nprofile1qy2hwumn8ghj7un9d3shjtnddaehgu3wwp6kyqpqknzsux7p6lzwzdedp3m8c3c92z0swzc0xyy5glvse58txj5e9ztqaufa4k (nprofile…fa4k) It turns out to be pretty straightforward to prove this in general, without worrying about the specific geometry of the (d-1)-dimensional faces of each regular polytope in R^d.
The Jacobian that gives the 2d-volume of the space of pairs of points on every line segment joining two parallel, congruent (d-1)-dimensional faces factors into three pieces, that when integrated give a product that contains the square of the (d-1)-volume of the faces, the square of the orthogonal distance between them, and some factors that only depend on d. But the total volume of the polytope, squared — which is the total 2d-volume in the denominator when computing the probability — also contains the first two factors, along with a factor of F^2.
When we multiply by F/2 to account for all the different pairs of opposite faces, we are left with k_d / F, where k_d depends only on the dimension.
Greg Egan
npub1qq…j5nhv
2025-08-25 23:26:05 UTC
in reply to nevent1q…eh3x
Author Public Key
npub1qqm7nu2qf25xd3mw6y6cyp4v2wr7ktf43ymp5wqz4u8jvz76ymtsjj5nhvPublished at
2025-08-25 23:26:05 UTCEvent JSON
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