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\title[PURE Math 2012 Residents' Program Week 3]
{PURE Math Residents' Program\\
Gr\"obner Bases and Applications\\
Week 3 Lectures}

\author
{John B. Little}

\institute[College of the Holy Cross]
{
  Department of Mathematics and Computer Science\\
  College of the Holy Cross\\
  $\ $\\
}

\date[June 2012]
{June 2012}

\begin{document}

\begin{frame}
  \titlepage
\end{frame}

\section{Day 1}

\begin{frame}
\frametitle{Overview of this week}
\begin{itemize}
\item<1-> The research project topics you will be working on 
will apply the algebra and geometry we have seen (e.g. Gr\"obner 
bases, varieties, etc.) to some interesting questions 
in \emph{celestial mechanics} (and possibly related question in
fluid mechanics
\item<2-> In this week's activities, we'll develop some 
background and look at basic cases to provide context
\item<3-> Fortunately, \emph{not} a lot of physics is 
\emph{required}, but if you \emph{have} seen some, that will help 
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Newtonian gravitation}
\begin{itemize}
\item<1-> Everyone should know about Newton's gravitational law
\item<2-> Says (in basic form) that \emph{magnitude} of gravitational force
exerted by one mass $m_1$ on another $m_2$ is proportional to the product
$m_1m_2$, and inversely proportional to the square of the distance:
\item<3->  $|F| = \frac{G m_1 m_2}{r^2}$
\item<4->  Suppose the masses are located at $q_1, q_2 \in \mathbb{R}^3$.
\item<5->  The gravitational force exerted by $m_1$ on $m_2$ is a vector
directed from $q_2$ back to $q_1$:
$$F = \frac{ - G m_1 m_2 (q_2 - q_1)}{r^3}$$
where $r = \|q_2 - q_1\|$ is the distance (note cancelation in magnitude formula!)
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Some history}
\begin{itemize}
\item<1-> Building on astronomical observations by Tycho Brahe (1546 - 1601), 
Johannes Kepler (1570 - 1630) developed three \emph{empirical laws} of 
planetary motion in a simplified planet, star (much larger mass) system:
\item<2->  \emph{Kepler 1}: The planet follows a planar, \emph{elliptical} orbit 
with the star at one focus
\item<3->  \emph{Kepler 2}: The radius vector from star to planet sweeps out equal areas in equal 
times as the planet orbits
\item<4->  \emph{Kepler 3}: The square of the period of orbit is proportional to the cube of 
the semimajor axis of the ellipse: $T^2 = k a^3$; see 
Wikipedia entry for Kepler's laws of planetary motion
\item<5->  Isaac Newton (1642-1727) basically developed his form of 
differential calculus to show that Kepler's laws follow from the 
inverse square law of gravitational attraction (and his own law $F = ma$) 
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{The Newtonian $n$-body problem}
\begin{itemize}
\item<1->  Basic problem in celestial mechanics is to understand what happens
when we have any number $n$ of masses $m_i$, each attracting all the others
according to Newtonian law.  Say the masses are located at $q_i$ (functions of $t$).
\item<2->  Then $F = ma$ for each mass leads to a \emph{system of 2nd order 
differential equations like this}
$$ m_i \frac{d^2 q_i}{dt^2} = \sum_{j \ne i} \frac{G m_i m_j (q_j - q_i)}{r_{ij}^3}$$
\item<3-> It's \emph{very difficult} to understand all properties of the 
solutions(!)
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{The Newtonian $n$-body problem, continued}
\begin{itemize}
\item<1->  For instance, even in the full 3-body problem, 
it is known that some solutions exhibit \emph{chaotic} behavior
\item<2->  Can be ``random-looking'' even though completely 
deterministic
\item<3->  Have \emph{sensitive dependence on initial 
conditions}
\item<4->  Virtually impossible to make long-term predictions in those
cases since we never know the initial conditions to 
arbitary precision in real-world measurements(!)
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Central Configurations}
\begin{itemize}
\item<1-> We will focus mostly on a special class of solutions of the
$n$-body problem called \emph{central configurations}
\item<2-> One way to define them:  A central configuration (with center
of mass at $\overline{q}$) is a configuration such that the acceleration 
vector of each body is proportional to the position vector from center of mass,
 all with the same (negative) proportionality constant $-\omega^2$
\item<3->  In other words, for all $i$, (setting $G = 1$)
$$\sum_{j \ne i} \frac{m_j (q_j - q_i)}{r_{ij}^3} + \omega^2 (q_i - \overline{q}) = 0$$
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Comments, and first properties}
\begin{itemize}
\item<1->  Note: the central configuration equations are \emph{algebraic}
equations on the coordinates of the position vector (\emph{not} differential equations)
\item<2->  The $q_i,q_j$ are now thought of as constant vectors, not functions 
of $t$ 
\item<3->  If $n$ masses are released from rest at positions satisfying the 
central configuration equations, they will collapse \emph{homothetically} to 
a complete collision in finite time
\item<4->  This means that at all times the configuration will be the
same as the original, up to scaling
\item<5->  \emph{Planar} central configurations are also \emph{relative equilibria} -- 
given the correct initial velocities, they can produce solutions of the $n$-body 
equations that undergo rigid rotation about the center of mass
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{An alternative definition}
\begin{itemize}
\item<1->  Let 
$$U = \sum_{1\le j < k \le n} \frac{G m_j m_k}{r_{jk}}$$
be the gravitational \emph{potential energy}, and
\item<2->  Let 
$$I = \frac{1}{2} \sum_{j=1}^n m_j \|q_j\|^2$$
be the \emph{moment of inertia}
\item<3->  Then the bodies form a central configuration if
$$\nabla U + \omega^2 \nabla I = 0$$
\item<4-> Lagrange multiplier
method  $\Rightarrow$ c.c.'s are critical points
of $U$ subject to $I = const$
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Practical applications}
\begin{itemize} 
\item<1-> Central configurations in the 3-body problem
\item<2-> At each given time a planet-sun (e.g. Earth-Sun) system has 5 ``Lagrange points'' 
which are the positions where a third mass could be placed to
form a central configuration:
\begin{center} 
\includegraphics[scale=.25]{LagPts.jpg}
\end{center}
(image credit: NASA Wilkinson Microwave Aniosotropy Probe project)
\end{itemize} 
\end{frame}

