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\begin{document}

\begin{center}
\textbf{Baton Rouge Lecture}
\end{center}

\begin{center}
Alexander Hahn, Notre Dame
\end{center}

Consider a group $G$ along with a set of generators $A$ that satisfies
$A^{-1} = A$. Many such situations exist: $G$ a
Weyl group (or more generally a Coxeter group) and $A$ the defining
hyperplane reflections; or $G$ a classical group and $A$ a
set of special elements coming from the underlying geometry (or from a
single conjugacy class of such elements); or $G =
SL_n(\mathbb{Z})$ with
$A$ the set of elementary  matrices; or in coding theory where interesting
codes are constructed from the Cayley graph arising from
certain $G$ and $A$ (this seems to be a hot topic currently).

\vspace{.1in}

\textbf{Question:} Given $G$ and $A$ and $\sigma \in G$ what is the length
of $\ell(\sigma)$ of $\sigma$? Or more precisely, are
there parameters arising from $\sigma$ from which $\ell(\sigma)$ be read off?

\vspace{.1in}

\textbf{Example 1.} Let $G$ be the symmetric group on $\{1, \dots, n\}$ and
let $A$ be the set of transpositions. Let $k(\sigma)$
be the the number of orbits of $\sigma$ (include the trivial orbits). Then
\[
\ell(\sigma) = n - k(\sigma)\ .
\]

\vspace{.1in}

\textbf{Example 2.} Let $ G $ be the alternating group on $\{1, \dots, n\}$
and let $ A $ be the set of three cycles, or equivalently,
the set of short commutators of transpositions. This time, let $ k(\sigma)
$ be the number of orbits of odd cardinality (again include
the trivial orbits). Then
$ n - k(\sigma) $ is even and
\[
\ell(\sigma) = \frac{1}{2}(n - k(\sigma)) \ .
\]

\vspace{.1in}

Now let $ V $ be a non-degenerate $ n-$dimensional quadratic space with
symmetric bilinear form $ B $ over a field $ F $
with $\hbox{char}(F) \neq 2 $. Let $O_n(V)$ be the orthogonal group of $V$.
For $\sigma \in O_n(V)$, let $ S $ be the subspace $ S =
(\sigma - 1_V)V
$ of
$ V
$.  This $ S $ is the \emph{space} of $ \sigma $.
Intuitively, this is where the "action" of $ \sigma $ is. In particular,
there is no action on the orthogonal complement $S^{\perp}$
of $S$; the fact is that $ S^{\perp} = \{x \in V \mid \sigma(x) = x\}.$ It
turns out that $\dim\, S$ is even if and only if $\sigma \in
O^+_n(V).$

For instance, $\sigma = 1_V$ if and only if $S = {0}.$ If $\dim \ S = 1$,
then $S$ is necessarily non-degenerate, and
$\sigma =  -1_S \perp 1_{S^{\perp}}$. These elements are the
\emph{hyperplane reflections} or \emph{symmetries}.
If $S = Fv$, denote $\sigma$  by $\tau_v.$  They are involutions, i.e.,
they satisfy $\sigma^2 = 1_V.$

We will define properties of $ \sigma $ by referring to $ S $. For example,
$ \sigma $ is
\emph{non-degenerate}, \emph{degenerate}, or \emph{totally degenerate} if $
S $ is non-degenerate, degenerate, or totally
degenerate, i.e., if  the radical $ \hbox{rad} \, S = S\, \cap\, S^{\perp}
$ of $ S $ is zero, non-zero, or $ S $. Similarly, $ \sigma $ is
\emph{anisotropic} if $ S $ is anisotropic. Symmetries are anisotropic. It
is easy to see that $\sigma$ is an involution
if and only if
$\sigma\,_{\rule{.15mm}{3mm} S} = -1_S.$ In particular, involutions are
non-degenerate. The degenerate elements $\sigma$ with
$\dim\, S = 2$ are the \emph{Eichler} transformations.

Notation: if an orthogonal transformation
$\sigma,
\mu,
\rho, \eta,$ etc. is under consideration, then then $S, U, R,$ and $E,$
etc. will automatically denote its space.

\vspace{.1in}

\textbf{Example 3.} Theorem (Cartan, Scherk, Dieudonn\'e). Let $ G $ be the
group $ O_n(V) $ and let $ A $ be the set of
symmetries. If $ \sigma $ is not totally degenerate, then
\[
\ell(\sigma) = \dim\, S \ .
\]
If $\sigma$ is totally degenerate, then $ \ell(\sigma) = \dim\, S\, +\, 2 $.

