Timeline for Rank gradient in free products amalgamating a finite subgroup
Current License: CC BY-SA 3.0
19 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Nov 17, 2015 at 9:27 | comment | added | HJRW | @Pablo, a (fg) group with infinitely many ends is precisely a group which splits over a finite group and is not virtually cyclic. This is a famous theorem of Stallings. | |
| Nov 17, 2015 at 7:51 | comment | added | Pablo | @YCor which examples of groups with infinitely many ends are there? | |
| Nov 17, 2015 at 0:09 | comment | added | YCor | Actually I had no doubt about the fact that fg groups with infinitely many ends have positive first $\ell^2$ Betti number, but the fact that fg groups with positive first $\ell^2$ Betti number have positive rank gradient is not exactly what is stated in the reference I gave. I guess it's known anyway. | |
| Nov 16, 2015 at 18:57 | comment | added | Pablo | @IanAgol this is true in general - see my previous comment. | |
| Nov 16, 2015 at 18:40 | comment | added | Ian Agol | This follows (assuming $A$ and $B$ are residually finite) exactly as in the referenced proof to my previous question. More generally, it holds when $C$ is separable in $A$ and $B$. | |
| Nov 16, 2015 at 18:08 | comment | added | Pablo | @YCor this probably follows from Proposition 3.1 in arxiv.org/pdf/0708.4327v2.pdf | |
| Nov 16, 2015 at 17:43 | comment | added | Pablo | @YCor you are right! But, can you show that the first $L^2$-Betti number is positive? | |
| Nov 16, 2015 at 17:40 | history | edited | Pablo | CC BY-SA 3.0 |
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| Nov 16, 2015 at 17:40 | comment | added | YCor | Sorry indeed I was misled by my intuition. From what I can read a few lines before Question 1.10 here (people.virginia.edu/~mve2x/Research/ershov-lueck120826.pdf), I'd expect that for groups with positive first $\ell^2$-Betti number, the rank gradient is positive. This would answer positively your question. | |
| Nov 16, 2015 at 17:30 | comment | added | Pablo | @YCor The rank gradient of this group is positive - this follows from the result of Lackenby which I have cited in my question... | |
| Nov 16, 2015 at 17:27 | comment | added | YCor | OK thanks for the correction, you're right. If $M$ is an infinite f.g. group with no nontrivial finite quotient and $K$ has order 2, do you know what the rank gradient of $K\ast (K\times M)$ is? | |
| Nov 16, 2015 at 17:03 | history | edited | Pablo | CC BY-SA 3.0 |
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| Nov 16, 2015 at 17:00 | comment | added | Pablo | @YCor this is the point I was trying to make in my previous comment. If the only finite quotient is trivial, then the rank gradient is just $d(A *_C B) - 1$ which is positive. | |
| Nov 16, 2015 at 16:56 | comment | added | YCor | If $A$ and $B$ both have no nontrivial finite quotient, then $A\ast B$ has no nontrivial finite quotient as well, so the rank gradient is zero. | |
| Nov 16, 2015 at 16:53 | comment | added | Pablo | @YCor - I have referenced a definition. I do not see which counterexamples arise if either $A$ or $B$ (or both) are not residually finite. Furthermore, I think that it is sufficient to assume only that $[A : C] > 2$ and not also $[B : C] > 2$. In the absence of residual finiteness, the resulting free product with amalgamation may have no finite index subgroups at all, but in this case, the rank gradient is positive by definition. | |
| Nov 16, 2015 at 16:50 | history | edited | Pablo | CC BY-SA 3.0 |
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| Nov 16, 2015 at 16:44 | comment | added | YCor | You should make a residual finiteness assumption to avoid trivial counterexamples (and also assume explicitly the implicit $[B:C]>1$). Also please give a definition, or link to another post, for the definition of rank gradient. | |
| Nov 16, 2015 at 16:38 | history | asked | Pablo | CC BY-SA 3.0 |