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% ---begin deftags---
\deftag{1}{sec-intro}
\deftag{1.1}{thm-Euclidean-domain}
\deftag{1.2}{thm-pid}
\deftag{2}{sec-triality}
\deftag{2.1}{thm-has-index-120}
\deftag{2.2}{thm-scalar-mults-generates-all}
\deftag{2.3}{thm-subgroup-of-index-120}
\deftag{3}{sec-main-theorem}
\deftag{3.1}{thm-main}
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\defcite{1}{ATLAS}
\defcite{2}{splag}
\defcite{3}{cox:integer-octaves}
\defcite{4}{dickson:new-simple-theory}
\defcite{5}{mahler:ideals-cayley-dickson-algebra}
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%-----------------Beginning-of-Document---------------------------
%
\def\O{{\bbb O}}
\def\k{{\cal K}}
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\def\spin{2^2G}
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\def\ideal{{\cal I}}
\author {Daniel Allcock\footnote*{Supported by an NSF
postdoctoral fellowship.}}
\title {Ideals in the Integral Octaves}
\address{Department of Mathematics\par
University of Utah\par
Salt Lake City, UT 84112.}
\email {allcock@math.utah.edu}
\homepage{http://www.math.utah.edu/$\sim$allcock}
\date {12 March 1998, revised 9 May 1998}
%\subject{}
%\note {}
\plaintitlepage
\abstract
We study the nonassociative ring of integral octaves
(or Cayley numbers or octonions)
discovered independently by Dickson and Coxeter. We prove
that every one-sided ideal is in fact two-sided, principal, and
generated by a rational integer.
\section{\Tag{sec-intro} Introduction}
The nonassociative ring $\k$ of integral octaves is a discrete
subring of the nonassociative field $\O$ of octaves---it is the
natural analogue of the the rational integers in $\Z$. Dickson
introduced $\k$ in \cite{dickson:new-simple-theory}; one can use
$\k$ to construct the finite simple groups now called $G_2(p)$
for $p$ a prime number. Much later, Coxeter
\cite{cox:integer-octaves} rediscovered the ring and obtained a
number of new results concerning it. The purpose of this note is
to obtain a complete description of the ideals in $\k$: every
one-sided ideal is actually two-sided and generated by a
rational integer. This substantially improves a result of Mahler
\cite{mahler:ideals-cayley-dickson-algebra}, who proved that any
one-sided ideal is generated by a rational integer multiple of
an element of one of three possible
norms. Our arguments rely on geometric properties of the $E_8$
lattice and its automorphism group, rather than the sort of
explicit computation that Mahler used.
Our definitions of $\O$ and $\k$ are those of
\ecite{ATLAS}{p.~85}. The nonassociative field $\O$ of octaves
is an 8-dimensional algebra over the real numbers $\R$, with
basis $e_\infty=1$, $e_0,\ldots,e_6$. Multiplication is defined
by the relations that for each $n=0,\ldots,6$, $e_n^2=-1$ and
the span of $e_\infty$, $e_n$, $e_{n+1}$ and $e_{n+3}$ is a copy
of the the (associative) field of quaternions, with
$e_ne_{n+1}e_{n+3}=-1$. Here the subscripts should be read
modulo 7. If $x=x_\infty e_\infty+x_0e_0+\cdots+ x_6e_6$ with
each $x_n\in\R$ then the conjugate of $x$ is $\bar x=x_\infty
e_\infty-x_0e_0-\cdots -x_6e_6$. The absolute value $|x|$ of $x$
is $\sqrt{x\bar x}$, which is a nonnegative real number; $|x|=0$
just if $x=0$. The identity $|xy|=|x|\,|y|$ (for any $x,y\in\O$)
is important for the proof of
theorem~\tag{thm-Euclidean-domain}. The real part of $x$ is
$x_\infty$; if this vanishes then we say that $x$ is imaginary.
