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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).
1
vote
Accepted
linking number and covering
The kind of argument you'd see Rolfsen make in his textbook would be to consider how one constructs the abelian cover explicitly via the Seifert surface of the trivial knot. This is a disc. Ensure i …
4
votes
A smooth analog of the mapping cylinder?
Generally the answer is no. For example, the 3-dimensional lens spaces $L_{7,2}$ and $L_{7,1}$ are homotopy-equivalent but not diffeomorphic. Let $f : L_{7,2} \to L_{7,1}$ be the homotopy equivalenc …
7
votes
Accepted
Dehn filling of hyperbolic 3-manifolds and Gromov volume
Regarding (1) and (2), yes, it's very possible. Moreover, provided $N$ is hyperbolic, I believe Thurston proves the Gromov norm of $N_1$ is strictly smaller than $N$'s volume. The only time Dehn fil …
22
votes
Accepted
What manifolds are boundaries of euclidian spaces ?
$N$ has to be a homotopy-sphere. So as long as it's dimension isn't $4$, there's a proof that it has to be the standard $S^{n-1}$.
These arguments appear in the Kosinski book on smooth manifolds. T …
2
votes
When does a CW-complex of dimension 2 embedd in $R^4$ ?
Shapiro's obstruction:
A. Shapriro, "Obstructions to the imbedding of a complex in Euclidean space, I. The first obstruction," Ann. of Math., 66 No. 2 (1957), 256--269.
4
votes
Isometry classification of spherical space forms
That the diffeomorphism and isometry problem is the same for spherical 3-manifolds (i.e. spherical space forms or they're also called elliptic manifolds) goes back to people like Reidemeister and Hein …
10
votes
Accepted
index of morse functions and homotopical dimension
Take a contractible $3$-manifold which is not homeomorphic to $\mathbb R^3$ -- like the Whitehead manifold.
If such a Morse function existed on the Whitehead manifold, it would be a Morse function w …
6
votes
The Freedman Dichotomies
One of Freedman's results is that a homology $3$-sphere admits a tame topological embedding into $\mathbb R^4$. So here is an odd fact:
Let $M$ be the Poincare dodecahedral space. There is an open …
20
votes
Accepted
Is there a 2 component link with full symmetry?
I think there is a non-hyperbolic link that does the job.
The link that I'm thinking of could be called the splice of two Bing doubles of a figure-8 knot. Another way to describe this link is to sta …
20
votes
Accepted
Pseudoisotopy in low dimensions
When $M$ is a compact $2$-manifold, with or without boundary, $P(M)$ is known. When $M$ is a 3-manifold there's bits and pieces known, especially once you get to more fine detail like pseudo-isotopy …
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 …
15
votes
Are non-PL manifolds CW-complexes?
Kirby and Siebenmann's paper "On the triangulation of manifolds and the Hauptvermutung" Bull AMS 75 (1969) is the standard reference for this, I believe.
The result is that compact topological manif …
24
votes
Examples of non-diffeomorphic smooth manifolds with diffeomorphic tangent bundle
For $k = \infty$ (a continuum, to be precise), the continuum of non-diffeomorphic smooth structures on $\mathbb R^4$ would suffice. The tangent bundle of any $\mathbb R^4$ is trivial (since $\mathbb R …
11
votes
Thurston geometries---the geometry of the universal cover of $SL(2, \mathbb{R})$
Yes, the answer is that you want a simply-connected manifold. All $SL_2(\mathbb R)$ manifolds are covered by the universal cover. Not all $SL_2(\mathbb R)$ manifolds are covered by $SL_2(\mathbb R) …
7
votes
Are knots determined by their complements within a homotopy class?
The answer is no.
It stems from the fact that links are not determined by their complements. If you take the Borromean rings (for example), and think of one component as being a knot in the exterio …