Kaloshin, Annals of Math 176, 2012. Finiteness (and ... · (1) Ideals, varieties, and algorithms: An introduction to computational algebraic geometry and commutative algebra, by Cox,

3

Nonlinear Multivariate:Bezout’s theorem:

A system of n polynomials in n variables of degree d1, . . . , dnhas at most

�ni=1 di isolated solutions in Cn.

References:

(1) Ideals, varieties, and algorithms: An introduction to computational algebraicgeometry and commutative algebra, by Cox, Little, and O’Shea, Springer.

(2) Using Algebraic Geometry, by Cox, Little, and O’Shea, Springer.

(3) Algorithmic Aspects of Grobner Fans and Tropical Varieties, Jensen, Ph.D.thesis 2007.

(4) Finiteness of central configurations of five bodies in the plane, Albouy andKaloshin, Annals of Math 176, 2012.

Finiteness (and enumeration) of relative equilibriain the N -body problem

Marshall Hampton

University of Minnesota Duluth

Albouy-Chenciner equations (symmetric):

n�

k=1

mk

�Sik(r

2jk − r2ik − r2ij) + Sjk(r

2ik − r2jk − r2ij)

�= 0

Asymmetric:

The Newtonian n-body problem

Central configurations

If we insist that the orbit of each particle is similar to that of theothers, they are forced into Keplerian orbits and they must form acentral configuration:

0 ≤ mi

xi ∈ Rd

− 8s12s13s14s23s24 + s12s13s14s23 − s13s214s23 + s12s13s14s24

− s213s14s24 + s12s13s23s24 + s12s14s23s24 + 8s13s14s23s24 − s14s

223s24 − s13s23s

224

−2m1r312r

313r

314r

323r

324−2m2r

312r

313r

314r

323r

324−2m3r

312r

313r

314r

323r

324−2m4r

312r

313r

314r

323r

324

+m4r312r

313r

314r

323−m4r12r

313r

514r

323+m4r12r

313r

314r

323r

224+m3r

312r

313r

314r

324−m3r12r

513r

314r

324

+m3r12r313r

314r

223r

324+m4r

312r

313r

323r

324+m4r12r

313r

214r

323r

324+m3r

312r

314r

323r

324+m3r12r

213r

314r

323r

324

+2m1r313r

314r

323r

324 + 2m2r

313r

314r

323r

324 −m3r12r

314r

523r

324 −m4r12r

313r

323r

524

1

(1) λ(xj − c) =�

i �=j

mi(xi − xj)

r3ij

(2) mjxj =�

i �=j

mjmi(xi − xj)

r3ij1 ≤ j ≤ n

Outline:

(1) Thanks; overview - cel mech and my path in it(2) Basic celestial mechanics - history, contributions to mathematics(3) light reading list(4) 2- and 3- body problem, Lagrange points(5) Chaos in the 3-body problem(6) Proof for 4 - requires more knowledge of ccs(7) four-body central configurations(8) finiteness. AC equations, polynomials(9) BKK theory(10) Reading list - Sottile, Cox, Little, and O’Shea(11) Future directions - Albouy-Kaloshin

Selected mathematics from the n-body problem

• Calculus - created by Newton for modelling planetary motion

• Convergence of power series - by Cauchy in studying Kepler’sequation

• Least squares error minimization - by Gauss to determine themotion of Ceres

• Dynamical systems and chaos - by Poincare for the n-bodyproblem

Fun reading in celestial mechanics

Diacu and Holmes

Standage

Peterson1

(1) λ(xj − c) =�

i �=j

mi(xi − xj)

r3ij

(2) mjxj =�

i �=j

mjmi(xi − xj)

r3ij1 ≤ j ≤ n

Here c is the location of the center of mass, and λ is a parameter.

Outline:








Diacu and Holmes

Standage

Peterson1

(1) λ(xj − c) =�

i �=j

mi(xi − xj)

r3ij

(2) Newton’s law: mjxj =�

i �=j

mjmi(xi − xj)

r3ij1 ≤ j ≤ n


Outline:








Diacu and Holmes

Standage

Peterson1

10

N-body problems of finiteness, enumeration and bounds

The Newtonian restricted four-body problem

Newtonian three-body problem (D = 3)

One equation from the Newtonian four-body problem

Albouy-Chenciner Equations for Central Configurations (1998)

(4) Sij =1

r3ij+ λ (i �= j) Sii = 0.

Valid for all force exponents, for configurations in every dimensionwith nonzero total mass.

Computational approaches with the Sage platform

Marshall Hampton

University of Minnesota, Duluth

Euler central configurations (1767)

Lagrange central configurations (1772)

(with masses = 1,2,3; on elliptical orbits)

The n-body problem


Orbits in which the shape of the configuration is constant (whichincludes the relative equilibria) are called central configurations andsatisfy the equation:







0 ≤ mi

xi ∈ Rd



223s24 − s13s23s

224

−2m1r312r

313r

314r

323r

324−2m2r

312r

313r

314r

323r

324−2m3r

312r

313r

314r

323r

324−2m4r

312r

313r

314r

323r

324

+m4r312r

313r

314r

323−m4r12r

313r

514r

323+m4r12r

313r

314r

323r

224+m3r

312r

313r

314r

324−m3r12r

513r

314r

324

+m3r12r313r

314r

223r

324+m4r

312r

313r

323r

324+m4r12r

313r

214r

323r

324+m3r

312r

314r

323r

324+m3r12r

213r

314r

323r

324

+2m1r313r

314r

323r

324 + 2m2r

313r

314r

323r

324 −m3r12r

314r

523r

324 −m4r12r

313r

323r

524

1

5

Concave equal mass four-body central configurations

Some four-body central configurations on elliptical orbits

(m1+m3)ρ5−(2m1+3m3)ρ

4+(m1+2m2+3m3)ρ3+(m1+3m2)ρ

2+(2m1+3m2)ρ−(m1+m2) = 0

Smale’s 18 Problems for the 21 Century:

(1) The Riemann hypothesis.(2) The Poincare conjecture.(3) Does P = NP?

. . .

(4)(5) sd

(6) Finiteness of relative equilibria of the n-body problem (for positivemasses)

Smale’s 6th problem for the 21st century:

Finiteness of relative equilibria of the n-body problem (for positive masses)

Introduced as a problem by Chazy in 1918; highlighted by Wintner in 1941.

Lagrange central configuration





0 ≤ mi

xi ∈ Rd



223s24 − s13s23s

224

−2m1r312r

313r

314r

323r

324−2m2r

312r

313r

314r

323r

324−2m3r

312r

313r

314r

323r

324−2m4r

312r

313r

314r

323r

324

+m4r312r

313r

314r

323−m4r12r

313r

514r

323+m4r12r

313r

314r

323r

224+m3r

312r

313r

314r

324−m3r12r

513r

314r

324

+m3r12r313r

314r

223r

324+m4r

312r

313r

323r

324+m4r12r

313r

214r

323r

324+m3r

312r

314r

323r

324+m3r12r

213r

314r

323r

324

+2m1r313r

314r

323r

324 + 2m2r

313r

314r

323r

324 −m3r12r

314r

523r

324 −m4r12r

313r

323r

524

1







0 ≤ mi

xi ∈ Rd



223s24 − s13s23s

224

−2m1r312r

313r

314r

323r

324−2m2r

312r

313r

314r

323r

324−2m3r

312r

313r

314r

323r

324−2m4r

312r

313r

314r

323r

324

+m4r312r

313r

314r

323−m4r12r

313r

514r

323+m4r12r

313r

314r

323r

224+m3r

312r

313r

314r

324−m3r12r

513r

314r

324

+m3r12r313r

314r

223r

324+m4r

312r

313r

323r

324+m4r12r

313r

214r

323r

324+m3r

312r

314r

323r

324+m3r12r

213r

314r

323r

324

+2m1r313r

314r

323r

324 + 2m2r

313r

314r

323r

324 −m3r12r

314r

523r

324 −m4r12r

313r

323r

524

1

3

2m3r313r

314r

323r

324 −m2r

313r

314r

523r34 + m2r

313r

314r

323r

224r34 + m2r

313r

314r

223r

324r34 −m1r

513r

323r

324r34+

m1r313r

214r

323r

324r34 + m1r

213r

314r

323r

324r34 −m1r

514r

323r

324r34 −m2r

313r

314r

524r34 + m2r

313r

314r

323r

334 + m2r

313r

314r

324r

334+

m1r313r

323r

324r

334 + m1r

314r

323r

324r

334 − 2m1r

313r

314r

323r

324r

334 − 2m2r

313r

314r

323r

324r

334 − 2m3r

313r

314r

323r

324r

334 = 0

mixed volume of aceqs, real solutions, positive solutions, bezout, ?

3body vortex: mxv: 4(subs), 32.

n=3 newt: mxv: 99, real: , pos:

n=4 vort: mxv: 80? (5120 without subs)

n=4 newt: mxv: 33201

10 equations for planar 4-bdy: mxv: 25380

aceqs for n=5: mxv: 133998561.

Dear Dr. Hampton: We have computed the mixed volume for the 5 body equations.

The mixed volume is: 133998561.

Problem Bezout bound Mixed volume Real Iso. Sol.s Pos. Real Iso. Sol.s

Min Max Min Max

d = 3, n = 3 729 99 7 7 4 4

d = 3, n = 4, m4 = 0 729 227 ≤ 23 ≥ 33 8 10

d = 3, n = 4 11390625 33201 ≥ 33 ∈ [50, 8472] 33 ∈ [50, 8472]

d = 3, n = 5 ≈ 1.6 · 1013

133998561∗

d = 2, n = 3, sij 27 4 4 4 2 4

d = 2, n = 3, rij 216 32

d = 2, n = 4, sij 15625 80 74

d = 2, n = 4, rij 1000000 5120 74

d = 2, n = 5, sij 282475249

d = 2, n = 5, rij ≈ 2.9 · 1011

∗: thanks to T. Y. Li for this computation.

