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# I wouldn’t give a fig for the simplicity on this side of complexity. But I would give my right arm for the simplicity on the far side of complexity. --

## Presentation on theme: "I wouldn’t give a fig for the simplicity on this side of complexity. But I would give my right arm for the simplicity on the far side of complexity. --"— Presentation transcript:

I wouldn’t give a fig for the simplicity on this side of complexity. But I would give my right arm for the simplicity on the far side of complexity. -- Oliver Wendell Holmes

The double bubble theorem in 3-space Every standard double bubble in R 3 has the least surface area required to separately enclose two volumes.

 Morgan  Foisy  Hass, Schlafly  Hutchings  Wichiramala  Ritore, Ros  Reichardt  Morgan  Foisy  Hass, Schlafly  Hutchings  Wichiramala  Ritore, Ros  Reichardt

A unified isoperimetric inequality Objects in R n having volume (or area) V and surface area (or perimeter) S satisfy the sharp inequality where r is the inradius of the minimizer in any of the following classes of objects:

Classes for which this inequality holds include: All bodies (with minimizer the round ball) All rectangular boxes (with minimizer the cube) All triangles (minimizer is equilateral) Cylindrical cans (popular calculus problem)

Classes for which this inequality holds include: All bodies (with minimizer the round ball) All rectangular boxes (with minimizer the cube) All triangles (minimizer is equilateral) Cylindrical cans (popular calculus problem)

Double and triple bubbles in the plane Double bubbles in 3-space Conjecturally: All multiple bubbles in the plane Triple to quintuple bubbles in 3-space (n+1)-fold bubbles in R n

Calibration is a great way to prove minimization. Find a progress monitor, in the form of a differential form or vector field Using a form of Stokes’ theorem to orchestrate the process, Make a (fully) local comparison between area and the integral of the monitor. The total monitor integral is the same for all competitors. Conclude the global comparison between competitors.

Metacalibration can be described as calibration combined with slicing, and enhanced by emulation. Slicing makes possible new variable types, and can average out a pointwise inequality requirement over a curve or sub-surface. Emulation guides and simplifies the statement and calculations.

Benefits of metacalibration are centered around the concepts of: Partial reduction Reallocation Emulation Differentiation of a measuring stick

Emulation 1. Start with two objects to compare: An ideal object I A competing object C 2. Match some aspect of C and I 3. Measure some aspect of I, based on step 2 4. Use that quantity to help measure C

To illustrate emulation, we offer the following isoperimetric proof, a-la-Schmidt: Theorem: For any body of volume V in R n, the surface area S satisfies where r is the radius of the round ball of volume V.

Theorem: For any body C of volume V in R n, the surface area S satisfies Proof: The theorem is true if n=1, in which case V = L = 2r and S = 2. Now take any n>1 and assume the theorem true for n-1.

Let C be a body of volume V in R n. Slice C with horizontal planes P t : {x n =t}. Let A max be the largest cross-sectional area. Let B be the round ball whose largest horizontal slice has area A max as well. Let V be the volume of B. C B

C B Now as the slicing plane P t passes upward through C, for every t find a plane Q t slicing through B so as to match the cross-sectional area A(t). Let z(t) be the z-coordinate of the plane Qt, with z=0 at the center of B.

C B Define G(t) = 2V(t) + V(t) - z(t) A(t) r Then G’ =2A + Az ’ - z ’A - z A’ r

Define G(t) = (n-1)V(t) + V(t) - z(t) A(t) r Then G’ = (n-1)A + Az ’ - z ’A - z A’ r (n-1)A - z A’ r =

Define G(t) = (n-1)V(t) + V(t) - z(t) A(t) r Then G’ = (n-1)A + Az ’ - z ’A - z A’ r (n-1)A - z A’ r =   P - z A’ r by the induction hypothesis, where  is the radius of the current slice of B.

Define G(t) = (n-1)V(t) + V(t) - z(t) A(t) r Then G’ = (n-1)A + Az ’ - z ’A - z A’ r (n-1)A - z A’ r = G’   P - z A’ r by induction, where  is the radius of the current slice of B. G’  (  - z) r 1 (P, A’)

Define G(t) = (n-1)V(t) + V(t) - z(t) A(t) r Then G’ = (n-1)A + Az ’ - z ’A - z A’ r (n-1)A - z A’ r = G’   P - z A’ r by induction, where  is the radius of the current slice of B. G’  (  - z) r 1 (P, A’)  S’

Define G(t) = (n-1)V(t) + V(t) - z(t) A(t) r G’  S’ G  S In the end, G = [(n-1)V + V ] / r =V r V r V r V r + + … + + which, by the AM-GM inequality, is minimized when V = V and thus r = r. So

G  S In the end, G = [(n-1)V + V ] / r =V r V r V r V r + + … + + which, by the AM-GM inequality, is minimized when V = V and thus r = r. So completing the proof by induction.

Comparison of methods for proving geometric minimization Deformation Variational methods Symmetrization Reduction Symmetrization Mod out by symmetry Directed slicing Calibration Equivalent problems Paired calibration

Slicing

Deformation Variational methods Symmetrization Reduction Symmetrization Mod out by symmetry Directed slicing Calibration Equivalent problems Paired calibration

Paired vector fields

Deformation Variational methods Symmetrization Reduction Symmetrization Mod out by symmetry Directed slicing Calibration Equivalent problems Paired calibration Metacalibration brings all these methods into one framework.

Deformation Variational methods more localized Symmetrization more versatile Reduction Symmetrization Mod out by symmetry Directed slicing more flexible Calibration applicable to more types Equivalent problems a central feature Paired calibration less rigid Metacalibration brings all these methods into one framework.

The double bubble theorem in 3-space 2 3

What is the question to which this piece is the answer? Metacalibration

What is the question to which this piece is the answer? Answer (i.e., question): Least “capillary surface area” for the given, fixed volumes

Divide and conquer Partition into pieces… Partition into pieces… Solve planar problems via Hutchings Solve planar problems via Hutchings Coordinate these results overall slices Coordinate these results over all slices

h1h1 h2h2 Slice competitor with horizontal planes Slice standard model with slanted planes, matching both volumes:

h1h1 h2h2 Proof. Slice competitor with horizontal planes Slice standard model, matching both volumes: Prove that such slicing planes exist and are unique Prove that S’ ≥ G’, where G is the calibration

h1h1 h2h2 Proof that S’ ≥ G’ uses variations equivalent problems calibration spherical inversion escorting Michael Hutchings’ planar method

h1h1 h2h2

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