presentations for complex involutory reflection groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T01:52:02Z http://mathoverflow.net/feeds/question/85654 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/85654/presentations-for-complex-involutory-reflection-groups presentations for complex involutory reflection groups Dima Pasechnik 2012-01-14T12:04:48Z 2012-03-04T06:39:01Z <p>It's a well-known result due to J.Tits that a finite-dimensional real reflection group has a faithful presentation, given by its Coxeter diagram (i.e. the linear group in question is isomorphic to the corresponding finitely presented abstract group). In our setting, we have an infinite subgroup $H\le GL(n)$ generated by finitely many complex involutory reflections, i.e. transformations that have the spectrum (1,1,..,1,-1).</p> <p>What can be said about a presentation for $H$? Does $H$ at least admit a finite presentation? If it matters, only one of our involutory generators has non-real entries.</p> http://mathoverflow.net/questions/85654/presentations-for-complex-involutory-reflection-groups/90182#90182 Answer by Misha for presentations for complex involutory reflection groups Misha 2012-03-04T06:39:01Z 2012-03-04T06:39:01Z <p>The answer is almost surely negative. Below are two examples of <em>discrete</em> non-finitely presentable subgroups of $GL(n,{\mathbb C})$ which are generated by finitely many finite order elements. In the first example generators are involutions but some have more than one eigenvalues $-1$, in the second example generators are "complex reflections," of order $5$, but where all but one eigenvalues are equal to $1$. By working harder one should be able to construct examples you are asking for, but I do not see sufficient motivation for doing so. My guess is that you were hoping for a positive result instead of counter-examples. </p> <p>Example 1. This example is a variation on the example in </p> <p>[1] M.Feighn and G. Mess, Conjugacy Classes of Finite Subgroups of Kleinian Groups, American Journal of Mathematics, Vol. 113, No. 1 (1991), 179-188. </p> <p>See also:</p> <p>G. Baumslag and J. E. Roseblade, Subgroups of direct products of free groups, J. London Math. Soc. (2) 30 (1984), 44–52</p> <p>where it is proven that subgroups of direct products of (virtually) free groups are "almost never" finitely presented. </p> <p>Let $F$ denote the group with the presentation $\langle x_1, x_2, x_3| x_1^2=x_2^2=x_3^2=1\rangle$. This group can be embedded as a discrete subgroup of $SU(1,1)$ (the 2-fold cover of $PU(1,1)$, the isometry group of the hyperbolic plane). Next, let $G=F\times F$. Then $G$ embeds discretely in $SU(1,1)\times SU(1,1)\subset SL(4, {\mathbb C})$. Consider homomorphism $\phi: F\to {\mathbb Z}$ sending $x_1, x_2$ to $0$ and $x_3$ to $1$. Let $\Phi=(\phi,\phi): G\to {\mathbb Z}$. Define $\Gamma$ to be the kernel of $\Phi$. Then $\Gamma$ is generated by the following elements: $$(x_1,1), (1,x_1), (x_2,1), (1,x_2), (x_3, x_3^{-1}).$$ Furthermore, $\Gamma$ is not finitely presentable. See [1] where a similar example is explained in detail. Now, every generator of $\Gamma$ is an involution. </p> <p>Example 2. This is a variation on the example in<br> <a href="http://front.math.ucdavis.edu/9808.5085" rel="nofollow">http://front.math.ucdavis.edu/9808.5085</a></p> <p>Start with the complex-hyperbolic lattice $G$ in $SU(2,1)\subset SL(3, {\mathbb C})$ which appears at number 9 on Thurston's list in [3], page 38. </p> <p>[3] <a href="http://front.math.ucdavis.edu/9801.5088" rel="nofollow">http://front.math.ucdavis.edu/9801.5088</a> </p> <p>Let $B^2$ be the complex 2-ball on which $G$ acts holomorphically. Geometry of the quotient was understood in detail by Hirzebruch et al. The quotient $B^2/G$, as an orbifold, admits a singular holomorphic fibration whose base is the projective line with three singular points of order $5$, while the generic fiber $\Sigma$ is the projective line with four singular points of order $5$. Now, let $\Gamma\subset G$ be the image of the fundamental group of the generic fiber $\Sigma$ in $G$. The group $\Gamma$ is generated by four complex reflections of order $5$ (each has two eigenvalues $1$ and one which is 5th root of unity). Then, using the main result of [2], one shows that $\Gamma$ is not finitely presentable. (The proof in [2] applies to torsion-free subgroup of finite index in $\Gamma$, but finite presentability is preserved by passing to finite index subgroups.) </p> <p>So, what is the difference with the result on real reflection groups? The key is that Tits-Vinberg theorem applies to groups $\Gamma$ generated by real reflections in faces of a convex cone $\Delta$ where "angles'' (whatever this means) have the form $\pi/m_i$. The cone $\Delta$ serves as a fundamental domain for the Coxeter group $\Gamma$ and one can read off the presentation of $\Gamma$ from the combinatorics and geometry of $\Delta$. Such a cone is completely missing in the case of groups generated by complex reflections. Even in the real case, once you drop the "$\pi/m_i$" assumption, the linear representation $\psi$ of the Coxeter group $\Gamma$ (whose presentation is defined by repeating the presentation which appears in the $\pi/m_i$ case) will fail to be faithful. By working harder, one can almost surely get examples of groups $\psi(\Gamma)$ which are not finitely-presented as well. </p>