# Tag Info

3

A tubular neighborhood of $S^1$ in $\mathbb{R}^2$ is an annulus (which is topologically just a cylinder). Making the one-point compactification is just collapsing the top and bottom borders of the cylinder together into a point, which is equivalently a sphere that's had two points collapsed:

6

Let $\mathcal S$ be a compact surface, possibly with boundary. Let $\text{Homeo}^+(\mathcal S)$ refer to the group of orientation-preserving homeomorphisms that fix the boundary pointwise with the compact-open topology. Throwing an $n$ in there means we add $n$ marked points in the interior, which I prefer over deleting points. (I'd be worried about the ...

1

Yes, these points are described by the open set $h^{-1}(0, \infty)$, where $h(x,y)= g(x,y)-f(x,y)$, and $h$ is smooth ( though continuous would be enough).

1

Jørgensen's inequality says that if two elements $A, B \in SL_2(\mathbb{C})$ generate a non-elementary discrete group, then $$|\mathrm{Tr}(A)^2-4| + |\mathrm{Tr}(ABA^{-1}B^{-1})-2| \ge 1$$ Assuming that $A=\begin{bmatrix} 1& 1 \\ 0 & 1\end{bmatrix}$ and $B=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ (with $a,b,c,d \in \mathbb{R}$ and ...

2

Contrary to the request of the OP, this is still a sketch. I put it here because it points to some relevant literature, hoping that somebody more expert than me will help to clarify it. The question reminded me of this paper by Fischer and Zastrow: the relevant part is the end of p. 12. See also this answer by George Lowther. You can assume that $\cup D_n$ ...

0

Here is an outline of an alternate method of proof: Regard $\mathbb R^2$ as $\mathbb C$ and write $D_n=\bar{B}(z_n,r_n), D^{\circ}_n=B(z_n,r_n)$. Step 1: $\mathbb C\setminus \{z_1,z_2,\ldots,\}$ is path-connected Step 2: There is a continuous surjection $f\colon \mathbb C\setminus \{z_1,z_2,\ldots,\}\to \mathbb C\setminus \bigcup D^{\circ}_n$ Proof of ...

1

The expression you wrote does not define a Riemannian metric for coordinates on $S^3$, because when $\eta=0$ or $\pi/2$ the metric degenerates. It does define coordinates for a portion of $S^3$, namely the portion where the coordinates satisfy the strict inequalities $0 < \eta < \pi/2$, $0 < \xi_1, \xi_2 < 2 \pi$. So perhaps you meant to use ...

0

I will add another answer, which depends on the assumption that the set $\mathbb{R}^2 -\bigcup_{n=1}^\infty D_n$ is uncountable and on the assumption that for every point $(x,y)$ in the plane there is a point in the set $\mathbb{R}^2 -\bigcup_{n=1}^\infty D_n$ that is at the positive distance(I mean strictly positive here and below in the proof where I use ...

0

The tangent Bundle of sphere is another example. For arbitrary vector bundle E, it seems that $E\otimes \mathbb{C}$ has a trivial line bundle, as a summand. http://mathoverflow.net/questions/209247/almost-complex-structure-and-nontrivial-idempotents Moreover, I guess that the complexification of canonical line bundle over $\mathbb{R}P^{n}\;\;n>1$ ...

1

So here is my attempt to clean up and clarify the other answer. We pick $A,B \in \Bbb R^2 \setminus S$ For some disks $D_n = B(c_n,r_n)$ we're going to find a "thickening" of $D_n$, $T_n$, such that its border $B_n$ is disjoint from every disk, and find a continuous projecti on $f_n : T_n \setminus \{ c_n\} \to B_n$. We also want the $T_n$ to be disjoint ...

0

The curve $\gamma= m_2 + 2 \ l_2$ bounds a Moibus strip $M$ inside the the solid torus $T_2\subset L(2,1)$ (can you see it?) and a two-disk $D$ inside $T_1 \subset L(2,1)$. The union $S= D \cup M$ is a closed non-orientable surface homeomorphic to $\mathbb{RP}^2$, and you can easily see that its complement in $L(2,1)$ is nothing but a 3-ball. Since ...