Note: The answer is complete and completely independent of $ F $ or $ V $.

This example parallels Example 1. What about the analogue of Example 2?

\vspace{.1in}

\textbf{Example 4.} Let $ G $ be the commutator subgroup $ \Omega_n(V) $ of
$ O_n(V) $ and let $ A $ be the set of short
commutators of  symmetries. What about $ \ell(\sigma) $ in this situation?

\vspace{.1in}

The short answer: Nothing until recently; much more difficult; dependence
on both $F$ and $V$.
The longer answer is the subject of this talk. Incidentally, I got
interested in this question about 5 years ago thanks to John Hsia
who was interested in the case $ F $ a  non-dyadic local field.  In this
case, a complete solution is possible. I'll describe what this is;
also discuss the dyadic case; then give some "global" insights.

\vspace{.1in}

Let $ \sigma \in \Omega_n(V) $. Let $\sigma = \tau_v \tau_w \tau_v \tau_w $
be a non-trivial short commutator of symmetries.
Then $S = Fv \oplus Fw.$ Also, $\sigma = \tau_v\tau_{\tau_w(v)} =
\tau_v\tau_{v'}$ with $ Q(v) = Q(v'),$ where $Q(x) = B(x, x).$
Conversely, any such product is a short commutator of symmetries. It is now
a direct consequence of CSD that
\[
\ell(\sigma) \geq \frac{1}{2} \dim\, S \ .
\]

Now let's call $ \sigma \in \Omega_n(V) $ \emph{short} if $ \ell(\sigma) =
\frac{1}{2} \dim\, S $, and \emph{long}
otherwise. Let $ \sigma $ be totally degenerate. Then  $ \sigma \in
\Omega_n(V) $ and by
CSD, $ \sigma $ is long.  CSD suggests that long ought to be the exception.
True?

\vspace{.2in}

\textbf{Goal:} The same as that of CSD. Namely, the complete description of
the long elements of $ \Omega_n(V) $ and the
determination of their length.

\vspace{.1in}

\textbf{Theorem 1}. Suppose that  \ card$\, \overset{\ast}{F}
/\overset{\ast}{F^2} \leq 2.$ Then the totally degenerate
elements $\sigma$ are the only long elements in $\Omega_n(V)$ and for
these, $\ell(\sigma) = \frac{1}{2} \dim \ S + 1$.

Easy consequence of CSD. Note that this result applies to $\mathbb{C},
\mathbb{R}$, and $\mathbb{F}_q.$

\vspace{.2in}

Kn\"uppel (1993) noticed the following as a result of his investigations
into of a different problem, namely the generation of the
orthogonal groups by symmetries from a fixed conjugacy class.

\ i) \ If $ \sigma $ is long and not an involution, then the quotient space
$ S/\hbox{rad}\,S $ is anisotropic, and

ii) \ If $ V $ is isotropic and $ \sigma $ is long, then $ \ell(\sigma) =
\frac{1}{2} \dim\, S\,+ \, 1.$

Kn\"uppel's observations are an important start towards the goal.

\vspace{.2in}

\textbf{The Tools:} The \emph{Zassenhaus} splitting; the \emph{Wall} form;
and \emph{Reduction mod the Radical}.

\vspace{.1in}

The \emph{Zasssenhaus splitting}. Let $ \sigma \in O_n(V).$ Consider the
subspace
\[
\{x \in V \mid (\sigma - 1_V)^kx = 0 \hbox{ \ some \ } k \}.
\]
This largest space on which $ \sigma $ acts as a \emph{unipotent}
transformation is non-degenerate. Let $ R $ be its orthogonal complement.
Note that $ \sigma R = R $ and $ \sigma R^{\perp} =
R^{\perp},$ and hence that
$\sigma = \sigma\,_{\rule{.15mm}{3mm} R^{\perp}} \perp
\sigma\,_{\rule{.15mm}{3mm} R} \ .$
Put $ \mu = \sigma\,_{\rule{.15mm}{3mm} R^{\perp}} \perp 1_R $ and $ \rho = 1_
{R^{\perp}} \perp \sigma\,_{\rule{.15mm}{3mm} R}.$ Then
\[
\sigma = \mu \,\cdot \,\rho \
\]
with $ \mu $ unipotent and $ \rho $ non-degenerate with space $ R$. This is the
\emph{Zassenhaus} splitting of $\sigma.$ Note that $ \mu $ and $ \rho $
commute. It turns out that
$\sigma$ is in $ \Omega_n(V) $ if and only if  $ \mu $ and $
\rho $ are both in $ \Omega_n(V).$ Non-trivial unipotent elements exist
only for isotropic $ V.$
Eichler transformations are unipotent.

