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The study of differentiable manifolds and differentiable maps. One fundamental problem is that of classifying manifolds up to diffeomorphism. Differential topology is what Poincaré understood as topology or “analysis situs”.
5
votes
Open book decompositions in dimension 4
Nice question; there's not much known in general. The signature obstruction you mention goes back to Winkelnkemper (Bull. Amer. Math. Soc. 79 (1973), 45–51) and is sufficient for simply connected $4n$ …
4
votes
Accepted
Index formula for elliptic operators acting on Sobolev sections vanishing on the boundary (s...
Local boundary conditions such as the Dirichlet condition you mention were considered in Atiyah-Bott, The index problem for manifolds with boundary. 1964 Differential Analysis, Bombay Colloq., 1964 pp …
2
votes
Accepted
An example of handle decomposition on modified $S^5$
I didn't really follow the details of your description, but here is a handle decomposition along the lines you suggest. Start with $((S^1 \times S^3)_a \coprod (S^1 \times S^3)_b) \times I$ where $I= …
3
votes
Accepted
A definition of linking number for knots in $S^3$ using chains in $D^4$
Here is a brief sketch. First, show that the intersection number between A and B is independent of the choice of specific chains. (In other words, if $A'$ and $B'$ are other 2-chains with the same pro …
7
votes
Accepted
What were the "questions unapproachable by other means" w.r.t. $KO$-invariants?
Taken in the context of the introduction, I would guess that at least in part they are referring to applications of the index theorem to questions about positive scalar curvature. Before Lichnerowicz' …
12
votes
The convex hull of a manifold whose cobordism class is trivial
Implicit in the other responses is the fact that if $M$ bounds a convex manifold $W$, then $W$ is contractible and so M has the homology of a sphere. So any null-cobordant manifold that is not a homol …
7
votes
Converse to Hopf degree theorem
A manifold $M$ of dimension $>1$ with $H^1(M) \neq 0$ does not have the self-Hopf property. For if so, then there is a map $f: M \to S^1$ that induces a non-trivial homomorphism $f_*: H_1(M) \to \math …
6
votes
In which dimensions is it true that every topological ball embedded by a smoothly embedded s...
Amended answer: It seems to me that Question 1 asks if a codimension 0 submanifold $B$ (not necessarily a ball) of a smooth submanifold $M$ whose boundary is smoothly embedded is a smooth submanifold …
7
votes
Accepted
Invertible 2-knots in $S^4$
Q1: This is true in the topological category and unknown in the smooth setting. In the topological setting, the fundamental group of $S^4 - K_1 \# K_2$ is $G_1 *_\mathbb{Z} G_2$ where $G_i$ are the fu …
4
votes
Embedding $S^1\sqcup S^2$ in $S^4$ and its variants
In dimensions at least 4, one has that homotopy of circles implies isotopy of circles. So if you fix an embedding of $S^2$ (eg the standard embedding), then you are really asking about embeddings of t …
9
votes
Godbillon–Vey invariant and leaf space of foliations
You should check out Thurston, William, Noncobordant foliations of $S^3$. Bull. Amer. Math. Soc. 78 (1972), 511–514.
He constructs foliations with arbitrary real-valued GV invariants. In this paper, h …
4
votes
Normal invariants
If you want specifically low-dimensional calculations, the Kirby-Taylor article A survey of 4-manifolds through the eyes of surgery does this for 4-manifolds. The discussion highlights the difference …
22
votes
Very particular kind of 4-manifolds. Classification
I would say no. If M is simply connected, then it is contractible and hence determined topologically by its boundary. But there's no current smooth classification; the case when the boundary is $S^3$ …
11
votes
Accepted
Smooth structure on direct product
Extending Michael Albanese's answer above, $M \times N^k$ will never be smoothable. For if it were then choose a point $p\in N$ and a chart U around $p$. Then $M \times U$ is an open subset of $M \tim …
7
votes
Accepted
Akbulut's cork involution
The boundary of W may be described as 0-framed surgery on both components of the link you drew. The link can be drawn in a more symmetric fashion, so that it is clear that there is an involution inter …