\hsize 6in 
\vsize9in 
%\font\rm=cmr12
%\rm
%\nopagenumbers
%\hsize 6.5truein
%\vsize 9truein
%\hoffset=1.0truein
%\voffset=1.0truein
%\baselineskip 12truept
\hfuzz 200 pt
\font\ninerm=cmr9
\font\sevenrm=cmr7
\font\fiverm=cmr5
% magic 1 cube 123
\setbox\strutbox=\hbox{\vrule height6pt depth 1pt width0pt}
\def\magiconecb#1{{
\vbox{\tabskip=0pt \offinterlineskip
\def\tablerule{\noalign{\hrule}}
\halign to#1{\strut##& \vrule##\tabskip=0em plus2em&
 ##& \vrule##&
 ##& \vrule##&
 ##& \vrule##\tabskip=0pt\cr\tablerule
&&   1&& 2&&  3&\cr\tablerule 
}
}
}}

\setbox\strutbox=\hbox{\vrule height6pt depth 1pt width0pt}
\def\magiconeca#1{{
\vbox{\tabskip=0pt \offinterlineskip
\def\tablerule{\noalign{\hrule}}
\halign to#1{\strut##& \vrule##\tabskip=0em plus2em&
 ##& \vrule##&
 ##& \vrule##\tabskip=0pt\cr\tablerule
&&   1&& 2&\cr\tablerule 
}
}
}}

\setbox\strutbox=\hbox{\vrule height6pt depth 1pt width0pt}
\def\magiconecc#1{{
\vbox{\tabskip=0pt \offinterlineskip
\def\tablerule{\noalign{\hrule}}
\halign to#1{\strut##& \vrule##\tabskip=0em plus2em&
 ##& \vrule##&
 ##& \vrule##&
 ##& \vrule##&
 ##& \vrule##&
 ##& \vrule##&
 ##& \vrule##\tabskip=0pt\cr\tablerule
&&   1&& 2&&  3&& 4&& 5&& 6&\cr\tablerule 
}
}
}}
\newcount\fncount \fncount=0
\def\Footnote#1{\global\advance\fncount by1 \footnote{${}^\the\fncount$}{#1}}
\newcount\scount \scount=-1
\newcount\eqcount \eqcount=1
\def\s#1{\global\advance\scount by1\eqcount=0
\par\vskip 12pt\noindent\S\the\scount
 \quad{{\bf #1}}\par}
\def\Leqno{\global\advance\eqcount by1\leqno{(\the\scount.\the\eqcount)}}
\newcount\thmcount \thmcount=0
\newcount\corcount \corcount=0
\def\Proclaim #1.  #2\par{\global\advance\eqcount by1\medbreak
 \noindent{\bf#1 (\the\scount.\the\eqcount): \enspace}{\sl#2}\par
 \ifdim\lastskip<\medskipamount \removelastskip\penalty55\medskip\fi}
\def\Thm#1{\global\advance\thmcount by1 \global\corcount=0 
\Proclaim Theorem {\the\thmcount}. {#1}\par }
\def\Cor#1{\global\advance\corcount by1
 \Proclaim Corollary \the\corcount. {#1}\par }
\def\author#1#2{\vskip 0.5truein\noindent\vbox{{\bf#1}\vskip 0.25truein{#2}}}
\def\x{{\phantom \ }}
\def\frame#1{(F_{#1},\sigma_{#1},\lambda_{#1})}
\def\SUM#1{\mathop{{\sum}^{#1}}}
\newdimen\tmpdimen

%% put arg in a box, keeping same baseline
\def\justboxit#1{{\setbox0\hbox{#1}\tmpdimen\dp0
  \advance\tmpdimen 0.4pt \lower\tmpdimen %% 0.4pt is the default 
					  %% rule thickness
  \vbox{\hrule\hbox{\vrule\unhbox0\vrule}\hrule}}}

%% add some extra space around arg, keeping same baseline
\def\justaddborder#1#2{{\setbox0\hbox{#1}\tmpdimen\dp0
  \advance\tmpdimen #2\lower\tmpdimen
  \vbox{\kern#2\hbox{\kern#2\unhbox0\kern#2}\kern#2}}}

%% put arg in a box with some extra space around it, keeping
%% same baseline
\newdimen\boxitmargin \boxitmargin = 1pt
\def\boxit#1{\justboxit{\justaddborder{#1}{\boxitmargin}}}

%% define binary op
\def\boxtimes{\mathbin{\boxit{$\times$}}}

%% Try an example
%$$L\timesquare L$$





{\nopagenumbers
\
\vskip 1 truein
\centerline{\bf Magic $N$-Cubes Form a Free Monoid}
\vskip .25truein
\centerline{Allan Adler}\par
\centerline{P.O. Box 20276, Cherokee Station, New York, NY 10021}\par
\vskip .5truein
\centerline{\bf Abstract}\bigskip

{\parindent 40pt\narrower\smallskip\noindent
In this paper we prove a conjecture stated in an earlier paper [A-L].
The conjecture
states that with respect to a rather natural operation, the set of 
$N$-dimensional magic cubes forms a free monoid for every integer $N>1$.
A consequence of this conjecture is a certain identity of formal Dirichlet
series. These series and the associated power series are shown to diverge.
Generalizations of the underlying ideas are presented. We also prove
variants of the main results for magic cubes with remarkable power sum
properties.}

\vskip .25 truein
\noindent {\bf(1980) AMS Classifications:}
 05A15, 05A17, 05A19, 05B15, 08A02, 08A05,
08B20, 10A05, 10A25, 10A50, 10H40, 10M20, 15A30, 15A33, 15A51, 20M05
\vskip .25truein
\noindent {\bf Key Words:} array, estimate, free monoid, generating function, 
inescapable submonoid, infinite dimensional variety, irreducible, left prime, 
magic N-cube, monoid, persuasive submonoid, semidirect product}
\pageno=1
%\headline{\tenrm ALLAN ADLER\hfil MAGIC $N$-CUBES FORM A FREE MONOID}
\font\smcp=cmcsc8
\headline={\ifnum\pageno>1 {\smcp the electronic journal of  
combinatorics 4 (1997), \#R15\hfill\folio} \fi}
%\vfill\eject
\vskip .5in

\s{Introduction}
According to the book of W.S.Andrews [An], the study of magic squares is quite
old and dates back to ancient Tibet, to 12th century China, to 9th century
Arab astrologers and perhaps much further. Indeed, Andrews speculates that
magic squares might even be prehistoric. Despite their antiquity, they
have often been regarded by serious mathematicians as a waste of time, although
many of the great mathematicians of the past, such as Euler [E1],[E2], 
Ramanujan [R1],(cf. [R2]) and
Fermat [F],  found it to their liking to waste 
quite a lot of time on them.
One possible reason for this double standard is that although magic squares
are a lot of fun, no one believes that it is possible to prove
general theorems about them. Indeed, until recently all work on magic squares
consisted in presenting methods of constructing 
magic $N$-cubes or of examining
the properties of particular magic $N$-cubes, with most attention being 
devoted to the case $N=2$. Since no explicit method has ever been given for
constructing all magic $N$-cubes, the work of past researchers has necessarily
been limited to special classes of magic $N$-cubes. Even so, their work
has in some cases also drawn on interesting mathematics such as the related 
studies of Latin squares and
finite geometries. One can also mention the paper [A-L], which used Prouhet
sequences\Footnote{Prouhet sequences have as a special case a sequence used
by Morse [Mo1] (cf. [Mo2], [M-H 1]) to give a counterexample to the
Poincar\'e-Bendixson theorem in higher genus and by Morse and Hedlund [M-H 2] 
to give an example of unending chess.} to construct an infinite class
of magic cubes with remarkable power sum properties, 
and the recent papers [A1],[A-W], in which 
$p$-adic L-functions [Iw] are used to construct an infinite class of
magic $p$-dimensional cubes.\Footnote{One can view this result as saying that
one can use the Riemann zeta function and certain Dirichlet L-functions
to construct magic $p$-cubes.}

In recent times, mathematicians concerned with areas
quite different from magic squares have been pleased and amused when they
could report that their researches have led them to objects which, if not
actually magic squares, at any rate exhibit some of the properties of magic
squares. Thus it was that Ax [Ax1], in his studies of quantum 
mechanics [Ax2], enjoyed referring to doubly stochastic matrices as 
``magic squares''; 
that Andr\'e Weil [W1] was able to point out properties of the period matrices 
of Fermat curves [W2] which made them like ``magic squares''; and that I.M.
Gelfand in his lectures
on generalized hypergeometric functions [Ge] was happy to report that some
of the objects of his theory are described by ``magic squares''. 
Some such ``magic squares'' then become legitimate objects of study
in their own right. For example, in his work on magic graphs, Richard Stanley 
[S1],[S2] used the 
Hilbert syzygy theorem to study the growth as $k\rightarrow\infty$ 
of the number of arrays of positive integers with entries less than $k$ and
having many of the properties of magic squares. Nevertheless, the study
of magic $N$-cubes themselves remains disreputable and probably always will.

