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Let $M$ be a compact codimension one submanifold of $\mathbb{R}^{n}$ which does not contaion $0$ and the origin lies in the bounded component of$\mathbb{R}^{n}-\{0\}$.

Is it true to say that:

For every $k>1$, there are (distinct) points $z_{1},z_{2},\ldots,z_{k}$ in M with $\sum_{i=1}^k z_{i}=0$

There are two motivations for this question:

1.For a closed curve $\gamma$ in the plane which surrounds origin, the geometric intuition says that this statement is likely true :a polygon with vertex on $\gamma$ with centroid at $0$.

  1. When the manifold $M$ is a retract of $\mathbb{R}^{n}-\{0\}$ with $\pi_{j}(M)\simeq \mathbb{Z}$ for some $j$, the statement is true for all $k$.

The reason is the following: Let $g:\mathbb{R}^{n}-\{0\}\to M$ be a retraction, then $g((z_{1}+z_{2}+\ldots +z_{k})/k)$ is a Mean(see the reference below) on $M$ which contradic to page $391$ of the following paper by B. Eckmann "Social Choice and Topology A Case of Pure and Applied Mathematics, Expo. Math 2004, 385-393

http://www.sciencedirect.com/science/article/pii/S0723086904800161

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    $\begingroup$ If true for a plane, just intersect your manifold with a general position plane and get your points. Maybe you want your points in general position, though? $\endgroup$
    – Jeff Strom
    Commented Apr 20, 2015 at 10:50
  • $\begingroup$ @JeffStrom No we do not require "general position". So you are using an inductive argument. But is it obvious that there is a hyperplane which global intersection with M is a manifold. It is intuitively obvious, but an exact reason? $\endgroup$ Commented Apr 20, 2015 at 14:33

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Maybe I am missing something, but aren't you answering your own question? By the Eckmann theorem you cite, the result is true for a plane curve (since the hypotheses obviously hold). By @JeffStrom's comment, once you intersect with a plane, the intersection will contain a plane curve surrounding the origin.

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  • $\begingroup$ Is it obvious that a closed curve surrounding 0 is globally a retract of punctured plane?Please see also my comment to Jeff Strom. Finally what about distinct $z_{i}$'s even for the plane case? $\endgroup$ Commented Apr 20, 2015 at 14:36

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