Towards a Proof of the Reinhardt Conjecture

Jacob Gross

June 31, 2017

Abstract

This document is the final write up for the author’s undergraduatesummer project. This project was completed during the summer of 2017and supervised by Professor Thomas C. Hales.

In 1934 Reinhardt conjectured that the hyperbolically smoothed oc-tagon gives a pessimal packing among optimal packings of centrally sym-metric convex bodies in the plane. In 2017 Hales proposed a program toprove Reinhardt’s conjecture with techniques in optimal control theory [5].In this project, we argue that splitting by cases on the homology classes ofthe phase space of Hales’ optimal control problem is a promising strategyfor a full proof of the Reinhardt conjecture. Any homology class not ofthe form (1, 3k + 1) violates boundary conditions. For k > 0, the regularsmoothed (6k + 2)-gon lies in class (1, 3k + 1) and a calibration-type ar-gument equates pessimality within its homology class to the existence ofa good Lagrangian submanifold of the phase space. For k < 0, computa-tional evidence suggests that the only extremals of class (1, 3k+1) are localmaxima and trajectories of class (1, 1) we conjecture to be unadmissible.

1 Introduction

Discrete geometers are often concerned with problems about optimal packings ofspheres, pentagons, etc. The Reinhardt problem is of a different ilk, more akin tothe Brachistochrone problem. The question is: What centrally symmetric convexbody, when optimally packed in the plane, yields the lowest packing density? Indimension 3, spheres are the least efficient packers. In dimension 2, however,circles are not the packing pessimal. Ulam’s packing conjecture is the conjecturethat heptagons are the packing density pessimum for convex bodies in the plane.Note that this problem is only interesting for convex bodies–otherwise one couldtake arbitrarily thin rings to get packings of arbitrarily small density.

Ulam’s packing conjecture is difficult because it is not yet known what theoptimal packing of heptagons is. So an easier problem would be to restrict toa class of shapes whose optimal packing is easy to find. Fejes Toth proved thatfor centrally symmetric convex bodies (e.g. convex bodies with a 180 degree ro-tational symmetry) optimal packings are always lattice packings [3]. And so itmakes good sense to try to find the centrally symmetric convex packing density

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pessimum.

Conjecture 1.1 [Reinhardt]. The hyperbolically smoothed regular octagon

along with its optimal packing, yields the least dense packing among all optimalpackings of centrally symmetric convex bodies in the plane R2.

For more informal articles on the Reinhardt conjecture see [2] or [4].

Hales’ program begins by stating Reinhardt’s conjecture as a problem inoptimal control theory. Optimal control problems are optimization problemswhose solutions are control functions. For example, we can ask how to park a carin minimal time. The solution to this problem is a time-dependent accelerationfunction that informs a driver how much to accelerate at each point in timeto obtain a time-minimal park. The solution to the time-minimal car parkingproblem is to accelerate as hard as the car can, say amax ≥ 0, until some timets (the switching time) and then slam on the break (acceleration of amin ≤ 0).This kind of solution is called bang-bang because it takes values only in extremalpoints of the control set (the domain of the control function).

From a certain point of view, smoothed polygons look like bang-bang solu-tions to optimal control problems. Consider the classical optimal control prob-lem Dubins Car Problem.

Problem 1.2. [Dubins Car Problem]. Given a car with initial positionvector x0 ∈ R2, how ought one drive the car so as to stop in terminal positionxf ∈ R2 in minimal time?

Clearly there are constraints on the possible acceleration and on the possibledeceleration of the car as well as on the possible curvature of turning arcs. Thesolution is bang-bang: always use maximal acceleration and always turn withmaximal curvature. Figure 1 shows an optimal Dubins car trajectory witha steering wheel that can only turn left and Figure 2 shows a portion of ahyperbolically smoothed polygon. This is the exact same figure as [5, figure 2].Indeed, smoothed polygons are drawn by abrupt changes in curvature.

The optimal control problem corresponding to the Reinhardt conjecture is ofa special kind; it satisfies the hypotheses the Filipov’s comapctness lemma. This

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Figure 1: A solution to the Dubins car problem with its circular arcs (left) anda polygon smoothed with hyperbolic arcs (right).

means that trajectories can be lifted to the cotangent bundle. Like the Euler-Lagrange equations in the calculus of variations, optimal control theory has afirst-order necessity test. This test is called Pontryagin’s Maximum Principle(PMP). A main result of [5] is that all smoothed (6k + 2)-gons are Pontryaginextremal, meaning they satisfy they PMP hypotheses. It remains therefore, toshow only that there are no other extremals.

