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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”.
9
votes
When is a manifold a tangent bundle?
Suppose $M$ is an open $2n$-manifold that is homotopy equivalent to a closed smooth $n$-manifold $N$, and suppose $n>2$. Then Haefliger's embedding theorem ensures that the homotopy equivalence $N\to …
4
votes
Euler class in the non-compact case
For the untwisted case see Dold's "Lectures on algebraic topology" section VIII.11. If $N$ is a oriented topological submanifold of an oriented manifold $M$ of codimension $k$, then one looks at the T …
11
votes
Quotient of arbitrary free involution on $S^n$
In every dimension $\ge 4$ there is a fake real projective space, i.e., a manifold that is homotopy equivalent but not homeomorphic to $RP^n$.
Here you can find a computation for the topological surg …
11
votes
Accepted
Obstruction to a general S^1-action
V. Puppe, in Simply connected manifolds without $S^1$-symmetry. Algebraic topology and transformation groups, Proc. Conf., Göttingen/FRG 1987, Lect. Notes Math. 1361, 261-268 (1988) proved the followi …
11
votes
Accepted
Piecewise linear Poincaré conjecture
For spheres of dimension $n>5$ the PL Poincare conjecture follows from the s-cobordism theorem. Indeed, removing disjoint two small open disks one gets an s-cobordism (this uses excision in homology, …
4
votes
Counting connected manifolds
There are countably many compact topological manifolds, as was shown by Cheeger-Kister in Counting topological manifolds, Topology 9 (1970) 149–151, https://doi.org/10.1016/0040-9383(70)90036-4. The p …
51
votes
Accepted
Can a topological manifold have different tangent bundles?
This is answered in [Crowley, Diarmuid J.; Zvengrowski, Peter D, On the non-invariance of span and immersion co-dimension for manifolds, Arch. Math. (Brno) 44 (2008), no. 5, 353–365], see here.
Spec …
10
votes
Accepted
The boundary of a domain whose interior is diffeomorphic to the ball
The case $n=4$ is open as far as I know.
The case $n=3$ follows since $\mathbb R^3$ is irreducible, so it contains no fake 3-disk, i.e. $\bar D$ must be the standard disk.
The case $n=5$ is equival …
6
votes
Existence of sections of the evaluation map for the diffeomorphism group
Evaluation maps are studied in rational homotopy theory, see e.g. Evaluation maps in rational homotopy by Félix and Lupton. Here is a sample result (see Corollary 1.9 of the above paper):
Let $\mat …
6
votes
Contractible manifold with boundary - is it a disc?
Sergei, there are lots of compact contractible smooth manifolds; see e.g. my answer
here.
I am a bit confused about what you say next. Are you claiming that any compact contractible manifold admits …
13
votes
Why is the first integral Pontryagin class a homeomorphism invariant?
For spin manifolds this is proved in Corollary 1.22 (p.17) of Kammeyer's Diploma. Kreck's claim that the "spin" assumption can be dropped is mentioned after the corollary, with the caveat that "the au …
4
votes
Accepted
Does every disc bundle come from a vector bundle?
Equip $N$ with a Riemannian metric, and prove that the normal exponential map to any compact submanifold is a diffeomorphism onto its image on some closed $\epsilon$-neighborhood of the the zero secti …
18
votes
Accepted
Does every smooth manifold of infinite topological type admit a complete Riemannian metric?
By Whitney embedding theorem any smooth manifold embeds into some Euclidean space as a closed subset. The induced metric is complete.
In fact, a good exercise is to show that any Riemannian metric is …
4
votes
Accepted
Homogeneous manifold deformation retracts onto compact submanifold
Mostow-Karpelevich theorem says that if $G/G^\prime$ is a homogeneous space where $G$, $G^\prime$ are Lie groups with finitely many connected components, and maximal compact subgroups $K\supset K^\pri …
26
votes
Accepted
Are there non-smoothable homotopy/homology spheres?
Since the Poincaré conjecture is known in all dimensions any homotopy $n$-sphere is homeomorphic to $S^n$ and hence admits a smooth structure.
Any manifold of dimension $\le 3$ admits a smooth struct …