PURE Math Residents' Program Gröbner Bases and

PURE Math Residents’ ProgramGröbner Bases and Applications

Week 3 Lectures

John B. Little

Department of Mathematics and Computer ScienceCollege of the Holy Cross

June 2012

John B. Little PURE Math 2012 Residents’ Program Week 3

Overview of this week

The research project topics you will be working on willapply the algebra and geometry we have seen (e.g.Gröbner bases, varieties, etc.) to some interestingquestions in celestial mechanics (and possibly relatedquestion in fluid mechanics

In this week’s activities, we’ll develop some backgroundand look at basic cases to provide contextFortunately, not a lot of physics is required, but if you haveseen some, that will help



The research project topics you will be working on willapply the algebra and geometry we have seen (e.g.Gröbner bases, varieties, etc.) to some interestingquestions in celestial mechanics (and possibly relatedquestion in fluid mechanicsIn this week’s activities, we’ll develop some backgroundand look at basic cases to provide context

Fortunately, not a lot of physics is required, but if you haveseen some, that will help



The research project topics you will be working on willapply the algebra and geometry we have seen (e.g.Gröbner bases, varieties, etc.) to some interestingquestions in celestial mechanics (and possibly relatedquestion in fluid mechanicsIn this week’s activities, we’ll develop some backgroundand look at basic cases to provide contextFortunately, not a lot of physics is required, but if you haveseen some, that will help


Newtonian gravitation

Everyone should know about Newton’s gravitational law

Says (in basic form) that magnitude of gravitational forceexerted by one mass m1 on another m2 is proportional tothe product m1m2, and inversely proportional to the squareof the distance:|F | = Gm1m2

r2

Suppose the masses are located at q1, q2 ∈ R3.The gravitational force exerted by m1 on m2 is a vectordirected from q2 back to q1:

F =−Gm1m2(q2 − q1)

r3

where r = ‖q2 − q1‖ is the distance (note cancelation inmagnitude formula!)



Everyone should know about Newton’s gravitational lawSays (in basic form) that magnitude of gravitational forceexerted by one mass m1 on another m2 is proportional tothe product m1m2, and inversely proportional to the squareof the distance:

|F | = Gm1m2r2


F =−Gm1m2(q2 − q1)

r3




Everyone should know about Newton’s gravitational lawSays (in basic form) that magnitude of gravitational forceexerted by one mass m1 on another m2 is proportional tothe product m1m2, and inversely proportional to the squareof the distance:|F | = Gm1m2

r2


F =−Gm1m2(q2 − q1)

r3





r2

Suppose the masses are located at q1, q2 ∈ R3.

The gravitational force exerted by m1 on m2 is a vectordirected from q2 back to q1:

F =−Gm1m2(q2 − q1)

r3





r2


F =−Gm1m2(q2 − q1)

r3



Some history

Building on astronomical observations by Tycho Brahe(1546 - 1601), Johannes Kepler (1570 - 1630) developedthree empirical laws of planetary motion in a simplifiedplanet, star (much larger mass) system:

Kepler 1: The planet follows a planar, elliptical orbit withthe star at one focusKepler 2: The radius vector from star to planet sweeps outequal areas in equal times as the planet orbitsKepler 3: The square of the period of orbit is proportionalto the cube of the semimajor axis of the ellipse: T 2 = ka3;see Wikipedia entry for Kepler’s laws of planetary motionIsaac Newton (1642-1727) basically developed his form ofdifferential calculus to show that Kepler’s laws follow fromthe inverse square law of gravitational attraction (and hisown law F = ma)


Some history

Building on astronomical observations by Tycho Brahe(1546 - 1601), Johannes Kepler (1570 - 1630) developedthree empirical laws of planetary motion in a simplifiedplanet, star (much larger mass) system:Kepler 1: The planet follows a planar, elliptical orbit withthe star at one focus

Kepler 2: The radius vector from star to planet sweeps outequal areas in equal times as the planet orbitsKepler 3: The square of the period of orbit is proportionalto the cube of the semimajor axis of the ellipse: T 2 = ka3;see Wikipedia entry for Kepler’s laws of planetary motionIsaac Newton (1642-1727) basically developed his form ofdifferential calculus to show that Kepler’s laws follow fromthe inverse square law of gravitational attraction (and hisown law F = ma)


Some history

Building on astronomical observations by Tycho Brahe(1546 - 1601), Johannes Kepler (1570 - 1630) developedthree empirical laws of planetary motion in a simplifiedplanet, star (much larger mass) system:Kepler 1: The planet follows a planar, elliptical orbit withthe star at one focusKepler 2: The radius vector from star to planet sweeps outequal areas in equal times as the planet orbits

Kepler 3: The square of the period of orbit is proportionalto the cube of the semimajor axis of the ellipse: T 2 = ka3;see Wikipedia entry for Kepler’s laws of planetary motionIsaac Newton (1642-1727) basically developed his form ofdifferential calculus to show that Kepler’s laws follow fromthe inverse square law of gravitational attraction (and hisown law F = ma)


Some history

Building on astronomical observations by Tycho Brahe(1546 - 1601), Johannes Kepler (1570 - 1630) developedthree empirical laws of planetary motion in a simplifiedplanet, star (much larger mass) system:Kepler 1: The planet follows a planar, elliptical orbit withthe star at one focusKepler 2: The radius vector from star to planet sweeps outequal areas in equal times as the planet orbitsKepler 3: The square of the period of orbit is proportionalto the cube of the semimajor axis of the ellipse: T 2 = ka3;see Wikipedia entry for Kepler’s laws of planetary motion