There are also sufficiency tests in optimal control theory. The most commonsufficiency test in optimal control theory is the Hamilton-Jacobi-Bellman (HJB)criterion. For example, the HJB technique solves the Dubins car problem [9].Unfortunately, the Hamilton-Jacobi-Bellman technique works on for problemswith fixed boundary conditions whereas the Reinhardt problem has periodicboundary conditions. In section 3, we prove a variant of this theorem for periodicboundary conditions. What is sacrificed, however, is that optimally is onlygotten within a fixed (first) homology class. This argument mimics a standardargument from calibrated geometry which shows that the volume of embeddedcalibrated submanifolds is stable under small perturbations of their embedding.Calibrated geometry and optimal control theory have notable overlap; perhapscalibrations will prove a fruitful way to attack the Reinhardt conjecture.

The homology of the phase space is Z⊕Z. Any class not of the form (1, 3k+1)violates boundary conditions. For k > 0, each class (1, 3k + 1) contains thesmoothed (6k + 2)-gon. Using Theorem 3.3. we eventually hope to prove thatthe trajectory corresponding to the smoothed (6k+2)-gon is a packing pessimumamong all trajectories class (1, 1 + 3k). For k < 0, computational evidencesuggests these trajectories yield local maxima. For (1, 1), heuristic reasoningstates these ought to correspond to ‘(homological) 2-gons.’ We conjecture thatso-called homological 2-gons (bodies whose corresponding trajectory has class(1, 1)) are not admissible.

Note that we do remove the singular locus of the phase space (as describedin [5, section 5]). Behaviour at this point Λsing will eventually have to be dealtwith.

An application of Theorem 3.3. requires the existence of a ‘good‘ Lagrangiansubmanifold L0 of T ∗M . The region R of optimality gotten in M is the projec-tion of the set traced out by L0 ∩H−1(0) under Hamiltonian flow (where H de-notes the Hamiltonian). By the Arnold-Louville theorem, completely integrablesystems on a 2n-dimensional symplectic manifold with n constants of motion(H = f1, f2, ..., fn) yield a Lagrangian fibration f = (f1, ..., fn) : M → Rn. The

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fiber over a regular value c ∈ Rn whose first coordinate is 0 is well-suited forTheorem 3.3.; in part because that way L0 ∩ H−1(0) = L0. The Hamiltonianof the Reinhardt control problem is SL2(R)-invariant and its action on T ∗Mis Hamiltonian. Noether’s theorem gives then f2, f3, and f4. We conclude byconjecturing the existence of a fifth first integral f5 making (H, f2, f3, f4, f5) acompletely integrable system on T ∗M and explaining why we could not find it.

2 The Optimal Control Problem

In this section, we summarize the results of [5].A convex body in Euclidean space Rd is a compact convex set with nonempty

interior. A centrally symmetric convex body D ⊂ Rd is a convex body such thatD is a translate of D. Reinhardt proved the existence of a convex centrally sym-metric disk Dmin in the plane with the property that the density of its densestlattice packing minimizes the density of the densest lattice packing among allconvex centrally symmetric disks in the plane and conjectured that Dmin is anoctagon whose corners are rounded out with hyperbolic arcs (note that a solu-tion only exists up to affine transformations of the plane, because density is anaffine transformation invariant quantity). He showed that the densest latticepacking is obtained by placing Dmin in a centrally symmetric hexagon of small-est area containing Dmin and tiling the plane with copies of the hexagon. Thisforces Dmin to admit a structure called hexagonal symmetry.

Definition 2.1. For each i = 0, ..., 5, let e∗i = (cos(2πi/6), sin(2πi/6)). A diskD said to be hexagonally symmetric if there exists a path g : I →2 (R) suchthat D is the union of arcs t 7→ g(t)e∗i .

Example 2.2 The circle is hexagonally symmetric, with g(t) =

(cos(t) sin(t)− sin(t) cos(t)

).

Example 2.3. Any hyperbolically smoothed (6k + 2)-gon is hexagonally sym-metric; the path g is computed in [5].

We consider now the following optimal control problem: Let

U = (u0, u1, u2) ∈ R3|∑

ui = 1, ui ≥ 0

be the control set and g : [0, tf ]→ SL2(R) be a C1 path such that g′ = gX, withX : [0, tf ] →2 (R) and det(X(t)) = 1, for all t ∈ [0, tf ]. We may also assumethat X21 > 0 [6]. The set of elements X ∈ sl2(R) with detX = 1 and X21 > 0

form the adjoint of J =

(0 1−1 0

). And that adjoint orbit is diffeomorphic to

the upper-half plane h.

