# Metrizability of the strong operator topology on the set of unitary operators of a separable Hilbert space

Let $H$ be a Hilbert space, and $\mathscr{U}(H)$ be the set of unitary operators on $H$, that is $T\in\mathscr{U}(H)\iff TT^*=T^*T=I$. Let $\{x_n\}$ be a countable dense subset of the unit ball, and define $$d(S,T):=\sum_n 2^{-n} \left( \|Tx_n-Sx_n\|+\|T^*x_n-S^*x_n\|\right)$$

I would like to show that the topology generated by this metric is the SOT topology. I've already shown that every set of the form

$$\{T\in\mathscr{U}(H):\|Tx-T_0x\|<\varepsilon\},$$ with $T_0\in\mathscr{U}(H)$, $x\in H$, and $\varepsilon>0$, is open with respect to $d$.

Now, given $\varepsilon>0$ and $T_0\in H$, I'd like to show that the ball $$B=\{ T\in \mathscr{U}(H):d(T,T_0)<\varepsilon\}$$ is SOT-open. Fix $S_0\in B$. Given $\delta>0$, we can choose $N$ such that $\sum_{n> N} 2^{-n}<\delta$, which allows us to control the tail of the series. Then we can consider the SOT-open set $$U=\bigcap_{n=1}^N \{T\in \mathscr{U}(H):\|Tx_n-S_0x_n\|<\delta\}\ni S_0.$$ which gives a bound for the summands $\|Tx_n-Sx_n\|$, but I'm at a loss as to how to bound the terms with the adjoints, that is, I need a bound for $\|T^*x_n-S^*x_n\|$, $n\le N$.

Notice that $$\|(T-S_0)^*x_n\|^2= \|T^*x_n\|^2 + \|S_0^*x_n\|^2 -2 \text{Re} \langle T^*x_n, S_0^*x_n \rangle \\ = 2\langle x_n,S_0S_0^*x_n\rangle -2\text{Re}\langle x_n, TS_0^*x_n\rangle = 2\text{Re}\langle x_n, (S_0-T)S_0^*x_n \rangle \\ \leq 2\|x_n\| \|(S_0-T)S_0^*x_n\|.$$
So if you slightly modify $U$ to the following SOT open set $$U = \bigcap_{n=1}^N\{T\in \mathscr{U}(H):\|Tx_n-S_0x_n\|<\delta\} \cap \{T\in \mathscr{U}(H):\|(T-S_0)S_0^* x_n\|<\frac{\delta^2}{2}\},$$ your proof should work.
• Thank you for your answer. Should the $S$ in the first real part be a $S_0$? Other than that, it seems great, didn't think to use that pseudo polarization identity. – Reveillark Nov 11 '17 at 19:34