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Can a positive curvature smooth metric on the disk be $C^0$ approximated by smooth flat metrics?

Similar question under extra assumption: the same conformal class.

If there are positive examples in both cases, then are there some relatively weak extra-assumptions that the answer is no? Also I am curious how the answer depends on the disk dimension.

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Anton Petrunin's paper, "Polyhedral approximations of Riemannian manifolds," may give some insights. PDF: math.psu.edu/petrunin/papers/poly.pdf –  Joseph O'Rourke Jun 21 '13 at 10:40
    
@Joseph, thank you for a link! –  Dmitri Scheglov Jun 21 '13 at 13:53

1 Answer 1

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No, because too many geometric quantities are continuous in the $C^0$ topology: e.g. length of curves, volume.

Take the standard metric on the radius $1$ sphere, remove a small open disk near the south pole, and push forward stereographically the metric to a metric on the disk. Then the length of the boundary in this metric is some $\ell\ll 1$, and the volume of the metric is close to $4\pi$. Any sufficiently $C^0$-close metric $\tilde g$ must have volume close to $4\pi$ and give to the boundary length much smaller than $1$. This prevents it to be flat, as it would violate the Euclidean isoperimetric inequality.

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What about a "piecewise flat" approximation, as in a many faced polyhedron? –  Carl Jun 21 '13 at 8:20
    
Then $C^0$ approximation is not possible, because your polyhedral metrics are not bounded (they are just $L^p$ for some $p$ depending on the angles at the vertex of the polyhedron. However, $C^0$ approximation of the distance is possible in this way, this is a quite old result that can be found in a book by Alexandrov and Zalghaler whose title I have forgot. –  Thomas Richard Jun 21 '13 at 9:03
    
@Benoit, clear enough, thank you! –  Dmitri Scheglov Jun 21 '13 at 13:54

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