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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).
4
votes
Translation distance in the curve complex
In the braid group, Ko and Lee have given a polynomial time test of reducibility using the Garside structure. (See http://arxiv.org/abs/math/0610746)
10
votes
Accepted
Are compact submanifolds of "S X (0,1)" with one boundary component handlebodies, where S is...
No. N could be homeomorphic to the exterior of any knot in the 3-sphere (i.e. a cube with a knotted hole).
7
votes
Accepted
Stallings fibration theorem
This should follow from geometrization. The fundamental group of the manifold is $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}$, and geometrization should tell you that the manifold is then $\mathrm{RP}^ …
19
votes
Fibers of fibrations of a 3-manifold over $S^1$
As you suspect, the answer is no. There are $3$-manifolds that fiber over the circle in infinitely many ways, with fibers of unbounded genera.
Thurston constructed a seminorm on $H_2(M,\partial M; \ …
4
votes
Teichmuller theory and moduli of Riemann surfaces
I think a good example is Kerckhoff's solution to the Nielsen Realization Problem, which asks if every finite subgroup of the mapping class group is realized as a group of isometries of some hyperboli …
3
votes
Normal subgroups of fuchsian groups
This doesn't say much about the elliptic case, but there is Hempel's theorem that the normal closure of an element $\alpha$ contains a (power of a) simple curve $\beta$ if and only $\alpha$ is a (powe …
17
votes
Accepted
Homology generated by lifts of simple curves
As far as I know, this is open.
In fact, I think the following weaker question is open.
Let $\Theta$ be the set of loops $\gamma$ in $\widetilde \Sigma$ such that the image of $\gamma$ in $\Sigma$ …
10
votes
Higher homotopy groups of slice disk complement
To add to Ryan's answer,
$2$--knots usually don't have aspherical complements, see Dyer & Vasquez, The sphericity of higher dimensional knots, Canad. J. Math. 25(1973), 1132-1136. This suggests a com …
6
votes
Why should I care about Heegaard-Floer theory?
Kronheimer, Mrowka, Ozsváth, and Szabó obtained a new proof of Gordon and Luecke's Knot Complement Theorem using monopole Floer homology. That's pretty good, I think. They also proved that $\mathbb{ …
12
votes
Accepted
Topological rigidity of compact manifolds in dimension three
Yes.
When the manifolds are Haken this is a theorem of Waldhausen. See Ian Agol's answer here.
Since your manifolds are aspherical, they are irreducible by the Poincaré conjecture. Since they have …
7
votes
Degrees of self-maps of aspherical manifolds
For $K(\pi, 1)$'s, the answer is:
Because Euler characteristic is multiplicative under covering spaces.
Edit: As pointed out in the comments, I was assuming the map was $\pi_1$--injective.
Here' …
29
votes
Fundamental group of 3-manifold with boundary
No. The Baumslag--Solitar groups $\langle a, b | ab^m a^{-1} = b^n \rangle$ are not three-manifold groups when $m \neq n$.
See:
Heil, Wolfgang H. Some finitely presented non-$3$-manifold groups. Proc …
15
votes
Accepted
Word problem for fundamental group of submanifolds of the 4-sphere
Update:
My memory was quite blurry about this when I originally answered.
See Gonzáles-Acuña, Gordon, Simon, ``Unsolvable problems about higher-dimensional knots and related groups,'' L’Enseigneme …
10
votes
Accepted
Seifert surfaces of torus knots
There's the usual description of the Seifert surface for a general cable obtained by taking copies of a Seifert surface for the knot and a fiber for the cable in the solid torus. See Ken Baker's disc …
4
votes
A problem/conjecture related to 4-manifolds that deserves a name. What name does it deserve?
In dimension 3, you have the sphere theorem, the torus theorem, the annulus theorem, and the disk theorem (which is the loop theorem and Dehn's lemma put together).
So, if you didn't require the sphe …