Isaac Newton (1642-1727) basically developed his form ofdifferential calculus to show that Kepler’s laws follow fromthe inverse square law of gravitational attraction (and hisown law F = ma)


Some history

Building on astronomical observations by Tycho Brahe(1546 - 1601), Johannes Kepler (1570 - 1630) developedthree empirical laws of planetary motion in a simplifiedplanet, star (much larger mass) system:Kepler 1: The planet follows a planar, elliptical orbit withthe star at one focusKepler 2: The radius vector from star to planet sweeps outequal areas in equal times as the planet orbitsKepler 3: The square of the period of orbit is proportionalto the cube of the semimajor axis of the ellipse: T 2 = ka3;see Wikipedia entry for Kepler’s laws of planetary motionIsaac Newton (1642-1727) basically developed his form ofdifferential calculus to show that Kepler’s laws follow fromthe inverse square law of gravitational attraction (and hisown law F = ma)


The Newtonian n-body problem

Basic problem in celestial mechanics is to understand whathappens when we have any number n of masses mi , eachattracting all the others according to Newtonian law. Saythe masses are located at qi (functions of t).

Then F = ma for each mass leads to a system of 2ndorder differential equations like this

mid2qi

dt2 =∑j 6=i

Gmimj(qj − qi)

r3ij

It’s very difficult to understand all properties of thesolutions(!)



Basic problem in celestial mechanics is to understand whathappens when we have any number n of masses mi , eachattracting all the others according to Newtonian law. Saythe masses are located at qi (functions of t).Then F = ma for each mass leads to a system of 2ndorder differential equations like this

mid2qi

dt2 =∑j 6=i

Gmimj(qj − qi)

r3ij




Basic problem in celestial mechanics is to understand whathappens when we have any number n of masses mi , eachattracting all the others according to Newtonian law. Saythe masses are located at qi (functions of t).Then F = ma for each mass leads to a system of 2ndorder differential equations like this

mid2qi

dt2 =∑j 6=i

Gmimj(qj − qi)

r3ij



The Newtonian n-body problem, continued

For instance, even in the full 3-body problem, it is knownthat some solutions exhibit chaotic behavior

Can be “random-looking” even though completelydeterministicHave sensitive dependence on initial conditionsVirtually impossible to make long-term predictions in thosecases since we never know the initial conditions to arbitaryprecision in real-world measurements(!)



For instance, even in the full 3-body problem, it is knownthat some solutions exhibit chaotic behaviorCan be “random-looking” even though completelydeterministic

Have sensitive dependence on initial conditionsVirtually impossible to make long-term predictions in thosecases since we never know the initial conditions to arbitaryprecision in real-world measurements(!)



For instance, even in the full 3-body problem, it is knownthat some solutions exhibit chaotic behaviorCan be “random-looking” even though completelydeterministicHave sensitive dependence on initial conditions

Virtually impossible to make long-term predictions in thosecases since we never know the initial conditions to arbitaryprecision in real-world measurements(!)



For instance, even in the full 3-body problem, it is knownthat some solutions exhibit chaotic behaviorCan be “random-looking” even though completelydeterministicHave sensitive dependence on initial conditionsVirtually impossible to make long-term predictions in thosecases since we never know the initial conditions to arbitaryprecision in real-world measurements(!)


Central Configurations

We will focus mostly on a special class of solutions of then-body problem called central configurations

One way to define them: A central configuration (withcenter of mass at q) is a configuration such that theacceleration vector of each body is proportional to theposition vector from center of mass, all with the same(negative) proportionality constant −ω2

In other words, for all i , (setting G = 1)∑j 6=i

mj(qj − qi)

r3ij

+ ω2(qi − q) = 0



We will focus mostly on a special class of solutions of then-body problem called central configurationsOne way to define them: A central configuration (withcenter of mass at q) is a configuration such that theacceleration vector of each body is proportional to theposition vector from center of mass, all with the same(negative) proportionality constant −ω2


mj(qj − qi)

r3ij

+ ω2(qi − q) = 0



We will focus mostly on a special class of solutions of then-body problem called central configurationsOne way to define them: A central configuration (withcenter of mass at q) is a configuration such that theacceleration vector of each body is proportional to theposition vector from center of mass, all with the same(negative) proportionality constant −ω2


mj(qj − qi)

r3ij

+ ω2(qi − q) = 0


Comments, and first properties

Note: the central configuration equations are algebraicequations on the coordinates of the position vector (notdifferential equations)

The qi , qj are now thought of as constant vectors, notfunctions of tIf n masses are released from rest at positions satisfyingthe central configuration equations, they will collapsehomothetically to a complete collision in finite timeThis means that at all times the configuration will be thesame as the original, up to scalingPlanar central configurations are also relative equilibria –given the correct initial velocities, they can producesolutions of the n-body equations that undergo rigidrotation about the center of mass



Note: the central configuration equations are algebraicequations on the coordinates of the position vector (notdifferential equations)The qi , qj are now thought of as constant vectors, notfunctions of t

If n masses are released from rest at positions satisfyingthe central configuration equations, they will collapsehomothetically to a complete collision in finite timeThis means that at all times the configuration will be thesame as the original, up to scalingPlanar central configurations are also relative equilibria –given the correct initial velocities, they can producesolutions of the n-body equations that undergo rigidrotation about the center of mass



Note: the central configuration equations are algebraicequations on the coordinates of the position vector (notdifferential equations)The qi , qj are now thought of as constant vectors, notfunctions of tIf n masses are released from rest at positions satisfyingthe central configuration equations, they will collapsehomothetically to a complete collision in finite time