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We impose the additional constraint that the following ODEs

x′ = f1(x, y;u) :=y(b+ 2ax− cx2 + cy2)

b+ 2ax− cx2 − cy2

y′ = f2(x, y;u) :=2(a− cx)y2

b+ 2ax− cx2 − cy2

are satisfied (here x, y denote upper-half plane coordinates), where

a :=u1 − u2√

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b :=u03− 2u1

3− 2u2

3c := u0.

The complete set of state equations is

x′ = f1(x, y;u) (1)

y′ = f2(x, y;u) (2)

g′ = gX. (3)

Definition 2.4. An admissible trajectory (g, z) : [0, tf ] → M := SL2(R) to bea solution of (1),(2), and (3) such that

1. the image of z lies in h∗

2. The endpoints of the trajectory are (g(0), z(0)) = (I, z0) and (g(tf ), z(tf )) =(R,R−1.z0), for some z0 ∈ h, where R := exp(Jπ/3).

3. the path g : [0, tf ]→textSL2(R) is homotopic in SL2(R) to the path given by rotation:

t 7→ exp(J/(3tf )), t ∈ [0, tf ].

Without loss of generality even further conditions can be imposed: the so-called star inequalities on sl2(R)

√3|X11| < X21, 3X12 +X21 < 0,

where Xij denotes the (i, j)th entry of X ∈ sl2(R). In upper-half plane coordi-nates, the star inequalities determine an open region

h∗ := (x, y) ∈ h| − 1√3< x <

1√3,

1

3< x2 + y2

pictured below in the the upper-half plane and disk models of hyperbolic geom-etry:

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Theorem 2.5 [5]. The cost functional of the Reinhardt optimal control problemis

g = areaD(g) = −3

2

∫ tf

0

trace(Jg−1g′) dt→ min

and its solution is given by an admissible trajectory.

This problem satisfies the hypotheses of Filipov’s compactness lemma whichallows trajectories to be lifted to the cotangent bundle T ∗M = SL2(R)×sl2(R)×h∗ × R2. Its Hamiltonian is

H(λ;u); = HLie(Λ, X) +Hh(ν, x, y;u),

HLie; = 〈Λ, X〉 − 3

2λcost〈J,X〉,

Hh(ν, x, y;u) := ν1f1(x, y;u) + ν2f2(x, y;u),

for costate variables λcost ∈ R,Λ ∈ sl2(R), ν = (ν1, ν2) ∈ R, and X = zJ z−1.z =z(x, y).

Optimal control theory has a first order necessity test: the Pontryagin Max-imum Principle (PMP). In our case it is equivalent to

1. the trajectory (g, z) satisfies ODEs (1), (2), and (3) for some measurablecontrol u : [0, tf ]→ U

2. the Hamiltonian H(λ, u) vanishes identically along the lifted control tra-jectory (λ, u)

3. the lifted trajectory λ : [0, tf ]→ T ∗M is Lipschitz continuous and satisfies

Λ′ = [λ,X],

ν′1 = −∂H+

∂x,

ν′2 =∂H+

∂y,

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where H+ denotes the pointwise maximum over the control simplex

4. for each t, the vector (λcost, λ(t)) ∈ R× T ∗M is nonzero

5. λcost is constant and λcost ≤ 0

6. transverality at the endpoints

Paths satisfying 1.− 6. are called Pontryagin extremal

Theorem 2.6 [5]. The smoothed regular (6k + 2)-gon lifts to a Pontryaginextremal trajectory.

3 Homologies of the Phase Space

The remaining task is to show that there are no other extremals. Knener’s higherorder maximum principle (HOMP) is a higher-order necessity test [8]. Perhapsit could eliminate every trajectory but those corresponding to smoothed regularpolygons and then we could just compare smoothed regular polygons showingthat the smoothed regular octagon, along with its optimal packing, gives anabsolute minimum.

A more direct approach would be to apply a sufficiency test. There is a suf-ficiency test in optimal control theory related to the Hamilton-Jacobi equationwhich we now state as Theorem 3.1.

