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\def\Aut{{\rm Aut}}
\proclaim Automorphism groups of Lorentzian lattices.
Journal of Algebra, Vol. 111, No. 1, Nov 1987, 133--153.
Richard E. Borcherds,
D.P.M.M.S.,
University of Cambridge,
16 Mill Lane,
Cambridge,
CB2 1SB,
England.
The study of automorphism groups of unimodular Lorentzian lattices
$I_{n,1}$ was started by Vinberg. These lattices have an infinite
reflection group (if $n\ge 2$) and Vinberg showed that the quotient of
the automorphism group by the reflection groups was finite if and only
if $n\le 19$. Conway and Sloane rewrote Vinberg's result in terms of
the Leech lattice $\Lambda$, showing that this quotient (for $n\le
19$) was a subgroup of $\cdot 0 = \Aut(\Lambda)$. In this paper we
continue Conway and Sloane's work and describe $\Aut(I_{n,1})$ for
$n\le 23$. In these cases there is a natural complex $U$ associated to
$I_{n,1}$, whose dimension is the virtual cohomological dimension of
the ``non-reflection part'' $G_n$ of $\Aut(I_{n,1})$, and which is a
point if and only if $n\le 19$. For $n=20,21$, and $22$ the group
$G_n$ is an amalgamated product of 2 subgroups of $\cdot 0$, while
$G_{23}$ is a direct limit of 6 subgroups of $\cdot 0$. The group
$G_{24}$ seems to be much more complicated (although it would probably
be just about possible to describe it). We also have a few results
about $G_n$ for large $n$; for example, if $n$ is at least 18 and
congruent to 2, 3, 4, 5, or 6 mod 8 then $\Aut(I_{n,1})$ is a
nontrivial amalgamated product. We find a few new lattices whose
reflection group has finite index in the automorphism group; for
example, the even sublattice of $I_{21,1}$ of determinant 4.
\proclaim 1.~Definitions.
In this section we summarize some standard definitions and results
that we will use. Many of the results can be found in Serre [5] or
Bourbaki [2].
We say that a group $G$ acting on a set fixes a subset $S$ if every
element of $G$ maps $S$ into $S$.
A {\it lattice} $L$ is a finitely generated free $Z$-module with an
integer valued bilinear form, written $(x,y)$ for $x$ and $y$ in $L$.
The {\it type} of a lattice is even (or II) if the {\it norm}
$x^2=(x,x)$ of every element $x$ of $L$ is even, and odd (or I)
otherwise. If $L$ is odd then the vectors in $L$ of even norm form an
even sublattice of index 2 in $L$. $L$ is called {\it positive
definite, Lorentzian, nonsingular,} etc. if the real vector space
$L\otimes R$ is.
If $L$ is a lattice then $L'$ denotes its {\it dual} in $L\otimes R$;
i.e. the vectors of $L\otimes R$ which have integral inner products
with all elements of $L$. The dual $L'$ contains $L$ and if $L$ is
nonsingular then $L'/L$ is a finite abelian group whose order is
called the {\it determinant} of $L$. (If $L$ is singular we say it has
determinant 0.) The lattice $L$ is called {\it unimodular} if its
determinant is 1. If $S$ is any subset of $L$ then $S^\perp$ is the
sublattice of elements of $L$ orthogonal to $S$.
Even unimodular lattices of a given signature and dimension exist if
and only if there is a real vector space with that signature and
dimension and the signature is divisible by 8. Any two indefinite
unimodular lattices with the same type, dimension, and signature are
isomorphic. $I_{m,n}$ and $II_{m,n}$ $(m\ge 1, n\ge 1)$ are the
unimodular lattices of dimension $m+n$, signature $m-n$, and type $I$
or $II$.
A vector $v$ in a lattice $L$ is called {\it primitive} if $v/n$ is
not in $L$ for any $n>1$. A {\it root} of a lattice $L$ is a primitive
vector $r$ of $L$ such that reflection in the hyperplane $r^\perp$
maps $L$ to itself. This reflection maps $v$ in $L$ to
$v-2r(v,r)/(r,r)$. Any vector $r$ in $L$ of norm 1 or 2 is a root.
If $L$ is unimodular then there is a unique element $c$ in $L/2L$ such
that $(c,v)\equiv v^2\bmod 2$ for all $v$ in $L$. The vector $c$ or
any inverse image of $c$ in $L$ is called a {\it characteristic
vector} of $L$, and its norm is congruent to the signature of $L\bmod
8$.
We now summarize some definitions and basic properties of finite root
systems. ``Root system'' will mean ``root system all of whose roots
have norm 2'' unless otherwise stated, so we only consider components
of type $a_n$, $d_n$, $e_6$, $e_7$, $e_8$. We use small letters $x_n$
to stand for spherical Dynkin diagrams. The types $e_3$, $e_4$, $e_5$
are the same as $a_2a_1$, $a_4$, and $d_5$. The types $d_2$ and $d_3$
are the same as $a_1^2$ and $a_3$.
The norm 2 vectors in a positive definite lattice $A$ form a root
system which we call the root system of $A$. The hyperplanes
perpendicular to these roots divide $A\otimes R$ into regions called
{\it Weyl chambers}. The reflections in the roots of $A$ generate a
group called the {\it Weyl group} of $A$, which acts simply
transitively on the Weyl chambers of $A$. Fix one Weyl chamber
$D$. The roots $r_i$ which are perpendicular to the faces of $D$ and
which have inner product at most 0 with the elements of $D$ are called
the {\it simple roots} of $D$. (These have opposite sign to what are usually
called the simple roots of $D$. This is caused by the irritating fact
that the usual sign conventions for positive definite lattices are not
compatible with those for Lorentzian lattices. With the convention
we use something is in the Weyl chamber if and only if it has inner
product at most 0 with all simple roots, and a root is simple if and
only if it has inner product at most 0 with all simple roots not equal
to itself.)
The {\it Dynkin diagram} of $D$ is the set of simple roots of $D$. It
is drawn as a graph with one vertex for each simple root of $D$ and
two vertices corresponding to the distinct roots $r,s$ are joined by
$-(r,s)$ lines. (If $A$ is positive definite then two vertices are
always joined by 0 or 1 lines. We will later consider the case that
$A$ is Lorentzian and then its Dynkin diagram may contain multiple
bonds, but these are not the same as the multiple bonds appearing in
$b_n$, $c_n$, $f_4$, and $g_2$.) The Dynkin diagram of $A$ is a union
of components of type $a_n$, $d_n$, $e_6$, $e_7$ and $e_8$. The {\it
Weyl vector} $\rho$ of $D$ is the vector in the vector space spanned
by roots of $A$ which has inner product $-1$ with all simple roots of
$D$. It is in the Weyl chamber $D$ and is equal to half the sum of the
positive roots of $D$, where a root is called positive if its inner
product with any element of $D$ is at least 0. A {\it tip} of a
spherical Dynkin diagram is one of the points of weight 1. The number
of tips of $a_n$, $d_n$ ($n\ge 3$), $e_6$, $e_7$, $e_8$ is $n$,
3,2,1,0. The tips of a connected Dynkin diagram $R$ are in natural 1:1
correspondence with the nonzero elements of the group $A'/A$, where
$A$ is the lattice generated by $R$.
The automorphism group of $A$ is a split extension of its Weyl group
by $N$, where $N$ is the group of automorphisms of $A$ fixing
$D$. This group $N$ acts on the Dynkin diagram of $D$ and $\Aut(A)$ is
determined by its Dynkin diagram $R$, the group $N$, and the action of
$N$ on $R$. There is a unique element $i$ of the Weyl group taking
$D$ to $-D$, and $-i$ is called the {\it opposition involution} of $D$
and is denoted by $\sigma$ or $\sigma(D)$. The element $\sigma$
fixes $D$ and has order 1 or 2. (Usually $-\sigma$ is called the
opposition involution.)
If $A$ is Lorentzian or positive semidefinite then we can still talk
about its root system and $A$ still has a fundamental domain $D$ for
its Weyl group and a set of simple roots. $A$ may or may not have a
Weyl vector.
We now describe the geometry of Lorentzian lattices and its relation to
hyperbolic space.
