68 votes
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Not all manifolds can be triangulated: In which dimensions?

In dimensions up to three, every manifold is triangulable (this is classical). In dimension 4, there are simply connected non-triangulable manifolds (such as the E8 manifold); in fact, a closed 4-...
mme's user avatar
  • 9,243
60 votes
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Do rings of smooth functions differ from rings of continuous functions?

No. In both the smooth function ring and the continuous function ring a maximal ideal $\frak m$ consists of the functions vanishing at some point. In the smooth case $\frak m/\frak m^2$ is the ...
Tom Goodwillie's user avatar
44 votes
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Questions on J. F. Nash's answer about his errors in the proof of embedding theorem

Igor already answered questions 1 and 2. What Nash wrote is an attempt to describe to a non-expert audience the solution scheme he had for non-compact manifolds. In the noncompact case he proceeds ...
Willie Wong's user avatar
  • 36.5k
36 votes

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

Here is a different proof which maybe clarifies a different aspect of the situation. The ring $C(X)$ of continuous functions on a compact Hausdorff space, as an abstract ring, actually knows its $C^{\...
Qiaochu Yuan's user avatar
35 votes
Accepted

Is $\mathbb{CP}^3$ minus two points the universal cover of a compact manifold?

More is true. Let $M^n$ be a closed connected simply connected manifold of dimension $\ge 3$. Let $p,q\in M$ be two distinct points. Suppose $M\setminus\{p,q\}$ is the universal cover of a closed ...
Vitali Kapovitch's user avatar
34 votes
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Is every rational realized as the Euler characteristic of some manifold or orbifold?

Products of 2-orbifolds with manifolds will do the trick. There are 2-orbifolds of Euler characteristic $1/n$ (take a quotient of $S^2$ by a rotation of order $2n$). Then take a product with a ...
Ian Agol's user avatar
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31 votes
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Suspension of a topological space

It’s not true. The Poincare sphere $P$ is a manifold, and its suspension is not. But its double suspension is homeomorphic to $S^5$ by Cannon’s “Double Suspension Theorem”. I learned about this from ...
Jeff Strom's user avatar
  • 12.5k
30 votes
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On a curious map from the complex projective plane into $S^5$

I think I can prove the following Claim There is no topological embedding of $\mathbb CP^2$ into $\mathbb R^6$. The proof uses the van Kampen obstruction. Let me review the idea. Suppose there is a ...
Gregory Arone's user avatar
29 votes

Can one determine the dimension of a manifold given its 1-skeleton?

You cannot hope to find the dimension exactly from the 1-skeleton alone. The complete graph on seven vertices is both the 1-skeleton of a triangulation of the two dimensional torus and of the five ...
Gjergji Zaimi's user avatar
27 votes

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

The answer for connected 2-dimensional orbifolds is no. Euler characteristic is $$\chi(O)=\chi(M)-\sum\left(1-\frac{1}{q}\right)-\frac{1}{2}\sum\left(1-\frac{1}{p}\right),$$ where $p,q\geq 2$ are ...
Alexandre Eremenko's user avatar
27 votes
Accepted

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

The answer is 'yes' by Britton's lemma (see wikipedia and, more generally, Serre's book Trees and Scott and Wall's article 'Topological methods in group theory'). Since $M$ and $U$ are smooth and ...
HJRW's user avatar
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26 votes
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Is symmetric power of a manifold a manifold?

$\newcommand{\Cone}{\operatorname{Cone}}$Let $d$ be the dimension of the manifold $M$. For $n \geq 2$, I will prove that the symmetric power $SP^n(M)$ is a manifold with boundary for $d=1$, a ...
Andy Putman's user avatar
  • 43.2k
25 votes
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Critical dimensions D for "smooth manifolds iff triangulable manifolds"

All smooth manifolds are triangulable, as you say. This follows from Morse theory, which dictates that you only need to know how to triangulate (PL) handle-attachments, which one can do by hand. The ...
mme's user avatar
  • 9,243
22 votes

Are there examples of non-orientable manifolds in nature?

Lanyards, strings used for holding name cards, often has the shape of a Mobius band. A picture of lanyard: Many more can be found on Google Image. The purpose of using this shape is probably to ...
22 votes

A manifold is a homotopy type and _what_ extra structure?