\begin{frame}
\frametitle{Meaning of Lagrange points}
\begin{itemize}
\item<1->  A satellite  placed at rest at one of these Lagrange points
will stay in a rotating orbit about the center of mass of the planet-sun system
(relative equilibrium property)
 due to gravitational attraction exerted on it by the planet and sun
\item<2->  Same phenomenon has been observed for many natural objects, too
\item<3->  For instance -- each cloud of ``Trojan'' asteroids has center of mass approx. at
a Lagrange point of Sun and Jupiter 
\item<4->  You will study the Lagrange points
in this week's lab assignment (via consideration of 
central configurations for the 3-body problem in general)
\item<5->  More generally, rings of Saturn are a central configuration with 
a large number of smaller masses orbiting the large mass of Saturn
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Saturn and its rings}
\begin{center}
\includegraphics[scale=.30]{saturn.jpg}
\end{center}
\end{frame}

\begin{frame}
\frametitle{The big problem}
\begin{itemize}
\item<1->  Major question about central configurations is
\item<2->  Given $n$ masses $m_1,\ldots,m_n$, at how many different locations
can these be placed to get central configurations?  In particular, is
the set of possible configurations \emph{finite}?  On Smale's 21st century problem list.
\item<3->  To make this meaningful, need some further
explanations or definitions, since we can translate, rotate, and scale
central configurations and the results are again central configurations
\item<4->  \emph{Convention:}  We will consider two central configurations
to be equivalent if there exists a mapping of $\mathbb{R}^k$ taking
one into the other that can be obtained as a composition of 
a rigid motion (translation, rotation) and a scaling
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Status of the ``big problem''}
\begin{itemize}
\item<1->  Answer is known to be yes for $n = 3, 4$
\item<2->  Some fairly strong results for $n = 5$
\item<3->  Only fairly limited special cases known in general
\item<4->  Extremely subtle problem in general(!)
\item<5->  For instance, it's known that there are 
collections of $n = 5$ ``masses,'' including one negative
value, for which there is a whole curve of central configurations
(not just a finite number, even taking rigid motions and
scaling into account)
\end{itemize}
\end{frame}

\section{Day 2}

\begin{frame}
\frametitle{Mutual distances as coordinates}
\begin{itemize}
\item<1-> Because of the way we want to consider equivalence, it
makes sense to try to set up the problem so that:
\item<2-> A fixed distance scale is used, and
\item<3-> The mutual distances $r_{ij}$ become the {\it coordinates}
for describing the configuration   
\item<4-> This leads to some subtleties that we will need to understand, 
though
\item<5-> For instance, what are all triples $(r_{12},r_{13},r_{23})$
that can represent the mutual distances of three points in $\mathbb{R}^2$?
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Some obvious restrictions}
\begin{itemize}
\item<1->  All $r_{ij} \in \mathbb{R}$ (they're distances!) 
and any nonreal solutions of the c.c. equations will not be meaningful
for the physical problem 
\item<2->  All $r_{ij} \ge 0$ and $ = 0$ only in a partial or total collision
case
\item<3->  By triangle inequalities, we also have $r_{12} + r_{23} - r_{13} \ge 0$, 
$r_{12} + r_{13} - r_{23} \ge 0$, and $r_{13} + r_{23} - r_{12} \ge 0$ with
equality in one if and only if the three points are \emph{collinear} 
\item<4->  Which one gives zero depends on which point is between the other two 
along the line
\item<5->  The set of all points in $\mathbb{R}^3$ satisfying these inequalities
is a triangular cone $\mathcal{C}$ with vertex at the origin  
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{The configuration space of triangles}
\begin{itemize}
\item<1-> We claim that every point in $\mathcal{C}$ represents a ``triangle''
(including ``degenerate'' cases in which the three points are collinear, and two
of them coincide)
\item<2-> For instance, consider the cases where $r_{13} \ge r_{12}$.  
\item<3-> Then we can place point 1 at the origin, point two at $(r_{12},0)$
\item<4-> Then point 3 can be any point on the circle 
$x^2 + y^2 = r_{13}^2$
\item<5-> Can see all $r_{23}$ with 
$$r_{13} - r_{12} \le r_{23} \le r_{13} + r_{12}$$
are attained (as in inequalities describing $\mathcal{C}$)  
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Dimensions}
\begin{itemize}
\item<1->  Note that $n$ points in $\mathbb{R}^k$ span at most an
$n-1$-dimensional space.  
\item<2-> For instance, three points always lie on a plane, and may
be collinear.  If they are not collinear, then they form the vertices
of a \emph{triangle} with nonzero area.
\item<3-> Four points always lie in a 3-dimensional space and may
be coplanar or collinear, etc.  If they are not coplanar, then
they form the vertices of a \emph{tetrahedron} with nonzero volume.
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Cayley-Menger determinants}
\begin{itemize}
\item<1->  In working with mutual distance coordinates there 
are certain expressions that come up repeatedly and have significant
geometric meaning.
\item<2->  For instance, consider configurations of $n = 3$ points. 
Then it can be shown (by you in today's discussion, for instance :) 
that
$$\Delta_{CM} = \det \left( \begin{matrix} 0 & 1 & 1 & 1\\
                1 & 0 & r_{12}^2 & r_{13}^2\\
                1 & r_{12}^2 & 0 & r_{23}^2\\
                1 & r_{13}^2 & r_{23}^2 & 0\end{matrix} \right) $$
is a constant multiple of the square of the area of the corresponding triangle 
if $(r_{12},r_{13},r_{23}) \in \mathcal{C}$, and is zero if the 
points are collinear.   So $\det \Delta_{CM} \ge 0$.
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Cayley-Menger determinants, continued}
In general, for a configuration of $n$ points, the CM determinant is
the $(n+1) \times (n+1)$ determinant 
$$\Delta_{CM} = \det\left( \begin{matrix} 
0 & 1 & 1 & \cdots & 1 & 1\\
1 & 0 & r_{12}^2 & \cdots& r_{1,n-1}^2 & r_{1n}^2\\
1 & r_{12}^2 & 0 & \cdots & r_{2,n-1}^2 & r_{2n}^2\\
\vdots & \vdots & \vdots & \ddots & \vdots &\vdots \\
1 & r_{1,n-1}^2 & r_{2,n-1}^2 &  \cdots & 0 & r_{n-1,n}^2\\
1 & r_{1n}^2 & r_{2n}^2 &  \cdots & r_{n-1,n}^2 & 0
                           \end{matrix}\right) $$ 
(Note: the matrix is symmetric.) 
\end{frame}