Equal mass four-body central configurations (Albouy 1996)

Warning: Some of these numbers have caveats -

for example, restrictions on the parameters


3

2m3r313r

314r

323r

324 −m2r

313r

314r

523r34 + m2r

313r

314r

323r

224r34 + m2r

313r

314r

223r

324r34 −m1r

513r

323r

324r34+

m1r313r

214r

323r

324r34 + m1r

213r

314r

323r

324r34 −m1r

514r

323r

324r34 −m2r

313r

314r

524r34 + m2r

313r

314r

323r

334 + m2r

313r

314r

324r

334+

m1r313r

323r

324r

334 + m1r

314r

323r

324r

334 − 2m1r

313r

314r

323r

324r

334 − 2m2r

313r

314r

323r

324r

334 − 2m3r

313r

314r

323r

324r

334 = 0











Min Max Min Max

d = 3, n = 3 729 99 7 7 4 4

d = 3, n = 4, m4 = 0 729 227 ≤ 23 ≥ 33 8 10

d = 3, n = 4 11390625 33201 ≥ 33 ∈ [50, 8472] 33 ∈ [50, 8472]

d = 3, n = 5 ≈ 1.6 · 1013

133998561∗

d = 2, n = 3, sij 27 4 4 4 2 4

d = 2, n = 3, rij 216 32

d = 2, n = 4, sij 15625 80 74

d = 2, n = 4, rij 1000000 5120 74

d = 2, n = 5, sij 282475249

d = 2, n = 5, rij ≈ 2.9 · 1011






3

2m3r313r

314r

323r

324 −m2r

313r

314r

523r34 + m2r

313r

314r

323r

224r34 + m2r

313r

314r

223r

324r34 −m1r

513r

323r

324r34+

m1r313r

214r

323r

324r34 + m1r

213r

314r

323r

324r34 −m1r

514r

323r

324r34 −m2r

313r

314r

524r34 + m2r

313r

314r

323r

334 + m2r

313r

314r

324r

334+

m1r313r

323r

324r

334 + m1r

314r

323r

324r

334 − 2m1r

313r

314r

323r

324r

334 − 2m2r

313r

314r

323r

324r

334 − 2m3r

313r

314r

323r

324r

334 = 0











Min Max Min Max

d = 3, n = 3 729 99 7 7 4 4

d = 3, n = 4, m4 = 0 729 227 ≤ 23 ≥ 33 8 10

d = 3, n = 4 11390625 33201 ≥ 33 ∈ [50, 8472] 33 ∈ [50, 8472]

d = 3, n = 5 ≈ 1.6 · 1013

133998561∗

d = 2, n = 3, sij 27 4 4 4 2 4

d = 2, n = 3, rij 216 32

d = 2, n = 4, sij 15625 80 74

d = 2, n = 4, rij 1000000 5120 74

d = 2, n = 5, sij 282475249

d = 2, n = 5, rij ≈ 2.9 · 1011





Smale’s 6th

problem for the 21st century:

3

2m3r313r

314r

323r

324 −m2r

313r

314r

523r34 + m2r

313r

314r

323r

224r34 + m2r

313r

314r

223r

324r34 −m1r

513r

323r

324r34+

m1r313r

214r

323r

324r34 + m1r

213r

314r

323r

324r34 −m1r

514r

323r

324r34 −m2r

313r

314r

524r34 + m2r

313r

314r

323r

334 + m2r

313r

314r

324r

334+

m1r313r

323r

324r

334 + m1r

314r

323r

324r

334 − 2m1r

313r

314r

323r

324r

334 − 2m2r

313r

314r

323r

324r

334 − 2m3r

313r

314r

323r

324r

334 = 0








Dear Dr. Hampton: We have computed the mixed volume for the 5 body equations. The mixed volume is: 133998561.


Min Max Min Max

d = 3, n = 3 729 99 7 7 4 4

d = 3, n = 4 11390625 33201 ≥ 33 ∈ [50, 8472] 33 ∈ [50, 8472]

d = 3, n = 5 ≈ 1.6 · 1013

133998561∗

d = 2, n = 3, sij 27 4 4 4 2 4

d = 2, n = 3, rij 216 32

d = 2, n = 4, sij 15625 80 74

d = 2, n = 4, rij 1000000 5120 74

d = 2, n = 5, sij 282475249

d = 2, n = 5, rij ≈ 2.9 · 1011


Smale’s 6th




Gareth Roberts found a continuum of solutions for the 5-body problem with one negative mass.

3

2m3r313r

314r

323r

324 −m2r

313r

314r

523r34 + m2r

313r

314r

323r

224r34 + m2r

313r

314r

223r

324r34 −m1r

513r

323r

324r34+

m1r313r

214r

323r

324r34 + m1r

213r

314r

323r

324r34 −m1r

514r

323r

324r34 −m2r

313r

314r

524r34 + m2r

313r

314r

323r

334 + m2r

313r

314r

324r

334+

m1r313r

323r

324r

334 + m1r

314r

323r

324r

334 − 2m1r

313r

314r

323r

324r

334 − 2m2r

313r

314r

323r

324r

334 − 2m3r

313r

314r

323r

324r

334 = 0










Min Max Min Max

d = 3, n = 3 729 99 7 7 4 4

d = 3, n = 4 11390625 33201 ≥ 33 ∈ [50, 8472] 33 ∈ [50, 8472]

d = 3, n = 5 ≈ 1.6 · 1013

133998561∗

d = 2, n = 3, sij 27 4 4 4 2 4

d = 2, n = 3, rij 216 32

d = 2, n = 4, sij 15625 80 74

d = 2, n = 4, rij 1000000 5120 74

d = 2, n = 5, sij 282475249

d = 2, n = 5, rij ≈ 2.9 · 1011


Smale’s 6th




Gareth Roberts found a continuum of solutions for the 5-body problem with one negative mass.4

Richard Moeckel and I proved finiteness in the 4-body case in 2002.

(1) N-body problem intro(2) .Euler and Lagrange(3) Chazy, Wintner, Smales 6th problem(4) Four body results. ”Deconstructive” method, eliminate exponents. Minkowski sum.(5) Vortex problem: easier, good test case(6) Summary of results

4


Warning: Some of these numbers have caveats -for example, restrictions on the parameters

From Grobli’s thesis, 1877




(1) The Riemann hypothesis.(2) The Poincar conjecture.(3) Does P = NP?

. . .

(4)(5) sd

(6) Finiteness of relative equilibria of the n-body problem (for positive masses)


5




(1) The Riemann hypothesis.(2) The Poincare conjecture.(3) Does P = NP?

. . .

(4)(5) sd

(6) Finiteness of relative equilibria of the n-body problem (for positivemasses)




In 1999, Gareth Roberts found a continuum of solutionsfor the 5-body problem with one negative mass.

8

(6) Finiteness of relative equilibria of the n-body problem (for positive

masses)

Smale’s 6th




In 1999, Gareth Roberts found a continuum of solutions for the 5-body

problem with one negative mass.


T.Y. Li computed the mixed volume (133998561) and mixed cells for the

5-body Albouy-Chenciner equations in 2005; in 2007 Anders Jensen

computed the tropical pre-variety.

Helmholtz

The theory of point vortices

Point vortex relative equilibria satisfy the Albouy-Chenciner

equations with a different exponent:

Some of the state of the art:

3

2m3r313r

314r

323r

324 −m2r

313r

314r

523r34 + m2r

313r

314r

323r

224r34 + m2r

313r

314r

223r

324r34 −m1r

513r

323r

324r34+

m1r313r

214r

323r

324r34 + m1r

213r

314r

323r

324r34 −m1r

514r

323r

324r34 −m2r

313r

314r

524r34 + m2r

313r

314r

323r

334 + m2r

313r

314r

324r

334+

m1r313r

323r

324r

334 + m1r

314r

323r

324r

334 − 2m1r

313r

314r

323r

324r

334 − 2m2r

313r

314r

323r

324r

334 − 2m3r

313r

314r

323r

324r

334 = 0










Min Max Min Max

d = 3, n = 3 729 99 7 7 4 4

d = 3, n = 4 11390625 33201 ≥ 33 ∈ [50, 8472] 33 ∈ [50, 8472]

d = 3, n = 5 ≈ 1.6 · 1013

133998561∗

d = 2, n = 3, sij 27 4 4 4 2 4

d = 2, n = 3, rij 216 32

d = 2, n = 4, sij 15625 80 74

d = 2, n = 4, rij 1000000 5120 74

d = 2, n = 5, sij 282475249

d = 2, n = 5, rij ≈ 2.9 · 1011


Smale’s 6th




Gareth Roberts found a continuum of solutions for the 5-body problem with one negative mass.

11

In 1999, Gareth Roberts found a continuum of solutions for the 5-body

problem with one negative mass.


Anders Jensen and I also proved a finiteness result for the 5-body spatial

central configurations in 2011.

Alain Albouy and Vadim Kaloshin proved finiteness in the 5-body case in

2012, apart from some special cases.

They also provided an alternate proof for the 4-body case which avoided the

large amount of symbolic computation in the 2002 result.

T.Y. Li computed the mixed volume (133998561) and mixed cells for the

5-body Albouy-Chenciner equations in 2005; in 2007 Anders Jensen computed

the tropical pre-variety.

Helmholtz

The theory of point vortices

Point vortex relative equilibria satisfy the Albouy-Chenciner

equations with a different exponent:

Some of the state of the art:

(2) Sij =1

r2ij+ λ (i �= j) Sii = 0.

(3)

n�

k=1

mk

�Sik(r



�= 0

4 MARSHALL HAMPTON AND RICHARD MOECKEL

matrix representing β. Choosing ki = − 12 |xi|2 shows that β is represented by the

matrix, B, whose entries are

(7) Bij = xi · xj − 12 |xi|2 − 1

2 |xj |2 = − 12r2

ij .

Multiplying both sides of (5) by Xt gives GA = 0. The matrix GA can beviewed as representing a (non-symmetric) bilinear form on P , in which case itis permissible to replace G by B. Taking the symmetric part gives the Albouy-Chenciner equations for central configurations:

(8) BA + AtB = 0.

Let ei denote the standard basis vectors in Rn and define eij = ei − ej . Then (8)is equivalent to the equations

(9) etij(BA + AtB)eij = 0 1 ≤ i < j ≤ n

To see this let γ(v, w) = vtCw be the symmetric bilinear form on P associated tothe matrix C = BA + AtB. Then (9) means that γ(eij , eij) = 0 for 1 ≤ i < j ≤ n.To show that γ = 0 it suffices to show that γ vanishes on the basis e1i, 2 ≤ i ≤ nof P . By the polarization identity

2γ(e1i, e1j) = γ(eij , eij)− γ(e1i, e1i)− γ(e1j , e1j),

this follows from (9).Equations (9) provide

�n2

�constraints on the

�n2

�mutual distances rij of a cen-

tral configuration. Conversely, it can be shown that if the quantities rij are themutual distances of some configuration in some Rd and if they satisfy (10), thenthe configuration is central [3]. It is remarkable that the equations themselves areindependent of d, so they determine the central configurations in all dimensions atonce.

To find the equations explicitly, note that

γ(eij , eij) = 2etijBAeij = 2(BAii + BAjj −BAij −BAji).

where BAij denotes the entries of the matrix BA. From (6) and (7) we find

(10)n�

k=1

mk

�Sik(r2

jk − r2ik − r2

ij) + Sjk(r2ik − r2

jk − r2ij)

�= 0

for 1 ≤ i < j ≤ n, where Sik, Sjk are given by (4).At this point we can normalize the equations to eliminate the dilation symmetry.

In fact, if we scale all the distances by setting rij = crij where c �= 0 is a constant,and if we further set λ� = c−3λ�, we obtain another solution. We will choose c toachieve the normalization

λ� = −1.

2.2. Dziobek’s Equations. For planar central configurations of n = 4 bodies,there is another set of equations which we will find useful. In this case, each ofthe equations in (2) contains three nonzero terms and the vectors xi − xj whichappear there are two-dimensional. Taking wedge products of the i-th equationwith one of the vectors xi − xk, yields an equation relating two of the areas ofthe triangles containing the point xi (the term involving xi − xk drops out). LetAl denote the oriented (signed) area of the triangle not containing xl, where we


Albouy-Chenciner Equations for Central Configurations

Valid for all force exponents, for configurations in every dimension with nonzero total mass.