The integral octaves $\k$ are the discrete subring of $\O$
consisting of those $x=\sum x_ne_n$ satisfying
$x_n\in{1\over2}\Z$ for all $n=\infty,0,\ldots,6$ and that the
set of $n$ for which $x_n\notin\Z$ is one of
$$
\displaylines{
\{0124\},\quad
\{0235\},\quad
\{0346\},\quad
\{0156\}, \cr
\{\infty013\},\quad
\{\infty026\},\quad
\{\infty045\},\quad
\{\infty0123456\},
\cr}
$$
or the complement of one of these. One may check \cite{cox:integer-octaves} that
$\k$ is closed under addition and multiplication. Furthermore, if
$x\in\k$ then $|x|^2\in\Z$, and under the metric on $\k$ induced
by the absolute value function, $\k$ is a scaled copy of the
$E_8$ lattice \ecite{splag}{ch.~4,~\S8.1}. In particular, the
minimal distance between elements of $\k$ is 1 and the covering
radius of $\k$ is $1/\sqrt2$. (This means that every element of
$\O$ lies within $1/\sqrt2$ of $\k$ and that $1/\sqrt2$ is the
smallest number for which this holds.) There are 240 elements of
$\k$ of absolute value 1, which are the units of $\k$. We denote
the set of these by $\units$.
The first step in applying the geometry of the $E_8$ lattice to
the study of ideals in $\k$ is the following theorem and its
corollary.
\beginproclaim Theorem
{\Tag{thm-Euclidean-domain}}
\cite{cox:integer-octaves} \cite{mahler:ideals-cayley-dickson-algebra}.
The function $x\mapsto|x|^2$ is a Euclidean norm on $\k$.
\endproclaim
\remark{Remark:}
By this we mean that for all $x,m\in\k$ there are $q,r\in\k$
such that $x=qm+r$ with $|r|^2<|m|^2$, and that there are also
$q',r'\in\k$ such that $x=mq'+r'$ with $|r'|^2<|m|^2$.
\beginproof{Proof:}
In the notation of the remark, let $q\in\k$ by such that $qm$ is
an element of $\k m$ nearest $x$, and let
$r=x-qm$. The right-multiplication map of $m$ increases
distances by a factor of $|m|$, so $\k m$ has covering radius
$|m|/\sqrt2$. Therefore $|r|\leq |m|/\sqrt2$. A similar argument
applied to $m\k$ completes the proof.
\endproof
\beginproclaim Corrollary
{\Tag{thm-pid}}.
Any left (resp. right) ideal in $\k$ has the form $\k m$
(resp. $m\k$) for some $m\in\k$.
\endproclaim
\beginproof{Proof:}
This follows from the usual argument that a Euclidean domain is
a principal ideal domain.
\endproof
Of course, since $\k$ is not associative, if $m$ is a random
integral octave then $\k m$ and $m\k$ might fail to be ideals.
Our main theorem is essentially the assertion that $\k m$ or
$m\k$ is an ideal if and only if $m$ is a product of a rational
integer and a unit of $\k$:
\def\statement{Any one-sided ideal $\ideal$ in $\k$ is two-sided,
principal, and has the form $\k n=n\k$ for some rational integer
$n$.}
\beginproclaim Theorem
{\tag{thm-main}}.
\statement
\endproclaim
\noindent
In order to prove the theorem we need to identify the group
generated by the left (or right) multiplication maps of units of
$\k$. We do this in
the next section, and prove the theorem in
section~\tag{sec-main-theorem}.
\section{\Tag{sec-triality}. Triality }
The best way to understand
the group generated by the left (or right) multiplication maps
of units of $\k$ is by considering the ``isotopy'' group of
$\k$ and its ``triality'' automorphism.
An isotopy is a triple $(A,B,C)$ of $\Z$-linear maps of $\k$
such that for all $x,y,z\in\k$, the equation $xyz=1$ holds just
if $A(x)B(y)C(z)=1$ holds. These equations make sense because if either
$xy\cdot z=1$ or $x\cdot yz=1$ then $x$, $y$ and $z$ all lie in
an associative algebra, so that $xyz$ is unambiguously defined
and equal to 1. The product of two isotopies is defined by
$$
(A,B,C)(A',B',C')=(A\circ A',B\circ B', C\circ C').
