\def\ssc{\bigskip\noindent{\it Some Specific Comments}\bigskip}
\def\finish{\bigskip\noindent {\it Paper: }
\bigskip\noindent{\it Presentation: } \bigskip\noindent
{\it Overall Project Grade: }\vfill\eject\noindent}
\nopagenumbers
\magnification=\magstep1
\noindent
{\it Natalie Cate and Becca Bates -- Buchberger's Criterion and Improvements to Buchberger's Algorithm}
\bigskip
This is a {\it very} well-written exposition of the proof of Buchberger's Criterion
and the improvements to Buchberger's algorithm based on the two criteria for
unnecessary $S$-polynomial reductions.  You have done an excellent and thorough
job with the details here.  The in-class presentation and the example you 
worked through were well-chosen and very illuminating.  So it's OK that you 
didn't really get to the specifics of Faug\`ere's alternate approach in the 
F4 algorithm.  
\ssc
\item{0)}  A somewhat random initial comment -- the Faug\`ere here is also the 
``F'' in the ``FGLM'' algorithm that Amy and Michelle worked on.
\bigskip
\item{1)}  It's not really a matter of ``right and wrong,'' of course.  
It be might be better to say something like "$h$'s with extra terms producing 
cancellations."  We often say things like this in lectures to make the explanations
more vivid, but they are best avoided in formal mathematical writing.
\bigskip
\item{2)}  Oops -- The underlined statement here {\it is the converse} of the 
Lemma, and that is the statement that is {\it not true in general}.  Namely, as we
saw in several calculations done in class, it is possible for an element
$f = a_1 f_1 + \cdots + a_s f_s$ to leave a nonzero remainder on division
by $F = [f_1, \ldots, f_s]$ when $F$ is not a GB for $I = \langle F\rangle$.  
(This can be caused, for instance, by the way the standard division algorithm
works -- it forces you to take the first element in the list $F$ whose
leading term divides the leading term of the intermediate dividend.  When
several choices are possible, some choices may lead to zero remainders,
showing $f \rightarrow_F 0$, while others may not lead to zero remainders.)
However, if $\overline{f}^F = 0$, then $f \rightarrow_F 0$, since then the 
expression $f = a_1 f_1 + \ldots a_s f_s$ from the division algorithm
does satisfy the condition ${\rm multideg}(a_if_i) \le {\rm multideg}(f)$
for all $i$ (again, by the way the division algorithm works). 
\bigskip
\item{3)}  This is just a comment for you about how the structure of the 
syzygies $S(F)$ you are describing here generalizes.  These properties lead
to the notion of what is called a {\it module} over the ring 
$k[x_1,\ldots,x_n]$.   Roughly speaking: a module is to a ring what
a vector space is to a field.  More precisely, a module over any commutative 
ring $R$ (with multiplicative identity element)
is a set $M$ with an addition operation and an operation of multiplication
by elements of $R$.  These operations must satisfy axioms that make $(M,+)$ 
into an abelian group, distributive laws $r(m + m') = rm + rm'$ and 
$(r+s)m = rm + sm$ for all $r,s \in R$ and $m,m' \in M$, and finally 
$1m = m$ for all $m \in M$ (where $1$ is the multiplicative identity in $R$).
What you are saying here is that $S(F)$ has this sort of structure.
It is a module over the ring $R = k[x_1,\ldots,x_n]$.  Later, when 
you mention ``linear combinations'' of syzygies, of course, you need
to allow for general polynomial coefficients, not just constant coefficients.
There is also a connection with the ideals that we studied in detail.
An ideal $I \subset R$ is just a module over $R$ that is a subset of $R$ itself.
More advanced treatments of commutative algebra treat ideals and modules
together, in fact, and there is also a theory of Gr\"obner bases for modules
over polynomial rings.
\finish
\noindent
{\it Ren\'ee Laverdi\`ere -- The Ring of Invariants and Molien's Theorem}
\bigskip
Very good presentation and paper on Molien's Theorem and the beginnings
of invariant theory for finite matrix groups.  There are a few 
specific comments I have about the content of the paper (see below).
The proof of Molien, in particular, is a bit rough still (although
I know that writing down the formulas you need there in \TeX\ is a
bit of a chore, and that might have caused the missing details!)