\begin{frame}
\frametitle{Configuration spaces in general}
\begin{itemize}
\item<1->  The mutual distance description for configurations of $n \ge 4$
points is more subtle
\item<2->  For instance with $n = 4$, we have $\binom{4}{2} = 6$
mutual distances
\item<3->  Can ask which 
$$(r_{12},r_{13},r_{14},r_{23},r_{24},r_{34}) \in \mathbb{R}_+^6$$
can be realized as mutual distances for configurations of $n = 4$ points.
\item<4->  Must have $r_{ij} \le r_{ik} + r_{kj}$ for all distinct triples
$\{i,j,k\} \subset \{1,2,3,4\}$, by triangle inequality
\item<5->  Must also have $\det \Delta_{CM} \ge 0$ (this is automatic in 
$n = 3$ case, but not here)
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Configuration spaces, continued}
\begin{itemize}
\item<1-> Example:  Let 
$$(r_{12},r_{13},r_{14},r_{23},r_{24},r_{34}) = \left(1,t,1,1,\frac{2}{t},1\right)$$
\item<2->  There is an interval of $t$ values for which all the triangle 
inequalities are satisfied, namely $1 < t < 2$, but
\item<3->  $\det \Delta_{CM} = \frac{-8(t^2 - 2)}{t^2} < 0$ unless
$t = \sqrt{2}$, which gives the unit square.  
\item<4->  In general a collection of distances $r_{ij} \in \mathbb{R}_+$
can be realized by a configuration if and only if all the triangle
inequalities hold, \emph{and}  $\det \Delta_{CM} \ge 0$.  
\end{itemize}
\end{frame}


