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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).
67
votes
Accepted
Poincaré Conjecture and the Shape of the Universe
In Einstein's theory of General Relativity, the universe is a 4-manifold that might well be fibered by 3-dimensional time slices. If a particular spacetime that doesn't have such a fibration, then it …
47
votes
Can every manifold be given an analytic structure?
There is an amazing theorem of Morrey and Grauert that says that not only does every (paracompact) smooth manifold have a real analytic structure, the real analytic structure is unique. Using Whitney …
37
votes
Accepted
What (if anything) happened to Intersection Homology?
Intersection homology and cohomology are still around, but as a topic they have just substantially been renamed. They are part of the theory of perverse sheaves, which are widely used in the Langland …
32
votes
Accepted
Why is Casson's invariant worth studying?
Wikipedia's description of the Casson invariant gives the first important reason to study it. As an invariant that comes from the $\text{SU}(2)$ representation variety of $\pi_1(M)$, it reveals in pa …
29
votes
Accepted
Manifolds are paracompact
Theorem: A countable atlas of charts for a Hausdorff $n$-manifold $M$ can be refined to a locally finite atlas. In fact, each chart only needs to be trimmed.
Proof: Let $U_1,U_2,\ldots$ be the chart …
24
votes
Are there any very hard unknots?
There are really two questions here: (1) Can you an untangle any unknot with relatively little work, say a polynomial number of geometric moves of some kind? (2) Given a knot, can you quickly figure …
24
votes
Accepted
Homotopies of triangulations
The standard name for this type of relation between two structures on $X$ is concordance rather than homotopy. If two structures on $X$ are isotopic (with the respect to the appropriate homeomorphism …
22
votes
Accepted
Riemannian metrics on non-paracompact manifolds
On the contrary, the long line does not have a Riemannian metric. Every countable subset of the long line has a least upper bound, so if it were Riemannian then a geodesic ray in the long direction w …
21
votes
Are there piecewise-linear unknots that are not metrically unknottable?
There is a survey paper on this general topic by Robert Connelly and Erik Demaine. As David Eppstein just posted, the answer is yes in 3D. However, it is a famous result of those two authors and Gun …
17
votes
Accepted
Embeddings of $S^2$ in $\mathbb{CP}^2$
The conjecture that every $S^2 \subseteq \mathbb{C}P^2$ is standard if it is homologous to flat is implied by the smooth Poincaré conjecture in 4 dimensions. It also implies a special of smooth Poinc …
17
votes
Intuition behind Alexander duality
I read Alexander duality as saying "If you have any type of doughnut in a sphere, then the outside must have handles or islands that fill the doughnut's holes." The proof that Ryan outlines exactly m …
15
votes
Classification of homology 3-spheres?
On the other hand, there is no particular classification of hyperbolic homology 3-spheres, much less hyperbolic links in homology 3-spheres, other than in general terms that they all come from hyperbo …
12
votes
Singular chains in Spivak's Calculus on Manifolds
It's clearly an elementary oversight that, on the other hand, doesn't matter for the real development of the material. Yes, the chain $c$ ought to be at least $C^1$ for the pullback to be continuous …
11
votes
Rugged manifold
To expand on Ryan's comment: A handle decomposition of a manifold is a restricted type of quotient of a disjoint union of handles. An $n$-dimensional $k$-handle is by definition the manifold $B^k \t …
10
votes
Utility of virtual knot theory?
My view of virtual knot theory: An R-matrix is a tensor with four indices that, in a natural sense, satisfies the Reidmeister relations. Actually it is enough to consider the third Reidemeister rela …