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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”.
13
votes
3
answers
3k
views
Contractible manifold with boundary - is it a disc?
I'm sure this is standard but I don't know where to look. Let $M$ be a contractible compact smooth $n$-manifold with boundary. Does it have to be homeomorphic to $D^n$? What about diffeomorphic?
[UPD …
17
votes
closed dual of vector fields
It is rarely possible.
First of all, if the manifold is simply connected, then there are no nowhere vanishing closed 1-forms at all. Indeed, every closed 1-form on such a manifold is a derivative of …
3
votes
Accepted
Are all $C^1$ arcs tame?
If $p$ is continuously differentiable up to and including the endpoints, then it has a $C^1$ extension to $(-\varepsilon,1+\varepsilon)$. Then it has a tubular neighborhood $U$ parametrized by a $C^1$ …
5
votes
Accepted
Upper bound on the number of intersections of algebraic manifolds with affine planes
Unless I misunderstood the question, here is a counter-example:
$d=3$, $k=2$, $f:\mathbb R^3\to\mathbb R^2$ is just a linear map, e.g. $f(x,y,z)=(x,y)$. Take $q=(0,0)$, then $Y=f^{-1}(q)$ is a straigh …
2
votes
volume growth of tubular neigbhorhood of critical values of an algebraic/differentiable map
EDIT: This answer is wrong: as pointed out by Alfonz, this map is not algebraic. The similar one with a rational approximation of $\sqrt2$ is, but then the degree is unbounded.
Consider the following …
9
votes
Accepted
Checking whether the image of a smooth map is a manifold
The specific $F(M)$ is not a smooth submanifold. Here is an argument.
To simplify formulas, I renormalize the sphere: let it be the set of $(z_1,z_2)\in\mathbb C^2$ such that $|z_1|^2+|z_2^2|=2$ rath …
15
votes
Accepted
Counting connected manifolds
Upper bound (assuming the manifolds are second countable): every manifold admits a complete metric, and the "set" of isometry classes of complete separable metric spaces is of cardinality continuum. I …