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Assume that

  • $\Omega$ is an open simply connected set in $\mathbb R^n$
  • (two-convexity) if 3 faces of a 3-simplex belong to $\Omega$ then whole simplex in $\Omega$.

Is it true that any component of intersection of $\Omega$ with any 2-plane is simply connected?

Comments:

  • Note that an open connected set $\Omega\subset\mathbb R^n$ is convex (in the usual sense) if together with two sides of triangle $\Omega$ contains whole triangle.

  • If the boundary of $\Omega$ is a smooth hypersurface then the answer is YES. [The above property implies that at most one principle curvature of the boundary is negative. Then the statement follows easily from a Morse-type argument; see Gromov's "Sign and geometric meaning of curvature", Lefschetz theorem in Section $\tfrac12$. In fact this argument, shows that in this case any component of intersection of $\Omega$ with any affine subspace (not necessary 2-dimensional) is simply connected.]

  • The answer is YES if $n=3$. [In this case I can mimic the Morse-type argument.]

  • The statement would follow if one could approximate any $\Omega$ by two-convex domains with smooth boundary. BUT the example below shows that such approximation does not exist for $n\ge 4$.

  • I was surprised that my question has a lot in common with this question

Example: We will construct two-convex simply connected open set $\Omega$ in $\mathbb R^4$ such that intersection $L_{t_0}$ of $\Omega$ with some hyperplane is not simply connected. [This NOT a counterexample, but it shows that it is impossible to prove it using smoothing, as indicated in the comments.]

Set $$\Pi=\{\,(x,y,z,t)\in\mathbb R^4\mid\,y< x^2\}.$$ Let $\Pi'$ be the image of $\Pi$ under a generic rotation of $\mathbb R^4$, say $(x,y,z,t)\mapsto(z,t,x,y)$.

Note that $\Pi$ is open and two-convex. Therefore $\Omega=\Pi\cap \Pi'$ is also open two-convex set. One can choose coordinates so that $\Omega$ is an epigraph for a function $f\colon\mathbb R^3\to\mathbb R$ like $$f=\max\{\alpha_1-\beta_1^2,\alpha_2-\beta_2^2 \},$$ where $\alpha_i$ and $\beta_i$ are linear functions. In particular $\Omega$ is contactable.

Let $L_{t_0}$ be the intersection of $\Omega$ with hyperplane $t=t_0$; it is a complement of two convex parabolic cylinders in general position. If these cylidners have a point of intersection then $\pi_1 L_{t_0}=\mathbb Z$.

[In particular, the function $f$ can not be approximated by smooth functions which Hessian has at most one negative eigenvalue value at all points.]

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For an open set $\Omega$, why is [ if 3 faces of a 3-simplex belong to $\Omega$ then whole simplex in $\Omega$ ] not the same as convex? –  Gerald Edgar Feb 18 '11 at 19:14
2  
Here's an article of mine where (a slight variant of) higher convexity is used: staff.science.uu.nl/~henri105/PDF/advgometry.pdf –  André Henriques Oct 1 '12 at 19:32
    
I don't understand the example. I'm visualizing pairs parabolic cylinders in general position and all of them have simply-connected complements. Could you describe a specific hyperplane that does the job? –  Will Sawin Jan 3 at 19:54
1  
@WillSawin, I guess your question is about last par, starting with "Let $L_{t_0}$...". We are in hypereplane, i.e., in $\mathbb R^3$. Imagine that your parabolic cylinders have exactly one point of intersection. Then the complement is homotoipically equivalent to $\mathbb{S}^1$. –  Anton Petrunin Jan 3 at 22:06
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