Theorem 3.1. Let L0 be the graph of a differential 1-form, H a Hamiltonianon T ∗M and W a domain in L0 ∩H−1(0)× R such that

π ΦW : W →M

is a diffeomorphism onto a domain of M , where Φ denotes Hamiltonian flowΦ(λ0, t) = etXH (λ0). Let

λt = etXH (λ0),

t ∈ [0, t1] be a normal extremal such that (λ0, t) ∈W for all t ∈ [0, t1]. Then thetrajectory q(t) = π(λt) (with the corresponding control u(t)) realizes a strictminimum of the cost

∫ τ0ϕ(q(t), u(t)) dt among all admissible trajectories such

that q(t) ∈ π Φ(W ) for all t ∈ [0, τ ], q(0) = q(0), q(τ) = q(t1), τ > 0.

Proof. [1, ch. 17]

Theorem 3.1. gives minimalilty among all admissible trajectories with thesame endpoints. The boundary conditions of the Reinhardt optimal controlproblem are

g(0) = I, g(tf ) := exp(Jπ/3)

andX(0) ∈ sl2(R), X(tf ) = R−1X(0)R ∈ sl2(R).

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[5, Section 3.5]. The terminal condition g(tf ) = exp(Jπ/3) is necessary becausethe six paths of the hexagonally symmetric disk must join together to givea closed curve e.g. g(tf )e∗j = g(0)e∗j+1 if and only if g(tf ) = R. For liftedtrajectories λ the boundary condition becomes

λ(0) = λ(tf ). (4)

Theorem 3.1. is proved by showing the cost is given by integration againstan exact form. So, by Stokes’ theorem, λ can be compared to any λ with thesame endpoints. Instead we integrate against a closed form. This accounts forperiodic boundary conditions but allows λ to be compared only to λ in the samehomology class.

Fix a Langrangian submanifold L0 ⊂ T ∗M . Write Lt = etXH (L0) to be itstranslations along Hamiltonian flow for each time t and define L = (λ, t)|λ ∈Lt, 0 ≤ t ≤ t1 ⊂ T ∗M × R.

Lemma 3.2. The form (s−Hdt)|L ∈ Ω1(T ∗M), where s denotes the Louville1-form, is closed.

Proof. First, note that d(s−Hdt)|L = (ω − dH ∧ dt)|L. Fix t. And

(ω − dH ∧ dt)Lt = ω|Lt

because t is constant along Lt. Because ˆetXHω = ω

ω|Lt= ( ˆetXHω)|L0

= ω|L0= 0

as L0 is Lagrangian with respect to ω. Thus (ω − dH ∧ dt) = 0.Note L is the image of the map (λ, t) 7→ (etXHλ, t) and so the tangent vector

to L which is transverse to Lt is

XH(λ) +∂

∂t∈ T(λ,t)L.

So

T(λ,t)L = T(λ,t)Lt ⊕ R(XH(λ) +∂

∂t).

Putting XH(λ) + ∂∂t into the first argument of ω − dH ∧ dt yields

ι(XH(λ) +∂

∂t)(ω − dH ∧ dt) = −dH + dH = 0.

Theorem 3.3. Let W be a domain in L0 ∩H−1(0)× R such that

π Φ|W : W →M

is a diffeomorphism onto a domain in M and let

λt = etXH (λ0)

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be a normal extremal such that (λ0, t) ∈W for all t ∈ [0, t1]. Then the extremaltrajectory q(t) = π(λt), with corresponding control u(t), realizes a strict mini-mum of the cost among all admissible trajectories λ, which are homologous toλt such that π λ ∈ π Φ(W ).

Proof. Write L = Φ(W ). Then π : L → π(L) is a diffeomorphism and (s−Hdt)|Lis closed. Moreover we may, without loss of generality, take (s −Hdt)|L = s|Lbecause H vanishes along extremals (as per the PMP).

Now, take a comparison trajectory λt that is homologous to λt and suchthat q(t) := π λt ∈ π(L). Because s|L is closed∫

λ

s =

∫λ

s.

And so ∫λ

s =

∫ t1

0

〈λt, q(t)〉 dt

=

∫ t1

0

〈λt, fu(t)(q(t))〉 dt

=

∫ t1

0

ϕ(q(t), u(t)) dt

because〈λt, fu(t)(q(t))〉 − ϕ(q(t), u(t)) = H(λt) = 0.

Also ∫λ

s =

∫ τ

0

〈λ(t), q(t)〉

=

∫ τ

0

hu(t)(λ(t)) dt+

∫ τ

0

ϕ(q(t), u(t)) dt

≤∫ τ

0

ϕ(q(t), u(t)) dt.

