Invertible matrices satisfying $[x,y,y]=x$. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T08:49:29Z http://mathoverflow.net/feeds/question/44737 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/44737/invertible-matrices-satisfying-x-y-yx Invertible matrices satisfying $[x,y,y]=x$. Mark Sapir 2010-11-03T22:09:01Z 2012-03-19T03:31:22Z <p>I have been thinking about this question for quite some time but now <a href="http://mathoverflow.net/questions/44680/norm-of-commutators-bis" rel="nofollow"> this </a> question by Denis Serre revived some hope. </p> <p><b> Question. </b> Let $x,y$ be invertible matrices (say, over $\mathbb C$) and $[x,y,y]=x$ where $[a,b]=a^{-1}b^{-1}ab$, $[a,b,c]=[[a,b],c]$. Does it follow that some power of $x$ is unipotent? </p> <p>The motivation is this. Consider the one-relator group $\langle x,y \mid [x,y,y]=x\rangle$. It is hyperbolic (proved by A. Minasyan) and residually finite (that is proved in my paper with A. Borisov). If the answer to the above question is "yes", then that group would be non-linear which would provide an explicit example of non-linear hyperbolic group. </p> <p><b> Update 1. </b> Can $x$ in the above be a diagonal matrix and not a root of 1?</p> <p><b> Update 2. </b> The group is residually finite, so it has many representations by matrices such that $x, y$ have finite orders (hence their powers are unipotents). </p> <p><b> Update 3. </b> The group has presentation as an ascending HNN extension of the free group: $\langle a,b,t \mid a^t=ab, b^t=ba\rangle$. So it is related to the Morse-Thue map. Properties of that map may have something to do with the question. See two quasi-motivations of the question as my comments below. </p> http://mathoverflow.net/questions/44737/invertible-matrices-satisfying-x-y-yx/44746#44746 Answer by Nick S for Invertible matrices satisfying $[x,y,y]=x$. Nick S 2010-11-03T22:38:13Z 2010-11-03T22:57:33Z <p><strong>Ignore this, it is wrong</strong></p> <p>I migth miss something simple, but</p> <p>$[a,b]^n=[a,b]$ for all $n$ hence $x^2=[x,y,y]^2=[x,y,y]=x$.</p> <p>Since $x$ is invertible and $x^2=x$ it follows $x=I$.</p> <p>using this it is easy to show that $[x,y,y]=x$ for x,y invertible if and only if $x=I$.</p> http://mathoverflow.net/questions/44737/invertible-matrices-satisfying-x-y-yx/44762#44762 Answer by David Speyer for Invertible matrices satisfying $[x,y,y]=x$. David Speyer 2010-11-04T01:09:39Z 2010-11-04T01:25:58Z <p>Here's a quick test which might disprove your hopes very quickly:</p> <p>Take $n$ to be small: Try $2$ first, and $5$ is probably near the limit of a computer algebra system. Choose $x$ to be a random $n \times n$ diagonal matrix with determinant $1$, for example, $\mathrm{diag}(17, 1/17)$. Write out your relation, leaving all the elements of $y$ as variables. After clearing denominators, you have $n^2$ simultaneuous homogenous equations in $n^2$ variables. (If I haven't made any dumb errors, they have degree $3n$.) Ask your favorite computer algebra system to solve them for you. If any of the roots are not on the hypersurface $\det y=0$, then you have a counterexample!</p>