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Results tagged with differential-topology
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user 318
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”.
37
votes
Accepted
All fiber bundles over $S^2$ extendable to $\mathbb{C}P^\infty$?
No it isn't, but I had to dig quite deep to get a counterexample. Let us look at smooth $(D^7, \partial D^7)$-bundles over $S^2$, i.e $D^7 \to E \overset{\pi} \to S^2$ with an identification $\partial …
- 18.2k
26
votes
Accepted
Approximation of homeomorphism by diffeomorphism
No. The space of homeomorphisms of a compact manifold is locally contractible:
A. V. Černavskiı̆. Local contractibility of the group of homeomorphisms of a manifold. Mat.
Sb. (N.S.), 79 (121):307–356 …
- 18.2k
23
votes
Accepted
Are homology spheres stably parallelisable?
Yes, they have stably trivial tangent bundles. A remark to this effect can be found on page 70 of
M. Kervaire "Smooth Homology Spheres and their Fundamental Groups"
but it is a little terse. It is e …
- 18.2k
18
votes
Accepted
Are the associative grassmannian and the quaternionic projective plane diffeomorphic?
According to
Characteristic Classes and Homogeneous Spaces, I
A. Borel and F. Hirzebruch,
Section 17, they are not even homotopy equivalent: $G_2 / SO(4)$ has mod $2$ homology in degree 2, whereas …
- 18.2k
13
votes
Examples of Self-Maps of E8-Manifold
Such a map $f : M \to M$ of degree $d >0$ satisfies, with respect to the cup-product pairing $\langle -, - \rangle$,
$$\langle f^*(x), f^*(y) \rangle = d \langle x, y \rangle.$$
Conversely, I claim t …
- 18.2k
11
votes
Accepted
Why is it true that if two 4-manifolds are homeomorphic then their squares are diffeomorphic?
It follows by smoothing theory. If $h : X \to Y$ is a homeomorphism between smooth 4-manifolds, one obtains two maps $X \to BO$ which become homotopic in $BTOP$. The difference between them is therefo …
- 18.2k
10
votes
Accepted
How does all of the bundles over a certain manifold characterize the homotopy class of the b...
If you are liberal enough as to what you consider to be "a bundle", then this is true. The reason is that the symmetric product
$$ G := SP^\infty(S^{n-1}) \simeq K(\mathbb{Z},n-1)$$
is a topological a …
- 18.2k
7
votes
Steenrod powers of Pontryagin classes
No, because there are not enough of them. The lowest Steenrod operation $\mathcal{P}^1$ raises degree by $2(q-1) = 4 \tfrac{q-1}{2}$, so $\mathcal{P}^1(p_1)$ has degree $4(\tfrac{q-1}{2}+1)$ and so yo …
- 18.2k
6
votes
Topological relationships between family of transversal intersections of manifolds
Let me rephrase the construction: you have a map
$$\varphi : [0,1] \times M \to \mathbb{R}^n$$
which for each $t \in [0,1]$ is an embedding ($\varphi(t,x) = x+a(t)$ in your notation) and is transverse …
- 18.2k
6
votes
Homologically trivial submanifolds
This was supposed to be a comment to Jeff's answer, but wouldn't fit.
If you can solve the bordism problem you get a smooth map $F: W \to M$, but $F$ is not homotopic to an immersion in general. The …
- 18.2k
5
votes
Is there an analog of compactified moduli spaces(/stacks) for smooth manifolds?
I have a piece of juvenilia on this topic, considering the case of zero-dimensional submanifolds. See Section 9 of
O. Randal-Williams, Embedded cobordism categories and spaces of submanifolds,
IMRN …
- 18.2k
4
votes
Accepted
cartesian product rigidity for the punctured open disc
Q1: No, see the third-from-last theorem of
R. H. Bing, The cartesian product of a certain nonmanifold and a line is $E^4$.
- 18.2k
4
votes
Accepted
Relative version of Whitney Immersion Theorem
No, you can't generally do this even with the added assumptions. The bundle $\tau = TS^{n-1} \to S^{n-1}$ is non-trivial for $n-1 \neq 1,3,7$, but it is stable after (one) trivialisation. Hence $\tau …
- 18.2k
3
votes
Accepted
When are bundles of odd and even differential forms isomorphic?
I will explain that as long as $n>2$ the real vector bundles $\Omega^{even}$ and $\Omega^{odd}$ over $M$ are isomorphic.
If $n>2$ then $dim(\Omega^{even}) = dim(\Omega^{odd}) = 2^{n-1} > n$ and so $\O …
- 18.2k
2
votes
Cobordism and finite sheeted covers of manifolds
Its not true in complex cobordism. In Quillen's paper ``Elementary proofs of some results of cobordism theory using Steenrod operations", Section 4, he computes the complex cobordism class of a princi …
- 18.2k