Let $L$ be an $(n+1)$-dimensional Lorentzian lattice (so $L$ has
signature $n-1$). Then the vectors of $L$ of zero norm form a double
cone and the vectors of negative norm fall into two components. The
vectors of norm $-1$ in one of these components form a copy of
$n$-dimensional hyperbolic space $H_n$. The group ${\rm Aut}(L)$ is a
product $Z_2\times {\rm Aut}_+(L)$, where $Z_2$ is generated by $-1$
and ${\rm Aut}_+(L)$ is the subgroup of ${\rm Aut}(L)$ fixing each
component of negative norm vectors. See Vinberg [8] for more
details.
If $r$ is any vector of $L$ of positive norm then $r^\perp$ gives a
hyperplane of $H_n$ and reflection in $r^\perp$ is an isometry of
$H_n$. If $r$ has negative norm then $r$ represents a point of $H_n$
and if $r$ is nonzero but has zero norm then it represents an infinite
point of $H_n$.
The group $G$ generated by reflections in roots of $L$ acts as a
discrete reflection group on $H_n$ so we can find a fundamental
domain $D$ for $G$ which is bounded by reflection hyperplanes. The
group ${\rm Aut}_+(L)$ is a split extension of this reflection group by
a group of automorphisms of $D$.
Finally we recall Conway's calculation of $\Aut_+(II_{25,1})$; see
Conway [3] or Borcherds [1]. If $\Lambda$ is the Leech lattice then
$II_{25,1}$ is isomorphic to the set of all points $(\lambda,m,n)$
with $\lambda$ in $\Lambda$, and $m$, $n$ integers, with the norm
given by $(\lambda,m,n)^2=\lambda^2-2mn$. If $w=(0,0,1)$ then the
roots of $II_{25,1}$ which have inner product $-1$ with $w$ are the
points $(\lambda,1,{1\over 2} \lambda^2-1)$ which form a set of simple
roots of a fundamental domain $D$ of the reflection group of
$II_{25,1}$, so that $\Aut_+(II_{25,1})$ is a split extension of the
reflection group by $\Aut(D)$, and $\Aut(D)$ is isomorphic to the
group $\cdot\infty$ of affine automorphisms of $\Lambda$, which is in
turn a split extension of $Z^{24}$ by Conway's group $\cdot
0=\Aut(\Lambda)$.
\proclaim 2.~Notation.
We define notation for the rest of this paper. $L$ is $II_{25,1}$,
with fundamental domain $D$ and Weyl vector $w$. We identify the
simple roots of $D$ with the affine Leech lattice $\Lambda$.
$R$ and $S$ are two sublattices of $II_{25,1}$ such that $R^\perp=S$,
$S^\perp=R$, and $R$ is positive definite and generated by a nonempty
set of simple roots of $D$. The Dynkin diagram of $R$ is the set of
simple roots of $D$ in $R$ and is a union of $a$'s, $d$'s, and
$e$'s. The finite group $R'/R$ is naturally isomorphic to $S'/S$ (in
more than one way) because $II_{25,1}$ is unimodular, and subgroups of
$R'/R$ correspond naturally to subgroups of $S'$ containing $S$ (in
just one way). We fix a subgroup $G$ of $R'/R$ and write $T$ for the
subgroup of $S'$ corresponding to it, so that an element $s$ of $S'$
is in $T$ if and only if it has integral inner product with all
elements of $G$. It is $T$ that we will be finding the automorphism
group of in the rest of this paper. For each component $R_i$ of the
Dynkin diagram of $R$ the nonzero elements of the group $\langle
R_i\rangle'/\langle R_i \rangle$ can be identified with the tips of
$R_i$, and $R'/R$ is a product of these groups. Any automorphism of
$D$ fixing $R$ acts on the Dynkin diagram of $R$ and on $R'/R$, and
these actions are compatible with the map from tips of $R$ to
$R'/R$. In particular we can talk of the automorphisms of $D$ fixing
$R$ and $G$. We will also write $R$ for the Dynkin diagram of $R$. We
write $x'$ for the projection of any vector $x$ of $II_{25,1}$ into
$S$.
{\sl Example.} $\Lambda$ contains a unique orbit of $d_{25}$'s; let
$R$ be generated by a $d_n$ $(n\ge 2)$ contained in one of these
$d_{25}$'s. (N.B.: $\Lambda$ contains two classes of $d_{16}$'s and
$d_{24}$'s.) Then $R^\perp=S$ is the even sublattice of $I_{25-n,1}$
and $R'/R$ has order 4. We can choose $G$ in $R'/R$ to have order 2 in
such a way that $T$ is $I_{25-n,1}$. We will use this to find ${\rm
Aut} (I_{m,1})$ for $m\le 23$.
We write $\Aut(T)$ for the group of automorphisms of $T$ induced by
automorphisms of $II_{25,1}$. This has finite index in the group of
all automorphisms of $T$ and is equal to this group in all the
examples of $T$ we give. This follows from the fact that if $g$ is an
automorphism of $T$ fixing $S$ and such that the automorphism of
$T/S$ it induces is induced by an automorphism of $R$ (under the
identification of $G$ with $S'/S$) then $g$ can be extended to an
automorphism of $II_{25,1}$.
\proclaim 3.~Some automorphisms of $T$.
The group of automorphisms of $D$ fixing $(R,G)$ obviously acts on
$T$. In this section we construct enough other automorphisms of $T$ to
generate $\Aut(T)$ and in the next few sections we find how these
automorphisms fit together. Recall that $\Lambda$ is the Dynkin diagram
of $D$ and $R$ is a spherical Dynkin diagram of $\Lambda$.
Let $r$ be any point of $\Lambda$ such that $r\cup R$ is a spherical
Dynkin diagram and write $R'$ for $R\cup r$. If $g'$ is any element of
$\cdot\infty$ such that $\sigma(R')g'$ fixes $r$ and $R$ then we
define an automorphism $g=g(r,g')$ of $II_{25,1}$ by
$g=\sigma(R)\sigma(R')g'$. (Recall that $\sigma(X)$ is the opposition
involution of $X$, which acts on the Dynkin diagram $X$ and acts as
$-1$ on $X^\perp$.) These automorphisms will turn out to be a sort of
generalized reflection in the sides of a domain of $T$.
\proclaim Lemma 3.1.
\item{(1)} If $g=g(r,g')$ fixes the group $G$ then $g$ restricted to
$T$ is an automorphism of $T$. $g$ fixes $R$ and $G$ if and only if
$\sigma(R')g'$ does.
\item{(2)} $g$ fixes the space generated by $r'$ and $w'$ and acts
on this space as reflection in $r'^\perp$.
\item{(3)} If $g'=1$ then $g$ acts on $T$ as reflection in $r'^\perp$.
($g'$ can only be 1 if $\sigma(R')$ fixes $r$.)
{\sl Proof.} Both $\sigma(R)$ and $\sigma(R')g'$ exchange the two cones
of norm 0 vectors in $L$ and fix $R$, so $g$ fixes both the cones of
norm 0 vectors and $R$ and hence fixes $R^\perp=S$. If $g$ also fixes
$G$ then it fixes $T$ as $T$ is determined by $G$ and $S$. $\sigma(R)$
acts as $-1$ on $G$ and fixes $R$, so $g$ fixes $R$ and $G$ if and
only if $\sigma(R')g'$ does. This proves (1).
$\sigma(R')g'$ fixes $r$ and $R$ and so fixes $r'$. $\sigma(R)$ acts
as $-1$ on anything perpendicular to $R$, and in particular on $r'$,
so $g(r')=[\sigma(R)\sigma(R')g'](r')=-r'$. If $v$ is any vector of
$L$ fixed by $g'$ and $v'$ is its projection into $T$, then
$g(v)=\sigma(R)\sigma(R')v$ so that $g(v)-v$ is in the space generated
by $R'$ and hence $g(v')-v'$ is in the space generated by $r'$. As
$g(r')=-r'$, $g(v')$ is the reflection of $v'$ in $r'^\perp$. In
particular if $v=w$ or $g'=1$ then $g'$ fixes $v$ so $g$ acts on $v'$
as reflection in $r'^\perp$. This proves (2) and (3). Q.E.D.
