Right Angled Artin Group Reference request - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T07:10:55Z http://mathoverflow.net/feeds/question/108848 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108848/right-angled-artin-group-reference-request Right Angled Artin Group Reference request Igor Rivin 2012-10-04T19:13:40Z 2012-10-05T19:11:35Z <p>The following should be true: every normal subgroup of a non-Abelian Right Angled Artin Group should contain a free group on two generators. Is there a standard reference one can cite for this?</p> http://mathoverflow.net/questions/108848/right-angled-artin-group-reference-request/108868#108868 Answer by Agol for Right Angled Artin Group Reference request Agol 2012-10-04T22:49:49Z 2012-10-04T22:49:49Z <p>A right-angled Artin group $G$ is the fundamental group of the compact non-positively curved cube complex called the Salvetti complex $X$. Thus $\tilde{X}$ is a CAT(0) space, and the action of $G$ on $\tilde{X}$ is proper and cocompact. </p> <p>Thus, by the <a href="http://books.google.com/books?id=3DjaqB08AwAC&amp;lpg=PA249&amp;ots=eNW6ZTW10t&amp;dq=solvable%2520subgroup%2520theorem&amp;pg=PA249#v=onepage&amp;q&amp;f=false" rel="nofollow">Solvable Subgroup Theorem</a>, every virtually solvable subgroup $H&lt; G$ contains a finitely generated abelian subgroup of finite index. Moreover, every element in $G$ will be semi-simple by Prop. 6.10 (2) from <a href="http://www.google.com/url?sa=t&amp;rct=j&amp;q=&amp;esrc=s&amp;source=web&amp;cd=1&amp;ved=0CCIQFjAA&amp;url=http%253A%252F%252Fwww.math.psu.edu%252Fpetrunin%252Fpapers%252Fscans%252Fbooks%252Fbridson.haefliger.pdf&amp;ei=qgpuUPTpFYSQiQLTyoCgCw&amp;usg=AFQjCNEhpIU1JkqEhAilaLlz3a5MaPeadg&amp;sig2=nGpi7dPEG8zvTkgSbd9qLA" rel="nofollow">Bridson-Haefliger</a>. By Corollary 7.2, there is an $H$-invariant closed convex subspace of $\tilde{X}$ isometric to a product $Y\times \mathbb{E}^n$, so that $H$ acts as the identity on $Y$ and acts cocompactly on the $\mathbb{E}^n$ factor. Any isometry of $\tilde{X}$ which normalizes $H$ preserves $Y\times \mathbb{E}^n$ and its splitting. We conclude that $\tilde{X}=Y\times\mathbb{E}^n$ when $H$ is normal in $G$. </p> <p>Now, I think for a CAT(0) cube complex $\tilde{X}$, the factorization $Y\times \mathbb{E}^n$ should actually be a product of cube complexes. This isn't quite right, since in the group $\mathbb{Z}^n$ there are clearly $\mathbb{R}^k$ factors of $\mathbb{R}^n$ which are not a factor of the cube complex structure. But I think if $H$ is assumed to be a virtually solvable normal subgroup of $G$ and maximal with respect to this property, then $\tilde{X}=Y\times \mathbb{E}^n$ should be a product of cube complexes. I think it should be possible to prove this by analyzing links of vertices of the Salvetti complex, and how the foliation by $\mathbb{E}^n$'s passes through them. </p> <p>But this implies that $G=G_1\times \mathbb{Z}^n$ where $G_1$ is a right-angled Artin group, since each edge of a cube of the Salvetti complex corresponds to a generator, so a cube complex factor isometric to $\mathbb{R}^n$ will give generators commuting with every other generator. </p> <p>As observed in one of the comments, $G$ satisfies the Tits alternative since it is linear, so a normal subgroup is either virtually solvable, and then $G$ splits off a $\mathbb{Z}^n$ factor, or the normal subgroup is not solvable and therefore must contain a free subgroup. However, one would need to fill in the above missing step to finish this argument. </p> http://mathoverflow.net/questions/108848/right-angled-artin-group-reference-request/108947#108947 Answer by Denis Osin for Right Angled Artin Group Reference request Denis Osin 2012-10-05T19:11:35Z 2012-10-05T19:11:35Z <p>Igor, I assume the question is about centerless RAAGs. Then it follows from the classical result of A. Baudisch (see MR0634562): every 2-generated subgroup of a RAAG is abelian or $F_2$. </p> <p>Indeed if $N$ is a non-trivial normal subgroup in a RAAG $G$, take any $x\in N$. Since $G$ is centerless there exists $g\in G$ that does not commute with $x$. Then $\langle x,g\rangle\cong F_2$ and hence $\langle x, x^g \rangle \cong F_2$.</p>