# First countability strong operator topology.

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?

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.
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 $$-1$$.
• 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 $\mathcal{H}$? – user672749 Sep 30 '19 at 18:59
• 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 $U(H)\times U(H)\to U(H)$. – s.harp Sep 30 '19 at 21:06