It is not the purpose of this article, however, to survey the subject of
magic $N$-cubes nor to create enthusiasm for the topic where it does not
already exist. It is instead to report on a theoretical development 
in the subject of magic $N$-cubes, and here I mean genuine magic $N$-cubes,
not the imitations which were mentioned in the previous paragraph. The simple
fact is that until the paper [A-L] appeared in 1977, no theorem had
ever been published concerning the set of all magic $N$-cubes as a whole.
What was new in [A-L], apart from a new construction of a remarkable class
of magic $N$-cubes, was the introduction ([A-L], p. 624) of an operation on the
set of all magic squares. This operation is associative and has an
identity element and therefore makes the set of magic squares into a monoid.
Furthermore, it was conjectured that with respect to this operation, the
set of magic squares is a free monoid.
It was also remarked in  [A-L], but not explicitly proved, that the operation 
makes sense for magic $N$-cubes for all $N$ and the conjecture was expected to 
hold for all $N>1$. In this paper, we show that the conjecture is 
true.\Footnote{For $N=1$, every permutation of the interval $[1,r]$ is
a magic $1$-cube, since there is only one ``side" and that side is also
the only ``diagonal". In this case, the monoid is definitely not free,
since for example, 
the magic $1$-cubes {\rm\magiconecb{0.5in}} and
{\rm\magiconeca{0.3in}} are irreducible
and commute with each other under the monoid operation, the product being
{\rm\magiconecc{0.9in}}. The generators of the monoid are the
irreducible elements of the monoid, but I don't know the relations among them.}
By adapting the techniques of the proof, we show that magic
$N$-cubes satisfying certain remarkable power sum properties also
form a free monoid. We then consider various generating functions for
the number of magic $N$-cubes and show that they diverge. Although pleasant
to know, the real significance of their divergence lies in the estimates
that one uses. For, what one really wants to know, either exactly or
approximately, is how many magic $N$-cubes there are of a given size.
We therefore summarize the estimates we obtain at the end of the article.
Undoubtedly one can get much better estimates even from the constructions
we have used. Whether one can get accurate asymptotic estimates for the
number of magic $N$-cubes is another question. It is conceivable that
the ``sporadic" magic squares which do not arise from general constructions
form an infinite set larger in the sense of its growth than the set
of magic squares that do arise from such constructions.
If this is the case, then the approach we have taken
to obtaining estimates cannot, even in principle, be made accurate through 
more careful study of general constructions. In order to test for this
possibility, it seems reasonable to suggest that in addition
to using computers to enumerate the magic squares of various sizes, as many
people are doing, one
should also be trying to see whether the magic squares produced by these
searches have natural generalizations to new infinite families.

The contents of the paper are as follows. In \S1, we prove (Theorem 1.26) that 
magic squares
form a free monoid. These ideas apply
as well in the case of magic $N$-cubes, but since they only involve
trivial modifications of the proof for magic squares, the details are left 
to the reader. In \S2, which is devoted to generalizations, 
we note that the frames of the magic
$N$-cubes form a monoid, as do the summation conditions to which they are
subjected, and we formulate these observations in a very general setting, 
motivated by the desire to understand the monoids ${\cal S}_N(t,g)$
(see \S3) from an
abstract point of view. In earlier draft of this article [A2], these 
generalizations were worked out in considerable detail including some
discussion of infinite dimensional varieties. In the present article, for
reasons of space, the discussion is reduced to a brief sketch. In \S3,
we consider classes of magic $N$-cubes that have certain remarkable power 
sum properties
for their entries. We denote these monoids ${\cal S}_N(t,g)$ and prove
in Theorem 3.1 that
they are free on infinitely many generators.
In \S4, we take up the question of the convergence of the generating functions
and, after showing that they diverge, conclude the article with a summary of
the estimates we obtained. These estimates can undoubtedly be greatly
improved by a more careful application of the methods we are using. However 
they are all based on the observation that certain methods of constructing 
magic
$N$-cubes are compatible with many permutations. More desirable than merely
having better estimates would be to have another observation of an essentially
different nature.

Throughout this article, $N$ will denote an integer greater than 1 and
${\bf Z}$ will denote the ring of integers. We will often regard ${\bf Z}$
as a monoid either under addition or under multiplication, as the situation
requires. If $a$ and $b$ are integers,
we will denote by $[a,b]$ the set of integers which are $\geq a$ and $\leq b$.
If $S$ is a monoid, we will denote the identity element of $S$ by $1_S$.

The author wishes to thank C.Cowsik for his help in proving the crucial
Lemma 1.25 and P.Sz\"usz for his help in proving the divergence of the 
generating functions. The author is also grateful to Roger Howe and
Walter Feit for making it possible for him to use facilities at Yale
University. Finally, the author is grateful to Michel Brou\'e for 
making it possible to use a computer at the \'Ecole Normale Sup\'erieure
in Paris, where final changes were made in the manuscript.\par\ \par

\s{Magic Squares}
% Definition 1.1
\Proclaim Definition. {\rm By a {\bf magic square of order ${\bf n}$}, also
called an {\bf ${\bf n\times n}$ magic square}, we mean an $n\times n$
square array $A=(a_{ij})$, $0\leq i,j\leq n-1$, of positive integers such that
{\parindent 40pt\item{(i)}each of the integers from 1 to $n^2$ inclusive occurs exactly once among the entries of $A$;
\item{(ii)}for $0\leq i\leq n-1$, the sum $\sum\limits_{j=0}^{n-1} a_{ij}$ is independent of $i$;
\item{(iii)}for $0\leq j\leq n-1$, the sum $\sum\limits_{i=0}^{n-1} a_{ij}$ is independent of $j$;
\item{(iv)}the sums $\sum\limits_{i=0}^{n-1} a_{ii}$ and $\sum\limits_{i=0}^{n-1} a_{i,n-i-1}$ are equal to the sums given in (ii)  (and therefore
to those in (iii) as well).}}\par
For example,\par\noindent
{\rm
% define \magicsqa to make a 3x3 magic square
%\setbox\strutbox=\hbox{\vrule height14pt depth 3.5pt width0pt}
\setbox\strutbox=\hbox{\vrule height7pt depth 2pt width0pt}
\def\magicsqa#1{{
\vbox{\tabskip=0pt \offinterlineskip
\def\tablerule{\noalign{\hrule}}
\halign to#1{\strut##& \vrule##\tabskip=0em plus2em&
 ##& \vrule##&
 ##& \vrule##&
 ##& \vrule##\tabskip=0pt\cr\tablerule
&&   8&& 1&&  6&\cr\tablerule 
&&   3&& 5&&  7&\cr\tablerule 
&&   4&& 9&&  2&\cr\tablerule 
}
}
}}
% define \magicsqb to make a full 4x4 magic square
\def\magicsqb#1{{
\vbox{\tabskip=0pt \offinterlineskip
\def\tablerule{\noalign{\hrule}}
\halign to#1{\strut##& \vrule##\tabskip=0em plus2em&
 ##& \vrule##&
 ##& \vrule##&
 ##& \vrule##&
 ##& \vrule##\tabskip=0pt\cr\tablerule
&&  \ 1&&15&&14&&\ 4&\cr\tablerule 
&&   12&&\ 6&&\ 7&&\ 9&\cr\tablerule 
&&   \ 8&&10&&11&&\ 5&\cr\tablerule 
&&   13&&\ 3&&\ 2&&16&\cr\tablerule 
}
}
}}
\vbox{
\centerline{
%\magicsqa{.8in},\ \boxit{1}\quad{\rm and}\quad\magicsqb{1.5in}}
\magicsqa{.5in},\ \boxit{1}\quad{\rm and}\quad\magicsqb{.8in}}
\vskip .25truein
\centerline{{\bf Figure 1}}
}
}
are magic squares.

