bio | website | |
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location | ||
age | ||
visits | member for | 5 years, 2 months |
seen | 9 mins ago | |
stats | profile views | 567 |
Mar 23 |
awarded | Enlightened |
Mar 23 |
awarded | Nice Answer |
Mar 13 |
answered | 2-bridge knots in the Rolfsen's table |
Mar 9 |
awarded | Nice Answer |
Jan 10 |
awarded | Yearling |
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 |