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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
deleted 124 characters in body