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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).
3
votes
Fundamental group of the complement of a codimension two submanifold
To your first question, the answer is yes.
Take a $k$-component trivial link in $S^n$, i.e. the boring, linear embedding
$$\sqcup_k S^{n-2} \to S^n$$
that is the boundary of a linear embedding
$$\sqcu …
2
votes
Handle decompositions subordinate to an open cover
If you call the standard $n$-simplex $\Delta^n$, i.e.
$$\Delta^n = \{ (x_0, x_1, \cdots, x_n) : x_i \geq 0 \forall i, \sum_i x_i = 1\}$$
then the function
$\phi : \Delta^n \to \mathbb R$
given by $\ph …
5
votes
Equivalence of knotted spheres in $S^4$
$\DeclareMathOperator{\Diff}{Diff}$ The answer is "yes modulo some small potatoes".
There is one case where the answer is a simple no: if $K$ and $K'$ are mirror images of each other you can have a di …
2
votes
Accepted
Numerical computation of the second Vassiliev invariant, and the permutation $(1 3 4 2)$
This is not a full answer to your question, but it gives some information.
The expression (1) is an integral version of the Polyak-Viro formula for the type-2 invariant, described here: https://academ …
9
votes
Why/does 'low-dimension' topology end with dimension 4?
There are of course ways to construct difficult problems in high-dimensional manifold theory. One of the ways high-dimensional manifold theory differs strongly from low-dimensional manifold theory i …
14
votes
Accepted
Isotopic diffeomorphisms of the sphere
This is Cerf's pseudoisotopy-implies-isotopy theorem.
Cerf's result is true in high dimensions, while it's independently known in a low-dimensional range. In dimension $2$ it goes back to the Earle–E …
3
votes
Lie group framing and framed bordism
Lie group framing is a reference to the group action. A Lie group $G$ acts on the left of $G$ by the map
$$(g,h) \longmapsto gh.$$
Similarly there are actions on the right, and conjugation actions, e …
9
votes
Generalization of the sphere theorem in dimension at least 4
I have a thread on the co-dimension $1$ generalization of Dehn's Lemma for 4-manifolds:
A problem/conjecture related to 4-manifolds that deserves a name. What name does it deserve?
My pair of papers w …
5
votes
Accepted
How to get a presentation of the mapping class group of the $n$-punctured sphere
$\DeclareMathOperator\Diff{Diff}\DeclareMathOperator\Emb{Emb}\DeclareMathOperator\fix{fix}$There's a variety of ways to do this. If you take "mapping class group" to mean "isotopy classes of diffeomo …
2
votes
Handle attachment information from Morse function and triangulation
Near a critical point $p$ of a genuine Morse function you have the local model
$$f(x) = f(p) + x_1^2 + \cdots + x_i^2 - x_{i+1}^2 - \cdots - x_n^2$$
Just below the height of $p$, your attaching sphere …
6
votes
Accepted
Is there a geometric interpretation of the second derivative of the Alexander polynomial at ...
Given a knot in $S^3$, think of it as an embedding
$$f : S^1 \to S^3.$$
The configuration space of $5$ distinct points in $S^3$ is denoted $C_5(S^3)$, this is a $15$-dimensional manifold and it consis …
4
votes
Accepted
Does every simply connected, orientable, non-compact, 3-manifold embed in $\mathbb{R}^3$?
When the manifold is the universal cover of a compact $3$-manifold $M$ (to begin with, lets say without boundary) then you construct the embedding by hands, using geometrization. In your question let …
2
votes
How to find the JSJ decomposition in the plumbing tree model of a graph manifold?
Given that all your ingredient manifolds are Seifert-fibered, your JSJ decomposition is going to be a subset of the separating tori in your plumbing construction, i.e. corresponding to your edges, if …
1
vote
applications of Sard's to differential topology
It's been a long time, but isn't your suggestion roughly Whitney's original approach to this problem?
I don't have Whitney's papers in front of me but this is roughly how I think his arguments went. …
7
votes
Accepted
Is a spin structure on a knot complement the same thing as an orientation of the knot?
Is the theorem true? There is an non-natural bijection. There is no natural bijection.
A link exterior is homotopy-equivalent to a $2$-complex, so a trivialization of the tangent bundle over the $2$ …