Frank Morgan

Here’s a sketch of a movie idea about an excellent scientist and citizen, which I had had for some time, but which took further form during a cathedral mass in Granada in 1999. (There is also my “A Mathematician at Heaven’s Gate.”)

On the very day our scientist, Prof. Blake, retires he dies, and to his pleasant surprise finds himself at the Pearly Gates.

“It’s incredible, religion was right about this.”

“Are you concerned at all?”

“No, I’ve led a good life, and I’m sure this God must be good and appreciative.”

“What good have you done?”

“I’ve led the advancements in my field and been an upstanding member of the community.”

“What is good about such work?”

“Well, the purpose of humanity is to understand the universe, and I’ve contributed to that.”

“Do you think you did it because it was important, or because it appealed to you?”

“No, it is important.”

“Yes, though some folks find it easy to rationalize whatever they want to do. Why do you think you’ve been so successful.”

“Well, I’ve worked hard I guess, made myself organized and persistent and energetic.”

“And what do you think we should do with others who have been less successful, less organized and energetic?”

“Well, I guess you can’t just pretend it doesn’t matter. Maybe you could let them try again, with a fresh start.”

“Actually, we were wondering whether you might go back and finish up their work for them.” Continue reading ‘Scientist in Heaven’ »

A new ingredient in Hales’s 2001 proof of the Hexagonal Honeycomb Conjecture* is his Chordal Isoperimetric Inequality (see Chapter 15 of my Geometric Measure Theory book). It provides a lower bound on the perimeter of a bulging curvilinear polygon in terms of the extra area X enclosed by the bulges. With some help from my new colleague Steven Miller, I’ve realized that a minor modification of the proof improves the constant π/8 to the optimal π/4 as the requisite upper bound on X.

            To modify the proof, add to the auxiliary, less constrained problem the hypothesis that the longest chord is no longer than the sum of the lengths of the others. In the delicate case 1 < L₀ ≤ 2, by the hypothesis the length of the longest chord is at most L₀/2 and a minimizer consists of arcs of equal curvature over two chords of length L₁ = L₂ = L₀/2. If one of the arcs were more than a semicircle, X would exceed the area of a circle of unit diameter, π/4, a contradiction. Hence we are in the convex range of the arc function and the old convexity argument kicks in.

            Similarly if L₀ > 2, a minimizer conists of m ≥ 1 arcs over chords of length L_i = 1 and perhaps one lower arc over a chord of length 0 ≤ L_m+1 < 1. If any of the first m arcs were more than a semicircle, again the area under two would exceed π/4, a contradiction, putting us back in the convex range of the arc function and the old proof.

*The Hexagonal Honeycomb Conjecture says that congruent regular hexagons provide a least-perimeter way to partition the plane into unit areas.