
> 
> Assume $\Gamma$ acts by isometries on a separable Hilbert space $H$, and 
>  $$\operatorname{diam} H/\Gamma\le 1.$$ 
>  Is it true that $H/\Gamma$ is compact?
> 
> 
> 




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**Stupid example.** Assume the action of $\Gamma$ on $H=\ell\_2$ is generated by coordinate translations $x\_n\mapsto x\_n+\epsilon\_n$. Then 
$$\operatorname{diam} H/\Gamma=\tfrac12\cdot\sqrt{\sum\_{n=1}^\infty\epsilon\_n^2}.$$ 
Thus, if $\operatorname{diam} H/\Gamma\le 1$ then $H/\Gamma$ is a quotient of Hilbert cube, and has to be compact.



Cobounded ⇒ cocompact?
The answer is "NO". To show this let us use the following:



> 
> **Lemma.** Let $L$ be a lattice in $\mathbb R^q$ ($q$ is any positive integer). 
>  Assume $$\operatorname{diam} \mathbb R^q/L>1000.$$ 
>  Then there is a midpoint $m$ of two points in $L$ such that $|m-x|>1$ for any $x\in L$.
> 
> 
> 


Modulo Lemma one can construct an action of parallel translations the following way:
Let us construct inductively a sequence of lattices $L\_q$ on $\mathbb R^q$ such that $\mathop{diam} \mathbb R^q/L\_q<1000$ and such that $|x|>1$ for any $x\in L$.
Start with standard $L\_1=\mathbb Z$ in $\mathbb R$.
To construct $L\_{q}$ take 
$$L\_{q}'=L\_{q-1}\times \mathbb Z\subset \mathbb R^{q-1}\times\mathbb R = \mathbb R^{q}.$$
If $\mathop{diam} \mathbb R^q/L'\_q < 1000$ set $L\_q = L'\_q$.
Othewise pass to the minimal lattice which contains $L'\_q$ and the midpoint provided by the Lemma.
Applying this construction finitely many times you get a lattice $L\_q$ with $\mathop{diam} \mathbb R^q/L\_q<1000$.


Continue the process, we get lattice $L\_\infty$ in $H$ which is a $1000$-net, its fundamental doamin contains a ball of radius 1; i.e. $H/L\_\infty$ is not compact.


*Proof of Lemma.*
For $z\in\mathbb R^q$, denote by $\rho(z)$ the minimal distance to a point in $L$.
Take a point $z\in\mathbb R^q$ which maximize distance to $L$.
So $\rho(z)\ge 1000$. Then there is a couple of points $x,y\in L$ such that
$\angle xzy\ge\pi/2$ and $|x-z|=|x-z|=\rho(z)$.
Let $m$ be the midpoint for $x$ and $y$.
Then 
$$|z-m|\le \frac{\rho(z)}{\sqrt{2}}$$ and therefore the distance from $m$ to any point of $L$ is at least $1000{\cdot}(1-\tfrac1{\sqrt{2}})>1$. $\square$


