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Topology of cell complexes and manifolds, classification of manifolds (e.g. smoothing, surgery), low dimensional topology (e.g. knot theory, invariants of 4-manifolds), embedding theory, combinatorial and PL topology, geometric group theory, infinite dimensional topology (e.g. Hilbert cube manifolds, theory of retracts).
1
vote
Question about maps of $S^{3}$-bundles
Yes, and you do not need the hypothesis on $H_2$. The projection map $\pi: X \to \Sigma$, having fibre $S^3$, is 3-connected, and hence we can attach cells of dimension $\geq 4$ to $X$ to form a CW-co …
34
votes
Manifold embedded in $R^{n+1}$ with a submanifold that doesn't embed in $R^n$
If I have understood the table at
http://www.lehigh.edu/~dmd1/immtable
correctly, then $\mathbb{RP}^{10}$ embeds into $\mathbb{R}^{17}$. But by
Mahowald, Mark On the embeddability of the real proje …
2
votes
generators of Out(F_n) and homology
For your second point, the map
$$d : \mathrm{Out}(F_n) \longrightarrow GL_n(\mathbb{Z}) \overset{\mathrm{det}}\longrightarrow \mathbb{Z}^\times$$
(given by the action on $\mathbb{Z}^n = H_1(F_n;\math …
8
votes
Accepted
Whitney sum formula for topological Pontryagin classes
Yes. A simple argument is that $BO \to BTOP$ is a rational equivalence and an H-space map (in fact even an infinite loop map), so it follows from the Whitney sum formula for vector bundles.
Edit: The …
10
votes
Accepted
Low degree cohomology of Eilenberg-MacLane space K(G,2)?
For a finite cyclic group G, in the range you ask for you get cohomology groups
$$\mathbb{Z}, 0, 0, G \cong Ext(G, \mathbb{Z}), 0.$$
One sees this by for example computing the Leray--Serre spectral se …
8
votes
Accepted
Topology of hypersurface of sphere fixed by homeomorphic involution
In "Smooth Homology Spheres and their Fundamental Groups" Kervaire proves that i) every 4-dimensional homology sphere bounds a contractible smooth manifold, and ii) for $d \geq 5$, every $d$-dimension …
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 …
14
votes
Accepted
Is it true that all sphere bundles are boundaries of disk bundles?
If you don't specify what structure group you want the disc bundle $D^{k+1} \to N \to M$ to have, then it is always true: you just take the fibrewise cone on the original family.
If you want to know …
80
votes
1
answer
3k
views
Topological cobordisms between smooth manifolds
Wall has calculated enough about the cobordism ring of oriented smooth manifolds that we know that two oriented smooth manifolds are oriented cobordant if and only if they have the same Stiefel--Whitn …
2
votes
Accepted
Smooth closed simply-connected $4$-manifold with $w_1 = w_2 = 0$ without a point admits symp...
By an application of Gromov's $h$-principle, an open manifold $W$ admits a symplectic structure precisely if $TW$ admits an almost complex structure. In the case you are considering it does, as $TW$ i …
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 …
2
votes
Is the hypersurface satisfying $\langle x-x_0,\nu\rangle>0$ diffeomorphic to sphere?
Let me suppose that $M$ is in addition compact, and for simplicity that $x_0=0$. Let $\phi : M \to S^n$ be the radial projection, which is a smooth map.
If $\phi$ is not an immersion, there is a tan …
5
votes
Almost free actions on simply-connected spaces
EDIT: This is wrong, as Jens explains below.
It is enough to show that the isotropy groups of the 0-cells are trivial: higher dimensional cells must have smaller isotropy.
We may suppose, by restr …
9
votes
Accepted
Intersection form of surface bundle over surface
Yes, such a thing exists, but I don't know an explicit example.
To see that it exists, it is clearest to me to consider the universal situation. For any $k \in \mathbb{Z}$ there is a space $\mathcal{S …
3
votes
Accepted
Density of compactly-supported homeomorphisms
I think this is true. It suffices to prove the
Lemma. Given an orientation-preserving homeomorphism $h$ of $\mathbb{R}^n$ there is a compactly-supported homeomorphsim $h_1$ which agrees with $h$ on th …