Timeline for Hyperbolic 3-manifolds with no geometrically finite structure
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
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| Feb 7, 2013 at 18:58 | comment | added | Sam Nead | [See page 14, paragraph (3) of Hatcher's notes.] Basically, if a boundary torus $T$ compresses, then there is a disk $D$ with various properties. Take a neighborhood $K$ of $T \cup D$ and consider the frontier of $K$ in the manifold $M$. This will be a sphere and so bound a ball in $M$, by irreducibility. Thus $M$ is a solid torus. | |
| Feb 7, 2013 at 18:53 | comment | added | Sam Nead | @Igor - Not quite! For example, the solid torus $S^1 \times B^2$ is irreducible and atoroidal (and hyperbolizable!) but the boundary torus is compressible. There are also Seifert fibered spaces which satisfy the hypothesis and conclusion, but that are of course not hyperbolic. | |
| Feb 7, 2013 at 18:45 | vote | accept | Igor Belegradek | ||
| Feb 7, 2013 at 18:45 | comment | added | Igor Belegradek | So that I do not forget: the boundary tori of an irreducible, atoroidal compact 3-manifold must be incompressible because in the geometrically finite structure they correspond to rank two cusps, which are incompressible. | |
| Feb 7, 2013 at 17:57 | history | edited | Sam Nead | CC BY-SA 3.0 |
Answering Igor's actual (?) question, linking to Hatcher.
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| Feb 7, 2013 at 17:53 | comment | added | Sam Nead | I will add a reference to my answer. | |
| Feb 7, 2013 at 17:26 | comment | added | Igor Belegradek | It is not a new question because after checking that an atoroidal, irreducible manifold is pared, you are done by Thurston (e.g. Theorem 1.43 of Misha's book which was initially stated in my question). Checking that it is pared is the only thing that takes some work. Anyway, thank you for the input. I will try filling in the details. | |
| Feb 7, 2013 at 17:13 | comment | added | Sam Nead | There is a purely geometric proof, as well. Suppose $N$ is hyperbolic, and $B_S$ and $B_T$ are disjoint horo-tori about $S$ and $T$. Suppose that $A$ is a compact annulus connecting $B_S$ to $B_T$. Lift everything to the universal cover. A component of the lift of $A$ is a strip (quasi-isometric to a line!) that fellow-travels lines in two distinct horospheres, a contradiction. | |
| Feb 7, 2013 at 17:08 | comment | added | Sam Nead | There are lots of copies of $Z^2$ in $\pi_1(K)$ and all of them have to be parabolic. This leads to contradictions. | |
| Feb 7, 2013 at 17:04 | comment | added | Sam Nead | @Igor - I suggest you ask this (new, to my eyes) question in a separate post. However, very briefly, a hyperbolic manifold is algebraically atoroidal - that is, any $Z^2$ subgroup is parabolic. [This is an exercise in hyperbolic geometry, using the discreteness of the group.] Let's now do just one of the many cases. Suppose that $S$ and $T$ are boundary tori, and $A$ is an annulus between them. Let $K$ be a small neighborhood of $S \cup A \cup T$. Then $K$ is homeomorphic to $P \times I$ where $P$ is a pair of pants. So $\pi_1(K)$ is a rank two free group, crossed with $Z$. | |
| Feb 7, 2013 at 15:42 | comment | added | Igor Belegradek | Sam, I would appreciate, if you could include in your answer a detailed explaination why a hyperbolizable manifold can contain no essential annulus joining two torus boundaries. This is the only thing I did not (and still do not) understand; what you wrote does not tell me anything new except stating that the answer is "no". | |
| Feb 7, 2013 at 15:06 | history | edited | Sam Nead | CC BY-SA 3.0 |
Cleaning up the logic.
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| Feb 7, 2013 at 14:54 | comment | added | Sam Nead | Because Theorem 19.6 is a different path to proving the hyperbolization theorem. See the remarks immediately after the proof of Theorem 15.3, at the top of page 372. | |
| Feb 7, 2013 at 14:44 | history | edited | Sam Nead | CC BY-SA 3.0 |
added 110 characters in body
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| Feb 7, 2013 at 14:22 | comment | added | Igor Belegradek | If what you say is true, why would anyone bother to prove the weaker statement 3 in my list (which is theorem 19.6 in Misha's book)? | |
| Feb 7, 2013 at 14:11 | comment | added | Sam Nead | You are correct - I was referring to the wrong thing. I have added a temporary fix, and will look for a precise reference. | |
| Feb 7, 2013 at 14:10 | history | edited | Sam Nead | CC BY-SA 3.0 |
FIxed large mistake!
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| Feb 7, 2013 at 12:49 | comment | added | Igor Belegradek | Sam, this is not an answer because it misses the most interesting case when $M$ has a boundary component of genus $>1$. In this case $N$ does not admit a finite volume hyperbolic metric. Of course, it might still admit a complete infinite volume hyperbolic metric. | |
| Feb 7, 2013 at 11:48 | history | answered | Sam Nead | CC BY-SA 3.0 |