% file  2lat1-5.tex   (has a .fr)
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\pageno=42
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\input ../louis.sty
\input ../l.sty
\input lattice.sty
\headline{Geometry of lattices I\s5 \hfill\er file 2lat1-5.tex.\qquad\today}

\entete {I \s5 }{The classification of two dimensional lattices}

\proclaim I\s5-1. Reduction of two variable quadratic forms. 

We can always choose an arbitrary visible vector to be a vector basis. 
For the 2-dimensional lattice $L$ we choose a shortest vector $\vec s\in 
S$. Then the other basis vector must be in one the two $\Sigma_\pm$ the 
nearest paralleles to $\lambda\vec s$ which contain lattice points 
\fnote{Define $\vec t\in E_2$ by $(\vec s,\vec t)=0$ and $\det(\vec s,\vec 
t)=\det L$ $\Rightarrow$ $N(\vec t)N(\vec s)=(\det L)^2$. This two 
paralleles are defined by $\pm\vec t+\lambda\vec s$; their equation in 
the basis $\vec s,\vec t$ is $(\vec t,\vec x)=\pm N(\vec t)$.}
of $o+L$; this means that the distance of $\Sigma_\pm$ to the the axis 
$\lambda\vec s$ is $(\det L)/\sqrt{N(\vec s)}$. One can choose as closure 
of the fundamental domain of the group $Z(\vec s)$ generated by the 
translation $\vec s$, the band 
\fnote{It is defined by $\Delta(\vec s)=\{\vec y\in E_2$, $|(\vec s,\vec y)|
\le\frac12 N(\vec s)\}$.}
$\Delta(\vec s)$ bounded by the two paralleles to the axis $\mu\vec t$ 
at a distance $\frac12\sqrt{N(\vec s)}$ of this axis. The intersection 
$\Delta\cap\Sigma_\pm$ contains the extremities of two lattice vector when they 
are in the interior of $\Delta(\vec s)$ and four when they are on its
boundary. We can chose one of these vectors such that its scalar product 
with $\vec s$ is negative; we denote it by $\vec b$. The corresponding 
quadratic form is $q_{11}=N(\vec s)\le q_{22}=N(\vec b)$ (remember that 
$\vec s\in S(L)$) and $-\frac12N(\vec s)\le q_{12}=q_{21}=(\vec s,\vec 
b)\le0$. To summarize the Lagrange reduced positive (symmetric) quadratic 
forms in $\Cal Q_2$ are defined by
\fnote{The sign $-$ in front of the $q_{12}$ term is arbitrary. We use 
it here because it has a natural generalisation to arbitrary $n$; e.g. 
see Lemma ??????\mgr.}:
$$0\le-2q_{12}\le q_{11}\le q_{22};\quad q_{11}>0. 
                           \eqno I5(1)              $$  
That domain in $\Cal C_+(\Cal Q_2)$ is a primitive domain of the action 
of $\glz2$: it represents the orbifold $\Cal L^o$.

\proclaim I\s5-2. The Bravais crystallographic systems.

We have first to recall the list and some notations for the finite subgroups 
of $O_2$. The matrices of $O_2$ of determinant 1 (respectively -1) are the 
rotations $r(\theta)$ by an angle $\theta$ around the origin (respectively, 
the reflections $s(\phi)$ through the axis of azimuth $\phi$):
$$r(\theta)={\cos\theta\ -\sin\theta \choose \sin\theta\ \ \cos\theta},\ 
\theta (\mod 2\pi)\quad s(\phi)= {\cos(2\phi)\ \ \sin(2\phi) \choose
\sin(2\phi)\ -\cos(2\phi)},\ \phi (\mod\pi).     \eqno I5(2)    $$
They satisfy the following relations:
$$ r(\theta)r(\theta')=r(\theta+\theta'),\ s(\phi)s(\phi')=
r\bigl(2(\phi-\phi')\bigr),
\ r(\theta)s(\phi)=s(\theta+\phi)=s(\phi)r(-\theta). \eqno I5(3)   $$
In particular:
$$ r(\theta)r(-\theta)=I,\quad s(\phi)^2=I,\quad s(\phi)r(\theta)s(\phi)=
r(-\theta),\quad 
r(\theta)s(\phi)r(\theta)\inv=s(\phi+2\theta).\eqno I5(4) $$
We denote by $C_n$ the $n$-element group formed by the rotations 
$r(2\pi k/n),\ 0\le k\le n-1$; it is isomorphic to $Z_n$ and 
$C_n\triangleleft O_2$. We denote by $C_s$ the conjugation class of 
subgroups $C_{s(\phi}\sim Z_2$ generated by the reflection $s(\phi)$. 
The two reflections $s(\phi)$ and $s(\phi+\pi/n)$ generates a $2n$ 
element group that we denote 
\fnote{The notations $C_n,\ C_s,\ C_{nv}$ are those of Sch\"onflies.}
by $C_{nv(\phi)}$, with the simple notation $C_{nv}$ for the conjugacy 
class $[C_{nv(\phi)}]_{O_2}$. These groups are ismorphic to the ``dihedral'' 
groups, defined by the generators and relations $s_1^2=s_2^2=1=(s_1s_2)^n$;
we denote them by $d_n$; they are non Abelian for $n>2$. The group $\cv 
n$ is the symmetry group of a regular polygon of $n$ vertices and edges.

