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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.2012-12-07
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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.2012-12-07
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    @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?2012-12-07
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    @jathd: I think we're saying the same thing.2012-12-07

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