Bicycle Polygons, Solitons, and the Darboux Transform · We say that two polygons P and Q are...

Bicycle Polygons, Solitons, and the DarbouxTransform

E. Tsukerman and I. Alevy

Stanford University and University of Chicago

August 8, 2012

E. Tsukerman and I. Alevy Bicycle Polygons, Solitons, and the Darboux Transform

Motivation behind the Problem

The Adventure of the Priory School by Arthur Conan Doyle

Figure: S. Tabachnikov, 2006


Definitions (S.Tabachnikov 2006)

Definition (Continuous Bicycle Curve)

Two planar curves γ(t) and Γ(t) are said to be bicycle curves ifthere exists L ∈ R such that Γ(t) = γ(t) + Lγ′(t).

Definition (Ambiguous Bicycle Curves)

We call the bicycle curve ambiguous if Γ(t) = γ(t)− Lγ′(t).

Definition (Rotation Number)

If (γ, Γ) are ambiguous bicycle curves parametrized so that theperimeter of Γ is 2π then we define the flotation number

ρ =

_

|xy |2π

.


Examples

Figure: F. Wegner, 2007

Two dimensional bodies which can float in all directions are given by !per =2"/n, thus for m = 1 and su!ciently small #. In this limit the $u can bedetermined from eq. (141) with

%(&! + $z) =n2

12(&! + $z) ! ni ! n

2tan(

n$z

2) + O(q), (199)

'(&! + $z) =2i

nq̂e"in!z cos(

n$z

2)en2("!+!z)2/24 + O(q̂), (200)

where eqs. (349) and (355) and #"3

= n, $3"3

= n2

12 have been used. Then eq.(141) yields

tan(n $u ) = n tan( $u ), (201)

in agreement with the results obtained in refs. [7, 12, 8], where $u correspondsto #

2 ! $0 and in ref. [6], where $u corresponds to "(.A few cross-sections of the bodies are shown in figs. 10 to 23. For odd n the

innermost envelope corresponds to density ( = 1/2.

Fig. 10 m/n = 1/3,# = 0.1

Fig. 11 m/n = 1/3,# = 0.2

Fig. 12 m/n = 1/3,# = 0.5

Fig. 13 m/n = 1/4,# = 0.1

Fig.# 14m/n = 1/4, # = 0.1

Fig. 15 m/n = 1/4,# = 0.2

31


Connection to Flotation Problem

For n odd the innermost envelope corresponds to density ρ = 1/2.

Question (Ulam’s Problem 19 from the Scottish Book)

Is the sphere the only solid of uniform density that floats inequilibrium in every position?

Answer

ρ = 1/2 there exist counterexamples (Auerbach, 1938)

The circle is the only such shape when ρ = 1/5, 2/5 (J.Bracho, L. Montejano, and D. Oliveros, 2001) and ρ = 1/4(S. Tabachnikov, 2006)


Definition

A Zindler carousel with n chairs is a system of curves {βi}ni=1 suchthat all βi are parametrizations of the same curve and

βi+n(t) = βi (t)

βi+1(t)− βi (t) = constant

mi (t) = βi (t)+βi+1(t)2 has tangent parallel to βi+1(t)− βi (t).

Theorem (J. Bracho, L. Montejano, D. Oliveros)

Zindler carousels solve the flotation problem.


Figure: Zindler Carousel with Five Chairs

14 j. bracho, l. montejano and d. oliveros

S�V�

��

R�V�

�

�

�

�

Figure 3

In the following, we will classify the figures that float in equilibrium in everyposition, according to their perimetral density.

The next Theorem establishes the main relation between carousels and thefloating body problem

Theorem 2. Let !(!) be a figure. Suppose that the system of interior chordswhich divide the area of !(!) in a fixed ratio ", {CA(t)}, exists and that the figure!(!) of density " floats in equilibrium in every position with perimetral density #,where # = q/n (irreducible fraction). Then there exist a Zindler carousel with n-chairs {$1(t), . . . , $n(t)}, such that ! = $1. Conversely, if {$1(t), . . . , $n(t)} is aZindler carousel, with the property that $i is a simple closed curve and the chords


Definition (Bicycle Polygon)

An n-gon with vertices given by {Vi}ni=0 is called an (n, k)-bicyclepolygon if it is equilateral and all the k diagonals are of equallength. The ratio k

n will be called the rotation number. It is closedif V0 = Vn and periodic if V0 + ~e1 = Vn.