It might also have been nice
to point out that the ring $k[x,y]^{C_2} = k[x^2,xy,y^2]$ has the following 
structure.  It is the image of the ring homomorphism:
$$\eqalign{\varphi : k[u,v,w] &\to k[x,y]\cr
                         u &\mapsto x^2\cr
                         v &\mapsto xy\cr
                         w &\mapsto y^2\cr}$$
By the First Isomorphism Theorem for rings, this says
$$k[x,y]^{C_2} \cong k[u,v,w]/\langle uw - v^2\rangle$$
(since $\ker(\varphi) = \langle uw - v^2\rangle$).  
This shows that the ring of invariants can be viewed as the {\it ring
of polynomial functions on a variety} V, namely $V = {\bf V}(uw - v^2)$
in ${\bf C}^3$.  This is a surface whose real points form a quadric cone.
(The origin is a singular point.)  It is this variety that 
gives the quotient space ${\bf C}^2/G$ in this case (see comment 4 below).
\ssc
\item{1)}  Saying that ${\rm GL}(n,k)$ is the set of units in $M(n,k)$
means that you are thinking of $M(n,k)$ as a ring under matrix addition
and multiplication.  Probably better to say that explicitly.
\item{2)}  It's true that a matrix $A$ can act on a vector field, but
the more basic action is the action on single vectors:  $v \mapsto Av$
for $v \in k^n$.  (Note:  a vector field would be something like 
a mapping $V : k^n \to k^n$ giving a vector $V(x)$ for each $x \in k^n$.)
\item{3)}  ``Indistinguishable'' in what sense?  (This means indistinguishable
from the point of view of polynomial functions.  Any polynomial vanishing 
on an orbit also vanishes on the closure of that orbit, since polynomials
are continuous functions.)
\item{4)}  What you wrote here doesn't quite make sense.  The two things
on either side of your $=$ sign are of two different types 
so they can't really be equal.   The $k^n/G$ should
mean the quotient space of $k^n$ under the action of $G$
(i.e. the set of orbits under the action of $G$).  For
finite groups, this set of orbits has the structure of a variety too, 
and the ring of polynomial functions on the quotient $k^n/G$ can 
be identified with the ring of invariants $k[x]^G$. 
\item{5)}  As we discussed in office hours one time, the real reason
this is true is just the fact that $x \mapsto A\cdot x$ is
homogeneous of degree 1 in the variables, so $A$ acts on the homogeneous 
components of each degree in polynomials ``separately''  (i.e. the 
homogenous component of degree $d$ in $f(A\cdot x)$ is just the 
image under $A$ of the homogenous component of degree $d$ in $f(x)$.
This observation makes the statement of your Theorem 3.4 almost ``obvious.''
\item{6)}  The Hilbert series itself makes sense over any field.  However
the statement of Molien's Theorem is not necessarily true (for some $G$), 
since some of the invariants might have complex coefficients.  That's 
the real reason for restricting to $k = {\bf C}$.  
\item{7)}  For completeness in this argument it would be good to 
say that you are using the linearity of the trace here
$$\dim(V^G) = {\rm trace}(P_G) = {1\over |G|} \sum_{A\in G} {\rm trace}(A)$$
which is the formula you are trying to show.  
\item{8)}  I think this proof got somewhat garbled (but I'm not sure if this was
a problem with the complicated \TeX\ or with your understanding of the details). 
Some of these formulas
should have sums over $d$, and/or sums over $A \in G$,
and you are missing $z^d$'s.  Should say:
$$\eqalign{
\sum_{d = 0}^\infty \dim({\bf C}[x]_d^G) z^d 
&= \sum_{d=0}^\infty \left({1\over |G|} \sum_{A\in G} {\rm trace}(A^d)\right) z^d\cr
&= \sum_{d=0}^\infty \left({1\over |G|} \sum_{A\in G} \left(\sum_{d_1 + \cdots + d_n = d}
\rho_{A,1}^{d_1} \cdots \rho_{A,n}^{d_n}\right) \right) z^d\cr
}$$
{\it Then}, you interchange the two outer sums to get 
$${1\over |G|} \sum_{A\in G} \left(\sum_{d=0}^\infty \left(\sum_{d_1 + \cdots + d_n = d}
\rho_{A,1}^{d_1} \cdots \rho_{A,n}^{d_n}\right) \right) z^d$$
{\it Finally}, you want to argue that the inner sum is the same as 
$${1\over \det(I - Az)}$$
This follows from some standard matrix indentities when you change basis
to put $A$ in diagonal form (formally the sum of a geometric
series, as you indicate).  Note that you also need $k = {\bf C}$ to 
say that the eigenvalues of your $A$'s exist in $k$ for this argument.  
