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

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

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 R3 has the least surface area required to separately enclose two volumes.

MorganMorgan FoisyFoisy Hass, SchlaflyHass, Schlafly HutchingsHutchings WichiramalaWichiramala Ritore, RosRitore, Ros ReichardtReichardt

MorganMorgan FoisyFoisy Hass, SchlaflyHass, Schlafly HutchingsHutchings WichiramalaWichiramala Ritore, RosRitore, Ros ReichardtReichardt

A unified isoperimetric inequalityObjects in Rn 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 Rn

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 I3. Measure some aspect of I, based on step 24. 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 Rn, the surface area S satisfies

where r is the radius of the round ballof volume V.

Theorem: For any body C of volume V in Rn, 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 Rn.

Slice C with horizontal planesPt: {xn=t}.

Let Amax be the largest cross-sectional area. Let B be the round ball whose largesthorizontal slice has area Amax as well.

Let V be the volume of B.

C

B

C

B

Now as the slicing plane Pt passesupward through C, for every t finda plane Qt slicing through B so asto match the cross-sectional area A(t).

Let z(t) be the z-coordinate of the planeQt, with z=0 at the center of B.

C

B

DefineG(t) =

2V(t) + V(t) - z(t) A(t)r

ThenG’ = 2A + Az ’ - z ’A - z A’

r

DefineG(t) =

(n-1)V(t) + V(t) - z(t) A(t)r

ThenG’ = (n-1)A + Az ’ - z ’A - z A’

r

(n-1)A - z A’r

=

DefineG(t) =

(n-1)V(t) + V(t) - z(t) A(t)r

ThenG’ = (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 theradius of the current slice of B.

DefineG(t) =

(n-1)V(t) + V(t) - z(t) A(t)r

ThenG’ = (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)r1 (P, A’)

DefineG(t) =

(n-1)V(t) + V(t) - z(t) A(t)r

ThenG’ = (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)r1 (P, A’) S’

DefineG(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

= Vr

Vr

Vr

Vr

+ + … + +

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

= Vr

Vr

Vr

Vr

+ + … + +

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 fieldsPaired vector fieldsPaired vector fieldsPaired 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

23

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 conquerDivide and conquer

Partition into pieces…Partition into pieces…Solve planar problems via Solve planar problems via

HutchingsHutchingsCoordinate these resultsCoordinate these results

overover all slicesall slices

h1

h2

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

h1

h2

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

h1

h2

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

h1

h2