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I have been trying to figure this problem out for a while, and while I believe someone must have figured it out hundreds of years ago, I still can't quite get it.

Suppose we have a 3-dimensional sphere and three distinct planes that pass through the center of the sphere. Each pair of two planes will cut a lune from the surface of the sphere. Consider the intersection of the three lunes, which is a triangle on the surface of the sphere. Now, suppose you can reflect the triangle across any of the three planes, take the union of both the triangle and its reflection and repeat the process with this new surface (reflect the 4-sided polygon on the sphere across any of the three planes, union the two surfaces together, etc.) The question is, is it possible to cover the entire sphere using this method? If not, is it possible to at least reach the diametrically opposed triangle (the reflection of the original triangle across the center of the sphere)? Is it possible to do so in a finite number of steps?

The motivation for this question can be thought of in terms of optics: if the three planes were mirrors, and we shine a laser pointer infinitely close to their intersection, would the light bounce back after a finite number of reflections? My intuition says yes, but alas I can't prove it.

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2 Answers 2


            Lunes
I believe this is what vlsd must mean by "the intersection of the three lunes." While I'm posting this clarifying (I hope!) image, I'll risk a heuristic argument to support why I think it is "possible to cover the entire sphere using this method":

(1) A region on the sphere is only a fixed point with respect to reflection in the three planes if the three great circles meet in one point. For example, a geodesic equilateral triangle is symmetric with respect to three circles meeting at its centroid, and would be fixed w.r.t. reflection in those three planes. In other words, a shape can only be fixed w.r.t. to three lines of reflection if those lines meet in a point. I assume that vlsd would exclude this degenerate situation (else the three-lune intersection is a point).

(2) Since the region is not a fixed point w.r.t. reflections (if (1) is correct), the only way the process could avoid covering the sphere is if repeated reflections and unions approached a limit shape. But the displacement of points in the reflecting region is lower-bounded by some function of the smallest angle between the planes, so I do not believe this can occur.

I am aware this is not a precise argument!

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Hi Prof O'Rourke. that's a very nice picture indeed. Also I think you hit a great point about using the angle to lower bound things. Can you make an argument for complete covering of the 2-plane with respect to reflection across three lines forming a triangle? –  John Jiang Oct 15 '11 at 0:00
    
@JohnJ: I think gnomonic projection (en.wikipedia.org/wiki/Gnomonic_projection) will render the two situations equivalent. –  Joseph O'Rourke Oct 15 '11 at 0:11
    
This is exactly what I meant. Thank you for the awesome clarification. I do also believe that there's a lower bound based on the angle between the planes, in fact for the 2D case (a circle and two lines) using such an argument gives an upper bound on the number of reflections needed to cover the circle, inversely proportional with the angle between the two lines. Thanks also for pointing out Gnomonic projections. –  Vlad Seghete Oct 17 '11 at 15:50

Do you really want 2-planes? In $\mathbb{R}^4$, the intersection of two $2$-planes is a point generically. So the intersection of that with the sphere will be empty. Say you are using $3$-planes, then the intersection of two such planes with the sphere will be a circle, generically. But then I think you just get the union of six circles for your reflection process. So certainly won't cover the sphere. I also don't understand what you mean by lune. In general the intersection will always be some lower dimensional spheres, since that's what you get by intersecting vector subspaces with the standard sphere. Maybe you are not thinking about standard spheres? Then again you would get subspheres as intersections. So I don't know where the lunes or triangles come from.

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I think when vlsd says a "3-dimensional sphere," she does not mean a 3-sphere, but rather a 2-sphere in $\mathbb{R}^3$. Under that interpretation, it is possible to make sense of the question, I think. –  Joseph O'Rourke Oct 14 '11 at 21:05
    
Thanks for bringing up that possibility. But that would even make less sense right? Because two 2-planes will intersect in a line, which will have two point intersection with the sphere. Then certainly one cannot cover the whole sphere in finite number of steps. So until further clarification I won't think about it for now. –  John Jiang Oct 14 '11 at 22:29
    
I think the intent of the construction is this: Take the $2$-sphere in $\mathbb{R}^3$ given by $x^2 + y^2 + z^2 = 1$. Given distinct planes $ax + by + cz = 0$ and $dx + ey + fz = 0$, the portion of the $2$-sphere lying in the intersection of $ax + by + cz \geq 0$ and $dx + ey + fz \geq 0$ is a digon. If you take three planes that don't have a mutual line of intersection, you can obtain a $3$-sided piece of the $2$-sphere. For the $xy$, $yz$, and $xz$ this is the portion lying in one of the octants. –  Chad Musick Oct 14 '11 at 22:43
    
Thanks to Chad for pointing out the correct interpretation of the question. It's indeed an interesting question. –  John Jiang Oct 14 '11 at 23:52

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