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Let $M$ be a compact orientable surface (manifold in $\mathbb R^3$) with boundary $S^1\times\{0\}$.Show that $M$ intersects the $z$-axis.

Some ideas:

$1)$Since $M$ is a compact orientable manifold with boundary, one way to solve this problem would be using Stokes.Assuming that $M$ doesn't intersect the $z$-axis, we should be able to integrate some differential form $\omega$ on $M$ or on $\partial{M}$ and get a contradiction.I just don't know what form I could integrate. Any suggestions?

$2)$When I first encountered this problem I tried this: by contradiction, if $M$ doesn't intersect the $z$-axis then ,if we take the projection $\pi:\mathbb R^3\rightarrow \mathbb R^2$, $\pi(x,y,z)=(x,y)$, the image of $M$ by $\pi$ is closed since $M$ is compact. Then if it doesn't intersect the $z$-axis, $(0,0)\notin \pi(M)$.Now,since the complement of $\pi(M)$ is open then there is a cylinder $B_{\epsilon}(0,0)\times \mathbb R$ that doesnt intersect $M$.I don't know what to do from here, it seems to be possible to finish the problem with the compacity of the surface, I don't think that we need the its orientability.I'm guessing this by intuition.Is there any way to finish this without the orientability, if not, is there any counterexample?

Thanks!

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Your first idea is a good one. Thinking of the xy-plane as $\mathbb C$, you've got the standard differential form $dz/z$ that gives the winding number around the origin. Integrating this around the circle gives $2\pi$ but it is locally exact, and if the circle bounded a surface, we could piece together the forms $\phi$ for which $d\phi =dz/z$ to show that the form is exact over the surface. But then Stokes Theorem gives a contradiction. –  Grumpy Parsnip Dec 7 '12 at 4:45
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Also, you definitely can remove the orientability restriction. If you perturb the surface so that it hits the $z$-axis in finitely many points transversely, then the number of intersections is the linking number mod 2 of the circle with the $z$-axis, which is 1. So the surface must hit the $z$-axis an odd number of times. –  Grumpy Parsnip Dec 7 '12 at 4:47
    
@JimConant I don't understand why you mention piecing together local primitives. Can't you just say that if $M$ doesn't meet the $z$ axis, the winding number form defines a closed $1$-form on $M$, whose integral on $\partial M$ is not $0$, contradicting Stokes' theorem? –  jathd Dec 7 '12 at 5:11
    
@jathd: I think we're saying the same thing. –  Grumpy Parsnip Dec 7 '12 at 14:23

1 Answer 1

Consider $M_{s}^{g}$ to be the manifold in question with genus $g$. Consider $M^{g}$ to be $M_{s}^{g}$ filled the circle into a region homeomorphic to $\mathbb{D}^{2}$ such that it has no intersection with $M^{g}_{s}$. This can always be done. By the classification theorem we have $M_{g}$ to be a closed oriented surface with $g$ handles.

I claim the following:

1) $M_{g}$ must contain at least one point in $z$-axis in its interior.

2) Any line passing through this point must intersect with $M_{g}$ at some point.

The second one is clear since $M_{g}$ are all compact.

The first one need a contradiction type argument. If $M_{g}$ has an intersection point with $z$-axis, then we are done. If $z$-axis is disjoint from $M_{g}$, then we can put a box such that $M_{g}$ is in the box, and $z$-axis is outside of the box. Now up to homeomorphism we can deform $M_{g}$ to the 'standard' type such that $\mathbb{S}^{1}$ remain a circle on one of the $g$ handles or on its surface area(hence contractible). In the former case the center of the circle is in $M_{g}$. In the later case it is clear the line must intersect $M_{g}$ at some point.

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