We let let ${\cal M}_2$ denote the set of all magic squares. We will define an
operation on ${\cal M}_2$. The operation will be denoted by $*$ and will be
referred to as {\bf multiplication}. Before defining our
operation on ${\cal M}_2$ formally, let us illustrate it with an example. 
Let $A$ be the $3\times3$ magic square in Figure 1 above and let $B$ be the
$4\times4$ magic square in Figure 1. To form the magic square $A\mathbin*B$,
note that $B$ is $4\times4$ in this case and make a big, empty $4\times4$ 
frame, as in Fig.2 below.\vskip .25in
\centerline{\hskip4.3truein\vbox{% redefine \magicsq to make big empty 4x4 frame
\setbox\strutbox=\hbox{\vrule height30pt depth 2pt width0pt}
\def\magicsq#1#2{\vbox{
\tabskip=0pt  \offinterlineskip
\def\tablerule{\noalign{\hrule}}
\halign to#1{\strut##& \vrule##\tabskip=0em plus2em&
 ##& \vrule##&
 ##\hfil& \vrule##&
 ##\hfil& \vrule##&
 ##& \vrule##\tabskip=0pt\cr\tablerule
&&    &&&&&&&\cr\tablerule
&&   &&&&&&&\cr\tablerule
&&   &&&&&&&\cr\tablerule 
&&   &&&&&&&\cr\tablerule 
}
\vskip .25in
\hskip0.35truein\hbox{\bf Fig. #2}
}}
\magicsq{2.0in}2
}\hskip0.5truein\hskip-3.8truein\vbox{\hbox{\phantom{eagle}}
\setbox\strutbox=\hbox{\vrule height9pt depth 1pt width0pt}
\def\magicsq#1#2{
\vbox{\tabskip=0pt  \offinterlineskip
\def\tablerule{\noalign{\hrule}}
\halign to#1{\strut##& \vrule##\tabskip=0em plus2em&
 ##& \vrule##&
 ##\hfil& \vrule##&
 ##\hfil& \vrule##&
 ##\hfil& \vrule##&
 ##\hfil& \vrule##&
 ##\hfil& \vrule##&
 ##\hfil& \vrule##&
 ##\hfil& \vrule##&
 ##\hfil& \vrule##&
 ##\hfil& \vrule##&
 ##\hfil& \vrule##& 
 ##& \vrule##\tabskip=0pt\cr\tablerule
&&   8&&1&&6&&\ &&\ &&\ &&\ &&\ &&\ &&\ &&\ &&\ &\cr\tablerule
&&   3&&5&&7&&&&&&&&&&&&&&&&&&&\cr\tablerule
&&   4&&9&&2&&&&&&&&&&&&&&&&&&&\cr\tablerule 
&&    && && &&&&&&&&&&&&&&&&&&&\cr\tablerule 
&&    && && &&&&&&&&&&&&&&&&&&&\cr\tablerule
&&    && && &&&&&&&&&&&&&&&&&&&\cr\tablerule
&&    && && &&&&&&&&&&&&&&&&&&&\cr\tablerule
&&    && && &&&&&&&&&&&&&&&&&&&\cr\tablerule
&&    && && &&&&&&&&&&&&&&&&&&&\cr\tablerule
&&    && && &&&&&&&&&&&&&&&&&&&\cr\tablerule
&&    && && &&&&&&&&&&&&&&&&&&&\cr\tablerule
&&    && && &&&&&&&&&&&&&&&&&&&\cr\tablerule
}}
\vskip .25in
\hskip0.4truein\hbox{\bf Fig. #2}}
{\rm
\magicsq{2in}3
}}\hfill}\par
Now locate the square in $B$ which contains the number 1 and place a copy of 
$A$ in the corresponding square of the frame we have just constructed,
as in Fig.3 above.
We can view this as a way of counting out 9 consecutive numbers. Now locate the
square in $B$ containing 2 and in the corresponding square of Fig. 3, count
out the next 9 numbers in the same pattern. It is the same to say that one
adds 9 to all of the entries of $A$ and places the result in the box 
corresponding to the position of the 2 in $B$, as in Fig.4 below.\bigskip
\hskip-2.1truein\hbox{\vbox{
\setbox\strutbox=\hbox{\vrule height9pt depth 1.5pt width0pt}
\def\magicsq#1#2{\vbox{\centerline{
\vbox{\tabskip=0pt  \offinterlineskip
\def\tablerule{\noalign{\hrule}}
\halign to#1{\strut##& \vrule##\tabskip=0em plus2em&
 ##& \vrule##&
 ##\hfil& \vrule##&
 ##\hfil& \vrule##&
 ##\hfil& \vrule##&
 ##\hfil& \vrule##&
 ##\hfil& \vrule##&
 ##\hfil& \vrule##&
 ##\hfil& \vrule##&
 ##\hfil& \vrule##&
 ##\hfil& \vrule##&
 ##\hfil& \vrule##& 
 ##& \vrule##\tabskip=0pt\cr\tablerule
&&   8&&1&&6&&\ \ &&\ \ &&\ \ &&\ \ &&\ \ &&\ \ &&\ \ &&\ \ &&\ \ &\cr\tablerule
&&   3&&5&&7&&&&&&&&&&&&&&&&&&&\cr\tablerule
&&   4&&9&&2&&&&&&&&&&&&&&&&&&&\cr\tablerule 
&&    && && &&&&&&&&&&&&&&&&&&&\cr\tablerule 
&&   &&&&&&&&&&&&&&&&&&&&&&&\cr\tablerule
&&   &&&&&&&&&&&&&&&&&&&&&&&\cr\tablerule
&&   &&&&&&&&&&&&&&&&&&&&&&&\cr\tablerule
&&   &&&&&&&&&&&&&&&&&&&&&&&\cr\tablerule
&&   &&&&&&&&&&&&&&&&&&&&&&&\cr\tablerule
&&    && && && && && &&17&&10&&15&& && && &\cr\tablerule
&&    && && && && && &&12&&14&&16&& && && &\cr\tablerule
&&    && && && && && &&13&&18&&11&& && && &\cr\tablerule
}
}}
\vskip .2truein
\centerline{\bf Fig. #2}
}}
{\rm
\magicsq{2.3in}4
}
}\vbox{\hbox{\phantom{squeeze}}\hskip-3.1truein
\vbox{
% redefine \magicsq to make a partial 5 x 5 magic square
%\setbox\strutbox=\hbox{\vrule height18pt depth 3.5pt width0pt}
\setbox\strutbox=\hbox{\vrule height9pt depth 2pt width0pt}
\def\magicsq#1#2{\vbox{\centerline{
\vbox{\tabskip=0pt  \offinterlineskip
\def\tablerule{\noalign{\hrule}}
\halign to#1{\strut\hfil##& \vrule\hfil##\tabskip=0em plus2em&
 \hfil##& \vrule\hfil##&
 \hfil##& \vrule\hfil##&
 \hfil##& \vrule\hfil##&
 \hfil##& \vrule\hfil##&
 \hfil##& \vrule\hfil##&
 \hfil##& \vrule\hfil##&
 \hfil##& \vrule\hfil##&
 \hfil##& \vrule\hfil##&
 \hfil##& \vrule\hfil##&
 \hfil##& \vrule\hfil##&
 \hfil##& \vrule\hfil##& 
 \hfil##& \vrule\hfil##\tabskip=0pt\cr\tablerule
&&    8&& 1&& 6&&134&&127&&132&&125&&118&&123&&35&&28&&33&\cr\tablerule
&&    3&& 5&& 7&&129&&131&&133&&120&&122&&124&&30&&32&&34&\cr\tablerule
&&    4&& 9&& 2&&130&&135&&128&&121&&126&&119&&31&&36&&29&\cr\tablerule 
&&   107&&100&&105&&53&&46&&51&&62&&55&&60&&80&&73&&78&\cr\tablerule 
&&   102&&104&&106&&48&&50&&52&&57&&59&&61&&75&&77&&79&\cr\tablerule
&&   103&&108&&101&&49&&54&&47&&58&&63&&56&&76&&81&&74&\cr\tablerule
&&   71&&64&&69&&89&&82&&87&&98&&91&&96&&44&&37&&42&\cr\tablerule
&&   66&&68&&70&&84&&86&&88&&93&&95&&97&&39&&41&&43&\cr\tablerule
&&   67&&72&&65&&85&&90&&83&&94&&99&&92&&40&&45&&38&\cr\tablerule
&&   116&&109&&114&&26&&19&&24&&17&&10&&15&&143&&136&&141&\cr\tablerule
&&   111&&113&&115&&21&&23&&25&&12&&14&&16&&138&&140&&142 &\cr\tablerule
&&   112&&117&&110&&22&&27&&20&&13&&18&&11&&139&&144&&137&\cr\tablerule
}
}}
\vskip .2truein
\centerline{\bf Fig. #2}
}}
{\rm
\magicsq{3.2in}5
}
}}}\par
Next one finds the 3 of $B$ and counts out the next 9 numbers in the 
corresponding place in the frame. Continuing in this way, we eventually get
the magic square in Fig.5 above.

The product square $A\mathbin*B$ has order 12,
the product of the orders of $A$ and $B$. It is convenient to have an
analytic expression for this operation and for that purpose we can work
in somewhat greater generality. 