Marshall Hampton







If we insist that the orbit of each particle is similar to that of the others, they are forced into Keplerian orbits and

they must form a central configuration:

0 ≤ mi

xi ∈ Rd



223s24 − s13s23s

224

1

Vol(λ1P1 + λ2P2) = a1λ21 + (mixed volume)λ1λ2 + a2λ

22

−m3r312r

313 + m3r12r

513 −m3r12r

313r

223 −m3r

312r

323 −m3r12r

213r

323 − 2m1r

313r

323−

2m2r313r

323 + 2m1r

312r

313r

323 + 2m2r

312r

313r

323 + 2m3r

312r

313r

323 + m3r12r

523 = 0

m2r512r13 −m2r

312r

313 −m2r

312r13r

223 − 2m1r

312r

323 − 2m3r

312r

323 −m2r

212r13r

323−

m2r313r

323 + 2m1r

312r

313r

323 + 2m2r

312r

313r

323 + 2m3r

312r

313r

323 + m2r13r

523 = 0

2m2r312r

313 + 2m3r

312r

313 −m1r

512r23 + m1r

312r

213r23 + m1r

212r

313r23 −m1r

513r23+

m1r312r

323 + m1r

313r

323 − 2m1r

312r

313r

323 − 2m2r

312r

313r

323 − 2m3r

312r

313r

323 = 0

−m3r312r

513r14r

324 + m3r

312r

313r

314r

324 + m3r

312r

313r14r

324r

234 −m2r

512r

313r14r

334 + m2r

312r

313r

314r

334+

m2r312r

313r14r

224r

334 + 2m1r

312r

313r

324r

334 + m3r

312r

213r14r

324r

334 + m2r

212r

313r14r

324r

334 + m3r

312r

314r

324r

334 + m2r

313r

314r

324r

334−

2m1r312r

313r

314r

324r

334 − 2m2r

312r

313r

314r

324r

334 − 2m3r

312r

313r

314r

324r

334 −m2r

313r14r

524r

334 −m3r

312r14r

324r

534 = 0

−m3r312r

314r

523r24 + m3r

312r

314r

323r

324 + m3r

312r

314r

323r24r

234 + 2m2r

312r

314r

323r

334 + m3r

312r

314r

223r24r

334−

m1r512r

323r24r

334 + m1r

312r

214r

323r24r

334 + m1r

212r

314r

323r24r

334 −m1r

514r

323r24r

334 + m3r

312r

314r

324r

334 + m1r

312r

323r

324r

334+

m1r314r

323r

324r

334 − 2m1r

312r

314r

323r

324r

334 − 2m2r

312r

314r

323r

324r

334 − 2m3r

312r

314r

323r

324r

334 −m3r

312r

314r24r

534 = 0

2m3r313r

314r

323r

324 −m2r

313r

314r

523r34 + m2r

313r

314r

323r

224r34 + m2r

313r

314r

223r

324r34 −m1r

513r

323r

324r34+

m1r313r

214r

323r

324r34 + m1r

213r

314r

323r

324r34 −m1r

514r

323r

324r34 −m2r

313r

314r

524r34 + m2r

313r

314r

323r

334 + m2r

313r

314r

324r

334+

m1r313r

323r

324r

334 + m1r

314r

323r

324r

334 − 2m1r

313r

314r

323r

324r

334 − 2m2r

313r

314r

323r

324r

334 − 2m3r

313r

314r

323r

324r

334 = 0

1

2

Albouy-Chenciner Equations for Central Configurations (1998)

(1) Sij =1

r3ij

+ λ (i �= j) Sii = 0.

Valid for all force exponents, for configurations in every dimension with

nonzero total mass.


Marshall Hampton







If we insist that the orbit of each particle is similar to that of the others, they

are forced into Keplerian orbits and they must form a central configuration:

0 ≤ mi

xi ∈ Rd

Four vortex problem equation versus Newtonian four-body

(1) λ(xj − c) =�

i�=j

mi(xi − xj)

rDij

(2) Sij =1

rDij

+ λ (i �= j) Sii = 0.


We eventually used the 6 Albouy-Chenciner equations together with theDziobek equations (1900):

For mixed volume bounds, we had to use a square system that implies theprevious one:

This has a mixed volume of 25380.

f0 = m1z1 + m2z2 + m3z3 + m4z4 = 0

f1 = m2z2r212 + m3z3r

213 + m4z4r

214 + k = 0

f2 = m1z1r212 + m3z3r

223 + m4z4r

224 + k = 0

f3 = m1z1r213 + m2z2r

223 + m4z4r

234 + k = 0

f4 = m1z1r214 + m2z2r

224 + m3z3r

234 + k = 0

Sij = zizj 1 ≤ i < j ≤ 4.

Sketch of our proof for the 4-body Newtonian and vortex problems:

For a candidate system of necessary equations for central configurations,we computed first the tropical prevariety from the Minkowski sum of theirNewton polytopes. The tropical prevariety is the set of normal cones to mixedfaces of the Minkowski sum.

1

(1) λ(xj − c) =�

i�=j

mi(xi − xj)

rDij

(2) Sij =1

rDij

+ λ (i �= j) Sii = 0.





f0 = m1z1 + m2z2 + m3z3 + m4z4 = 0

f1 = m2z2r212 + m3z3r

213 + m4z4r

214 + k = 0

f2 = m1z1r212 + m3z3r

223 + m4z4r

224 + k = 0

f3 = m1z1r213 + m2z2r

223 + m4z4r

234 + k = 0

f4 = m1z1r214 + m2z2r

224 + m3z3r

234 + k = 0

Sij = zizj 1 ≤ i < j ≤ 4.



1

2

0 ≤ mi

xi ∈ Rd



223s24 − s13s23s

224

− 2(m1 + m2 + m3 + m4)r312r

313r

314r

323r

324 + m4r

312r

313r

314r

323 −m4r12r

313r

514r

323

+ m4r12r313r

314r

323r

224 + m3r

312r

313r

314r

324 −m3r12r

513r

314r

324 + m3r12r

313r

314r

223r

324

+ m4r312r

313r

323r

324 + m4r12r

313r

214r

323r

324 + m3r

312r

314r

323r

324 + m3r12r

213r

314r

323r

324

+ 2(m1 + m2)r313r

314r

323r

324 −m3r12r

314r

523r

324 −m4r12r

313r

323r

524 = 0


22

−m3r312r

313 + m3r12r

513 −m3r12r

313r

223 −m3r

312r

323 −m3r12r

213r

323 − 2m1r

313r

323−

2m2r313r

323 + 2m1r

312r

313r

323 + 2m2r

312r

313r

323 + 2m3r

312r

313r

323 + m3r12r

523 = 0

m2r512r13 −m2r

312r

313 −m2r

312r13r

223 − 2m1r

312r

323 − 2m3r

312r

323 −m2r

212r13r

323−

m2r313r

323 + 2m1r

312r

313r

323 + 2m2r

312r

313r

323 + 2m3r

312r

313r

323 + m2r13r

523 = 0

2m2r312r

313 + 2m3r

312r

313 −m1r

512r23 + m1r

312r

213r23 + m1r

212r

313r23 −m1r

513r23+

m1r312r

323 + m1r

313r

323 − 2m1r

312r

313r

323 − 2m2r

312r

313r

323 − 2m3r

312r

313r

323 = 0

−m3r312r

513r14r

324 + m3r

312r

313r

314r

324 + m3r

312r

313r14r

324r

234 −m2r

512r

313r14r

334 + m2r

312r

313r

314r

334+

m2r312r

313r14r

224r

334 + 2m1r

312r

313r

324r

334 + m3r

312r

213r14r

324r

334 + m2r

212r

313r14r

324r

334 + m3r

312r

314r

324r

334 + m2r

313r

314r

324r

334−

2m1r312r

313r

314r

324r

334 − 2m2r

312r

313r

314r

324r

334 − 2m3r

312r

313r

314r

324r

334 −m2r

313r14r

524r

334 −m3r

312r14r

324r

534 = 0

−m3r312r

314r

523r24 + m3r

312r

314r

323r

324 + m3r

312r

314r

323r24r

234 + 2m2r

312r

314r

323r

334 + m3r

312r

314r

223r24r

334−

m1r512r

323r24r

334 + m1r

312r

214r

323r24r

334 + m1r

212r

314r

323r24r

334 −m1r

514r

323r24r

334 + m3r

312r

314r

324r

334 + m1r

312r

323r

324r

334+

m1r314r

323r

324r

334 − 2m1r

312r

314r

323r

324r

334 − 2m2r

312r

314r

323r

324r

334 − 2m3r

312r

314r

323r

324r

334 −m3r

312r

314r24r

534 = 0

One of six Albouy-Chenciner equations

for the Newtonian 4-body problem

A rich source of polynomial problems:

Solutions, bounds, and finiteness

of polynomial systems in Sage

We eventually used the 6 Albouy-Chenciner equations together with the Dziobek equations(1900):

A hierachy of problems for polynomial systems

(1) Computing/characterizing solutions:

Groebner bases (Magma, Singular, Macaulay2)

Homotopy continuation methods (Phcpack, Bernoulli, Hom4ps)

Isolation methods

(2) Counting solutions, upper and lower bounds:

Mixed volumes, fewnomial methods

(3) Finiteness/dimension of solution set:

Tropical methods (Gfan)1


for the Newtonian 4-body problem

A rich source of polynomial problems:

Solutions, bounds, and finiteness

of polynomial systems in Sage

We eventually used the 6 Albouy-Chenciner equations together with the Dziobek equations(1900):

A hierachy of problems for polynomial systems

(1) Computing/characterizing solutions:

Groebner bases (Magma, Singular, Macaulay2)

Homotopy continuation methods (Phcpack, Bernoulli, Hom4ps)

Isolation methods

(2) Counting solutions, upper and lower bounds:

Mixed volumes, fewnomial methods

(3) Finiteness/dimension of solution set:

Tropical methods (Gfan)1

What do we know about a system of polynomials?

Univariate:We know many things, in particular the

Fundamental Theorem of Algebra(an nth-degree polynomial has exactly

n solutions in C when counted with multiplicity).

Linear Multivariate:The system Ax = b has a unique solution if A is invertible.

Nonlinear Multivariate:Bezout’s theorem:

A system of n polynomials in n variables of degree d1, . . . , dnhas at most

�ni=1 di isolated solutions in Cn.

References:

(1) Ideals, varieties, and algorithms: An introduction to computational algebraicgeometry and commutative algebra, by Cox, Little, and O’Shea, Springer.




Finiteness of relative equilibria in the N -body problem

Marshall Hampton

University of Minnesota Duluth1

10

Problem with Bezout’s Theorem:

Consider the eigenvalue problem: Av = λv

with a normalization condition�

v2i = 1

with a given A ∈ Mn(C) and unknowns λ ∈ C, v ∈ Cn.

Bezout’s theorem says that if there are finitely many solutions,there are at most (2)n+1.

But the true maximum of isolated solutions is at most 2n.

Puiseux series: x(t) =�∞

i=i0ait

iq , q ∈ N, i0 ∈ Z

Simple example:Consider two polynomials in variables x and y, with a parameter b:

f1 = x3 + xy2 − x− y + by

= x3y0 + x1y2 − x1y0 − x0y1 + bx0y1

= x(x2 + y2 − 1) + y(m− 1)

f2 = x2 + y2 − 1

= x2y0 + x0y2 − x0y0

N(f2) = convex hull of {(1, 0), (0, 1), (0, 0)}

N(f1) = convex hull of {(3, 0), (1, 2), (1, 0), (0, 1)}

Grobner bases

A powerful approach to studying any polynomial systemis to compute a Grobner basis for the ideal generated by the polynomials.

