Skip to main content
22 events
when toggle format what by license comment
Dec 22, 2012 at 17:58 comment added Deane Yang Looks nice. Too bad I'm only allowed to upvote once.
Dec 22, 2012 at 17:37 history edited Liviu Nicolaescu CC BY-SA 3.0
deleted 1 characters in body; Post Made Community Wiki
Dec 22, 2012 at 17:35 comment added Liviu Nicolaescu @ Will The more precise Normalization condition (se Edit 2) prevents this pathology from happening.
Dec 22, 2012 at 17:16 history edited Liviu Nicolaescu CC BY-SA 3.0
added 872 characters in body
Dec 22, 2012 at 17:01 comment added Will Sawin That axiom is preserved by any bijection.
Dec 22, 2012 at 16:55 history edited Liviu Nicolaescu CC BY-SA 3.0
added 937 characters in body
Dec 22, 2012 at 10:29 comment added Liviu Nicolaescu @A.Rex How about the condition that the center of mass of a single point of mass 1 is that point?
Dec 21, 2012 at 22:20 comment added aorq As far as I can tell, you are not using very many properties of $\mathbb{R}$ at all. To be more explicit, pick any continuous bijection $f$ of $\mathbb{R}$ with itself. Then it seems like your center of mass axioms are satisfied by $f^{-1}$ applied to the (usual) center of mass of $f(p)$ for $p$ in your divisor.
Dec 21, 2012 at 17:41 comment added Liviu Nicolaescu I think that the problem is more challenging with positive integral weights. Also I'm not requiring a $GL$-equivariance.
Dec 21, 2012 at 17:36 history edited Liviu Nicolaescu CC BY-SA 3.0
added 343 characters in body
Dec 21, 2012 at 17:31 comment added Liviu Nicolaescu @Deane & alvarezpaiva With the modification of axiom 3 it is almost line a valuation. Namely the new axiom requires that the centroid keeps track of both the location of the centroid and the total mass of the systems of particles. Of corse the masses could be positive real numbers. Also instead of a discrete system of points one could consider positive measures.
Dec 21, 2012 at 17:03 comment added Deane Yang Juan-Carlos, I've been thinking of making a comment like that but forgot who did it. Also, is the centroid really a valuation? I believe that the volume times the centroid (i.e., $\int_K x\,dx$) is a valuation.
Dec 21, 2012 at 16:53 comment added Deane Yang Why are you restricting $m_D$ to the positive integers? Why not allow it to be positive real-valued?
Dec 21, 2012 at 16:44 comment added alvarezpaiva @Deane: I'm guessing the question would be more or less solved by Schneider's characterization of the centroid (or the moment vector) as GL(n)-invariant vector valuation.
Dec 21, 2012 at 16:23 history edited Liviu Nicolaescu CC BY-SA 3.0
added 68 characters in body
Dec 21, 2012 at 15:28 comment added Liviu Nicolaescu I don't yet see even simpler things, why the the mass of the center of mass is equal to the mass of the corresponding divisor. Maybe this should also be part of the requirements. I thought that playing with tricks like $C(\delta_0+k\delta_t)= C( C(\delta_0+delta_t)+(k-1)\delta_t)$ will yield something. In any case, it is intriguing, because the above 5 conditions impose a lot of restrictions, maybe not enough of them.
Dec 21, 2012 at 15:05 comment added Deane Yang Sorry for the running commentary but even if you replace your axiom 5 by something like axiom 3 above, I don't see how to find the center of mass of two points with unequal masses. I do like this approach, though.
Dec 21, 2012 at 14:53 comment added Deane Yang I don't see how to use your axioms to get the center of mass of the divisor consisting of two distinct points with equal mass. Any chance you could explain? How about adding an axiom like 3) in the original question?
Dec 21, 2012 at 14:24 history edited Liviu Nicolaescu CC BY-SA 3.0
added 36 characters in body
Dec 21, 2012 at 12:31 history edited Liviu Nicolaescu CC BY-SA 3.0
added 405 characters in body
Dec 21, 2012 at 11:22 history edited Liviu Nicolaescu CC BY-SA 3.0
added 58 characters in body
Dec 21, 2012 at 11:16 history answered Liviu Nicolaescu CC BY-SA 3.0