Timeline for Invertible matrices satisfying $[x,y,y]=x$
Current License: CC BY-SA 4.0
21 events
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| Aug 31, 2018 at 10:07 | history | edited | user6976 | CC BY-SA 4.0 |
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| Mar 29, 2018 at 17:51 | comment | added | user6976 | @AlexanderChervov: See my answer below. | |
| Mar 28, 2018 at 8:02 | vote | accept | CommunityBot | moved from User.Id=6976 by developer User.Id=69903 | |
| Mar 28, 2018 at 7:34 | answer | added | user6976 | timeline score: 13 | |
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Jun 7, 2012 at 13:00 | comment | added | user6976 | @Alexander: There was no progress so far, any new information would be very nice. | |
| Jun 7, 2012 at 10:06 | comment | added | Alexander Chervov | Is there some progress on this amasing question ? If I have time I can do numerical MatLab experiments for the following fix "x"; try to find numerically y. I.e. minimize | [x y y]-x|^2 | |
| Mar 19, 2012 at 3:31 | history | edited | user6976 | CC BY-SA 3.0 |
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| Mar 18, 2012 at 21:02 | comment | added | user6976 | @Terry: Yes, this was another quasi-reason. Consider $G=\langle x,y\mid x^y=x^2\rangle$. Then in every linear representation of $G$, conjugating $x$ by powers of $y^{-1}$ will produce matrices that are closer and closer to 1. So if $x,y$ are matrices $\lim_{n\to\infty} x^{y^{-n}}=1$. This means that $x$ is a unipotent element "of $y$" in Margulis' terminology, hence $x$ is unipotent. Now we have a similar presentation $\langle x,z,y\mid x^y=xz, z^y=zx\rangle$, so the idea was to show that some power of $x$ satisfies the limit property above. | |
| Mar 18, 2012 at 19:40 | comment | added | Terry Tao | A trivial observation: setting $z := [x,y]$, the condition $[x,y,y]=x$ is equivalent to the assertion that the pair $(x,z)$ is conjugate to $(xz,zx)$ after conjugation by $y$. So the question is equivalent to the question of whether a pair of matrices $(x,z)$ which has the property of being conjugate to $(xz,zx)$ is such that all the eigenvalues of $x$ (or equivalently, $z$, which is necessarily conjugate to $x$) are roots of unity. Unfortunately, I got stuck after this observation: the conjugacy does give a number of trace identities involving various words in z,x, but not enough of them... | |
| Mar 18, 2012 at 14:04 | comment | added | user6976 | Here is one of the quasi-reasons why I think the answer is positive. If $G=\langle x,y \mid [x,y,y]=x\rangle$ is linear, then it has a representation over a number field, hence over $\mathbb{Q}$. Therefore the sequence of indexes of subgroups of finite index of $G$ must grow polynomially (take congruence subgroups). This would imply that certain polynomial maps over finite fields have many quasi-fixed points with long orbits (see our paper with Borisov). The latter seems to be impossible. | |
| Mar 18, 2012 at 3:06 | history | edited | user6976 | CC BY-SA 3.0 |
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| Jul 25, 2011 at 1:30 | history | edited | user6976 | CC BY-SA 3.0 |
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| Jul 25, 2011 at 0:26 | history | edited | user6976 | CC BY-SA 3.0 |
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| Apr 5, 2011 at 17:27 | history | edited | user6976 | CC BY-SA 2.5 |
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| Apr 5, 2011 at 16:51 | history | edited | user6976 | CC BY-SA 2.5 |
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| Nov 4, 2010 at 1:09 | answer | added | David E Speyer | timeline score: 4 | |
| Nov 3, 2010 at 23:12 | comment | added | user6976 | @Lukasz: Yes, there are even non-residually finite ones: $BS(2,3)=\langle x,y \mid y^{-1}x^2y=x^3\rangle$. There are also residually finite 1-related groups which are not linear. Those were constructed in our paper with Cornelia Drutu (in J. Algebra). The point is that this group is hyperbolic. There is an example of a non-linear hyperbolic group due to M. Kapovich (which easily follows from the super-rigidity of certain rank 1 lattices and a Gromov-Olshanskii theorem). But that example has no explicit presentation. This one would be the first explicit example. | |
| Nov 3, 2010 at 23:06 | comment | added | Łukasz Grabowski | Are there any one-relator groups known not to be linear? | |
| Nov 3, 2010 at 22:38 | history | edited | user6976 |
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| Nov 3, 2010 at 22:09 | history | asked | user6976 | CC BY-SA 2.5 |