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Let $\Omega$ be a round disc of radius $\alpha<\pi/2$ on the unit sphere $\mathbb{S}^2$. 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 $\mathbb{S}^2$ with the same area as $\Omega$ also admits a $(1,\tfrac{\alpha}{\sin\alpha})$-bi-Lipschitz map to the plane?


  • 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. (To prove the statement appy the idea from "A New Area Maximization Proof for the Circle" by Lawlor.)

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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
@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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