Kaloshin, Annals of Math 176, 2012. Finiteness (and ... · (1) Ideals, varieties, and algorithms:...
Transcript of Kaloshin, Annals of Math 176, 2012. Finiteness (and ... · (1) Ideals, varieties, and algorithms:...
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:
(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) Newton’s law: 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:
(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
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
Central configurations
Orbits in which the shape of the configuration is constant (whichincludes the relative equilibria) are called central configurations andsatisfy the equation:
Euler central configurations (1767)
Lagrange central configurations (1772)
(with masses = 1,2,3; on elliptical orbits)
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
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
(with masses = 1,2,3; on elliptical orbits)
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
Euler central configurations (1767)
Lagrange central configurations (1772)
(with masses = 1,2,3; on elliptical orbits)
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
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
Some four-body central configurations on elliptical orbits
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
Concave equal mass four-body central configurations
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.
Warning: Some of these numbers have caveats -
for example, restrictions on the parameters
Some four-body central configurations on elliptical orbits
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
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 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.
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.
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
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 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.
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.
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
Equal mass four-body central configurations (Albouy 1996)
Warning: Some of these numbers have caveats -for example, restrictions on the parameters
From Grobli’s thesis, 1877
Concave equal mass four-body central configurations
Some four-body central configurations on elliptical orbits
Smale’s 18 Problems for the 21 Century:
(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)
Smale’s 6th problem for the 21st century:
5
Concave equal mass four-body central configurations
Some four-body central configurations on elliptical orbits
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.
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
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.
In 1999, Gareth Roberts found a continuum of solutions for the 5-body
problem with one negative mass.
Richard Moeckel and I proved finiteness in the 4-body case in 2002.
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
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 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.
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.
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.
Richard Moeckel and I proved finiteness in the 4-body case in 2002.
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
2jk − r2ik − r2ij) + Sjk(r
2ik − r2jk − r2ij)
�= 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
N-body problems of finiteness, enumeration and bounds
Albouy-Chenciner Equations for Central Configurations
Valid for all force exponents, for configurations in every dimension with 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 Newtonian n-body problem
Central configurations
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
− 8s12s13s14s23s24 + s12s13s14s23 − s13s214s23 + s12s13s14s24
− s213s14s24 + s12s13s23s24 + s12s14s23s24 + 8s13s14s23s24 − s14s
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.
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 Newtonian n-body problem
Central configurations
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.
Newtonian three-body problem (D = 3)
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.
Newtonian three-body problem (D = 3)
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
2
0 ≤ mi
xi ∈ Rd
− 8s12s13s14s23s24 + s12s13s14s23 − s13s214s23 + s12s13s14s24
− s213s14s24 + s12s13s23s24 + s12s14s23s24 + 8s13s14s23s24 − s14s
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
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
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
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
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.
(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 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
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
Groebner bases of linear systems
= Reduced echelon form
Consider the linear system:
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
Groebner bases of linear systems
= Reduced echelon form
Consider the linear system:
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
Groebner bases of linear systems
= Reduced echelon form
Consider the linear system:
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
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
Marshall Hampton () Geometric Visualization of Algebraic Information November 1, 2008 5 / 6
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
Marshall Hampton () Geometric Visualization of Algebraic Information November 1, 2008 5 / 6
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
Marshall Hampton () Geometric Visualization of Algebraic Information November 1, 2008 5 / 6
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
Marshall Hampton () Geometric Visualization of Algebraic Information November 1, 2008 5 / 6
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
Marshall Hampton () Geometric Visualization of Algebraic Information November 1, 2008 5 / 6
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
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
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.
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
Term orderings and weight vectors
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.
Marshall Hampton () The polyhedral geometry of polynomial systems November 5, 2009 2 / 30
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.
