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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”.
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 $C^1$ …
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 …
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 …
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 …
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 …
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 …
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 …