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Results tagged with differential-topology
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user 3460
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”.
23
votes
Accepted
What can we say about the Cartesian product of a manifold with its exotic copy?
Your question seems to be about simply connected exotic 4-manifolds, for which the answer is yes. That's because $M$ and $M^E$ are h-cobordant (by Wall), say via an h-cobordism W. Then $M \times W$ is …
- 19.4k
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 $4n$ …
- 19.4k
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 …
- 19.4k
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= …
- 19.4k
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 …
- 19.4k
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' …
- 19.4k
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 …
- 19.4k
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 …
- 19.4k
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 …
- 19.4k
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 …
- 19.4k
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 …
- 12.9k
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 …
- 12.9k
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 …
- 19.4k
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 $S^3$ …
- 19.4k
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 …
- 19.4k