\proclaim Lemma 3.2.
\item{(0)} The subgroup of $\cdot\infty=\Aut(D)$
fixing $(R,G)$ maps onto the subgroup of $\Aut_+(T)$ fixing $w'$.
\item{(1)} If $\sigma(R')$ fixes $(R,G)$ then $g(r,1)$ acts on $T$
as reflection in $r'^\perp$ and this is an automorphism of $T$.
\item{(2)} If $\sigma(R')$ does not fix $(R,G)$ then we define a map $f$ from
a subset of $\Aut(D)$ to $\Aut(T)$ as follows:
\itemitem{}If $h$ in $\Aut(D)$ fixes all points of $R'$ then we put $f(h)=h$
restricted to $T$.
\itemitem{} If $h'$ in $\Aut(D)$ acts as $\sigma(R')$ on $R'$ then we put
$f(h')=g(r,h')$ restricted to $T$. (This is a sort of twisted reflection
in $r'^\perp$.)
\item{} Then the elements on which we have defined $f$ form a subgroup of
$\Aut(D)$ and $f$ is an isomorphism from this subgroup to its image in
$\Aut(T)$. $f(h)$ fixes $r'^\perp$ and $w'$ while $f(h')$ fixes $r'^\perp$
and acts on $w'$ as reflection in $r'^\perp$. (So $f(h)$ fixes the two
half-spaces of $r'^\perp$ while $f(h')$ exchanges them.)
{\sl Proof.} Parts (0) and (1) follow from 3.1.
If $h$ in $\Aut(D)$ fixes all points of $R'$ then it certainly fixes
all points of $R$ and $G$ and so acts on $T$. $h$ also fixes $w$ and
$w'$.
If $h'$ acts as $\sigma(R')$ on $R'$ then $\sigma(R')g'$ fixes all
points of $R$, so by 3.1(1), $g(r,h')$ is an automorphism of $T$, and
by 3.1(2), $g(r,h')$ maps $w'$ to the reflection of $w'$ in
$r'^\perp$. (There may be no such automorphisms $g'$, in which case the
lemma is trivial.) It is obvious that $f$ is defined on a subgroup of
$\Aut(D)$, so it remains to check that it is a homomorphism.
We write $h$, $i$ for elements of $\Aut(D)$ fixing all points of $R'$
and $h'$, $i'$ for elements acting as $\sigma(R')$. All four of these
elements fix $R'$ and so commute with $\sigma(R')$. $\quad h$, $i$,
$\sigma(R')h'$, and $\sigma(R')i'$ fix $R$ and so commute with
$\sigma(R)$. $\sigma(R)^2=\sigma(R')^2=1$. Using these facts it follows
that
$$\eqalign{
f(hi)&= hi=f(h)f(i)\cr
f(hi')&=\sigma(R)\sigma(R')hi' = h\sigma(R)\sigma(R')i'=f(h)f(i')\cr
f(h'i)&= \sigma(R)\sigma(R')h'i=f(h')f(i)\cr
f(h'i')&= h'i' \cr
&=\sigma(R)^2\sigma(R')^2h'i'\cr
&=\sigma(R)^2\sigma(R')h'\sigma(R')i'\cr
&=\sigma(R)\sigma(R')h'\sigma(R)\sigma(R')i'\cr
&=f(h')f(i')\cr
}$$
so $f$ is a homomorphism. Q.E.D.
\proclaim 4.~Hyperplanes of $T$.
We now consider the set of hyperplanes of $T$ of the form $r'^\perp$,
where $r$ is a root of $II_{25,1}$ such that $r'$ has positive
norm. These hyperplanes divide the hyperbolic space of $T$ into
chambers and each chamber is the intersection of $T$ with some chamber
of $II_{25,1}$. We write $D'$ for the intersection of $D$ with $T$.
In the section we will show that $D'$ is often a sort of fundamental
domain with finite volume. It is rather like the fundamental domain of
a reflection group, except that it has a nontrivial group acting on
it, and the automorphisms of $T$ fixing sides of $D'$ are more
complicated than reflections.
\proclaim Lemma 4.1. $D'$ contains $w'$ in its interior and, in particular, is
nonempty.
{\sl Proof.} To show that $w'$ is in the interior of $D'$ we have to
check that no hyperplane $r^\perp$ of the boundary of $D$ separates
$w$ and $w'$, unless $r$ is in $R$. $r$ is a simple root of $D$ with
$-(r,w)=1$ so it is enough to prove that $(r,\rho)\ge 0$, where
$\rho=w-w'$ is the Weyl vector of the lattice generated by
$R$. $-\rho$ is a sum of simple roots of $R$, so $(r,\rho)\ge 0$
whenever $r$ is a simple root of $D$ not in $R$ because all such
simple roots have inner product $\le 0$ with the roots of $R$. This
proves that $w'$ is in the interior of $D'$. Q.E.D.
\proclaim Lemma 4.2.
The faces of $D'$ are the hyperplanes $r'^\perp$, where $r$ runs
through the simple roots of $D$ such that $r\cup R$ is a spherical
Dynkin diagram and $r$ is not in $R$. In particular $D'$ has only a
finite number of faces because $R$ is not empty.
{\sl Proof.} The faces of $D'$ are the hyperplanes $r'^\perp$ for the
simple roots $r$ of $D$ such that $r'^\perp$ has positive norm, and
these are just the simple roots of $D$ with the property in 4.2. $D'$
has only a finite number of faces because the Leech lattice
(identified with the Dynkin diagram of $II_{25,1}$) has only a finite
number of points at distance at most $\sqrt 6$ from any given point in
$R$. Q.E.D.
\proclaim Lemma 4.3.
If $R$ does not have rank 24 (i.e., $T$ does not have dimension 2)
then $D'$ has finite volume.
{\sl Proof.} $D'$ is a convex subset of hyperbolic space bounded by a
finite number of hyperplanes, and this hyperbolic space is not one
dimensional as $T$ is not two dimensional, so $D'$ has finite volume
if and only if it contains only a finite number of infinite
points. $R$ is nonempty so it contains a simple root $r$ of $D$. The
points of $D'$ at infinity correspond to some of the isotropic
subspaces of $II_{25,1}$ in $r^\perp$ and $D$. The hyperplane
$r^\perp$ does not contain $w$ as $(r,w)=-1$, so the fact that $D'$
contains only a finite number of infinite points follows from 4.4
below (with $V=r^\perp$). Q.E.D.
\proclaim Lemma 4.4.
If $V$ is any subspace of $II_{25,1}$ not containing $w$ then $V$
contains only a finite number of isotropic subspaces that lie in $D$.
{\sl Proof.} Let $II_{25,1}$ be the set of vectors $(\lambda,m,n)$
with $\lambda$ in $\Lambda$, $m$ and $n$ integers, with the norm given
by $(\lambda,m,n)^2=\lambda^2-2mn$. We let $w$ be $(0,0,1)$ so that
the simple roots are $(\lambda,1,\lambda^2/2-1)$. As $V$ does not
contain $w$ there is some vector $r=(v,m,n)$ in $V^\perp$ with
$(r,w)\ne 0$, i.e., $m\ne 0$. We let each norm 0 vector $z=(u,a,b)$
which is not a multiple of $w$ correspond to the point $u/a$ of
$\Lambda\otimes Q$. If $z$ lies in $V$ then $(z,r)=0$, so
$(u/a-v/m)^2=(r/m)^2$, so $u/a$ lies on some sphere in $\Lambda\otimes
Q$. If $z$ is in $D$ then $u/a$ has distance at least $\sqrt 2$ from
all points of $\Lambda$ (i.e., it is a ``deep hole''), but as
$\Lambda$ has covering radius $\sqrt 2$ these points form a discrete
set so there are only a finite number of them on any sphere. Hence
there are only a finite number of isotropic subspaces lying in $V$ and
$D$. Q.E.D.
{\sl Remark.} If $w$ is in $V$ then the isotropic subspaces of
$II_{25,1}$ in $V$ and $D$ correspond to deep holes of $\Lambda$ lying
on some affine subspace of $\Lambda\otimes Q$. There is s universal
constant $n_0$ such that in this case $V$ either contains at most
$n_0$ isotropic subspaces in $D$ or contains an infinite number of
them. If $w$ is not in $V$ then $V$ can contain an arbitrarily large
number of isotropic subspaces in $D$.
