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Jan 24, 2020 at 15:41 history closed Ben McKay
user44191
Luc Guyot
Max Horn
David Handelman
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Jan 24, 2020 at 11:53 comment added Gerald Edgar Even without $n+1$ linearly independent points, there exists an isometry such that its restriction is $f$; but in that case it is not unique.
Jan 24, 2020 at 10:08 history edited asv CC BY-SA 4.0
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Jan 24, 2020 at 9:35 review Close votes
Jan 24, 2020 at 15:41
Jan 24, 2020 at 8:12 comment added asv @LevSoukhanov: Thanks, you are right. This is a final answer in fact.
Jan 24, 2020 at 8:03 comment added Lev Soukhanov convexity is not needed: non-empty interior is enough, moreover, having $n+1$ linearly independent points is enough. distances to these points determine the point in a space uniquely, so knowing images of these points you determine the images of every point of space.
Jan 24, 2020 at 7:41 history asked asv CC BY-SA 4.0