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## 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.

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 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 2010 at 0:07 @Anton: Of course you did not say what $\Gamma$ is. – Mark Sapir Oct 22 2010 at 0:09 @Mark: (1) $diam=\infty$ (2) Γ is a group, what else can act? – Anton Petrunin Oct 22 2010 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 2010 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 2010 at 7:30
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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 $1000{\cdot}(1-\tfrac1{\sqrt{2}})>1$. $\square$

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