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
All of the possible Groebner bases
(via Anders Jensen’s Gfan and Sage)
Marshall Hampton () Geometric Visualization of Algebraic Information November 6, 2008 6 / 19
All of the possible Groebner bases
(via Anders Jensen’s Gfan and Sage)
Marshall Hampton () Geometric Visualization of Algebraic Information November 6, 2008 6 / 19
All of the possible Groebner bases
(via Anders Jensen’s Gfan and Sage)
Marshall Hampton () Geometric Visualization of Algebraic Information November 6, 2008 6 / 19
Groebner fan colored by degree
Marshall Hampton () Geometric Visualization of Algebraic Information November 4, 2008 5 / 12
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
Marshall Hampton () Geometric Visualization of Algebraic Information November 4, 2008 5 / 13
Groebner fan colored by degree
Marshall Hampton () Geometric Visualization of Algebraic Information November 4, 2008 5 / 12
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
Marshall Hampton () Geometric Visualization of Algebraic Information November 4, 2008 5 / 13
tropical line
Tropical Geometry
The study of the algebraic geometry of the tropical semiring (R,⊕,⊗) where
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
Problem with Bezout’s Theorem:
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).
Marshall Hampton () Geometric Visualization of Algebraic Information November 3, 2008 4 / 9
Newton polytopes and amoeba
50x3+83x2y +24xy2+y3+392x2+414xy +50y2−28x +59y−100 = 0
Marshall Hampton () Geometric Visualization of Algebraic Information November 5, 2008 10 / 16
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.
Marshall Hampton () The polyhedral geometry of polynomial systems November 4, 2009 10 / 25
Newton polytopes and amoeba
50x3+83x2y +24xy2+y3+392x2+414xy +50y2−28x +59y−100 = 0
Marshall Hampton () Geometric Visualization of Algebraic Information November 5, 2008 10 / 16
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)}
Marshall Hampton () Geometric Visualization of Algebraic Information November 6, 2008 2 / 20
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)}
Marshall Hampton () Geometric Visualization of Algebraic Information November 6, 2008 2 / 20
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)}
Marshall Hampton () Geometric Visualization of Algebraic Information November 6, 2008 2 / 20
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)}
Marshall Hampton () Geometric Visualization of Algebraic Information November 6, 2008 2 / 20
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)}
Marshall Hampton () Geometric Visualization of Algebraic Information November 6, 2008 2 / 20
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)}
Marshall Hampton () Geometric Visualization of Algebraic Information November 6, 2008 2 / 20
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
Title: Finiteness and enumeration of polynomial systems
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.
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, lo-
cation
(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
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
Title: Finiteness and enumeration of polynomial systems
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.
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, lo-
cation
(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
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
1
Title: Finiteness and enumeration of polynomial systems
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.
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, lo-
cation
(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
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
Amoeba of x2+ y2 − 1
1
Title: Finiteness and enumeration of polynomial systems
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.
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, lo-
cation
(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
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
1
Title: Finiteness and enumeration of polynomial systems
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.
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, lo-
cation
(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
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
Amoeba of x2+ y2 − 1
1
Newton polytopes and amoeba
50x3+83x2y +24xy2+y3+392x2+414xy +50y2−28x +59y−100 = 0
Marshall Hampton () Geometric Visualization of Algebraic Information November 5, 2008 10 / 16
Newton polytope and amoeba
50x3 +y3−100 = 0
Marshall Hampton () Geometric Visualization of Algebraic Information November 3, 2008 5 / 9
Newton polytopes and amoeba
50x3+83x2y +24xy2+y3+392x2+414xy +50y2−28x +59y−100 = 0
Marshall Hampton () Geometric Visualization of Algebraic Information November 5, 2008 10 / 16
Newton polytopes and amoeba
50x3+83x2y +24xy2+y3+392x2+414xy +50y2−28x +59y−100 = 0
Marshall Hampton () Geometric Visualization of Algebraic Information November 5, 2008 10 / 16
Newton polytopes and amoeba
50x3+83x2y +24xy2+y3+392x2+414xy +50y2−28x +59y−100 = 0
Marshall Hampton () Geometric Visualization of Algebraic Information November 5, 2008 10 / 16
The Minkowski Sum
A lot of information about the relationships between a collection of
polytopes is encoded in the Mink
Marshall Hampton () The polyhedral geometry of polynomial systems November 4, 2009 13 / 25
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}.
Marshall Hampton () The polyhedral geometry of polynomial systems November 4, 2009 13 / 26
Minkowski Sums in Two Dimensions
Marshall Hampton () Geometric Visualization of Algebraic Information November 4, 2008 5 / 11
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
(2) Newton’s law: mjxj =�
i �=j
mjmi(xi − xj)
r3ij1 ≤ j ≤ n
Here c is the location of the center of mass, and λ is a parameter.
Lagrange points in a rotating frame.
