# Tag Info

## Hot answers tagged gt.geometric-topology

17

I presume by "acyclic" you are referring to homology with $\mathbb{Z}$ coefficients. There are many such examples. For instance, you can take two elements $u,v$ in the free group $F_2$ of rank 2 that satisfy the $C'(1/6)$ small-cancellation condition, and also such that $u,v$ together generate the abelianisation $\mathbb{Z}^2$. Explicit examples are easy ...

15

The Higman group with presentation $$\langle{a,b,c,d}\mid{aba^{-1}b^{-2}},~bcb^{-1}c^{-2},~cdc^{-1}d^{-2},~ dad^{-1}a^{-2}\rangle$$ is perfect, and the 2-complex associated to this presentation has Euler characteristic 0. Hence this complex is acyclic. It is in fact aspherical, but it may be simpler to observe that Higman's group is also an iterated ...

9

Here is a proof with pictures. Observe that blowing up and blowing own doesn't change the 3 manifold,i.e, the boundary. So all these above pictures have same boundary. As a 3 manifolds all of these are isomorphic.

9

This answer is courtesy of Sam Corson who kindly pointed out the following. Theorem: The Hawaiian earring group $\pi_1(\mathbb{H})$ is essentially freely indecomposable, i.e. if $\pi_1(\mathbb{H})\cong G_1\ast G_2$, then one of $G_1$ or $G_2$ must be a finitely generated free group. The key is to apply a special case of Theorem 1.3 in K. Eda, Atomic ...

9

As Golla pointed out that since every smooth $4$-manifold has a handle decomposition, you can draw a Kirby diagram. See the following pretty nice picture from Akbulut's lecture notes (now it is a published book titled as $4$-manifolds). As Golla listed that lots of Brieskorn spheres are known to be bound integral or rational homology balls, i.e., they are ...

7

There are many examples of the sort, in effect. As far as I know, Akbulut and Kirby (Mazur manifolds, Michigan Math. J. 26 (1979)) proved that $\Sigma(2,5,7)$, $\Sigma(3,4,5)$, and $\Sigma(2,3,13)$ bound contractible 4-manifolds; their work was then extended by Casson and Harer (Some homology lens spaces which bound rational homology balls, Pacific Math. J. ...

4

I'm obliged to add "The topology, geometry, and dynamics of free groups. Part I: Outer space, fold paths, and the Nielsen/Whitehead problems" by Lee Mosher which I find very helpful (looking forward to Parts II & III).

4

Yes, this is true. It is shown (either in the paper you cite or the other McShane-Rivin paper) that the length of a simple closed geodesic is quasi-the-same as the combinatorial length ($m+n$) (this is easy, because a simple geodesic stays away from the cusp), and that, in turn, is easily seen to be quasi-the-same as the Euclidean length.

4

In fact, more is true and you do not need separate arguments for rank 1 and higher rank. The following is Theorem 13.1(i) in the book of Ballmann, Gromov and Schroeder "Manifolds of nonpositive curvature": Suppose that $(M,g)$ is a complete real-analytic Riemannian manifold of nonpositive curvature and finite volume. Then $M$ is tame: It is diffeomorphic ...

4

I'll try to summarize YCor's comments into an answer (using big guns): Let $G$ be the real points of an algebraic group (a restriction by the OP in the comments) and assume $\Gamma$ irreducible. Then Raghunathan shows that the answer is "yes" if $\Gamma$ is arithmetic. Margulis (Discrete subgroups of semisimple Lie groups) says that $\Gamma$ will be ...

3

The cited (early) work by Cerf proves that, given a submanifold Y in a manifold X, the obvious map Diff(X)->Emb(Y,X) is a locally trivial fibration. I guess that Budney and Gabai mean the following. By Palais, all embeddings D^3->S^4 are isotopic. Hence, for i=0, 1, the complement C_i of a small open tubular neighborhood U_i of Delta_i in S^4 is ...

2

Ian's argument of mean curvature is wonderfully simple. Here is another one. Rotate your surface to put it in generic position with respect to the heigth function z; then, the preimage of z is a Morse function f on RP^2, which has no critical point of index 1 (saddle point) since the surface is locally convex. Hence, every critical point of f has index 0 or ...

1

Such a homotopy exists and in fact you can assume that it is an isotopy. This is a "standard fact" in the theory of mapping class groups. See Proposition 2.2 of the "Primer" by Farb and Margalit.

1

You may simply find the surgery diagram of Brieskorn spheres. This is from Özbağcı's lecture notes. The small Seifert fibered $3$-manifold $M(r_1, r_2, r_3)$ is defined by the following rational surgery diagram. Then $- \Sigma(2,3,5) \cong M \left(\frac{-1}{2}, \frac{1}{3},\frac{1}{5} \right) \cong \partial E_8 \cong S^3_1(3_1)$, \$- \Sigma(2,3,4) \...

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