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visits | member for | 4 years, 11 months |
seen | 14 hours ago | |
stats | profile views | 559 |
Dec 19 |
awarded | Necromancer |
Nov 5 |
comment |
Fantastic properties of Z/2Z
So I guess that the correct reference should be Gromov's "Asymptotic invariants of infinite groups" but it is ~300 pages and I haven't read it. Danny gave a talk at Cornell and in the first ~15 mins he covers the density model and gives a sketch of this result. It's available here: cornell.edu/video/danny-calegari-random-groups-diamonds-glass |
Nov 4 |
answered | Fantastic properties of Z/2Z |
Oct 13 |
awarded | Notable Question |
Sep 24 |
revised |
Is there a table of (fibred knot) monodromies?
Added link to more data. |
Jul 2 |
awarded | Curious |
May 9 |
comment |
Does small Perron-Frobenius eigenvalue imply small entries for integral matrices?
Thanks, Do you know of any reference for this result? |
May 9 |
accepted | Does small Perron-Frobenius eigenvalue imply small entries for integral matrices? |
May 9 |
asked | Does small Perron-Frobenius eigenvalue imply small entries for integral matrices? |
May 9 |
awarded | Informed |
Apr 1 |
awarded | Necromancer |
Feb 27 |
comment |
Is there a table of (fibred knot) monodromies?
For anyone looking for more data like this, significantly more data can now be found at: bitbucket.org/Mark_Bell/bundle-censuses/overview |
Feb 27 |
comment |
from Dehn twists to surgery diagram
Neil, because of the orientation of $b$, the left Dehn twist $d_\beta$ sends $a$ to $ab^{-1}$ not $ab$. Therefore its matrix in $SL(2, \mathbb{Z})$ is $(1, 0, -1, 1)$. Hence $w$ corresponds to $(0, 1, -1, 1)^6 = (1, 0, 0, 1)$. So $M$ should have Euclidean geometry. By using Alexander's trick on $a \cup b$ you can even check that $(d_\alpha d_\beta)^6 = id$ by hand. Or if you prefer, software packages such as Twister can build the bundle for you and you can check that it is $T^3$. |
Jan 10 |
awarded | Yearling |
Nov 7 |
awarded | Popular Question |
Oct 3 |
awarded | Caucus |
Aug 31 |
revised |
Is the normalised Kauffman bracket more powerful than the Kauffman bracket?
title clearer |
Aug 31 |
asked | Is the normalised Kauffman bracket more powerful than the Kauffman bracket? |
Apr 25 |
awarded | Nice Question |
Feb 28 |
comment |
Knot database including text names
Thanks for the suggestion. Just to let you know you can actually skip the volume filtering, Dr. Weeks built it directly into the snappea kernel! Line 90 of isometry.c defines "CRUDE_VOLUME_EPSILON" to be 0.01. Later, around line 140, when testing if two manifolds are isometric if they differ by more than this amount the entire calculation is aborted and the function returns "not isometric". |