Today, 02 September 2015, we worked out the Coulomb energy for the 
4-point configuration U_T=(u_T_1, u_T_2, u_T_3, u_T_4) in S^2 defined 
by the vertices of the regular tetrahedron (say u_T_1=(0, 0, 1) and
the other u_T_i=(*, *, -1/3), all with |u_i-u_j|^2=2(1-u_i•u_j)=8/3). 
And we began proving that U_T minimizes the f-energy

    		 E_f(U)=\Sum_{i<j}f(|u_i-u_j|^2)

among all 4-point configurations U = (u_1, ... , u_4) in S^2, for any
non-negative, decreasing, convex function f:(0,4] -> R (the Coulomb
energy with f(t) = 1/√t is a special case).  More generally (we'll see
09 September) the same holds for the regular (k-1)-simplex in S^{k-2}.


PROJECT 1.  Compare the Coulomb energy (or the general E_f) for U_T
and for the U_4 consisting of 4 equally spaced points around the
equator on S^2.  If we go down a dimension to the circle S^1, it turns
out that k equally spaced points is also a minimizer, for any k (of
course k=3 is a special case of the above, the points at the vertices
of the regular 2-simplex, a.k.a. equilateral triangle); although the
corresponding equatorial configuration on S^2 is not a minimizer for
k>3, show it is critical (in a sense we'll make precise soon).  Can
you prove the 6-point configuration defined by the vertices of the
octahedron, or the 12-point configuration defined by the vertices of
the icosahedron, are also E_f minimizers (also try comparing their
energies, say for Coulomb, with the energy of 6 or 12 equally spaced
points around an equator)?  Cohn and Kumar prove this for any f that
is "completely monotonic" which is stronger than non-negative,
decreasing, and convex - how much about f do you need to assume?  Can
you at least show these (and the 20-point configuration defined by
dodecahedron vertices) are critical?  It is unknown whether there are
any other "universal minimizers" on S^2 for k other than 4, 6 or 12.

      	    		 *	*	*

I expect to have explained (by 09 September) what it means for U to be
a critical configuration, but its physical interpretation is that the
net force on each point exerted by the other points of U must be
normal to the sphere - in other words, the tangential gradient (which
we'll also explain 09 September) of E_f vanishes at U.

      	    		 *	*	*

Here's an excerpt from the Cohn & Kumar paper with some implicit
calculus exercises you should try to check too:

Recall that a C^∞ [∞'ly differentiable or "smooth"] function f: I → R
on an interval I is "completely monotonic" if (−1)^k f^(k)(x) ≥ 0 for
all x ∈ I and all k ≥ 0 (see [Wid, p. 145]) and "strictly completely
monotonic" if strict inequality always holds in the interior of I. Of
course, derivatives at endpoints of intervals denote one-sided
derivatives. In this paper, endpoints are irrelevant, because we use
half-open intervals closed on the right, and one can show using the
mean value theorem that complete monotonicity on the interior implies
that it holds also at the right endpoint.

If a completely monotonic function fails to be strictly completely
monotonic, then it must be a polynomial: if f^(k)(x) = 0, then
complete monotonicity implies that f^(k)(y) = 0 for all y > x, in
which case f^(k) is identically zero because it is analytic (Theorem 3a
in [Wid, p. 146]).  

All inverse power laws f(r) = 1/r^p with p > 0 are strictly completely
monotonic on (0,∞), and they are the most important cases for our
purposes. Other examples include f(r) = e^{−cr} with c > 0.  

It might seem preferable to use completely monotonic functions of
distance, rather than squared distance, but squared distance is in
fact more natural than it first appears. It simplifies formulas
appearing in later sections of this paper, it fits into the general
framework described in Section 8, and it is more general than using
distance: it is not hard to prove that if r → f(r^2) is completely
monotonic on a subinterval (a,b) of (0,∞), then f is completely
monotonic on (a^2,b^2), but not vice versa.

[Wid] D. V. Widder, The Laplace Transform, Princeton University Press,
Princeton, New Jersey, 1941. MR0005923 (3:232d)