$$
By \ecite{ATLAS}{p.~85}, the group of isotopies is isomorphic to an
extension $\spin$ of the simple group
$\pso=O_8^+(2)$. Furthermore, $\spin$ is generated by
the triples $(L_u,R_u,B_u)$ for $u\in\units$, where $L_u$
(resp. $R_u$) is the map given by left (resp. right) multiplication by
$u$, and $B_u$ is the ``bimultiplication'' map $x\mapsto
u^{-1}xu^{-1}$. The fact that $(L_u,R_u,B_u)$ is an isotopy
follows from the Moufang identity: $u(xy)u=(ux)(yu)$ for all
$u,x,y\in\O$. For each $u\in\O\setminus\{0\}$ the maps $L_u$, $R_u$ and $B_u$
are orientation-preserving maps of $\O$, since this is obviously
true for $u=1$ and $\O\setminus\{0\}$ is connected. This implies
that there are three different maps $\pi_0$, $\pi_1$ and $\pi_2$
from $\spin$ to the rotation group of $\k$; these carry
$(A,B,C)$ to $A$, $B$ and $C$, respectively. By the rotation
group of $\k$ we mean the full group of orientation-preserving
isometries of $\k$; this is the commutator subgroup of the $E_8$
Weyl group. It is a central extension $\so$ of
$O_8^+(2)$. Finally, the description of $\spin$ in terms of
isotopies makes visible the ``triality'' automorphism
$\tau:(A,B,C)\mapsto(B,C,A)$. It is obvious that
$\pi_m\circ\tau^n=\pi_{m+n}$, where the subscripts should be
read modulo 3.
We define $H$ to be the subgroup of $\spin$ generated by those
triples $(L_u,R_u,B_u)$ with $u\in\im\units$, and we set
$H_\ell=\pi_0(H)$, $H_r=\pi_1(H)$ and $H_b=\pi_2(H)$. That is,
$H_\ell$, $H_r$ and $H_b$ are respectively the subgroups of
$\so$ generated by the left, right and bimultiplication maps
of elements of $\im\units$.
\beginproclaim Lemma
{\Tag{thm-has-index-120}}.
$H_b$ is a maximal subgroup of $\so$ and preserves $\{\pm1\}$.
\endproclaim
\beginproof{Proof:}
For $u\in\im\units$ it is easy to check that $B_u$ negates $1$
and $u$ and fixes $u^\perp\cap\im\O$ pointwise. (This uses the
fact that orthogonal imaginary octaves anticommute.) Therefore
$H_b$ acts on $\im\k$ as the $E_7$ Weyl group, since $\im\k$ is
a copy of the $E_7$ lattice, being the orthogonal complement of
a minimal vector in a copy of the $E_8$ lattice. Furthermore,
an element of $H_b$ exchanges $1$ and $-1$ just if it reverses
orientation on $\im\O$. Therefore $H_b$ is isomorphic to the
$E_7$ Weyl group, and a computation reveals that
$|\so|/|H_b|=120$. The maximality follows because by
\ecite{ATLAS}{p.~85}, $\pso$ has no proper subgroups of index $<120$.
\endproof
\beginproclaim Lemma
{\Tag{thm-scalar-mults-generates-all}}.
The maps $L_u$ (or alternately, the $R_u$ or the $B_u$), for
$u\in\units$, generate all of $\so$.
\endproclaim
\beginproof{Proof:}
By lemma~\tag{thm-has-index-120}, $H_b$ is a maximal subgroup of
$\so$. Since there are $u\in\units$ such that
$B_u(1)\notin\{\pm1\}$, the group generated by all the
$B_u$ is all of $\so$. This is just the assertion that
$\pi_2(\spin)=\so$. By triality, we see that each of $\pi_0(\spin)$ and
$\pi_1(\spin)$ is also all of $\so$. The lemma follows.