Before briefly describing Grobner bases, here is a summary of their properties:

≈ 1% of the time, your computation will finish in less than 5 minutes, and you candetermine everything you want about the solutions to polynomial system.

≈ 99% of the time, your computation will take more than a hundred years, andrequire at least 128 GB of memory.

Fewnomials:

Bihan-Sottile bound (improving a previous result of Khovanskii):

A system of n polynomials in n variables having a total of n+ k + 1 distinctmonomials has fewer than

e2 + 3

42

k2

nk

non-degenerate solutions in the positive orthant.

Another result is that of Li, Rojas, and Wang, who showed thata system of two trinomials has at most 5 positive real solutions.

Tropical Geometry

The study of the algebraic geometry of the tropical semiring (R,⊕,⊗) where

x⊕ y = min{x, y}1

Groebner bases of linear systems

= Reduced echelon form

Consider the linear system:

x +y +z−1 = 0,

2∗x −y +3z−2 = 0,

−x +4y +z +1 = 0

Marshall Hampton () Geometric Visualization of Algebraic Information November 1, 2008 4 / 5




x +y +z−1 = 0,

2∗x −y +3z−2 = 0,

−x +4y +z +1 = 0





x +y +z = 1,

2x−y +3z = 2,

−x +4y +z =−1

The corresponding augmented coefficient matrix can be reduced (by

Gauss-Jordan reduction) to the reduced echelon form:

1 1 1 1

2 −1 3 2

−1 4 1 −1

→

1 0 0 1

0 1 0 0

0 0 1 0

which corresponds to the Groebner basis {x−1,y ,z}Marshall Hampton () Geometric Visualization of Algebraic Information November 1, 2008 4 / 5




x +y +z = 1,

2x−y +3z = 2,

−x +4y +z =−1



1 1 1 1

2 −1 3 2

−1 4 1 −1

→

1 0 0 1

0 1 0 0

0 0 1 0





x +y +z = 1,

2x−y +3z = 2,

−x +4y +z =−1



1 1 1 1

2 −1 3 2

−1 4 1 −1

→

1 0 0 1

0 1 0 0

0 0 1 0


Lexicographic Groebner basis example

For the nonlinear system

x2 +y +z = 1,2x2−y2 +3z = 2,−x +4y +z2 =−1

The lexicographic Groebner basis with x > y > z is

z8−12z6−30z5 +18z4 +164z3 +381z2 +253z,

y − 8704

2598655z7 +

312

112985z6 +

79424

2598655z5 +

195639

2598655z4− 143608

2598655z3

− 6966

44045z2− 2552233

2598655z,

x− 34816

2598655z7 +

1248

112985z6 +

317696

2598655z5 +

782556

2598655z4− 574432

2598655z3

− 71909

44045z2− 10208932

2598655z−1




x2 +y +z = 1,2x2−y2 +3z = 2,−x +4y +z2 =−1


z8−12z6−30z5 +18z4 +164z3 +381z2 +253z,

y − 8704

2598655z7 +

312

112985z6 +

79424

2598655z5 +

195639

2598655z4− 143608

2598655z3

− 6966

44045z2− 2552233

2598655z,

x− 34816

2598655z7 +

1248

112985z6 +

317696

2598655z5 +

782556

2598655z4− 574432

2598655z3

− 71909

44045z2− 10208932

2598655z−1




x2 +y +z = 1,2x2−y2 +3z = 2,−x +4y +z2 =−1


z8−12z6−30z5 +18z4 +164z3 +381z2 +253z,

y − 8704

2598655z7 +

312

112985z6 +

79424

2598655z5 +

195639

2598655z4− 143608

2598655z3

− 6966

44045z2− 2552233

2598655z,

x− 34816

2598655z7 +

1248

112985z6 +

317696

2598655z5 +

782556

2598655z4− 574432

2598655z3

− 71909

44045z2− 10208932

2598655z−1




x2 +y +z = 1,2x2−y2 +3z = 2,−x +4y +z2 =−1


z8−12z6−30z5 +18z4 +164z3 +381z2 +253z,

y − 8704

2598655z7 +

312

112985z6 +

79424

2598655z5 +

195639

2598655z4− 143608

2598655z3

− 6966

44045z2− 2552233

2598655z,

x− 34816

2598655z7 +

1248

112985z6 +

317696

2598655z5 +

782556

2598655z4− 574432

2598655z3

− 71909

44045z2− 10208932

2598655z−1




x2 +y +z = 1,2x2−y2 +3z = 2,−x +4y +z2 =−1


z8−12z6−30z5 +18z4 +164z3 +381z2 +253z,

y − 8704

2598655z7 +

312

112985z6 +

79424

2598655z5 +

195639

2598655z4− 143608

2598655z3

− 6966

44045z2− 2552233

2598655z,

x− 34816

2598655z7 +

1248

112985z6 +

317696

2598655z5 +

782556

2598655z4− 574432

2598655z3

− 71909

44045z2− 10208932

2598655z−1


To generalize Gauss-Jordan reduction to nonlinear systems, we need

a way to systematically order terms of a polynomial.

The highest-order (leading) terms in pairs of polynomials are can-

celled to produce ‘S-polynomials’:

L1 = the leading term of p1 , L2 = the leading term of p2

Spoly(p1, p2) =LCM(L1, L2)

L1p1 −

LCM(L1, L2)

L2p2

Example: For p1 = x3 + y3 + x2y − 1 and p2 = y4 + x2 + x, if theleading terms are x3 and x2 then

Spoly(p1, p2) =x3

x3p1 −

x3

x2p2 = −xy4 + x2y + y3 − x2 − 1

The restricted three-body problem (m3 = 0)

Outline of proving finitely many solutions to a polynomial system:

(1) Compute the tropical prevariety - directions where the amoeba

tentacles can overlap. (Polyhedral problem)

(2) Compute (often using Grobner bases) any solutions to the initial

forms of the tropical prevariety. (Algebraic problem)

(3) Investigate in more detail, using Puiseaux series, directions which

have nontrivial solutions. (Mostly unsolved problem)

The behavior of the amoeba tentacles suggests the approach of

studying the polynomial solutions out at the tips. This idea has many

names: the logarithmic limit, the Bergman fan, BKK theory, and

tropical geometry.

Albouy-Chenciner polynomials for the

three-body problem central configurations

−2(m1 +m2 +m3)r312r

313r

323m1 + 2r313r

323m1 + 2r313r

323m2 + r312r

313m3−

r12r513m3 + r12r

313r

223m3 + r312r

323m3 + r12r

213r

323m3 − r12r

523m3 = 0

−2(m1 +m2 +m3)r312r

313r

323m1 + 2r312r

323m1 − r512r13m2 + r312r

313m2+

r312r13r223m2 + r212r13r

323m2 + r313r

323m2 − r13r

523m2 + 2r312r

323m3 = 0

−2(m1 +m2 +m3)r312r

313r

323m1 − r512r23m1 + r312r

213r23m1 + r212r

313r23m1−

r513r23m1 + r312r323m1 + r313r

323m1 + 2r312r

313m2 + 2r312r

313m3 = 0







L1p1 −

LCM(L1, L2)

L2p2


Spoly(p1, p2) =x3

x3p1 −

x3

x2p2 = −xy4 + x2y + y3 − x2 − 1












tropical geometry.



−2(m1 +m2 +m3)r312r

313r

323m1 + 2r313r

323m1 + 2r313r

323m2 + r312r

313m3−

r12r513m3 + r12r

313r

223m3 + r312r

323m3 + r12r

213r

323m3 − r12r

523m3 = 0

−2(m1 +m2 +m3)r312r

313r

323m1 + 2r312r

323m1 − r512r13m2 + r312r

313m2+

r312r13r223m2 + r212r13r

323m2 + r313r

323m2 − r13r

523m2 + 2r312r

323m3 = 0

−2(m1 +m2 +m3)r312r

313r

323m1 − r512r23m1 + r312r

213r23m1 + r212r

313r23m1−

r513r23m1 + r312r323m1 + r313r

323m1 + 2r312r

313m2 + 2r312r

313m3 = 0

Term orderings and weight vectors

Elimination

Term orderings are closely related to gradings by weight vectors. For

a n-dimensional polynomial ring, a vector ω ∈ Rn≥0 defines a grading

by:

xα ≺ω xβ ⇐⇒ �ωα� < �ωβ�

Example: if p1 = x3 + y3 + x2y − 1 and ω = (2, 1) then x3 is the

highest-graded term.







L1p1 −

LCM(L1, L2)

L2p2


Spoly(p1, p2) =x3

x3p1 −

x3

x2p2 = −xy4 + x2y + y3 − x2 − 1












tropical geometry.




Term orderings are closely related to gradings by weight vectors. For a

n-dimensional polynomial ring, a vector ω ∈ Rn≥0

defines a grading by:

xα ≺ω xβ ⇐⇒ �ωα�< �ωβ �

Example: if p1 = x3 +y3 +x2y −1 and ω = (2,1) then x3 is the


Marshall Hampton () The polyhedral geometry of polynomial systems November 5, 2009 2 / 30


Elimination

Term orderings are closely related to gradings by weight vectors. For

a n-dimensional polynomial ring, a vector ω ∈ Rn≥0 defines a grading

by:

xα ≺ω xβ ⇐⇒ �ωα� < �ωβ�

Example: if p1 = x3 + y3 + x2y − 1 and ω = (2, 1) then x3 is the








L1p1 −

LCM(L1, L2)

L2p2


Spoly(p1, p2) =x3

x3p1 −

x3

x2p2 = −xy4 + x2y + y3 − x2 − 1












tropical geometry.



All of the possible Groebner bases

(via Anders Jensen’s Gfan and Sage)








Groebner fan colored by degree



−6s12s13s23 +s12s13−s2

13 +s12s23 +6s13s23−s2

23 = 0,

−6s12s13s23−s2

12 +s12s13 +6s12s23 +s13s23−s2

23 = 0,

−6s12s13s23−s2

12 +6s12s13−s2

13 +s12s23 +s13s23 = 0





−6s12s13s23 +s12s13−s2

13 +s12s23 +6s13s23−s2

23 = 0,

−6s12s13s23−s2

12 +s12s13 +6s12s23 +s13s23−s2

23 = 0,

−6s12s13s23−s2

12 +6s12s13−s2

13 +s12s23 +s13s23 = 0


tropical line

Tropical Geometry


x⊕ y = min{x, y}

x⊗ y = x+ y

This is a sort of algebraic encoding of polyhedral propertiesof normal fans and amoeba.

A network of ideas:

Polyhedral methods (Newton polytopes, polyhedral homotopies)

Tropical geometry and tropical varieties

BKK (Bernstein, Kushnirenko, and Khovansky) theory

Puiseux series and initial forms

Fewnomial theory


Consider the eigenvalue problem

Av = λv

with a normalization condition

|v| = 1

with A ∈ Mn(C) and v ∈ Cn.1

Newton polytopes and amoebae

The Newton polytope of a polynomial is the convex hull of the

exponent vectors of its monomials. For polynomials in one variable,

these are simply line segments. For example, for

p(x) = x +x2 +x6

the Newton polytope is the convex hull of (1), (2), and (6), which is the

line segment from (1) to (6).