\proclaim 5.~A complex.
We have constructed enough automorphisms to generate $\Aut(T)$, and
the problem is to fit them together to give a presentation of
$\Aut(T)$. We will do this by constructing a contractible complex
acted on by $\Aut(T)$. For example, if this complex is one dimensional
it is a tree, and groups acting on trees can often be written as
amalgamated products.
{\sl Notation.} We write $\Aut_+(T)$ for the group of automorphisms of
$T$ induced by $\Aut_+(II_{25,1})$. $D_r$ is a fundamental domain of
the reflection subgroup of $\Aut_+(T)$ containing $D'$. The hyperbolic
space of $T$ is divided into chambers by the conjugates of all
hyperplanes of the form $r'^\perp$ for simple roots $r$ of $D$.
\proclaim Lemma 5.1.
Suppose that for any spherical Dynkin diagram $R'$ containing $R$ and
one extra point of $\Lambda$ there is an element of $\cdot\infty$
acting as $\sigma(R')$ on $R'$. Then $\Aut_+(T)$ acts transitively on
the chambers of $T$ and $\Aut(D_r)$ acts transitively on the chambers
of $T$ in $D_r$.
{\sl Proof. } By lemma 3.2 there is an element of $\Aut_+(T)$ fixing
any face of $D'$ corresponding to a root $r$ of $D$ and mapping $D'$
to the other side of this face. Hence all chambers of $T$ are
conjugates of $D'$. Any automorphism of $T$ mapping $D'$ to another
chamber in $D_r$ must fix $D_r$, so $\Aut(D_r)$ acts transitively on
the chambers in $D_r$. Q.E.D.
$D_r$ is decomposed into chambers of $T$ by the hyperplanes
$r'^\perp$ for $r$ a root of $II_{25,1}$. We will write $U$ for the
dual complex of this decomposition and $U'$ for the subdivision of
$U$. $U$ has a vertex for each chamber of $D_r$, a line for each pair
of chambers with a face in common, and so on. $U'$ is a simplicial
complex with the same dimension as $U$ with an $n$-simplex for each
increasing sequence of $n+1$ cells of $U$. $U$ is not necessarily a
simplicial complex and need not have the same dimension as the
hyperbolic space of $T$; in fact it will usually have dimension 0, 1,
or 2. For example, if $D_r=D'$ then $U$ and $U'$ are both just points.
\proclaim Lemma 5.2. $U$ and $U'$ are contractible.
{\sl Proof. } $U$ is contractible because it is the dual complex of
the contractible space $D_r$. ($D_r$ is even convex.) $U'$ is
contractible because it is the subdivision of $U$. Q.E.D.
\proclaim Theorem 5.3.
Suppose that $\Aut(D_r)$ acts transitively on the maximal simplexes of
$U'$, and let $C$ be one such maximal simplex. Then $\Aut(D_r)$ is the
sum of the subgroups of $\Aut(D_r)$ fixing the vertices of $C$
amalgamated over their intersections.
{\sl Proof.} By 5.2, $U'$ is connected and simply connected. $C$ is
connected and by assumption is a fundamental domain for $\Aut(D_r)$
acting on $U'$. By a theorem of Macbeth (Serre [6, p. 31]) the group
$\Aut(D_r)$ is given by the following generators and relations:
{\sl Generators:} An element $\hat g$ for every $g$ in $\Aut(D_r)$
such that $C$ and $g(C)$ have a point in common.
{\sl Relations:} For every pair of elements $(s,t)$ of $\Aut(D_r)$
such that $C$, $s(C)$, and $u(C)$ have a point in common (where
$u=st$) there is a relation $\hat s\hat t=\hat u$.
Any element of $\Aut(D_r)$ fixing $C$ must fix $C$ pointwise. This
implies that $C$ and $g(C)$ have a point in common if and only if $g$
fixes some vertex of $C$, i.e., $g$ is in one of the groups $C_0$,
$C_1,\ldots$ which are stabilizers of the vertices of $C$, so we have
a generator $\hat g$ for each $g$ that lies in (at least) one of these
groups. There is a point in all of $C$, $s(C)$, and $u(C)$ if and only
if some point of $C$, and hence some vertex of $C$, is fixed by $s$
and $t$. This means that we have a relation $\hat s\hat t=\hat u$
exactly when $s$ and $t$ both lie in some group $C_i$. This is the
same as saying that $\Aut(D_r)$ is the sum of the groups $C_i$
amalgamated over their intersections. Q.E.D.
{\sl Example.} If $C$ is one dimensional then $\Aut(D_r)$ is the free
product of $C_0$ and $C_1$ amalgamated over their intersection. If the
dimension of $C$ is not 1 then $\Aut(D_r)$ cannot usually be written
as an amalgamated product of two nontrivial groups.
\proclaim 6.~Unimodular lattices.
In this section we apply the results of the previous section to find
the automorphism group of $I_{m,1}$ for $m\le 23$.
\proclaim Lemma 6.1.
Let $X$ be the Dynkin diagram $a_n$ ($1\le n\le 11$), $d_n$ ($2\le n\le
11$), $e_n$ ($3\le n\le 8$), or $a_2^2$ which is contained in
$\Lambda$. Then any automorphism of $X$ is induced by an element of
$\cdot\infty$. If $X$ is an $a_n$ ($1\le n\le 10$), $d_n$ ($2\le n\le
25$, $n\ne 16$ or $24$), $e_n$ ($3\le n\le 8$), or $a_2^2$ then
$\cdot\infty$ acts transitively on Dynkin diagrams of type $X$ in
$\Lambda$.
{\sl Proof.} A long, unenlightening calculation. See section 9. Q.E.D.
{\sl Remark.} $\cdot\infty$ acts simply transitively on ordered
$a_{10}$'s in $\Lambda$. There are two orbits of $d_{16}$'s and
$d_{24}$'s (see section 9 and example 2 of section 8) and many orbits
of $a_n$'s for $n\ge 11$.
{\sl Notation.} We take $R$ to be a $d_n$ contained in a $d_{25}$ for
some $n$ with $2\le n\le 23$. If $R$ is $d_4$ we label the tips of $R$
as $x$, $y$, $z$ in some order, and if $R$ is $d_n$ for $n\ne 4$ we
label the two tips that can be exchanged by an automorphism of $R$ as
$x$ and $y$. If $n=3$ or $n\ge 5$ we label the third tip of $R$ as
$z$. We let $G$ be the subgroup of $\langle R'\rangle/\langle R
\rangle $ of order 2 which corresponds to the tip $z$ if $n\ge 3$ and
to the sum of the elements $x$ and $y$ if $n=2$. An automorphism of
$R$ fixes $G$ if $n\ne 4$ or if $n=4$ and it fixes $z$. The lattice
$S=R^\perp$ is isomorphic to the even sublattice of $I_{25-n,1}$. We
let $T$ be the lattice corresponding to $G$ that contains $S$, so that
$T$ is isomorphic to $I_{25-n,1}$.
\proclaim Lemma 6.2.
Any root $r$ of $V$ such that $r\cup R$ is a spherical Dynkin diagram
is one of the following types:
\itemitem{Type $a$:} $r\cup R$ is $d_na_1$ (i.e., $r$ is not joined to any
point of $R$.) $r'$ is then a norm 2 vector of $T$.
\itemitem{Type $d$:}
$r$ is joined to $z$ if $n\ge 3$ or to $x$ and $y$ if $n=2$, so that
$r\cup R$ is $d_{n+1}$. $r'$ has norm 1.
\itemitem{Type $e$:}
$r$ is joined to just one of $x$ or $y$, so that $r\cup R$ is $e_{n+1}$.
$2r'$ is then a characteristic vector of norm $8-n$ in $T$.
{\sl Proof.} Check all possible cases. Q.E.D.