\vspace{.2in}

The \emph{Wall} form. For $\sigma \in O_n(V),$ define
\[
(\ , \ )_\sigma : S \times S \longrightarrow F
\]
by the equation
$(\sigma x - x, \sigma y - y)_\sigma = B(\sigma x - x, y)$ for all  $\sigma
x - x$ and $\sigma y - y$
in $S.$ This form is a non-degenerate, bilinear form on $S$ (but it
is almost never symmetric).  Note that the space $ S $ is now equipped both
with
$( \ , \ )_\sigma $ and the restriction of  $ B.$  When  $(\  , \ )_\sigma$
is under consideration we will denote $ S $ by $ S_\sigma.$

One can check that $\sigma$ is an involution if and only if $( \ , \
)_\sigma $ is symmetric, and that in this case,
$( \ , \ )_\sigma = -\frac{1}{2}B.$ Also, $\sigma$ is totally degenerate if
and only if $S_{\sigma}$ is
alternating. Let $\sigma = \mu \rho$ be the Zassenhaus splitting of
$\sigma$ with $\mu$ totally degenerate and $\rho$
an involution. What can you say about $S_{\sigma}$?

The key facts are these. Let $W_1$
be a non-degenerate subspace of  $S_{\sigma}$. Then there is a unique
$\sigma_1 \in O_n(V)$ - the transformation belonging to
$W_1$ - such that $(S_1)_{\sigma_1} = W_1$. If
$S_{\sigma} = W_1 \perp W_2$ ($W_2$ is the right complement of $W_1$), then
$\sigma = \sigma_1 \cdot \sigma_2 $, where
$\sigma_2$ belongs to $W_2$. Relevant to the current context is that
$\sigma$ is short if and only if
$S_{\sigma}$ is a (right) orthogonal sum of planes of discriminant one that
are non-alternating. (As an aside, the Wall form supplies a
useful definition of the spinor norm via disc $S_{\sigma}$).

\vspace{.2in}

The \emph{Reduction mod the Radical} construction.
Let $ M $ be any subspace of $ V$.  The quotient space $ M/\hbox{rad} \,M $
becomes a non-degenerate quadratic space with bilinear form $ B' $ defined by
\[
B'(x + \hbox{rad } M, y + \hbox{rad } M) = B(x, y) \hbox{ \ for all } \ x,
y \in M\ .
\]
Let $ O[M] $ be the subgroup of $ O_n(V) $ defined by
\[
O[M] = \{\eta \in O_n (V) \ \mid \ E \subseteq M\}.
\]
Let $ \eta \in O[M].$ Since $E^{\perp} \supseteq M^{\perp} \supseteq
\hbox{rad } M,$ we see that
$ \eta\,_{\rule{.15mm}{3mm}\hbox{\scriptsize{rad}} \,M} =
1_{\hbox{\scriptsize{rad}} \,M}.$
So we can define
 \[
^{\sim} \ : \ O[M] \longrightarrow O(M/\hbox{rad} \ M)
\]
by $\overset{\sim}{\eta}(x + \hbox{rad} \, M) = \eta x + \hbox{rad} \, M.$
For $ v \in M $ anisotropic, $ \tau_v \in O[M]
$  and $ \overset{\sim}{\tau}_v = \tau_{v + {\hbox{\scriptsize{rad}}
\,M}}.$ Check that
\[
\hbox{ker }  ^{\sim} \ = \{ \eta \in O[M] \ | \ (\eta - 1_V)M
\subseteq {\hbox{rad}} \, M \}.
\]
If $\eta$ is in the kernel then,
\[
(\eta - 1_V)^3V \subseteq  (\eta - 1_V)^2M \subseteq (\eta -
1_V)(\hbox{rad} \,M) = {0} .
\]
In particular, $ \eta $ is unipotent.

