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Results tagged with differential-topology
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user 8103
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
Projective span of a manifold
With apologies for the self-promotion, my student Baylee Schutte and I now have a paper on this topic called Projective span of Wall manifolds. We calculate the projective span of a family of manifold …
- 35.9k
3
votes
A question on the manifold $ \{n\otimes n-m\otimes m:n,m\in S^2,(n,m)=0\} $
I'm not too comfortable with the description of $N$ in terms of matrix tensor products, but the manifold $M/(\mathbb{Z}_2\times\mathbb{Z}_2)$ you describe in your comment is the manifold of pairs of o …
- 35.9k
7
votes
Realizing integral homology classes on non-orientable manifolds by embedded orientable subma...
Here are some comments that don't really answer the question, but are too long for the comment box.
Firstly, the Poincaré dual of $\nu\in H_n(M;\mathbb{Z})$ is a twisted integer class $D\nu\in H^{m-n} …
- 35.9k
14
votes
Integral homology classes that can be represented by immersed submanifolds but not embedded ...
This is a great question, and I don't have an answer but this is too long for a comment.
Working mod $2$, a codimension $k$ homology class $z\in H_{m-k}(M;\mathbb{Z}/2)$ is realizable by an embedding …
- 35.9k
9
votes
Accepted
Integral homology classes of which no multiples admit embedded representatives with trivial ...
With the trivial normal bundle condition, it's fairly easy to produce non-realizable examples using Théorème II.2 of Thom's paper. Namely, a class $z\in H_l(M^n;\mathbb{Z})$ is realizable by an embedd …
- 35.9k
7
votes
Accepted
Meaning of the first Chern class of the unit tangent bundle of a surface
The Poincaré-Hopf holds for non-orientable manifolds also. For a closed non-orientable surface $\Sigma$ the structure group of the sphere tangent bundle does not reduce to $SO(2)=S^1$, so it's classif …
- 35.9k
3
votes
Accepted
Equivariant Whitney approximation
This is Corollary 1.12 in
Wasserman, A. G., Equivariant differential topology, Topology 8, 127-150 (1969). ZBL0215.24702.
The proof is essentially the same as the one given by Peter Michor in his answ …
- 35.9k
7
votes
Accepted
on second cohomology of $S^1$-manifold
Yes. This follows from the Leray-Serre spectral sequence of the fibre bundle
$$
M\to M_{S^1} \to BS^1
$$
which has $E_2^{p,q}=H^p(BS^1;H^q(M;\mathbb{Z}))$ and converges to (the associated graded of th …
- 35.9k
16
votes
Can the nth projective space be covered by n charts?
It seems worth giving the cup-length argument, as it's relatively short and sweet.
Suppose $\mathbb{R}P^n=U_1\cup\cdots\cup U_n$, with each $U_i\approx\mathbb{R}^n$, and let $c\in H^1(\mathbb{R}P^n;\m …
- 35.9k
6
votes
Accepted
Künneth formula and induced map in homologies
Here is an example which will not make you very happy. There is a degree one map $f:S^2\times S^1 \to S^3$ which just collapses the complement of an embedded open disk. Take $a\in H_2(S^2;\mathbb{Z})$ …
- 35.9k
22
votes
Is there a $4$-manifold which Immerses in $\mathbb{R}^6$ but doesn't Embed in $\mathbb{R}^7$?
Edit: The answer below is incorrect. In fact, $\bar{w}_3(\mathbb{R}P^2\times\mathbb{R}P^2)=0$ (thanks to Rafal Walczak for pointing this out) so by the cited result $\mathbb{R}P^2\times\mathbb{R}P^2$ …
- 35.9k
16
votes
Accepted
Non orientable, closed manifold covered by two contractible charts
No. Recall that the Lusternik-Schnirelmann category of a space $X$, denoted $\operatorname{cat}(X)$, is the minumum $k$ such that $X$ may be covered by open sets $U_0,U_1,\ldots, U_k$ such that each i …
- 35.9k
1
vote
Non-transverse intersection of submanifolds
There is a generalisation of transversality called clean intersection, used by Bott, Quillen and others. I gave some references in my answer to Reference for base change of cohomology pull-push for cl …
- 35.9k
6
votes
Intersection modulo 2 theory for infinite dimensional manifolds?
This was supposed to be a comment but got too long.
The general result which encompasses both your examples in finite dimensions is the following: If $Y\subseteq Z$ is a submanifold of codimension $k$ …
- 35.9k
3
votes
Accepted
Is identification of double points of an immersion smooth?
I think the answer to the first 2 questions is yes. Most of the details are in the thesis of Ralph Herbert:
Herbert, Ralph J., Multiple points of immersed manifolds, Mem. Am. Math. Soc. 250, 60 p. (19 …
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