As we have seen, to be the holohedry $P_L$ of a lattice in dimension $n$ a 
subgroup of $O_n$ has to be finite and conjugate (in $\glr n$) to a subgroup 
of $\glz n$. Since the trace of a matrix is invariant by conjugation, a 
necessary condition is $\forall g\in P_L,\ \tr g\in Z$. In dimension 2 
this condition is sufficient and restrict the choice to $1=C_1,\ C_s,\  
C_n\ ,C_{nv}$ with $n=2,3,4,6$. As we have also seen, another condition is 
$-I_n\in P_L$. In dimension 2, the symmetry through the origin $-I_2$ is 
the rotation by $\pi$ and this restrict the possible holohedries to 
$C_2,\cv2,C_4,\cv4,C_6,\cv6$. 

The group $C_4$ cannot be a holohedry. Indeed, consider $\vec s_1$, one 
of the shortest vectors of $L$, its $C_4.\vec s_1$ orbit: $\vec s_1,\vec s_2,
-\vec s_1,-\vec s_2,$ with $(\vec s_1.\vec s_2)=0$, and denote by $L'\le L$ 
the lattice it generates. If there exists $\vec\ell\in L$, $\vec\ell\notin L'$, 
then it is possible to find integers $\lambda_1,\lambda_2$ such that $\vec w=
\vec\ell+\lambda_1\vec s_1+\lambda_2\vec s_2$ be in a fundamental domain of 
$L'$, e.g. the closed square of vertices $\vec o,\vec s_1,\vec s_2,\vec 
s_3=\vec s_1+\vec s_2$. From its construction, $\vec w$ is 
different from the vectors forming the four vertices of the square; 
there is at least one vertex (let us denote its vector $\vec x$), such that 
$|\vec w-\vec x|$ is smaller that the side of the square; this is absurd since $\vec s$ 
was chosen as shortest vectors. So $L'=L$; this lattice, generated by 
$C_4.\vec s$ has the symmetry of this orbit: it is $C_{4v}$. \hb
By a similar proof one shows that $C_6$ cannot be a lattice holohedry. 

So we are left with four possible holohedries: $\cv4$ for the square 
lattices, $\cv6$ for the hexagonal ones, $\cv2$ for instance for the 
rectangular ones and $C_2$ for the generic lattices. These groups label 
the two dimensional Bravais crystallographic systems. Hence the list: 
$$\{BCS\}_2=\{ C_2,\ \cv2,\ \cv4,\ \cv6\}=\{2,\ mm,\ 4m,\ 6m\}
\eqno I5(5)       $$
where the last notations are those of {\er [ITC83]} used by all 
crystallographers: the groups are given by generators, $m$ meaning a 
reflection, and the figure $k$ a rotation by $2\pi/k$. We recall that 
the groups $\cv n$ are generated by two reflections. The two maximal 
crystal systems correspond to the holohedries $\cv4$ and $\cv6$ whose 
orders are 8 and 12; we check that smc(8,12)=24 is equal to the 
Minkowski number $\overline3\!|$ (see I4(32)\mge).

\proclaim I\s5-3. The Bravais classes.

