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Let $\mathcal{H}$ be a Hilbert space and let $\mathcal{U}(\mathcal{H})$ be the group of unitary automorphisms, endowed with the strong operator topology. Now I want to show $\mathcal{U}(\mathcal{H})$ is a topological group. If $\mathcal{H}$ is separable, the strong operator topology is first countable and I can work with sequence $T_k,S_k\to T,S$ and show that $T_kS_k\to TS$. If $\mathcal{H}$ is not separable, is the strong operator topology still first countable?

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If $H$ is separable then SOT on $B(H)$ is first countable and so SOT on $U(H)$ is also first countable. Infact SOT on $U(H)$ is even metrisable! However, if $H$ is not separable, then neither $B(H)$ nor $U(H)$ are first countable.

In order to show this we note that in first countable spaces the sequential closure and topological closure agree. So to show that the spaces are not first countable we construct a set whose sequentially closure differs from its SOT-closure.

For $B(H)$ we let $P$ denote the space of finite rank orthogonal projections. Note that any sequence $p_{V_n}$ can converge (in SOT) at most to the projection onto $\bigcup_{n=0}^\infty V_n$, that is has image a separable Hilbert space. However the finite-dimensional subspaces of $H$ form a directed set, hence we may consider the net $p_V$ where $V$ ranges over the finite-dimensional subspaces. It is immediate that this net converges in SOT to the identity operator, hence the SOT-closure of $P$ is strictly bigger than the sequential closure of $P$.

For $U(H)$ use the same kind of idea. Choose some Hilbert-basis $\{e_i\}_{i\in J}$ and let $X$ denote the diagonal operators in this basis where all but finitely many diagonal elements are $+1$ and the others are $-1$. As before you can check that any sequential SOT-limit of such operators can have at most countably many $-1$'s on the diagonal, but you can construct a net (indexed by the finite subsets of $J$) converging to the operator $-1$.

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  • $\begingroup$ Thank you, this is very clear, I see why. If $\mathcal{H}$ happens to be inseparable, is it still true that multiplication in $\mathcal{U}(\mathcal{H})$ is continuous? This was easy to show in the case of first countability, but how can I show this for general $\mathcal{H}$? $\endgroup$ – user672749 Sep 30 '19 at 18:59
  • $\begingroup$ Note that if $H$ is infinite dimensional then multiplication $B(H)\times B(H)\to B(H)$ is _not _ a SOT continuous map. However, if $X\subseteq B(H)$ is a bounded subset (for example $X=U(H)$), then multiplication is continuous on $X\times B(H)\to B(H)$ regardless of the size of $H$. This follows from $\|(AB-A_\alpha B_\alpha) x \|≤ \|A_\alpha\|\,\|(B-B_\alpha)x\| + \|(A-A_\alpha)Bx\|$ and the right-hand side going to $0$ if $A_\alpha\to A$ and $B_\alpha\to B$ in SOT (remember $\|A_\alpha\|$ is bounded). It then follows that multiplication is continuous as a map on $U(H)\times U(H)\to U(H)$. $\endgroup$ – s.harp Sep 30 '19 at 21:06

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