The inequality holds because maxu∈Uhu(λ(t)) = H(λ(t)) = 0. Finally, the in-equality is actually strict if the curve t 7→ λ(t) is not a solution of λ = H(λ) e.g.if it is not t 7→ λ(t).

It is worth pointing out that this argument echoes a standard one in thetheory of calibrated geometry stating that calibrated submanifolds have volumestable under perturbations of their embedding (see Harvey and Lawson [7]).Calibrated submanifolds are, therefore, a kind of minimal submanifold.

Returning to our specific case, observe that the trajectory in h∗ correspond-ing to the regular smoothed octagon consists of four arcs. It looks like a trianglearound i ∈ h∗ with its first side retraced (Figure 5). Any regular smoothed

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(6k+2)-gon looks like a triangle around i ∈ h∗ with 1+3k arcs [5]. We quotientout T ∗M by the group of rotations 〈R = e2πi/3〉. This has no isotropy exceptat i ∈ h∗. We remove the “orbifold singularity” and take H1.

Theorem 3.4. H1(T ∗Mo/〈R〉) = Z⊕Z, where T ∗Mo = T ∗(SL2(R)× h∗/i).

Proof. The cotangent bundle projection

(SL2(R)× sl2(R))× (h∗/i × R2)/〈R〉 (SL2(R)× h∗/i)/〈R〉π

has contractible fibers and so induces an isomorphism on holomogy. The punc-tured star region h∗/i deformations retracts to S1 so

H1((SL2(R)× sl2(R))× (h∗/i × R2)/〈R〉) ∼= Z⊕ Z.

The Reinhardt problem can be approached by case analysis on homologyclasses.

Theorem 3.5. Let 0 < k ∈ N. Then the homology class (1, 3k + 1) ∈H1(T ∗Mo/〈R〉) contains the trajectory corresponding to the regular smoothed(6k + 2)-gon.

Proof. Explicit trajectories for the smoothed regular (6k + 2)-gon are given in[5].

Theorem 3.6. Let γ be a trajectory of homology class (a, b). Suppose thatb 6= 1 (mod 3) or a 6= 1. Then γ is not a solution of the Reinhardt problem.

Proof. If b 6= 1 (mod 3), boundary condition (4) is not satisfied.

What remains are the cases (1, 3k + 1) for k = 0 and for k < 0. For k < 0,computational evidence suggests that the only extremal in class (1, 1 + 3k) arelocal maxima [5].

Conjecture 3.7. Let γ be a trajectory of homology class (1, 3k + 1) for k < 0and let λ(6k+2)-gon be the trajectory for the smoothed regular (6k+2)-gon. Thenthe cost of λ(6k+2)-gon is strictly larger than the cost of γ.

The case k = 0 is stranger. As the homology class (1, 3k + 1) contains thesmoothed regular (6k + 2)-gon, by this pattern, the class (1, 1) should containa “homological 2-gon.”

Conjecture 3.8. There are no admissible trajectories of homology class (1, 1).

Note that pessimality is only gotten within π(L). We need therefore a La-grangian L0

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Figure 2: Hamiltonian vector field of H = x2+y2

2 along with the Lagrangian(x, y)|x = y.

i. the initial conditions of the smoothed octagon are contained in it λoct(0) ∈L0

ii. its projection covered the whole base space π(L) = M .

Even in a simple case, there can be great variation in the size of π(L).

Example 3.9. Consider M = R, H = x2+y2

2 , and L0 = (x, y) ∈ R|x = y ⊂R2 = T ∗R. The Hamiltonian vector field of H is depicted in Figure 2. Any pointon the x-axis can be obtained by moving some point in L0 along Hamiltonianflow. The image of projection onto the x-axis π : T ∗R → R is all of R. This isthe “best case scenario.”

Example 3.10. Consider M = R, H = −y, and L0 = (0, x)|x ∈ R. TheHamiltonian vector field of H is depicted in Figure 3. Any point in L0 is movedbelow the x-axis. After flowing for any time ε, L is a line distance ε below thex-axis. The image of the projection π : T ∗R → R is ∅. This is the “worst casescenario.” We get pessimality nowhere.

4 Langrangian Fibrations

Proving pessimality of smoothed regular polygons within their homology classreduces now to the problem of finding a “good” Lagrangian (in the sense de-scribed in Section 3). In particular, this Lagrangian must contain the initialconditions of the smoothed regular (6k + 2)-gon. A tempting option is just totake the graph of the unique left-invariant 1-form generated by the lie algebraportion of Λ(6k+2)-gon(0). But because SL2(R) is semi-simple, this form is notclosed and closedness is required to get a Lagrangian graph. So instead, we willhunt for Langragian submanifolds via Lagrangian fibrations.

