Timeline for Units in the group ring over fours group after Gardam
Current License: CC BY-SA 4.0
56 events
| when toggle format | what | by | license | comment | |
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| Sep 10, 2023 at 16:40 | review | Suggested edits | |||
| Sep 11, 2023 at 14:59 | |||||
| Mar 23, 2021 at 12:03 | comment | added | ARG | @VilleSalo correct, and on the index... | |
| Mar 23, 2021 at 12:03 | comment | added | ARG | @Ycor & VilleSalo I don't know if there are guidelines as to when one should stop sharing "research ideas" on MO (or perhaps you guys should start a paper). But I sure would like to see those matrices. | |
| Mar 23, 2021 at 12:02 | comment | added | Ville Salo | ...and on the index | |
| Mar 23, 2021 at 11:54 | comment | added | ARG | @VilleSalo Yes I think computing commutators will only give you that the group is not solvable (but then it's important to have the bound on the solvable length) | |
| Mar 23, 2021 at 9:13 | comment | added | Ville Salo | (I didn't mean to promise I'll do it, anyone feel free to produce them.) | |
| Mar 23, 2021 at 9:11 | comment | added | Ville Salo | Yes I guess I thought without thinking that it's enough to check commutators of generators up to some depth by commutator formulas, but I don't think that's true. The matrices should be easy to produce, I can do it (but no easier than anyone else). | |
| Mar 23, 2021 at 8:57 | comment | added | ARG | @Ycor & VilleSalo yes that was exactly what I was fishing for: a computable criterion to decide for solvability (perhaps it can only conclude non-solvability). Given that it seems feasible to compute relatively large elements in the group (at least VilleSalo can do it, my computing skills are too thin for this), one should be able to get something. But it would be nice to know where to stop! And it's true too, the matrices are probably computable. Looks like this might open some original research! | |
| Mar 23, 2021 at 8:04 | comment | added | Ville Salo | FWIW, I computed 5000 elements of the Gardam group (I mean the one generated by $a,b,p$), always multiplying previous ones by $a,b,p$ and always keeping only the 5000 with minimal ($> 1$) length and secondarily minimal total length (with some arbitrary convention), until this converged. I found $5000$ elements of support $21$, nothing with less. (This is about the exchange between Abdollahi and Gardam, maybe Gardam went further already?) | |
| Mar 23, 2021 at 6:59 | comment | added | Ville Salo | I mean, it would be cool to know an absolute upper bound on the amount of computation that needs to be done. | |
| Mar 23, 2021 at 6:35 | comment | added | Ville Salo | Of course, but are there abstract bounds known for the index so that we can immediately find some explicit $n$ such that this group is nonamenable iff it contains a free $n$-ball? | |
| Mar 23, 2021 at 6:23 | comment | added | YCor | At this point I think one should first compute explicitly this embedding into $4\times 4$ matrices, and make explicit these matrices corresponding to the units $p,q$ (and also to the trivial units $a,b$). The knowledge of these matrices, their characteristic polynomial etc should provide useful information on the Zariski closure of the group they generate, say much information about their powers, etc. | |
| Mar 23, 2021 at 5:26 | comment | added | Ville Salo | We need a bound on the index of the solvable subgroup too, but assuming we have those I think this gives that both amenability and nonamenability can be decided from a finite ball. I guess that was @YCor's idea. I don't know where the strongest bounds are given but I can fire up my script... | |
| Mar 23, 2021 at 5:03 | comment | added | Ville Salo | Ooh that's a great idea | |
| Mar 22, 2021 at 23:46 | comment | added | ARG | @Ycor my memory is murky... what's the bound on the length (if solvable)? one could eventually check commutators of the unit $p$, its powers $p^n$ and their conjugates... | |
| Mar 22, 2021 at 20:50 | comment | added | YCor | This group is naturally a subgroup of $\mathrm{GL}_4(R)$, $R=\mathbf{F}_2[a^{\pm 1}, b^{\pm 1}, c^{\pm 1}])=\mathbf{F}_2[\mathbf{Z}^3]$ and in particular satisfies the Tits alternative: has a free subgroup, or virtually solvable of small length. See my comment to mathoverflow.net/a/387186/14094 | |
| Mar 22, 2021 at 9:58 | comment | added | Ville Salo | Maybe the correct thing to do is to compare with the growth of the submonoid generated by the support of $p$. By amenability, assuming this monoid is a group (I didn't check this though) I would expect that the boundary behavior eventually disappears and the density goes to about a half. Calculated like this I get relative densities 1:1.000, 2:0.648, 3:0.588, 4:0.500, 5:0.499, 6:0.475, 7:0.489, 8:0.462, 9:0.455, 10:0.458, 11:0.466, 12:0.458 for $p^n$. | |