\item{9)}  Although the result of this computation is correct, note that
you're using $\det(A - zI)$ rather than $\det(I - Az)$ here.  This happens
to work for this group since
$$A - zI = \pmatrix{-z & -1\cr 1 & -z} \Rightarrow \det(A - zI) = z^2 + 1$$
and
$$I - zA = \pmatrix{1 & z \cr -z & 1} \Rightarrow \det(I - zA) = z^2 + 1$$
too.  But as we discussed, you should be using $\det(I - zA)$ (the ``reversed
version'' of the characteristic polynomial).
\item{10)}  $f_3$ should be $f_3 = x^3y - xy^3$ (you're missing the $x$ in the 
second term, which makes this nonhomogeneous!)
You could also say more here!  Where do the {\it three}
linearly independent invariants of degree 4 come from?  You have $f_2, f_3$ and 
the third one is $f_1^2 = (x^2 + y^2)^2$.  
\finish
\noindent
{\it Amy Borbely and Michelle Hedrick -- Gr\"obner Basis Conversion; Adaptation of the 
FGLM Algorithm}
\bigskip
Very good project on the FGLM algorithm and Gr\"obner Basis conversion.  
The presentation was quite good too (though slightly rough in a couple of spots
corresponding to the specific comments below).  The example was well-chosen
and very illuminating, though.  For the paper, it would have been nice to
include an example in more than two variables.  We discussed that, so
I was slightly surprised not to see that.  I'm slightly curious about
this -- was there some problem with
the procedures in that case?  I don't think there should have been 
provided that you gave the input of the correct grevlex staircase.  
(I didn't see any potential problems, though, so I'm guessing the 
difficulty was just seeing what was in the grevlex staircase in 
a case with more variables -- it can be hard to visualize!).  

For your general knowledge about this area -- there are other GB conversion
algorithms that are not limited to the case of zero-dimensional ideals, and
a lot of additional work has been done since the original FGLM paper.  
One method, called the ``Gr\"obner Walk'' makes use of the construction
of the Gr\"obner fan that Katie worked on for her project.  The 
algorithm ``walks through the fan'' and converts every time you cross
from one cone into another.  This sounds like a lot of work, but the
conversion is usually not too difficult when the cones are adjacent.  
Another (maybe better) method is called the ``Hilbert-function-driven 
method.''  It uses some of the same ideas about counting dimensions of
subspaces of ideals of different homogeneous degrees as in Renee's talk.
 
\ssc
\item{1)}  Actually, although in practice you would always make the input
GB a reduced GB in the starting order $M_1$, there isn't any reason it has 
to be reduced.  The reason that isn't necessary is that remainders are well-defined
and unique when you divide by any GB for the ideal with respect to the $M_1$
order (recall Exercise 1 in Section 2.6 of ``IVA'').
\bigskip
\item{2)}  Not exactly.  The monomials in $Blex$ are actually the 
``{\it lexicographic staircase}'' monomials when the algorithm terminates.  
You're not using them for the remainder calculations.  They correspond
to the rows in your $A$-matrix.
\bigskip
\item{3)}  Normally, you say that an {\it ideal} is zero-dimensional,
not a Gr\"obner basis.  This is just a matter of how the terminology is
defined.  But
one point that is not too well-explained here in the paper (or in
the presentation) was just what this means.  Let ${\bf Q} \subset k$
Saying an ideal $I \subset k[x_1,\ldots,x_n]$ is zero-dimensional 
means (by definition) that in ${\bf C}^n$, ${\bf V}(I)$ is a finite set.  
