Questions tagged [manifolds]

A manifold is a topological space that locally resembles Euclidean space near each point. More precisely, each point of an n-dimensional manifold has a neighbourhood that is homeomorphic to the Euclidean space of dimension n.

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23 votes
2 answers
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Uniqueness of compactification of an end of a manifold

Let $M$ be an $n$-dimensional manifold (smooth or topological). I call $\bar{M}$ a compactification of $M$ if it is an $n$-dimensional compact manifold with boundary $\partial \bar{M}$, an $(n-1)$-...
Igor Khavkine's user avatar
7 votes
2 answers
1k views

G-spaces and manifolds

In his book "The geometry of geodesics" H. Busemann defines the notion of a G-space to be a space which satisfies the following axioms: The space is metric The space is finitely compact, i.e., a ...
Dror Atariah's user avatar
39 votes
1 answer
6k views

Not all manifolds can be triangulated: In which dimensions?

I know that Ciprian Manolescu has settled the triangulation conjecture in the negative: Not all manifolds can be triangulated. I've only read secondary literature on this result, which did not detail ...
Joseph O'Rourke's user avatar
41 votes
4 answers
4k views

When is a submanifold of $\mathbf R^n$ given by global equations?

Let $M \subset \mathbf R^n$ be a (smooth) submanifold of dimension $d$. Under which conditions does there exist global equations defining $M$? By global equations I mean : does there exist a smooth ...
Oblomov's user avatar
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31 votes
1 answer
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Is there a manifold with fundamental group $\mathbb{Q}$?

It is known that the fundamental group of a locally path connected, path connected compact metric space is finitely presented or uncountable. Furthermore the fundamental group of every manifold is ...
123...'s user avatar
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14 votes
1 answer
476 views

3-fold of general type homeomorphic to rational 3-fold

Is there a smooth (complex projective) 3-fold of general type which is homeomorphic (in the complex topology) to a rational $3$-fold? I am aware of such examples in complex dimension $2$, for ...
Nick L's user avatar
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13 votes
4 answers
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When is a Riemannian metric equivalent to the flat metric on $\mathbb R^n$?

I'm looking for an easily-checked, local condition on an $n$-dimensional Riemannian manifold to determine whether small neighborhoods are isometric to neighborhoods in $\mathbb R^n$. For example, for ...
Theo Johnson-Freyd's user avatar
10 votes
1 answer
771 views

rational homotopy of a manifold

Given a finite dim rational homotopy type satisfying Poincaré duality, what is the best reference to when it is the rational homotopy type of a fin dim manifold?
Jim Stasheff's user avatar
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143 votes
20 answers
23k views

Are there examples of non-orientable manifolds in nature?

Whilst browsing through Marcel Berger's book "A Panoramic View of Riemannian Geometry" and thinking about the Klein bottle, I came across the sentence: "The unorientable surfaces are never discussed ...
59 votes
7 answers
7k views

Status of PL topology

I posted this question on math stackexchange but received no answers. Since I know there are more people knowledgeable in geometric and piecewise-linear (PL) topology here, I'm reposting the question. ...
55 votes
8 answers
9k views

Is there a Whitney Embedding Theorem for non-smooth manifolds?

For smooth $n$-manifolds, we know that they can always be embedded in $\mathbb R^{2n}$ via a differentiable map. However, is there any corresponding theorem for the topological category? (i.e. Can ...
Jake's user avatar
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46 votes
3 answers
8k views

Connected sum of topological manifolds

A definition of the connected sum of two $n$-manifolds $M$ and $M'$ begins by considering two $n$-balls $B$ in $M$, $B'$ in $M'$, and glueing the varieties $M\setminus \mathring B$ and $M'\setminus \...
ACL's user avatar
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26 votes
1 answer
4k views

Classification of 1-dimensional manifolds (not second-countable)

It is easy to see that every connected $1$-dimensional second-countable manifold (that is, what is often called just a manifold) is either homeomorphic to $\mathbb{R}$ or to $S^1$. Now let's drop the ...
Martin Brandenburg's user avatar
22 votes
6 answers
3k views

Does every vector bundle allow a finite trivialization cover?

Suppose there is a vector bundle (smooth, with constant rank finite-dimensional fibres) over a (smooth, second-countable, Hausdorff, not necessarily connected) manifold $B$ of dimension $n$. (a) Is ...
Fiktor's user avatar
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21 votes
2 answers
4k views

Topological $n$-manifolds have the homotopy type of $n$-dimensional CW-complexes

I search for a chain of clean references, which lead the fact of topological manifolds of dimension $n$ having the homotopy type of a CW of dimension $n$. Milnor's On spaces having the homotopy type ...
Tom's user avatar
  • 489
18 votes
3 answers
3k views

Algebraic varieties which are topological manifolds

Inspired by this thread, which concludes that a non-singular variety over the complex numbers is naturally a smooth manifold, does anyone know conditions that imply that the topological space ...
David Steinberg's user avatar
15 votes
3 answers
2k views