\begin{frame}
\frametitle{The Albouy-Chenciner equations}
\begin{itemize}
\item<1->  In mutual distance coordinates, the equations for 
central configurations can be written in the following form (\emph{asymmetric} Albouy-Chenciner):
\item<2->  For each pair $1 \le i, j \le n$:
$$G_{ij} = \sum_{k=1}^n m_k S_{ki}(r_{jk}^2 - r_{ik}^2 - r_{ij}^2) = 0$$
where 
$$S_{ki} = \begin{cases} r_{ki}^{-3} - 1 & \text{\ if\ } k \ne i\\
                          0 & \text{\ if\ } k = i \end{cases}$$
\item<3-> Note: $r_{\ell m} = r_{m \ell}$ for all $\ell,m$. $S_{ki} = S_{ik}$.)  
One nontrivial equation for each pair $1 \le i,j \le n$ with $i \ne j$ 
(everything cancels out if $i = j$).  
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{A comment about AC equations}
\begin{itemize}
\item<1->  The form of the AC equations we are using here 
comes by setting $\omega^2 = 1$ in the definition
\item<2->  This has the effect of imposing a particular
distance scale so that 
\item<3->  We \emph{never} get two central configurations
that are homothetic.  
\item<4->  In particular, for instance if three masses
$m_1,m_2,m_3$ at vertices of a triangle form a central 
configuration, then there are \emph{no similar triangles} that
will also give solutions of the AC equations.
\item<5-> OK for our purposes!
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{What the AC equations look like}
\begin{itemize}
\item<1->  Let's see what the AC equations look like for 4 masses
\item<2->  The AC equation for $i,j = 1,2$ is 
$$-2 m_2 S_{12} r_{12}^2 +m_3 S_{13}(r_{23}^2-r_{13}^2-r_{12}^2)+
m_4 S_{14}(r_{24}^2-r_{14}^2-r_{12}^2) = 0$$ 
where $S_{jk} = \frac{1}{r_{jk}^3} - 1$ as before
\item<3-> Note: can clear denominators to get a polynomial
\end{itemize}
\end{frame}



\section{Days 3-4}

\begin{frame}
\frametitle{An example}
\begin{itemize}
\item<1-> Say we want to study \emph{collinear} cc's
of $n = 4$ bodies. Use original form of cc equations (not AC). 
\item<2-> \begin{picture}(200,30)
          \put(20,15){\circle*{3}}
          \put(60,15){\circle*{5}}
          \put(120,15){\circle*{2}}
          \put(180,15){\circle*{6}}
          \put(20,15){\line(1,0){160}}
          \put(38,3){$a$}
          \put(88,3){$b$}
          \put(148,3){$c$}
          \put(16,22){$m_1$}
          \put(56,22){$m_2$}
          \put(116,22){$m_3$}
          \put(176,22){$m_4$}
          \end{picture}
\item<3->  Fix the first mass at $q_1 = (0,0)$, then 
place the others at $q_2 = (a,0)$, $q_3 = (a + b, 0)$, $q_4 = (a + b + c,0)$
(with $a,b,c > 0$).  
\item<4->  This gives $r_{12} = a$, $r_{13} = a+b$, $r_{14} = a+b+c$, 
$r_{23} = b$, $r_{24} = b + c$, and $r_{34} = c$.
\item<5->  The center of mass is at 
$$\overline{q} = \frac{\sum_i m_i q_i}{\sum_i m_i}$$
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Example, continued}
\begin{itemize}
