Cobounded ⇒ cocompact?

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?

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.

• Just a thought: Let $H=\ell_2$. Let $\Gamma$ be the subgroup of the additive group of $H$ generated by sequences with integral norms. Then $\Gamma$ acts on $H$ by translations. What is the diameter of $H/\Gamma$ and is the quotient compact? I do not have time to check it myself now. But it should not be that difficult. – Mark Sapir Oct 22 '10 at 0:07
• @Anton: Of course you did not say what $\Gamma$ is. – Mark Sapir Oct 22 '10 at 0:09
• @Mark: (1) $diam=\infty$ (2) Γ is a group, what else can act? – Anton Petrunin Oct 22 '10 at 1:55
• @Anton: Since you have computed the diameter of the quotient, could you then give an example of two orbits arbitrary far apart? I thought of the following argument. Suppose that $f\in \ell_2$ has norm $M$. Subtract $[M]/M \cdot f$ (with integral norm $[M]$), get norm $<1$. Hence the diam. is at most 2. If I do not see something here, did you consider already all subgroups $\Gamma$ of the additive group of $H$ acting by translations? – Mark Sapir Oct 22 '10 at 3:05
• @Mark: If the dimension of $H$ is at least 2, the group $\Gamma$ generated by vectors of integral norms is transitive: Given $f \in H, f\neq 0$, connect $v:=\frac{f}{||f||}$ to $-v$ via an arc $\phi\colon [0,1]\to H$ on the unit sphere and consider $w_t:=\phi(t)+f$. By the intermediate value theorem, $||w_{t_0}||$ will be integral for some $t_0$, so $f$ is the difference of two vectors of integral norm. – Guntram Oct 22 '10 at 7:30

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 $$1000{\cdot}(1-\tfrac1{\sqrt{2}})>1$$. $$\square$$