In particular if $r$ is of type $a$ or $d$, or of type $e$ with $n=6$
or 7, then $r'$ (or $2r'$) has norm 1 or 2 and so is a root of
$T$. Note that in these cases $\sigma(R\cup r)$ fixes $R$ and $z$ and
therefore $G$, so by 3.2(2), $r'^\perp$ is a reflection of $T$. In the
remaining four cases ($r$ of type $e$ with $2\le n\le 5$)
$\sigma(r\cup R)$ does not fix both $R$ and $G$.
\proclaim Corollary 6.3.
If $n\ge 6$ then then reflection group of $T=I_{25-n,1}$ has finite
index in $\Aut(T)$. Its fundamental domain has finite volume and a
face for each root $r$ of $\Lambda$ such that $r\cup R$ is a spherical
Dynkin diagram.
{\sl Proof.} This follows from the fact that all walls of $D'$ give
reflections of $T$ so $D'$ is a fundamental domain for the reflection
group. By 4.3, $D'$ has finite volume. Q.E.D.
{\sl Remarks.} The fact that a fundamental domain for the reflection
group has finite volume was first proved in Vinberg [8] for $n\ge 8$
and in Vinberg and Kaplinskaja [9] for $n=6$ and $7$ (i.e., for
$I_{18,1}$ and $I_{19,1}$). Conway and Sloane [4] show implicitly that
for $n\ge 6$ the non-reflection part of $\Aut(T)$ is the subgroup of
$\cdot\infty$ fixing $R$ and their description of the fundamental
domain of the reflection group in these cases is easily seen to be
equivalent to that in 6.3. $I_{18,1}$ and $I_{19,1}$ have a ``second
batch'' of simple roots of norm 1 or 2; from 6.2 we see that this
second batch consists of the roots which are characteristic vectors
and they exist because of the existence of $e_8$ and $e_7$ Dynkin
diagrams. The non-reflection group of $I_{20,1}$ is infinite because of
the existence of $e_6$ Dynkin diagrams and the fact that the
opposition involution of $e_6$ acts non-trivially on the
$e_6$. ($\dim(I_{20,1})=1+\dim(II_{25,1})+\dim(e_6)$.)
From now on we assume that $n$ is 2, 3, 4, or 5. Recall that $D_r$ is
the fundamental domain of the reflection group of $T$ containing $D$
and $U$ is the complex which is the dual of the complex of conjugates
of $D$ in $D_r$.
\proclaim Lemma 6.4. $\Aut(D_r)$ acts transitively on the vertices of $U$.
{\sl Proof. } This follows from 6.1 and 5.1. Q.E.D.
\proclaim Lemma 6.5.
If $n=3$, $4$, or $5$ then $U$ is one dimensional and if $n=2$, then $U$
is two dimensional. (By the remarks after 6.3, $U$ is zero
dimensional for $n\ge 6$.)
{\sl Proof.} Let $s$ and $t$ be simple roots of $\Lambda$ such that
$s\cup R$ and $t\cup R$ are $e_{n+1}$'s, so that $s'^\perp$ and
$t'^\perp$ give two faces of $D'$ inside $D_r$. Suppose that these
faces intersect inside $D_r$. We have $s'^2=t'^2=2-n/4$, $(s',t')\le
0$, and $r=s'+t'$ lies in $T$ (as $2s'$ and $2t'$ are both
characteristic vectors of $T$ and so are congruent mod $2T$). $r'^2$
cannot be 1 or 2 as then the intersection of $s'^\perp$ and $t'^\perp$
would lie on the reflection hyperplane $r^\perp$, which is impossible
as we assumed that $s'^\perp$ and $t'^\perp$ intersected somewhere in
the interior of $D_r$. Hence
$$ 3\le r^2=(s'+t')^2\le 2(2-n/4)\le 2(2-2/4)=3$$
so $r^2=3$, $n=2$, and $(s',t')=0$.
If $n=3$, 4, or 5 this shows that no two faces of $D'$ intersect in
the interior of $D_r$, so the graph whose vertices are the conjugates
of $D'$ in $D_r$ such that two vertices are joined if and only if the
conjugates of $D'$ they correspond to have a face in common is a
tree. As it is the 1-skeleton of $U$, $U$ must be one dimensional.
If $n=2$ then it is possible for $s'^\perp$ and $t'^\perp$ to
intersect inside $D_r$. In this case they must intersect at right
angles, so $U$ contains squares. However, in this case $s$ and $t$
cannot be joined to the same vertex of $R=a_1^2$, and in particular it
is not possible for three faces of $D'$ to intersect inside $D_r$, so
$U$ is two dimensional. (Its two-dimensional cells are squares.)
Q.E.D.
If $X$ is a Dynkin diagram in $\Lambda$ we will write $G(X)$ for the
subgroup of $\cdot\infty$ fixing $X$. If $X_1\subset X_2\subset
X_3\cdots$ is a sequences of Dynkin diagrams of $\Lambda$ we write
$G(X_1\subset X_2\subset X_3\cdots)$ for the sum of the groups
$G(X_i)$ amalgamated over their intersections.
\proclaim Theorem 6.6.
The non-reflection part of $\Aut_+(T)=\Aut_+(I_{25-n,1})$ is given by
$$
\eqalign{
G(d_5\subset e_6) & \hbox{ if } n=5,\cr
G(d_4^*\subset d_5) & \hbox{ if } n=4
\hbox{ (where $d_4^*$ means that one of the tips of the $d_4$ is labeled)},\cr
G(a_3\subset a_4) & \hbox{ if } n=3,\cr
G(a_1^2\subset a_1a_2\subset a_2^2) & \hbox{ if } n=2.\cr
}
$$
{\sl Proof.} By 6.4, $\Aut(D_r)$ acts transitively on the vertices of
$U$. Using this and 6.1 it is easy to check that it acts transitively
on the maximal flags of $U$, or equivalently on the maximal simplexes
of the subdivision $U'$ of $U$. For example, if $n=5$ this amounts to
checking that the group $G(d_5)$ acts transitively on the $e_6$'s
containing a $d_5$.
By 5.3 the group $\Aut(D_r)$ is the sum of the groups fixing each
vertex of a maximal simplex $C$ of $U'$ amalgamated over their
intersections. By 3.2 the subgroup of $\Aut(D_r)$ fixing the vertex
$D'$ of $U$ can be identified with $G(R)=G(d_n)$ and the subgroup
fixing a face of $D'$ can be identified with $G(e_{n+1})$ for the
$e_{n+1}$ corresponding to this face. These groups are the groups
fixing two of the vertices of $C$, and if $n=2$ it is easy to check
that the group fixing the third vertex can be identified with
$G(a_2^2)$. Hence $\Aut(D_r)$ is $G(d_n\subset e_{n+1})$ if $n=3$, 4,
or 5 (where if $n=4$ we have to use a subgroup of index 3 in
$G(d_4)$), and $G(d_2\subset e_3\subset a_2^2)$ if $n=2$. Q.E.D.
{\sl Examples.} $n=5$: $\Aut(I_{20,1})$ The domain $D'$ has 30 faces of
type $a$, 12 of type $d$, and 40 of type $e$. $\Aut(D')$ is
$\Aut(A_6)$ of order 1440 (where $A_6$ is the alternating group of
order 360). $\Aut(D_r)$ is
$$\Aut(A_6)*_H(S_3\wr Z_2),$$
where $S_3\wr Z_2=G(e_6)$ is a wreathed product and has order
72. $H$ is its unique subgroup of order 36 containing an element of
order 4. (Warning: $\Aut(A_6)$ contains two orbits of subgroups
isomorphic to $H$. The image of $H$ in $\Aut(A_6)$ is not contained in
a subgroup of $\Aut(A_6)$ isomorphic to $S_3\wr Z_2$.) $\Aut(D_r)$ has Euler
characteristic $1/1440+1/72-1/36=-19/1440$.