\vspace{.2in}

\textbf{Theorem 2.} Let $ \sigma \in \Omega_n(V)$ be long with $ \sigma $
neither totally degenerate nor an involution.  Let
$ \sigma = \mu\rho $ be the Zassenhaus splitting of $ \sigma$. Then

\begin{enumerate}
\item[i)] The space of $\mu$ satisfies $U = \hbox{rad} \,U \perp T$ with
$T$ anisotropic and the space of $\sigma$ satisfies $S =
\hbox{rad} \,U \perp (T \perp R)$ with $T \perp R$ anisotropic.

\item[ii)] The unipotent element $\mu$ is \emph{special}. This means that $
\mu $ is a product
of $ \frac{1}{2}(\dim \,U) $ commuting Eichler transformations and that $
(\mu - 1_V)^3 = 0.$ (The totally
degenerate elements are precisely the unipotents with $ (\mu - 1_V)^2 = 0.)$

\item[iii)] The non-degenerate element $ \rho $ is anisotropic and long.
\end{enumerate}


\vspace{.1in}

Part (i) follows from the Zassenhaus splitting and the insight of
Kn\"uppel. That $(\mu - 1_V)^3 = 0$ is a quick consequence of
reduction with $U/\hbox{rad} \,U.$ Namely, because $\mu$ is unipotent,
$\overset{\sim}{\mu} \in O(U/\hbox{rad} \,U)$ is
unipotent. But this quotient is anisotropic, so $\mu \in \hbox{ker }
^{\sim}.$ That $\rho$ is anisotropic follows from
(i); the rest of (ii) and (iii) are labor intensive.

\vspace{.1in}

This Theorem - it holds for any $ F $ - reduces the problem to the
following two questions:

\vspace{.1in}

\begin{enumerate}
\item[\textbf{A)}] \ Determine which of the elements in Theorem 2 are
actually long and compute their lengths. (Recall that if $ V $ is
isotropic, then the length of any long element $ \sigma $ is $
\frac{1}{2}\dim\, S \, + 1$.)

\vspace{.1in}

\item[\textbf{B)}] \ Classify all long anisotropic elements $ \rho $ in $
\Omega_n(V)$.
\end{enumerate}

\vspace{1.4in}

Let's see what the answers are in case $ F $ is a local field. We begin
with Question A:

\textbf{Proposition 3.} Suppose that $ F $ is a local field and consider
any element $ \sigma = \mu\rho \in \Omega_n(V)$ that
satisfies conditions (i) - (iii) of Theorem 2. Then $ T = {0}$, $\mu$ is
totally degenerate,  $ \dim\, R = 4$, and $\sigma$ is long.
Finally, $\ell(\sigma) = \frac{1}{2} \dim \, S \, + 1$. In particular, all
long elements of $\Omega_n(V)$ (excluding involutions and
totally degenerate elements) can be constructed by splicing totally
degenerate and long anisotropic elements together.

That $ T = {0} $ and $ \dim\, R = 4$, follows easily from
the fact that $ 4 \le \dim \ R \le \dim \ (T \perp R) \le 4$.
That $\sigma$ is long comes from a combination of reduction mod the radical
with properties of the Wall form.
In view of Kn\"uppel's result,
$\ell(\sigma) = \frac{1}{2} \dim \, S \, + 1$ only needs verification for
$\dim \,V = 4.$

\vspace{.2in}

In reference to Question B, there is a sharp dichotomy between the
non-dyadic case and the dyadic case.

\textbf{Proposition 4.} If $ F $ is non-dyadic, then anisotropic long
elements occur only for $ n \ge 5 $ and they are the following:
\[
\rho = 1 \perp \rho\,_{\rule{.15mm}{3mm} R} \ \ \ \hbox{with} \ \ \
\rho\,_{\rule{.15mm}{3mm} R} \in O'_4(R) - \Omega_4(R) \ .
\]
Such elements exist because the index of $\Omega_4(R)$ in $O'_4(R)$ is two
and they are all long. To prove the proposition, it is
enough to verify that all elements in $\Omega_4(V)$ are short.

