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While doing some research on polytopes I came to the following question. Maybe it's already somewhere but anyway I'll post it here.

Let $X\subset \mathbb{R}^3$ be such that, for every plane $P$, $P\cap X$ is simply connected. Is $X$ convex?

I'm not sure, but I think maybe it's necesary to assume some well behaving like local simple connectedness. Anyway I think this is true with the apropiate asumptions. I would not be surprised if it was true just as stated.

Probably this is true even in greater dimensions.

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By simply connected, you implicitely mean connected and simply connected, right? –  Benoît Kloeckner Jun 21 '10 at 16:40
    
For dimensions $n$ greater than 3, are you assuming that you intersect planes with $X$ or that you intersect $n-1$-planes with $X$? –  supercooldave Jun 21 '10 at 16:40
    
n-1 planes and yes, connected and simply connected –  Cristos A. Ruiz Jun 21 '10 at 22:41
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5 Answers

up vote 5 down vote accepted

How about some tomography? This should work if $X$ is open. Assume $X\subset \mathbb R^3$ is nonempty and for every plane $H$ the intersection $H\cap X$ is either contractible or empty. (Note that an open subset of the plane is contractible if it is ((nonempty,) connected, and) simply connected.)

Claim 1: $X$ is contractible.

Proof: Consider the space of all pairs $(x,H)$, $x\in X$ and $H$ a plane containing $x$.

Fiber this by $(x,H)\mapsto x$. It's a trivial bundle over $X$ with fiber $P^2$.

Fiber the same thing by $(x,H)\mapsto H$. The nonempty fibers are contractible, so the domain is homotopy equivalent to the image, which in turn fibers over $P^2$ with contractible fibers.

So $X\times P^2$ has the same homotopy type as $P^2$, and upon further inspection $X$ is contractible.

Claim 2: $X$ is convex.

Proof: Let $L$ be any line whose intersection with $X$ is nonempty, and play the same game again with pairs $(x,H)$, but now the plane $H$ is constrained to contain $L$. Call the space of all such pairs $Y$. On the one hand, $Y$ is equivalent to the circle $P^1$ by the same kind of argument as before. On the other hand, $Y$ is the blowup of the $3$-manifold $X$ along the $1$-manifold $X\cap L$. The complement $Y'$ of $(X\cap L)\times P^1$ in $Y$ is the same as the complement of $X\cap L$ in $X$, so it is connected. Therefore all of the group $H_1(Y,Y')$ comes from $H_1(Y)$. But the latter group is of rank $1$ while the former is isomorphic to $H_0( (X\cap L)\times P^1)$. (I am using mod $2$ coefficients.) It follows that $X\cap L$ is connected.

Maybe this generalizes to $\mathbb R^n$. The hypothesis I am thinking of is that every nonempty hyperplane section is contractible, or equivalently $(n-2)$-connected. I don't believe that every hyperplane section simply connected is enough if $n>3$

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I don't think that Claim 2 is true if $X$ is the interior of my modification of Diego Matessi's example: $X=\{ (x,y,z)\in\mathbb{R}^2\times\mathbb{R}_+: x<0 \text{ or } x^2<y^2+z^2\}.$ –  Victor Protsak Jun 21 '10 at 22:34
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Your $X$ intersects the plane $x=z+1$ in a disconnected set. –  Tom Goodwillie Jun 21 '10 at 23:51
    
Yes, you are right, thank you! The vertex of the parabola is below the $xy$-plane, so there are two pieces. Once I drew it, it became obvious. –  Victor Protsak Jun 22 '10 at 0:38
    
I'm chosing this one as the correct answer. There was also Matessi's correct counterexample and Pak's reference (by the way, that book looks really good). But I think there's more merit in this one, and it looks correct to me, also I don't know if the case of $X$ open was known, and the method used to solve it is very appealing to me. Thanks very much for your responses! –  Cristos A. Ruiz Jun 24 '10 at 18:28
    
Thanks. I don't swear that it's correct, though. –  Tom Goodwillie Jun 27 '10 at 1:13
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Edit Corrected.

Here is an elementary statement with a similar flavor (this does not answer the question).

