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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 $C^1$ …
Sergei Ivanov's user avatar
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 …
Sergei Ivanov's user avatar
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 …
Sergei Ivanov's user avatar
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 …
Sergei Ivanov's user avatar
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 …
Sergei Ivanov's user avatar
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 …
Sergei Ivanov's user avatar
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 …
Sergei Ivanov's user avatar