This means that at all times the configuration will be thesame as the original, up to scalingPlanar central configurations are also relative equilibria –given the correct initial velocities, they can producesolutions of the n-body equations that undergo rigidrotation about the center of mass



Note: the central configuration equations are algebraicequations on the coordinates of the position vector (notdifferential equations)The qi , qj are now thought of as constant vectors, notfunctions of tIf n masses are released from rest at positions satisfyingthe central configuration equations, they will collapsehomothetically to a complete collision in finite timeThis means that at all times the configuration will be thesame as the original, up to scaling

Planar central configurations are also relative equilibria –given the correct initial velocities, they can producesolutions of the n-body equations that undergo rigidrotation about the center of mass



Note: the central configuration equations are algebraicequations on the coordinates of the position vector (notdifferential equations)The qi , qj are now thought of as constant vectors, notfunctions of tIf n masses are released from rest at positions satisfyingthe central configuration equations, they will collapsehomothetically to a complete collision in finite timeThis means that at all times the configuration will be thesame as the original, up to scalingPlanar central configurations are also relative equilibria –given the correct initial velocities, they can producesolutions of the n-body equations that undergo rigidrotation about the center of mass


An alternative definition

Let

U =∑

1≤j<k≤n

Gmjmk

rjk

be the gravitational potential energy, and

Let

I =12

n∑j=1

mj‖qj‖2

be the moment of inertiaThen the bodies form a central configuration if

∇U + ω2∇I = 0

Lagrange multiplier method ⇒ c.c.’s are critical points of Usubject to I = const



Let

U =∑

1≤j<k≤n

Gmjmk

rjk

be the gravitational potential energy, andLet

I =12

n∑j=1

mj‖qj‖2

be the moment of inertia

Then the bodies form a central configuration if

∇U + ω2∇I = 0




Let

U =∑

1≤j<k≤n

Gmjmk

rjk


I =12

n∑j=1

mj‖qj‖2


∇U + ω2∇I = 0




Let

U =∑

1≤j<k≤n

Gmjmk

rjk


I =12

n∑j=1

mj‖qj‖2


∇U + ω2∇I = 0



Practical applications

Central configurations in the 3-body problem

At each given time a planet-sun (e.g. Earth-Sun) systemhas 5 “Lagrange points” which are the positions where athird mass could be placed to form a central configuration:

(image credit: NASA Wilkinson Microwave AniosotropyProbe project)


Practical applications

Central configurations in the 3-body problemAt each given time a planet-sun (e.g. Earth-Sun) systemhas 5 “Lagrange points” which are the positions where athird mass could be placed to form a central configuration:

(image credit: NASA Wilkinson Microwave AniosotropyProbe project)


Meaning of Lagrange points

A satellite placed at rest at one of these Lagrange pointswill stay in a rotating orbit about the center of mass of theplanet-sun system (relative equilibrium property) due togravitational attraction exerted on it by the planet and sun

Same phenomenon has been observed for many naturalobjects, tooFor instance – each cloud of “Trojan” asteroids has centerof mass approx. at a Lagrange point of Sun and JupiterYou will study the Lagrange points in this week’s labassignment (via consideration of central configurations forthe 3-body problem in general)More generally, rings of Saturn are a central configurationwith a large number of smaller masses orbiting the largemass of Saturn



A satellite placed at rest at one of these Lagrange pointswill stay in a rotating orbit about the center of mass of theplanet-sun system (relative equilibrium property) due togravitational attraction exerted on it by the planet and sunSame phenomenon has been observed for many naturalobjects, too

For instance – each cloud of “Trojan” asteroids has centerof mass approx. at a Lagrange point of Sun and JupiterYou will study the Lagrange points in this week’s labassignment (via consideration of central configurations forthe 3-body problem in general)More generally, rings of Saturn are a central configurationwith a large number of smaller masses orbiting the largemass of Saturn



A satellite placed at rest at one of these Lagrange pointswill stay in a rotating orbit about the center of mass of theplanet-sun system (relative equilibrium property) due togravitational attraction exerted on it by the planet and sunSame phenomenon has been observed for many naturalobjects, tooFor instance – each cloud of “Trojan” asteroids has centerof mass approx. at a Lagrange point of Sun and Jupiter

You will study the Lagrange points in this week’s labassignment (via consideration of central configurations forthe 3-body problem in general)More generally, rings of Saturn are a central configurationwith a large number of smaller masses orbiting the largemass of Saturn



A satellite placed at rest at one of these Lagrange pointswill stay in a rotating orbit about the center of mass of theplanet-sun system (relative equilibrium property) due togravitational attraction exerted on it by the planet and sunSame phenomenon has been observed for many naturalobjects, tooFor instance – each cloud of “Trojan” asteroids has centerof mass approx. at a Lagrange point of Sun and JupiterYou will study the Lagrange points in this week’s labassignment (via consideration of central configurations forthe 3-body problem in general)

More generally, rings of Saturn are a central configurationwith a large number of smaller masses orbiting the largemass of Saturn



A satellite placed at rest at one of these Lagrange pointswill stay in a rotating orbit about the center of mass of theplanet-sun system (relative equilibrium property) due togravitational attraction exerted on it by the planet and sunSame phenomenon has been observed for many naturalobjects, tooFor instance – each cloud of “Trojan” asteroids has centerof mass approx. at a Lagrange point of Sun and JupiterYou will study the Lagrange points in this week’s labassignment (via consideration of central configurations forthe 3-body problem in general)More generally, rings of Saturn are a central configurationwith a large number of smaller masses orbiting the largemass of Saturn


Saturn and its rings


The big problem

Major question about central configurations is

Given n masses m1, . . . , mn, at how many differentlocations can these be placed to get centralconfigurations? In particular, is the set of possibleconfigurations finite? On Smale’s 21st century problem list.To make this meaningful, need some further explanationsor definitions, since we can translate, rotate, and scalecentral configurations and the results are again centralconfigurationsConvention: We will consider two central configurations tobe equivalent if there exists a mapping of Rk taking oneinto the other that can be obtained as a composition of arigid motion (translation, rotation) and a scaling


The big problem

Major question about central configurations isGiven n masses m1, . . . , mn, at how many differentlocations can these be placed to get centralconfigurations? In particular, is the set of possibleconfigurations finite? On Smale’s 21st century problem list.

