# Tag Info

## Hot answers tagged gt.geometric-topology

17

Take a non-orientable $S^n$ bundle over $S^1$ with $n \geq 2$ (*), (sometimes called generalised Klein bottles) then covering $S^1$ by two intervals and taking preimages should work. (*) Let $\tau : S^n \rightarrow S^{n}$ be a non-orientable diffeomorphism of $S^n$ given by $(x_{1},\ldots,x_{n+1}) \mapsto (x_1,\ldots,-x_{n+1})$. Then such a bundle is given ...

14

The first part of my argument is borrowed from Iosif Pinelis, and the second part is different. Every open set contains a closed cube, so it suffices to show the closed cube does not have measure $0$. In fact, we will show that for a cube of side length $r$, the total volume of open cubes covering it is at least $r^n$. To do this, assume that it is covered ...

9

Since $c\subset H_g$ intersects an essential disc $D$ in a single point, the boundary of a regular neighbourhood of $D\cup c$ is another disc $D'$, which splits $H_g$ into a solid torus containing $D\cup c$ and the rest. You can forget about the rest (this is a $\partial$-connected sum) and consider the solid torus alone. Here, if you push $c$ inside the ...

8

Let $X$ be any paracompact space. Then Hilbert vector bundles over $X$ are classified by homotopy classes of maps $[X, BU(\mathcal H)]$. But when $\mathcal H$ is infinite-dimensional, the group $U(\mathcal H)$ is contractible (this is Kuiper's theorem), and hence every infinite-dimensional Hilbert bundle over a paracompact space is trivializable. Hilbert ...

7

If I understand the question correctly, the lens space $L(5,2)$ provides a counterexample. Let $f$ be the identity, and let $S \subset L$ be a circle that generates the first homology. For concreteness sake, take $L$ to be the union of two solid tori of the form $S^1 \times D^2$, and let $S = S^1 \times pt$ in one of the solid tori. So the restriction of $f$...

7

The $\tilde{\rho}$-invariants contain the $\tilde{\rho}_G$-invariants, at least. The map $\Sigma_g \to \Sigma_b$ defines a pullback map $H^1( \Sigma_b, \mathbb Q) \to H^1(\Sigma_g, \mathbb Q)$ (i.e. cohomology is a contravariant functor). The image of this pullback map is the $G$-invariants in $H^1(\Sigma_g, \mathbb Q)$. The image of this pullback map ...

7

The answer is that the outer automorphism group is $\mathbb{Z}_2$. (Compare Ian Agol's answer to What is the order of the isotopy group of the Brieskorn homology 3-sphere?). This is obtained by cobbling together several standard facts. Much of this can be found in Boileau and Otal, Scindements de Heegaard et groupe des homéotopies des petites variétés de ...

7

The details of the result slightly stronger than that oulined in Tom Goodwillie's answer appears as Proposition 3.5 of this paper of Bustamante, Krannich, and myself. Let me state a version of this result that uses a bit less notation that the one in the paper: Let $M$ be a compact smooth manifold of dimension $d$ and $N \subset \mathrm{int}(M)$ be a ...

6

You should check out Thurston, William, Noncobordant foliations of $S^3$. Bull. Amer. Math. Soc. 78 (1972), 511–514. He constructs foliations with arbitrary real-valued GV invariants. In this paper, he described the GV invariant as "measuring the helical wobble" of a foliation.

6

The answer is negative. Already in dimension 4 there are fake real-projective spaces, which are smooth 4-manifolds homotopy-equivalent but not homeomorphic to $RP^4$. These correspond to smooth free involutions $\sigma: S^4\to S^4$ which are not topologically conjugate to orthogonal transformations. Similar examples exist in higher dimensions. See manifold ...

5

You don't say what the morphisms in $C$ are supposed to be. I'll take them to be isotopy classes of orientation-preserving diffeomorphisms. Let $\mathbb{N}$ be the category with object set $\mathbb{N}$ and only identity morphisms. Then the genus gives a functor $\pi\colon C\to\mathbb{N}$, and by choosing a surface $X_g$ of genus $g$ for all $g\geq 0$ we ...

5

Here is more direct and elementary argument. Lemma: $\mathrm{Aut}(W_3)$ and $\mathrm{Aut}(\mathbb{F}_2)$ are isomorphic, where $W_3$ denotes the free product $\mathbb{Z}_2 \ast \mathbb{Z}_2 \ast \mathbb{Z}_2$. Sketch of proof. Let $F \leq W_3$ denote the kernel of the morphism $W_3 \twoheadrightarrow \mathbb{Z}_2$ sending all the generators $a,b,c$ to $1$. ...

5

As Mike notes Hilbert bundles are trivial over most spaces. The paper MR2481802 (2010e:46083) Reviewed Abbondandolo, Alberto (I-PISA); Majer, Pietro (I-PISA) Infinite dimensional Grassmannians. (English summary) J. Operator Theory 61 (2009), no. 1, 19–62. 46T05 (47A53 58B15) https://arxiv.org/pdf/math/0307192.pdf Studies the homotopy type of the relative ...