Then Theorem 6 in Section 5.3 of ``IVA'' gives a collection of equivalent
statements, including the one you used regarding the 
leading terms of any GB for $I$ (for each $i$ the existence of some $g \in GB$
such that $LM(g) = x_i^{m_i}$ for some $m_i \ge 0$).  
\finish
\noindent
{\it Katie Fitzpatrick -- The Gr\"obner Fan of an Ideal}
\bigskip
Very good presentation and paper on Gr\"obner fans.  I can tell you
probably have not had too much experience giving presentations like this.  
What you did was very good and the twisted cubic example was very instructive; 
the only thing I would suggest is to perhaps try to write a bit more on 
the board (even at the expense of that slowing you down somewhat).  
Sometimes what you wrote was only a (slightly cryptic) part of what 
you were saying and that made it difficult for a listener
to get back into the flow if he or she had to think about a 
point you were making or his/her attention was distracted momentarily.  

The last example is also good (the + sign in the second generator
is actually OK, I guess).  I was actually thinking of a different
way to generalize the twisted cubic.  The ideal defining any parametric 
curve of the form $\alpha(t) = (t,t^m,t^n)$ will also give an ideal whose
Gr\"obner fan has a ray common to all cones:  namely the 
homogeneous weight vector $(1,m,n)$.  
\ssc
\item{1)}  The fact that matrix orders represent all monomial orders
is actually a theorem of an Italian mathematician named Lorenzo Robbiano. 
This result was actually a ``big deal'' in the subject when it first appeared.
\bigskip
\item{2)}  An interesting aspect of the matrices defining some orders
like the {\it grevlex} is that the same order is represented if you 
add scalar multiples of ``higher'' rows (i.e. smaller row index) to ``lower''
row (i.e. larger row index) in the matrix.  (Think about how the comparison
works!)  This means that you actually never need to use negative entries
in vectors either (all rows of the matrix can be vectors in the positive
orthant).  For instance, the {\it grevlex} order, $x > y > z$, in $k[x,y,z]$ 
can also be defined by 
the matrix obtained by adding row 1 to row 2 and then the resulting row
2 to row 3:
$$M_2' = \pmatrix{1 & 1 & 1 \cr 1 & 1 & 0 \cr 1 & 0 & 0 \cr}$$
This observation is useful, for instance, since some software restricts
the weight matrices to have nonnegative integer entries in each row.
\bigskip
\item{3)}  Of course, there are several sets inequalities defining $C_G$ in
general -- one set from each of the elements of $G$.  So to make that perfectly
clear it would be good to say explicitly:  ``for all $i$, $1 \le i \le t$.''
\bigskip
\item{4)}  In the proof that ${\rm Mon}(I)$ is finite for all ideals,
the one point that wasn't completely clear was that when you find the 
set $\Sigma_1$ and the monomial ideal $N_1$, you then actually 
want to argue that the given set of generators
$\{f_1,\ldots,f_s\}$ for $I$ {\it cannot be a Gr\"obner basis} of $I$ with 
respect to any of the orders in $\Sigma_1$.  (Reason: if it is for any one, then
it has to be for all of them.  But the starting set $\Sigma$ is chosen so
that the ideals $\langle LT_{>}(I) \langle$ for $>$ in $\Sigma$ were all
{\it distinct}.)  That's what allows you to say there must be some 
other $f_{s+1}$ and some leading term that is not divisible by any of the
leading terms of $f_i$, $i = 1,\ldots,s$.  Then you repeat the same argument
to produce the chains 
$$\Sigma \supset \Sigma_1 \supset \Sigma_2 \supset \cdots $$
and c
$$N_1 \subset N_2 \subset \cdots $$
\finish
\noindent
{\it Julianne Kulevich -- Computer Algebra Methods for Studying and Computing 
Molecular Conformations}
\bigskip
This was a good project and presentation, although I have some specific comments
and suggestions about things you could have added or done differently.  The 
main comment here is 1 below.  It's sometimes important in working on applied
problems like this to make sure you really understand where the equations
you end up solving come from in some detail.  
