Timeline for Units in the group ring over fours group after Gardam
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Dec 4, 2021 at 7:18 | history | edited | ARG | CC BY-SA 4.0 |
updated to take "news" into account
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| Mar 21, 2021 at 17:41 | history | edited | ARG | CC BY-SA 4.0 |
removed tentative proof and wrote informal discussion instead
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| Mar 15, 2021 at 20:01 | history | edited | ARG | CC BY-SA 4.0 |
corrected missing hypothesis about zero-divisors; rewrote second part
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| Mar 15, 2021 at 15:27 | comment | added | ARG | @VilleSalo sorry I completely misunderstood what you meant with the $2\times 2$ monoid. Now I see your point (at least for the sets $F_{i,k}$). For the sets $F_{i,k,n}$ I'm not sure yet (since $F_{i,k,n}$ is not closed under addition). I'll need to edit the answer at some point anyway (since there are hypothesis missing in Lemma 1!). | |
| Mar 15, 2021 at 10:28 | comment | added | ARG | @VilleSalo I decided to post it as a separate question because the status of this "lemma 2" should really have been settled a long time ago (in the 70's if not even before). | |
| Mar 15, 2021 at 8:46 | comment | added | ARG | @VilleSalo So there is clearly one thing which does not go through for the monoid of $2\times2$ matrices. Namely look at $\big( \begin{smallmatrix} m_{11} & m_{12} \\ m_{21} & m_{22} \end{smallmatrix} \big)$ then, even if the support of these elements is contained in $B_i$, by multiplying with a simple matrix like $m=\big(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix} \big) $ one gets that the resulting elements have support in $(B_i)^2$ and not $B_i S$ for a set $S$ which only depends on the element $m$. So I'm absolutely sure my arguments fails dramatically for this monoid. | |
| Mar 15, 2021 at 6:01 | comment | added | Ville Salo | Quite possibly I just have a point in every open set, but I'm not really seeing these. One overarching problem is that everything you say seems to work just as well for the monoid of $2$-by-$2$-matrices over this group ring, and I know that one is not amenable. | |
| Mar 14, 2021 at 22:15 | comment | added | ARG | @VilleSalo Come to think of it, there might be another variant which is easier to compute. Let $\Pi_{k,n}$ to be the elements which can be written as products of the $k$ first in your enumeration, with at most $n$ uses of each of these elements. Then let $\Phi_{i,k,n}$ to be the translates of these elements so that the support lies in $B_i$. Then $|\Phi_{i,k,n} \cap \Phi_{i,k,n} m| \geq |$all those products with at most $n-1$ appearances of $m|$ and the boundary effect is also negligible. | |
| Mar 14, 2021 at 21:32 | comment | added | ARG | Indeed the estimates are quite hard to go through ... (and I do not claim to have done them). I'd be really surprised if $F_{i,k}$ would coincide with $F_i$ (especially since $i >> k$). But if the boundary effect is what bothers you here is another variant. Take $F_{i,j,k}$ to be the all the translates (in the sense of the usual action/embedding of $\Gamma$ in $R[\Gamma]$) of the elements of $F_{j,k}$ so that the support still lies in $B_i$. If $i >> j$ and $j$ is much larger than the support of any of the first $k$ elements in your enumeration, it "should" be fine. | |
| Mar 14, 2021 at 19:01 | comment | added | Ville Salo | I don't understand Lemma 2. For instance if $F_{i,k}$ happens to be just the set of all elements with support contained in $B_i$ (is that possible?) then it seems the "boundary effect" is quite big: almost all elements will have something on the support (even if only a small proportion of the "entropy" is generated there). | |
| Mar 12, 2021 at 11:37 | history | answered | ARG | CC BY-SA 4.0 |