Universal Lower Bounds for Potential Energy of Spherical Codes

Abstract

We derive and investigate lower bounds for the potential energy of finite spherical point sets (spherical codes). Our bounds are optimal in the following sense—they cannot be improved by employing polynomials of the same or lower degrees in the Delsarte–Yudin method. However, improvements are sometimes possible, and we provide a necessary and sufficient condition for the existence of such better bounds. All our bounds can be obtained in a unified manner that does not depend on the potential function, provided the potential is given by an absolutely monotone function of the inner product between pairs of points, and this is the reason we call them universal. We also establish a criterion for a given code of dimension n and cardinality N not to be LP-universally optimal; e.g., we show that two codes conjectured by Ballinger et al. to be universally optimal are not LP-universally optimal.

Appendix

In this appendix, we give the proof of Theorem 4.4(b), which has appeared so far only in the dissertation of Boumova [7]. We set $P_i^{(\frac{n-1}{2}, \frac{n-3}{2})}(t) := P_i^{1,0}(t)$ for short and use the Christoffel–Darboux formula (see [27, Equation (5.65)])

$$\begin{aligned} P_i^{1,0}(t)\sum _{j=0}^i r_j=\sum _{j=0}^i r_jP_j^{(n)}(t)={n+i-2 \atopwithdelims ()i}\frac{P_i^{(n)}(t)-P_{i+1}^{(n)}(t)}{1-t}. \end{aligned}$$

(37)

We also set $P_i^{(n)}(t)=\sum _{j=0}^i a_{i,j}t^j$.

For the proof of Theorem 4.4(b), we show for each $n\ge 3$ and s in the interior of $I_{2k-1}$ that $Q_{2k+3}(n,s)<0$ for k sufficiently large, and thus, by Theorem 4.1, the bound $R_{2k-1}(n,N;h)$ can be improved. For this purpose, we present several lemmas.

Lemma 4.14

Let $n \ge 3$, $s \in \left( t_{k-1}^{1,1},t_k^{1,0}\right)$, and $\alpha _i=\alpha _i(s)$, $i=1,\ldots , k$ be the associated Levenshtein quadrature nodes, see (22). Then $Q_{2k+3}(n,s)<0$ if and only if

$$\begin{aligned} \sum _{i=1}^{k} \alpha _i^2 < \frac{2k^2+k+1-n}{n+4k+2}. \end{aligned}$$

Proof

This is Corollary 4.4(a) from [11]. $\square$

To find the sum $\sum _{i=1}^{k} \alpha _i^2$ appearing in Lemma 4.14, we express the sums $\sum _{i=1}^{k} \alpha _i$ and $\sum _{1 \le i< j\le k} \alpha _i\alpha _j$ as functions of n, k, and s.

Lemma 4.15

For every $s \in I_{2k-1}$, $k \ge 2$, the numbers $\alpha _1,\alpha _2,\ldots ,\alpha _k$ satisfy the equalities

$$\begin{aligned} \sum _{i=1}^{k} \alpha _i= & {} -\frac{k}{n+2k-2} X , \end{aligned}$$

(38)

$$\begin{aligned} \sum _{1 \le i< j \le k} \alpha _i\alpha _j= & {} - \frac{k^2-k}{2(n+2k-4)} + \frac{k(k-1)}{(n+2k-2)(n+2k-4)} X, \end{aligned}$$

(39)

where

$$\begin{aligned} X = 1 - \frac{(n+2k-1)(n+k-2)}{k(n+2k-3)} \cdot \frac{P_k^{1,0}(s)}{P_{k-1}^{1,0}(s)}. \end{aligned}$$

Proof

The numbers $\alpha _1,\alpha _2,\ldots ,\alpha _{k}$ are the roots of the equation (24)

$$\begin{aligned} q(t) = P_k^{1,0}(t)P_{k-1}^{1,0}(s) - P_k^{1,0}(s)P_{k-1}^{1,0}(t)=0. \end{aligned}$$

Comparing the coefficients in (37), we obtain

$$\begin{aligned} q(t)= & {} \frac{a_{k,k} r_k P_{k-1}^{1,0}(s)}{\sum _{i=0}^k r_i} t^k \\&+ ~a_{k-1,k-1} r_{k-1} \left( \frac{P_{k-1}^{1,0}(s)}{\sum _{i=0}^k r_i} - \frac{P_k^{1,0}(s)}{\sum _{i=0}^{k-1} r_i} \right) t^{k-1}\\&+ ~\left[ \frac{( a_{k,k-2} r_k + a_{k-2,k-2} r_{k-2}) P_{k-1}^{1,0}(s)}{\sum _{i=0}^k r_i} - \frac{ a_{k-2,k-2} r_{k-2} P_k^{1,0}(s)}{\sum _{i=0}^{k-1} r_i} \right] t^{k-2}+\cdots \end{aligned}$$