\item<1-> For instance, if the masses are all equal (so we can set $m_i = 1$, all $i$),
and we make $\omega^2 = 1$ (this fixes the distance scale)
the central configuration equations become:
\item<2-> 
\begin{eqnarray*}
0 &=& {a}^{-2} + (a+b)^{-2}+ (a+b+c)^{-2}\\
  &&- 3/4\,a -1/2\,b - 1/4\,c \\
0 &=& -{a}^{-2}+{b}^{-2}+ (b+c)^{-2}\\
&&+1/4\,a-1/2\,b-1/4\,c\\
0 &=& -(a+b) ^{-2} - {b}^{-2}+{c}^{-2}\\
&&+1/4\,a+1/2\,b-1/4\,c\\
0 &=& -(a+b+c)^{-2} - (b+c)^{-2} - {c}^{-2}\\
&&+1/4\,a+1/2\,b+3/4\,c
\end{eqnarray*} 
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Example, continued}
\begin{itemize}
\item<1->  Clear denominators to get polynomial equations
\item<2->  When we do that, get factors of $a+b, b+c, a + b + c$
in some polynomials; since we want $a,b,c > 0$, those factors
cannot be zero either
\item<3-> (``Trick'') to find only solutions with $a,b,c,a+b,b+c,a+b+c \ne 0$
we can add a variable $t$ and a new equation 
$$1 - tabc(a+b)(b+c)(a+b+c) = 0$$
\item<4->  Compute a Gr\"obner basis for the ideal generated by these with 
respect to a monomial order set up to eliminate $t$, break ties with 
grevlex.  [Demo with {\tt Sage}]
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{How many solutions?}
\begin{itemize}
\item<1-> Note that the leading terms of the Gr\"obner basis
polynomials contain powers of each of the variables:  $t, a^5, b^8, c^{10}$
\item<2-> This implies that the set of all complex solutions of the
system is \emph{finite} (hence the same is true for the real solutions!)
\item<3->  
\begin{theorem}[Th. 6 in Chapter 5, \S 3 of IVA] 
Let $I$ be an ideal in $\mathbb{C}[x_1,\ldots,x_n]$ 
and $>$ a monomial order.  Then TFAE:
\begin{enumerate}
\item[i.]  $V(I)$ is a finite set
\item[ii.]  Let $G$ be a GB of $I$ wrt $>$.  For each $i$, 
there is $g_i \in G$ with $LT(g_i) = x_i^{m_i}$ for some $m_i$
\end{enumerate}
\end{theorem}
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Sketch of proof of this theorem}
\begin{proof}
\begin{itemize}
\item<1-> ii $\Rightarrow$ i: Think about the pictures we drew in the proof
of Dickson's Lemma.  If ii is true, there are only finitely many monomials
that are not in $\langle LT(I)\rangle$.  Hence for each $i$, the remainders
of the powers $x_i^n$, $n \ge 0$ on division by $G$ must be linearly dependent.  
This implies that there is some $h_i(x_i) \in I$ for each $i$.  It follows 
that $V(I)$ is finite.  
\item<2-> i $\Rightarrow$ ii:  If $V$ is a finite set, then for each $i$
there is a univariate polynomial $h_i(x_i) \in I(V(I))$.  By a big theorem
from Chapter 4 of IVA (the ``Nullstellensatz''), $h_i \in \sqrt{I}$ so some power
$h_i^\ell \in I$.  The leading term of $h_i^\ell$ is a power  
$x_i^{\ell k_i}$.  By the definition of a GB, there must be some $g_i \in G$ whose
leading term divides $x_i^{\ell k_i}$, so $LT(g_i)$ is a power of $x_i$.
\end{itemize}
\end{proof}
\end{frame}