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Figure 3: Hamiltonian vector field of H = −y along with the Lagrangian(x, 0).

Definition 4.1. Let (M,ω) be a symplectic manifold. A Lagrangian fibration isa surjective map π : M → B whose regular fibers are Langrangian submanifoldsof M .

Example 4.2. Let T ∗M be a contanget bundle endowed with the canonicalsymplectic form. Then the cotangent bundle projection T ∗M → M is a La-grangian fibration.

Perhaps, in the spirit of Example 4.2. we could take the fiber over theiniital condition of the smoothed octagon. This is difficult to deal with howeverbecause the region W of Theorem 3.3. is L0 ∩H−1(0)×R and it is not obvioushow to describe the fiber over the smoothed regular octagons initial conditionsintersected with the fiber of the Hamiltonian over 0.

Fortunately, Lagrangian fibrations can also be obtained from completely in-tegrable systems. And in that case Lagrangians will lie in H−1(0).

Definition 4.3. Let (M,ω) be a symplectic manifold of dimension 2n. Acompletely integrable system (or just integrable system) is a n smooth real-valuedfunctions f1, ..., fn such that

1. they are in involution fi, fj = 0, where −,− denotes the Poissonbracket on C∞(M).

2. ∇f1, ...,∇fn are linearly independent at each point p ∈M

Theorem 4.4. (Arnold-Louville). Let (M,ω) be an integrable system of di-mension 2n with integrals of motion f1 = H, f2, . . . , fn. Then f = (f1, ..., fn) :M → Rn is a Lagrangian fibration.

In sum, if we can find an integrable system H = f1, ..., f5 on the phasespace of the Reinhardt optimal control problem T ∗Mo/〈R〉 we can look at preim-

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age of regular values c ∈ Rn whose first coordinate is 0 to obtain L0∩H−1(0) =π−1(c) ∩H−1(0) = L0. Also the other coordinates of c can be chosen so as toinclude the initial conditions of regular smoothed polygons.

Note the our H is SL2(R)-invariant. This allows us to find first integralsby Noether’s theorem. This is how conserved quantities arise in physics; thesystem’s Hamiltonian is invariant under the Hamiltonian action of a Lie group.

Theorem 4.5 [Noether]. Consider a Hamiltonian action of a Lie group G on asymplectic manifold (M,ω). Let γ : g→ C∞(M), γ : X → ϕX be a correspond-ing anti-Lie algebra map. Let H ∈ C∞(M) be a G-invariant smooth function.Then for any X ∈ g the function γ(X) = ϕX is a constant of motion for H.

For example O(3) acts by left multiplication on R3. This action is Hamil-tonian and preserves the Hamiltonian of a particle in a central force field. Thecorresponding constants of motion are the particle’s angular momenta. Theevident action of SL2(R) on T ∗Mo/〈R〉 is Hamiltonian.

Theorem 4.6. The action by of SL2(R) be left multiplication on T ∗Mo/〈R〉 isHamiltonian and H is SL2(R)-invariant.

Proof. SL2(R) acts on itself by left multiplication. Lifted actions are Hamil-tonian. There is an explicit formula for H and it is independent of SL2(R)coordinates.

Corollary 4.7. The lifted of SL2(R) on T ∗Mo/〈R〉 yields three independentfunctions f1, f2, f3 with H, fi = 0, i = 1, 2, 3.

Proof. Choose a basis e1, e2, e3 for sl2R and apply Noether’s theorem; ∇f1,∇f2and ∇f3 are independent because e1, e2, and e3 are.

As a corollary, there are three independent constants of motion for H corre-sponding to a basis of sl2(R).Because sl2(R) is only 3-dimensional another con-stant of motion is needed. A natural idea is lifting the action of PSL2(R) on hand use the two functions of motion corresponding to a basis of the two dimen-sion lie algebra psl2(R). But PSL2(R) does not act on h∗. The one-dimensionalgroup R acts on h∗ by strictly upward translations, but it does not act on h∗/i.Its existence, therefore, is left as a conjecture.

Conjecture 4.8. There exists a function f5 making H, f2, f3, f4, f5 intoa completely integrable system such that the fiber of θ = (H, f2, f3, f4, f5) :T ∗Mo/〈R〉 → Rn, for some c ∈ Rn, contains λoct(0), and π(Φ(θ−1(c) × R)) =Mo/〈R〉.

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