Does such a subgroup exist? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T13:58:45Z http://mathoverflow.net/feeds/question/6635 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/6635/does-such-a-subgroup-exist Does such a subgroup exist? Arnaud Brot 2009-11-24T00:05:40Z 2009-11-24T16:07:20Z <p>I am looking for a certain masa in a $II_1$ factor which is singular and has nontrivial Takesaki invariant. For this I am looking for an example of an inclusion of groups $H\subset G$ such that:</p> <ul> <li>$G$ is a countable icc (infinite conjugacy class) group</li> <li>$H$ is abelian</li> <li>$\forall g\in G-H,\{ hgh^{-1} |h\in H \}$ is infinite</li> <li>$|H\backslash G/H| \geq 3$</li> <li>there exists $g\in G-H$ and $h_1\neq h_2\in H$ with $h_1 g=gh_2$.</li> </ul> <p>Does such an example exist?</p> http://mathoverflow.net/questions/6635/does-such-a-subgroup-exist/6642#6642 Answer by Yemon Choi for Does such a subgroup exist? Yemon Choi 2009-11-24T01:08:25Z 2009-11-24T02:26:04Z <p>This is off the cuff, and I'm very much a dilettante when it comes to group theory, so I hope there isn't an error in what follows. Corrections welcome, of course.</p> <p><b>[EDIT: it has been pointed out below that the group given below doesn't quite work. I'm leaving the bulk of this "answer" here, in case it suggests a correct solution or warns people off the same mistake I made.]</b></p> <p>I think the group $G$ with presentation $\langle g, h | hg =gh^n \rangle$, where $n\geq 2$, will do the job, with $H$ being the group generated by $h$. [Conditions 2,5]</p> <p>Elements of this group have a normal form with all the $g$s on the left and all the $h$s on the right. <b>[EDIT: this is not quite right, one has to take care over negative powers of $g$.]</b> Multiplying on the left or on the right by an element of $h$ should, once we bring to normal form, not change the index of $g$ in the normal form, and so there are infinitely many double $H$-cosets, taking care of Condition 4.</p> <p>Also, given an element of the form $g^ah^b$ where $a\neq 0$, then some back-of-the-envelope scribbling indicates that repeated conjugation by $h$ ought to increase the absolute value of the index of $h$ in the resulting normal form, so that conjugation by $h$ cannot be an operation of finite order. <strike>That would take care of Condition 3.</strike> <b>[EDIT: this is incorrect/insufficient, see comments below.]</b></p> <p>Finally, I think Condition 1 should follow from some further case-by-case analysis (given a non-identity element in $H$, conjugate repeatedly by $g$; and all the elements in $G-H$ are taken care of by condition 3).</p> <p>(The group $G$ is an example of a <a href="http://eom.springer.de/B/b130070.htm" rel="nofollow">Baumslag-Solitar group</a>, and these beasts have been quite well studied over the years, I'm told. I don't know if you can do similar games with other B-S groups.)</p> http://mathoverflow.net/questions/6635/does-such-a-subgroup-exist/6648#6648 Answer by Agol for Does such a subgroup exist? Agol 2009-11-24T03:46:00Z 2009-11-24T03:55:50Z <p>I think a lattice in the rank 3 solvable Lie group Sol works. For any 2x2 matrix A &isin; SL<sub>2</sub> &#8484; with tr(A) > 3, take the extension G of H=&#8484;<sup>2</sup> by &#8484;, where 1 acts by A on &#8484;<sup>2</sup>. We may write elements of G as (k, h), k &isin; &#8484;, h &isin; &#8484;<sup>2</sup>. The subgroups (k,0) and (0,h) are additive in the coordinates, and (k,0)(0,h)=(k,h). We have the relation (0,h)(1,0) =(1,0)(0, A(h)) (so the 5th condition holds). For example, the matrix </p> <p>$$\begin{pmatrix} 2 &amp; 1 \\ 1&amp; 1\end{pmatrix}$$</p> <p>gives rise to the fundamental group of 0-framed surgery on the figure 8 knot complement. G is countable icc, and H=&#8484;<sup>2</sup> is a normal subgroup, G/H = &#8484;, so the 2nd and 4th conditions are satisfied. The 3rd condition is satisfied, since for h &isin; H= &#8484;<sup>2</sup>, (0,h)(k,g)(0,-h) = (k, g+A<sup>k</sup>(h) -h), which one can see is infinite as one varies h (for k &isin; &#8484; - 0 ). </p> http://mathoverflow.net/questions/6635/does-such-a-subgroup-exist/6682#6682 Answer by Steven Deprez for Does such a subgroup exist? Steven Deprez 2009-11-24T13:09:56Z 2009-11-24T13:09:56Z <p>[generalization of Agol's answer]</p> <p>Take $H$ a group and let $K$ act on $H$ by isomorphisms (write the action as $\sigma$) and consider $G=H\rtimes_\sigma K$. Then</p> <ul> <li>condition 2 is satisfied if $H$ is abelian</li> <li>condition 4 is satisfied if $K$ contains at least 3 elements</li> <li>condition 5 is satisfied if $K$ acts non-trivially</li> <li>condition 3 is satisfied if ${h^{-1}\sigma_k(h) : h\in H}$ is infinite for all $k \in K$</li> <li>condition 1 is satisfied if $K$ acts with infinite orbits on $H$ and condition 3 is satisfied.</li> </ul> http://mathoverflow.net/questions/6635/does-such-a-subgroup-exist/6700#6700 Answer by Steven Deprez for Does such a subgroup exist? Steven Deprez 2009-11-24T16:07:20Z 2009-11-24T16:07:20Z <p>A more singular example:</p> <p>Take an infinite index inclusion of abelian groups $K\subset H$. Let a non-trivial group $L$ act on $K$ by automorphisms. Then the amalgamated free product $G=H\underset{K}{\ast} (K\rtimes L)$ satisfies the conditions. Moreover, $L(H)\subset L(G)$ $\,$is a singular masa. On can use the results of Ioana, Peterson and Popa for this, but maybe there are more elementary ways to see this.</p>