| Mar 22, 2021 at 9:49 | comment | added | Ville Salo | In the post I represent $p$ over the generators $a$, $b$, and the radius is $7$. So I used that to calculate. The support of $p$ is size $21$ and the $7$-ball in $P$ has cardinality $363$. The density is thus 5.8 percent. Taking powers and considering the density in the $7n$-ball, we get approximately 1:0.058, 2:0.050, 3:0.056, 4:0.052, 5:0.054, 6:0.053, 7:0.056, i.e. the density doesn't seem to change. (This doesn't seem very surprising to me, it's a complicated cellular automaton, but again I do recommend someone checks these numbers.) | |
| Mar 22, 2021 at 9:36 | comment | added | Ville Salo | You can check what you suggest (the "dimension" of the support), but the dimension could be full (so the limit is three) and you could even have positive density. That's what I would guess happens; except for the boundaries you fill about half the ball. | |
| Mar 22, 2021 at 9:31 | comment | added | Ville Salo | One could of course ask whether we reach the (cubic) growth of this group. I can check the densities. | |
| Mar 22, 2021 at 9:31 | comment | added | YCor | Oh, of course :) one should rather check if $\log(u_n)/\log(n)$ seems to converge to a positive limit. | |
| Mar 22, 2021 at 9:29 | comment | added | Ville Salo | Well, we don't have exponential growth, it's bounded by the growth of the group? (The list of pairs $n:k$ was the answer to what I guessed you were asking, so the size $k$ of the support of the polynomial $p^n$ where $p$ is the polynomial of Giles.) | |
| Mar 22, 2021 at 9:29 | comment | added | YCor | By the way $P$ embeds into a split abelian-by-finite group and such questions might be natural to ask in this setting (allowing torsion), in which case the first test case would be describing units in $\mathbf{F}_2D_\infty$. | |
| Mar 22, 2021 at 9:28 | comment | added | YCor | Still the successive quotients decrease which doesn't clearly show an exponential growth. | |
| Mar 22, 2021 at 9:18 | comment | added | Ville Salo | Support sizes according to my script 1:21, 2:129, 3:469, 4:1005, 5:2029, 6:3421, 7:5673, 8:8111, 9:11469, 10:15961, 11:21767, 12:27905, 13:35759, 14:44707, 15:55863, 16:65999, 17:75527, 18:90003, 19:110467, 20:128909. I could go further but I feel that's disrespectful to my computer, I really doubt that suddenly collapses to one. | |
| Mar 22, 2021 at 9:13 | comment | added | Ville Salo | I'm not following, is $a$ the generator or is it Giles' polynomial $p$? I'll check powers of $p$ after the $8$-ball finishes. (Ok it finished right this second, seems the $8$-ball is free. Do recheck, I wrote very messy Python code without careful checking because I wasn't planning to do original research.) | |
| Mar 22, 2021 at 9:10 | comment | added | YCor | What about powers? could one describe powers of $a$ and show that no power of $a$ is a group element? computing some small powers might be doable. | |
| Mar 22, 2021 at 9:05 | comment | added | Ville Salo | @YCor: This was somehow what my last paragraph was meant to be about. I checked more carefully and I retract my claim about a small identity satisfied by $p$ and $b$ (I don't know where I got that). I calculated some small balls and the pair $p,a$ seems to generate a free $7$-ball, i.e. all 4373 products are different. I wouldn't say this is very convincing of anything yet, but at least it's not immediately obvious even that this subgroup is not free. | |
| Mar 21, 2021 at 20:39 | answer | added | YCor | timeline score: 7 | |
| Mar 21, 2021 at 18:12 | comment | added | YCor | Have you thought about any question in the direction "showing that the group of units is not too small"? For instance, is it immediate that $P$ has infinite index? Can a lower bound on the (possibly infinite Hirsch length) be given? can it be shown that the group of units is not virtually abelian? etc. The amenability question sounds a bit elaborate to start with, since amenability is not purely algebraic (well it is, but for quite elaborate very recent reasons). | |
| Mar 21, 2021 at 18:05 | history | edited | YCor | CC BY-SA 4.0 |
added tags, added conjecture, avoided Z_2 ambiguous esp in ring context, etc
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| Mar 19, 2021 at 15:53 | comment | added | Giles Gardam | @AlirezaAbdollahi 21 is the smallest that I know of. Maybe 21 is minimal for this group but not for all groups? | |
| Mar 14, 2021 at 21:36 | answer | added | ARG | timeline score: 4 | |
| Mar 13, 2021 at 18:15 | comment | added | Alireza Abdollahi | @GilesGardam Do you think that 21 is the least length for a possible counterexample? Have you done any work on elements of length 19? | |
| Mar 12, 2021 at 13:13 | comment | added | ARG | @VilleSalo This is why math is so fun: you have mysterious things that can be so elegantly solved (I'm referring to Giles' Answer) | |
| Mar 12, 2021 at 11:37 | answer | added | ARG | timeline score: 4 | |
| Mar 11, 2021 at 18:25 | answer | added | Giles Gardam | timeline score: 19 | |
| Mar 11, 2021 at 6:45 | comment | added | Ville Salo | This torsion question reminds me of my other question mathoverflow.net/questions/339692/… (which is still completely open). These are such mysterious things. | |
