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6 votes

Is there a contractible hyperbolic 3-orbifold of finite volume?

As pointed out by Moishe Kohan in the comments below, the following doesn't answer the question as asked, because my group $\Gamma$ is not contained in $SO(3,1)$. Anyway, here is an easy description ...
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4 votes

Ideal triangulations of $3$-manifolds with "cusps" of genus $\ge 2$

A nice class of examples are the (generalized) triangulations with only one edge. The manifolds obtained by removing an open neighborhood of the vertex have totally geodesic boundary (or a cusp in the ...
  • 62.8k
13 votes
Accepted

Is there a contractible hyperbolic 3-orbifold of finite volume?

Yes. For example, let $M$ be the figure-eight knot complement. So $M$ is a hyperbolic manifold with volume a bit more than 2. The manifold $M$ has a two-fold symmetry $\tau$ that fixes, pointwise, a ...
  • 21.1k
2 votes

Centre of orbifold fundamental group of torus (Klein bottle) with one cone point

The centre of any non-cyclic one-relator group with torsion is trivial. Moreover, the centraliser of any non-identity element of a one-relator group with torsion is cyclic. This is B. B. Newman’s ...
6 votes

Homology of spherical $3$-manifold group

The attaching map has to kill $\pi_3(S^3/G)$, and the map $\mathbb Z =\pi_3(S^3) \to \pi_3 (S^3/G) $ induced by the covering is an isomorphism, so $\pi_3$ is generated by the class of the covering 3-...
  • 122k
7 votes

Centre of orbifold fundamental group of torus (Klein bottle) with one cone point

The groups have trivial center, as pointed out by Sam Nead. Another, more combinatorial, way to show this is to apply the algorithm from [1], which decides whether any given one-relator group has a ...
8 votes
Accepted

Centre of orbifold fundamental group of torus (Klein bottle) with one cone point

These groups have trivial centers. As one proof, they are both fuchsian and so embed in $\mathrm{PSL}(2, \mathbb{R})$ (well, the orientation preserving subgroups do). However, elements of $\mathrm{...
  • 21.1k
8 votes

How to prove that Lie group framing on S^1 represents the Hopf map in framed cobordism

I wrote out careful proofs of all of this and more in my note "Homotopy groups of spheres and low-dimensional topology", available here.
  • 40.6k
0 votes
Accepted

Finitely generated groups with Hölder-exotic space of ends?

The question is solved positively in the paper The Hausdorff dimension of the harmonic measure for relatively hyperbolic groups by Matthieu Dussaule, Wenyuan Yang, which appeared on arXiv on October ...
16 votes

Topology of the space of embedded genus $g$ surfaces in $S^3$

The first observation is that $\mathcal{E}_g$ is not connected when $g > 0$. This is due to the existence of "knotting". For example, in genus one, let $K$ and $K'$ be smooth knots. ...
  • 21.1k
2 votes

Question about and good reference for Kahn and Markovic result

I think that a good reference could be the following: Proposition ([1, Proposition 4.1 (or 5.1 in the arXiv version)]): Let $M$ be a closed hyperbolic 3-manifold, and regard $\pi_1(M)$ as acting on $\...
  • 674
1 vote

Ideal triangulations of $3$-manifolds with "cusps" of genus $\ge 2$

I think you mean, in the first paragraph “the link of each ideal vertex is a torus” and in the second paragraph “the link of an ideal vertex is instead a surface of genus two”. Such triangulations are ...
  • 21.1k
1 vote

Existence of covering isomorphism

I suppose that "non-compact complex algebraic curve" means complex affine curve. The following counterexample was proposed by my friend Fedor Pakovich. Let $D=\mathbf{C}\backslash\{-1,1\}$. ...
13 votes
Accepted

Stable torus that is not a torus

Suppose $M\times S^1$ is homeomorphic to $T^{n+1}$. Then $\pi_1(M\times S^1) \cong \pi_1(T^{n+1})$, so $\pi_1(M)\oplus\mathbb{Z} \cong \mathbb{Z}^{n+1}$, and hence $\pi_1(M) \cong \mathbb{Z}^n$. ...
2 votes
Accepted

Spaces satisfying a strong Cartan-Hadamard theorem

Note that Hilbert spaces (of all dimensions finite or infinite) are the only geodesic spaces with extendable geodesics which are flat in the sense of Alexandrov. Therefore $X$ has to have extendable ...
9 votes
Accepted

Are any embeddings $[0,1]\to\mathbb{R}^3$ topologically equivalent?

As igorf pointed out in the comment, the answer to the first question is 'no'. A quick counterexample is by looking into the complement of Fox-Artin arc, which is not simply connected. See figure ...
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3 votes

Action of noncentral mapping classes on curves or arcs on a surface

If there are infinitely many (isotopy classes of) curves, then yes. Here is sketch of a proof. Let $S$ be the surface and let $\mathcal{C}(S)$ be the curve complex. The diameter of the curve complex ...
  • 21.1k
5 votes
Accepted

Action of noncentral mapping classes on curves or arcs on a surface

Yes, this is true - there are many ways to prove it, but I'll hit it with a hammer. Let $C(\Sigma)$ denote the curve complex of $\Sigma$. Suppose not, then $f$ would map every curve $[c]$ in $\mathcal{...
  • 62.8k
6 votes
Accepted

Knot concordance, hyperbolicity and amphichirality

Neither of these properties are preserved by concordance. As was pointed out in the comments, any hyperbolic (for the first question) or chiral (for the second question) knot which is concordant to ...
10 votes
Accepted

Two surfaces in a 4-manifold whose algebraic intersection number is zero

Yes, this can be done by tubing one surface along the other. Suppose that you have two intersection points $p_+, p_- \in \Sigma_1 \cap \Sigma_2$ of opposite signs. Suppose also that $\Sigma_1$ and $\...
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