55 votes
Accepted

Why can't Antoine's necklace fall apart?

if we remove a segment of one of the tori the object still cannot be separated by a sphere, and yet can fall apart macroscopically. Well, sure it can. The key observation is that "sphere" means "...
Alexander Gruber's user avatar
  • 26.5k
20 votes

Intuition behind Khinchin's constant

Here is what I think of as the standard explanation. I learned this from The Art of Computer Programming, Section 4.5.3. Start with some number $z$ whose continued fraction we want to find. I'll ...
David E Speyer's user avatar
13 votes

What are some applications of these specific pure math areas?

In most fields of pure mathematics, there aren't a lot of industry options that rely directly on your expertise or involve continuing with your research program. Instead there are jobs which use the ...
Noah Snyder's user avatar
  • 9,785
13 votes

$4$-manifold with fundamental group $\Bbb Z/4\Bbb Z$

As pointed out, every finitely presented group arises as the fundamental group of a closed smooth orientable four-manifold. The same is not true of non-orientable manifolds as $\mathbb{Z}/3\mathbb{Z}$ ...
Michael Albanese's user avatar
12 votes
Accepted

$4$-manifold with fundamental group $\Bbb Z/4\Bbb Z$

A non-orientable example: consider the automorphism $f : S^2 \times S^2$ given by $(x, y) \mapsto (y, -x)$ where $-$ denotes the antipode map. This map has order $4$ and gives a free action of $\...
Qiaochu Yuan's user avatar
11 votes

What's true in $\mathbb{R}^4$, false in $\mathbb{R}^3$ and uninteresting in $\mathbb{R}^5$?

Due to the theory of quaternions, due to Hamilton, $\bf R^4$ has a structure of a of non commutative field. The only dimension for which $\bf R^n$ is a field are $n=1,2, 4$ . As an application, the ...
Thomas's user avatar
  • 7,255
11 votes

Why does $\mathbb{RP}^2$ not continuously embed in $\mathbb{R}^3$?

The answers I've found in MathSE (1, 2) indeed appear to assume smoothness. Let $A \subset \mathbb{R}^3$ be the embedding of a compact hypersurface on $\mathbb{R}^3$. Since it is embedded in $\...
Aloizio Macedo's user avatar
  • 33.6k
10 votes

What's true in $\mathbb{R}^4$, false in $\mathbb{R}^3$ and uninteresting in $\mathbb{R}^5$?

Lets bump knot theory up a dimension. In general, an $n$-sphere can be non-trivially knotted in $\mathbb{R}^{n+2}$. Obviously, an $n$-sphere can be embedded in $\mathbb{R}^{n+1}$, where it is ...
N. Owad's user avatar
  • 6,662
10 votes

Fundamental group of the total space of an oriented $S^1$ fiber bundle over $T^2$

Let me use $T$ to denote the torus. Oriented circle bundles over $T$ (indeed over any closed, connected surface) are classified up to orientation preserving bundle isomorphism by an integer known as ...
Lee Mosher's user avatar
  • 115k
10 votes
Accepted

Fundamental group of the total space of an oriented $S^1$ fiber bundle over $T^2$

Here's a geometric view point which computes (a presentation of) $\pi_1(M)$. As a byproduct, we establish The group $\pi_1(M)$ is nilpotent of at most $2$-steps. That is, $[\pi_1(M),\pi_1(M)]$ need ...
Jason DeVito - on hiatus's user avatar
9 votes

Prove by elementary methods: the plane cannot be covered by countably many copies of the letter "Y"

An idea that occured to me: Let there be some countable covering $G$ of the plane through "Y letters". Pick some circle $S$ in the plane. Clearly $G$ covers $S$. But each "Y letter" has at most 6 ...
John P's user avatar
  • 1,176
8 votes
Accepted

What closed 3-manifolds have fundamental group $\Bbb Z$?

The first argument of your proof sounded a little familiar to me, and I realized I saw it in Hempel. So for a maybe more citeable reference and also some other infinite non-$\mathbb Z$ examples, let ...
Daniel Valenzuela's user avatar
8 votes
Accepted

Covering Dimension of a Subset of the Plane

This paper by Mardesic shows: Let $X \subseteq \mathbb{R}^n$ which is compact and $k$-dimensional. Then there are coordinates $1 \le i_1 < i_2 \ldots i_k \le n$ such that $\operatorname{int}(\...
Henno Brandsma's user avatar
8 votes
Accepted

Non-orientable cover of a non-orientable surface

If $M$ is a connected manifold, let $nM$ denote the connected sum of $n$ copies of $M$. First note that if $X \to Y$ is a $k$-sheeted covering, then $\chi(X) = k\chi(Y)$. So if $X$ is a four-sheeted ...
Michael Albanese's user avatar
8 votes
Accepted

The 2 skeleton of a 3 manifold is the 2 skeleton of a $K(\pi, 1)$

I think you misunderstood what was being claimed. Since the $2$-skeleton on $M$ is the $2$-skeleton of a $K(\pi_1(M), 1)$ ... It is not claimed that the $2$-skeleton of $M$ is a $K(\pi_1(M), 1)$. ...
Michael Albanese's user avatar
8 votes
Accepted

A wild knot and its complement

Both (blue and red) curves represent non-trivial elements in the fundamental group of the complement. That is, none of them can be shrunk to a point. To see this, consider the linking number of these ...
Arnaud Mortier's user avatar
8 votes

Can the connected sum of three copies of $\mathbb{RP}^3$ be the total space of a fiber bundle?

Let $3\mathbb{RP}^3$ denote the connected sum of three copies of $\mathbb{RP}^3$. If $F \to 3\mathbb{RP}^3 \xrightarrow{\pi} B$ is a fiber bundle with manifold fiber and base, then $F$ and $B$ are ...
Michael Albanese's user avatar
7 votes
Accepted

Embedding 3-manifolds in Euclidean space

You can find some discussion and references on page 136 of the book by Lee "Introduction to smooth manifolds". In particular one of the result quoted is that Wall showed in 1965 that every 3-manifold ...
GFR's user avatar
  • 5,291
7 votes
Accepted

Bridge Presentation of a Knot

The most common definition of bridge presentation is the minimum number of local maxima over all diagrams of a knot. Similarly, there will the same number of local minima. With this, we can think of ...
N. Owad's user avatar
  • 6,662
7 votes
Accepted

To what extent are homeomorphisms just deformations?

Here's a theorem. Let $X$ be a closed subset of $\newcommand{\R}{\Bbb R}\R^m$ and $Y$ be a closed subset of $\R^n$. Let $X$ and $Y$ be homeomorphic. Embed $\R^m$ and $\R^n$ into $\R^{m+n}$ by $(x_1,\...
Angina Seng's user avatar
6 votes

Intuition behind Khinchin's constant

If you want a geometric explanation of the result, the existence of Khinchin's constant is a consequence of the erodicity property of the geodesic flow on the modular surface (more precisely, on the ...
Moishe Kohan's user avatar
  • 92.2k
6 votes

Existence of a vector field with one singularity on a surface

The statement actually holds in any dimension! There are references to this appearing in Hopf and Alexandroff's Topologie (1935), but I'm sure it was known earlier. Claim. Every compact, connected, ...
Kyle's user avatar
  • 5,963

Only top scored, non community-wiki answers of a minimum length are eligible