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This is a followup to this question, where I learned that curvature is invariant to rotations.

I have learned of a version of curvature that is invariant under affine transformations.

I am wondering if there a is a form of curvature between the two. Invariant under uniform scaling and rotation but not all affine transformations?

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up vote 4 down vote accepted

I don't know if this would suit you, but one thing you can consider (much more naive than the notion of affine curvature) is to fix a point P_0 on your curve, and then consider the function on the curve given by sending a point P to the quantity

curvature(P)/curvature(P_0) .

This is a kind of relative curvature, where you measure how much everything is curving in comparison to the curvature at P_0, and is invariant under scaling and rotation.

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Interesting. Perhaps it is obivious but do you what transformations this construction is invariant under? Or is it only invariant under rotation and uniform scaling? – Jonathan Fischoff Aug 1 '10 at 3:25
Curvature is essentially a Euclidean concept (we are approximating a curve by a circle of best possible radius at a given point), if you like, it incorporates concepts both of shape and size. Thus it is invariant under Eulcidean transformations: rotations, reflections, translations. The "relative curvature" in my answer involves shape, but not size, and so is invariant under Euclidean transformations, but also scaling. If you want further invariance, you need to consider more estoric measures of your curve, such as the special affine and affine curvatures discussed in the link you gave. – Matt E Aug 1 '10 at 3:45
(cont'd from above) But these measures will be less intuitive, since they won't explicitly be measuring shape or size, but something less familiar. (E.g. the special affine curvature measures something which is held in common by all ellipses of a given area, no matter what their shape. What this "something" is is less apparent to the eye than curvature or relative curvature.) – Matt E Aug 1 '10 at 3:47
Thanks Matt. :) – Jonathan Fischoff Aug 1 '10 at 4:12
You're very welcome! – Matt E Aug 1 '10 at 4:32

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