5

Let me mention another proof of the $\sigma$-additivity of the lenght of intervals, that uses transfinite induction instead of HB compactness. Say $n=1$, and we wish to prove that the set-function $$\lambda:[a,b)\mapsto b-a$$ is $\sigma$-additive on the family $\mathcal J$ of all left-closed, right-open intervals of $\mathbb{R}$ The key observation is that ...

5

Any nonempty open set in $\mathbb R^n$ contains a compact cube $C$ of volume $v:=|C|>0$. (All our cubes will be assumed to have edges parallel to the coordinate axes.) So, it is enough to show that $C$ cannot be covered by a set $S$ of (say) open cubes with total volume $<v$. By compactness, without loss of generality the cardinality (say $k$) of the ...

4

The simplest construction I can think of is as follows. Let $X$ be the complement of the Borromean rings in the three-sphere. Then an $n$-fold cover of $X$ (unwrapping only one boundary torus) will again be hyperbolic and will have $2n + 1$ torus boundary components.

4

The group $\mathrm{Aut}(F_2)$ is not relatively hyperbolic. This is contained in (the proof of) Theorem 8.1 of Behrstock-Drutu-Mosher. We first pass to $\mathrm{Aut}^+(F_2)$, the preimage of $\mathrm{SL}(2, \mathbb{Z})$. By Remark 7.2 of Behrstock-Drutu-Mosher it is enough to consider this index two subgroup. Let $S$ be a copy of the two-torus. Let $Z = \{... 3 This question for a product of one Gromov hyperbolic space is a famous open problem (see for example Bestvina's problem list) and as far as I am aware the general case is also open. 3 This is a bit of a long comment and perhaps a useful complement to the other answers. Having quickly read the question, I almost proceeded to downvote. But on second thought, I think it is not as trivial as it seems and, in fact, lies at the very heart of the standard construction of Lebesgue measure even for$n=1$using Carathéodory's Theorem. After some ... 3 A generalization of a barycentric subdivision is the notion of a stellar subdivision, where one performs a barycentric subdivison of a simplex$s$and subdivides all simplices accordingly that contain$s$as a face. The inverse of a stellar subdivision is called a stellar weld. Now let$K$and$L$be two abstract simplicial complexes. If the topological ... 3 Regarding the second question. This reduces to the homeomorphism problem for three manifolds, which is known due to geometrisation. Here is a sketch. Fix$M$as well as the knots$K$and$L$. Let$X_K$and$X_L$be the knot complements. We mark the boundary of each with the meridional slope of$K$and$L$, respectively. We ask our solution to the ... 2 I think the answer is no. There is a metrizable version of the Tangent Disc Topology, namely where instead of extending the Euclidean topology in the upper half-plane to all of$\mathbb{R}$you extend it to a countable subset of$\mathbb{R}$. This will be locally path-connected and connected (see Counterexamples in Topology by Steen/Seebach), but not ... 2 For (a),$C$does bound a disk in$B^+$which is later called$\Delta'$, but by assumption$C$does not bound a disk in$F\cap B^+$. A priori, the components of$F\cap B^+$are subsurfaces of the sphere$F$; they could be disks, annuli, pairs of pants, etc. For (b), the surface$F$could intersect a bubble in any number of saddles. These saddles inherit an ... 2 The statement is false. Here is a counter-example. Let$T$be an ideal triangle (say in the unit disk model). Let$S$be the surface obtained by doubling$T$across it’s boundary: that is, take two copies and glue by the identity on the boundary. Let$x$be the centre of$T$. The Dirichlet domain based at$x$has six vertices with three ideal and three ... 2 It should also be mentioned that$[0,1]$being not a null Lebesgue set can also be derived by the Lebesgue-Vitali characterization theorem for Riemann integrable function. If$[0,1]$were negligible, any bounded function on it would be Riemann integrable, but we know that for instance the Dirichlet function$\chi_{[0,1]\cap\mathbb{Q}}$is not. 2 You seem to be correct. I was wrong, probably because this result appears in Ciesielski's book ("Set theory for the working mathematician" CUP, 1997; Theorem 6.1.3) near a result of Sierpinski. I corrected my question accordingly. 2 As for conformal moduli of the tori (more precisely, Teichmuller parameters) that appear: It is hard to tell, afaik, there is no explicit description. We know that these will be elements of$\bar{{\mathbb Q}}$(this is essentially due to Selberg, 1960) and that they form a dense subset in${\mathbb H}^2\$ in the case of one boundary torus. See Nimershiem, ...

1

Though it does not answer this question, the paper https://arxiv.org/abs/1509.03748 contains a weaker notion of nonpositive curvature than CAT(0) which all groups acting geometrically on product of Gromov hyperbolic groups satisfy and such that there are examples of non-CAT(0)-groups which also satisfy this weaker notion.

1

One can show that a closed cube is not even negligible using (hyper)rectangles in your definition of "negligible". Indeed, given a cover with rectangles (w.l.o.g. initially open up to enlarging them a bit, w.l.o.g. finite by compactness of the cube, and w.l.o.g. contained in the cube), up to subdividing them you can assume that whenever a side of a ...

1

Theorem 4 in the paper "Branched polynomial covering maps" by Vagn Lundsgaard Hansen (Topology and its Applications 125(2002) 63-72).

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