Let ${\cal G}$ be an abelian group and choose an element $u$ of ${\cal G}$
once and for all. Denote by ${\cal A}({\cal G})$ the set of all square arrays
of elements of ${\cal G}$, the size of the arrays being arbitrary. If
$A=(a_{ij})$, $0\leq i,j\leq m-1$,
 and $B=(b_{kl})$, $0\leq k,l \leq n-1$
 are elements of ${\cal A}({\cal G})$ of sizes
$m\times m$ and $n\times n$ respectively, their product $A\mathbin*B$
will be the $mn\times mn$ matrix $E=(e_{\alpha\beta})$ whose 
$(\alpha,\beta)$-th entry is given
by
% equation 1.2
$$e_{\alpha\beta}=m^2\cdot(b_{kl}+u)+a_{ij}\Leqno$$
where
% equation 1.3
$$ (\alpha,\beta)=m\cdot(k,l)+(i,j).\Leqno$$
Here one should note that as $i,j$ run over the integers from 0 to $m-1$ and
$k,l$ run over the integers from 0 to $n-1$, the numbers $\alpha,\beta$ 
will run over
the integers from from 0 to $mn-1$. So we have defined all the entries of 
$E$. A moment's thought will show that if we take ${\cal G}$ to be the
group ${\bf Z}$ of rational integers and $u=-1$ and if we let $A$ and $B$ be
magic squares, then we recover the operation we have defined for magic squares.
It is also easy to see that in general the $1\times1$ array $\boxit{u}$ is a
2-sided identity element for the operation $*$. Furthermore, the reader will
easily verify that the operation $*$ is associative (see for example [A-L]).
A set $S$ closed under an associative operation is called a {\bf semigroup}.
A semigroup with an identity element is
called a {\bf monoid}. We therefore have the following result.
% Lemma 1.11
\Proclaim Lemma. Let ${\cal G}$ be an abelian group.
The set ${\cal A}({\cal G})$ of all square arrays with entries
in ${\cal G}$ is a monoid with identity
element $\boxit{u}$ with respect to the operation $*$ defined by equations
(1.2) and (1.3).\par
If $S$ is a monoid with identity element $s$ with respect to an associative
operation $\circ$ then a subset $T$ of $S$ is called a {\bf submonoid} of
$S$ if $T$ contains the identity element $s$ of $S$ and is closed under
the operation $\circ$.
Returning to the original construction involving magic squares, we have
the following result, whose straightforward proof is left to the reader
(cf.[A-L]).
% Proposition 1.12
\Proclaim Proposition. Let ${\cal M}_2$ denote the set of all magic squares. 
Let 
${\cal A}({\bf Z})$ denote the monoid obtained by taking $G$ to be the additive
group of
integers and $u=-1$ in Lemma (1.4). Then ${\cal M}_2$ is a submonoid of 
${\cal A}({\bf Z})$.\par\noindent 
If we modify condition (i) of Definition 1.1 so that the entries of a magic
square of order $n$ run from 0 to $n^2-1$ instead of from 1 to $n^2$ and
modify the product $*$, as given in Figures 2-6, accordingly, then the
analytic expression for the product becomes simpler. It is the product
inherited from ${\cal A}({\bf Z})$ when we take $u=0$ instead of $u=-1$.
\par\noindent The notation ${\cal M}_2$ and ${\cal A}({\bf Z})$ introduced
in Proposition (1.5) will be retained throughout this section.
% Definition 1.22
\Proclaim Definition. {\rm Let $S$ be a monoid with identity element $s$
with respect to an operation $\circ$. We say that $S$ is
is {\bf left cancellative} if for any elements $a,b,c$ of $S$, the identity
% equation 1.23
$$a\circ b=a\circ c\Leqno$$
implies $b=c$. Similarly we say that $S$ is {\bf right cancellative}
if for any elements $a,b,c$ of $S$, the identity 
% equation 1.24
$$a\circ c=b\circ c\Leqno$$
implies $a=b$.}\par
% Lemma 1.25
\Proclaim Lemma. Let ${\cal G}$ be an abelian group and let $u$ be an element
of ${\cal G}$. Then the monoid $({\cal A}({\cal G}),*,u)$ is right 
cancellative. \par\noindent
Proof: Let $A$, $B$ and $C$ be elements of 
${\cal A}({\cal G})$. Suppose that $A$ is $m\times m$, $B$ is $n\times n$ and
$C$ is $p\times p$. If $A\mathbin*C=B\mathbin*C$ then we must have $mp=np$
and therefore $m=n$. We therefore have
% equation 1.26
$$ m^2\cdot(c_{kl}-1)+a_{ij}=m^2\cdot(c_{kl}-1)+b_{ij}\Leqno$$
for $0\leq i,j\leq m-1$ and $0\leq k,l\leq n-1$, which implies that
$a_{ij}=b_{ij}$ for all $i,j$. Therefore $A\mathbin*C=B\mathbin*C$ implies
$A=B$ and we are done.
% Lemma 1.27
\Proclaim Lemma. Let ${\cal G}$ be an abelian group and let $u$ be an element
of ${\cal G}$. Then the monoid $({\cal A}({\cal G}),*,u)$ is left cancellative 
if and only if the group ${\cal G}$ is torsion-free.\par\noindent
Proof:  Let $A,B,C$ be elements of ${\cal A}({\cal G})$ and suppose that $A$ 
is $m\times m$, $B$ is $n\times n$ and $C$ is $p\times p$. If 
$A\mathbin*B=A\mathbin*C$ then we must have $mn=mp$ and therefore $n=p$. 
It follows that
% equation 1.28
$$m^2\cdot(b_{kl}-1)+a_{ij}=m^2\cdot(c_{kl}-1)+a_{ij}\Leqno$$
for $0\leq i,j\leq m-1$ and $0\leq k,l \leq n-1$, which implies that
$m^2\cdot (b_{kl}-c_{kl})=0$ for all $k,l$. Since the group $G$ is torsion
free, it follows that $b_{kl}=c_{kl}$ for all $k,l$. Therefore, the equation
$A\mathbin*B=A\mathbin*C$ implies $B=C$ and we are done.
% Definition 1.29
\Proclaim Definition. {\rm Let $S$ be a monoid with identity element $s$ with
respect to the law of composition $\circ$. Let $T$ be a submonoid of $S$.
We say that $T$ is {\bf persuasive} in $S$ if the following condition is
satisfied: for any elements $a,b,c$ of $S$ such that $a\circ b=c$, if
two of the elements $a,b,c$ belong to $T$ then so does the third.
We say that $T$ is {\bf inescapable} in $S$ if whenever the identity
$a\circ b=c$ holds in $S$ with $c$ in $T$, the elements $a,b$ must also
belong to $T$. It is easy to see that an inescapable submonoid is 
persuasive.}\par
% Lemma 1.30
\Proclaim Lemma. The submonoid ${\cal M}_2$ of ${\cal A}({\bf Z})$
is persuasive. The arrays which satisfy conditions (ii),(iii) of
Definition 1.1 form an inescapable submonoid of ${\cal A}({\bf Z})$
\par\noindent
Proof: Let $A,B,C$ be elements of ${\cal A}({\bf Z})$, where $A$ is 
$m\times m$, $B$ is $n\times n$ and $C$ is $mn\times mn$, and suppose that
$A\mathbin*B=C$. Then we have
% equation 1.31
$$\matrix{
\sum\limits_{r=0}^{mn-1}c_{rs}&=&\sum\limits_{i=0}^{m-1}\sum\limits_{k=0}^{n-1}
\left(m^2\cdot(b_{kl}-1)+a_{ij}\right)\cr
&&\cr
&&\cr
&=&\left(m^3\sum\limits_{k=0}^{n-1}b_{kl}\right)-m^3n+\left(n\sum\limits_{i=0}^{m-1}a_{ij}\right),\cr}\Leqno$$
where $(r,s)=m\cdot(k,l)+(i,j)$. If in the above equations, we replace
the row or column of $A$ we are using by another and keep the row or
column of $B$ fixed, we do not change the value of the expression.
That is because $C$ belongs to ${\cal M}_2({\bf Z})$ and all we are
doing is choosing another row or column of $C$. Since we have kept the
row or column of $B$ fixed, its contribution is also unchanged. It follows
that the sum over the row or column of $A$ is likewise unchanged, which
proves that $A$ satisfies conditions (ii) and (iii) of Definition 1.1.
Similarly, if one fixes the row or column of $A$ and varies the row
or column of $B$, one proves that $B$ satisfies conditions (ii) and (iii)
of Definition 1.1. This proves that arrays satisfying conditions (ii) and
(iii) form an inescapable submonoid. As for condition (iv), the sum 
$\sum c_{ii}$ over the main diagonal of $C$ is $m^2\sum (b_{jj}-1)$ plus
$\sum a_{kk}$. If we change to the other diagonal, the sum for $C$ doesn't
change and therefore it doesn't change for $A$ if and only if it doesn't
change for $B$. This shows that condition (iv) is persuasive. 
It remains to verify condition (i) of Definition (1.1). If $A$ and $B$ are
magic squares then by Proposition (1.5) so is $C$. Therefore if two of 
$A,B,C$ are magic squares then $C$ is a magic square and therefore satisfies 
condition
(i) of Definition (1.1). We can write
% equation 1.31
$$c_{rs}=m^2\cdot(b_{kl}-1)+a_{ij}\Leqno$$
where
% equation 1.32
$$(r,s)=m\cdot(k,l)+(i,j)\Leqno$$
and $0\leq i,j\leq m-1$ and $0\leq k,l\leq n-1$. If $A$ satisfies condition
(i) of Definition (1.1) then in equation (1.16) the elements $a_{ij}$ are
determined by the condition that $c_{rs}-a_{ij}$ must be divisible by
$m^2$ and therefore we have
% equation 1.33
$$b_{kl}=\left[{c_{rs}+m^2-1\over m^2}\right]\Leqno$$
which implies that
$$1\leq b_{kl}\leq n^2.\Leqno$$
Since the $(mn)^2$ numbers $c_{rs}$ in equation (1.16) must be distinct,
the same must be true of the $n^2$ numbers $b_{kl}$. Therefore, $B$
satisfies condition (i) of Definition (1.1). Conversely, suppose that $B$
satisfies condition (i) of Definition (1.1). By equation (1.16), the $m^2$
elements $a_{ij}$ must be a full set of representatives for the residue
classes modulo $m^2$ and for each $c_{rs}$ there is a unique $a_{ij}$ such
that $c_{rs}-a_{ij}$ is divisible by $m^2$. If some $a_{ij}$ is less than 1,
choose $(r,s)$ such that $c_{rs}-a_{ij}$ is divisible by $m^2$ and such
that $c_{rs}$ is as large as possible. Then we have
% equation 1.34
$$c_{rs}-a_{ij}\geq c_{rs}\geq (mn)^2\Leqno$$
which by equation (1.16) implies that
% equation 1.35
$$b_{kl}\geq n^2+1,\Leqno$$
contradicting the fact that $B$ satisfies condition (i) of Definition (1.1).
Therefore we must have $a_{ij}\geq1$ for all $i,j$. A similar argument,
which we leave to the reader, shows that $a_{ij}\leq m^2$ for all $i,j$.
Since the $a_{ij}$ represent all of the $m^2$ residue classes modulo $m^2$,
it follows that $A$ satisfies condition (i) of Definition (1.1). This
proves the lemma.
\Proclaim Remark. {\rm I don't know whether ${\cal M}_2$ is an
inescapable submonoid of ${\cal A}({\bf Z})$.}\par\noindent

