compact manifold

Question

I am study for Ph.D. qualification examination in geometry, i don't know solve this question it's from Utah university 1999 test:$$ \\$$ "Let $M$ be a surface compact without boundary embedded in $\mathbb R^3$ . Show there is a two-plane $P$ in $\mathbb R^3$ which intersects $M$ such that $P \cap M$ consist of finitely many disjoint smooth closed loops. " $$ $$ First i tried to define the height function $f:M\to\mathbb R $,such that $f(x,y,z)=z$.The vector field in $M$, $X_{p}=\text{grad}. f(p)$, will be normal to the level set of a height , which coincides with a tangent hyperplane parallel to the xy plane. I take the flow of $X$. The integral curves of X will be the curves in the problem?How i proof that curve in level set of $f$(i.e. the hyperplane)? I know threre is a problem in lee manifolds book which afirm the integral curves are difeomorfics to a $S^1$,in any cituations . I am very grateful for solution or sugestion.

Answer 1

I assume that you want $P \cap M \neq \emptyset$ or else the problem is trivial. You have the good intuition but I don't think you really need to consider the flow of $X$. You want to find a $z_0$ such that all the connected components of $\{z = z_0\} \cap M$ are diffeomorphic to $\mathbb{S}^1$.

First of all, avoid critical points of $X$. Notice that $S = \{z \in \mathbb{R}|\exists (x,y,z) \in M, X(x,y,z) = 0\}$ is by definition the set of critical values of $f : M \rightarrow \mathbb{R}$. By Sard's theorem, it is negligible so you can choose a $z_0$ in the image of $f$ (thus $\{z = z_0\} \cap M \neq \emptyset$) such that for all $(x,y,z_0) \in M$, $X(x,y,z_0) \neq 0$.

Now, you can show that $\{z = z_0\} \cap M$ is a $1$-dimensional manifold (use transversality). Then, it is compact because $M$ is and the only connected compact $1$-dimensional manifold is $\mathbb{S}^1$. It proves the result you wanted.