Let n be a positive integer not less than 2. Does anyone know of a theorem stating that- for each n- there exists a simple closed curve c(n), which (1) is a subset of n-dimensional Euclidean space E(n) and (2) does not contain n+1 pairwise distinct points all belonging to the same (n-1)-dimensional Hyperplane of E(n)? For n=2 there is such a theorem. This theorem is almost "trivial". We have merely to picture the circumference of a circle. I am asking whether, and if so, to what extent this "trivial" theorem has been generalized. What is the situation when n is greater than 2?
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asked May 2, 2015 at 18:50
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2$\begingroup$ If $n$ is even, I think an appropriate real form of a normal rational curve will do (more or less by definition). If $n$ is odd, that gives you a curve in the projective space only, so more work is needed. $\endgroup$– Alex DegtyarevCommented May 2, 2015 at 19:01
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Extending my comment, I can give an almost complete answer. If $n$ is even, an example is an appropriate normal rational curve, e.g., $t\mapsto(t/p,t^2/p,\ldots,t^n/p)$, where $p(t)$ is any real polynomial of degree $n$ and without real roots. I conjecture that, for $n$ odd, such a curve doesn't exist. Here is a proof for $n=3$. The curve is obviously not planar. Consider the pencil of planes through a generic tangent. Apart from the tangency point, a generic plane intersects the curve at an even number of points, hence, one can find one with at least two points of intersection. Now, move this plane slightly off the tangency point to get $4$ coplanar points on the curve. I think that this proof can be generalized to any $n$ by considering a pencil of sufficiently osculating hyperplanes.
Added: My proof for $n=3$ was too complicated. Here is a cleaner proof for any odd $n$, not using any smoothness. Take $n$ generic points on the curve and consider the hyperplane $H$ through these points. For homological reason, the intersection of $H$ and the curve must be $0\bmod2$; hence, there is one more point.
answered May 2, 2015 at 20:08
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$\begingroup$ I guess you meant for $p$ to be degree $n$. (If the degree is smaller, the curve goes off to infinity, and if the degree is bigger, the number of intersections could go up.) $\endgroup$ Commented May 2, 2015 at 22:16
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$\begingroup$ @HughThomas Yes, of course; this is implied by the name "normal rational curve" $\endgroup$ Commented May 2, 2015 at 23:08
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$\begingroup$ Many thanks for such a complete answer. I am not too familiar with homology theory so it may take me awhile to fully understand your proof. $\endgroup$ Commented May 3, 2015 at 18:17
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1$\begingroup$ @GarabedGulbenkian In this context, "homological reason" is just lack of time to explain. Since the curve is closed, it must intersect a hyperplane in an even number of points. In plane English, once it enters a subspace, it should also exit it the same number of times. $\endgroup$ Commented May 3, 2015 at 18:28