The following definition is motivated by the corresponding notions from
elementary number theory.
% Definition 1.36
\Proclaim Definition. {\rm Let $S$ be a monoid with identity element $s$ with
respect to an operation $\circ$. We will say that element $t$ of $S$ is
{\bf irreducible} if $t\neq s$ and if for any elements $u,v$ of $S$ such 
that $t=u\circ v$
we have either $u=s$ or $v=s$. An element $t$ of $S$ is said to be 
{\bf left prime} if for any elements $u,v,w$ of $S$ such that $t\circ u
=v\circ w$, there is an element $x$ of $S$ such that $v=t\circ x$.
This definition of irreducibility is more restrictive than the one generally
accepted. However since the monoids in which we wish to study irreducible
elements have no units other than the identity element, our definition will in 
practice coincide with the usual one.}\par

The following lemma follows by an obvious induction on the order of the
order of the magic square.
% Lemma 1.38
\Proclaim Lemma. The set of all irreducible magic squares generates the
monoid ${\cal M}_2$ of all magic squares. In other words, every magic
square can be written as a product of irreducible magic squares.\par\noindent
In order to prove that the monoid ${\cal M}_2$ is free, we only have to
prove that every magic square is {\it uniquely} a product of magic squares.
Indeed, a monoid is free if and only if it is freely generated by its
irreducible elements. That this is the case for ${\cal M}_2$ follows,
in analogy with the proof of the fundamental theorem of arithmetic,
by a simple induction as soon as one has proved that an irreducible
magic square is left prime. Therefore, the freeness of ${\cal M}_2$ is a
consequence of the following simple lemma. 
% Lemma 1.44
\Proclaim Lemma. {\rm\bf (Irreducible Implies Prime)} Let $A,B,C,D$ be
magic squares such that $A*B=C*D$.
 Suppose $A$ is $m\times m$, $B$ is $n\times n$, $C$ is 
$p\times p$ where $m\leq p$. If $A$ is irreducible then there is a
magic square $E$ such that $C=A\mathbin*E$.\par\noindent
Proof: We will first show that $m$ must divide $p$. Here we prefer to give
a geometric argument since the key insight is geometrical and is lost in
analytic formulations of the proof. We will therefore refer to the geometric
description of the operation on magic squares, as illustrated in Figures 2-6.
Let $F=A\mathbin*B=C\mathbin*D$. If we view $F$ is the product $C\mathbin*D$
then $F$ is partitioned into $p\times p$ squares. Let $C_1$ denote the
$p\times p$ square which contains the number 1 in $F$. Then $C_1$ is an
exact replica of the magic square $C$ and in particular contains all
of the entries of $F$ which are $\geq1$ and $\leq p^2$ and none of the
entries of $F$ which are $>p^2$. If we now view $F$ as the product
$A\mathbin*B$ then $F$ is instead partitioned into $m\times m$ squares.
The first such square contains the numbers from 1 to $m^2$ and is denoted
$A_1$. The second such square contains the numbers from $m^2+1$ to $2m^2$
and is denoted $A_2$ and so on. If $m$ does not divide $p$ then in particular
$m<p$ and $m^2$ does not divide $p^2$. Let $h$ be the smallest integer such
that $h\cdot m^2\geq p^2$. Then the square $A_h$ lies partly in the square 
$C_1$
and partly outside of it. Let $g>0$ be an integer. If $g<h$ then the square 
$A_g$ lies entirely inside $C_1$ and if $g>h$ then $A_g$ lies entirely outside
of $C_1$. Since $A_h$ is $m\times n$ and $C_1$ is $p\times p$ with
$m<p$, there must be a column of $C_1$ which does not meet
the square $A_h$. Since that column must be covered by disjoint squares
$A_g$ lying entirely inside of $C_1$, it follows that $m$ does divide $p$
after all. Retaining the definition of $h$, we see that $h\cdot m^2=p^2$
and therefore $h=t^2$ for some positive integer $t$. The square $C_1$ is
therefore covered by squares $A_g$ and is therefore of the form $A\mathbin*E$
where $E$ is a $t\times t$ array of nonnegative integers. Since $A$ and $C$
are magic squares, it now follows from Lemma (1.14) that $E$ is a magic
square and we are done.
% Theorem 1.45
\Proclaim Theorem. The monoid ${\cal M}_2$ of all magic squares is freely
generated by the set of all irreducible magic squares.\par\noindent
Proof: In view of the Irreducibility Implies Prime Lemma and the remarks
preceding it, the proof may safely be entrusted to the reader.
\s{Generalizations}
In the preceding section, we have stated our result in the case of magic
squares because of the highly geometric and visual nature of the operation
and proof and the comparatively simple notation in that case. One can also
define {\bf magic $N$-dimensional cubes} (also known as {\bf magic $N$-cubes})
in a similar way to that given in Definition 1.1 and the reader is invited
to provide the definition herself, the only caution being that the condition
(iv) on the sum over the diagonals is generalized in $N$ dimensions to the
sum over the {\it longest} diagonals, the ones joining opposite corners of
the $N$-cube. Similarly one can, with only minor changes, define the operation
$*$ on the set ${\cal M}_N$ of all magic $N$-cubes. The same arguments then 
apply with only minor modifications to prove that the set ${\cal M}_N$
of all magic 
$N$-cubes forms a free monoid. We omit the details of this generalization, 
whose only complications are notational.

