Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?


share|cite|improve this question
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
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

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.

share|cite|improve this answer

Your Answer


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.