All Questions
29 questions
56
votes
2
answers
3k
views
How to add essentially new knots to the universe?
A knot is an embedding of a circle $S^{1}$ in $3$-dimensional Euclidean space, $\mathbb{R}^3$. Knots are considered equivalent under ambient isotopy. There are two different types of knots, tame and ...
25
votes
2
answers
844
views
Which homotopy classes $S^3 \to S^2$ lift to embeddings $S^3 \to S^2 \times D^3$?
The question is, for a smooth embedding
$$f : S^3 \to S^2 \times D^3$$
one can compose the map $f$ with projection $\pi : S^2 \times D^3 \to S^2$, giving the map $\pi \circ f : S^3 \to S^2$.
Which ...
- 44.4k
24
votes
5
answers
3k
views
Can surfaces be interestingly knotted in five-dimensional space?
It's possible this question is trivial, in which case it will be answered quickly. In any case, I realized that it's a basic question the answer to which I should know but do not.
Everybody loves ...
- 54.6k
15
votes
2
answers
968
views
Semidirect product decomposition of the Borromean rings group
Let $X=S^3\setminus B$ be the link complement of the Borromean rings.
(source)
Then $G=\pi_1(X)$ has a presentation of the form
$$
G = \langle \; a,b,c \mid [a,[b^{-1},c]],\; ...
- 35.9k
12
votes
2
answers
2k
views
The Alexander polynomial of a slice knot, Reidemeister tosion, Whitehead group
My question is about the Alexander polynomial of a slice knot.
For a slice knot $K$,
Fox-Milnor and Terasaka proved that
$$ \Delta_{K}(t) \doteq f(t) f(t^{-1})$$
for some polynomial $f(t) \in \...
- 391
11
votes
4
answers
1k
views
Knot diagrams, sets of moves and equivalence relations
Short version: Does anyone study equivalence classes generated by a given set of "moves" (in the sense of, but not limited to, Reidemeister moves) on the set of knot diagrams?
Yes, I ...
- 17.6k
10
votes
3
answers
843
views
Invariants of high-dimensional knots
In the study of knots in three dimensions, it can be shown that the fundamental group together with a specification of a meridian and longitude form a complete invariant for knots. What is known about ...
- 1,025
8
votes
2
answers
518
views
High-dimensional ribbon knots
Let us suppose that we have a ribbon embedding $S^n \rightarrow S^{n+2}$ for $n\geq 3$. Call this knot $K$. By a theorem of Levine (and Trotter for $n=3$ I believe) we know $K$ is unknotted if the ...
- 1,025
7
votes
1
answer
640
views
Length of shortest possible knot
Consider a line L in R^3 in the shape of a trefoil knot. Consider the surface S that is the union of all unit circles that have centers on this line and whose tangent vectors are all perpendicular to ...
- 453
7
votes
1
answer
448
views
Computing $\pi_2$ of the complement of a 2-knot or spaces with aspherical splittings?
As far as I know, it is an open question if the complement of a ribbon disk $D^2 \subset B^4$ is aspherical. In reading "Some remarks on a problem of J.H.C Whitehead" by Howie, it is noted that there ...
- 5,349
6
votes
3
answers
555
views
Given a link $L\subset S^3$ how to construct a link $L'$ whose complement have hyperbolic structure?
Thurston claimed that almost all closed 3-manifolds are hyperbolic. To support this, he said that every closed 3-manifold is obtained by Dehn surgery along some link whose complement is hyperbolic. ...
- 3,751
6
votes
1
answer
324
views
Computation of $\pi_1$ for a Mazur manifold and its boundary
If we attach a $4$-dimensional $1$-handle $D^1 \times D^3$ to a $4$-dimensional $0$-handle $B^4$, we obtain $S^1 \times D^3$. The null homologous knot in $S^1 \times S^2$ indicated in the picture ...
- 606
6
votes
0
answers
211
views
$\mathbb{Z}/2\mathbb{Z}$ coefficients in gysin sequence
I am reading the article "Signature of links" by Kauffman and Taylor. Here they show that it is possible to calculate the nullity of a link $L\subset S^3$ by knowing the first betti number of the ...
- 521
5
votes
1
answer
285
views
unlinking when relaxing the homeomorphism condition
Say that we have two knots $K_1$ and $K_2$ in $S^3$ linked together in $S^3$ and forming the Hopf link. Usually, we can prove that we cannot unlink them by using a link invariant that shows that the &...
