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 ...
14
votes
Accepted
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 $\...
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}...
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) $...
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\...
10
votes
Accepted
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}{...
10
votes
Accepted
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 ...
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
...
4
votes
Accepted
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 ...
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 ...
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 [...
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 ...
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 ...
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 ...
1
vote
Accepted
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 $M$ ...
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