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Homotopy theory, homological algebra, algebraic treatments of manifolds.

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 …
Oscar Randal-Williams's user avatar
58 votes

Examples where it's useful to know that a mathematical object belongs to some family of objects

This is similar to Dan Petersen's answer, but more elementary. A fact I always mention when talking to students about matrix groups is that the diagonalisable matrices (over $\mathbb{C}$) are dense in …
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
32 votes
Accepted

Fundamental group of a topological pullback

There is a ``Mayer--Vietoris" sequence $$\cdots \to \pi_2(Z, z) \to \pi_1(E, e) \to \pi_1(X, x) \times \pi_1(Y, y) \to \pi_1(Z,z) \to \pi_0(E) \to \cdots$$ that can be developed by fitting together th …
Oscar Randal-Williams's user avatar
31 votes

Does a "Chern character" exist for any generalized cohomology theory?

For any (connective) spectrum $E$ one may rationalise it to get a rational spectrum $E_\mathbb{Q}$, and a map $E \to E_\mathbb{Q}$. Now rational spectra split as wedges of Eilenberg-Mac Lane spectra, …
Oscar Randal-Williams's user avatar
26 votes
Accepted

Mumford conjecture: Heuristic reasons? Generalizations? ... Algebraic geometry approaches?

All current proofs of Mumford's conjecture in fact prove a far stronger result, the "Strong Mumford conjecture", first formulated by Ib Madsen. This says the following (where by "moduli space" in the …
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
26 votes
Accepted

Homology theory represented by Madsen-Tillmann spectra

This is an exercise in understanding the Pontrjagin--Thom correspondence. The group $\pi_k(MTO(n) \wedge X_+)$ is in bijection with tuples of a $(n+k)$-manifold $M$, an $n$-dimensional vector bund …
Oscar Randal-Williams's user avatar
22 votes
Accepted

Group completion theorem

The statement that $M \to \Omega BM$ is a weak equivalence when $M$ is a group-like topological monoid is indeed easier: the map $EM = B(M \wr M) \to BM$ is then a quasi-fibration, has geometric fibre …
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
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
19 votes
Accepted

Product-like structures on spheres

Your condition determines the map $a_1 \vee a_2 : S^n \vee S^n \to S^n$ on the $n$-skeleton, so the question is when does this extend over the $2n$-cell of $S^n \times S^n$. The $2n$-cell is by defini …
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
18 votes
Accepted

Loop spaces and infinite braids

Yes, $B \beta_\infty$ is homology equivalent to $\Omega^2_0 S^2$, the zero component of the double loop space of $S^2$. The map $B_\infty \to S_\infty$ induces the obvious stablisation map $\Omega^2_0 …
Oscar Randal-Williams's user avatar
18 votes
Accepted

Is every ''group-completion'' map an acyclic map?

I think I have been able to reproduce the "argument by Wagoner" (perhaps it was removed from the published version?). It certainly holds in more generality that what I have written below, using the no …
Oscar Randal-Williams's user avatar

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