To make this meaningful, need some further explanationsor definitions, since we can translate, rotate, and scalecentral configurations and the results are again centralconfigurationsConvention: We will consider two central configurations tobe equivalent if there exists a mapping of Rk taking oneinto the other that can be obtained as a composition of arigid motion (translation, rotation) and a scaling


The big problem

Major question about central configurations isGiven n masses m1, . . . , mn, at how many differentlocations can these be placed to get centralconfigurations? In particular, is the set of possibleconfigurations finite? On Smale’s 21st century problem list.To make this meaningful, need some further explanationsor definitions, since we can translate, rotate, and scalecentral configurations and the results are again centralconfigurations

Convention: We will consider two central configurations tobe equivalent if there exists a mapping of Rk taking oneinto the other that can be obtained as a composition of arigid motion (translation, rotation) and a scaling


The big problem

Major question about central configurations isGiven n masses m1, . . . , mn, at how many differentlocations can these be placed to get centralconfigurations? In particular, is the set of possibleconfigurations finite? On Smale’s 21st century problem list.To make this meaningful, need some further explanationsor definitions, since we can translate, rotate, and scalecentral configurations and the results are again centralconfigurationsConvention: We will consider two central configurations tobe equivalent if there exists a mapping of Rk taking oneinto the other that can be obtained as a composition of arigid motion (translation, rotation) and a scaling


Status of the “big problem”

Answer is known to be yes for n = 3, 4

Some fairly strong results for n = 5Only fairly limited special cases known in generalExtremely subtle problem in general(!)For instance, it’s known that there are collections of n = 5“masses,” including one negative value, for which there is awhole curve of central configurations (not just a finitenumber, even taking rigid motions and scaling into account)



Answer is known to be yes for n = 3, 4Some fairly strong results for n = 5

Only fairly limited special cases known in generalExtremely subtle problem in general(!)For instance, it’s known that there are collections of n = 5“masses,” including one negative value, for which there is awhole curve of central configurations (not just a finitenumber, even taking rigid motions and scaling into account)



Answer is known to be yes for n = 3, 4Some fairly strong results for n = 5Only fairly limited special cases known in general

Extremely subtle problem in general(!)For instance, it’s known that there are collections of n = 5“masses,” including one negative value, for which there is awhole curve of central configurations (not just a finitenumber, even taking rigid motions and scaling into account)



Answer is known to be yes for n = 3, 4Some fairly strong results for n = 5Only fairly limited special cases known in generalExtremely subtle problem in general(!)

For instance, it’s known that there are collections of n = 5“masses,” including one negative value, for which there is awhole curve of central configurations (not just a finitenumber, even taking rigid motions and scaling into account)



Answer is known to be yes for n = 3, 4Some fairly strong results for n = 5Only fairly limited special cases known in generalExtremely subtle problem in general(!)For instance, it’s known that there are collections of n = 5“masses,” including one negative value, for which there is awhole curve of central configurations (not just a finitenumber, even taking rigid motions and scaling into account)


Mutual distances as coordinates

Because of the way we want to consider equivalence, itmakes sense to try to set up the problem so that:

A fixed distance scale is used, andThe mutual distances rij become the coordinates fordescribing the configurationThis leads to some subtleties that we will need tounderstand, thoughFor instance, what are all triples (r12, r13, r23) that canrepresent the mutual distances of three points in R2?



Because of the way we want to consider equivalence, itmakes sense to try to set up the problem so that:A fixed distance scale is used, and

The mutual distances rij become the coordinates fordescribing the configurationThis leads to some subtleties that we will need tounderstand, thoughFor instance, what are all triples (r12, r13, r23) that canrepresent the mutual distances of three points in R2?



Because of the way we want to consider equivalence, itmakes sense to try to set up the problem so that:A fixed distance scale is used, andThe mutual distances rij become the coordinates fordescribing the configuration

This leads to some subtleties that we will need tounderstand, thoughFor instance, what are all triples (r12, r13, r23) that canrepresent the mutual distances of three points in R2?



Because of the way we want to consider equivalence, itmakes sense to try to set up the problem so that:A fixed distance scale is used, andThe mutual distances rij become the coordinates fordescribing the configurationThis leads to some subtleties that we will need tounderstand, though

For instance, what are all triples (r12, r13, r23) that canrepresent the mutual distances of three points in R2?



Because of the way we want to consider equivalence, itmakes sense to try to set up the problem so that:A fixed distance scale is used, andThe mutual distances rij become the coordinates fordescribing the configurationThis leads to some subtleties that we will need tounderstand, thoughFor instance, what are all triples (r12, r13, r23) that canrepresent the mutual distances of three points in R2?