Therefore, we have

$$\begin{aligned} \sum _{i=1}^{k} \alpha _i= & {} -~\frac{a_{k-1,k-1} r_{k-1} \left( \frac{P_{k-1}^{1,0}(s)}{\sum _{i=0}^k r_i} - \frac{P_k^{1,0}(s)}{\sum _{i=0}^{k-1} r_i} \right) }{\frac{a_{k,k} r_k P_{k-1}^{1,0}(s)}{\sum _{i=0}^k r_i}} \\= & {} -~\frac{a_{k-1,k-1}r_{k-1} }{a_{k,k}r_k} \left( 1 - \frac{\sum _{i=0}^k r_i}{\sum _{i=0}^{k-1} r_i} \cdot \frac{P_k^{1,0}(s)}{P_{k-1}^{1,0}(s)} \right) \\= & {} -\frac{k}{n+2k-2} \left( 1 - \frac{(n+2k-1)(n+k-2)}{k(n+2k-3)} \cdot \frac{P_k^{1,0}(s)}{P_{k-1}^{1,0}(s)} \right) . \end{aligned}$$

Here, we used that

$$\begin{aligned} \frac{a_{k-1,k-1}}{a_{k,k}}=\frac{n+k-3}{n+2k-4} \end{aligned}$$

from (1), the constants $r_i$ as in (2), and $\sum _{i=0}^j r_i$, which is equal to the Delsarte–Goethals–Seidel bound D(n, 2j) (this is for $j=k$ and $j=k-1$).

Similarly, we obtain

$$\begin{aligned} \sum _{1 \le i< j \le k} \alpha _i\alpha _j= & {} \frac{\frac{( a_{k,k-2} r_k + a_{k-2,k-2} r_{k-2}) P_{k-1}^{1,0}(s)}{\sum _{i=0}^k r_i} - \frac{ a_{k-2,k-2} r_{k-2} P_k^{1,0}(s)}{\sum _{i=0}^{k-1} r_i} }{\frac{a_{k,k} r_k P_{k-1}^{1,0}(s)}{\sum _{i=0}^k r_i}}\\= & {} \frac{ a_{k,k-2}}{a_{k,k} } + \frac{ a_{k-2,k-2} r_{k-2}}{a_{k,k} r_k} \left( 1 - \frac{\sum _{i=0}^k r_i}{\sum _{i=0}^{k-1} r_i} \frac{P_k^{1,0}(s)}{P_{k-1}^{1,0}(s)} \right) \\= & {} - \frac{k^2-k}{2(n+2k-4)} + \frac{k(k-1)}{(n+2k-2)(n+2k-4)} \\&\times \left( 1 - \frac{(n+2k-1)(n+k-2)}{k(n+2k-3)} \frac{P_k^{1,0}(s)}{P_{k-1}^{1,0}(s)} \right) . \end{aligned}$$

$\square$

It follows from Lemma 4.14 and equalities (38) and (39) that we have to investigate the sign of the function

$$\begin{aligned} G(n,k,s)= & {} \sum _{i=1}^{k} \alpha _i^2 - \frac{2k^2+k+1-n}{n+4k+2}\\= & {} \left( \sum _{i=1}^{k} \alpha _i \right) ^2 - 2\sum _{1 \le i< j \le k} \alpha _i\alpha _j - \frac{2k^2+k+1-n}{n+4k+2} \\= & {} \frac{k^2}{(n+2k-2)^2}X^2 - \frac{2k(k-1)}{(n+2k-2)(n+2k-4)}X \\&+ \frac{k(k-1)}{n+2k-4} - \frac{2k^2+k+1-n}{n+4k+2}, \end{aligned}$$

where X is as in Lemma 4.15 and $s \in I_{2k-1}=\left[ t_{k-1}^{1,1},t_k^{1,0}\right]$.

Lemma 4.16

For fixed n and k, the function G(n, k, s) is decreasing for $s \in I_{2k-1}$.

