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Complex, contact, Riemannian, pseudo-Riemannian and Finsler geometry, relativity, gauge theory, global analysis.

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 …
Oscar Randal-Williams's user avatar
3 votes

Spin structure on mapping torus

You can do this iff the spin structures $\mathfrak{s}$ and $f^*(\mathfrak{s})$ are isomorphic. When $X$ is the 2-torus the set of Spin structures is naturally in bijection with $\mathbb{Z}/2 \oplus \ …
Oscar Randal-Williams's user avatar
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 …
Oscar Randal-Williams's user avatar
20 votes

Can one give an immersion of exotic sphere $S^7$ in a standard sphere $S^8$ of radius $1$?

Yes. By Smale-Hirsch theory it is enough to find a bundle injection $T\Sigma \to \epsilon^8$, so it is enough to find a trivialisation of $T\Sigma \oplus \epsilon^1$. It is a theorem of Kervaire and M …
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
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 …
Oscar Randal-Williams's user avatar
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 …
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
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
4 votes

Restrictions of diffeomorphisms

This is closely related to (and can be proved by the same methods as) the fact that if we fix another manifold $N$ then the restriction map $$\mathrm{Emb}(M, N) \to \mathrm{Emb}(S, N)$$ is a locally t …
Oscar Randal-Williams's user avatar
4 votes

Is every $S^3$ block bundle over $S^4$ a fiber bundle?

In my paper ``Generalised Miller--Morita--Mumford classes for block bundles and topological bundles" with Johannes Ebert, we construct a block $\mathbf{HP}^2$-bundle $\pi: E^{20} \to S^{12}$ which can …
Oscar Randal-Williams's user avatar
20 votes
Accepted

Uniqueness of compactification of an end of a manifold

Suppose $\overline{M}_i$, $i=0,1$, are compact smooth manifolds with boundary whose interiors are diffeomorphic: let $\psi$ be such a diffeomorphism, and $M$ for either interior (identified via $\psi$ …
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 …
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