28
votes
Accepted
Can I wrap a suitcase with hair ties
This configuration should work:
Edit (to provide credit/context): Michael Freedman's solution (see Ian Agol's post) is the original one. Ian directed me to this problem and gave me the hint that ...
- 376
14
votes
Accepted
Unlinked interlocking planar polygons
As Sam Hopkins commented, 8 vertices are enough. Let $Q$ be the pentagon from the picture and let $\pi$ be the plane containing it. Now we can define the triangle $P$ as a triangle of less diameter ...
- 10.6k
13
votes
Does there exist a discrete gauge theory as a TQFT detecting the figure-8 knot?
Short answer: Untwisted Dijkgraaf-Witten theories with non-abelian gauge groups (e.g. $S_3$) distinguish most knots from the unknot and from each other.
Longer answer:
Here's my understanding of ...
- 12.8k
11
votes
Can I wrap a suitcase with hair ties
I shared this question with Michael Freedman who came up with a solution similar to Larsen’s and asked me to post on his behalf.
He made a physical realization too.
I’ll address Anton’s question of ...
10
votes
Accepted
Infinitely many Brunnian links with bounded crossings
There are simpler diagrams of Brunnian links that have simple limits.
If you make an infinite chain of the C-shaped components, say periodically, then removing any one will let you ambient-isotope ...
- 28k
7
votes
Accepted
Amenable link groups
No. Suppose that $K$ is a non-trivial knot. Then its knot genus is at least one. Thus $X = S^3 - K$ contains a $\pi_1$-essential surface (with boundary) of genus at least one. Thus $\pi_1(X)$ ...
- 28.1k
7
votes
In knot theory, what is this link property and how to detect it: "linkings between components separate nicely"
Nice question--I'm not sure if this already has a name.
Here is one way to show that a link is not a necklace, that applies to the Borromean rings. Suppose that the link has components $(R,G,B)$, and ...
- 19.4k
7
votes
Is there a "simplest" way to embed a graph in 3-space?
I agree with the comments that there will be no simplest embedding in general. Nevertheless in certain cases there are embeddings with nicer properties.
One way to embed a simplicial graph (ie no ...
- 68.8k
6
votes
Tangled random triangles: One giant component?
Here is @fedja's clever example: The magenta triangle is topologically loose
but metrically "stuck":
fedja: "(link two large ....
6
votes
Accepted
Revisiting Gordon-Luecke theorem
I think that the correct theorem should be the following:
If $L$ does not contain any split unknot or any two coaxial components, the image of the map is finite
Following Cameron Gordon, two coaxial ...
- 10.5k
6
votes
Unlinked interlocking planar polygons
It is not possible with 7 (i.e., with a triangle $T$ and a quadrilateral $Q$). I write a rough proof.
First, any quadrilateral $Q$ lying in a plane $\pi$ can be partitioned in two triangles $Q_1$ and $...
- 380
6
votes
Accepted
How to prove the product of Whitehead manifold and $\mathbb{R}$ is homeomorphic to $\mathbb{R}^4$?
I would suggest a classical proof showing that the one-point compactification of $W$ is a manifold factor. See Wild wild whitehead manifold. The proof is originally due to J. Andrews and L. Rubin, ...
- 2,083
6
votes
Accepted
Are all Torus Links in fact Lorenz links or not?
The point is that the two papers use slightly different definitions of "Lorenz Links".
The newer paper defines Lorenz links as links on the Lorenz template. With this definition all torus links are ...
- 10.4k
5
votes
Accepted
Wrapping a suitcase with large rotational symmetry
Edit: my previous answer was incorrect
No, this one you cannot do. If one had such an arrangement of circles in the solid torus inside $R^3$, and quotiented by a rotation, then by the equivariant Dehn'...
- 68.8k
5
votes
Do triple-linked graphs exist?
If you restrict to straight-line embeddings (where edges are line segments), then the answer is yes: using the result in Erdos-Szekeres in high dimensions there exists some $n$ such that if you have $...
