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I have discovered the following results about the sums of squared distances between points on an $n$-sphere (and proved them). To the best of my knowledge (and my advisor's knowledge), these results are new. However, since this is classical geometry and the proofs are not hard, I wanted to ask if anyone knows of their existence in the literature.

I have searched the literature, and I do find authors considering similar questions (e.g., what distribution of points on an $n$-sphere maximizes the sum of distances between the points) but I do not see anyone considering the sum of squared distances.

I want to emphasize that I am not asking for proofs of these results. I have already proven them myself.

First, a bit of terminology:

  • An $n$-sphere is the set of all points equidistance from a fixed point in $\mathbb{R}^n$.
  • A set of points $\mathcal{V}$ is centrally symmetric if it is closed under the antipodal map, i.e., for every $P \in \mathcal{V}$, the point $-P$ is in $\mathcal{V}$.
  • A set of points $\mathcal{V}$ is transitive if for each $P, Q \in \mathcal{V}$, there exists a symmetry of $\mathcal{V}$ such that $P$ is mapped to $Q$.

Result 1. Let $\mathcal{V}$ be a set of $V$ points on a unit $n$-sphere, and let $\mathcal{C}$ be the multiset of the lengths of all the chords between them. Then: $$\sum_{c \, \in \, \mathcal{C}}c^2= V^{2}(1-d^2)$$ where $d$ is the distance between the centroid of $\mathcal{V}$ and the center of the unit $n$-sphere.

This leads to a nice corollary: The centroid of $\mathcal{V}$ coincides with the center of the $n$-sphere if and only if $\sum_{c \, \in \, \mathcal{C}}c^2= V^{2}$.

Result 2. Let $\mathcal{V}$ be a transitive, centrally symmetric set of $V$ points on a unit $n$-sphere, and let $\mathcal{L}$ be the set of distinct lengths of all the chords between them. Then: $$\sum_{l \, \in \, \mathcal{L}}l^2 = 2k+2$$ where $k$ is the cardinality of $\mathcal{L}$.

Question: Has anyone seen these results in the literature? If so, where? (I suppose reference to any similar results would be helpful as well...)

Note: My research was motivated by the following two facts found in the literature. My results subsume these (if one thinks of the vertices of a regular polygon as a point configuration).

  1. For a regular polygon inscribed in a unit circle, $\sum_{c \, \in \, \mathcal{C}}c^2= V^{2}$.
  2. For a regular polygon inscribed in a unit circle, $\sum_{l \, \in \, \mathcal{L}}l^2 \in \mathbb{Z}$.

Fact 1 is in an unpublished paper by S. Mustonen at https://www.survo.fi/papers/Roots2013.pdf. Fact 2 is in a popular math book by J. Kappraff: "Beyond Measure: A Guided Tour through Nature, Myth, and Number." (World Scientific Publishing, River Edge, NJ, 2002). A related question was stated as a 1923 MAA Monthly problem (Morley, F. V., Harding, A. M.: 2925. Am. Math. Mon. 30(1), 44 (1923)).

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  • $\begingroup$ I don't understand the definition of transitive (but that's OK, since you never use it). $\endgroup$ – Gerry Myerson Mar 20 at 2:00
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    $\begingroup$ @GerryMyerson. It's easier to understand if you think about the convex hull of the point set--i.e., a convex polytope. The polytope has symmetries, and it is (vertex) transitive if any vertex of the polytope can be mapped to any other vertex via a symmetry of the polytope... Does that help? $\endgroup$ – Jessie Copher Mar 20 at 13:56
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    $\begingroup$ Result 2 seems surprising. My combinatorial intuition would be that counting distinct chord lengths should be much messier than that. Does anyone have any intuition as to why this should have come out as nice as it did? $\endgroup$ – Nathan Reading Mar 22 at 13:26
  • $\begingroup$ @GerryMyerson Thanks for pointing out that I hadn't used "transitive." It was supposed to be a hypothesis for Result 2. I have fixed this in my question, and also reworded the definition of "transitive" to hopefully be clearer. $\endgroup$ – Jessie Copher Mar 22 at 13:38
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There are two basic formulae for the moment of inertia, which I formulate for the system of unit masses: given $V$ points $p_1,\dots,p_V\in \mathbb{R}^n$, define by $J(q)=\sum |q-p_i|^2$ the moment of intertia of a point $q$. Then

1) (Lagrange formula) $J(q)=J(z)+V|q-z|^2$, where $z$ is the centroid of our $V$ points.

2) (Jacobi formula) $J(z)=V^{-1}\sum_{i<j} |p_i-p_j|^2$.

These formulae are known under different names, I follow the book of Balk and Boltyanskiy "Geometry of masses".

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  • $\begingroup$ To clarify: $n$ is the dimension of the space and $V$ is the number of points, right? Since you say "our $n$ points" in the first formula, I'm wondering which of the $n$'s in the formula should be $V$'s. $\endgroup$ – Nathan Reading Mar 20 at 16:14
  • $\begingroup$ Sorry. Corrected $\endgroup$ – Fedor Petrov Mar 20 at 16:20
  • $\begingroup$ @FedorPetrov I'm familiar with moment of inertia around an axis, but not moment of inertia at a point. Could you explain what you mean by this? (I tried to look at "Geometry of masses," but my Russian is not good.) $\endgroup$ – Jessie Copher Apr 1 at 17:03
  • $\begingroup$ In general it is $\sum_i m_i |q-p_i|^2 $, where $m_i$ are masses in $p_i$. For axis the same, but you take the squares of distances to the axis, not to the point. $\endgroup$ – Fedor Petrov Apr 1 at 20:26
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The sum of squares of chords has proven to be particularly useful in studying arrangements of lines for redundant linear encodings of data.

Suppose $n=\binom{d+1}{2}$. In this case, we may identify $\mathbb{R}^n$ with the vector space of real symmetric $d\times d$ matrices, and $\mathbb{R}\mathbf{P}^{d-1}$ with the subset of matrices that represent orthogonal projections of rank 1:

$$ \mathbb{R}\mathbf{P}^{d-1} \cong\{xx^\top:x\in S^{d-1}\}. $$

Notice that these matrices have unit Frobenius norm, and so they reside in the unit sphere in this space of matrices. The square of the chord length between $xx^\top$ and $yy^\top$ is given by

$$ \|xx^\top-yy^\top\|_F^2 =2-2(x^\top y)^2. $$

As such, the sum of the squared chord lengths between pairs in $\mathcal{V}=\{x_ix_i^\top\}_{i=1}^V$ is given by

$$ \sum_{1\leq i<j\leq V}\|x_ix_i^\top-x_jx_j^\top\|_F^2 =V^2-\sum_{i=1}^V\sum_{j=1}^V(x_i^\top x_j)^2. $$

The double sum on the right-hand side is the frame potential of $\{x_i\}_{i=1}^V$, studied extensively by Benedetto and Fickus in "Finite Normalized Tight Frames." (It's a beautiful paper!)

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