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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”.
24
votes
Accepted
Examples of odd-dimensional manifolds that do not admit contact structure
According to a well known result of Martinet, every compact orientable $3$-dimensional manifold has a contact structure [2], see also [1] for various proofs. On the other hand we have
Theorem. For …
14
votes
Accepted
Can $C^1$ mappings with derivative of low rank be approximated by smooth maps?
There is a counterexample.
Example. There is $f\in C^1(\mathbb{R}^5,\mathbb{R}^5)$ with ${\rm rank}\, Df\leq 3$ that cannot be approximated in the supremum norm by mappings
$g\in C^2(\mathbb{R}^5,\ …
13
votes
Accepted
Is there a sensible notion of a winding number of a closed curve in $\mathbb{R}^n$, $n\geq 3...
No, you cannot define a winding number if $n\geq 3$ since, as pointed out in a comment by Anthony Carapetis, any two curves in $\mathbb{R}^n\setminus\{ p\}$ are homotopic, and a winding number should …
13
votes
Checking that the image of a curve is not contained in a hyperplane
A curve $\alpha$ in $\mathbb{R}^3$ is called non-degenerate if $\alpha'$ and $\alpha''$ are linearly independent at every point.
A curve parametrized by arc-length is a Frenet curve if $\alpha''\neq 0 …
10
votes
Books in advanced differential topology
I highly recommend an amazing and highly underestimated trilogy Modern Geometry.
It covers not only differential geometry, but also differential and algebraic topology of manifolds.
Dubrovin, B. A.; F …
9
votes
Accepted
Existence of a certain foliation of $\mathbb R^n$
EDIT: Originally I could prove that there is such a foliation by topological manifolds:
Clearly, if $\mathbb{Q}^n$ is the set if points with all rational coordinates, you can have a foliation by paral …
9
votes
Whitney embedding theorem for Hölder manifolds
Every $C^1$ manifold admits a compatible $C^\infty$ structure. You can find a proof in Hirsch's "Differential topology". It is actually quite easy and based on a fact that smoothing a $C^1$ diffeomorp …
7
votes
Accepted
Map between manifolds and open dense subsets
Yes. I think you do not even have to assume that the preimages are finite. $f$ can be any $C^1$ mapping from $X$ onto $Y$. Let $y\in Y$ be a regular value of the mapping $f$. Then for some $\epsilon>0 …
7
votes
Topological obstructions to existence of immersion
I believe that the following Riemannian manifold $M$ cannot be embedded to a compact manifold $N$:
Since a compact manifold $N$ has a finite atlas consisting of domains diffeomorphic to balls, and …
6
votes
Accepted
Is $L^1$ strong convergence of Jacobians valid for maps between manifolds?
You actually do not need to assume that the mappings are Lipschitz as it is true for general $W^{1,n}$ mappings
Theorem. If $\mathcal{M}$ and $\mathcal{N}$ are smooth compact and oriented manifolds, …
6
votes
Accepted
Smooth functions on subsets of $\mathbb{R}^n$
The answer is yes for functions defined on closed sets $X\subset\mathbb{R}^n$.
In Section 1.5.5 in [1] we have a necessary and a sufficient condition of the existence of an extension to a $C^m$ functi …
6
votes
Accepted
$C^1$ perturbation of diffeomorphism is diffeomorphism?
Assuming that $M$ is a compact manifold, the answer is yes. Indeed, $\det Df(x)\neq 0$ for $x\in M$ and if $|Df(x)-Df_\epsilon(x)|$ is small, then $\det Df_\epsilon(x)\neq 0$, because the set of inver …
6
votes
Global diffeomorphisms of $\mathbb R^n$
$\kappa$ must be an affine isometry. If $\gamma:[0,1]\to\mathbb{R}^n$ is a smooth curve and $L(\gamma)$ denotes its length, then
$$
L(\kappa\circ\gamma)=\int_0^1|D(\kappa\circ\gamma)(t)|\, dt=
\int_0 …
5
votes
Thom's gradient conjecture and analyticity
The result quoted by the author of the questions is not correct.
Suppose we have an analytic function $f: U \to {\mathbb R}$, where $U\subset {\mathbb R}^n$ is an open subset and $0 \in U$ is a c …
5
votes
Accepted
Homogeneous regular (= polynomial component) maps with odd degree and their being global hom...
The answer is yes.
If a homogeneous polynomial map $F:\mathbb{R}^m\to\mathbb{R}^m$ is a local homeomorphism, then it is a global homeomorphism.
We say that a map is proper if preimages of compact s …