Newton polytopes and amoeba

50x3+83x2y +24xy2+y3+392x2+414xy +50y2−28x +59y−100 = 0


Newton polytopes and amoebae

The Newton polytope of a polynomial is the convex hull of the

exponent vectors of its monomials. For polynomials in one variable,

these are simply line segments. For example, for

p(x) = x +x2 +x6

the Newton polytope is the convex hull of (1), (2), and (6), which is the

line segment from (1) to (6).

Note that the length of the Newton polytope is equal to the number of

non-zero solutions.



50x3+83x2y +24xy2+y3+392x2+414xy +50y2−28x +59y−100 = 0


f1 = x3 +xy2−x−100y

= x3y0 +x1y2−x1y0−100x0y1

N(f1) = convex hull of {(3,0),(1,2),(1,0),(0,1)}

f2 = x2 +y2−2

= x2y0 +x0y2−2x0y0

N(f2) = convex hull of {(1,0),(0,1),(0,0)}


f1 = x3 +xy2−x−100y

= x3y0 +x1y2−x1y0−100x0y1

N(f1) = convex hull of {(3,0),(1,2),(1,0),(0,1)}

f2 = x2 +y2−2

= x2y0 +x0y2−2x0y0

N(f2) = convex hull of {(1,0),(0,1),(0,0)}


f1 = x3 +xy2−x−100y

= x3y0 +x1y2−x1y0−100x0y1

N(f1) = convex hull of {(3,0),(1,2),(1,0),(0,1)}

f2 = x2 +y2−2

= x2y0 +x0y2−2x0y0

N(f2) = convex hull of {(1,0),(0,1),(0,0)}


f1 = x3 +xy2−x−100y

= x3y0 +x1y2−x1y0−100x0y1

N(f1) = convex hull of {(3,0),(1,2),(1,0),(0,1)}

f2 = x2 +y2−2

= x2y0 +x0y2−2x0y0

N(f2) = convex hull of {(1,0),(0,1),(0,0)}


f1 = x3 +xy2−x−100y

= x3y0 +x1y2−x1y0−100x0y1

N(f1) = convex hull of {(3,0),(1,2),(1,0),(0,1)}

f2 = x2 +y2−2

= x2y0 +x0y2−2x0y0

N(f2) = convex hull of {(1,0),(0,1),(0,0)}


f1 = x3 +xy2−x−100y

= x3y0 +x1y2−x1y0−100x0y1

N(f1) = convex hull of {(3,0),(1,2),(1,0),(0,1)}

f2 = x2 +y2−2

= x2y0 +x0y2−2x0y0

N(f2) = convex hull of {(1,0),(0,1),(0,0)}


Title: Finiteness and enumeration of polynomial systems

Abstract: The study of solutions of polynomial systems is a fundamental

topic in mathematics. In this talk I will discuss the finiteness and enumeration

of solutions to families of polynomial systems that arise in celestial mechanics

and vortex dynamics - however, the methods used can be applied in greater

generality. I will also briefly introduce the Sage computational platform which

is used for this work. Sage is free, open-source, and builds on many other suc-

cessful projects. For this talk, I will focus on the components Gfan, PHCpack,

Singular, and some Sage-native code.

Finiteness and enumeration of polynomial systemscomputational approaches with the Sage platform

Necessary pieces/outline

(1) Basic intro to cel mech and vortices - equations, pictures, people.

(2) Relative equilibria and central configurations

(3) Smale’s problem and generalizations

(4) polynomial solutions - finiteness, enumeration, classification, location

(5) poly ideas and BKK

(a) 1-D Newton polytopes

(b) Minkowski sums and mixed volume - Annette’s work

(c) phcpack and visualization

A brief introduction to Sage

Amoebae

Amoeba of x3+ xy2 − x− y + by

2x− x2+ x3 − 3x4

x + x4

1


Abstract: The study of solutions of polynomial systems is a funda-

mental topic in mathematics. In this talk I will discuss the finiteness

and enumeration of solutions to families of polynomial systems that

arise in celestial mechanics and vortex dynamics - however, the meth-

ods used can be applied in greater generality. I will also briefly intro-

duce the Sage computational platform which is used for this work. Sage

is free, open-source, and builds on many other successful projects. For

this talk, I will focus on the components Gfan, PHCpack, Singular, and

some Sage-native code.






(4) polynomial solutions - finiteness, enumeration, classification, lo-

cation






Amoebae

Definition: the amoeba of a curve defined by f(x, y) = 0 is the set

{(log |x|, log |y|) : f(x, y) = 0, x ∈ C− 0, y ∈ C− 0}

Amoeba of x3+ xy2 − x− y + by

1

















cation






Amoebae



Amoeba of x3+ xy2 − x− y

1

















cation






Amoebae




Amoeba of x2+ y2 − 1

1

















cation






Amoebae




1

















cation






Amoebae




Amoeba of x2+ y2 − 1

1


50x3+83x2y +24xy2+y3+392x2+414xy +50y2−28x +59y−100 = 0


Newton polytope and amoeba

50x3 +y3−100 = 0



50x3+83x2y +24xy2+y3+392x2+414xy +50y2−28x +59y−100 = 0



50x3+83x2y +24xy2+y3+392x2+414xy +50y2−28x +59y−100 = 0



50x3+83x2y +24xy2+y3+392x2+414xy +50y2−28x +59y−100 = 0


The Minkowski Sum

A lot of information about the relationships between a collection of

polytopes is encoded in the Mink


The Minkowski Sum

A lot of information about the relationships between a collection of

polytopes is encoded in their Minkowski sum:

The Minkowski sum of polyhedra P1,P2, . . . ,Pm ∈ Rn is the polyhedron

m

∑i=1

Pi = {m

∑i=1

pi : (p1, . . . ,pm) ∈ P1×P2× . . .×Pm}.


Minkowski Sums in Two Dimensions


Minkowski sum of the Newton polytopes of

x3 + xy

2 − x− y and x2 + y

2 − 1:

MV (N(f1), N(f2))

= Area(N(f1) + N(f2))− Area(N(f1))− Area(N(f2))

= 3n− n/2− n/2 = 2n = Bernstein Bound

Theorem. D. N. Bernstein’s Theorem (1973)

If a system of n polynomials in n variables has finitely manynon-zero solutions, then the number of non-zero solutions isbounded by the mixed volume of the Newton polytopes of thepolynomials.

The Mixed Volume and Bernstein’s TheoremA set of n polytopes in n dimensions determines a quantity

called the mixed volume. One formula for the mixed volume is:

MV (P1, . . . , Pn) =

n�

k=1

(−1)n−k

�

I∈{1,...,n},|I|=k

V oln(

�

i∈I

Pi)

For two polytopes in R2, this becomes

MV (P1, P2) = Area(P1 + P2)− Area(P1)− Area(P2)

Poyhedral homotopy continuation

Polyhedral homotopy Hi = tgi + eic(1− t)fi

Example: the following two polynomials each have total de-

gree 2n, so there are at most 4n2 complex solutions to the system

f1 = f2 = 0 if the solution set is finite.

f1 = 1 + x + xn−1

yn−1

+ 2xny

n

f2 = 1 + y + xn−1

yn−1

+ 2xny

n

Bezout bound is (2n)2 = 4n2

We will see the actual number of solutions in (C∗)2 is 2n.

1

+ =

=

Amoeba, Minkowski sum, and normal fan for

x3 + xy2 − x− y and x2 + y2 − 1:

(1) λ(xj − c) =�

i �=j

mi(xi − xj)

r3ij


i �=j

mjmi(xi − xj)

r3ij1 ≤ j ≤ n


Lagrange points in a rotating frame.

Part II: Polynomials and Polytopes

Outline:







1

tropical line

Tropical Geometry


x⊕ y = min{x, y}

x⊗ y = x+ y


A network of ideas:



BKK (Bernstein, Kushnirenko, and Khovansky) theory

Puiseux series and initial forms


Consider the eigenvalue problem

Av = λv

with a normalization condition

|v| = 1

with A ∈ Mn(C) and v ∈ Cn.

Bezout’s theorem says that if there are finitely many solutions,there are at most (n+ 1)2.

1


i=i0ait

iq , q ∈ N, i0 ∈ Z

f1 = x3+ xy2 − x− y + by

= x3y0 + x1y2 − x1y0 − x0y1 + bx0y1

= x(x2+ y2 − 1) + y(m− 1) = 0

f2 = x2+ y2 − 1

= x2y0 + x0y2 − x0y0

N(f2) = convex hull of {(1, 0), (0, 1), (0, 0)}

N(f1) = convex hull of {(3, 0), (1, 2), (1, 0), (0, 1)}

Our initial form system for the (−1,−1) inward pointing normal then consists of

the highest total-degree terms:

{x3+ xy2 = 0, x2

+ y2 = 0}We can choose x = 1/t and then y = ±i/t+ . . ..For simplicity, lets consider the case where y = i/t+ . . ..The next nonzero term in y(t) will be linear: y = i/t+ a1t+ . . .This gives rise the to the initial form system:

−1− i+ im+ 2ia1 = 0

−1 + 2ia1 = 0

which has the solution a1 = −i/2 only if the parameter m = 1.

1

10


Consider the eigenvalue problem: Av = λv

with a normalization condition�

v2i = 1

with a given A ∈ Mn(C) and unknowns λ ∈ C, v ∈ Cn.

Bezout’s theorem says that if there are finitely many solutions,there are at most (2)n+1.

But the true maximum of isolated solutions is at most 2n.


i=i0ait

iq , q ∈ N, i0 ∈ Z

Simple example:Consider two polynomials in variables x and y, with a parameter m:

f1 = x3 + xy2 − x− y +my

= x3y0 + x1y2 − x1y0 − x0y1 +mx0y1

= x(x2 + y2 − 1) + y(m− 1)

f2 = x2 + y2 − 1

= x2y0 + x0y2 − x0y0

N(f2) = convex hull of {(1, 0), (0, 1), (0, 0)}

N(f1) = convex hull of {(3, 0), (1, 2), (1, 0), (0, 1)}

2

f1 = x3+ xy2 − x− y + by = 0

f2 = x2+ y2 − 1 = 0



{x3+ xy2 = 0, x2

+ y2 = 0}

We can choose x = 1/t and then y = ±i/t+ . . ..

For simplicity, lets consider the case where y = i/t+ . . ..

The next nonzero term in y(t) will be linear: y = i/t+ a1t+ . . .

This gives rise the to the initial form system:

−1− i+ im+ 2ia1 = 0

−1 + 2ia1 = 0



Univariate:

We know many things, in particular the

Fundamental Theorem of Algebra

(an nth-degree polynomial has exactly


Linear Multivariate:

The system Ax = b has a unique solution if A is invertible.

Nonlinear Multivariate:

Bezout’s theorem:

A system of n polynomials in n variables of degree d1, . . . , dn

2

f1 = x3+ xy2 − x− y + by = 0

f2 = x2+ y2 − 1 = 0



{x3+ xy2 = 0, x2

+ y2 = 0}





−1− i+ im+ 2ia1 = 0

−1 + 2ia1 = 0



Univariate:








Bezout’s theorem:


2

f1 = x3+ xy2 − x− y + by = 0

f2 = x2+ y2 − 1 = 0



{x3+ xy2 = 0, x2

+ y2 = 0}





−1− i+ im+ 2ia1 = 0

−1 + 2ia1 = 0



Univariate:








Bezout’s theorem:


2

f1 = x3+ xy2 − x− y + by = 0

f2 = x2+ y2 − 1 = 0



{x3+ xy2 = 0, x2

+ y2 = 0}





−1− i+ im+ 2ia1 = 0

−1 + 2ia1 = 0


Minkowski sum of N(f1) and N(f2).


