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Results tagged with gt.geometric-topology
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user 13119
Topology of cell complexes and manifolds, classification of manifolds (e.g. smoothing, surgery), low dimensional topology (e.g. knot theory, invariants of 4-manifolds), embedding theory, combinatorial and PL topology, geometric group theory, infinite dimensional topology (e.g. Hilbert cube manifolds, theory of retracts).
28
votes
Accepted
The Jones polynomial at specific values of $t$
The evaluation of the Jones polynomial at $e^{i\pi/3}$ is related to the number of 3-colourings $tri(K)$ of $K$ (see also here) as well as to the topology of the branched double cover $\Sigma(K)$:
$$t …
- 10.9k
22
votes
Manifold embedded in $R^{n+1}$ with a submanifold that doesn't embed in $R^n$
If you are interested in an example in codimension 2 examples (which also happens to involve two orientable manifolds), according to the table on page 2 in the survey
Davis, Donald M. Embeddings o …
- 10.9k
20
votes
Accepted
Does there exist a closed manifold with vanishing reduced rational cohomology but nonvanishi...
As mme noted in the comments, such examples cannot exist in odd dimensions, for Euler characteristic reasons. They can't exist in dimension 2 either, by classification. I claim that in all other dimen …
- 10.9k
20
votes
Accepted
Example of homeomorphism of $3$-manifolds
Here is a sequence of moves that gets you from left to right. (Pictures are done with Frenk Swenton's Kirby calculator.)
First, we convert from rational to integral surgery.
Then we blow up twice (th …
- 10.9k
19
votes
Accepted
Obtain 4-manifolds by repeating surgeries of submanifolds in $S^4$
If an $n$-dimensional smooth manifold $X'$ is obtained from $X$ by doing some surgeries, then there is an $(n+1)$-dimensional smooth cobordism $W$ from $X$ to $X'$; vice-versa, any handle decompositio …
- 10.9k
17
votes
Accepted
Can Khovanov homology have arbitrarily large torsion?
This paper from earlier this year (Jan 18, to be precise) proves the existence of $\mathbb{Z}/n\mathbb{Z}$-torsion for $n\le 8$ and $\mathbb{Z}/2^s\mathbb{Z}$-torsion for $s\le23$. It also states at t …
- 10.9k
17
votes
Accepted
4-dimensional cohomology $\mathbb{CP}^2$'s
No. If $\Sigma$ is any homology 4-sphere with non-trivial fundamental group, $\mathbb{CP}^2 \# \Sigma$ is a homology $\mathbb{CP}^2$ with non-trivial fundamental group. (Here $\#$ denotes connected su …
- 10.9k
13
votes
Accepted
Higher homotopy groups of irreducible 3-manifolds
An irreducible 3-manifold $M$ is aspherical if and only if it's not a finite quotient of $S^3$, which in turn is equivalent to having infinite fundamental group. Essentially you've already outlined th …
- 10.9k
12
votes
Accepted
Distance between two knots
The Gordian distance measures precisely the number of crossings you need to change to turn a knot into another.
MathWorld gives the reference:
Murakami, H. Some Metrics on Classical Knots, Math. Ann …
- 10.9k
12
votes
Behavior of genus function on a 4-manifold for sums
Sometimes the function $G$ can be constantly 0: consider the class $x = [S^2\times\{p\}]$ in $H_2(S^2\times F)$, where $F$ is a surface.
Then $G(nx)$ can be realised by an embedded sphere for all $n$: …
- 10.9k
11
votes
How to show whether a given knot and its mirror image are the same or not?
There are several ways to try and tell apart a knot from its mirror
Probably the most "classic" way to do this is using the signature;
You can compute Jones polynomial and check that it's not symmet …
- 10.9k
11
votes
Does every embedded 2-sphere in $\mathbb{R}^n$ bound an embedded ball?
Haefliger proved that $S^k$ in $\mathbb{R}^n$ is unknotted if $2n > 3(k+1)$. Therefore, any 2-sphere embedded in $\mathbb{R}^n$ with $n\ge 5$ is unknotted. On the other hand, as Mark Grant points out …
- 10.9k
10
votes
Is the Lisca-Matic bound (aka slice-Bennequin bound) strictly stronger than the Bennequin bo...
To answer the question in the title: yes, the slice Bennequin inequality is stronger than the Bennequin inequality. For instance, it shows that all slice knots have negative maximal self-linking numbe …
- 10.9k
10
votes
Accepted
Two surfaces in a 4-manifold whose algebraic intersection number is zero
Yes, this can be done by tubing one surface along the other.
Suppose that you have two intersection points $p_+, p_- \in \Sigma_1 \cap \Sigma_2$ of opposite signs. Suppose also that $\Sigma_1$ and $\S …
- 10.9k
10
votes
Presentations of exotic 4-manifolds
I guess that this is as explicit and low-tech as it gets: if $X$ is a K3 surface (i.e. a non-singular quartic hypersurface in $\mathbb{CP}^3$, with the complex orientation), then $X \# \overline{\math …
- 10.9k