Proof

The function G(n, k, s) is quadratic with respect to X. Since $s \in \left[ t_{k-1}^{1,1},t_k^{1,0}\right] \subset \left( t_{k-1}^{1,0},t_k^{1,0}\right]$, the ratio $P_k^{1,0}(s)/P_{k-1}^{1,0}(s)$ increases in $I_{2k-1}$ (see [26, Corollary 2.1]). Therefore X decreases in s in the same interval, and we need to determine the numbers

$$\begin{aligned} X_1= & {} 1 - \frac{(n+2k-1)(n+k-2)}{k(n+2k-3)} \cdot \frac{P_k^{1,0}\left( t_{k-1}^{1,1}\right) }{P_{k-1}^{1,0}\left( t_{k-1}^{1,1}\right) } \end{aligned}$$

and

$$\begin{aligned} X_2= & {} 1 - \frac{(n+2k-1)(n+k-2)}{k(n+2k-3)} \cdot \frac{P_k^{1,0}\left( t_k^{1,0}\right) }{P_{k-1}^{1,0}\left( t_k^{1,0}\right) } \end{aligned}$$

(the end points of the interval of variation of X). We obtain

$$\begin{aligned} X_1 = 1 + \frac{(n+2k-1)(n+k-2)(n+2k-3)}{k(n+2k-3)(n+2k-1)}= \frac{n+2k-2}{k} \end{aligned}$$

using (30) and

$$\begin{aligned} X_2 = 1, \end{aligned}$$

because $P_k^{1,0}\left( t_k^{1,0}\right) = 0$.

We can already locate the numbers $X_1$ and $X_2$ with respect to the minimum of the graph of the quadratic function

$$\begin{aligned} g(X)= & {} G(n,k,s) \\= & {} \frac{k^2}{(n+2k-2)^2}X^2~ -~\frac{2k(k-1)}{(n+2k-2) (n+2k-4)}X\\&~ + ~\frac{k(k-1)}{n+2k-4}~-~\frac{2k^2+k-n+1}{n+4k+2}. \end{aligned}$$

The minimum of g(X) is attained at the point

$$\begin{aligned} X_0 = \frac{(k-1)(n+2k-2)}{k(n+2k-4)}. \end{aligned}$$

We have

$$\begin{aligned} X_0 - X_2 = X_0-1= -\frac{n-2}{k(n+2k-4)} < 0 \end{aligned}$$

for every $n \ge 3$ and $k \ge 2$. This shows that $X_0 < 1 = X_2 < X_1$; i.e., $X_2$ and $X_1$ lie on the left side of $X_0$. Hence g(X) decreases from $g(X_1)$ to $g(X_2)$ when X decreases from $X_1$ to $X_2$. This means that G(n, k, s) decreases in s in the whole interval $I_{2k-1}$. This completes the proof. $\square$

Thus we need to consider the sign of the function G(n, k, s) in the end points of the interval $I_{2k-1}=[t_{k-1}^{1,1},t_k^{1,0}]$. Define the functions

$$\begin{aligned} \varphi _1(n,k) = G\left( n,k,t_{k-1}^{1,1}\right) =g(X_1) \end{aligned}$$

and

$$\begin{aligned} \varphi _2(n,k) = G\left( n,k,t_k^{1,0}\right) =g(X_2). \end{aligned}$$

From the above, we have

$$\begin{aligned} \varphi _1(n,k) > G(n,k,s) > \varphi _2(n,k) \end{aligned}$$

for all $s \in \left( t_{k-1}^{1,1},t_k^{1,0}\right)$. We calculate $\varphi _1(n,k)$ and $\varphi _2(n,k)$.

Lemma 4.17

For every $n \ge 3$ and $k \ge 2$, we have

$$\begin{aligned} \varphi _1(n,k)= & {} \frac{ (4-n)k^2 + 4(n-2)k + 2n^2-5n}{(n+2k-4)(n+4k+2)}, \nonumber \\ \varphi _2(n,k)= & {} \frac{(n-2)(n+2k-1)(n-k^2-2)}{(n+4k+2) (n+2k-2)^2}. \end{aligned}$$

(40)

Proof

Plug $X_1 = \frac{n+2k-2}{k}$ and $X_2=1$ into g(X). $\square$

Theorem 4.18

Let $n\ge 3$ and $k \ge 9$ satisfy

$$\begin{aligned} n \le \frac{k^2-4k+5 + \sqrt{k^4-8k^3-6k^2+24k+25}}{4}. \end{aligned}$$

Then $Q_{2k+3}(n,s)<0$ for all $s \in \left( t_{k-1}^{1,1},t_k^{1,0} \right)$.

Proof

For $k \ge 9$, all values of n such that

$$\begin{aligned} 3 \le n \le \frac{k^2-4k+5 + \sqrt{k^4-8k^3-6k^2+24k+25}}{4} \end{aligned}$$

are solutions of the inequality $\varphi _1(n,k)<0$ (see (40)). Therefore $G(n,k,s)<0$ for all $s \in \left( t_k^{1,0},t_k^{1,1}\right)$ in this case. This means that $Q_{2k+3}(n,s)<0$ for $s \in \left( t_{k-1}^{1,1}, t_k^{1,0} \right)$. $\square$