\begin{frame}
\frametitle{Back to collinear 4-body cc's}
\begin{itemize}
\item<1->  In our case, the ideal contains a univariate
polynomial of degree $105$ in $a$.   
\item<2->  This factors; the smaller factor has this form (all terms
have exponent divisible by $3$ !)
$$16 a^{21} - 160 a^{18} + \cdots + 6856 a^9 - 8880 a^6 + 5936 a^3-1568$$
\item<3->  The method of Sturm sequences (see this week's lab) 
can be used to show this has exactly one real root, which is positive.  
\item<4->  Note that it must have \emph{at least} one real root since it has odd degree.  
\item<5->  The other factor has even degree and no real roots.
\end{itemize}
\end{frame}

\begin{frame}
\frametitle{Collinear 4-body cc's, continued}
\begin{itemize}
\item<1->  The ideal also contains univariate polynomials in $b,c$
\item<2->  These are very similar to the one for $a$.   (In fact, the polynomial for $c$
is \emph{exactly the same as} the polynomial for $a$; can you see why this
might be true?)
\item<3->  It follows with a bit of computation 
that there is \emph{exactly one real solution} of the cc equations in this case,
\item<4->  In fact it has been proved (by other methods!) that for \emph{all}
choices of $m_i > 0$, there is exactly one collinear cc of the form we 
are studying (\emph{Moulton's theorem}). 
\end{itemize}
\end{frame}
\end{document}