In every dimension the generic lattices form only one Bravais class: 
by their definition, their holohedry is $\{I_n,-I_n\}$, the center of 
$\glr n$; they are no distinct element of $\glr n$ conjugate to $-I_n$. 
In every dimension $n\ge2$, while the reflections through an hyperplane are 
all conjugate in $\glq n$ (a fortiori in $\glr n$), this is not true in 
$\glz n$. From the matrices introduced in I4(10)\mge :
$$pm=\sigma_3=\pmatrix{1&\phantom-0\cr 0&-1\cr},\quad cm=\sigma_1=
\pmatrix{0&1\cr 1&0\cr}        \eqno I4(10)                      $$
we build two reflection matrices $m_i=\sigma_i\oplus I_{n-2}$. They 
cannot be conjugated in $\glz n$; if they were, this would also be true 
of the two matrices $I_n+m_3$ and $I_n+m_1$; that is not possible:
indeed, the greatest common divisor of all elements of each matrix is  
2 for the former and 1 for the latter but the conjugation by an element 
of $\glz n$ cannot change the greatest commun divisor of all elements of 
the matrix to be conjugated. So there are at least two conjugacy classes 
of reflections  in $\glz n$ and one can prove that there are only two.

We give here a direct proof for $n=2$. A reflection in 
$\glz2$ has trace $0$ and determinant $-1$; so its general form is 
$$s=\pmatrix {a&b\cr c&-a\cr},\ a,b,c\in Z,\ a^2+bc=1. \eqno I5(6) $$
In particular, $\sigma_3$ and $\sigma_1$ are two reflections and we know 
that they are not conjugate in $\glz2$. If $s\ne\pm\sigma_3$, it is not 
diagonal. We may not be able to diagonalize it by conjugation in $\glz2$, 
but we can make it upper triangular. Indeed, corresponding to the eigenvalue 
$1$, it has, up to a sign, a unique integral, visible eigenvector 
$v={\alpha \choose \beta }$ with $\alpha = b/k$, $\beta = (1-a)/k$ where 
$k=\gcd(b,1-a)$ is the largest common divisor of $b$ and $1-a$. 
Then (as we saw in I\s2-2), we can choose a pair $\alpha ',\beta '$ 
of relatively prime integers such that $\alpha \beta '-\beta \alpha'=1$,
to complete a conjugating matrix:
$$\pmatrix {\beta '&-\alpha '\cr -\beta &\alpha\cr}\pmatrix {a&b\cr c&-a\cr}
\pmatrix {\alpha &\alpha '\cr \beta &\beta '\cr}=\pmatrix {1&x\cr 0&-1\cr}
=t,\ x=2a\alpha'\beta'+b\beta'^2-c\alpha'^2.\eqno I5(7)        $$
Depending wether $x$ is even ($=2y$), or odd ($x=2y+1)$, the 
matrix $t$ can be conjugate to $\sigma_3$ or $\sigma_1$ by the 
matrices $\pmatrix {1&y\cr 0&1\cr}$ and $\pmatrix{1&y\cr 1&1+y\cr}$ 
respectively. This ends the proof of the existence of exactly two conjugacy 
classes of reflections in $\glz2$. These two classes can be labelled by the 
class $(\mod2)$ of $x$ in I5(7\mge); using that $\alpha'$ and $\beta'$ are 
relatively prime, this class can be expressed as function of the 
matrix elements $a,b,c$ of the reflection matrix $s$:
$$x\equiv b+c+bc \equiv a+b+c+1\ (\mod2)  \eqno I5(8)         $$
The two arithmetic classes of reflections corresponding to the classes 
$0,1$ of $x\mod2$ are labelled in {\er [ITC83]}, $pm$, $cm$ respectively. A 
reflection from either class and $-I_2$ generate a holohedry of the 
orthorhombic crystallographic system. The corresponding two Bravais classes 
are denoted respectively by $pmm,\ cmm$ by the crystallographers.\hb
Figure I4-4a shows an example of lattice for each Bravais class. That of 
the class $cmm$ can be oftained from one of the class $pmm$ as a 
sublattice of index two with $\Cal K=\Cal N$ in equation I2(12)\mge.
\midinsert
\medskip\vskip 7truecm
%\special{psfile=lat1-5f3.ps} 
Fig. I5-3. Lattices of the two Bravais class of the $mm$=orthorhombic  
crystallographic system: (i) belongs to $pmm$ and (ii) to $cmm$. 
\medskip
\endinsert
Two reflections $s$ and $s'\ne-s$ of $\glz2$ generate in general an infinite 
subgroup. We know that the only finite group they can generate are conjugate 
in $\glr2$ to $\cv n$, $n=4,3,6,$. By conjugation in $\glz2$, the 
reflection $s'$ can be transformed either in $\sigma_3$ or $\sigma_1$; what 
are the possible $s$ of I5(6\mge) such that $\tr ss'=$ either 0 (rotation by 
$\pi/2$), or 1 (rotation by $\pi/3$)? We have four cases to consider: \hb
i) $s'=\sigma_3$, $\tr ss'=0$ $\Leftrightarrow $ $2a=0$ $\Leftrightarrow $
$s=\sigma_1$;\hb
ii) $s'=\sigma_1$, $\tr ss'=0$ $\Leftrightarrow $ $b+c=0$; with 
$a^2+bc=1$ this implies $a^2=1+b^2$; the only integer solutions $b=0$, 
$a=\pm1$ so $s=\sigma_3$.\hb
These two cases prove that every $\cv4$ subgroups of $\glz2$ are conjugate 
to the one generated by $\pm\sigma_3,\pm\sigma_1$.\hb
iii) $s'=\sigma_3$, $\tr ss'=1$ $\Leftrightarrow $ $2a=1$, impossible: 
$a\notin Z$; so all reflections of $\cv6$ subgroups of $\glz2$ can be 
transformed by conjugation in to $\pm s'$ and $\pm s$ such that\hb
iv) $s'=\sigma_1$, $\tr ss'=1$ $\Leftrightarrow$ $b+c=1$; with 
$a^2+bc=1$ this implies $(b-c)^2=4a^2-3$ $\Leftrightarrow$ 
$(2a+b-c)(2a-b+c)=3$ $\Rightarrow$ $b-c=\pm1$ and the two solutions for 
$s$ are (up to a sign) $b=0,c=1$ or $b=1,c=0,$. With all compatible 
choice of signs, one can form two contragredient $\cv6$ subgroups which 
are conjugate by $\sigma_1$. That concludes the determination of the 
five Bravais classes in dimension 2. They are denoted by the 
crystallographers as:
$$\{BC\}_2=\{p2,\ pmm,\ cmm,\ p4m,\ p6m\}. \eqno I5(9) $$
 