One can in fact generalize much further with no
essential difficulty. Since some of these generalizations might be of
independent interest, and since to some extent they facilitate the
discussion of \S3, we will indicate some of them briefly,
leaving the verification of the elementary details to the reader.
\par\noindent
\item{(1)} Let $S,H$ be monoids and suppose that $H$ is an $S$-module, i.e.
we have a homomorphism from $S$ into the monoid of endomorphisms of the
monoid $S$. Denote by ${\cal F}(S,H)$ the set of all pairs $(F,\sigma)$
consisting of a finite subset $F$ of $H$ and an element $\sigma$ of $S$.
The pair $(F,\sigma)$ is called a {\bf frame}, $F$ is called the {\bf turf}
of the frame and $\sigma$ is called the {\bf nominal order} of the frame.
Then ${\cal F}(S,H)$ forms a monoid  in which the product of $(F_1,\sigma_1)$
and $(F_2,\sigma_2)$ is the frame $(F_3,\sigma_1\sigma_2)$, where $F_3$
is the set of all elements of the form $\sigma_1(f_2)f_1$ with $f_1\in F_1$
and $f_2\in F_2$. If we view a magic $N$-cube as an array, hence a function
defined on a certain cubical subset of ${\bf Z}^N$, this definition allows
us to consider domains which instead are finite subsets $F$ of arbitrary
monoids $H$. It should be clear that the monoid ${\cal F}(S,H)$ is simply
the semidirect product of $S$ and the monoid ${\cal P}(H)$ of finite subsets
of $H$.
\item{(2)} Let $S,H$ be as in (1) and let $G$ be a commutative $S$-module.
Let $u$ be an element of $G$ which will be fixed throughout the discussion.
If $(F,\sigma)\in{\cal F}(S,H)$, then by an {\bf array of type $(F,\sigma)$
with entries in $G$}, we mean a triple $(F,\sigma,\lambda)$ where $\lambda$
is a function from $F$ to $G$. If we do not wish to specify the type
$(F,\sigma)$ of the array, we will refer to it as an {\bf array framed
in ${\cal F}(S,H)$}. We call $\sigma$ the {\bf order} of the array
$(F,\sigma,\lambda)$. We denote by ${\cal A}(S,H,G)$ the set of all arrays
framed in ${\cal F}(S,H)$. Then ${\cal A}(S,H,G)$ is naturally a semigroup
in which the product of $(F_1,\sigma_1,\lambda_1)$ and
$(F_2,\sigma_2,\lambda_2)$ is the array $(F_3,\sigma_3,\lambda_3)$,
where $(F_3,\sigma_3)$ is the product of $(F_1,\sigma_1)$ and
$(F_2,\sigma_2)$ in ${\cal F}(S,H)$ and where the function $\lambda_3$
is defined by
$$\lambda_3(h_3)={\sum}'(\sigma_1(\lambda_2(h_2)+u)+\lambda_1(h_1))$$
where the summation runs over all elements $(h_1,h_2)\in F_1\times F_2$
such that $h_3=\sigma_1(h_2)h_1$. If $g$ is an element of $G$, 
we will denote by $\boxit{g}$ the framed array 
$\boxit{g}=(\{e\},s,\lambda:\{e\}\rightarrow G)$ defined by 
$\lambda(e)=g$,
where $e$ is the identity element of $H$. This notation generalizes and
supersedes the notation $\boxit{u}$ introduced in \S1. If $g$ is taken
to be an element of $G$ such that $g+u=1_G$, where $1_G$ is the identity
element of $G$, then ${\cal A}(S,H,G)$ is actually a monoid whose identity
element is $\boxit{g}$. Assume that such an element $g$ exists. The monoid 
${\cal A}(S,H,G)$ may then be described as the semidirect product of $S$
with the monoid ${\cal P}(H;G)$ of all finite partial functions from
$H$ to $G$, this being the kernel of the homomophism $(F,\sigma,\lambda)
\mapsto\sigma$ of ${\cal A}(S,H,G)$ onto $S$.
\item{(3)} In order to introduce the notion of a magic frame, we have to
be able to specify the summation conditions.
To do so in our present generality, we observe that the 
summation conditions traditionally considered in the study of magic $N$-cubes
also form a monoid and use that insight in order to formulate our
generalization. More precisely, let $S,H,G$ be as in (1) and let $n$ be a 
positive integer. We introduce the following monoids.
$$\matrix{
M_1^{(1)}&=&{\cal F}(S,H)&\phantom{gleep}&M_1^{(n)}&=&\prod\limits_{i=1}^nM_1^{(1)}\cr
&&\cr
M_2^{(1)}&=&{\cal F}(S,{\cal P}(H))&\phantom{gleep}&
M_2^{(n)}&=&\prod\limits_{i=1}^nM_2^{(1)}.\cr}$$
In the definition of $M_1^{(n)}$ and $M_2^{(n)}$ we use the product notation
to denote the fibre product over $S$, not the cartesian product.
In practice, an element of $M_1^{(1)}$ will be a subset of $H$ over which
we wish to sum the entries of an array. In the case of magic $N$-cubes, such
a subset might be an orthogonal or a great diagonal of the $N$-cubes,
but one considers other subsets as well. In general, we will want to sum over 
various subsets
and we wish to handle all such subsets simultaneously. For example, 
if $N$ is fixed and $H={\bf Z}^N$, we can let 
$n=2^{N-1}$ and view the set of $n$ great diagonals of the $N$-cube as an 
element of $M_1^{(n)}$. In that case, if $(F_i,\sigma_i)$ is an $N$-cube
for $i=1,2$ and if $A_i$ is the element of $M_1^{(n)}$ corresponding to
the set of great diagonals of $(F_i,\sigma_i)$, then the product in $M_1^{(n)}$
of $A_1$ and $A_2$ will be the set of great diagonals of the product of
$(F_1,\sigma_1)$ and $(F_2,\sigma_2)$. This is quite satisfying but does
not suffice to describe the orthogonals. The reason is that the number
of orthogonals of an $N$-cube depends on the size of the $N$-cube and not
just on $N$. For this reason we introduced the monoid $M_2^{(1)}$. The
orthogonals parallel to a particular axis of an $N$-cube form an element of
$M_2^{(1)}$. If for $i=1,2$, $B_i$ is the element of $M_2^{(1)}$ corresponding 
to the orthogonals of an $N$-cube $(F_i,\sigma_i)$ parallel to a particular 
axis (the same axis for $i=1,2$), then the product in $M_2^{(1)}$ of $B_1$
and $B_2$ is the element corresponding to the set of orthogonals in that
same direction in the product of $(F_1,\sigma_1)$ and $(F_2,\sigma_2)$.
This is also satisfying, but since there are $N$ directions in which to 
choose the orthogonals, the system of all orthogonals is described by an
element of $M_2^{(N)}$. Thus, the set of summation conditions which a 
magic $N$-cube is expected to satisfy is an element of the monoid
$$M_1^{(n)}\times M_2^{(N)},$$
where $n=2^{N-1}$. In order to simplify the notation, we observe that the
monoid $M_1^{(1)}$ may be embedded in the monoid $M_2^{(1)}$ by associating
to a subset $F$ of $H$ the singleton $\{F\}$. Therefore, we may regard the
set of summation conditions as an element of $M_2^{(d)}$ where $d=N+2^{N-1}$.
These considerations motivate the following definition:
let $(F,\sigma,\lambda)$ be an element of
${\cal A}(S,H,G)$, let $d$ be a positive integer and let $C=(C_1,\dots,C_d)$ be
an element of $M_2^{(d)}$ such that for $1\leq i\leq d$, every element of $C_i$
is a subset of $F$. We say that the array $(F,\sigma,\lambda)$ is 
{\bf ${\bf C}$-stochastic} if for $1\leq i\leq d$ and for $E$ in $C_i$, the 
summation $\sum_{h\in E}\lambda(h)$ is independent of $i$ and $E$. 
Suppose that $\phi$ is a homomorphism from
a submonoid ${\cal F}$ of the monoid ${\cal F}(S,H)$ to the monoid $M_2^{(d)}$
such that for every $(F,\sigma,\lambda)$ in ${\cal F}$ the order of 
$\phi(F,\sigma,\lambda)$ is $\sigma$.
We will say that an array $(F,\sigma,\lambda)$ is {\bf ${\bf phi}$-stochastic}
if $(F,\sigma)$ belongs to ${\cal F}$ and if $(F,\sigma,\lambda)$ is
$\phi(F,\sigma)$-stochastic. If $\cal F$ and $\phi$ are such,
we denote by ${\cal M}(\phi)$ the subset of ${\cal A}(S,H,G)$ consisting
of all $\phi$-stochastic arrays. Then ${\cal M}(\phi)$ is a submonoid of 
${\cal A}(S,H,G)$. Now suppose that the only pair $(\sigma,g)\in S\times G$ 
with $\sigma(g)=1_G$ is $(1_S,1_G)$, where $1_S$ is the identity element
of $S$. Then then the monoid ${\cal M}(\phi)$ is a persuasive submonoid
of ${\cal A}(S,H,G)$.

By placing some restrictions on the monoid $G$, one can generalize
further the notions of stochasticity introduced in (3). 
More precisely, one assumes that $G$ is the additive group of an
algebraically closed field $k$. In that case, one can introduce
monoids of summation conditions by means of certain classes of
infinite dimensional
varieties over $k$ and prove that they are free. Space does not permit the 
detailed description of this general construction 
and we presently know only one interesting example of it, namely in
connection with the monoids ${\cal S}_N(t,g)$ defined in the next section.
So we will confine our attention to that example, which we describe in the
next section without reference to infinite dimensional varieties.

\s{Magic $N$-cubes with remarkable power sum properties}
Let $n\in[1,N]$. An $n$-cube contained in an $N$-dimensional cubical array
$f$ of order $d$
is then the intersection of that array with a certain number of hyperplanes
parallel to the faces of the array. As such it is completely described
by selecting a subset $J$ of cardinality $N-n$ of $[1,N]$ and prescribing
a function $h:J\rightarrow[0,d-1]$. Two $n$-cubes having the same set $J$
are said to be {\bf conjugate}.

Let $g:[1,N]\rightarrow {\bf Z}$
be any nonnegative function and let $t>1$ be an integer. 
We denote by ${\cal S}_N(t,g)$
the set of all magic $N$-cubes $A$ satisfying  the following two conditions:
\item{(a)}The order of $A$ is a power of $t$, say $t^s$.
\item{(b)}For $1\leq n\leq N$, conjugate $n$-dimensional subcubes have
equal sums of the $r$-th powers of their entries, for $0\leq r\leq g(n)$.