0

EDIT: This answer has a flaw in itself. It seems that the set $\mathbb{R}^2 -\bigcup_{n=1}^\infty D_n$ will not always be path connected, under the assumption that my line of reasoning is correct an that there is no flaw in the proof below. So let us construct the countable set of pairwise disjoint disks in the following way: Let $A_{11}$ be the disk with ...

2

Your idea of using a triangulation each of whose simplices lies in one of the $U_i$ is a good one. There exists such a covering of the form $V_0 \cup \ldots \cup V_D$ where $D$ is the dimension of $M$, such that if $M^{(k)}$ denotes the $k$-skeleton of the triangulaion then $$M^{(k)} \subset V_0 \cup \ldots \cup V_k \quad\text{for each k=1,\ldots,D.}$$ ...

4

The four non-orientable non-compact Euclidean 3-manifolds are: $M\times \mathbb{R}$, $M\times S^1$, $K\times \mathbb{R}$, and $K \times S^1$, where $M$ denotes the open Mobius band (i.e. the quotient of $\mathbb{R}^2$ by a glide reflection) and $K$ denotes the Klein bottle. Note that these are non-homeomorphic, since their fundamental groups are ...

1

Here is another solution, somewhat in the same spirit as Olivier Bégassat's answer. In fact, we will prove a stronger result : any bijection $f : \mathbb{Z}^2 \to \mathbb{Z}^2$ comes from a smooth isotopy $h : \mathbb{R}^2 \times [0,1] \to \mathbb{R}^2$ such that $h(x,0) = x$ and $h(-,1) = f : \mathbb{Z}^2 \to \mathbb{Z}^2$. As in Hagen von Eitzen's answer, ...

2

I think it is path-connected. Let $S = \{ D_n \, | \, n \in \mathbb{N} \}$ be the set of closed disjoint disks under consideration. For any integer $m \ge 1$, let $N_m = \{ n \in \mathbb{N} \, | \, \mathrm{radius}(D_n) \in [1/m, 1/m-1) \}$ (with the convention $1/0 = \infty$) and $S_m = \{ D_n \, | \, n \in N_m \}$. Let $P, Q \in \mathbb{R}^2 \backslash S$ ...

3

Here's a sort of diagrammatic argument from an old homework assignment: Construct $\mathbb{R}P^3$ as the quotient of $B^3 \subset \mathbb{R}^3$ under the antipodal map $a: \partial B^3 \to \partial B^3$. Let $K$ be the knot in $\mathbb{R}P^3$ obtained as the quotient of the vertical segment $V=\{(0,0,z) \in \mathbb{R}^3 : -1 \leq z \leq 1\}$ in $B^3$. As ...

3

Here's an idea on how to construct a solution for the bijection you describe. It appears feasible by a diffeomorphism of the plane. First we solve the following problem : extend the bijection \begin{array}{RCL} \phi:\Bbb Z\times\lbrace-1,0\rbrace & \longrightarrow & \Bbb Z\times\lbrace-1,0\rbrace\\ (n,\epsilon) & \longmapsto & \begin{cases} ...

18

Yes. We can consider a sequence of disks $D_1,D_2,\ldots$ around $(\sqrt 2,\pi)$ such that the $n$th disk has precisely $n$ lattice points in its interior (and none on its boundary). Let $z_n$ be the lattice point in $D_n\setminus D_{n-1}$. Construct accordingly a sequence $C_1, C_2, \ldots$ of simply connected compact sets with smooth boundary such that ...

0

Let $S$ be the surface. I will be using the following facts: 1) There are two types of simple closed curves, separating and non-separating. Separating closed curves are zero in $H^1(S)$. 2) $H^1(S)=\frac{\pi_1(S)}{[\pi_1(s),\pi_1(S)]}$, (Theorem 2A.1 of Hatcher). 3) For any two non-separating curves $a$ and $b$, there exists a homeomorphism $\phi$ such ...

Top 50 recent answers are included