| Mar 11, 2021 at 6:25 | comment | added | ARG | @GilesGardam & VilleSalo a silly remark: when $n$ is prime, then over $\mathbb{F}_n$ there cannot be $n$ torsion, since $0=p^n -1 = (p-1)^n$ (so one of the powers of $(p-1)^k$ times $(p-1)$ form a zero divisor. This makes it unlikely (but not impossible) to have torsion over $\mathbb{Z}$... | |
| Mar 10, 2021 at 22:27 | comment | added | ARG | @GilesGardam (& VilleSalo) It was probably careless speculation from my part to extend the idempotent conjecture to all $n$. So the "correct" statement of the idempotent conjecture is with $n=2$, and in this case $(p-1)(p+1)$ is clearly a zero divisor (even over $\mathbb{F}_2$) [note that the variation of idempotent with $p^n=p$ works for $n=2$ or $3$]. But indeed, a torsion element of odd order $>2$ might not be an actual contradiction... One can always hope that for an explicit element $p^{n-1} + \cdots + p + 1$ can be shown to be non-zero, but that might be a tall order. | |
| Mar 10, 2021 at 20:21 | comment | added | Ville Salo | @GilesGardam: Isn't $0 = p^n -1 = (p-1)(p^{n-1} + \cdots + p + 1)$ a perfectly fine proof? Oh wait, why isn't the latter term $0$? I suppose I didn't understand the proof. | |
| Mar 10, 2021 at 19:36 | comment | added | Giles Gardam | @ARG this is a very theoretical question at this point, but can you get a contradiction to the zero divisor conjecture from a torsion element if the field isn't $\mathbb{F}_2$? | |
| Mar 10, 2021 at 11:39 | comment | added | Ville Salo | The group $GL(2, \mathbb{F}_2[P])$ is not amenable, so be careful. | |
| Mar 10, 2021 at 10:40 | comment | added | ARG | yes, I wrote garbage. I'll try to write things properly (if it works) sometime in the not too far future. | |
| Mar 10, 2021 at 5:19 | comment | added | Ville Salo | I mean I believe $|pF_i|/|F_i|=1$ so the limit is 1 too, but that's not amenability. As for the correct formula, I don't see why it holds. | |
| Mar 10, 2021 at 5:12 | comment | added | Ville Salo | If it works it's a good answer. But do check the formulas, at least one looks iffy to me (maybe a typo in the limit). | |
| Mar 9, 2021 at 23:07 | comment | added | ARG | here would be an argument for "of $G$ is a subgroup of an amenable monoid $M$, then $G$ is amenable": the subgroup $G$ partitions $M$ in right cosets (this uses that $G$ is a group); take a bunch $m_i$ of representatives for the $G$-cosets; given a function $f$ in $\ell^{\infty}G$ extend it to $\ell^{\infty}M$ by $\tilde{f}(m_ig) = f(g)$; define the invariant mean on $\ell^{\infty}G$ by $\mu(f) = \mu_M(\tilde{f})$ where $\mu_M$ is an invariant mean on the monoid $M$. | |
| Mar 9, 2021 at 19:18 | comment | added | ARG | As for amenability, the monoid $\mathbb{F}_2[P]$ is surely amenable. Look at $F_i =$ elements supported on the ball of radius $i$, then for any element $p \in \mathbb{F}_2[P]$, $\lim_i \tfrac{|pF_i| }{|F_i|} =1$. Is there a result to conclude that a subgroup of an amenable monoid is amenable? A submonoid of a amenable group is not always amenable. (and yes, the capital letters are very convenient to copy in code!) | |
| Mar 9, 2021 at 15:10 | comment | added | Ville Salo | You are also correct about the capital notation being confusing, it's just that I imagined some people will find it useful for the formula to be copy-pasteable for quick checking, and I figured the $A$ & $B$ convention is better for that than $^{-1}$. I copied and pasted it from my little Python script. | |
| Mar 9, 2021 at 15:09 | comment | added | Ville Salo | Ok I did not connect the dots there. So on $P$, we can immediately say that the group of units of the group ring is torsion-free. | |
| Mar 9, 2021 at 14:44 | comment | added | ARG | Maybe I'm missing a point in your last question, but the zero divisor conjecture (which holds for the group $P$ since it is solvable) implies the idempotent conjecture. If $p$ had finite order, that would contradict the zero divisor conjecture, since $0 = p^n-1 = (p-1)(p^{n-1}+p^{n-2} + \ldots + p^2 + p + 1)$. (Also the convention that a capital letter is an inverse is very misleading there, since $p$ is an element and $P$ is either the inverse of $p$ or the group $P$; but again it's just me being pointy) | |
| Mar 9, 2021 at 12:52 | comment | added | Ville Salo | I realized that some time after posting but I figured it's not worth bumping the post over. Inverses included in presentation now. | |
| Mar 9, 2021 at 12:50 | history | edited | Ville Salo | CC BY-SA 4.0 |
A and B added in presentation
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| Mar 9, 2021 at 10:04 | comment | added | ARG | it might be nice to mention what are $A$ and $B$ (although I'll admit it's a fairly common shortcut) | |
| Mar 6, 2021 at 10:59 | history | edited | Ville Salo | CC BY-SA 4.0 |
added 2 characters in body
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| Mar 6, 2021 at 10:42 | history | asked | Ville Salo | CC BY-SA 4.0 |