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visits | member for | 4 years, 3 months |
seen | yesterday | |
stats | profile views | 531 |
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". |
Feb 15 |
answered | Interesting mathematical documentaries |
Jan 10 |
awarded | Yearling |
Oct 15 |
accepted | Manifolds with prescribed fundamental group and finitely many trivial homotopy groups |
Oct 14 |
revised |
Manifolds with prescribed fundamental group and finitely many trivial homotopy groups
added 25 characters in body |
Oct 14 |
comment |
Manifolds with prescribed fundamental group and finitely many trivial homotopy groups
Sorry, despite my efforts to make sure I wrote "finitely presented" I ended up writing finitely generated. I'll edit the question. |
Oct 14 |
awarded | Nice Question |
Oct 12 |
asked | Manifolds with prescribed fundamental group and finitely many trivial homotopy groups |
Sep 28 |
comment |
Problems about the Estimate the Unknotting Number
Yes, all knots have unknotting number less than $\lfloor n/2 \rfloor$. Suppose that making $k > \lfloor n/2 \rfloor$ crossing flips results in the unknot, then you can check that flipping all unflipped crossings instead also does. This clearly requires $\leq \lfloor n/2 \rfloor$ flips and so the unknotting number is at most $\lfloor n/2 \rfloor$. An equivalent bound also holds for unlinking links. |
Sep 4 |
comment |
Determine if a matrix is unimodular
In fact, thanks to LU decomposition, computing the determinant is at least as fast as computing a matrix product. So we can compute the determinant exactly in $O(n^{2.376}). See en.wikipedia.org/wiki/LU_decomposition#Theoretical_complexity. Similarly, in Storjohann's paper "Near Optimal Algorithms for Computing Smith Normal Forms of Integer Matrices" he shows a similar inequality. That is, for an integer square matrix, computing its SNF is at least as fast as computing a matrix product. |
Sep 3 |
revised |
Determine if a matrix is unimodular
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