Univariate:







The Mixed Volume and Bernstein’s TheoremA set of n polytopes in n dimensions determines a quantity called

the mixed volume. One formula for the mixed volume is:

MV (P1, . . . , Pn) =

n�

k=1

(−1)n−k

�

I∈{1,...,n},|I|=k

V oln(

�

i∈I

Pi)



Example: the following two polynomials each have total degree 2n,

so there are at most 4n2 complex solutions to the system f1 = f2 = 0

if the solution set is finite.

f1 = 1 + x + xn−1

yn−1

+ 2xny

n

f2 = 1 + y + xn−1

yn−1

+ 2xny

n



We can obtain better estimates of the number of solutions to a

polynomial system by studying the Newton polytopes of the

polynomials.

Non-zero solutionsLet us focus on the non-zero solutions to a polynomial equation.

If we factor out the zero solutions,

p(x) = amxm

+am+1xm+1

+. . .+anxn

= xm

(am+am+1x+. . .+anxn−m

)

we get a corollary of the fundamental theorem of algebra: the num-

ber of nonzero complex solutions to a nth-degree polynomial is n−m

where m is the degree of the lowest nonzero term.

For example, p(x) = x3 + x

7 = x3(1 + x

4) has 7− 3 = 4 non-zero

roots.

Univariate Polynomials

What can we know about polynomial solutions?

1





MV (P1, . . . , Pn) =

n�

k=1

(−1)n−k

�

I∈{1,...,n},|I|=k

V oln(

�

i∈I

Pi)








f1 = 1 + x + xn−1

yn−1

+ 2xny

n

f2 = 1 + y + xn−1

yn−1

+ 2xny

n





polynomials.

Non-zero solutionsLet us focus on the non-zero solutions to a polynomial equa-

tion. If we factor out the zero solutions,

p(x) = amxm

+am+1xm+1

+. . .+anxn

= xm


)

1

A set of n polytopes in n dimensions determines a quantity called


MV (P1, . . . , Pn) =

n�

k=1

(−1)n−k

�

I∈{1,...,n},|I|=k

V oln(

�

i∈I

Pi)






f1 = 1 + x + xn−1

yn−1

+ 2xny

n

f2 = 1 + y + xn−1

yn−1

+ 2xny

n





polynomials.



p(x) = amxm

+am+1xm+1

+. . .+anxn

= xm


)





7 = x3(1 + x


roots.



1



MV (P1, . . . , Pn) =

n�

k=1

(−1)n−k

�

I∈{1,...,n},|I|=k

V oln(

�

i∈I

Pi)








f1 = 1 + x + xn−1

yn−1

+ 2xny

n

f2 = 1 + y + xn−1

yn−1

+ 2xny

n





polynomials.



p(x) = amxm

+am+1xm+1

+. . .+anxn

= xm


)





7 = x3(1 + x


roots.

Univariate Polynomials1



MV (P1, . . . , Pn) =

n�

k=1

(−1)n−k

�

I∈{1,...,n},|I|=k

V oln(

�

i∈I

Pi)






f1 = 1 + x + xn−1

yn−1

+ 2xny

n

f2 = 1 + y + xn−1

yn−1

+ 2xny

n





polynomials.



p(x) = amxm

+am+1xm+1

+. . .+anxn

= xm


)





7 = x3(1 + x


roots.



1



MV (P1, . . . , Pn) =

n�

k=1

(−1)n−k

�

I∈{1,...,n},|I|=k

V oln(

�

i∈I

Pi)






f1 = 1 + x + xn−1

yn−1

+ 2xny

n

f2 = 1 + y + xn−1

yn−1

+ 2xny

n





polynomials.



p(x) = amxm

+am+1xm+1

+. . .+anxn

= xm


)





7 = x3(1 + x


roots.



1

Minkowski sum interpolation

A polytope P1 can be continuously deformed into another

polytope P2 by the Minkowski sum homotopy (1 − t)P1 + tP2,

t ∈ [0, 1].

The following animation shows a sequence of such homotopies

between the Platonic solids, in order to illustrate the Minkowski

sum in three dimensions.

The name ‘mixed volume’ comes from another description.

In the expansion of the volume of a scaled Minkowski sum with

λi ∈ R+,

V oln(λ1P1 + λ2P2 . . . + λnPn) =

λn1v(n,0,...,0) + λn−1

1 λ2v(n−1,1,0,...,0) + . . .

+λ1λ2 · · · λnv(1,...,1) + . . .

+λn1v(0,...,0,n)

the coefficient v(1,...,1) = MV (P1, . . . , Pn).

Minkowski sum of the Newton polytopes of

x3 + xy2 − x− y and x2 + y2 − 1:

MV (N(f1), N(f2))







MV (P1, . . . , Pn) =

n�

k=1

(−1)n−k

�

I∈{1,...,n},|I|=k

V oln(

�

i∈I

Pi)

1



Example: the following two polynomials each have total degree 2n, so

there are at most 4n2 complex solutions to the system f1 = f2 = 0 if the

solution set is finite.

f1 = 1 + x + xn−1

yn−1

+ 2xny

n

f2 = 1 + y + xn−1

yn−1

+ 2xny

n


Non-zero solutionsLet us focus on the non-zero solutions to a polynomial equation. If we

factor out the zero solutions,

p(x) = amxm

+am+1xm+1

+ . . .+anxn

= xm

(am +am+1x+ . . .+anxn−m

)

we get a corollary of the fundamental theorem of algebra: the number

of nonzero complex solutions to a nth-degree polynomial is n−m where

m is the degree of the lowest nonzero term.

For example, p(x) = x3 +x

7 = x3(1+x

4) has 7−3 = 4 non-zero roots.



For high-degree polynomials, or systems of polynomials, we cannot

expect nice symbolic formulae for the solutions.

So we need to consider other sorts of information.

One thing we might be able to do is determine the exact number of

solutions. For polynomials in one variable, we have:

The Bezout Bound

There is a generalization of the fundamental theorem of algebra to

multivariate polynomials, known as Bezout’s Theorem. For complex

solutions to multivariate polynomials it implies what we call

The Bezout Bound:

1



Example: the following two polynomials each have total degree 2n, so

there are at most 4n2 complex solutions to the system f1 = f2 = 0 if the

solution set is finite.

f1 = 1 + x + xn−1

yn−1

+ 2xny

n

f2 = 1 + y + xn−1

yn−1

+ 2xny

n


Non-zero solutionsLet us focus on the non-zero solutions to a polynomial equation. If we

factor out the zero solutions,

p(x) = amxm

+am+1xm+1

+ . . .+anxn

= xm

(am +am+1x+ . . .+anxn−m

)

we get a corollary of the fundamental theorem of algebra: the number

of nonzero complex solutions to a nth-degree polynomial is n−m where

m is the degree of the lowest nonzero term.

For example, p(x) = x3 +x

7 = x3(1+x

4) has 7−3 = 4 non-zero roots.



For high-degree polynomials, or systems of polynomials, we cannot

expect nice symbolic formulae for the solutions.

So we need to consider other sorts of information.

One thing we might be able to do is determine the exact number of

solutions. For polynomials in one variable, we have:

The Bezout Bound

There is a generalization of the fundamental theorem of algebra to

multivariate polynomials, known as Bezout’s Theorem. For complex

solutions to multivariate polynomials it implies what we call

The Bezout Bound:

1



MV (P1, . . . , Pn) =

n�

k=1

(−1)n−k

�

I∈{1,...,n},|I|=k

V oln(

�

i∈I

Pi)






f1 = 1 + x + xn−1

yn−1

+ 2xny

n

f2 = 1 + y + xn−1

yn−1

+ 2xny

n





polynomials.



p(x) = amxm

+am+1xm+1

+. . .+anxn

= xm


)





7 = x3(1 + x


roots.



1

+ =

MV (N(f1), N(f2))







MV (P1, . . . , Pn) =

n�

k=1

(−1)n−k

�

I∈{1,...,n},|I|=k

V oln(

�

i∈I

Pi)








f1 = 1 + x + xn−1

yn−1

+ 2xny

n

f2 = 1 + y + xn−1

yn−1

+ 2xny

n





polynomials.

1

Fewnomials:

Bihan-Sottile bound (improving a previous result of Khovanskii):

A system of n polynomials in n variables having a total of n+ k + 1 distinctmonomials has fewer than

e2 + 3

42

k2

nk

non-degenerate solutions in the positive orthant.

Another result is that of Li, Rojas, and Wang, who showed thata system of two trinomials has at most 5 positive real solutions.

Tropical Geometry


x⊕ y = min{x, y}

x⊗ y = x+ y


A network of ideas:



BKK (Bernstein, Kushnirenko, and Khovanskii) theory1

−2(m1 +m2 +m3)r312r

313r

323m1 + 2r313r

323m1 + 2r313r

323m2 + r312r

313m3−

r12r513m3 + r12r

313r

223m3 + r312r

323m3 + r12r

213r

323m3 − r12r

523m3 = 0

−2(m1 +m2 +m3)r312r

313r

323m1 + 2r312r

323m1 − r512r13m2 + r312r

313m2+

r312r13r223m2 + r212r13r

323m2 + r313r

323m2 − r13r

523m2 + 2r312r

323m3 = 0

−2(m1 +m2 +m3)r312r

313r

323m1 − r512r23m1 + r312r

213r23m1 + r212r

313r23m1−

r513r23m1 + r312r323m1 + r313r

323m1 + 2r312r

313m2 + 2r312r

313m3 = 0

R3.¡r12,r13,r23¿ = QQ[] n3eqms = [R3(q.subs(m1:1,m2:-102/100,m3:1000)) forq in n3eqs] from sage.interfaces.phc import phc s3 = phc.blackbox(n3eqms,R3)

Recommended reading on Grobner basesand BKK theory


i=i0ait

iq , q ∈ N, i0 ∈ Z

x3 + xy2 − x− y and x2 + y2 − 1:

(1) λ(xj − c) =�

i �=j

mi(xi − xj)

r3ij


i �=j

mjmi(xi − xj)

r3ij1 ≤ j ≤ n




Outline:

(1) Thanks; overview - cel mech and my path in it(2) Basic celestial mechanics - history, contributions to mathematics(3) light reading list(4) 2- and 3- body problem, Lagrange points(5) Chaos in the 3-body problem(6) Proof for 4 - requires more knowledge of ccs(7) four-body central configurations(8) finiteness. AC equations, polynomials(9) Groebner bases(10) BKK theory

1

Albouy-Chenciner polynomials for thethree-body problem central configurations

−2(m1 +m2 +m3)r312r

313r

323m1 + 2r313r

323m1 + 2r313r

323m2 + r312r

313m3−

r12r513m3 + r12r

313r

223m3 + r312r

323m3 + r12r

213r

323m3 − r12r

523m3 = 0

−2(m1 +m2 +m3)r312r

313r

323m1 + 2r312r

323m1 − r512r13m2 + r312r

313m2+

r312r13r223m2 + r212r13r

323m2 + r313r

323m2 − r13r

523m2 + 2r312r

323m3 = 0

−2(m1 +m2 +m3)r312r

313r

323m1 − r512r23m1 + r312r

213r23m1 + r212r

313r23m1−

r513r23m1 + r312r323m1 + r313r

323m1 + 2r312r

313m2 + 2r312r

313m3 = 0

R3.¡r12,r13,r23¿ = QQ[] n3eqms = [R3(q.subs(m1:1,m2:-102/100,m3:1000)) forq in n3eqs] from sage.interfaces.phc import phc s3 = phc.blackbox(n3eqms,R3)

