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Dec
18 |
awarded | Good Answer |
Oct
16 |
awarded | Yearling |
Jul
30 |
comment |
Surgery along an arc connecting the components of a $2$-component link gives the unknot
It's certainly used in the paper by Bao. I didn't see how the other papers use it to prove their results about band sums, although Eudave-Muñoz does invoke it afterward to apply his result to the cabling conjecture. |
Jul
30 |
answered | Surgery along an arc connecting the components of a $2$-component link gives the unknot |
Jul
25 |
comment |
contact manifolds dimension five
Done, unless you had any other developments in mind. |
Jul
25 |
revised |
contact manifolds dimension five
Added links to recent developments |
Jul
23 |
comment |
Legendrian knot in 3-sphere
@nikita: The adjunction inequality says that $tb(L)\le 2g_*(L)-1$, so equality is still possible, but either way if you have a symplectic surface then it actually satisfies an adjunction formula $\langle c_1(\omega), \Sigma\rangle + \Sigma\cdot\Sigma = 2g(\Sigma)-2$. In response to your second comment, if the surface is symplectic then its branched double cover is naturally a symplectic manifold and so the contact structure which comes from taking the branched double cover of the transverse knot in S^3 is in fact symplectically fillable. |
Jul
23 |
answered | Legendrian knot in 3-sphere |
Jul
9 |
comment |
Concrete examples of covering from the 3-torus to the 3-sphere
The 3-torus can't be a branched double cover because the triple cup product on $H^1$ would then be zero. |
Feb
6 |
comment |
Regularity of the taut foliation
See "Approximating C^0-foliations" by Kazez and Roberts (arxiv.org/abs/1404.5919), which addresses exactly this issue. |
Jan
12 |
answered | Are there spaces in which there are no fibered knots? |
Oct
16 |
awarded | Yearling |
Jul
8 |
answered | Do Heegaard Floer homology detect fibred knot in general oriented 3-manifold? |
Jun
13 |
comment |
Heegaard Floer Homology of double branched cover
There are actually lots of non-QA knots whose branched double covers are L-spaces. See section 6.1 of arxiv.org/pdf/1205.5261.pdf for some discussion and explicit examples, including the $P(p_1,\dots,p_n,-q) $ pretzel knots where $p_i,q>0$ and $q = \min(p_1,\dots,p_n)$. |
Mar
26 |
awarded | Guru |
Dec
17 |
awarded | Enlightened |
Dec
17 |
awarded | Nice Answer |
Nov
21 |
comment |
Link surgery on $S^2\times S^1$
For n=2 you do have a product link. You can see it by writing the pages $(S^1\times I)$ of the open book collectively as $(S^1 \times I) \times S^1$, and then gluing in two solid tori $S^1\times D^2$ by using the disks $\{*\} \times D^2$ to cap off the annuli $(\{*\} \times I) \times S^1$ (which consist of one arc from each page). Thus in the surgered manifold the annuli $(\{*\}\times I)\times S^1$ can be closed up to form spheres $\{*\}\times S^2$, one for each point of $S^1$, and the link $L_2$ contributes a pair of points to each of those spheres. |
Nov
21 |
comment |
Link surgery on $S^2\times S^1$
The argument only works for $n>0$ because when $n=0$ it isn't really an open book decomposition anymore, and then you really do have $S^2\times S^1$. You can check the n=2 case directly by Kirby calculus, though: it's 0-surgery on an unknot and on two of its meridians, and if you blow up one of those meridians you get a chain of unknots with surgery coefficients -1,-1,0,0. Blow down the second to get a chain with coefficients 0,1,0, and then the middle one, and then either of the remaining ones and you're left with a single 0-framed unknot, which gives $S^1\times S^2$. |
Nov
21 |
answered | Link surgery on $S^2\times S^1$ |