Timeline for Gromov's pseudogroups and Tao's approximate groups
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Jan 12, 2018 at 5:21 | history | edited | Martin Sleziak | CC BY-SA 3.0 |
minor typos
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| Oct 21, 2016 at 9:48 | comment | added | Mikhail Katz | @NoahSchweber, Igor Belegradek's comment below my answer is a good illustration of what I have in mind. | |
| Oct 20, 2016 at 22:38 | comment | added | Noah Schweber | @mikhail based on what? Do you see a meaningful relation between them? | |
| Oct 20, 2016 at 16:54 | comment | added | Mikhail Katz | I think you may be underestimating the relation between the two notions. | |
| Oct 20, 2016 at 16:04 | comment | added | john mangual | yes now you get it. Gromov clearly had something in mind but his definition doesn't capture any essense. Another problem is that Tao-Green-Breullard require an ancient group. an approximate group lives inside another group (integers, permutations, matrices, etc). Gromov's definition is intrinsic. Any set with a sometimes-working binary operation will do, such as a collection of paths satisfying some length constraint. These can be embedded into the original fundamental group, maybe. | |
| Oct 20, 2016 at 15:55 | comment | added | Joël | Neither will I. I have no issue with your main point that the two notions are not related in an interesting way. In fact, Gromov's notion seems to me a very general, rather uninteresting, concept (which is not to say, of course, that Gromov doesn't do extremely interesting things with it). If you take any set $S$ with a marked point $e$, you can make it a pseudo-group in the sense of the definition you gave by defining $ae = ea =a$ and $a^2=e$ for any $a \in S$ and leaving all the other multiplications undefined. Clearly, this pseudo-group structure has no interest whatsoever. | |
| Oct 20, 2016 at 15:44 | comment | added | john mangual | the definition of approximate group emphasizes covering by translations eg this nilprogression $A$ of integers $\lfloor n \sqrt{2} \lfloor n \sqrt{3} \rfloor\rfloor $ for $-N < n < N$ is a 100-approximate subgroup since $A+A$ can be covered by 100 translates $A+ b$ for various integers $b$ - you are certainly welcome to show (or fail to show) that Gromov's definition and Breullard's are the same. I will not do so here. | |
| Oct 20, 2016 at 15:22 | comment | added | Joël | "These are closed under multiplication (when a multiplication exists) , while Approximate groups are definitely not closed". I am not really convinced by this argument. If you re-define the multiplication of an approximate group by saying "undefined" when the result is not in the approximate group, you transform your "closed but not defined everywhere" multiplication into a non-closed multiplication. | |
| Oct 20, 2016 at 14:54 | history | answered | john mangual | CC BY-SA 3.0 |