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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
Why almost every geodesic arc is generic?
The fact that the union of all simple geodesics has Hausdorff dimension 1 is a classic result of Joan Birman and Caroline Series
Birman, Joan S., and Caroline Series. "Geodesics with bounded intersec …
3
votes
Is a generic closed orientable hyperbolic 3-manifold Haken?
Firstly, the random Heegard splitting model and the random fibering model are discussed at length in my preprint. (I believe the OP is familiar with it). I believe the question is asked there, though …
2
votes
Maximum order of MCG finite order elements
The answers to all the questions can be found in that fond of wisdom Farb-Margalit:
Farb, Benson; Margalit, Dan, A primer on mapping class groups, Princeton Mathematical Series. Princeton, NJ: Princ …
4
votes
Regarding the Thurston norm and the ways that a three-manifold can fiber over the circle
See Autumn Kent's answer to this question.
0
votes
Action of U(1) on a sphere bundle, non-vanishing vector fields on odd-dimensional manifolds
A very closely related question is discussed here:
nowhere vanishing vector field on a manifold
22
votes
Accepted
Which closed orientable $4$-dimensional manifolds cannot be embedded in $6$-space?
This is true if and only if $X^4$ is spin and its signature vanishes. This is on p. 345 in Gompf/Stipsicz (4-manifolds and Kirby calculus), who cite Ruberman: Imbedding four-manifold and slicing links …
2
votes
If a polyhedron is homeomorphic to a simplex, is it piecewise-linear homeomorphic?
I am not sure if this is what you are asking, but check out http://en.wikipedia.org/wiki/Exotic_R4 (note that in dimension four, PL is the same as smooth).
3
votes
Sufficient conditions for a 3D tetrahedral complex to be homeomorphic to a 3D ball
First you check that your complex is a PL manifold with boundary (this is easy, the hardest part is checking that the links of interior vertices are 2-spheres (which is an euler characteristic argumen …
8
votes
extension of surface homeomorphism
Really a long comment on @Agol's answer:
One does not actually need the fact (which I was not aware of) to show what the OP wants from what @Agol says, since by Casson-Bleiler, any mapping class who …
2
votes
extension of surface homeomorphism
I accidentally found this Rice University thesis (never published, it seems) which studies the question in some depth:
Author Jamie Bradley Jorgensen
Title Surface homeomorphisms that do not extend t …
25
votes
Thurston's "tinker toy" problem
The result comes by way of Nash's theorem which states that every smooth manifold is a component of a real algebraic variety.
Nash, John, Real algebraic manifolds, Ann. Math. (2) 56, 405-421 (1952). …
5
votes
Accepted
Counterexample to high dimensional Nielsen realization problem
This is discussed at great length (with copious references) in Jonathan Block's and Shmuel Weinberger's paper. The title of the paper is the suggestive "On the generalized Nielsen realization problem" …
8
votes
regular tiling of a surface of genus 2 by heptagons
A picture is in the comments to John Baez' Blog Post. EDIT For the second part of the question, no there is no obstruction, as long as the implied angle ($2\pi/v$) is smaller than the angle of a Eucli …
6
votes
Determine if an $n$-dimensional mesh of simplices is a non-manifold
In all dimensions, something is a manifold if the link of every cell is a sphere. This, sadly, is undecidable if the dimension of your complex is at least five. It is decidable (but not quickly) for c …
1
vote
homotopy groups of an orbifold
As a topological space, this is homotopy equivalent to $\mathbb{D}^3,$ so the homotopy groups are whatever they are for $\mathbb{D}^3.$ As an orbifold, the fundamental group is $\mathbb{PSL}(2, \mathb …