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Let $\Omega$ be a round disc of radius $\alpha<\pi/2$ on the standard sphere. It is easy to construct a $(1,\tfrac{\alpha}{\sin\alpha})$-bi-Lipschitz map from $\Omega$ to the plane.

Is it true that any convex domain $\Omega'$ on $S^2$ with the same area as $\Omega$ also admits a $(1,\tfrac{\alpha}{\sin\alpha})$-bi-Lipschitz map to the plane?

Comments:

  • This problem appears in Milnor's A problem in cartography. Amer. Math. Monthly 76 1969 1101--1112.

  • I spent quite a bit of time to solve it, but without success. I only noticed that if one exchange "area" above to "perimeter" then the answer is YES.

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In the non-convex case: Are you defining distance as spherical distance, or distance along paths that stay in your domain? (For spherical distance and a global lipschitz condition, I think there are quick counterexamples.) –  Martin M. W. Nov 14 '09 at 15:33
    
Right, the condition is local. –  Anton Petrunin Nov 14 '09 at 15:58
    
Could it be that for the nonconvex problem you still need that the set be simply-connected or something? Otherwise you can take a small-area neighborhood of a "triangulation by tiny triangles"'s 1-skeleton and I think it makes a counterexample, since if none of the grid's 1-cycles expands a huge amount, then the inverse map has a huge lipschitz constant. –  Mircea Mar 22 '12 at 12:39
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@Mircea, well, let's do convex first. You are definetely right if the Lipschitz condition is global; if it is only local then I am not sure. –  Anton Petrunin Mar 22 '12 at 19:14
    
@Anton Petrunin, I agree, the maps could still "crumple" the skeleton while keeping the local condition. Can I ask you how it all works in the case of perimeter? I thought that if one takes the (globally)1-lipschitz map defined just on the boundary whose image encloses maximum area, then that should extend to the wanted map (as the projection did for the disk), but I got stuck in proving that. –  Mircea Mar 24 '12 at 11:55

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