Some obvious restrictions

All rij ∈ R (they’re distances!) and any nonreal solutions ofthe c.c. equations will not be meaningful for the physicalproblem

All rij ≥ 0 and = 0 only in a partial or total collision caseBy triangle inequalities, we also have r12 + r23 − r13 ≥ 0,r12 + r13 − r23 ≥ 0, and r13 + r23 − r12 ≥ 0 with equality inone if and only if the three points are collinearWhich one gives zero depends on which point is betweenthe other two along the lineThe set of all points in R3 satisfying these inequalities is atriangular cone C with vertex at the origin



All rij ∈ R (they’re distances!) and any nonreal solutions ofthe c.c. equations will not be meaningful for the physicalproblemAll rij ≥ 0 and = 0 only in a partial or total collision case

By triangle inequalities, we also have r12 + r23 − r13 ≥ 0,r12 + r13 − r23 ≥ 0, and r13 + r23 − r12 ≥ 0 with equality inone if and only if the three points are collinearWhich one gives zero depends on which point is betweenthe other two along the lineThe set of all points in R3 satisfying these inequalities is atriangular cone C with vertex at the origin



All rij ∈ R (they’re distances!) and any nonreal solutions ofthe c.c. equations will not be meaningful for the physicalproblemAll rij ≥ 0 and = 0 only in a partial or total collision caseBy triangle inequalities, we also have r12 + r23 − r13 ≥ 0,r12 + r13 − r23 ≥ 0, and r13 + r23 − r12 ≥ 0 with equality inone if and only if the three points are collinear

Which one gives zero depends on which point is betweenthe other two along the lineThe set of all points in R3 satisfying these inequalities is atriangular cone C with vertex at the origin



All rij ∈ R (they’re distances!) and any nonreal solutions ofthe c.c. equations will not be meaningful for the physicalproblemAll rij ≥ 0 and = 0 only in a partial or total collision caseBy triangle inequalities, we also have r12 + r23 − r13 ≥ 0,r12 + r13 − r23 ≥ 0, and r13 + r23 − r12 ≥ 0 with equality inone if and only if the three points are collinearWhich one gives zero depends on which point is betweenthe other two along the line

The set of all points in R3 satisfying these inequalities is atriangular cone C with vertex at the origin



All rij ∈ R (they’re distances!) and any nonreal solutions ofthe c.c. equations will not be meaningful for the physicalproblemAll rij ≥ 0 and = 0 only in a partial or total collision caseBy triangle inequalities, we also have r12 + r23 − r13 ≥ 0,r12 + r13 − r23 ≥ 0, and r13 + r23 − r12 ≥ 0 with equality inone if and only if the three points are collinearWhich one gives zero depends on which point is betweenthe other two along the lineThe set of all points in R3 satisfying these inequalities is atriangular cone C with vertex at the origin


The configuration space of triangles

We claim that every point in C represents a “triangle”(including “degenerate” cases in which the three points arecollinear, and two of them coincide)

For instance, consider the cases where r13 ≥ r12.Then we can place point 1 at the origin, point two at (r12, 0)

Then point 3 can be any point on the circle x2 + y2 = r213

Can see all r23 with

r13 − r12 ≤ r23 ≤ r13 + r12

are attained (as in inequalities describing C)



We claim that every point in C represents a “triangle”(including “degenerate” cases in which the three points arecollinear, and two of them coincide)For instance, consider the cases where r13 ≥ r12.

Then we can place point 1 at the origin, point two at (r12, 0)



r13 − r12 ≤ r23 ≤ r13 + r12




We claim that every point in C represents a “triangle”(including “degenerate” cases in which the three points arecollinear, and two of them coincide)For instance, consider the cases where r13 ≥ r12.Then we can place point 1 at the origin, point two at (r12, 0)



r13 − r12 ≤ r23 ≤ r13 + r12







r13 − r12 ≤ r23 ≤ r13 + r12







r13 − r12 ≤ r23 ≤ r13 + r12



Dimensions

Note that n points in Rk span at most an n − 1-dimensionalspace.

For instance, three points always lie on a plane, and maybe collinear. If they are not collinear, then they form thevertices of a triangle with nonzero area.Four points always lie in a 3-dimensional space and maybe coplanar or collinear, etc. If they are not coplanar, thenthey form the vertices of a tetrahedron with nonzerovolume.


Dimensions

Note that n points in Rk span at most an n − 1-dimensionalspace.For instance, three points always lie on a plane, and maybe collinear. If they are not collinear, then they form thevertices of a triangle with nonzero area.

Four points always lie in a 3-dimensional space and maybe coplanar or collinear, etc. If they are not coplanar, thenthey form the vertices of a tetrahedron with nonzerovolume.


Dimensions

Note that n points in Rk span at most an n − 1-dimensionalspace.For instance, three points always lie on a plane, and maybe collinear. If they are not collinear, then they form thevertices of a triangle with nonzero area.Four points always lie in a 3-dimensional space and maybe coplanar or collinear, etc. If they are not coplanar, thenthey form the vertices of a tetrahedron with nonzerovolume.


Cayley-Menger determinants

In working with mutual distance coordinates there arecertain expressions that come up repeatedly and havesignificant geometric meaning.

For instance, consider configurations of n = 3 points. Thenit can be shown (by you in today’s discussion, for instance:) that

∆CM = det

0 1 1 11 0 r2

12 r213

1 r212 0 r2

231 r2

13 r223 0

is a constant multiple of the square of the area of thecorresponding triangle if (r12, r13, r23) ∈ C, and is zero if thepoints are collinear. So det ∆CM ≥ 0.


Cayley-Menger determinants

In working with mutual distance coordinates there arecertain expressions that come up repeatedly and havesignificant geometric meaning.For instance, consider configurations of n = 3 points. Thenit can be shown (by you in today’s discussion, for instance:) that

∆CM = det

0 1 1 11 0 r2

12 r213

1 r212 0 r2

231 r2

13 r223 0

is a constant multiple of the square of the area of thecorresponding triangle if (r12, r13, r23) ∈ C, and is zero if thepoints are collinear. So det ∆CM ≥ 0.


