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”.
6
votes
Accepted
Computing the Euler characteristic of the complex projective plane using differential topology
There is a canonical way to construct holomorphic vector fields on $\mathbb{C}P^2,$ and that way is described in Zoladek's "Monodromy Group", page 335. If you read the description, it will be pretty c …
5
votes
Computing the Euler characteristic of the complex projective plane using differential topology
By the way, a very cool (to my mind) way of computing the Euler characteristic of $\mathbb{C}P^n$ is to treat it as the $n$-fold symmetric product of $\mathbb{C}P^1 = \mathbb{S}^2$ with itself. Then, …
1
vote
on product of some spaces
This question (or a close relative) is discussed and answered in
When factors may be cancelled in homeomorphic products?
4
votes
Extension of a group action beyond the boundary
If you are asking for a topological action, then it would seem you can just double $M$ along the boundary. I am not sure this works in the smooth category..
3
votes
Circle Bundles of surfaces
See this wikipedia article on Seifert Fibered Spaces.
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
Structural stability on the circle
You can read Jaco Palis' PhD thesis. The result is presumably also in deMelo/vanStrien
http://w3.impa.br/~demelo/tablecon.html
EDIT
Here is the link to the published version of Palis's thesis:
http:// …
9
votes
Extending a diffeomorphism of the sphere $S^2$ to the ball $D^3$
A combinatorial proof of Smale's theorem (actually a stronger theorem, which at the time of the paper cited was a conjecture of N. Kuiper) is given in the paper
The space of simplexwise linear homeom …
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 …
4
votes
Accepted
Set of density matrices
This is a symmetric space when equipped by the metric discussed in this question (it is Riemannian, when $p=2$). The metric is clearly invariant under the action; its other properties can be checked ( …
0
votes
When a homeomorphism is a diffeomorphism w.r.t to a suitable smooth structure?
Just a slight addition to Danny Ruberman's answer: more example in the same vein are discussed in David Gay's note "4-manifolds which are homeomorphic but not diffeomorphic".
11
votes
Accepted
is there a diffeomorphism with only finite orbits but of infinite order?
No, there is no such animal. This is due to Montgomery (1938), see my answer to this question: Nonperiodic points of homeomorphisms of a ball
4
votes
Diffeomorphisms of a surface in terms of generators.
If your question is: can you present the homeotopy group in terms of generators and relations, the answer is "yes", following the work of Hatcher-Thurston, Wajnryb, and most recently M. Korkmaz, who g …
10
votes
Accepted
Is a compact, connected, orientable 3-manifold with $\mathbb{Z}^K$ fundemental group uniquel...
NO, since the three-torus $T^3$ does not have this form.
EDIT if the OP really means a free product of $\mathbb{Z}$s, so the free group $F_k,$ then the answer is YES. It is a fact (see Hempel's book, …
2
votes
Reference request: gluing manifolds along pieces of boundary
This seems to be discussed here. (particularly page 6)