Characterizing Hessians among symmetric bilinear tensors

I apologize in advance if this is somewhat elementary, but: Let $(M,g)$ be a compact Riemannian manifold. Is there a "characterization" of which symmetric bilinear tensors $B\in Sym^2(M)$ ...
Renato G. Bettiol's user avatar
15 votes
3 answers
3k views

Proving the existence of good covers

Usually one proves the existence of good covers in compact manifolds by Riemannian methods: we pick an arbitrary Riemannian metric, prove that geodesically convex neighborhoods exist, that they are ...
Mariano Suárez-Álvarez's user avatar
14 votes
2 answers
2k views

Reference Request: Compact manifolds with boundary have the homotopy type of a CW-complex

Let $M$ be a compact manifold (possibly non-smooth) manifold with boundary $\partial M$. Is the inclusion $\partial M\hookrightarrow M$ homotopy equivalent to the inclusion of a subcomplex into a CW-...
archipelago's user avatar
  • 2,964
8 votes
1 answer
499 views

Universal covers of non-prime 3-manifolds

Let $M$ be a closed, connected, oriented 3-manifold. If $M$ is prime, then we know what the universal cover of $M$ looks like: it is either $S^3, \mathbb{R}^3$ or $S^2 \times \mathbb{R}$ depending on ...
Minkowski's user avatar
  • 571
7 votes
1 answer
538 views

What manifolds can have a (non-piecewise) linear structure?

By the definition I'm using, all manifolds are Hausdorff and second countable. For all non-negative integers $n$, I define $B_n$ to be $\bigl\{ \mathbf{v} \in \mathbf{R}^n : \lVert\mathbf{v}\rVert <...
user avatar
7 votes
2 answers
1k views

Any 3-manifold can be realized as the boundary of a 4-manifold

We know "Any closed, oriented $3$-manifold $M$ is the boundary of some oriented $4$-manifold $B$." See this post: Elegant proof that any closed, oriented 3-manifold is the boundary of some ...
wonderich's user avatar
  • 10.3k
6 votes
0 answers
207 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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5 votes
1 answer
485 views

Every unorientable 4-manifold has a $Pin^c$, $Pin^{\tilde c+}$ or $Pin^{\tilde c-}$ Structure

The precise statement on J. W. Morgan's "The Seiberg-Witten Equations and Applications to the Topology of Smooth Four-Manifolds (MN-44)" that 4-manifold $X$ admits a Spinc structure (Lemma 3.1.2) ...
wonderich's user avatar
  • 10.3k
2 votes
1 answer
298 views

Density of continuous functions to interior in set of all continuous functions

Let $M$ be an $m$-dimensional manifold and $N$ be an $n$-dimensional manifold with boundary. Suppose also that the topology on $N$ can be described by a metric. Thus, the set $C(M,N)$ can be endowed ...
ABIM's user avatar
  • 5,019
107 votes
8 answers
15k views

What do heat kernels have to do with the Riemann-Roch theorem and the Gauss-Bonnet theorem?

I know the following facts. (Don't assume I know much more than the following facts.) The Atiyah-Singer index theorem generalizes both the Riemann-Roch theorem and the Gauss-Bonnet theorem. The ...
Qiaochu Yuan's user avatar
51 votes
3 answers
11k views

What is the difference between holonomy and monodromy?

What is the difference between holonomy and monodromy? And what is the simplest example in which one is trivial and the other is not?
James Propp's user avatar
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43 votes
4 answers
3k views

Do rings of smooth functions differ from rings of continuous functions?

Let $M$, $N$ be connected nondiscrete compact smooth manifolds. Can the ring of continuous functions on $M$ be isomorphic to the ring of smooth functions on $N$?
Arshak Aivazian's user avatar
40 votes
1 answer
2k views

Are there only countably many compact topological manifolds?

Up to homeomorphism, there are 2 one-dimensional topological manifolds and countably many 2- and 3-dimensional compact manifolds, respectively, since each manifold in these dimensions can be ...
Dominik's user avatar
  • 3,007
35 votes
2 answers
4k 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
  • 31k
30 votes
2 answers
2k views

A manifold is a homotopy type and _what_ extra structure?

Motivation: Surfaces Closed oriented 2-manifolds (surfaces) are "classified by their homotopy type". By this we mean that two closed oriented surfaces are diffeomorphic iff they're homotopy ...
Manuel Bärenz's user avatar
25 votes
2 answers
4k views

Questions on J. F. Nash's answer about his errors in the proof of embedding theorem

In the interview of John Nash taken by Christian Skau and Martin Gaussen, in EMS Newsletter, September, 2015 when asked Is it true, as rumours have it, that you started to work on the embedding ...
trequartista's user avatar
24 votes
1 answer
1k views

Non-regular Connected Hausdorff Banach Manifold

After reading this MO post, I am wondering: Is every (connected) Hausdorff Banach manifold a regular space? Though unjustified, page 53 of this paper nonchalantly states: "Note that a Hausdorff ...
Benjamin Dickman's user avatar
24 votes
2 answers
3k views

Are topological manifolds homotopy equivalent to smooth manifolds?