Let $M$ be a compact manifold and $\mathcal{F}$ a family of functions $f: M\to \mathbb{R}^{n_f}$ with the following properties: a. $\mathcal{F}$ is closed under linear projections b. The pre-image $f^{-1}(c)\subset M$ under $f\in\mathcal{F}$ of any point $c\in \mathbb{R}^{n_f}$ is either empty or connected. Then the image $f(M)$ of $M$ under any $f\in\mathcal{F}$ is convex.

Using Morse theory, Atiyah proved that these hypotheses hold for the moment maps of Hamiltonian torus actions on a compact symplectic manifold $M$1 and concluded that the image $\mu(M)$ is a convex polytope whose vertices are the images of the fixed points.

Atiyah, Convexity and commuting Hamiltonians, Bull. London Math. Soc. 14 (1982), no. 1, 1–15


Footnote

1 More precisely, families of commuting Hamiltonians which generate a compact subgroup in $Diff(M).$

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There is something wrong: the continuous image of a connected set is connected, so that the right-hand side is satisfied for all connected $K$. –  Benoît Kloeckner Jun 22 '10 at 12:26
    
Hmm... you are right, I need to check the paper. It's possible that the condition is really in terms of intersections with hyperplanes. –  Victor Protsak Jun 22 '10 at 13:18
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A version of this is well known:

Let $X\subset \Bbb R^3$ be a compact set and suppose every intersection of $X$ by a plane is contractible. Then $X$ is convex.

This is due to Schreier (1933) in $\Bbb R^3$, and Aumann (1936) generalized this to higher dimensions. See this and other related results in Ju.D. Burago and V.A. Zalgaller, Sufficient criteria of convexity, J. Math. Sci. 10 (1978). 395–435. Incidentally, this a (hard) exercise in my book (Exc. 1.25).

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It is not true for non locally-connected sets. Take a path $\gamma$ in $\mathbb{R}^2$ with one end that accumulates to a subsegment of $\gamma$ (in a $\sin(1/x)$ way). Take the product of $\gamma$ with $\mathbb{R}$, and add the interior of $\gamma$ times $\{0\}$.

If you want it to be compact, take $\gamma\times\{0\}$ union with a cone based on $\gamma$.

Note that $\gamma$ is also a weird example of a compact set of the plane that is pathwise connected, simply connected, and whose complement has two connected components.

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Re "weird example of a compact set": what you mean is probably a "quasicircle" (Hatcher, Alg Top, Sec 1.3, Ex 7), because the complement of the closure of the graph of $\sin(1/x), x>0$ is clearly contractible. –  Victor Protsak Jun 21 '10 at 23:17
    
Yes, that is what I tried to describe. –  Benoît Kloeckner Jun 22 '10 at 12:19
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In coordinates $(x,y,z)$, consider { $z \geq 0$} and remove from it the half line { $(t,0,0) | t > 0 $}. You get something non convex, but it seems to me that it also satisfies your property. Maybe you need some boundedness?

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Nice example, that can be made closed by removing a semi-infinite open cylinder instead of a half-line. –  Benoît Kloeckner Jun 21 '10 at 16:56
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@Benoit: Would it not then be possible to choose $P$ slanting so that it clips a bit of the base of the cylinder near the origin, and so creating a hole in $P \cap X$? Maybe I misunderstand what you intend. –  Joseph O'Rourke Jun 21 '10 at 18:49
    
Joseph: That's right! It can be fixed by removing a half-cone $\{ (x,y,z): x>0, x^2>y^2+z^2\}$ instead. Then $P\cap X$ is $P_+\setminus Q$, where $P_+$ is a half-plane and $Q$ is empty, a point, a wedge, a region above a parabola or hyperbola, or a an elliptical segment with the chord along the boundary line of $P_+.$ –  Victor Protsak Jun 21 '10 at 22:15
    
As Tom Goodwillie pointed out, the intersection with $x=z+1$ is disconnected (it is the region above a parabola whose vertex is below the boundary of the half-plane), so this doesn't work, either. –  Victor Protsak Jun 22 '10 at 0:42
    
Unboundedness is not the issue. Diego Matessi's example can be modified by intersecting with max(|x|,|y|,z)<1. –  Tom Goodwillie Jun 22 '10 at 2:04
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