\ssc
\item{1)}  I was hoping to see some more details about how these equations
are derived, both in the paper and when I asked you about this in 
the presentation.  That's what you need to make the connection between
the solutions $(t_1,t_2,t_3)$ of the transformed equations and the 
actual geometry of the molecules they represent.   These equations 
can be derived like this.  For instance, introduce coordinates so 
$P_6$ is the origin and $\overline{P_6P_4}$ is the $x$-axis . 
Then from the geometry of the triangles $\Delta P_4 P_5 P_6$ and
$\Delta P_2 P_4 P_6$, 
the coordinates of $P_5$ are
$$(L_6 \cos(a_1), L_6\cos(a_1) \cos(\theta_3), L_6\sin(a_1)\sin(\theta_3))$$
where $a_1$ is the angle at $P_6$ in the ``flap triangle'' $\Delta P_4 P_5 P_6$.
Similarly, but with somewhat more work you get the coordinates of $P_1$ are 
$$\eqalign{
(&L_1 \left(\cos(a_3)\cos(a_2) + \sin(a_3)\sin(a_2)\cos(\theta_1)\right),\cr
 &L_1 \left(\sin(a_2)\cos(a_3) - \sin(a_3)\cos(a_2)\cos(\theta_1\right),\cr
 &L_1 \sin(a_3)\sin(\theta_1) )\cr}$$
where $a_2$ is the angle in the base triangle at $P_6$ and $a_3$ is the 
angle in the flap triangle $\Delta P_1P_2P_6$.
Then the first equation comes from the fact that you know the bond angle
at $P_6$ so the dot product of the vectors $P_1$ and $P_5$ satisfies:
$$P_1 \cdot P_5 - \|P_1\| \|P_5\| \cos(\psi_6) = 0$$
You can work out this equation and get the explicit forms of the $\alpha_{2j}$
in terms of $L_6, L_1$, and the sines and cosines of the angles $a_1, a_2, a_3$.
Similarly, the other two equations come from the angles at $P_2$ and $P_4$.
\bigskip
\item{2)}  I'm not 100\% certain about the relative magnitudes of the 
improvements from a) better implementation of GB's in Maple 10, and 
b) faster computer hardware on our radius server (and more memory space
for intermediate computations) versus what Emiris 
and Mourrain had to work with in 1998.  I would guess that both
had an impact, but perhaps the software improvements had more effect.
\bigskip
\item{3)}  The large coefficients in the final result are just the 
``tip of the iceberg.''  In the computation of the {\it lex}
GB you are actually building up more and more polynomials with these
huge coefficients as you go along.  The final reduced GB has only 
three elements, but intermediate stages could have many more, with
even more terms than the polynomials here.
In practical terms, this means that the amount of information 
you need to store in the course of the calculation can easily
fill up the whole amount of memory available (and more).  When
that happens, programs like Maple will actually write polynomials
(or parts of them) to temporary storage on the main hard disk 
of the computer system, and read them back in as those polynomials
are used.  This ``swapping'' information in and out of memory
ends up taking up so much time and computing power to manage
that the computation grinds to a halt.  
\bigskip
\item{4)}  It's not entirely clear from the way you set up the 
solving why there are only four different solution vectors $(x,y,z) = (t_1,t_2,t_3)$.
It would have been better to use the full {\it lex} GB and the Extension
Theorem.  (Note that the other two GB polynomials have the form
$y - p_{15}(z), x - q_{15}(z)$, where the $p_{15}$ and $q_{15}$
are polynomials in $z$ alone, each of degree 15.  This shows directly that each value
of $z$ gives only one $x$ and one $y$.  If you plug one of the four real
solutions of the $z$-equation into the other polynomials in the 
Gr\"obner basis you get one value of $y$ and one value of $x$ in 
each case.)  Your way makes it look
like there might be more than one solution vector with each $z$-value. 
It's not easy to tell which $x$'s go with which $y$'s to give full 
solutions.
\bigskip
\item{5)}  If you had done the derivation of the equations in more detail, 
you could have worked backwards to solve for the actual locations of the 
$P_i$ and plotted them yourself using Maple(!)
\bigskip
\item{6)}  We didn't study Cayley-Menger matrices in this course,
so to include this discussion, it would have been much better to 
say more about this.  What does the matrix represent?  What do
you do with it?
\finish

 
\bye
