthonormal sequence in $H$, and consider the operators $T_n:\xi\mapsto (e_n|\xi) e_0$ for $n\geq 1$. Again, the compact operator $T_n$ has spectrum $\{0\}$ and so $1+T_n$ is invertible. By Bessel's Inequality $T_n\rightarrow 0$ strongly, and $T_n^*\rightarrow 0$ weakly, hence $\sigma$-weakly, as $(T_n)$ is bounded. So $1+T_n, 1+T_n^*\rightarrow 1$ $\sigma$-weakly. However, $$ T_nT_n^*(\xi) = (e_0|\xi) T_n(e_n) = (e_0|\xi) e_0 $$ so $(1+T_n)(1+T_n^*) \rightarrow 1 + p_0$ weakly, where $p_0$ is the projection onto $\mathbb Ce_0$.

(I do not know a reference. However, looking in e.g. Dixmier's book or Stratila-Zsido, there are exercises which motivate the 2nd counter-example. Once you have this, the 1st counter-example is not so hard to think of.)

Question 2

Let $G$ be a bounded subgroup of $GL(H)$, say $K=\sup\{ \|T\| :T\in G \}$.

This has an affirmative answer for the strong topologies. Let $(S_i), (T_i)$ be nets in $G$ converging strongly to $1$. For $\xi\in H$, the estimate $$ \| T_i S_i\xi - \xi\| \leq \|T_iS_i\xi - T_i\xi\| + \|T_i\xi-\xi\| \leq K \|S_i\xi - \xi\|+ \|T_i\xi-\xi\| $$ shows that $T_iS_i\rightarrow 1$ strongly. The estimate $$ \|T_i^{-1}\xi-\xi\| = \|T_i^{-1} (\xi-T_i\xi)\| \leq K \|T_i\xi-\xi\| $$ shows that $T_i^{-1}\rightarrow 1$ strongly.

As $G^*:=\{T^*:T\in G \}$ is also a bounded subgroup of $GL(H)$, the same arguments applied to $G^*$ and $G$ at the same time show the results for the strong$^*$-topology.

By replacing $H$ with $H\otimes\ell^2$ and letting $T\in G$ act as $T\otimes 1$, shows the result for the $\sigma$-strong and $\sigma$-strong$^*$-topologies.

Above, we found bounded sequences $(S_n), (T_n)$ which converge $\sigma$-weakly to $1$, but with $S_nT_n\not\rightarrow 1$ weakly. However, a little thought will show that the subgroup these operators generate is not bounded. As I do not have a good source of examples of bounded (but not unitarizable) subgroups of $GL(H)$, I leave this open question:

If $G\subseteq GL(H)$ is a bounded subgroup, must it necessarily be a topological group for the weak or $\sigma$-weak topologies?