Part II: Polynomials and Polytopes
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
1
tropical line
Tropical Geometry
The study of the algebraic geometry of the tropical semiring (R,⊕,⊗) where
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
Problem with Bezout’s Theorem:
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
Puiseux series: x(t) =�∞
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
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 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
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.
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, . . . , dn
2
f1 = x3+ xy2 − x− y + by = 0
f2 = x2+ y2 − 1 = 0
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.
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, . . . , dn
2
f1 = x3+ xy2 − x− y + by = 0
f2 = x2+ y2 − 1 = 0
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.
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, . . . , dn
2
f1 = x3+ xy2 − x− y + by = 0
f2 = x2+ y2 − 1 = 0
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.
Minkowski sum of N(f1) and N(f2).
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.
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)
Poyhedral homotopy continuation
Polyhedral homotopy Hi = tgi + eic(1− t)fi
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
Bezout bound is (2n)2 = 4n2
We will see the actual number of solutions in (C∗)2 is 2n.
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
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.
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 equa-
tion. If we factor out the zero solutions,
p(x) = amxm
+am+1xm+1
+. . .+anxn
= xm
(am+am+1x+. . .+anxn−m
)
1
A 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)
Poyhedral homotopy continuation
Polyhedral homotopy Hi = tgi + eic(1− t)fi
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
Bezout bound is (2n)2 = 4n2
We will see the actual number of solutions in (C∗)2 is 2n.
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
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 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
Bezout bound is (2n)2 = 4n2
We will see the actual number of solutions in (C∗)2 is 2n.
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 Polynomials1
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)
Poyhedral homotopy continuation
Polyhedral homotopy Hi = tgi + eic(1− t)fi
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
Bezout bound is (2n)2 = 4n2
We will see the actual number of solutions in (C∗)2 is 2n.
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
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)
Poyhedral homotopy continuation
Polyhedral homotopy Hi = tgi + eic(1− t)fi
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
Bezout bound is (2n)2 = 4n2
We will see the actual number of solutions in (C∗)2 is 2n.
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
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))
= 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)
1
Poyhedral homotopy continuation
Polyhedral homotopy Hi = tgi + eic(1− t)fi
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
Bezout bound is (2n)2 = 4n2
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.
Univariate Polynomials
What can we know about polynomial solutions?
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
Poyhedral homotopy continuation
Polyhedral homotopy Hi = tgi + eic(1− t)fi
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
Bezout bound is (2n)2 = 4n2
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.
Univariate Polynomials
What can we know about polynomial solutions?
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
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)
Poyhedral homotopy continuation
Polyhedral homotopy Hi = tgi + eic(1− t)fi
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
Bezout bound is (2n)2 = 4n2
We will see the actual number of solutions in (C∗)2 is 2n.
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 (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.
We can obtain better estimates of the number of solutions to a
polynomial system by studying the Newton polytopes of the
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
The study of the algebraic geometry of the tropical semiring (R,⊕,⊗) where
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 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
Puiseux series: x(t) =�∞
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
(2) Newton’s law: mjxj =�
i �=j
mjmi(xi − xj)
r3ij1 ≤ j ≤ n
Here c is the location of the center of mass, and λ is a parameter.
Lagrange points in a rotating frame.
Part II: Polynomials and Polytopes
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
Puiseux series: x(t) =�∞
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
(2) Newton’s law: mjxj =�
i �=j
mjmi(xi − xj)
r3ij1 ≤ j ≤ n
Here c is the location of the center of mass, and λ is a parameter.
Lagrange points in a rotating frame.
Part II: Polynomials and Polytopes
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.
Newtonian three-body problem (D = 3)
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
N-body problems of finiteness, enumeration and bounds
Albouy-Chenciner Equations for Central Configurations
Valid for all force exponents, for configurations in every dimension with 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 Newtonian n-body problem
Central configurations
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
− 8s12s13s14s23s24 + s12s13s14s23 − s13s214s23 + s12s13s14s24
− s213s14s24 + s12s13s23s24 + s12s14s23s24 + 8s13s14s23s24 − s14s
223s24 − s13s23s
224
1
2
− 8s12s13s14s23s24 + s12s13s14s23 − s13s214s23 + s12s13s14s24
− s213s14s24 + s12s13s23s24 + s12s14s23s24 + 8s13s14s23s24 − s14s
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
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
N-body problems of finiteness, enumeration and bounds
The Newtonian restricted four-body problem
Newtonian three-body problem
One equation from the Newtonian four-body problem
Albouy-Chenciner Equations for Central Configurations
(1) Sij =1
r3ij
+ λ (i �= j) Sii = 0.
