Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've read in a paper that if $M$ is a prime, reducible $3$-manifold, then $\pi_{1}(M) \cong \mathbb{Z}$. Can anyone explain why this is true?

Thanks in advance.

share|improve this question
1  
en.wikipedia.org/wiki/… –  Brandon Carter Jul 10 '12 at 19:30
add comment

1 Answer

up vote 2 down vote accepted

I'm probably assuming $M$ to be orientable.

Reducible means that there is an essential sphere $S \subseteq M$. Two cases happen : either that sphere disconnects $M$, and $M$ is decomposable as a nontrivial connected sum (so it is not prime) or $M$ is homeomorphic to $S^2 \times S^1$.

So, $S^2 \times S^1$ is the only prime reducible $3$-manifold.

share|improve this answer
2  
The same technique of proof works in the non-orientable case. Either way, take a loop intersecting this non-separating sphere once transversally, then an open neighborhood is either a (punctured) $S^2\times S^1$ or its nonorientable cousin. –  user641 Jul 10 '12 at 20:18
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.