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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Apr 14, 2011 at 16:07 comment added Igor Belegradek Incidentally, applying the "geometric proof" to the plane (instead of the disk) we get the amusing conclusion that if $\phi$ is harmonic and non-constant, and $g_0$ is the standard metric on the plane, then $e^\phi g_0$ is not complete. This is because conformal automorphisms of the plane are affine maps, and pulling back the standard metric $g_0$ on the plane by an affine map yields the rescaling of $g_0$ by the conformal factor of the map.
Apr 14, 2011 at 15:53 history answered Igor Belegradek CC BY-SA 3.0