- 12.4k
5
votes
Accepted
Knotted concordances of slice links
I think this is likely an unknown question. Namely, the negation of 3) would follow from 1) and 2) if
strongly slice links are strongly ribbon (which seems to be open)
ribbon disks bounding the ...
- 68.8k
5
votes
Accepted
Is there a "simplest" way to embed a graph in 3-space?
Another way to generate a topologically simple embedding into $\mathbb{R}^3$ is to embed the graph in a surface of minimal genus which in turn embeds into $\mathbb{R}^3$. One can generate the ...
- 687
5
votes
Accepted
Why is this Brieskorn manifold a rational homology sphere?
The key is pointed out by HJRW in his comment: there's a missing piece in your explanation, which is the genus of the base $B$ of the Seifert fibration. Némethi writes:
$$
2g-2 = (n−2)A/a−\sum q_i,
$$
...
- 10.9k
4
votes
Unlinked interlocking planar polygons
Here is another example with 8 vertices: a short fat Star Trek symbol and a square in orthogonal planes.
Since the distance between the base points of the red figure is greater than its height, one ...
- 7,149
4
votes
Accepted
Links and non-orientable surfaces
Yes, the surface is orientable. To simplify the LaTex and the exposition, I will change the notation and setting a small amount.
Suppose that $F$ is a compact connected embedded surface in three-...
- 28.1k
4
votes
Accepted
Determine if a closed braid is a link/unlink
A braid gives a braid closure. This can be drawn as a knot (or link) diagram. There are then various approaches to solve the unknot (or unlink) recognition problem given a diagram. This begins with ...
- 28.1k
3
votes
An equivalence relation on knots similar to concordance
As far as I understand your question, the answer is "no". Consider the Whitehead link.
https://en.wikipedia.org/wiki/Whitehead_link
The two components of the Whitehead link are unknots (...
- 28.1k
3
votes
Unlinked interlocking planar polygons
I think that 8 might be possible, by interlocking two Star Trek symbols as shown below.
Adendum: This candidate may not work, as quarague points out, but I leave it as a potential "how not to&...
- 7,149
3
votes
Accepted
Embedding linklessly embeddable graphs without Borromean rings
Isn't it that linklessly and flatly embeddable are the same family, and that a flat embedding can not contain a Borromean ring?
upd - clarification:
From the wiki article on linkless embeddings: "...
3
votes
Accepted
Reference request: Can iterated torus links be mutated?
I don't think that it's that difficult to deduce this for prime iterated cable links. The point being that the preimage of a Conway sphere is an essential torus, so must be isotopic into a Seifert ...
- 68.8k
3
votes
Links defined by link-severance tableau
If I'm reading your notation correctly, the symbol $\{l_1, l_2\}$ means the two element link with components labelled $l_1, l_2$ together with the additional information that they are unlinked? ...
- 44.3k
3
votes
Do triple-linked graphs exist?
I have since come across the following paper which seems to answer the question affirmatively in a very strong sense:
E. Flapan, B. Mellor, R. Naimi, "Intrinsic linking and knotting are ...
- 13.6k
3
votes
Accepted
A uniform upper bound for the linking number of periodic orbits of algebraic vector fields
The Lorenz equations are quadratic, and already have an infinite number of distinct knotted and linked orbits. An answer to one of your other questions has good references on Lorenz orbits, but I also ...
- 6,571
2
votes
Accepted
Link invariants distinguishing components
The multivariable Alexander polynomial has the potential to do this. It has one variable $t_i$ for each link component, and if there is an isotopy interchanging the $i^{th}$ and $j^{th}$ component ...
- 19.4k
2
votes
Non-zero winding number on a space curve implies a linked curve in the zero set?
Now the statement is indeed correct. What follows below is and answer to the previous version of the question which I'll keep for the moment.
Note that the condition for $f(C)$ to have non-zero ...
- 28.9k
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