Recommended reading on Grobner basesand BKK theory


i=i0ait

iq , q ∈ N, i0 ∈ Z

x3 + xy2 − x− y and x2 + y2 − 1:

(1) λ(xj − c) =�

i �=j

mi(xi − xj)

r3ij


i �=j

mjmi(xi − xj)

r3ij1 ≤ j ≤ n




Outline:

(1) Thanks; overview - cel mech and my path in it(2) Basic celestial mechanics - history, contributions to mathematics(3) light reading list(4) 2- and 3- body problem, Lagrange points(5) Chaos in the 3-body problem(6) Proof for 4 - requires more knowledge of ccs(7) four-body central configurations(8) finiteness. AC equations, polynomials(9) Groebner bases

1

(1) λ(xj − c) =�

i�=j

mi(xi − xj)

rDij

(2) Sij =1

rDij

+ λ (i �= j) Sii = 0.





f0 = m1z1 + m2z2 + m3z3 + m4z4 = 0

f1 = m2z2r212 + m3z3r

213 + m4z4r

214 + k = 0

f2 = m1z1r212 + m3z3r

223 + m4z4r

224 + k = 0

f3 = m1z1r213 + m2z2r

223 + m4z4r

234 + k = 0

f4 = m1z1r214 + m2z2r

224 + m3z3r

234 + k = 0

Sij = zizj 1 ≤ i < j ≤ 4.



1





Marshall Hampton







If we insist that the orbit of each particle is similar to that of the others, they are forced into Keplerian orbits and

they must form a central configuration:

0 ≤ mi

xi ∈ Rd



223s24 − s13s23s

224

1

2



223s24 − s13s23s

224

− 2(m1 + m2 + m3 + m4)r312r

313r

314r

323r

324 + m4r

312r

313r

314r

323 −m4r12r

313r

514r

323

+ m4r12r313r

314r

323r

224 + m3r

312r

313r

314r

324 −m3r12r

513r

314r

324 + m3r12r

313r

314r

223r

324

+ m4r312r

313r

323r

324 + m4r12r

313r

214r

323r

324 + m3r

312r

314r

323r

324 + m3r12r

213r

314r

323r

324

+ 2(m1 + m2)r313r

314r

323r

324 −m3r12r

314r

523r

324 −m4r12r

313r

323r

524 = 0


22

−m3r312r

313 + m3r12r

513 −m3r12r

313r

223 −m3r

312r

323 −m3r12r

213r

323 − 2m1r

313r

323−

2m2r313r

323 + 2m1r

312r

313r

323 + 2m2r

312r

313r

323 + 2m3r

312r

313r

323 + m3r12r

523 = 0

m2r512r13 −m2r

312r

313 −m2r

312r13r

223 − 2m1r

312r

323 − 2m3r

312r

323 −m2r

212r13r

323−

m2r313r

323 + 2m1r

312r

313r

323 + 2m2r

312r

313r

323 + 2m3r

312r

313r

323 + m2r13r

523 = 0

2m2r312r

313 + 2m3r

312r

313 −m1r

512r23 + m1r

312r

213r23 + m1r

212r

313r23 −m1r

513r23+

m1r312r

323 + m1r

313r

323 − 2m1r

312r

313r

323 − 2m2r

312r

313r

323 − 2m3r

312r

313r

323 = 0

−m3r312r

513r14r

324 + m3r

312r

313r

314r

324 + m3r

312r

313r14r

324r

234 −m2r

512r

313r14r

334 + m2r

312r

313r

314r

334+

m2r312r

313r14r

224r

334 + 2m1r

312r

313r

324r

334 + m3r

312r

213r14r

324r

334 + m2r

212r

313r14r

324r

334 + m3r

312r

314r

324r

334 + m2r

313r

314r

324r

334−

2m1r312r

313r

314r

324r

334 − 2m2r

312r

313r

314r

324r

334 − 2m3r

312r

313r

314r

324r

334 −m2r

313r14r

524r

334 −m3r

312r14r

324r

534 = 0

−m3r312r

314r

523r24 + m3r

312r

314r

323r

324 + m3r

312r

314r

323r24r

234 + 2m2r

312r

314r

323r

334 + m3r

312r

314r

223r24r

334−

m1r512r

323r24r

334 + m1r

312r

214r

323r24r

334 + m1r

212r

314r

323r24r

334 −m1r

514r

323r24r

334 + m3r

312r

314r

324r

334 + m1r

312r

323r

324r

334+

m1r314r

323r

324r

334 − 2m1r

312r

314r

323r

324r

334 − 2m2r

312r

314r

323r

324r

334 − 2m3r

312r

314r

323r

324r

334 −m3r

312r

314r24r

534 = 0

2m3r313r

314r

323r

324 −m2r

313r

314r

523r34 + m2r

313r

314r

323r

224r34 + m2r

313r

314r

223r

324r34 −m1r

513r

323r

324r34+

m1r313r

214r

323r

324r34 + m1r

213r

314r

323r

324r34 −m1r

514r

323r

324r34 −m2r

313r

314r

524r34 + m2r

313r

314r

323r

334 + m2r

313r

314r

324r

334+

m1r313r

323r

324r

334 + m1r

314r

323r

324r

334 − 2m1r

313r

314r

323r

324r

334 − 2m2r

313r

314r

323r

324r

334 − 2m3r

313r

314r

323r

324r

334 = 0



Newtonian three-body problem



(1) Sij =1

r3ij

+ λ (i �= j) Sii = 0.



Marshall Hampton







If we insist that the orbit of each particle is similar to that of the others, they are forced into Keplerian orbits andthey must form a central configuration:

0 ≤ mi1



Newtonian three-body problem



(1) Sij =1

r3ij

+ λ (i �= j) Sii = 0.



Marshall Hampton







If we insist that the orbit of each particle is similar to that of the others, they are forced into Keplerian orbits andthey must form a central configuration:

0 ≤ mi1

Celest Mech Dyn Astr (2011) 109:321–332DOI 10.1007/s10569-010-9328-9

ORIGINAL ARTICLE

Finiteness of spatial central configurationsin the five-body problem

Marshall Hampton · Anders Jensen

Received: 12 August 2010 / Revised: 22 November 2010 / Accepted: 8 December 2010 /Published online: 8 January 2011© The Author(s) 2011. This article is published with open access at Springerlink.com

Abstract We strengthen a generic finiteness result due to Moeckel by showing that thenumber of spatial central configurations of the Newtonian five-body problem with positivemasses is finite, apart from some explicitly given special cases of mass values.

Keywords Central configurations · n-Body problem · Tropical geometry ·Polyhedral fan · Albouy–Chenciner equations

1 Introduction

In this paper we present a computer-assisted proof of the finiteness of the spatial centralconfigurations of the Newtonian five-body problem with positive masses, with the exceptionof some explicit special cases of mass values.

By the Newtonian spatial n-body problem we mean the dynamical system given by

m j x j =!

i != j

mi m j (xi " x j )

r3i j

1 # j # n (1)

where xi $ R3 is the position of particle i , ri j is the distance between xi and x j , and mi isthe mass of particle i (Newton 1687).

M. Hampton (B)Department of Mathematics and Statistics, University of Minnesota, 1117 University Drive,Duluth, MN 55812, USAe-mail: [email protected]

A. JensenMathematisches Institut, Georg-August-Universität Göttingen, Bunsenstr. 3-5,37073 Göttingen, Germanye-mail: [email protected]

123

Author's personal copy

Celest Mech Dyn Astr (2011) 109:321–332DOI 10.1007/s10569-010-9328-9

ORIGINAL ARTICLE

Finiteness of spatial central configurationsin the five-body problem

Marshall Hampton · Anders Jensen

Received: 12 August 2010 / Revised: 22 November 2010 / Accepted: 8 December 2010 /Published online: 8 January 2011© The Author(s) 2011. This article is published with open access at Springerlink.com

Abstract We strengthen a generic finiteness result due to Moeckel by showing that thenumber of spatial central configurations of the Newtonian five-body problem with positivemasses is finite, apart from some explicitly given special cases of mass values.

Keywords Central configurations · n-Body problem · Tropical geometry ·Polyhedral fan · Albouy–Chenciner equations

1 Introduction

In this paper we present a computer-assisted proof of the finiteness of the spatial centralconfigurations of the Newtonian five-body problem with positive masses, with the exceptionof some explicit special cases of mass values.

By the Newtonian spatial n-body problem we mean the dynamical system given by

m j x j =!

i != j

mi m j (xi " x j )

r3i j

1 # j # n (1)

where xi $ R3 is the position of particle i , ri j is the distance between xi and x j , and mi isthe mass of particle i (Newton 1687).

M. Hampton (B)Department of Mathematics and Statistics, University of Minnesota, 1117 University Drive,Duluth, MN 55812, USAe-mail: [email protected]

A. JensenMathematisches Institut, Georg-August-Universität Göttingen, Bunsenstr. 3-5,37073 Göttingen, Germanye-mail: [email protected]

123

Author's personal copy

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Annals of Mathematics 176 (2012), 1–54http://dx.doi.org/10.4007/annals.2012.176.1.

Finiteness of central configurationsof five bodies in the plane

By Alain Albouy and Vadim Kaloshin

Abstract

We prove there are finitely many isometry classes of planar central con-

figurations (also called relative equilibria) in the Newtonian 5-body prob-

lem, except perhaps if the 5-tuple of positive masses belongs to a given

codimension 2 subvariety of the mass space.

1. Introduction and statements

Let (xk, yk) ∈ R2, k = 1, . . . , n, be the positions of n points in the plane R2.We call these points the bodies. Body k has a mass mk > 0. We will studythe system

Çx1y1

å= m2r

−312

Çx21y21

å+m3r

−313

Çx31y31

å+ · · ·+mnr

−31n

Çxn1yn1

å(1)

Çx2y2

å= m1r

−312

Çx12y12

å+m3r

−323

Çx32y32

å+ · · ·+mnr

−32n

Çxn2yn2

å

· · ·Çxnyn

å= m1r

−31n

Çx1ny1n

å+ · · ·+mn−1r

−3(n−1)n

Çx(n−1)n

y(n−1)n

å,

where xkl = xl − xk, ykl = yl − yk and rkl =Äx2kl + y2kl

ä1/2> 0. Some short

notation will be useful. We call fk ∈ R2, k = 1, . . . , n, the right-hand sides ofthe equations. System (1) is

(2) qk = fk, k = 1, . . . , n, with qk =

Çxkyk

å∈ R2.

Let us recall the meaning of this system. Newton’s equations of the n-bodyproblem are the 3-dimensional version of qk = −fk, k = 1, . . . , n. Newton’sequations possess few “simple” solutions if n ≥ 3. They are called homo-graphic or self-similar solutions. In these solutions the configuration remainsin the same similarity class, and each of the n bodies behaves as a body ina 2-body problem. Laplace [16], [17] remarked that if there is a λ > 0 such

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Annals of Mathematics 176 (2012), 1–54http://dx.doi.org/10.4007/annals.2012.176.1.

