Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
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”.
4
votes
Closed simple curves in $\mathbb{R}\mathbb{P}^2$
Well, if you take the double cover, under your assumptions the lift is two simple closed curves in $S^2,$ the complement of which will be two disks and an annulus, so the original curve bounds a disk …
3
votes
Torus action implying infinite fundamental group
By taking products, it seems clear that something like $d\leq 2n/3$ is possible without infinite fundamental group, and this is, in some metaphysical sense, sharp: see Torus actions on rationally elli …
3
votes
Any 3-manifold can be realized as the boundary of a 4-manifold
In this question it is shown how to see that two smooth manifolds are topologically cobordant if and only if they are smoothly cobordant, which answers some subset of the questions (the second manifol …
11
votes
What is the weakest negative curvature condition ensuring a manifold is a $K(G,1)$?
For 1-3, yes, by the Cartan-Hadamard Theorem.5-... No. For example, every 3-manifold admits a metric of negative scalar curvature (I think this is actually true for any manifold, due to Lohkamp).
6
votes
Obstruction to a general S^1-action
It is a result of Atiyah-Hirzebruch (1970) that the $\hat{A}$ genus of a spin manifold with a nontrivial $S^1$ action vanishes, and a result of Herrera and Herrera that the same result, if the manifo …
10
votes
Accepted
A question on continuous maps from Möbius to itself
I might be missing something, but for $f = g\circ h,$ where $h$ is the retraction onto the core circle (see this question), and $g$ is the rotation of the circle by, say, 1 radian there appears to be …
-1
votes
Smale's theorem for $C^1$ diffeomorphisms of the sphere
I am quite sure it follows from:
Bloch, Ethan D.; Connelly, Robert; Henderson, David W., The space of simplexwise linear homeomorphisms of a convex 2-disk, Topology 23, 161-175 (1984). ZBL0547.57016. …
2
votes
Can a finite group action by homeomorphisms of a three-manifold be approximated by a smooth ...
I am a little confused. It is a theorem of Bing that there are periodic homeomorphisms of $S^3$ which are not conjugate to orthogonal actions (one of these has the Alexander horned sphere as the fixed …
3
votes
Accepted
How to understand this isomorphism?
The question is terribly put , but the answer is: $S_{0, 4}$ is the four times punctured sphere. You can think of this sphere as the ideal simplex in $\mathbb{H}^3$ (it is a theorem of mine that this …
22
votes
Accepted
A manifold is a homotopy type and _what_ extra structure?
You are talking about the (much studied) Poincare duality spaces. For a survey, see the very nice one by John Klein: (seems to be unpublished, but dates to April 2010).
1
vote
Accepted
Fixed-point-free action and cohomology of a finite group
You may want to check out Alex Adem's paper:
Adem, Alejandro, Cohomological restrictions on finite group actions, J. Pure Appl. Algebra 54, No.2-3, 117-139 (1988). ZBL0686.57023.
If you look at the p …
1
vote
Most general version for the Gauss-Bonnet theorem for polygons
See Pressley's Elementary Differential Geometry book, Chapter 13, to be specific, for an exhaustive discussion.
12
votes
Were 3-manifolds with $\sec>0$ known to be space forms before Ricci flow?
No, this was not known before Hamilton's paper. I think a Ricci-flow-free argument would still be of interest, so if you know how to do it, by all means...
18
votes
Homeomorphisms of $S^n\times S^1$
This is false for $n=1.$ The mapping class group of the torus is $SL(2, Z),$ of which the homeomorphisms you describe are but a small part - the parabolic matrices $\begin{pmatrix}1 & n\\ 0 &1\end{pma …
3
votes
Can we convert any non-vanishing vector field into geodesic field by changing metric?
The magic word is "geodesible", and the question in full generality is open. For surfaces, it has been answered by H. Gluck:
Gluck, Herman. "Dynamical behavior of geodesic fields." Global theory of d …