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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

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 …
Oscar Randal-Williams's user avatar
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 …
Oscar Randal-Williams's user avatar
2 votes

Injectivity of the $\alpha$-genus

No. See for example the work of Anderson, Brown, and Peterson https://projecteuclid.org/euclid.bams/1183527786
Oscar Randal-Williams's user avatar
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 …
Oscar Randal-Williams's user avatar
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 …
CuriousUser's user avatar
  • 1,452
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 …
Dario's user avatar
  • 683
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 …
Oscar Randal-Williams's user avatar
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 …
Oscar Randal-Williams's user avatar
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 …
Oscar Randal-Williams's user avatar
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 …
Oscar Randal-Williams's user avatar
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 $E^4$.
Oscar Randal-Williams's user avatar
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 …
Oscar Randal-Williams's user avatar
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 …
Oscar Randal-Williams's user avatar
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 …
Oscar Randal-Williams's user avatar
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 …
Oscar Randal-Williams's user avatar

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