There exist topological manifolds which don't admit a smooth structure in dimensions > 3, but I haven't seen much discussion on homotopy type. It seems much more reasonable that we can find a smooth ...
Jake's user avatar
  • 795
23 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
  • 18.3k
23 votes
2 answers
2k views

Are homeomorphic open subsets of $\mathbb{R}^n$ also diffeomorphic?

Let $U_1, U_2$ be open subsets of $\mathbb{R}^n$. Both are naturally differentiable submanifold, getting the differentiable structure from $\mathbb{R}^n$. Further, both are natural topological ...
Mark Ullmann's user avatar
23 votes
2 answers
1k views

Is every rational realized as the Euler characteristic of some manifold or orbifold?

Let me first ask the question for two-dimensional compact, connected manifolds and orbifolds. Then, if the answer is No, one can remove various conditions on the dimension, and allow non-compact ...
Joseph O'Rourke's user avatar
22 votes
2 answers
2k views

What is an example of an orbifold which is not a topological manifold?

In Thurston's book The Geometry and Topology of Three-Manifolds it is proven that the underlying space of a two-dimensional orbifold is always a topological surface. Are there any easy examples of ...
user88649's user avatar
  • 271
18 votes
4 answers
3k views

When is a finite cw-complex a compact topological manifold?

I think the statement of the question is pretty straightforward. Given a finite $n$-dimensional CW complex, are there necessary and sufficient conditions for determining that it is also a compact $n$-...
William's user avatar
  • 712
18 votes
5 answers
2k views

Is every real n-manifold isomorphic to a quotient of $\mathbb{R}^n$?

I'm curious about the following: Is every real $n$-manifold isomorphic to a quotient of $\mathbb{R}^n$? Thanks. EDIT: As Tilman points out, the manifold should be connected. Also, yes, I'm thinking ...
Eivind Dahl's user avatar
17 votes
1 answer
1k views

Homology spheres and fundamental group

I have a curiosity about homology spheres: I was wondering if they were uniquely characterized by their fundamental group. I.e. given two $n-$dimensional (integral) homology spheres with isomorphic ...
Dario's user avatar
  • 643
17 votes
3 answers
2k views

Is symmetric power of a manifold a manifold?

A Hausdorff, second-countable space $M$ is called a topological manifold if $M$ is locally Euclidean. Let $SP^n(M): = \left(M \times M \times \cdots \times M \right)/ \Sigma_m$, where product is done $...
JE2912's user avatar
  • 359
17 votes
1 answer
2k views

Proofs of Rohlin's theorem (an oriented 4-manifold with zero signature bounds a 5-manifold)

A celebrated theorem of Rohlin states the following An oriented closed 4-manifold $M^4$ bounds an oriented 5-manifold if and only if the signature of $M^4$ is zero. Simple homological arguments ...
Bruno Martelli's user avatar
16 votes
3 answers
1k views

Does injectivity of $\pi_1(\partial U) \to \pi_1(M)$ imply injectivity of $\pi_1(U) \to \pi_1(M)$?

Let $M$ be a smooth compact manifold of dimension $n$, and let $U$ be a smooth compact manifold with boundary, of the same dimension $n$, embedded in $M$. The embedding induces maps on $\pi_1$. If $...
Yaniv Ganor's user avatar
  • 1,873
16 votes
2 answers
2k views

homotopy type of embeddings versus diffeomorphisms

Previously, I asked a question on mathoverflow comparing smooth embeddings and diffeomorphisms, which received a very interesting and somewhat unexpected answer by Agol. I now ask a further question ...
Ricardo Andrade's user avatar
16 votes
1 answer
588 views

If all balls around two points are isometric... -- manifold version

This question is a natural follow-up of this other question, asked earlier today by wspin. Let's say that a metric space $(X,d)$ has two poles if: there are two distinct points $x$, $y$ such that ...
Marco Golla's user avatar
  • 10.4k
14 votes
1 answer
565 views

Obstruction of spin-c structure and the generalized Wu manifods

Bockstein homomorphim and obstruction of spin-c structure: Let $w_2$ be the Stiefel Whintney class of manifold $M$. Let the Bockstein homomorphim $\beta$ be the $$ H^2(\mathbb{Z}_2,M) \to H^3(\mathbb{...
annie marie cœur's user avatar
14 votes
2 answers
2k views

Tubular Neighborhood Theorem for $C^1$ Submanifold

Can anyone reference/disprove the theorem in the case where the embedded submanifold is merely $C^1$ instead of smooth? I have a compact $C^1$ embedded submanifold of $\mathbb{R}^n$ without boundary ...
L P's user avatar
  • 323
13 votes
3 answers
963 views

Closed manifolds with the fixed point property

The real projective plane ${\bf P}^2({\bf R})$ is the only closed surface with the fixed point property: all continuous maps from the plane to itself has a fixed point. The projective spaces ${\bf P}^{...
coudy's user avatar
  • 18.5k
13 votes
1 answer
582 views

Are there examples of Einstein manifolds with unbounded curvature?

Are there any known examples of Einstein manifolds $(M, g)$ such that $$\sup_{x \in M} \|\text{Rm}(x) \| = \infty$$ I'm looking for these examples because they might provide a counter-example to a ...
Matheus Andrade's user avatar