Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Following is an argument given by Hempel where I am unable to understand his comment about choosing a loop close enough to a surface. Can somebody please elucidate this:

Lemma: If $F$ is a compact connected surface properly embedded in a $3$-manifold $M$ and if $image(i_*:H_1(F;Z/2Z)\rightarrow{H_1(M;Z/2Z)})=0$, then $F$ is 2-sided in $M$.

Proof: By regular neighborhood theory, it suffices to show that $F$ separates some connected neighborhood of $F$. If this is not the case then there is a loop $J\subset{M}$ such that $J\cap{F}$ is a single point, with transverse intersection. We may choose $J$ close enough to $F$ so that $J$ is homologous to zero (mod 2) in $M$. This contradicts homological invariance (mod 2) of intersection numbers. QED.

Doubt: Everything else is clear to me except the bold part in the proof. I don't think we will be able to bring $J$ close to $F$ unless it already bounds a disc. So I can see the proof only in the simply connected $M$ case where after suitably perturbing the loop and using the loop theorem to bound, I can apply an ambient isotopy to bring the disc itself closer to $F$. But how do we see this for the non-simply connected $M$? A complicated loop in $M$ might go all around it.

Reference: Lemma 2.1, Chapter 2 (Heegard Splittings), 3-manifolds(book) by Hempel.

share|improve this question
add comment

2 Answers

up vote 3 down vote accepted

We want to prove that in the case F is not one-sided, we may replace J by a curve J' that is contained in a small neighborhood of F and interesects F in the same way as J. By assumtion F is one sided. Consider the boundary B of a small neighborhood N of $F$. Since F is one-sided, B is connected. Now, conisder the intersection of J with B. There are even number of intersections, since B is the boundary. So you can throw the part of J that does not belong to the neighbohood N an close it to a connected curve J' by segments in B (we assumed that B is connected). This explanes the words written in bold.

share|improve this answer
    
Thanks Dmitri. I think this works! –  Maharana Dec 24 '09 at 21:23
add comment

We have to prove that $F$ separates some connected neighborhood. Choose a neighbourhood $U$ that can be contracted onto $F$ (e.g. a tubular one) and suppose $F$ does not separate $U$. Then there is a curve $J$ in $U$ that intersects $F$ transversally at one point.

Every homological class in $U$ becomes zero mod 2 in $M$ (since the image of the homology mod 2 of $U$ = the image of the homology mod 2 of $F$ is zero).

In other words, for tubular neighborhoods "close enough" is not necessary, any curve would do.

Sorry, got the notation wrong the first time around.

share|improve this answer
    
thanks algori, both yours and Dmitri's idea is same but his is somewhat more explicit so I am accepting it! You get my up vote! cheers! –  Maharana Dec 25 '09 at 9:50
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.