Questions tagged [topological-manifolds]

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"Homotopy homomorphisms" of homeomorphisms of Euclidean space

For a topological group $G$, an older term for a map $BG \to BG$ is a "homotopy homomorphism". If $G$ is connected, taking based loops shows that a homotopy class of such a map is the same ...
skupers's user avatar
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4 votes
1 answer
253 views

"Thickening" an arc on a 2-manifold

What is the argument for the fact that each arc in the interior of a 2-manifold can be "thickend" to obtain a 2-cell containing the arc in its interior and being disjoint from any ...
Palina's user avatar
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7 votes
0 answers
269 views

When does the tangent microbundle of a closed orientable topological $4k$-manifold have a trivial rank 2 subbundle?

$\DeclareMathOperator{\Top}{Top} \DeclareMathOperator{\co}{H}$Let $M$ be a closed orientable connected topological manifold of dimension $4k$ with $k > 1$. It is known (David Frank, On the index of ...
Cihan's user avatar
  • 1,546
1 vote
0 answers
103 views

Bijective continuous map from subset of $\mathbb{R}^n$ to a manifold of dimension $n$

I'd like to know if the following assertion is true or not (if true I'd like an example): There exists a positive integer $n$, and a manifold $M$ of dimension $n$ such that there is no subset $X \...
Rahul Sarkar's user avatar
6 votes
1 answer
318 views

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

In Lurie's "Higher Algebra", Remark 5.4.5.2 towards the end, there is the following statement: "It follows that $\mathbb{E}_M$ can be identified with the colimit of a diagram of $\infty$...
Markus Zetto's user avatar
5 votes
1 answer
192 views

Stable smoothing of topological manifolds relative to an embedding

Let $M$ be a topological manifold. We know that $M$ is stably smoothable if and only its tangent microbundle, up to stabilization, admits a reduction to vector bundle. Now I wonder if there is a ...
UVIR's user avatar
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6 votes
0 answers
162 views

Can $X_4 \times S^1$, $X_4 \times I^1$, or $X_4 \times \mathbb{R}^1$ be a triangulable, PL or DIFF manifold, if $X_4$ is a non-triangulable manifold? [duplicate]

Question: If $X_4$ is a non-triangulable topological (TOP) manifold, can $X_4 \times S^1$, $X_4 \times I^1$, or $X_4 \times \mathbb{R}^1$ be a triangulable manifold? can $X_4 \times S^1$, $X_4 \...
wonderich's user avatar
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6 votes
0 answers
206 views

If $X_d$ is a non-triangulable manifold, can $X_d \times T^k$, $X_d \times I^k$, or $X_d \times \mathbb{R}^k$ be a triangulable manifold?

If $X_d$ is a non-triangulable manifold, can $X_d \times T^k$, $X_d \times I^k$, or $X_d \times \mathbb{R}^k$ always be a triangulable manifold? Let $X_d$ be a $d$-manifold which is NOT a ...
wonderich's user avatar
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13 votes
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Structures between PL and smooth

Let $X$ be a topological manifold of dimension at least five. The Kirby-Siebenmann invariant $ks(X)\in H^4(X,\mathbb{Z}_2)$ is an obstruction to the existence of a PL structure on $X$. If it vanishes, ...
Philip Engel's user avatar
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11 votes
0 answers
213 views

Torus trick without surgery theory

It follows from surgery theory that in dimension $\geq 5$ every closed PL manifold homotopy equivalent to a torus has a finite cover which is PL homeomorphic to a torus. This is an important ...
user124543's user avatar
10 votes
1 answer
538 views

Is every open topological $d$-manifold homotopy equivalent to a CW-complex of dimension $\leq d-1$?

Let $M$ be a connected open topological $d$-manifold (without boundary). Whitehead showed that if $M$ has a PL structure, there exists a subcomplex of dimension $\leq d-1$ onto which $M$ deformation ...
Cihan's user avatar
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2 votes
1 answer
997 views

Homology of topological manifolds

Let $X$ be a topological manifold of dimension $n$ (assuming perhaps that there is a countable basis of open sets). Do NOT assume that $X$ is compact, or oriented, or triangulable (so do not assume it ...
Pierre's user avatar
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7 votes
0 answers
173 views

Existence of codimension 1 topological foliations

One of the many famous theorems proven by William Thurston is that a closed connected smooth manifold $M$ admits a codimension 1 smooth foliation if and only if $\chi(M)=0$: W.P. Thurston, Existence ...
Moishe Kohan's user avatar
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4 votes
0 answers
157 views

(Non-)Orientability of non-triangulable manifolds

We heard and learned from Mike Miller's answer to Not all manifolds can be triangulated: In which dimensions? that "All orientable 5-dimensional manifolds are triangulable. In dimensions at least ...
annie marie cœur's user avatar
10 votes
3 answers
1k views

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

I will just repeat the title: Is there a closed non-smoothable 4-manifold with zero Euler characteristic? I am guessing yes simply based on other existence theorems I have seen for 4-manifolds.
Cihan's user avatar
  • 1,546
14 votes
3 answers
899 views

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

Recall that the Stiefel-Whitney classes of a smooth manifold are defined to be those of its tangent bundle - this definition doesn't extend to topological manifolds as they don't have a tangent bundle....
Michael Albanese's user avatar
5 votes
0 answers
116 views

Is there a non-smoothable punctured manifold?

Does there exist a connected topological manifold $M$ such that $M-\{pt\}$ is non-smoothable? My understanding is that Quinn showed that these are always smoothable in dimension 4 (in fact in ...
Cihan's user avatar
  • 1,546
20 votes
2 answers
781 views

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

Let $M^3$ be a smooth three-manifold, and let $\gamma:G\to\operatorname{Homeo}(M)$ be a finite group action on $M$ by homeomorphisms. Can $\gamma$ can be $C^0$-approximated by smooth group actions $...
John Pardon's user avatar
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11 votes
0 answers
172 views

"The TOP h-cobordism theorem without surgery??"

Kirby and Siebenmann's book on topological manifolds contains the following intriguing passage on page 141: I believe no such proof has been discovered, though I'd be happy to be corrected on that. ...
skupers's user avatar
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6 votes
0 answers
374 views

Do all topological manifolds admit locally flat embeddings into R^n?

In his 1969 paper “Locally flat imbeddings of topological manifolds” Lees proved that a closed oriented second countable topological manifold admits a locally flat embedding into some R^n. Does the ...
Dmitri Pavlov's user avatar
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0 answers
49 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(...
Lars Pettersen's user avatar
7 votes
0 answers
324 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, ...
Michael Albanese's user avatar
25 votes
2 answers
758 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 ...
Eric S.'s user avatar
  • 596
3 votes
1 answer
260 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 $Y$ ...
Guangbo Xu's user avatar
  • 1,197
2 votes
1 answer
120 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 $^{\mathbb S^n}/_{\mathcal D}$ ...
Mark Hagen's user avatar
12 votes
1 answer
717 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 ...
ACL's user avatar
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12 votes
0 answers
365 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^\...
skupers's user avatar
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8 votes
1 answer
411 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 ...
skupers's user avatar
  • 7,823
35 votes
2 answers
3k 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 ...
Misha's user avatar
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7 votes
2 answers
955 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 ...
Alex M.'s user avatar
  • 5,067
17 votes
1 answer
1k 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 ...
Yuji Tachikawa's user avatar
22 votes
3 answers
2k 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=...
John Pardon's user avatar
  • 18.1k
11 votes
1 answer
526 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?
Janesis's user avatar
  • 113
29 votes
1 answer
2k 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 ...
Andrew Ranicki's user avatar