Igor is giving a good reference to the topic. For completeness, my education, and satisfaction of other reader's laziness, I'm going to give a rough outline here. A Poincaré complex is, very ...
Manuel Bärenz's user avatar
22 votes

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

In this answer, let's assume that all functions are real-valued. In the ring of continuous functions, the following are equivalent: $f\geq 0$ there is some $g$ with $g^2=f$. For each $n>0$, ...
Joseph Van Name's user avatar
21 votes
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A manifold is a homotopy type and _what_ extra structure?

You are talking about the (much studied) Poincare duality spaces. For a survey, see the very nice one by John Klein: (seems to be unpublished, but dates to April 2010).
Igor Rivin's user avatar
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21 votes
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Smooth curve in $\mathbb{R}^3$ not contained in real analytic surface?

Here is an example: Let $\gamma:\mathbb{R}\to\mathbb{R}^3$ be defined by $\gamma(t) = \bigl(t,\exp(-1/t^2),0\bigr)$ for $t<0$, $\gamma(0) = (0,0,0)$ and $\gamma(t) = \bigl(t,0,\exp(-1/t^2)\bigr)$ ...
Robert Bryant's user avatar
20 votes

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

"Excess dimensions": If a manifold can be topologically embedded in $\mathbb{R}^N,$ and you can prove that it can be isometrically embedded in $\mathbb{R}^M,$ then the quantity he is talking about is $...
Igor Rivin's user avatar
  • 95.5k
19 votes

Closed manifold with non-vanishing homotopy groups and vanishing homology groups

As suggested by Lennart Meier, the connected sum $M=P\#P$ of two copies of the Poincaré homology sphere gives an example. The homotopy groups $\pi_n(M)$ are nonzero for all $n>1$ because the ...
Allen Hatcher's user avatar
19 votes
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Manifolds as Cauchy completed objects

(Expanding on Phil Tosteson’s comment.) No: the Cauchy-completion characterisation doesn’t hold for the PL, topological, or complex-analytic cases. The key technical point is that split idempotents ...
Peter LeFanu Lumsdaine's user avatar
18 votes
Accepted

A question on connected sum of compact manifolds

Let us suppose that $dim(M)\geq 3$ then we have that: $\pi_1(M \# M)\cong \pi_1(M)*\pi_1(M)$, $H_*(M;\mathbb{Z})\cong H_*(M;\mathbb{Z})\oplus H_*(M;\mathbb{Z})$ when $*< dim(M)$. As $M$ is ...
David C's user avatar
  • 9,792
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
  • 9,243
18 votes
Accepted

Transitive embedding of the projective plane $\Bbb R P^2$ into the $4$-sphere

Yes. You take each vector $v \in \mathbb{R}^3$ to the vector $v \cdot v \in \operatorname{Sym}^2\mathbb{R}^3=\mathbb{R}^6$. This takes each unit vector $v$ to the same place as $-v$. So it descends to ...
Ben McKay's user avatar
  • 25.3k
18 votes
Accepted

Is there a closed manifold whose universal cover is $\mathbb{R}^n\setminus\{x_1, \dots, x_k\}$ for some $k > 1$?

If we demand that the universal cover is homeomorphic / diffeomorphic to $\mathbb{R}^n \setminus \{x_1,\ldots,x_k\}$ with $k>1$ the answer is no, there are no such closed manifolds. Each missing ...
Geva Yashfe's user avatar
  • 1,356
18 votes
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Converse to Hopf degree theorem

See the second half of the answer for a complete characterisation of closed orientable manifolds with the Hopf property. Note that $X$ having the Hopf property is equivalent to the injectivity of $\...
Michael Albanese's user avatar
17 votes

Self-covering spaces

Here are some pretty examples of self covering manifolds: Suppose that $F$ is a manifold and $f \colon F \to F$ is a periodic homeomorphism of period $k$. We define $M_f = F \times I \,/\, f$ to be ...
Sam Nead's user avatar
  • 25.4k
17 votes

Existence of non-null-homotopic map from $M^n$ to $S^{n-1}$

Theorem. If $M$ is a compact connected oriented $3$-manifold, then there is always a homotopically non-trivial map $f:M\to \mathbb{S}^2$. Proof. Let $g:M\to\mathbb{S}^3$ be a map of degree one. Such ...
Piotr Hajlasz's user avatar
17 votes
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Classification of closed 3-manifolds with finite first homology group?

The answer is no by Yves' comments. Let me add that there are plenty of explicit constructions of closed hyperbolic 3--manifolds with finite homology, and this is a generic phenomenon (for example ...
Jean Raimbault's user avatar

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