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Let $M$ be prime closed compact orientable 3-manifold with infinite fundamental group $G=\pi_1(M)$. Can there be finite order elements in $G$ ? What are the possibilities for $H_1(M)$ - can it be any abelian finitely generated group ?

If we have finite cover $N\to M$. In this case there is monomorphism $\pi_1(N)\to\pi_1(M)$. Is there also monomorphism $H_1(N)\to H_1(M)$ ?

From the "virtually Haken conjecture" I conclude that there exists finite cover $N$ which is Haken. It means it contains incompressible surface. It means $H_2(N)\neq0$.

If $M$ is closed compact orientable 3-manifold with finite fundamental group then $M$ is spherical. It means it is quotient of $S_3$ by finite subgroup of $SO_4$. This subgroup is fundamental group of $M$.

Anubhav Mukherjee
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mmm
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It cannot be $S^1\times S^2$, so in fact it is irreducible. So $\pi_1$ infinite implies its universal cover is contractible and moreover is $\mathbb R^3$ [Geometrization Theorem]. To see it's universal cover is contractible, observe that irreducibility implies $\pi_2$ is zero. So infact it's universal cover has $\pi_2$ zero. Now $\pi_1$ infinite implies it's universal cover is non-compact, so $H_3=0$ and thus by Hurewich theorem $\pi_3=0$. And with similar arguments it's all homotopy groups are zero. So it is contractible. And Thurston-Parelman's geometrization says that it is infact diffeomorphic with $\mathbb R^3$

Now if it has a finite order element in $\pi_1$ then infact we can have a finite free group action on $\mathbb R^3$ (WHY?). But it is not possible. So contradiction.

Anubhav Mukherjee
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  • Thank you. How do you deduce that universal cover is contractible ? Then $H_1$ is just sum of few copies of $Z$... or maybe not . I am not familiar with infinite group properties. – mmm Jun 30 '18 at 13:44
  • Thurston's-Parelman's Geometrization conjecture(theorem). Since you sited vertual Haken, I thought you must be familiar with this as well – Anubhav Mukherjee Jun 30 '18 at 17:34
  • @MarekMitros see the edit – Anubhav Mukherjee Jun 30 '18 at 21:31
  • Thank you for more details. I know about geometrization conjecture. I am trying to understand how decomposition of prime manifold along incompressible tori help classify 3-manifolds. Can we construct list of all prime manifolds this way ? I should post new question about it. – mmm Jul 01 '18 at 19:36
  • @MarekMitros what do you mean by 'this way' here in your last sentence? – Anubhav Mukherjee Jul 01 '18 at 20:57
  • Yes, I used shortcut. Once we decomposed given prime manifold $M$ on components having tori boundary we should been able to glue up the components back to obtain $M$. Therefore I though that it is possible to list all possible components $C_1,C_2,...$ each of them having tori boundary (and no more incompressible torus) and have recipe of gluing such that any prime manifold can be obtained this way. This procedure should be so good, so for each prime $M$ there should be different pair (components, gluing recipe). Obviously we have already such procedure - it is surgery on link but .... – mmm Jul 02 '18 at 04:57
  • (continued) it is surgery on link embedded in $S^3$, but is this 3-manifolds classification ? Maybe my knowledge is little here. Do we know when prime manifold is obtained as result of surgery and when two links give the same prime manifold ? If yes, then it remains to list all possible links in $S^3$. – mmm Jul 02 '18 at 05:03
  • @mark your procedure is not clear to me...moreover what will you do in case of hyperbolic closed manifold. And also if you are fully satisfied with my answer, then may accept it :) – Anubhav Mukherjee Jul 02 '18 at 17:48
  • I accepted your answer. I have asked new question how decomposition along incompressible tori help in 3-manifold classification. https://math.stackexchange.com/questions/2837854/how-decomposition-of-prime-3-manifold-along-incompressible-tori-help-in-classifi – mmm Jul 03 '18 at 09:22