# Tag Info

### 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 ...
• 2,999

### 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
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

### 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 ...
• 3,732

### 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

### 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 ...
Accepted

• 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 $D=\mathbf{C}\backslash\{-1,1\}$. ...
• 81.6k
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 $\pi_1(M) \cong \mathbb{Z}^n$. ...
• 17.7k
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 ...
• 39.4k
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 ...
• 1,345

### 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