\vspace{.1in}

Suppose that $ F $ is dyadic. Here the matter is much more subtle. In this
case, $ O'_4(R) = \Omega_4(R).$ So there are no long
elements of the type above.  It follows that all anisotropic long elements
have the form
\[
\rho = 1 \perp \rho\,_{\rule{.15mm}{3mm} R} \ \ \ \hbox{with} \ \ \
\rho\,_{\rule{.15mm}{3mm} R} \hbox{ \ long in \ }
\Omega_4(R)
\ .
\]

\vspace{.1in}
So we need to classify the long elements in $ \Omega_4(V) $ with $ V $
anisotropic.  Milnor [8] provides a strategy:

\vspace{.1in}

Let $ V $ be any non-degenerate quadratic space over a local field $ F$ (of
characteristic not $2$). Let $ m(X) $ be an irreducible
monic polynomial in $ F[X].$ Then $m(X)$ is the minimal polynomial of an
element of $O_n(V)$ if and only if its degree $ k $ divides
$ n $, it is symmetric, and disc\, $V = (m(1)m(-1))^\frac{n}{k}
\overset{\ast}{F^2}.$ Given such a polynomial $ m(X),$ then -
and this is one of the main results of Milnor's paper -  there is precisely
one conjugacy class of elements in $ O_n(V) $ with minimal
polynomial $ m(X).$ Notice that if
$
\rho
$ is long and anisotropic in
$ \Omega_n(V) $ then the entire conjugacy class of $ \sigma $ (in $O_n(V)$)
consists of long and anisotropic elements of $
\Omega_n(V).$ Thus when looking for long elements, we are looking for
conjugacy classes of them.

\vspace{.2in}

Turn to the study of the long elements in $ \Omega_4(V) = O'_4(V) $ (with $
F $ dyadic and $ V $ anisotropic). Let
$m(X)$ be the minimal polynomial of a long element $\sigma \in
\Omega_4(V)$. The factor
$X + 1$ is the only linear factor that $m(X)$ can have.

\vspace{.1in}

\ \ 1) Suppose $ \deg\, m(X) = 1$. Then $ m(X) = X + 1 $. So $\sigma = -1_V
$. This element is in $\Omega_4(V),$ and it is long
if and only if $ -1 \in \overset{\ast}{F^2}.$
\vspace{.1in}

\ \ 2) Suppose $ \deg\, m(X) = 2$. If $ m(X) $ is reducible, then $m(X) =
(X + 1)^2$. But this means that $-\sigma$ is unipotent.
But $V$ anisotropic means that $-\sigma = 1_V,$ impossible. So $m(X)$ is
irreducible, hence $ m(X) =
X^2 - cX + 1.$ It turns out that the unique corresponding conjugacy class
is long if and only if
$ c - 2 \in \overset{\ast}{F^2}$. If $\sigma$ has a minimal polynomial of
this form, then $\sigma$ is in $\Omega_4(V).$
\vspace{.1in}

\ \ 3) Suppose $ \deg\, m(X) = 3$. By Milnor, $ m(X) $ is reducible. So
$m(X) = (X + 1)(X^2 - cX + 1)$ with $X^2 - cX + 1$
irreducible. (The quadratic factor must be irreducible by the "unipotent"
argument above.) In this case, $ -1 $ and $c - 2$ must both
be in $\overset{\ast}{F^2}.$ This case arises.
\vspace{.1in}

\ \ 4) Suppose $ \deg\, m(X) = 4$. In this case, either

\ \ \ \ \ \ \ a) $m(X) = (X^2 - cX + 1)(X^2 - dX + 1)$ \ with both factors
irreducible, or

\ \ \ \ \ \ \ b) $m(X) = X^4 - cX^3 - dX^2 - cX + 1$ \ is irreducible.

I know that (a) arises. Most probably, (b) does too.

\vspace{.1in}

\textbf{The "Long" Criterion}: Let $\sigma \in \Omega_4(V).$ Then $\sigma$
is long if and only if
\[
Q(\sigma x - x) = -\epsilon_x^2Q(x) \ \hbox{ for all  } \ x \in V \hbox{
and some } -\epsilon_x \in  \overset{\ast}{F}.
\]

The Long Criterion immediately provides the conclusion in (1). It turns out
that case (2) is precisely the situation where
all $\epsilon_x^2$ are equal, namely to $c - 2$. What about (3) and (4a)?
These situations are similar. In each case,
the factorization $m(X) = p_1(X)p_2(X)$ provides two unique planes, namely
$U = p_1(\sigma)V$ and $W = p_2(\sigma)V$
on which $\sigma$ acts. Check that $V = U \perp W.$ In order for $\sigma$
to be long, the Long Criterion that has to hold
for both $U$ and $W$ it must be extendable to all of $V$.