Cayley-Menger determinants, continued

In general, for a configuration of n points, the CM determinantis the (n + 1)× (n + 1) determinant

∆CM = det

0 1 1 · · · 1 11 0 r2

12 · · · r21,n−1 r2

1n1 r2

12 0 · · · r22,n−1 r2

2n...

......

. . ....

...1 r2

1,n−1 r22,n−1 · · · 0 r2

n−1,n1 r2

1n r22n · · · r2

n−1,n 0

(Note: the matrix is symmetric.)


Configuration spaces in general

The mutual distance description for configurations of n ≥ 4points is more subtle

For instance with n = 4, we have(4

2

)= 6 mutual distances

Can ask which

(r12, r13, r14, r23, r24, r34) ∈ R6+

can be realized as mutual distances for configurations ofn = 4 points.Must have rij ≤ rik + rkj for all distinct triples{i , j , k} ⊂ {1, 2, 3, 4}, by triangle inequalityMust also have det ∆CM ≥ 0 (this is automatic in n = 3case, but not here)



The mutual distance description for configurations of n ≥ 4points is more subtleFor instance with n = 4, we have

(42


Can ask which

(r12, r13, r14, r23, r24, r34) ∈ R6+





(42


Can ask which

(r12, r13, r14, r23, r24, r34) ∈ R6+

can be realized as mutual distances for configurations ofn = 4 points.

Must have rij ≤ rik + rkj for all distinct triples{i , j , k} ⊂ {1, 2, 3, 4}, by triangle inequalityMust also have det ∆CM ≥ 0 (this is automatic in n = 3case, but not here)




(42


Can ask which

(r12, r13, r14, r23, r24, r34) ∈ R6+

can be realized as mutual distances for configurations ofn = 4 points.Must have rij ≤ rik + rkj for all distinct triples{i , j , k} ⊂ {1, 2, 3, 4}, by triangle inequality

Must also have det ∆CM ≥ 0 (this is automatic in n = 3case, but not here)




(42


Can ask which

(r12, r13, r14, r23, r24, r34) ∈ R6+



Configuration spaces, continued

Example: Let

(r12, r13, r14, r23, r24, r34) =

(1, t , 1, 1,

2t, 1

)

There is an interval of t values for which all the triangleinequalities are satisfied, namely 1 < t < 2, but

det ∆CM = −8(t2−2)t2 < 0 unless t =

√2, which gives the unit

square.In general a collection of distances rij ∈ R+ can be realizedby a configuration if and only if all the triangle inequalitieshold, and det ∆CM ≥ 0.



Example: Let

(r12, r13, r14, r23, r24, r34) =

(1, t , 1, 1,

2t, 1

)There is an interval of t values for which all the triangleinequalities are satisfied, namely 1 < t < 2, but






Example: Let

(r12, r13, r14, r23, r24, r34) =

(1, t , 1, 1,

2t, 1




square.

In general a collection of distances rij ∈ R+ can be realizedby a configuration if and only if all the triangle inequalitieshold, and det ∆CM ≥ 0.



Example: Let

(r12, r13, r14, r23, r24, r34) =

(1, t , 1, 1,

2t, 1






The Albouy-Chenciner equations

In mutual distance coordinates, the equations for centralconfigurations can be written in the following form(asymmetric Albouy-Chenciner):

For each pair 1 ≤ i , j ≤ n:

Gij =n∑

k=1

mkSki(r2jk − r2

ik − r2ij ) = 0

where

Ski =

{r−3ki − 1 if k 6= i

0 if k = i

Note: r`m = rm` for all `, m. Ski = Sik .) One nontrivialequation for each pair 1 ≤ i , j ≤ n with i 6= j (everythingcancels out if i = j).



In mutual distance coordinates, the equations for centralconfigurations can be written in the following form(asymmetric Albouy-Chenciner):For each pair 1 ≤ i , j ≤ n:

Gij =n∑

k=1

mkSki(r2jk − r2

ik − r2ij ) = 0

where

Ski =

{r−3ki − 1 if k 6= i

0 if k = i




In mutual distance coordinates, the equations for centralconfigurations can be written in the following form(asymmetric Albouy-Chenciner):For each pair 1 ≤ i , j ≤ n:

Gij =n∑

k=1

mkSki(r2jk − r2

ik − r2ij ) = 0

where

Ski =

{r−3ki − 1 if k 6= i

0 if k = i



A comment about AC equations

The form of the AC equations we are using here comes bysetting ω2 = 1 in the definition

This has the effect of imposing a particular distance scaleso thatWe never get two central configurations that arehomothetic.In particular, for instance if three masses m1, m2, m3 atvertices of a triangle form a central configuration, thenthere are no similar triangles that will also give solutions ofthe AC equations.OK for our purposes!



The form of the AC equations we are using here comes bysetting ω2 = 1 in the definitionThis has the effect of imposing a particular distance scaleso that

We never get two central configurations that arehomothetic.In particular, for instance if three masses m1, m2, m3 atvertices of a triangle form a central configuration, thenthere are no similar triangles that will also give solutions ofthe AC equations.OK for our purposes!



The form of the AC equations we are using here comes bysetting ω2 = 1 in the definitionThis has the effect of imposing a particular distance scaleso thatWe never get two central configurations that arehomothetic.

In particular, for instance if three masses m1, m2, m3 atvertices of a triangle form a central configuration, thenthere are no similar triangles that will also give solutions ofthe AC equations.OK for our purposes!



