Existence of certain sequences defined by matrices which are quasi-isometric to $\mathbb R^d$. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T13:36:16Z http://mathoverflow.net/feeds/question/74046 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/74046/existence-of-certain-sequences-defined-by-matrices-which-are-quasi-isometric-to Existence of certain sequences defined by matrices which are quasi-isometric to $\mathbb R^d$. Roland Bacher 2011-08-30T08:38:55Z 2011-08-30T09:06:14Z <p>Given a strictly positive integer $d$ and $k$ square-matrices $A_1,\dots,A_k$ of size $d$, we define $S(A_1,\dots,A_k)=\lbrace n^{1/d}A_1^nA_2^n\cdots A_k^nV\ \vert\ n\in \mathbb N\rbrace$ where $V=(1,0,0,\dots,0)^t$.</p> <p>Given $d\geq 1$ does there always exist an integer $k$ and $k$ orthogonal matrices $A_1,\dots,A_k$ such that $S(A_1,\dots,A_k)$ is quasi-isometric to the Euclidean space $R^d$? (Quasi-isometry means simply that there exists a real number $B$ such that balls of radius $B$ centered at all points of $S(A_1,\dots,A_k)$ cover $R^d$.)</p> <p>The answer is yes for $d=1,2$ with $k=1$ (take $A_1=-1$ for $d=1$ and a rotation of angle $2\pi/\tau$ where $\tau=(1+\sqrt{5})/2$ or any irrational number with bounded continued fractions developpement for $d=2$).</p> <p>$d=3$ is the first dimension where I am not sure of the answer. ($k=1$ is clearly impossible for $d\geq 3$.)</p>