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0
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0answers
33 views

A C(B)-module structure on the function algebra of the total space of a vector bunlde $\pi:V \to B$

For a continuous vector bundle $\pi:V \to B$ vector bundle over a compact Hausdorff space $B$, and $C(B)$, $C(V)$ the continuous complex valued functions on $B$ and $V$ respectively, we can give $C(...
2
votes
1answer
139 views

Stiefel-Whitney classes of closed topological manifolds with no smooth structure

If $M$ is a smooth manifold, the total Stiefel-Whitney class of $M$ is defined to by the total Stiefel-Whitney class of the tangent bundle, i.e. $w(M) := w(TM)$. If $M$ is a closed smooth manifold, ...
17
votes
1answer
437 views

Are compact topological $n$-manifolds recursively enumerable?

Earlier this year it was asked on MO, "Are there only countably many compact topological manifolds?" Thanks to Cheeger and Kister, the answer is yes. On the other hand, Manolescu recently debunked ...
3
votes
1answer
115 views

Existence of tubular neighborohoods of locally flat topological embeddings

Suppose $X$ is a topological manifold and $Y \subset X$ is a locally flat submanifold. We know that $Y$ doesn't necessarily have a tubular neighborhood. My definition of a tubular neighborhood of $Y$ ...
2
votes
1answer
59 views

Shrinkable decompositions with uncountably many non-degenerate elements?

Let $\mathcal D$ be an upper semicontinuous decomposition of $\mathbb S^n$ and let $\mathcal D'\subset\mathcal D$ be the set of non-singletons. The decomposition space $^{\mathbb S^n}/_{\mathcal D}$ ...
10
votes
1answer
318 views

Piecewise linear (PL) structures on $\mathbf R^4$

One can read in Wikipedia that the 4-dimensional affine space $\mathbf R^4$ has uncountably many piecewise linear structures (in contrast with other dimensions, where it has exactly one). A reference ...
11
votes
0answers
216 views

Is the quotient map of the action of homeomorphisms on embeddings well-behaved?

It is well known that if $M$ and $N$ are smooth manifolds, the diffeomorphisms $Diff(M)$ act continuously on the smooth embeddings $Emb^{C^\infty}(M,N)$ by precomposition, if both are given the $C^\...
6
votes
1answer
239 views

A homological criterion for collapsibility?

On page 256 of Kirby and Siebenmann one finds the following lemma (its proof an "elementary exercise", so they only give a hint): Taking $A$ to be a point and iterating this collapsing lemma, this ...
28
votes
2answers
886 views

good covers of manifolds

It is well-known and easy to prove (see for instance this post) that every smooth manifold admits a "good cover", i.e. a locally finite cover by open balls such that all nonempty intersections of the ...
3
votes
2answers
382 views

The group of diffeomorphisms with compact support

Let $M$ be a topological/differentiable manifold. Is there any topology on the group of homeomorphisms/diffeomorphisms with compact support, turning it into a (locally-)compact topological group? (My ...
13
votes
1answer
785 views

How can gauge theory techniques be useful to study when topological manifolds can be triangulated?

I was reading a review article arXiv:1310.7644 and it was explained there that in the last few years it was proven that there are topological manifolds of dimension greater than four that cannot be ...
14
votes
3answers
931 views

Does a *topological* manifold have an exhaustion by compact submanifolds with boundary?

If $M$ is a connected smooth manifold, then it is easy to show that there is a sequence of connected compact smooth submanifolds with boundary $M_1\subseteq M_2\subseteq\cdots$ such that $M=\bigcup_{i=...
10
votes
1answer
246 views

orbit space of a topological manifold

Given a compact Lie group G acting freely on a topological manifold M, is it true that the orbit space M/G is also a topological manifold? If so, why?
23
votes
1answer
1k views

When is a compact topological 4-manifold a CW complex?

Freedman's $E_8$-manifold is nontriangulable, as proved on page (xvi) of the Akbulut-McCarthy 1990 Princeton Mathematical Notes "Casson's invariant for oriented homology 3-spheres". Kirby showed that ...