$n=4$: $\Aut(I_{21,1})$. $D'$ has 42 faces of type $a$, 56 of type
$d$, and 112 of type $e$. $\Aut(D')$ is $L_3(4)\cdot 2^2$. $\Aut(D_r)$
is
$$L_3(4).2^2 *_{M_{10}}\Aut(A_6).$$
$\Aut(A_6)$ has 3 subgroups of index 2, which are $S_6$, $PSL_2(9)$, and
$M_{10}$. $\Aut(D_r)$ has Euler characteristic $-11/2^8.3^2.7.$
$n=3$: $\Aut(I_{22,1})$. $D'$ has 100 faces of type $a$, 1100 of type
$d$, and 704 of type $e$. $\Aut(D')$ is $HS.2$, where $HS$ is the
Higman-Sims simple group, and $\Aut(D_r)$ is
$$HS.2*_HH.2$$
where $H$ is $PSU_3(5)$ of order $2^4.3^2.5^3.7$. Its Euler
characteristic is $3.13/2^{10}.5^5.7.11$.
$n=2$: $\Aut(I_{23,1})$. $D'$ has 4600 faces of type $a$, 953856 of
type $d$, and 94208 of type $e$. $\Aut(D')$ is $\cdot 2\times2$, where
$\cdot 2$ is one of Conway's simple groups. $\Aut(D_r)$ is the direct
limit of the groups
$$
\matrix
{
&&&\cdot2\times 2&=&G(a_1^2)\cr
&&\nearrow&&\nwarrow\cr
&McL&\leftarrow&PSU_4(3)&\rightarrow&PSU_4(3).2\cr
&\downarrow&&\downarrow&&\downarrow\cr
G(a_1a_2)=&McL.2&\leftarrow&PSU_4(3).2
&\rightarrow&PSU_4(3)\cdot D_8&=G(a_2^2)\cr
}
$$
$McL$ is the McLaughlin simple group and $D_8$ is the dihedral group
of order 8. The direct limit is generated by $\cdot 2\times 2$ and an
outer automorphism of $McL$; the $PSU_4(3)$'s are there to supply one
additional relation. The Euler characteristic of $\Aut(D_r)$ is the
sum of the reciprocals of the orders of the groups in the center and
the vertices of the diagram above minus the sum of the reciprocals of
the groups on the edges (see Serre [7]), which is
$3191297/2^{20}.3^6.5^3.7.11.23$.
{\sl Remark.} For $n=2$, 3, 4, 5, or 7 the number of faces of $D'$ of
type $e$ is $(24/(n-1)-1)2^{12/(n-1)}$, which is $23\cdot2^{12}$,
$11\cdot2^6$, $7\cdot2^4$, $5\cdot 2^3$, or $3\cdot 2^2$. For $n=6$
this expression is $20.06\ldots$ and there are 20 faces of type
$e$. Table 3 of Conway and Sloane [4] gives the number of faces of
type $a$ and $d$ of $I_{m,1}$ for $m\le 23$.
\proclaim 7.~More about $I_{n,1}$.
Here we give more information about $I_{n,1}$ for $20\le n\le 23$. In
the tables in Conway and Sloane [4] the heights of the simple roots
they calculate appear to lie on certain arithmetic progressions; we
prove that they always do. We then prove that the dimension of the
complex $U$ is the virtual cohomological dimension of the
non-reflection part of $\Aut(I_{n,1})$.
{\sl Notation.} $D$ is a $d_n$ of $\Lambda$ contained in a
$d_{25}$. Let $C$ be the sublattice of all elements of $T=I_{25-n,1}$
which have even inner product with all elements of even norm. $C$
contains $2T$ with index 2 and the elements of $C$ not in $2T$ are the
characteristic vectors of $T$. $w'$ is the projection of $w$ into $T$
and $D_r$ is the fundamental domain of the reflection group of $T$
containing $w'$.
\proclaim Lemma 7.1.
Suppose $2\le n\le 5$ Then all conjugates of $w'$ in $D_r$ are
congruent $\bmod 2^{n-2}C$. (If $n\ge 6$ then $D_r=D'$ so there are no
other conjugates of $w'$ in $D_r$.)
{\sl Proof.} It is sufficient to prove that any two conjugates of $w'$
which are joined as vertices of the graph of $D_r$ are congruent
$\bmod 2^{n-2}C$ because this graph is connected, and we can also
assume that one of these vertices is $w'$ because $\Aut(D_r)$ acts
transitively on its vertices. Let $w''$ be a conjugate of $w'$ joined
to $w'$.
$w''$ is the reflection of $w'$ in some hyperplane $e^\perp$, where
$e$ is a characteristic vector of $T$ of norm $8-n$, so $e$ is in $C$.
Therefore
$$w''=w'-2(w',e)e/(e,e).$$
$e=2r'$ for a vector $r$ of $\Lambda$ such that $r\cup R$ is an
$e_{n+1}$ diagram. The projections of $r$ and $w$ into the lattice
$I^n$ containing $R$ are $({1\over 2},{1\over 2},\ldots,{1\over 2})$
and $(0,1,\ldots,n-1)$, which have inner product $(n-1)n/4$. Hence
$(w',r')=(w,r)-(\hbox{inner product of projections of $w$ and $r$ into
$\langle R\rangle$})=-1-(n-1)n/4$, so
$$-2(w',e)/(e,e)=(4+(n-1)n)/(8-n).$$
For $2\le n\le 5 $ the expression on the right is equal to $2^{n-2}$, so
$w'$ and $w''$ are congruent ${} \bmod 2^{n-2}C$. Q.E.D.
\proclaim Theorem 7.2.
Suppose $n=2$, $3$, $4$, or $5$ and let $r$ be a simple root of the
fundamental domain $D_r$ of $T=I_{25-n,1}$.
\item{} If $r^2=1$ then $(r,w')\equiv -n\bmod 2^{n-2}$.
\item{} If $r^2=2$ then $(r,w')\equiv -1\bmod 2^{n-1}$.
{\sl Proof.} $(r,w')=(s',w'')$ for some conjugate $w''$ of $w$ in
$D_r$ and some simple root $s'$ of $I_{25-n,1}$ that is the projection
of a simple root $s$ of $D$ into $I_{25-n,1}$. By 7.1, $w'$ is
congruent to $w'' \bmod 2^{n-2}C$ so $(r,w')$ is congruent to
$(s',w')\bmod 2^{n-2}$ if $r^2=1$ and ${} \bmod 2^{n-1}$ if $r^2=2$
(because elements of $C$ have even inner product with all elements of
norm 2). $(s',w')$ is equal to $(s,w)-$(inner product of projections
of $s$ and $w$ into $R$), which is $-1$ if $s'$ has norm 2 and
$-1-(n-1)$ if $s'$ has norm 1. Q.E.D.
This explains why the heights of the simple roots for $I_{m,1}$ with
$20\le n\le 23$ given in table 3 of Conway and Sloane [4] seem to lie
on certain arithmetic progressions.
Now we show that the dimension of the complex $U$ is the virtual
cohomological dimension of $\Aut(D_r)$.
\proclaim Lemma 7.3.
The cohomological dimension of any torsion free subgroup of
$\Aut(D_r)$ is at most $\dim(U)$.
{\sl Proof.} Any such subgroup acts freely on the contractible complex $U$.
Q.E.D.
\proclaim Corollary 7.4.
The virtual cohomological dimension of $\Aut(D_r)$ is equal to $\dim(U)$.
{\sl Proof.} $\Aut(D_r)$ contains torsion-free subgroups of finite
index, so by 7.3 the v.c.d. of $\Aut(D_r)$ is at most $\dim(U)$. For
$n\le 5$, $\Aut(D_r)$ is infinite and so has v.c.d. at least 1, while
for $n=2$. $\Aut(D_r)$ contains subgroups isomorphic to $Z^2$ (because
there are 22-dimensional unimodular lattices whose root systems
generate a vector space of codimension 2) so $\Aut(D_r)$ has v.c.d. at
least 2. Q.E.D.
Lemma 7.3 implies that $\Aut(D_r)$ contains no subgroups of the form
$Z^i$ with $i>\dim(U)$, and this implies that if $L$ is a
$(24-n)$-dimensional unimodular lattice then the space generated by
roots of $L$ has codimension at most $\dim(U)$. This can of course
also be proved by looking at the list of such lattices. (Vinberg used
this in reverse: he showed that the non-reflection part of
$\Aut(I_{m,1})$ was infinite for $m\ge 20$ from the existence of
19-dimensional unimodular lattices with root systems of rank 18. There
are two such lattices, with root systems $a_{11}d_7$ and $e_6^3$; they
are closely related to the two Niemeier lattices $a_{11}d_7e_6$ and
$e_6^4$ containing an $e_6$ component.)