Given a regular n-gon we can construct a (2n, k) bicycle polygonfor any odd k < n (S. Tabachnikov, 2006).

Question

Are there other constructions?


Theorem (S. Tabachnikov, 2006)

Every convex (n, k)-bicycle polygon is regular if

k = 2

n odd and k = 3

n = 2k + 1

n = 3k

Proposition (T., A.)

If n is odd then every convex (4n, n)-bicycle polygon is regular. If nis even then all odd vertices lie equally-spaced on one circle and alleven vertices lie on a concentric circle at equal-spacing.

Proof.

Darboux transform of a Rhombus.


Darboux Transformation

Definition (Darboux Transformation)

We say that two polygons P and Q are Darboux transforms ofeach other with parameter l if each quadrilateral PiQiPi+1Qi+1 isan isosceles trapezoid with side length l :

|PiQi | = |Pi+1Qi+1| = l and PiQi+1 ‖ QiPi+1.


Given a vector ~ov and a vector ~ou, a calculation shows that theimage of u under the trapezoidal rule is given by

D ~ov (u) =| ~ov |2 − | ~ou|2| ~uv |2 ~uv .


Darboux Transform of a Parallelogram


Definition (Trapezoidal Condition)

We will say that a discrete (n, k)-path satisfies the trapezoidalcondition if ViVi+k+1 ‖ Vi+1Vi+k for each i ∈ N..

A discrete (n, k) path satisfying the trapezoidal condition may beinterpreted in terms of the Darboux Transform.


- Consider a periodic framework of n bars of equal length joined attheir endpoints E0,E1, ...,En in a linear arrangement. - Apply theDarboux Transformation k times so that Ei becomes Ei+1. - If thelinkage satisfies these conditions, we may label each Ei as Vik andDl(Vik+m) = Vik+m+1 for m = 0, 1, ..., k − 1. The result is an(n, k)-path satisfying the trapezoidal condition.


Consider a polygonal line P with vertices V0,V1, ...,Vn−1. Let v0be a vector with its origin at V0. Having a vector vi at vertex Vi ,we obtain a vertex vi+1 of the same length at Vi+1 via thetrapezoidal rule.


For a fixed length of v0, we may view the map taking v0 to vj as aself-map of the circle of radius |v0| = |vj | by identifying the circleat V0 with circle at Vj .

Definition ((Monodromy for the Darboux Transformationn)

In the n-periodic and closed cases, the monodromy map is the mapacting on the identified circles at V0 and Vn which takes v0 to vn.


Theorem

The Darboux transformation is area and perimeter preserving onclosed polygons.


Darboux Transformations Preserve Area


(Darboux Transform of a Triangle) Let T = V0V1V2 be a triangleand let d be the diameter of its circumcircle. The monodromymap is hyperbolic on (0, d), parabolic at d and elliptic on (d ,∞).The Darboux transforms of T are rotations of T inside of itscircumcircle and these are precisely those induced by vectors fromV0 to the circumcircle.


Corollary

The Darboux transformation of a rhombus R is a rhombuscongruent to R.

Proof.

Let s be the side length of R and let α be one of its angles. Thearea of R is

s2 sinα.

Therefore D(R) must have angle α or π − α. In both cases D(R)is congruent to R.


Open Problems

Darboux transform for quadrilaterals

Dynamics of the Darboux transform

Connections to group theory?

Interpretation of the Darboux transform in hyperbolic andspherical geometry

Vector flow for odd linkages


Continuous Vector Flow on 3-Linkages


Bicycle Polygons, Solitons, and the Darboux Transform · We say that two polygons P and Q are...

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