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A three-manifold is a space that locally looks like Euclidean three-dimensional space

4 votes

Attaching a thickened annulus between two 3-manifold

Let me try to give a more elementary argument. Assume there is a compression disk. One can assume it is transversal to $A\times\left\{\frac{1}{2}\right\}$, thus it intersects $A\times\left\{\frac{1}{ …
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4 votes

Tight vs. overtwisted contact structure

Perhaps I'm wrong but I thought that historically the notions were considered first in the work of Bennequin who proved the existence of contact structures not equivalent to the standard contact struc …
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3 votes

Does Dehn filling always decrease Gromov norm?

In fact, for manifolds with torus boundary you have $$\parallel M,\partial M\parallel_0=\parallel M,\partial M\parallel$$ so that the two inequalities are plainly equivalent. (More generally, this eq …
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1 vote

Does Dehn filling always decrease Gromov norm?

There is also an explicit argument which shows $$\parallel M_1\cup_{A}M_2\parallel \le \parallel M_1,\partial M_1\parallel + \parallel M_2,\partial M_2\parallel$$ whenever $A=\partial M_1=\partial M_2 …
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9 votes
Accepted

3-dimensional h-cobordisms

For $M_0=M_1=S^2$, this follows from the 3-dimensional Poincaré conjecture: glueing in 3-balls, you get a simply connected $3$-manifold, that has to be $S^3$, so $W=S^2\times\left[0,1\right]$. For sur …
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11 votes

3-manifolds with isomorphic fundamental groups

(Edit: Everything what follows is about closed and orientable $3$-manifolds.) Non-spherical geometric $3$-manifolds are determined by their fundamental group. This is proved in Peter Scott's paper "T …
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7 votes

Can bilipschitz models of hyperbolic 3-manifolds be made effective?

See Bowditch: link text Systems of bands in hyperbolic 3-manifolds with an approach to the Brock-Canary-Minsky Theorem (though not through their model manifold) that is, in principle, effective. Thou …
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0 votes

what is the meaning of a curve $C$ representing Identity in fundamental group?

Dehn's Lemma (= Papakyriakopoulos' Theorem) asserts: if C represents 0 in $\pi_1M$ and if C is a simple closed curve, then C bounds an embedded disk. The assumption on C being a simple closed curve i …
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6 votes

braid representation of a Montesinos link

No, the braid index of a Montesinos link can be as large as one wishes. One way to see this is as follows. There is a lower bound for the braid index in terms of the Homfly polynomial due to Morton …
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4 votes
1 answer
423 views

Hyperbolic Dehn surgeries and SU(2)-representations

Let $S^3-K$ be the complement of the figure eight knot complement. Thurston, in his Lecture Notes, constructed a hyperbolic structure, which comes from a discrete, faithful representation $\pi_1(S^3-K …
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6 votes

Gluing two 3 manifolds along their boundary

Ad (2): the mapping class group of surfaces is isomorphic to the outer automorphism group of its fundamental group - this is a theorem of Baer-Dehn-Nielsen. In particular there is even a finite set of …
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4 votes
2 answers
296 views

Quadratic cusp shape

Which hyperbolic $3$-manifolds are known to have quadratic cusp shape? Explanations: Cusps of hyperbolic $3$-manifolds are products torus x interval. They lift to horoballs in hyperbolic $3$-space, wh …
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5 votes

Mapping Class Groups and torus (JSJ) decomposition of closed 3-manifolds

The natural definition of a higher Dehn twist at a surface $F$ is to identify a neighborhood of $F$ homeomorphic to $F\times\left[0,1\right]$, take a loop $\phi_t$ in $Homeo(F)$ based at $id$, and con …
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7 votes
Accepted

Reconciling Sullivan's theorem with the hyperbolic structure of the Figure–8 knot complement

The statement is only true if you restrict to geometrically finite hyperbolic metrics (possibly of infinite volume) and ignore parabolic elements, which basically means that you ignore the boundary …
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9 votes

Homeomorphic but Non-Conjugate Mapping Tori

McMullen and Taubes 4-manifolds with inequivalent symplectic forms and 3-manifolds with inequivalent fibrations constructs 3-manifolds $N$ with different fibrations, whose Euler classes do not lie in …
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