\proclaim 8.~Other examples.
We list some more examples of (not necessarily unimodular) Lorentzian
lattices with their automorphism groups.
{\sl Example 1.} $R$ is an $e_8$ in $\Lambda$ so that $T$ is
$II_{17,1}$. The fundamental domain $D'$ of the reflection group has
finite volume and its Dynkin diagram is the set of points of $\Lambda$
not connected to $e_8$, which is a line of 17 points with 2 more
points joined onto the 3rd and 15th points. This diagram was found in
Vinberg [8].
{\sl Example 2.} Similarly if $R$ is one of the $d_{16}$'s of
$\Lambda$ not contained in a $d_{17}$ then $T$ is $II_{9,1}$ and the
points of $\Lambda$ not joined to $R$ form an $e_{10}$ which is the
Dynkin diagram of $II_{9,1}$.
{\sl Example 3.} All $e_7$'s of $\Lambda$ are conjugate; if $R$ is one
of them then $T$ is the 19-dimensional even Lorentzian lattice of
determinant 2. There are 3+21 roots $r$ for which $r\cup R$ is a
spherical Dynkin diagram and these 24 points are arranged as a ring of
18 points with an extra point joined on to every third point. The
three roots joined to the $e_7$ correspond to norm 2 roots $r$ of $T$
with $r^\perp$ unimodular, while the other 21 roots correspond to norm
2 roots of $T$ such that $r^\perp$ is not unimodular. The
non-reflection part of $\Aut_+(T)$ is $S_3$ of order 6 acting in the
obvious way on the Dynkin diagram.
{\sl Example 4.} Let $R$ be the unique orbit of $e_6$'s in
$\Lambda$. Then $T$ is the 20-dimensional even Lorentzian lattice of
determinant 3. $D'$ is a fundamental domain for the reflection group of
$T$ and has $12+24$ faces, coming from 12 roots of norm 6 and 24 of
norm 2. The non-reflection group of $\Aut_+(T)$ is a wreath product
$S_3\wr Z_2$ of order 72. (Remark added in 1998: this example was first
found by Vinberg in ``The two most algebraic $K3$ surfaces'',
Math. Ann. 265 (1983), no. 1, 1--21.)
{\sl Example 5.} $R$ is $d_4$ and $T$ is the even sublattice of
$I_{21,1}$ so that $T$ has determinant 4. The domain $D'$ is a
fundamental domain for the reflection group of $T$ and has 168 walls
corresponding to roots of norm 4 and 42 walls corresponding to roots
of norm 2. $\Aut(D)$ is isomorphic to $L_3(4).D_{12}$ of order
$2^8.3^3.5.7$. $T$ is a 22-dimensional Lorentzian lattice whose
reflection group has finite index in its automorphism group; I do not
know of any other such lattices of dimension $\ge 21$. (Remark added
in 1998: Esselmann recently proved in ``\"Uber die maximale Dimension
von Lorentz-Gittern mit coendlicher Spiegelungsgruppe'', Number Theory
61 (1996), no. 1, 103--144, that the lattice $T$ is essentially the
only example of a Lorentzian lattice of dimension at least 21 whose
reflection group has finite index in its automorphism group.) $T$ is
contained in three lattices isomorphic to $I_{21,1}$ each of whose
automorphism groups has index 3 in $\Aut(T)$. However the reflection
groups of these lattices do not have finite index in their
automorphism groups.
{\sl Example 6.} $R$ is $a_1$ and $T$ is the 25-dimensional even
Lorentzian lattice of determinant 2. This time $D'$ is not a
fundamental domain for the reflection group. It has 196560 faces
corresponding to norm 2 roots and 16773120 faces perpendicular to norm
6 vectors (which are not roots). However, the simplicial complex of $T$
is a tree so $\Aut(D')$ is $(\cdot 0)*_{(\cdot 3)}(2\times \cdot 3)$,
i.e., it is generated by $\cdot 0$ and an element $t$ of order 2 with
the relations that $t$ commutes with some $\cdot 3$ of $\cdot 0$. $D'$
has finite volume but if any of its 16969680 faces are removed the
resulting polyhedron does not!
\proclaim 9.~The automorphism groups of high-dimensional Lorentzian lattices.
{\sl Notation.} $L$ is $II_{8n+1,1}$ $(n\ge 1)$ and $D$ is a
fundamental domain of the reflection group of $L$.
$X$ is the Dynkin diagram of $D$. In this section we will show that if
$8n\ge 24$ then $\Aut(L)$ acts transitively on many subsets of $X$,
and use this to generalize some of the results of the previous
sections to higher dimensional lattices.
\proclaim Lemma 9.1.
Let $R$ be a spherical Dynkin diagram. Suppose that whenever $R'$ is a
spherical Dynkin diagram in $X$ which is isomorphic to $R$ plus one
point $r$ there is an element $g$ of $\Aut(D)$ such that $g\sigma(R')$
fixes $R$ (resp. fixes all points of $R$).
\item{} For any map $f:R\mapsto X$ we construct $(M,f',C)$, where
\item{$M$}
is the lattice $f(R)^\perp$,
\item{$f'$}
is the map from $R'/R$ to $M'/M$ such that $f(r)\equiv-f'(r)\bmod L$
for $r$ in $R'/R$,
\item{$C$}
is the cone of $M$ contained in the cone of $L$ containing $D$.
\item{} If $f_1$, $f_2$ are two such maps then the images $f_1(R)$,
$f_2(R)$ (resp. $f_1$ and $f_2$) are conjugate under $\Aut(L,D)$ if
the two pairs $(M_1,f_1',C_1)$, $(M_2,f_2',C_2)$ are isomorphic.
{\sl Proof.} It is sufficient to show that a triple $(M,f',C)$
determines $f(R)$ (resp. $f$) up to conjugacy under
$\Aut(L,D)$. Given $M$ and $f'$ we can recover $L$ as the lattice
generated by $R\oplus M$ and the elements $r\oplus f'(r)$ for $r$ in
$R'$. We have a canonical map from $R$ to this $L$, so we have to show
that the Weyl chamber $D$ of $L$ is determined up to conjugacy by
elements of the group fixing $R$ and $M$ (resp. fixing $M$ and fixing
all points of $R$.) This Weyl chamber is determined by its
intersection with $R$ and $M$, and its intersection with $R$ is just
the canonical Weyl chamber of $R$. Its intersection with $M$ is in the
cone $C$ and is in some Weyl chamber of the norm 2 roots of $M$. All
such Weyl chambers of $M$ in $C$ are conjugate under automorphisms of
$L$ fixing $M$ and all points of $R$, so we can assume that the
intersection with $M$ is contained in some fixed Weyl chamber $W$ of
$M$.
By 3.1 and the assumption on $R$ all the Weyl chambers of $L$ whose
intersection with $M$ is in $W$ are conjugate under the group of
automorphisms of $L$ fixing $R$ (resp. fixing all points of $R$) and
hence $(M,f',C)$ determines $f(R)$ (resp. $f$). Q.E.D.
\proclaim Corollary 9.2.
If $R$ is $e_6$, $e_7$, $e_8$, $d_4$, or $d_m$ $(m\ge 6)$ then two copies of
$R$ in $X$ are conjugate under $\Aut(D)$ if and only if their
orthogonal complements are isomorphic lattices.
{\sl Proof. } If $R'$ is any Dynkin diagram containing $R$ and one
extra point then $\sigma(R')$ fixes $R$. The result now follows from
9.1. Q.E.D.
{\sl Remark.} If $L$ is $II_{9,1}$ or $II_{17,1}$ then $\Aut(D)$ is
not transitive on $d_5$'s. $\sigma(e_6)$ does not fix the $d_5$'s in
$e_6$.
\proclaim Lemma 9.3.
$\Aut(D)$ is transitive on $e_8$'s. The simple roots of $D$
perpendicular to an $e_8$ form the Dynkin diagram of $II_{8n-7,1}$ and
the subgroup of $\Aut(D)$ fixing the $e_8$ is isomorphic to the
subgroup of $Aut(II_{8n-7,1})$ fixing a Weyl chamber.
