Cobounded ⇒ cocompact? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T04:19:11Z http://mathoverflow.net/feeds/question/43110 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/43110/cobounded-cocompact Cobounded ⇒ cocompact? Anton Petrunin 2010-10-21T23:24:14Z 2010-10-27T23:20:38Z <blockquote> <p>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?</p> </blockquote> <hr> <p><strong>Stupid example.</strong> 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.</p> http://mathoverflow.net/questions/43110/cobounded-cocompact/43862#43862 Answer by Anton Petrunin for Cobounded ⇒ cocompact? Anton Petrunin 2010-10-27T20:02:30Z 2010-10-27T23:20:38Z <p>The answer is "NO". To show this let us use the following:</p> <blockquote> <p><strong>Lemma.</strong> 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 $x\in L$?</p> </blockquote> <p>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 $\mathop{diam} \mathbb R^q/L_q&lt;1000$ 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 $\mathop{diam} \mathbb R^q/L'_q &lt; 1000$ 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 $\mathop{diam} \mathbb R^q/L_q&lt;1000$.</p> <p>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.</p> <p><em>Proof of Lemma.</em> 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 $1000{\cdot}(1-\tfrac1{\sqrt{2}})>1$. $\square$</p>