Let's consider case (3). Note that $\sigma\,_{\rule{.15mm}{3mm}W} = -1_W.$
By the Long Criterion applied to $U$ and $W$ we get
$c - 2$ and $ -1$ both in $\overset{\ast}{F^2}.$ So put $c - 2 = s^2$ and
$-1 = i^2.$ Notice that
$\overset{\ast}{Q(U)} \cap \overset{\ast}{Q(W)}$ is empty, otherwise $V$
would contain a plane of discriminant
$\overset{\ast}{F^2} = -\overset{\ast}{F^2};$ not possible because $V$ is
anisotropic. Put
\[
s = -2it^{-1} \hbox{ \ for some \ } t  \in \overset{\ast}{F} \hbox{  \ and
set \ } B = \frac{Q(\overset{.}{U})}{Q(\overset{.}{W})} \ ,
\]
where $\overset{.}U $ and $\overset{.}W$ denote the non-zero elements of
$U$ and $W$. The extendability of the
Long Criterion to
$ V $ translates into the question of the existence - and precise
description - of the elements
$t \in \overset{\ast}{F}$ such that

\[
(*)   \ \ 1 + \frac{t^2 - 1}{1 + \beta} \ \in  \overset{\ast}{F^2} \ \hbox{
for all    } \ \beta \in  B \ .
\]

Do such $t$ exist?  YES! To see this rewrite the above as
\[
\frac{1 + \beta - \beta + t^{2\beta}}{1 + \beta} \ = 1 + \frac{t^2 - 1}{1 +
\beta^{-1}} \ .
\]

Now observe that $\{ |1 + \beta| \mid \beta \in \overset{\ast}{Q(W)} \}$ is
bounded below by $|4|$. For suppose
that $|1 + \beta| \leq |4\pi|$ for some $\beta$. Then $1 + \beta = 4\alpha
\pi$ for some local integer $\alpha$; but this means that
$-\beta = 1 - 4\alpha \pi \in \overset{\ast}{F^2}$ by the Local Square
Theorem.  Because $-\beta \in \overset{\ast}{Q(W)}$
(notice that $-1 \in \overset{\dot}{Q(U)}$), we have a contradiction. Now
choose $t$ such that $|t^2 - 1|$ is small enough and apply
the Local Square Theorem again to get (*).

\vspace{.1in}
$i^2 =$Then $\overset{\ast}{Q(U)}$ is a
subgroup of index $2$ of $\overset{\ast}{F}.$ If
Global Fields?

\vspace{.2in}

\centerline{\bf Bibliography}
\bigskip
\noindent
[1] E. Artin, Geometric Algebra, Wiley Interscience, New York, 1966.
\smallskip

[2]  E. W. Ellers and J. Malzan, Products of reflections in the kernel of the
spinorial Norm, {\it Geom. Ded.} 36 (1990), 279-285.
\smallskip

[3]  A. J. Hahn, Unipotent elements and the spinor norms of Wall and
Zassenhaus, {\it  Archiv der Math.},
32 (1979), 114-122.
\smallskip

[4] A. J. Hahn, The elements of the orthogonal group $\Omega_n(V)$ as
products of commmutators of symmetries, J. Alg., 1995
\smallskip

[5]  A. J. Hahn and O. T. O'Meara, The Classical Groups and $K$-Theory,
{\it Grundlehren der Mathematik}, Springer-Verlag, 1989.
\smallskip

[6]  F. Kn\"uppel, Products of simple isometries of given conjugacy types,
{\it Forum Math.} 5 (1993), 441-458.
\smallskip

[7]  T. Y. Lam, The Algebraic Theory of Quadratic Forms, Benjamin, Reading
MA, 1973.
\smallskip

[8] J. Milnor, On Isometries of Inner Product Spaces, Inventiones Math. 8
(1969), 83-97.
\smallskip

[9] O. T. O'Meara, Introduction to Quadratic Forms,
 {\it Grundlehren der Mathematik}, Springer-Verlag, 1971.
\smallskip



\end{document}




\it{Mathematics Subject Classification } 20G15, 20G25, 20F05}} \smallskip



Department of Mathematics \par
University of Notre Dame \par
Notre Dame, IN  46556-5683 \par
United States \par
E-mail:Alexander.J.Hahn.$1\@$nd.edu

\end{document}