- 504
5
votes
2
answers
463
views
Zero surgery on a Seifert fiber space
I have a problem with understanding what is a neighbourhood of a singular fiber in a Seifert fibered space coming from the zero surgery. For me a 3-manifold $Y$ is a SFS if it has a decomposition into ...
- 152
4
votes
2
answers
374
views
Knot theory in handlebodies of arbitrary genus
It is well known that not all graphs embed on the plane (e.g. the graph $K_{3,3}$). However, every finite graph embeds into a surface of some genus. One can think of this procedure as starting with a ...
4
votes
2
answers
360
views
Chirality and Anti-Chirality of links in 3 and in 5 dimensions
We know there is a chiral knot which is a knot that is not equivalent to its mirror image. It is well known in the mathematical field of knot theory:
https://en.wikipedia.org/wiki/Chiral_knot
My ...
- 10.5k
4
votes
1
answer
264
views
Invariance of Khovanov homology under first Reidemester move
I am studying Khovanov homology from five lectures on Khovanov homology
and I want to try to show Khovanov homology is invariant under first Reidemester move but I cannot understand how we can write
...
- 141
3
votes
2
answers
409
views
Alexander duality and homology equivalence
While reading the paper of Kauffman and Taylor "Signature of links" I found the following situation.
In the proof of Theorem 2.6 they suppose that two links $L_1, L_2\in \mathbb{S}^3$ are ...
- 521
3
votes
0
answers
75
views
Approximative extension of the autohomeomorphism of the complement of the trivial knot?
Let $S^1\subset \mathbb{R}^3$ be the unit circle and suppose $h\colon \mathbb{R}^3\setminus S^1\to \mathbb{R}^3\setminus S^1$ is a homeomorphism. Clearly it might be that $h$ cannot be extended to $S^...
- 396
3
votes
0
answers
173
views
More questions about high-dimensional knot invariants
In a question yesterday I asked about the existence of algebraic invariants for embeddings of n-manifolds into n+2-spheres. The answers in the positive dimension all made certain assumptions about ...
- 1,025
2
votes
1
answer
130
views
Space of the trivial long knot in the thickened surface
Let $F$ be a compact oriented surface and $x_0\in F$ a basepoint. Consider the set $\mathcal E=Emb_0(I,F\times I)$ of embeddings $\sigma\colon I\hookrightarrow F\times I$, $\sigma(\partial I)=\{x_0\}\...
- 357
1
vote
1
answer
279
views
Ways to prove that $n$-component Brunnian link is nontrivial
The attached image shows a way to construct an $n$-component Brunnian link for any $n\geq 3$. That is, this link is not trivial, but deleting any of its components makes the new link trivial. The ...
- 214
1
vote
1
answer
255
views
Questions about a few terminologies in "Knots and Links" by Rolfsen
In "Knots and Links" by Rolfsen, he mentioned words like *"the collar of a boundary", "bicollared boundary", "a bicollar on the boundary". I just wonder what ...
- 15
1
vote
1
answer
326
views
Addition of two homology classes is zero in construction of Poincare Sphere
I ask here the question since it hasn't been answered in
Math Stack Exchange.
I am working through Greenberg and Harper, Lecture notes on Algebraic Topology, and I am having trouble with one ...
- 909
1
vote
1
answer
2k
views
Homology and homotopy type for knot complements
I'm trying to understand a paper by Kawauchi and Matumoto, 'An estimate of infinite cyclic coverings and knot theory.' In one of their proofs they have a manifold $E$ which is the complement of a ...
- 1,025
1
vote
0
answers
160
views
Higher dimensional Seifert surfaces and link numbers of higher knots
In 3-manifold topology, the notion of Seifert surface is well known. It is then used to define link numbers of knots.
Question: Consider embedding $N^n \rightarrow M^{2n+1}$ of n-dimensional manifold $...
- 593
0
votes
1
answer
205
views
Space-time trajectory that cannot be straightened and its braid form
Considering we have the space-time trajectory of multiple particles (or any objects) in the X-Y-Time coordination system. Given a projection direction, we can obtain the braid form of the space-time ...
- 73
0
votes
0
answers
119
views
Nullity of the linking matrix of a framed link $L$ equals the first betti number of the manifold obtained by surgery on $L$
I have asked this on mathstackexchange as well. I'm not necessarily asking for a proof, just a hint or a point to the right direction (although I'm not saying that a proof isn't welcome). I'm studying ...
- 101