Is the unitary group $U(H)$ with the strong operator topology locally compact? Suppose that $H$ is a complex Hilbert space. Endow the unitary group $U(H)$ with the strong operator topology (SOT) - that is, $u_{i}\to u$ in $U(H)$ if and only if $u_{i}x\to ux$ in $H$ for all $x\in H$. One can show that $U(H)$ is a topological group.
Is $U(H)$ locally compact?
My intuition says that this is not true for general $H$. However, this is true for $H=\mathbb{C}^{n}$. In fact, if $H$ is finite dimensional, then $U(H)$ is even compact.
 A: Assume that $U(H)$ is locally compact. Then the identity operator admits a compact neighborhood. This compact neighborhood must contain a SOT-basic neighborhood, which is of the form $$\Omega:=\{V\in U(H): \|Vx_i-x_i\|<\varepsilon\text{ for all }i=1,\dots,n\} $$
for some $\varepsilon>0$ and $x_1,\dots,x_n\in H$. We will construct a sequence of unitaries in this set that does not have any convergent subnet and this will be a contradiction.
Take the subspace $K=\text{span}\{x_1,\dots,x_n\}$ and find a (finite) orthonormal basis for this subspace, say $E_0=\{e_1,\dots,e_l\}$. We can then extend $E_0$ to an ONB of $H$, say $E$, which is a countable set, $E=\{e_1,\dots,e_l,e_{l+1},e_{l+2},\dots\}$.
Let $\sigma:\{l+1,l+2,\dots,\}\to\{l+1,l+2,\dots\}$ be any bijection. We define $V_\sigma:H\to H$ by $V_\sigma e_j=e_j$ for all $j=1,\dots,l$ and $V_\sigma e_j=e_{\sigma(j)}$ for all $j>l$. The operators $V_\sigma$ are all unitaries since they map an ONB onto an ONB in a bijective way and actually $V_\sigma\in\Omega$.
We now take a sequence of bijections: $\sigma_1$ is defined by permuting $l+1$ and $l+2$ and leaving everything else fixed. $\sigma_2$ is defined by permuting $l+1$ and $l+3$ and leaving everything else fixed. $\sigma_3$ is defined by permuting $l+1$ and $l+4$ and leaving everything else fixed. In general, $\sigma_n$ is defined by permuting $l+1$ and $l+n+1$ and leaving everything else fixed. Set $V_n:=V_{\sigma_n}$. We will show that $\{V_n\}$ does not have any convergent subnet (in the SOT topology).
Let $\Lambda$ be a directed set and $\{V_{k_\lambda}\}$ be a subnet of $\{V_n\}$. If the net $\{V_{k_\lambda}\}_{\lambda\in\Lambda}$ converges in the SOT topology, then $\{V_{k_\lambda}(e_{l+1})\}_{\lambda\in\Lambda}$ is a convergent net in $H$ and thus is a Cauchy net. So, for $\varepsilon=1$, there must exist some $\lambda_0\in\Lambda$, such that for all $\lambda_1,\lambda_2\geq\lambda_0$ we have $$\|V_{k_{\lambda_1}}(e_{l+1})-V_{k_{\lambda_2}}(e_{l+1})\|<1. \;\;\;\;\;(\star)$$
But $V_{k_\lambda}(e_{l+1})=e_{\sigma_{k_\lambda}(l+1)}=e_{l+k_\lambda+1}$. Now
$$\|V_{k_\lambda}(e_{l+1})-V_{k_{\lambda'}}(e_{l+1})\|^2=\|e_{l+k_\lambda+1}-e_{l+k_{\lambda'}+1}\|^2=2$$
for all $k_\lambda\neq k_{\lambda'}$.
Now since we have a subnet, the map $\Lambda\to\mathbb{N}$, $\lambda\mapsto k_\lambda$ is final, i.e. it is monotone and given any integer $n_0$ we can find $\lambda\in\Lambda$ such that $k_\lambda\geq n_0$.
For the integer $n_0:=k_{\lambda_0}+1$, we find $\lambda_1\in\Lambda$ such that $k_{\lambda_1}>k_{\lambda_0}+1$. Since $\Lambda$ is directed, find $\lambda_2\in\Lambda$ such that $\lambda_2\geq\lambda_0$ and $\lambda_2\geq\lambda_1$, so $k_{\lambda_2}\geq k_{\lambda_1}>k_{\lambda_0}$. In particular, $k_{\lambda_2}\neq k_{\lambda_0}$, so
$$\|V_{k_{\lambda_2}}(e_{l+1})-V_{k_{\lambda_0}}(e_{l+1})\|^2=2$$
which is a contradiction by $(\star)$
Edit: As pointed out in the comment section, the ONB need not be countable if $H$ is non-separable. In that case, one should adjust the bijections to fix everything in the basis and simply permute one specific vector of $E$ minus the finite part $E_0$ with a different vector of the basis every time. The proof then works verbatim.
A: Here's a short argument.
(0) For a finite subset $F\subset H$ and $\varepsilon>0$ let $V(F,\varepsilon)$ be the basis of closed neighborhoods of identity (as in the previous answer) $\{s:\sup_{v\in F}\|sv-v\|\le\varepsilon\}$.
(1) the topological group $U(H)$ is not compact if $H$ is infinite-dimensional. Indeed, let $(e_n)_{n\in\mathbf{Z}}$ be orthonormal and complete it to an orthonormal basis. Let $T$ be the shift $e_n\mapsto e_{n+1}$, fixing other basis elements. Then the subgroup $\langle T\rangle$ has trivial intersection with $V(\{e_0\},1)$. So it is discrete infinite, and hence $U(H)$ is not compact.
(2) If by contradiction $U(H)$ is locally compact, there exists $F$ and $\varepsilon>0$ such that $V(F,\varepsilon)$ is compact. But then its closed subset $V(F,0)$ is compact. But the latter is topologically isomorphic to $U(H')$ where $H'$ is the orthogonal of $F$, hence is also infinite-dimensional. This contradicts (1).
