Metrizability of space of unitary operators on Hilbert space Let $\mathbb{H}$ be a (complex) Hilbert space. Define $\mathbb{P}$ as a projective space over $\mathbb{H}$, i.e. $\mathbb{P}=\left(\mathbb{H}\setminus\{0\}\right)/\sim$, where $f\sim g$ iff there exists $\lambda \in \mathbb{C}^{\times}$ s.th. $f=\lambda g$.
Define $U(\mathbb{H})$ the space of unitary transformations, i.e.
$U(\mathbb{H}):=\{U:\mathbb{H}\rightarrow \mathbb{H}, \ \mathbb{C}\mathrm{-linear, bijective}, \ \langle Uf,Ug\rangle=\langle f,g\rangle \ \forall f,g\in\mathbb{H}\}$ 
Similary we define $U(\mathbb{P})=\hat\gamma(U(\mathbb{H}))$, where $\hat\gamma$ is induced by $\gamma$, i.e. $\hat\gamma(U)(\varphi):=\gamma(U(f))$, for $U\in U(\mathbb{H}), \ \varphi=\gamma(f)\in\mathbb{P}$.
By strong topology I mean topology generated by subbasis
$\{U\in U(\mathbb{H}) \ | \ \|U_0(f)-U(f)\|<r \}$, for $U_0\in U(\mathbb{H}),\ r>0, \ f\in \mathbb{H}$.
If $\mathbb{H}$ is separable one can show that $U(\mathbb{P})$ and $U(\mathbb{H})$ are both metrizable (connected) topological groups. 
My question : I'm looking for an examples for such nonseparable Hilbert spaces such that: 
a) $U(\mathbb{H})$ and $U(\mathbb{P})$ are non-metrizable topological groups
b) $U(\mathbb{H})$ and $U(\mathbb{P})$ are non-metrizable and at least one of them is not a topological group
c) one of $U(\mathbb{H}), \ U(\mathbb{H})$ is a metrizable topological group but the other one is not.
 A: My answer is mainly on $U(\Bbb H)$. It should be a topological group. More precisely: 


*

*This is claimed in the first paragraph of [Usp], where (probably) is not specified is the space $\Bbb H$ real or complex.

*Prop. 1 of [Schot] (with a very inaccurate proof) claims this for a complex space $\Bbb H$ and a paragraph after it contains comments of references. The Remark on p.4 probably claims that $U(\Bbb P)$ is also a topological group.
I have some heuristic suggesting that is if $\Bbb H$ is not separable then the group $U(\Bbb H)$ is non-metrizable. Indeed, assume 
that the group $U(\Bbb H)$ has a countable base $\mathcal B$ of the unit $e$. Therefore there exists a countable set $C\in\Bbb H$ such that for each neighborhood $N\in\mathcal B$ there exist a finite subset $F$ of $C$ and a number $r>0$ such that the intersection of the subbase elements of $U_0=e$ defined by $f\in F$ and $r$ is contained in $N$. Since the space $\Bbb H$ is not separable there should exist a vector $f_0\in\Bbb H$ which is not contained in the closed linear subspace of $\Bbb H$ generated by $C$. Then the subbase element of $U_0=e$ defined by $f_0$ and $r=1$ should not contain any element of the base $\mathcal B$.
References
[Schot] Martin Schottenloher, The Unitary Group In Its Strong Topology.
[Usp] V. V. Uspenskij, On unitary representations of groups of isometries.