The form of the AC equations we are using here comes bysetting ω2 = 1 in the definitionThis has the effect of imposing a particular distance scaleso thatWe never get two central configurations that arehomothetic.In particular, for instance if three masses m1, m2, m3 atvertices of a triangle form a central configuration, thenthere are no similar triangles that will also give solutions ofthe AC equations.

OK for our purposes!



The form of the AC equations we are using here comes bysetting ω2 = 1 in the definitionThis has the effect of imposing a particular distance scaleso thatWe never get two central configurations that arehomothetic.In particular, for instance if three masses m1, m2, m3 atvertices of a triangle form a central configuration, thenthere are no similar triangles that will also give solutions ofthe AC equations.OK for our purposes!


What the AC equations look like

Let’s see what the AC equations look like for 4 masses

The AC equation for i , j = 1, 2 is

−2m2S12r212+m3S13(r2

23−r213−r2

12)+m4S14(r224−r2

14−r212) = 0

where Sjk = 1r3jk− 1 as before

Note: can clear denominators to get a polynomial



Let’s see what the AC equations look like for 4 massesThe AC equation for i , j = 1, 2 is

−2m2S12r212+m3S13(r2

23−r213−r2

12)+m4S14(r224−r2

14−r212) = 0





Let’s see what the AC equations look like for 4 massesThe AC equation for i , j = 1, 2 is

−2m2S12r212+m3S13(r2

23−r213−r2

12)+m4S14(r224−r2

14−r212) = 0




An example

Say we want to study collinear cc’s of n = 4 bodies. Useoriginal form of cc equations (not AC).

r t q ua b c

m1 m2 m3 m4

Fix the first mass at q1 = (0, 0), then place the others atq2 = (a, 0), q3 = (a + b, 0), q4 = (a + b + c, 0) (witha, b, c > 0).This gives r12 = a, r13 = a + b, r14 = a + b + c, r23 = b,r24 = b + c, and r34 = c.The center of mass is at

q =

∑i miqi∑

i mi


An example


r t q ua b c

m1 m2 m3 m4


q =

∑i miqi∑

i mi


An example


r t q ua b c

m1 m2 m3 m4

Fix the first mass at q1 = (0, 0), then place the others atq2 = (a, 0), q3 = (a + b, 0), q4 = (a + b + c, 0) (witha, b, c > 0).

This gives r12 = a, r13 = a + b, r14 = a + b + c, r23 = b,r24 = b + c, and r34 = c.The center of mass is at

q =

∑i miqi∑

i mi


An example


r t q ua b c

m1 m2 m3 m4

Fix the first mass at q1 = (0, 0), then place the others atq2 = (a, 0), q3 = (a + b, 0), q4 = (a + b + c, 0) (witha, b, c > 0).This gives r12 = a, r13 = a + b, r14 = a + b + c, r23 = b,r24 = b + c, and r34 = c.

The center of mass is at

q =

∑i miqi∑

i mi


An example


r t q ua b c

m1 m2 m3 m4


q =

∑i miqi∑

i mi


Example, continued

For instance, if the masses are all equal (so we can setmi = 1, all i), and we make ω2 = 1 (this fixes the distancescale) the central configuration equations become:

0 = a−2 + (a + b)−2 + (a + b + c)−2

−3/4 a − 1/2 b − 1/4 c0 = −a−2 + b−2 + (b + c)−2

+1/4 a − 1/2 b − 1/4 c0 = −(a + b)−2 − b−2 + c−2

+1/4 a + 1/2 b − 1/4 c0 = −(a + b + c)−2 − (b + c)−2 − c−2

+1/4 a + 1/2 b + 3/4 c


Example, continued

For instance, if the masses are all equal (so we can setmi = 1, all i), and we make ω2 = 1 (this fixes the distancescale) the central configuration equations become:

0 = a−2 + (a + b)−2 + (a + b + c)−2

−3/4 a − 1/2 b − 1/4 c0 = −a−2 + b−2 + (b + c)−2

+1/4 a − 1/2 b − 1/4 c0 = −(a + b)−2 − b−2 + c−2

+1/4 a + 1/2 b − 1/4 c0 = −(a + b + c)−2 − (b + c)−2 − c−2

+1/4 a + 1/2 b + 3/4 c


Example, continued

Clear denominators to get polynomial equations

When we do that, get factors of a + b, b + c, a + b + c insome polynomials; since we want a, b, c > 0, those factorscannot be zero either(“Trick”) to find only solutions witha, b, c, a + b, b + c, a + b + c 6= 0 we can add a variable tand a new equation

1− tabc(a + b)(b + c)(a + b + c) = 0

Compute a Gröbner basis for the ideal generated by thesewith respect to a monomial order set up to eliminate t ,break ties with grevlex. [Demo with Sage]


Example, continued

Clear denominators to get polynomial equationsWhen we do that, get factors of a + b, b + c, a + b + c insome polynomials; since we want a, b, c > 0, those factorscannot be zero either

(“Trick”) to find only solutions witha, b, c, a + b, b + c, a + b + c 6= 0 we can add a variable tand a new equation

1− tabc(a + b)(b + c)(a + b + c) = 0



Example, continued

Clear denominators to get polynomial equationsWhen we do that, get factors of a + b, b + c, a + b + c insome polynomials; since we want a, b, c > 0, those factorscannot be zero either(“Trick”) to find only solutions witha, b, c, a + b, b + c, a + b + c 6= 0 we can add a variable tand a new equation

1− tabc(a + b)(b + c)(a + b + c) = 0



Example, continued

Clear denominators to get polynomial equationsWhen we do that, get factors of a + b, b + c, a + b + c insome polynomials; since we want a, b, c > 0, those factorscannot be zero either(“Trick”) to find only solutions witha, b, c, a + b, b + c, a + b + c 6= 0 we can add a variable tand a new equation

1− tabc(a + b)(b + c)(a + b + c) = 0



How many solutions?