Valid for all force exponents, for configurations in every dimension with 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 Newtonian n-body problem
Central configurations
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
N-body problems of finiteness, enumeration and bounds
The Newtonian restricted four-body problem
Newtonian three-body problem
One equation from the Newtonian four-body problem
Albouy-Chenciner Equations for Central Configurations
(1) Sij =1
r3ij
+ λ (i �= j) Sii = 0.
Valid for all force exponents, for configurations in every dimension with 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 Newtonian n-body problem
Central configurations
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.)
One of six Albouy-Chenciner equations
for the Newtonian 4-body problem1
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.)
One of six Albouy-Chenciner equations
for the Newtonian 4-body problem1
The mixed volume of the Albouy-Chenciner equations for N = 3 is
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.
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!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.)
One of six Albouy-Chenciner equations
for the Newtonian 4-body problem1
The mixed volume of the Albouy-Chenciner equations for N = 3 is
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.
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!1
The mixed volume of the Albouy-Chenciner equations for N = 4 is
MV (4) = D6 + 48D5 + 216D4 + 416D3 − 1008D2 + 384D
The mixed volume of the Albouy-Chenciner equations for N = 3 is
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.
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 as1
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.)
One of six Albouy-Chenciner equations
for the Newtonian 4-body problem1
The mixed volume of the Albouy-Chenciner equations for N = 3 is
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.
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!1
The mixed volume of the Albouy-Chenciner equations for N = 4 is
MV (4) = D6 + 48D5 + 216D4 + 416D3 − 1008D2 + 384D
The mixed volume of the Albouy-Chenciner equations for N = 3 is
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 2.
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.
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 as1
The mixed volume of the Albouy-Chenciner equations for N = 4 is
MV (4) = D6 + 48D5 + 216D4 + 416D3 − 1008D2 + 384D
The mixed volume of the Albouy-Chenciner equations for N = 3 is
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.
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 as1
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.)
One of six Albouy-Chenciner equations
for the Newtonian 4-body problem1
The mixed volume of the Albouy-Chenciner equations for N = 3 is
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.
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!1
The mixed volume of the Albouy-Chenciner equations for N = 4 is
MV (4) = D6 + 48D5 + 216D4 + 416D3 − 1008D2 + 384D
The mixed volume of the Albouy-Chenciner equations for N = 3 is
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 2.
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.
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 as1
The mixed volume of the Albouy-Chenciner equations for N = 3 is
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.
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.)
1
The mixed volume of the Albouy-Chenciner equations for N = 4 is
MV (4) = D6 + 48D5 + 216D4 + 416D3 − 1008D2 + 384D
The mixed volume of the Albouy-Chenciner equations for N = 3 is
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.
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 as1
The mixed volume of the Albouy-Chenciner equations for N = 4 is
MV (4) = D6 + 48D5 + 216D4 + 416D3 − 1008D2 + 384D
The mixed volume of the Albouy-Chenciner equations for N = 3 is
MV (3) = D3 + 12D2 − 12D
Conjecture:
These are simply all choices of rij = e2πim/D, m ∈ {0, 1, . . . , D− 1}
Surprisingly, six of these choices are degenerate with multiplicity 2.
For positive masses, there are four positive real solutions to theAlbouy-Chenciner equations for N = 3.
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 curves1
The mixed volume of the Albouy-Chenciner equations for N = 4 is
MV (4) = D6 + 48D5 + 216D4 + 416D3 − 1008D2 + 384D
The mixed volume of the Albouy-Chenciner equations for N = 3 is
MV (3) = D3 + 12D2 − 12D
Conjecture:
These are simply all choices of rij = e2πim/D, m ∈ {0, 1, . . . , D− 1}
Surprisingly, six of these choices are degenerate with multiplicity 2.
For positive masses, there are four positive real solutions to theAlbouy-Chenciner equations for N = 3.