We also denote by ${\cal M}_N(t)$ the monoid of all magic $N$-cubes satisfying
only condition (a) above. In the proof, we will use notation introduced
in (1) and (2) of \S2.
\Proclaim Theorem. Let $t,g$ be as above. Then ${\cal S}_N(t,g)$ is an
inescapable submonoid
of the monoid ${\cal M}_N$ of all magic $N$-cubes. Furthermore, 
${\cal S}_N(t,g)$ is a free monoid on infinitely many generators.\par\noindent
Proof: Let $A,B$ be elements of ${\cal S}_N(t,g)$ and let $C$ be
their product. Any $n$-dimensional subcube $K$ of $C$ is the product in 
${\cal F}({\bf Z},{\bf Z}^N)$ of the corresponding subcubes $K',K''$ of $A$ and
$B$ respectively. Normally we think of the $n$-subcubes as subsets 
of $A,B$ and $C$ but we can view them as elements of 
${\cal F}({\bf Z},{\bf Z}^N)$ by replacing, e.g. $K'$ by the pair $(K',t^s)$,
where $t_s$ is the order of $C$. Here the action of ${\bf Z}$ on ${\bf Z}^N$
is the usual ${\bf Z}$-module structure on ${\bf Z}^N$.
Since  we care about the {\it entries} of these $n$-cubes, we are concerned
with the elements of ${\cal A}(S,{\bf Z}^N,G)$, where $S,G$ are 
${\bf Z}$ under multiplication and addition respectively and where $G$ is 
viewed as an $S$-module via $(s,g)\mapsto s^Ng$. Denote by $f,f',f''$
the functions that give the entries of the subcubes 
$K,K',K''$ respectively. We then have
$$\eqalign{\sum_{v\in K}f(v)^r&=
\sum_{v'\in K'}\sum_{v''\in K''}(t^{sN}f''(v'')+f'(v'))^r\cr
&=\sum_{j=0}^r\left({r\atop j}\right)
\left(\sum_{v'\in K'}f'(v')^{r-j}\right)\cdot
\left(\sum_{v''\in K''}t^{jsN}f''(v'')^j\right)\cr}\Leqno$$
Since the sums over $K'$ and $K''$ are independent of the choices of the
$n$-cubes $K',K''$ within their conjugacy classes, 
the sum over $K$ is independent of the choice of the $n$-cube $K$
within its conjugacy class. This proves that ${\cal S}_N(t,g)$ is a submonoid 
of ${\cal M}_N$. Furthermore, the submonoid ${\cal S}_N(t,g)$ is
an inescapable submonoid of ${\cal M}_N$. Indeed, suppose $A,B,C$ are magic
$N$-cubes such that $A*B=C$ and suppose that $C$ belongs to ${\cal S}_N(t,g)$.
Choose $K,K',K''$ as before and use the above
identity for the sum of the $r$-th powers of the entries of $K$,
which we know is independent of the choice of the $n$-cube $K$
within its conjugacy class. If we fix the $n$-cube $K''$ and allow the
$n$-cube $K'$ to vary in its conjugacy class, it follows by an easy induction 
on $r$ that, for $1\leq r\leq g(n)$,
the sum of the $r$-th
powers of the entries of the entries of $K'$ are independent of the
choice of $K'$ within its conjugacy class. Similarly, by fixing $K'$
and letting $K''$ vary in its conjugacy class we see that
the sum of the $r$-th powers of the 
entries of $K''$ is independent of the choice of the $n$-cube $K''$
within its conjugacy class. This proves that ${\cal S}_N(t,g)$ is
an inescapable submonoid of ${\cal M}_N$. It follows that
an irreducible element of ${\cal S}_N(t,g)$ is also irreducible when
viewed as an element of ${\cal M}_N$. 
To see that ${\cal S}_N(t,g)$ is free, it suffices to prove the analogue
of the Irreducibility Implies Prime Lemma for this situation. So suppose
that $A,B,C,D$ are elements of ${\cal S}_N(t,g)$ of orders $m,n,p,q$
where $m\leq p$, and suppose that $A*B=C*D$ with $A$ irreducible.
In the proof of the
freeness of ${\cal M}_N$, which we left to the reader, one proves along
the way that ${\cal M}_N$ satisfies the analogue of the Irreducibility
Implies Prime Lemma. Since $A$ is also irreducible as a magic $N$-cube,
it follows that we can write $C$ as the product $A*E$ of $A$ and
another magic $N$-cube $E$. It then follows that $E$ belongs to 
${\cal S}_N(t,g)$. Therefore, ${\cal S}_N(t,g)$ is free. Finally,
to see that ${\cal S}_N(t,g)$ has infinitely many generators, we will
show that it contains infinitely many irreducible magic $N$-cubes. 
In Theorem 1 on p.620 of [A-L], it is shown how to construct a magic
$N$-cube of order $t^M$ for every positive integer $M$ such that
$t$ divides $MN$ and such that $M>1$ when $t$ is even. We will call that 
construction the {\bf method of Prouhet sequences}.
Hence infinitely many elements of ${\cal M}_N(t)$
are constructed by the method of Prouhet sequences.
Furthermore, in Theorem 5 on p.626 of [A-L]
it is shown that for  $1\leq n\leq N$ every magic $N$-cube $A$ constructed in
this way has the following property:
for all $n$, conjugate $n$-dimensional subcubes of $A$ have equal sums of
 the $k$-th powers of their entries for $k=0,1,\dots, nr-1$. It follows that
all but finitely many of the magic $N$-cubes in ${\cal M}_N(t)$ constructed
by the method of Prouhet sequences belong to ${\cal S}(t,g)$. Therefore we will
be done as soon as prove the following lemma.
% Lemma 7.5
\Proclaim Lemma. If  $M+1$ is divisible by $t$ then 
the magic $N$-cube of order $t^M$ constructed by the method of 
Prouhet sequences is irreducible.
\par\noindent
Proof: We refer the reader to Theorem 1 of [A-L] or to [A1], pp.16-18,
for details of the method
of Prouhet sequences. 
Let $A=(a_{i_1,\dots,i_N})$ be a magic $N$-cube of order $t^M$
constructed by the method of Prouhet sequences. Let
$(i)=(i_1,\dots,i_N)$ and $(j)=(j_1,\dots,j_N)$ be the
$N$-tuples such that $a_{(i)}=0$, $a_{(j)}=1$. 
Then we have $(i)=(0,\dots,0)$ and $(j)=(t^M-1,\dots,t^M-1,0)$.
It follows that the entries $0$ and $1$ are on opposite faces of the
$N$-cube and cannot lie in any proper subcube of dimension $N$.
Therefore $A$ is irreducible.\par\noindent