Finiteness of central configurationsof five bodies in the plane

By Alain Albouy and Vadim Kaloshin

Abstract

We prove there are finitely many isometry classes of planar central con-

figurations (also called relative equilibria) in the Newtonian 5-body prob-

lem, except perhaps if the 5-tuple of positive masses belongs to a given

codimension 2 subvariety of the mass space.

1. Introduction and statements

Let (xk, yk) ∈ R2, k = 1, . . . , n, be the positions of n points in the plane R2.We call these points the bodies. Body k has a mass mk > 0. We will studythe system

Çx1y1

å= m2r

−312

Çx21y21

å+m3r

−313

Çx31y31

å+ · · ·+mnr

−31n

Çxn1yn1

å(1)

Çx2y2

å= m1r

−312

Çx12y12

å+m3r

−323

Çx32y32

å+ · · ·+mnr

−32n

Çxn2yn2

å

· · ·Çxnyn

å= m1r

−31n

Çx1ny1n

å+ · · ·+mn−1r

−3(n−1)n

Çx(n−1)n

y(n−1)n

å,

where xkl = xl − xk, ykl = yl − yk and rkl =Äx2kl + y2kl

ä1/2> 0. Some short

notation will be useful. We call fk ∈ R2, k = 1, . . . , n, the right-hand sides ofthe equations. System (1) is

(2) qk = fk, k = 1, . . . , n, with qk =

Çxkyk

å∈ R2.

Let us recall the meaning of this system. Newton’s equations of the n-bodyproblem are the 3-dimensional version of qk = −fk, k = 1, . . . , n. Newton’sequations possess few “simple” solutions if n ≥ 3. They are called homo-graphic or self-similar solutions. In these solutions the configuration remainsin the same similarity class, and each of the n bodies behaves as a body ina 2-body problem. Laplace [16], [17] remarked that if there is a λ > 0 such

1

The mixed volume of the Albouy-Chenciner equations for N = 3 is

MV (3) = D3 + 12D2 − 12D

Bernstein’s theorem: the mixed volume determines the maximum number ofisolated solutions to a system of polynomials in (C∗)n, and generically this anexact count.

The common refinement of two fans F1 and F2 is defined as

F1 ∧ F2 = {C1 ∩ C2}(C1,C2)∈F1×F2

Mixed volumes, normal fans, and Bernstein’s theorem

Addition on elliptic curves

Sage Wiki has a growing collection of @interact examples

Thanks!

Highways in the solar system (thanks to Maciej and co.)


for the Newtonian 4-body problem1


MV (3) = D3 + 12D2 − 12D



F1 ∧ F2 = {C1 ∩ C2}(C1,C2)∈F1×F2




Thanks!





MV (3) = D3 + 12D2 − 12D

Conjecture:

These are simply all choices of rij = e2πim/D, m ∈ {0, 1, . . . , D − 1}

For positive masses, there are always four positive real solutions tothe Albouy-Chenciner equations for N = 3.



F1 ∧ F2 = {C1 ∩ C2}(C1,C2)∈F1×F2




Thanks!1


MV (3) = D3 + 12D2 − 12D



F1 ∧ F2 = {C1 ∩ C2}(C1,C2)∈F1×F2




Thanks!





MV (3) = D3 + 12D2 − 12D

Conjecture:





F1 ∧ F2 = {C1 ∩ C2}(C1,C2)∈F1×F2




Thanks!1


MV (4) = D6 + 48D5 + 216D4 + 416D3 − 1008D2 + 384D


MV (3) = D3 + 12D2 − 12D

Conjecture:

Again, these are all the choices of rij = e2πim/D,m ∈ {0, 1, . . . , D − 1}

Some of these can be degenerate for particular mass values.

Surprisingly, six of these choices are degenerate with multiplicity 2when D = 3.

The generalized Albouy-Chenciner equations for N = 3 have finitelymany solutions for any integer D > 0 (result by Michael Cook).

Currently there is not a finiteness result for just theAlbouy-Chenciner equations for N = 4, but it should be possible to

prove this for all D ≥ 2.

For positive masses, there are four positive real solutions to theAlbouy-Chenciner equations for N = 3.


The common refinement of two fans F1 and F2 is defined as1


MV (3) = D3 + 12D2 − 12D



F1 ∧ F2 = {C1 ∩ C2}(C1,C2)∈F1×F2




Thanks!





MV (3) = D3 + 12D2 − 12D

Conjecture:





F1 ∧ F2 = {C1 ∩ C2}(C1,C2)∈F1×F2




Thanks!1


MV (4) = D6 + 48D5 + 216D4 + 416D3 − 1008D2 + 384D


MV (3) = D3 + 12D2 − 12D

Conjecture:



Surprisingly, six of these choices are degenerate with multiplicity 2.








MV (4) = D6 + 48D5 + 216D4 + 416D3 − 1008D2 + 384D


MV (3) = D3 + 12D2 − 12D

Conjecture:











MV (3) = D3 + 12D2 − 12D



F1 ∧ F2 = {C1 ∩ C2}(C1,C2)∈F1×F2




Thanks!





MV (3) = D3 + 12D2 − 12D

Conjecture:





F1 ∧ F2 = {C1 ∩ C2}(C1,C2)∈F1×F2




Thanks!1


MV (4) = D6 + 48D5 + 216D4 + 416D3 − 1008D2 + 384D


MV (3) = D3 + 12D2 − 12D

Conjecture:











MV (3) = D3+ 12D2 − 12D

Conjecture:

For positive masses, there are always four positive real solutions to

the Albouy-Chenciner equations for N = 3.

Bernstein’s theorem: the mixed volume determines the maximum number of

isolated solutions to a system of polynomials in (C∗)n, and generically this an

exact count.


F1 ∧ F2 = {C1 ∩ C2}(C1,C2)∈F1×F2




Thanks!


1


MV (4) = D6 + 48D5 + 216D4 + 416D3 − 1008D2 + 384D


MV (3) = D3 + 12D2 − 12D

Conjecture:











MV (4) = D6 + 48D5 + 216D4 + 416D3 − 1008D2 + 384D


MV (3) = D3 + 12D2 − 12D

Conjecture:

These are simply all choices of rij = e2πim/D, m ∈ {0, 1, . . . , D− 1}





F1 ∧ F2 = {C1 ∩ C2}(C1,C2)∈F1×F2


Addition on elliptic curves1


MV (4) = D6 + 48D5 + 216D4 + 416D3 − 1008D2 + 384D


MV (3) = D3 + 12D2 − 12D

Conjecture:






F1 ∧ F2 = {C1 ∩ C2}(C1,C2)∈F1×F2




MV (4) = D6 + 48D5 + 216D4 + 416D3 − 1008D2 + 384D


MV (3) = D3 + 12D2 − 12D

Conjecture:






F1 ∧ F2 = {C1 ∩ C2}(C1,C2)∈F1×F2




MV (4) = D6 + 48D5 + 216D4 + 416D3 − 1008D2 + 384D


MV (3) = D3 + 12D2 − 12D

Conjecture:






F1 ∧ F2 = {C1 ∩ C2}(C1,C2)∈F1×F2




MV (4) = D6 + 48D5 + 216D4 + 416D3 − 1008D2 + 384D


MV (3) = D3 + 12D2 − 12D

Conjecture:






F1 ∧ F2 = {C1 ∩ C2}(C1,C2)∈F1×F2




MV (4) = D6 + 48D5 + 216D4 + 416D3 − 1008D2 + 384D


MV (3) = D3 + 12D2 − 12D

Conjecture:







F1 ∧ F2 = {C1 ∩ C2}(C1,C2)∈F1×F2

Mixed volumes, normal fans, and Bernstein’s theorem1


MV (4) = D6 + 48D5 + 216D4 + 416D3 − 1008D2 + 384D


MV (3) = D3 + 12D2 − 12D

Conjecture:






F1 ∧ F2 = {C1 ∩ C2}(C1,C2)∈F1×F2




MV (4) = D6 + 48D5 + 216D4 + 416D3 − 1008D2 + 384D


MV (3) = D3 + 12D2 − 12D

Conjecture:






F1 ∧ F2 = {C1 ∩ C2}(C1,C2)∈F1×F2




MV (4) = D6 + 48D5 + 216D4 + 416D3 − 1008D2 + 384D


MV (3) = D3 + 12D2 − 12D

Conjecture:







F1 ∧ F2 = {C1 ∩ C2}(C1,C2)∈F1×F2

Mixed volumes, normal fans, and Bernstein’s theorem1


MV (4) = D6 + 48D5 + 216D4 + 416D3 − 1008D2 + 384D


MV (3) = D3 + 12D2 − 12D

Conjecture:




Currently there is not a finiteness result for just theAlbouy-Chenciner equations for N = 4, but it should be possible

to prove this for all D ≥ 2.




F1 ∧ F2 = {C1 ∩ C2}(C1,C2)∈F1×F2

1

PROPERTIES OF GENERALIZED ALBOUY-CHENCINEREQUATIONS

MARSHALL HAMPTON

1. Three-body problem

For D = 3, mixed volume is 99. Six of the 27 third-root of unity solutions havemultiplicity two - those which sum to 0.For positive masses, there seem to always be the same number (4) of positive real

solutions regardless of D. The number of real solutions is 7 for odd D and 32 for evenD.

2. Jacobian structure

Let

Sij = r−Dik − 1, Aijk = r2jk − r2ik − r2ij

qijk = mkSikAijk = (r−Dik − 1)(r2jk − r2ik − r2ij)

fijk = qijk + qjik

fij =n�

k=1

fijk

pij =

�rD−2ij

n�

k �=i,j

rDikrDjk

�fij

= 2(−MrDij +mi +mj)n�

k �=i,j

rDikrDjk

+mk

n�

k �=i,j

�rDij

�n�

l �=i,j,k

rDil

n�

l �=i,j

rDjl +n�

l �=i,j,k

rDjl

n�

l �=i,j

rDil

�

+ rD−2ij

�r2jk

n�

l �=i,j

rDil

n�

l �=i,j,k

rDjl + r2ik

n�

l �=i,j

rDjl

n�

l �=i,j,k

rDil

−(r2+Dik + r2+D

jk )n�

l �=i,j,k

rDjl rDil

��

1

In some ways, the complexity of the Albouy-Chenciner equationsdoes not increase much after N = 4.


MV (4) = D6 + 48D5 + 216D4 + 416D3 − 1008D2 + 384D


MV (3) = D3 + 12D2 − 12D

Conjecture:









1

References:(1) Ideals, varieties, and algorithms: An introduction to computational algebraic

geometry and commutative algebra, by Cox, Little, and O’Shea, Springer.




Finiteness of relative equilibria in the N -body problem

Marshall Hampton

University of Minnesota Duluth

Albouy-Chenciner equations (symmetric):

n�

k=1

mk

�Sik(r



�= 0

Asymmetric:

n�

k=1

mkSik(r2jk − r2ik − r2ij) = 0

Sij = r−3ij + λ = r−3

ij − 1

(we choose to set λ = −1).

1

Documents

Kaloshin, Annals of Math 176, 2012. Finiteness (and ... · (1) Ideals, varieties, and algorithms: An introduction to computational algebraic geometry and commutative algebra, by Cox,