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 curves1
The mixed volume of the Albouy-Chenciner equations for N = 4 is
MV (4) = D6 + 48D5 + 216D4 + 416D3 − 1008D2 + 384D
The mixed volume of the Albouy-Chenciner equations for N = 3 is
MV (3) = D3 + 12D2 − 12D
Conjecture:
Again, these are all the choices of rij = e2πim/D,m ∈ {0, 1, . . . , D − 1}
Surprisingly, six of these choices are degenerate with multiplicity 2.
For positive masses, there are four positive real solutions to theAlbouy-Chenciner equations for N = 3.
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 curves1
The mixed volume of the Albouy-Chenciner equations for N = 4 is
MV (4) = D6 + 48D5 + 216D4 + 416D3 − 1008D2 + 384D
The mixed volume of the Albouy-Chenciner equations for N = 3 is
MV (3) = D3 + 12D2 − 12D
Conjecture:
These are simply all choices of rij = e2πim/D, m ∈ {0, 1, . . . , D− 1}
Surprisingly, six of these choices are degenerate with multiplicity 2.
For positive masses, there are four positive real solutions to theAlbouy-Chenciner equations for N = 3.
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 curves1
The mixed volume of the Albouy-Chenciner equations for N = 4 is
MV (4) = D6 + 48D5 + 216D4 + 416D3 − 1008D2 + 384D
The mixed volume of the Albouy-Chenciner equations for N = 3 is
MV (3) = D3 + 12D2 − 12D
Conjecture:
Again, these are all the choices of rij = e2πim/D,m ∈ {0, 1, . . . , D − 1}
Surprisingly, six of these choices are degenerate with multiplicity 2.
For positive masses, there are four positive real solutions to theAlbouy-Chenciner equations for N = 3.
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 curves1
The mixed volume of the Albouy-Chenciner equations for N = 4 is
MV (4) = D6 + 48D5 + 216D4 + 416D3 − 1008D2 + 384D
The mixed volume of the Albouy-Chenciner equations for N = 3 is
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 2.
For positive masses, there are four positive real solutions to theAlbouy-Chenciner equations for N = 3.
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 theorem1
The mixed volume of the Albouy-Chenciner equations for N = 4 is
MV (4) = D6 + 48D5 + 216D4 + 416D3 − 1008D2 + 384D
The mixed volume of the Albouy-Chenciner equations for N = 3 is
MV (3) = D3 + 12D2 − 12D
Conjecture:
These are simply all choices of rij = e2πim/D, m ∈ {0, 1, . . . , D− 1}
Surprisingly, six of these choices are degenerate with multiplicity 2.
For positive masses, there are four positive real solutions to theAlbouy-Chenciner equations for N = 3.
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 curves1
The mixed volume of the Albouy-Chenciner equations for N = 4 is
MV (4) = D6 + 48D5 + 216D4 + 416D3 − 1008D2 + 384D
The mixed volume of the Albouy-Chenciner equations for N = 3 is
MV (3) = D3 + 12D2 − 12D
Conjecture:
Again, these are all the choices of rij = e2πim/D,m ∈ {0, 1, . . . , D − 1}
Surprisingly, six of these choices are degenerate with multiplicity 2.
For positive masses, there are four positive real solutions to theAlbouy-Chenciner equations for N = 3.
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 curves1
The mixed volume of the Albouy-Chenciner equations for N = 4 is
MV (4) = D6 + 48D5 + 216D4 + 416D3 − 1008D2 + 384D
The mixed volume of the Albouy-Chenciner equations for N = 3 is
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 2.
For positive masses, there are four positive real solutions to theAlbouy-Chenciner equations for N = 3.
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 theorem1
The mixed volume of the Albouy-Chenciner equations for N = 4 is
MV (4) = D6 + 48D5 + 216D4 + 416D3 − 1008D2 + 384D
The mixed volume of the Albouy-Chenciner equations for N = 3 is
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 2.
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.
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
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.
The mixed volume of the Albouy-Chenciner equations for N = 4 is
MV (4) = D6 + 48D5 + 216D4 + 416D3 − 1008D2 + 384D
The mixed volume of the Albouy-Chenciner equations for N = 3 is
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.
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.
1
References:(1) Ideals, varieties, and algorithms: An introduction to computational algebraic
geometry 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 of relative equilibria in 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:
n�
k=1
mkSik(r2jk − r2ik − r2ij) = 0
Sij = r−3ij + λ = r−3
ij − 1
(we choose to set λ = −1).
1