18 votes
Accepted

Is there a closed non-smoothable 4-manifold with zero Euler characteristic?

The Kirby-Siebenmann invariant in $H^4(M;\Bbb Z/2)$, an obstruction to smoothability, is additive under connected sum in dimension 4. In even dimensions, $\chi(M \# N) = \chi(M) + \chi(N) -2$. To ...
mme's user avatar
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14 votes
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Is the top Stiefel-Whitney number of a topological manifold the Euler characteristic mod two?

As you say, we define $w_n = \sum \text{Sq}^i \nu_{n - i}$, where $\nu_{n-i}$ is the Wu class, the class such that $\nu_{n-i} \cup c = \text{Sq}^{n-i} c$ for $c \in H^{i}$. So as a corollary we have $\...
mme's user avatar
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13 votes

Is the top Stiefel-Whitney number of a topological manifold the Euler characteristic mod two?

It has been proved in the preprint (page 6) by Renee Hoekzema that the vanishing of the $w_{n}(M)$ implies $\chi(M)$ is even. The proof uses the fact that a symplectic vector space over $\mathbb{F}_{2}...
Bombyx mori's user avatar
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12 votes

Is there a closed non-smoothable 4-manifold with zero Euler characteristic?

Alternatively, one can start with the $E_8$-manifold and connect sum with (five copies of) $S^1\times S^3$, and appeal to Donaldson's diagonalisation theorem instead. More precisely, the (negative) $...
Marco Golla's user avatar
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11 votes

Is there a closed non-smoothable 4-manifold with zero Euler characteristic?

You can also get this from Donaldson's theorem by a similar device. Take a non-diagonalizable definite form with even rank $2n$, and realize it (Freedman again) by a simply connected manifold. Then $W\...
Danny Ruberman's user avatar
10 votes
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Is every open topological $d$-manifold homotopy equivalent to a CW-complex of dimension $\leq d-1$?

$\DeclareMathOperator{\co}{H} \DeclareMathOperator{\ch}{C} \newcommand{\zz}{\mathbb{Z}} \newcommand{\nn}{\mathbb{N}} \newcommand{\A}{\mathcal{A}} \newcommand{\B}{\mathcal{B}} \DeclareMathOperator{\lf}{...
Cihan's user avatar
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10 votes
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Are compact topological $n$-manifolds recursively enumerable?

In a note of Freedman and Zuddas, they show that this is true for dimensions $\geq 4$. In the "Background" section of the paper, they describe the solution in the higher dimensional case ...
Ian Agol's user avatar
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8 votes

Can a finite group action by homeomorphisms of a three-manifold be approximated by a smooth action?

See Corollary 3.1 of the following paper by Robert Francis Craggs for the case of any involution of $S^3$ whose fixed set is homeomorphic to $S^2$: http://www.ams.org/mathscinet-getitem?mr=257966 ...
Ian Agol's user avatar
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4 votes
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Stable smoothing of topological manifolds relative to an embedding

I think the answer is yes, it follows from this: if $M$ is a triangulable manifold of dimension $n$ greater than 4, and if the total space $E$ of a vector bundle over $M$ is smoothable, then the ...
Connor Malin's user avatar
  • 5,191
4 votes

Homology of topological manifolds

It's already been noted in the comments that 1) is false: take for example an infinite genus surface. 2) is true. For an oriented manifold you have $H_i(M)\cong H^{n-i}_c(M)$, the cohomology with ...
Greg Friedman's user avatar
4 votes

"Thickening" an arc on a 2-manifold

Suppose that $S$ is the surface, $\alpha$ is the arc, and $K$ is the compact set. Proof 1: Find an arc $\alpha'$ so that $\gamma = \alpha \cup \alpha'$ is a Jordan curve (+). The Jordan/Schoenflies [...
Sam Nead's user avatar
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3 votes

Is the top Stiefel-Whitney number of a topological manifold the Euler characteristic mod two?

Elaborating on Mark's comment, associated to a manifold of type CAT=PL,DIFF,TOP or a menagerie of others we have a $\mathbb{R}^n$ bundle with structure group $CAT$. In all these cases, we can form the ...
Connor Malin's user avatar
  • 5,191
2 votes

Can a finite group action by homeomorphisms of a three-manifold be approximated by a smooth action?

I am a little confused. It is a theorem of Bing that there are periodic homeomorphisms of $S^3$ which are not conjugate to orthogonal actions (one of these has the Alexander horned sphere as the fixed ...
Igor Rivin's user avatar
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2 votes
Accepted

$\mathbb{E}_M$ as colimit of little cubes operads

I agree this is sort of scattered around as remarks (e.g. HA.2.3.3.4, say). Let's try to spell it out more cleanly. I claim that, if $X$ is a groupoid, then the category of $X$-families of operads is ...
Dylan Wilson's user avatar
  • 13.2k
1 vote
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A sufficient condition for a collection of open sets of a manifold to contain all open sets

Yes, it is true that $\mathcal{U}$ contains all open sets of $M$. The proof is a minor modification of Weiss's argument, and proceeds in several steps. Step1 We show that every open set of $M$ ...
Ken's user avatar
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