\s{Divergence of generating functions}
It is well known that the freeness of these monoids implies
identities of formal Dirichlet series of the form
$$\sum_{n=1}^\infty \displaystyle{a_n\over n^s}
=\left({1-\sum_{n=3}^\infty \displaystyle{b_n\over n^s}}\right)^{-1}$$
where $a_n$ denotes the number of elements of size $n$ in the monoid
and $b_n$ denotes the number of irreducible elements of size $n$
 of the monoid. We will now consider the convergence properties of
these formal Dirichlet series and their associated power series.
% Theorem 8.2
\Proclaim Theorem. Let $N$, $a_n$ (resp. $b_n$) be the number of
magic $N$-cubes (resp. irreducible magic $N$-cubes) of order $n$. Then the 
formal power series 
$$\sum_{n=1}^\infty a_nt^n,\ \sum_{n=3}^\infty b_nt^n$$
both have radius of convergence equal to 0. In particular, the formal Dirichlet
series introduced above are everywhere divergent for this monoid.\par\noindent
Proof: Let $X=(X_{k\ell})$,
 $0\le k,\ell\le N-1$ denote the $N\times N$ matrix 
whose
$(k,\ell)$-th entry is $1+N\delta_{k\ell}$. If $M$ is a positive integer 
relatively prime to $(2N-1)!$ and
if we view $X$ as having entries in the ring ${\bf Z}/M{\bf Z}$ then
$X$ satisfies the following conditions:
(1) $X$ is invertible; (2) Every entry of $X$ is a unit of
${\bf Z}/M{\bf Z}$; (3) If $e$ is a vector whose entries are
all $\pm1$ then the entries 
of the vector $Xe$ are units of ${\bf Z}/M{\bf Z}$.\par
 For every such $M$, we can construct a magic $N$-cube 
$${}_MA=({}_MA_{i_1,\dots,i_N})$$ of order $M$ whose 
entries are defined by
$${}_MA_{i_1,\dots,i_N}=\sum_{k=0}^{N-1}j_kM^k,$$
where the digits $j_k$ are determined by the conditions $0\le j_k\le M-1$
and
$$j_k\equiv \sum_{k=0}^{N-1}X_{k\ell}i_\ell\ ({\rm mod}\ M).$$
For a discussion of the ideas behind this construction, see \S1 of the
author's paper [A1]. We will merely remark that: 
%\halign{\quad#\hfil&\quad#\hfil\cr
%(I)&condition (1) guarantees that each of the numbers $0,1,\dots,M^N-1$ occurs\cr
%& exactly once in ${}_MA$;\cr
%(II)&condition (2) guarantees that for any fixed $k$, the $k$-th digits  of\cr
%&the entries of any orthogonal of ${}_MA$ are $0,1,\dots,M-1$ in some order;\cr
%(III)&condition (3) guarantees that for any fixed $k$, the $k$-th digits of\cr
%& the entries of any diagonal of ${}_MA$ are $0,1,\dots,M-1$ in some order.\cr}
\itemitem{(I)}condition (1) guarantees that each of the numbers 
$0,1,\dots,M^N-1$ occurs exactly once in ${}_MA$;
\itemitem{(II)}condition (2) guarantees that for any fixed $k$, the $k$-th 
digits  of the entries of any orthogonal of ${}_MA$ are $0,1,\dots,M-1$ in some
order;
\itemitem{(III)}condition (3) guarantees that for any fixed $k$, the $k$-th 
digits of the entries of any diagonal of ${}_MA$ are $0,1,\dots,M-1$ in some 
order.\par\noindent
Peter Sz\"usz pointed out to the author that the observations (I),(II),(III) 
imply that 
if $\sigma$ is any permutation of the numbers $0,1,2,\dots M-1$ then the
array ${}_MA^\sigma$ defined by
$${}_MA^\sigma_{i_1,\dots,i_N}=\sigma({}_MA_{i_1,\dots,i_N})$$
is also a magic $N$-cube. Therefore $a_M\ge M!$ whenever $M$ is relatively
prime to $(2N-1)!$, and that already suffices to prove that the summation
$\sum_{n=1}^\infty a_nt^n$ diverges for $t\neq0$. To show that 
$\sum_{n=3}^\infty b_nt^n$
also diverges, we first prove that the magic $N$-cube ${}_MA$ is irreducible.
Indeed, the number $0$ is placed in the center of ${}_MA$ and the number $1$
is placed in the box indexed by $(i)=(i_0,\dots,i_{N-1})$ where $(i)$ satisfies
$0\leq i_k\leq M-1$ and
$$\sum_{\ell=0}^{N-1}(1+N\delta_{k\ell})\cdot i_k\equiv \delta_{0k}\ 
({\rm mod}\ M)\Leqno$$
for $0\leq k\leq N-1$.
It follows that $i_1=i_2=\dots=i_{N-1}$ and 
$$\eqalign{
(N+1)\cdot i_0+(N-1)\cdot i_1&\equiv 1\ ({\rm mod}\ M)\cr
i_0+(2N-1)\cdot i_1&\equiv 0\ ({\rm mod}\ M).\cr}$$
Solving these equations modulo $M$ we obtain
$$i_0\equiv{2N-1\over 2N^2}\ ({\rm mod}\ M)\ {\rm and}\ 
i_1\equiv{-1\over2N^2}\ ({\rm mod}\ M).$$
In particular, since $M$ is relatively prime to $(2N-1)!$, both $i_0$ and 
$i_1$ are units modulo $M$.
If ${}_MA$ were reducible, we could tile it with subcubes of order $R$ with
$0$ and $1$ lying in a common $N$-cube $B$
of 
order $R$. Here $R<M$ is a factor of $M$. The $N$ cube $B$ would have to
contain all of the nonnegative integers less than $R^N$. For $0\leq\nu\leq 
R-1$, the integer $\nu$ occurs at the point $\nu\cdot(i)$. Therefore,
for $0\le \nu\le R-1$ the $N$-cube $B$ would contain the multiple 
$\nu\cdot(i)$ of $(i)$.
% For each such $\nu$ we could then
%find an integer $j_\nu$ such that $0\leq j_\nu\leq R-1$ and such that
%$$\nu\cdot a_0\equiv j_\nu\ ({\rm mod}\ M).$$
In particular, for $0\leq\nu\leq R-1$, both $\nu i_0$ and $\nu i_1$ must
lie among the elements $0,1,\dots,R-1$ of ${\bf Z }/M{\bf Z}$. 
Since $i_0$ and $i_1$ are units modulo $M$, they cannot be divisible by $R$. 
Therefore the only way all the multiples $\nu i_0$ and $\nu i_1$
with $0\leq\nu\leq R-1$ can be
$<R$ is if $i_0=i_1=1$. But then
$1/N\equiv i_0-i_1\equiv0$ modulo $M$, which is impossible.
Therefore $0$ and $(i)$ cannot
lie in a common subcube, which proves that ${}_MA$ is irreducible. 
Furthermore, the proof that ${}_MA$ is irreducible depends only on the 
locations
of the numbers $0,1,\dots,D-1$, where $D$ is the largest divisor of $M$
other than $M$ itself. Therefore any permutation $\sigma$ that fixes
these numbers
will map ${}_MA$ to an irreducible magic $N$-cube ${}_MA^\sigma$.
Since $M$ is relatively prime to $(2N-1)$, we have $D=M/p$, where $p$
is the smallest prime dividing $M$ and $p\geq(2N+1)$. Therefore, we have
$$b_M\ge (M-D)!>\Gamma\left({2MN\over2N+1}\right)\geq\Gamma(4M/5)$$
which implies that the series $\sum_{n=3}^\infty b_nt^n$ diverges for $t\neq0$.

We have a similar result for the monoids ${\cal S}_N(t,g)$, but unfortunately
somewhat weaker: we can only prove that the Dirichlet series diverge.
% Theorem 8.3
\Proclaim Theorem. Let $N>1$ be an integer and let $g:[1,N]\rightarrow S$
be any function.
For every positive integer $n$ denote by $c_n$ (resp. $d_n$) the number
of magic $N$-cubes (resp. irreducible magic $N$-cubes) of order $n$ 
belonging to the monoid $S_N(t,g)$. Then the Dirichlet series
$$\sum_{n=1}^\infty {c_n\over n^s},\ \sum_{n=1}^\infty {d_n\over n^s} $$
diverge for every complex number $s$.\par\noindent
Proof: Let $n$ run over all positive integers $M$ such that $t$ divides
$MN$ and such that $M>1$ if $t$ is even. If $\sigma$ is a permutation on
the integers $0,1,\dots,M-1$ and if $x$ is an integer such that
$0\leq x\leq t^M-1$, then we write $\sigma(x)$ to denote the 
integer obtained by permuting the digits of $x$ in the base $t$ according
to the permutation $\sigma$. If $(i)=(i_1,\dots,i_N)$ is an $N$-tuple
of integers such that $0\leq i_\nu\leq t^M-1$ for $1\leq \nu\leq N$
then we write $\sigma\cdot(i)$ to denote the $N$-tuple obtained from
$(i)$ by replacing $i_N$ by $\sigma(i_N)$. Let $A=(a_{i_1,\dots,i_N})$
be the magic $N$-cube of order $t^M$ constructed by the method of
Prouhet sequences (cf. [A-L],Theorem 1). As noted in the proof of Theorem
3.1, for all but at most finitely many $M$ such that $t$ divides $MN$,
the magic $N$-cube lies in ${\cal S}_N(t,g)$. If $\sigma$ is a permutation
on $0,\dots,M-1$ then we define the array $A^\sigma=(a^\sigma_{i_1,\dots,i_N})$
by the rule 
$$a^\sigma_{(i)}=a_{\sigma\cdot(i)}.$$
It is easy to see that for $1\leq k\leq N$ the mapping 
$(i)\mapsto \sigma\cdot(i)$ maps $k$-subcubes to $k$-subcubes and preserves
the relation of conjugacy of subcubes. It follows that $A^\sigma$ also
belongs to ${\cal S}_N(t,g)$. This shows that for $n=t^M$ with $M$ sufficiently
large and such that $t$ divides $MN$, we have $c_n\geq M!$. Furthermore,
since the proof of Lemma 3.3 shows that $A^\sigma$ is also irreducible,
we actually have $d_n\geq M!$. Therefore we have,
letting $M$ run over sufficiently large multiples of $t$,
$$\sum_{n=1}^\infty{c_n\over n^s} \geq
\sum_{n=1}^\infty{d_n\over n^s} \geq
\sum {M!\over t^{Ms}}$$
and the summation on the right diverges by the ratio test.

Since the problem of determining the numbers $a_n,b_n,c_n,d_n$, and
others like them, or at least giving good estimates for them, is
a matter of independent interest, we summarize in the following
theorem the estimates we have obtained. Undoubtedly one can easily
do much better.
\Proclaim Theorem. We have the following inequalities.
\item{(i)} $a_M\geq M!$ if $M$ is relatively prime to $(2N-1)!$.
\item{(ii)}$b_M\geq (M-D)!$ if $M$ is relatively prime to $(2N-1)!$,
where $D$ is the largest proper divisor of $M$; 
\item{(iii)} $c_n\geq d_n\geq M!$ if $n=t^M$.
\par\noindent
The estimates (i) and (ii) could be improved by studying the question of
how 
many matrices one could have used instead of the matrix $X$ we actually chose
for the proof of Theorem 4.1. This would only improve the estimate by
at most a factor which is polynomial in $M$. We haven't investigated this
in any detail since we believe that the actual growth of $b_M$ is much larger
than any polynomial multiple of $M!$.\bigskip
\centerline{\bf References}\par
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\vfill\eject
\end