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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).
6
votes
Accepted
Embedding theory for contractible manifolds
(This is by far not a complete answer, just an example.) In dimension 4, a paper of Livingstone (build on previous work of Lickorish) constructs some (compact with boundary) contractible 4-manifold wh …
3
votes
Accepted
Geometry of a manifold after Dehn filling, in terms of geometry pre-filling
The general principle is that a generic filling of a geometric manifold belongs to the same geometry of the original manifold. This holds notably in hyperbolic geometry by Thurston's Dehn filling theo …
5
votes
Accepted
Are Seifert fibered spaces with a horizontal surface exactly the surface bundles over the ci...
This is not quite true, because after splitting along the horizontal surface you may get interval bundles over non-orientable surfaces, and these are not products (I am assuming your 3-manifold is ori …
9
votes
Accepted
embeddings of graphs into surfaces
The answer in general is no. One obstruction is that a 4-valent vertex has three local resolutions as a pair of trivalent vertices, and only two of them may be realized in a given surface.
Edit I hav …
5
votes
Some mid-sized ¿hyperbolic? manifolds and SnapPea
That's just a long comment about tri13. I don't have Regina now, but you can get some informations by looking only at the 1-skeleton $G$ of the closed triangulation for tri13. Among the (experimentall …
13
votes
Is the topological concept of collapsible useful?
There is a simple reason for appreciating collapsible objects: a collapsible (PL) n-manifold is always (PL) homeomorphic to a disc! (Although a contractible one may not, for instance in dimension 4.) …
6
votes
Accepted
a special type of 2 component link complement
I suppose that you are asking whether such links are determined by their complement. If this is the case, the answer is yes and it is a consequence of Gordon-Luecke's "knots are determinent by their c …
8
votes
Accepted
whether a kind of surgery can go on infinitely many steps?
The answer is "no", although it seems that a homological argument is not enough as Kevin's and Bin's examples show. I describe an argument which uses geometrization.
There is a quantity which decreas …
2
votes
Accepted
Twisting equivalent links and the isotopy type of the resulting links
The result does not depend on the diagram, because the twisting operation does not depend on the disc $D$ with $\partial D = K$ that you choose. You can see this $n$-th twist as a self-diffeomorphism …
4
votes
Accepted
Can one construct a regular neighborhood without an ambient space?
One might think of the abstract regular neighbourhood as a canonical way of thickening $K$ to a handle decomposition of sufficiently high dimension.
If $\dim K = 1$, then $K$ has a canonical $n$-dim …
3
votes
Non-collapsible complexes
If you either collapse or de-collapse (the term expand is used here) a 2-dimensional polyhedron that contains a Bing's house, using only strata of dimension <=2, you end up with another 2-dimensiona …
9
votes
Accepted
Existence of fibered surfaces in arbitrary 4-manifolds?
I suppose that, for a $n$-manifold $M$, containing a "fibered codimension-2 manifold" $N\subset M$ means that $N$ has trivial normal neighbourhood $\nu N = N\times D^2$ and its complement $M \setminus …
20
votes
fundamental group and complete invariant of irreducible 3-manifolds
Perelman has proved Thurston's geometrization conjecture, which says that every irreducible 3-manifold decomposes along its canonical decomposition along tori into pieces, each admitting a geometric s …
7
votes
Compelling evidence that two basepoints are better than one
If you study the set of ideal triangulations of a fixed punctured surface you find out that:
("generators") by using sequences of flips you can relate any pair of triangulations,
("relators") two su …
4
votes
Accepted
How to distinguish Pretzel links with the same coefficients?
Richard Bedient has proved in 1984 that two Pretzel knots $P$ and $\sigma P$ are equivalent if and only if $\sigma$ is a cyclic permutation, an order reversing permutation, or a composition of both. H …