{\sl Proof.} The transitivity on $e_8$'s is in 9.2. The rest of 9.3
follows easily. Q.E.D.
\proclaim Lemma 9.4.
If $8n\ge 24$ then for any $e_6$ in $X$ there is an element of
$\Aut(D)$ inducing $\sigma(e_6)$ on it.
{\sl Proof. } By 9.2, $\Aut(D)$ is transitive on $e_6$'s so it is
sufficient to prove it for one $e_6$. It is true form $8n=24$ by
calculation and using 9.3 it follows by induction for $8n>24$. Q.E.D.
\proclaim Theorem 9.5. Classification of $d_m$'s in $X$.
\item{(1)}
$\Aut(D)$ acts transitively on $e_6$'s $e_7$'s, and $e_8$'s in $X$.
if $8n\ge 24$ then for any $e_6$ there is an element of $\Aut(D)$
inducing the nontrivial automorphism of this $e_6$.
\item{(2)}
For any $m$ with $4\le m\le 8n+1$ there is a unique orbit of $d_m$'s in
$X$ such that $d_m^\perp$ is not unimodular, unless $m=5$ and $8n=8$
or $16$. Any automorphism of such a $d_m$ is induced by an element of
$\Aut(D)$ if and only if $m\le 8n-13$.
\item{(3)}
For any $m$ with $16\le 8m\le 8n$ there is a unique orbit of
$d_{8m}$'s such that $d_{8m}^\perp$ is unimodular, and these are the
only $d$'s whose orthogonal complement is unimodular. There is no
element of $\Aut(D)$ inducing the nontrivial automorphism of $d_{8m}$.
{\sl Proof. } Part (1) follows from 9.2 and 9.4 because there is only
one isomorphism class of lattices of the form $e_i^\perp$ for
$i=6,7,8$. From 9.4, 9.2, and 9.1 it follows that two $d_m$'s of $X$
are conjugate under $\Aut(D)$ if and only if their orthogonal
complements are isomorphic, unless $m=5$ and $n\le 2$. $d_m^\perp$ is
either the even sublattice of $I_{8n+1-m,1}$, or $m$ is divisible by 8
and $d_m^\perp$ is $II_{8n+1-m,1}$. In the second case we must have
$m\ge 16$ because if $m$ was 8 the Dynkin diagram of
$(II_{8n+1-m,1})^\perp$ would be $e_8$ and not $d_8$. This shows that
there is one orbit of $d_m$'s unless $8|m$, $m\ge 16$ or $m=5$, $n\le
2$ in which case there are two orbits.
If $d_m^\perp$ is unimodular then $d_m$ is contained in an even
unimodular sublattice of $L$, and there are no automorphisms of this
lattice acting non-trivially on $d_m$, so there are no elements of
$\Aut(D)$ inducing a nontrivial automorphism of $d_m$.
If $d_m^\perp$ is not unimodular then there is an element of $\Aut(D)$
inducing a nontrivial automorphism of $d_m$ if and only if there is an
automorphism of the Dynkin diagram of $I_{8n+1-m,1}$ acting
non-trivially on $M'/M$, where $M$ is the sublattice of even elements
of $I_{8n+1-m,1}$. There is no such automorphism of $I_k$ for $k\le
13$ and there is such an automorphism for $k=14$, so there is an
element of $\Aut(D)$ inducing a nontrivial automorphism of $d_m$ for
$m=8n-13$ and there is no such element if $m\ge 8n-12$. If $m<8n-13$
then our $d_m$ is contained in a $d_{8n-13}$ so there is still a
nontrivial automorphism of $d_m$ induced by $\Aut(D)$. Finally, if
$m=4$ and $8n\ge 24$ then as in the proof of 9.4 we see that there is
some $d_4$ such that $\Aut(D)$ induces all automorphisms of $d_4$. As
$\Aut(D)$ is transitive on $d_4$'s, this is true for any $d_4$. Q.E.D.
\proclaim Lemma 9.6.
If $2\le m\le 11$ and $8n\ge 24$ then for any $a_m$ in $L$ there is an
element of $\Aut(D)$ inducing the nontrivial automorphism of $a_m$.
{\sl Proof. } This is true for $8n=24$ by calculation. There is an
element of $\Aut(D)$ acting as $\sigma(a_m)$ on $a_m$ if and only if
there is an automorphism of the Weyl chamber of the lattice
$M=a_m^\perp$ which acts as $-1$ on $M'/M$. The lattice $M$ is isomorphic to
$N\oplus e_8^{n-3}$, where $N$ is $a_m^\perp$ for some $a_m$ in
$II_{25,1}$ and $N$ has an automorphism of its Weyl chamber, so $M$
has one too. Hence for $m\le 11$ there is an element of $\Aut(D)$
inducing $\sigma(a_m)$ on $a_m$. Q.E.D.
\proclaim Corollary 9.7.
If $8n\ge 24$ and $m\le 10$ then $\Aut(D)$ is transitive on $a_m$'s in $D$.
{\sl Proof.} It follows from 9.6 and 9.5 that if $R'$ is any spherical
Dynkin diagram in $X$ containing $a_m$ and one extra point (so $R'$ is
$a_{m+1}$, $d_{m+1}$, $e_{m+1}$, or $a_ma_1$) then there is an element
$g$ of $\Aut(D)$ such that $g\sigma(R')$ fixes $a_m$. Hence by 9.1,
$\Aut(D)$ is transitive on $a_m$'s. Q.E.D.
\proclaim Corollary 9.8.
If $n\ge 20 $ and $n\equiv 4,5$ or $6\bmod 8$ then the non-reflection
part of $\Aut(I_{n,1})$ can be written as a nontrivial amalgamated
product.
{\sl Proof. } The results of this section show that the analogue of
6.1 is true for $II_{8i+1,1}$ for $8i\ge 24$. This is all that is
needed to prove the analogue of 6.6. Q.E.D.
(If $n\ge 23$ then the group cannot be written as an amalgamated
product of finite groups.)
{\sl Remark.} If $n\ge 10$ and $n\equiv 2$ or $3\bmod 8$ and $G$ is
the subgroup of $\Aut(I_{n,1})$ generated by the reflections of
non-characteristic roots, then $\Aut(I_{n,1})/G$ is a nontrivial
amalgamated product.
\proclaim References.
\item{1.} R. E. Borcherds, The Leech lattice. Proc. Roy. Soc. London Ser.
A 398 (1985), no. 1815, 365--376.
\item{2.}
N. Bourbaki, ``Groupes et alg\`ebres de Lie'', Chaps. 4--6, Hermann, Paris,
1968.
\item{3.} J. H. Conway,
The automorphism group of the $26$-dimensional even unimodular Lorentzian
lattice. J. Algebra 80 (1983), no. 1, 159--163.
\item{4.} J. H. Conway, N. J. A. Sloane,
Leech roots and Vinberg groups. Proc. Roy. Soc. London Ser. A
384 (1982), no. 1787, 233--258.
\item{5.} J.-P. Serre, A course in arithmetic. Graduate Texts in Mathematics,
No. 7. Springer-Verlag, New York-Heidelberg, 1973.
\item{6.} J.-P. Serre, ``Trees'', Springer-Verlag,
Berlin-New York, 1980. ISBN: 3-540-10103-9
\item{7.} J.-P. Serre, Cohomologie des groupes discrets.
Prospects in mathematics (Proc.
Sympos., Princeton Univ., Princeton, N.J., 1970), pp. 77--169.
Ann. of Math. Studies, No. 70, Princeton Univ.
Press, Princeton, N.J., 1971.
\item{8.} \`E. B. Vinberg,
Some arithmetical discrete groups in Loba\v cevski\u\i ~spaces.
Discrete subgroups
of Lie groups and applications to moduli
(Internat. Colloq., Bombay, 1973), pp. 323--348. Oxford Univ. Press,
Bombay, 1975.
\item{9.} Vinberg, Kaplinskaja, On the groups $O_{18,1}(Z)$ and
$O_{19,1}(Z)$, Soviet Math. 19, No 1 (1978) 194-197.
\bye