Note that the leading terms of the Gröbner basispolynomials contain powers of each of the variables:t , a5, b8, c10

This implies that the set of all complex solutions of thesystem is finite (hence the same is true for the realsolutions!)

Theorem 1 (Th. 6 in Chapter 5, §3 of IVA)

Let I be an ideal in C[x1, . . . , xn] and > a monomial order. ThenTFAE:

i. V (I) is a finite set

ii. Let G be a GB of I wrt >. For each i, there is gi ∈ G withLT (gi) = xmi

i for some mi


How many solutions?







i for some mi


How many solutions?







i for some mi


Sketch of proof of this theorem

Proof.ii ⇒ i: Think about the pictures we drew in the proof ofDickson’s Lemma. If ii is true, there are only finitely manymonomials that are not in 〈LT (I)〉. Hence for each i , theremainders of the powers xn

i , n ≥ 0 on division by G mustbe linearly dependent. This implies that there is somehi(xi) ∈ I for each i . It follows that V (I) is finite.

i ⇒ ii: If V is a finite set, then for each i there is aunivariate polynomial hi(xi) ∈ I(V (I)). By a big theoremfrom Chapter 4 of IVA (the “Nullstellensatz”), hi ∈

√I so

some power h`i ∈ I. The leading term of h`

i is a power x`kii .

By the definition of a GB, there must be some gi ∈ Gwhose leading term divides x`ki

i , so LT (gi) is a power of xi .


Sketch of proof of this theorem

Proof.ii ⇒ i: Think about the pictures we drew in the proof ofDickson’s Lemma. If ii is true, there are only finitely manymonomials that are not in 〈LT (I)〉. Hence for each i , theremainders of the powers xn

i , n ≥ 0 on division by G mustbe linearly dependent. This implies that there is somehi(xi) ∈ I for each i . It follows that V (I) is finite.i ⇒ ii: If V is a finite set, then for each i there is aunivariate polynomial hi(xi) ∈ I(V (I)). By a big theoremfrom Chapter 4 of IVA (the “Nullstellensatz”), hi ∈

√I so

some power h`i ∈ I. The leading term of h`

i is a power x`kii .

By the definition of a GB, there must be some gi ∈ Gwhose leading term divides x`ki

i , so LT (gi) is a power of xi .


Back to collinear 4-body cc’s

In our case, the ideal contains a univariate polynomial ofdegree 105 in a.

This factors; the smaller factor has this form (all terms haveexponent divisible by 3 !)

16a21 − 160a18 + · · ·+ 6856a9 − 8880a6 + 5936a3 − 1568

The method of Sturm sequences (see this week’s lab) canbe used to show this has exactly one real root, which ispositive.Note that it must have at least one real root since it hasodd degree.The other factor has even degree and no real roots.



In our case, the ideal contains a univariate polynomial ofdegree 105 in a.This factors; the smaller factor has this form (all terms haveexponent divisible by 3 !)

16a21 − 160a18 + · · ·+ 6856a9 − 8880a6 + 5936a3 − 1568





16a21 − 160a18 + · · ·+ 6856a9 − 8880a6 + 5936a3 − 1568

The method of Sturm sequences (see this week’s lab) canbe used to show this has exactly one real root, which ispositive.

Note that it must have at least one real root since it hasodd degree.The other factor has even degree and no real roots.




16a21 − 160a18 + · · ·+ 6856a9 − 8880a6 + 5936a3 − 1568

The method of Sturm sequences (see this week’s lab) canbe used to show this has exactly one real root, which ispositive.Note that it must have at least one real root since it hasodd degree.

The other factor has even degree and no real roots.




16a21 − 160a18 + · · ·+ 6856a9 − 8880a6 + 5936a3 − 1568



Collinear 4-body cc’s, continued

The ideal also contains univariate polynomials in b, c

These are very similar to the one for a. (In fact, thepolynomial for c is exactly the same as the polynomial fora; can you see why this might be true?)It follows with a bit of computation that there is exactly onereal solution of the cc equations in this case,In fact it has been proved (by other methods!) that for allchoices of mi > 0, there is exactly one collinear cc of theform we are studying (Moulton’s theorem).



The ideal also contains univariate polynomials in b, cThese are very similar to the one for a. (In fact, thepolynomial for c is exactly the same as the polynomial fora; can you see why this might be true?)

It follows with a bit of computation that there is exactly onereal solution of the cc equations in this case,In fact it has been proved (by other methods!) that for allchoices of mi > 0, there is exactly one collinear cc of theform we are studying (Moulton’s theorem).



The ideal also contains univariate polynomials in b, cThese are very similar to the one for a. (In fact, thepolynomial for c is exactly the same as the polynomial fora; can you see why this might be true?)It follows with a bit of computation that there is exactly onereal solution of the cc equations in this case,

In fact it has been proved (by other methods!) that for allchoices of mi > 0, there is exactly one collinear cc of theform we are studying (Moulton’s theorem).



The ideal also contains univariate polynomials in b, cThese are very similar to the one for a. (In fact, thepolynomial for c is exactly the same as the polynomial fora; can you see why this might be true?)It follows with a bit of computation that there is exactly onereal solution of the cc equations in this case,In fact it has been proved (by other methods!) that for allchoices of mi > 0, there is exactly one collinear cc of theform we are studying (Moulton’s theorem).


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