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Since the concept of distance in Euclidean space is not invariant in projective space, that is , distance is invariant under Euclidean transformations but not under projective transformations, is it possible to define a distance from point to lines in 3D projective space which is invariant under projective transformations?

LCFactorization
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    Maybe relevant: http://en.wikipedia.org/wiki/Fubini%E2%80%93Study_metric – Qiaochu Yuan Jun 16 '14 at 06:49
  • thank you 翘楚; can we obtain algebraic formula from such definitions? that is, give the homogeneous coordinates of both the 3D point, and the line (maybe Pl"{u}cker coordiante of line), how to give the formula of such a metric/distance? – LCFactorization Jun 16 '14 at 07:12

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Without additional reference objects, this is impossible: any pair of a point and a line can be mapped to any other such pair by a projective transformation, so any projectively invariant quantity you assign to that has to be constant, i.e. be the same for all pairs, in which case I wouldn't call it a distance.

If you introduce additional reference objects, then you can measure with respect to these, but you'll have to ensure that these reference objects remain fixed under (your subset of) projective transformations. One common example is using a conic as the fundamental domain of a Cayley-Klein metric.

MvG
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  • If the requirement be reduced to such a projectively invariant "metric/distance": the "distance" (e.g., sum of squares) of one point to at least three(probably two distinct lines would be enough?) lines, how about it? – LCFactorization Jun 16 '14 at 07:04
  • @LCFactorization: Three lines are defined by their three points of intersection, so in a sense your comment is asking about an invariant of four non-collinear points. Since any four non-collinear points can be mapped to any other four, there can't be a non-constant invariant of a point and three lines either. – MvG Jun 16 '14 at 08:23
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    @LCFactorization: I missed the fact that you're talking about 3D, so my last comment doesn't exactly apply. But a line in space corresponds to 4 degrees of freedom, a point to 3, so three lines and one point amounts to 15 degrees of freedom, which again just equals the 15 degrees of freedom a projective transform in 3D has. So you can still map almost any such configuration to any other, although it's a bit more difficult to see. – MvG Jun 16 '14 at 10:10