\endproof
We remark the the groups $\spin$ and $H$ are respectively ${\rm
Spin}_8^+(\F_2)$, the universal central extension of $O_8^+(2)$,
and ${\rm Pin}_7(\F_2)$, the direct product of $\Z/2$ and the
universal central extension of $O_7(2)$. Furthermore, the groups
$H_b$, $H_\ell$ and $H_r$ are isomorphic to $2\times O_7(2)$,
${\rm Spin}_7(\F_2)$ and ${\rm Spin}_7(\F_2)$. One may also
consider the isotopy group of $\O$ and the obvious analogues of
$H$, $H_b$, $H_\ell$ and $H_r$. These five groups (after
discarding Euclidean factors) are
respectively isomorphic to ${\rm Spin}_8(\R)$, ${\rm
Pin}_7(\R)$, $O_7(\R)$, ${\rm Spin}_7(\R)$ and ${\rm
Spin}_7(\R)$. It is remarkable that the exceptional phenomena
(triality, etc.) arising in the $\F_2$-versions of these groups
can be ``embedded'' in the real versions. In particular, the
triality automorphism $\tau$ acts in both cases as the
order-three automorphism of the Dynkin diagram $D_4$, the
Dynkin diagram of an orthogonal group in 8 dimensions.
\section{\Tag{sec-main-theorem}. The main theorem}
The proof of our main theorem is now quite short. We continue
using the notation of section~\tag{sec-triality}.
\beginproclaim Theorem
{\Tag{thm-main}}.
\statement
\endproclaim
\beginproof{Proof:}
The result for right ideals follows formally from that for left
ideals, since the identity $\overline{xy}=\bar y\bar x$ (for any
$x,y\in\O$) implies that the conjugate of a right ideal is a
left ideal. So suppose $\ideal$ is a nonzero left ideal. By
corollary~\tag{thm-pid} we have $\ideal=\k m$ where $m$ is some minimal
(nonzero) element of $\ideal$. Since $\ideal$ is a left ideal it is
preserved by the group generated by the maps $L_u$ for
$u\in\units$. If $\k$ were associative then this group would
have order 240. But $\k$ is not, and by
lemma~\tag{thm-scalar-mults-generates-all}, the group is
the full rotation group $\so$ of $\k$. Observe that $\ideal$ has 240
minimal vectors, namely the vectors $um$ for $u\in\units$. It is
obvious that group generated by the $L_u$ acts transitively on
these, so the subgroup preserving $m$ has index
240 and order 1,451,520. The only elements of the $E_8$ lattice
with a stabilizer this large are the rational integral multiples
of minimal lattice vectors. One sees this by considering the
full isometry group of $\k$, the $E_8$ Weyl group. The
stabilizer of any vector is (conjugate to) a Weyl group
corresponding to a subdiagram of the $E_8$ Dynkin diagram. The
only subdiagram whose associated Weyl group is large enough is
the (unique) $E_7$ subdiagram. The conjugates of this subgroup
are just the stabilizers of the various minimal vectors $u$ of
the lattice. For each such $u$, the only vectors stabilized by
the stabilizer of $u$ are the real multiples of $u$. Therefore
$m=un$ for some $u\in\units$, $n\in\Z$. Then $\ideal=\k
m=\k\cdot un=\k n$ and the proof is complete.
\endproof
\section{References}
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% ---begin bibentries---
\bibitem{ATLAS}
J.~H. Conway, R.~T. Curtis, S.~P. Norton, R.~A. Parker, and R.~A. Wilson.
{\it {A}{T}{L}{A}{S} of Finite Groups}.
Oxford, 1985.
\bibitem{splag}
J.~H. Conway and N.~J.~A. Sloane.
{\it Sphere Packings, Lattices, and Groups}.
Springer-Verlag, 1988.
\bibitem{cox:integer-octaves}
H.~S.~M. Coxeter.
Integral {C}ayley numbers.
{\it Duke Math. J.}, 13:561--78, 1946.
\bibitem{dickson:new-simple-theory}
L.~E. Dickson.
A new simple theory of hypercomplex integers.
{\it Journal de Math\'ematiques Pures et Appliqu\'ees}, 2:281--326,
1923.
\bibitem{mahler:ideals-cayley-dickson-algebra}
K.~Mahler.
On ideals in the {C}ayley-{D}ickson algebra.
{\it Proc. Roy. Irish Acad.}, 48:123--133, 1943.
% ---end bibentries---
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\bye