\proclaim I\s5-4. The $\glz2$ orbits on $\overline{\Cal C_+}(\Cal Q_2)$.
 
In the three dimensional space $\Cal Q_2$ of real symmetrical matrices we 
take as basis the three matrices $I_2,\sigma_3,\sigma_1$:
$$\Cal Q_2\ni q=tI_2+x\sigma_3+y\sigma_1=\pmatrix{t+x&y\cr y&t-x\cr},
\quad \det q=t^2-x^2-y^2 \eqno I5(10)                     $$
We knew from I\s3-3 that $\tr q^2=2(t^2+x^2+y^2)$ corresponds to the 
Euclidean metric; but as we saw in I\s3-4, the action $q\mapsto 
mqm^\top$ of $\splr2$ (the group of the elements of $\glr2$ with 
determinant $\pm1$) and of its subgroup $\glz2$ on $\Cal Q_2$ preserves 
$\det q$; as I5(10) shows, that is the pseudo-euclidean metric of Lorentz
\fnote{The trace of the characteristic equation of $q$ yields 
$\det q=\frac12((\tr q)^2$. By polarization one obtains the $\glz n$ 
invariant bilinear form $b(q_1,q_2)=\frac12(\tr q_1\tr q_2-\tr q_1q_2)$}.
So we have the interesting geometric situation:\hb
i) the affine space of $\Cal Q_2$ is a three dimensional space time and 
$\splr2$ is its orthochronous Lorentz group (i.e. the subgroup preserving the 
orientation of time of index two in the full pseudo-euclidean group). 
$\Cal Q_2$ itself can be compared to the energy-momentum vector space; in it\hb
ii) the cone $\Cal C_+(\Cal Q_2)$
is the interior of the future light cone which is itself the $\glr2$ orbit 
$\partial_1\overline{\Cal C_+}(\Cal Q_2)$ which is also, with the origin, 
the set of extremal points of the closure $\overline{\Cal C_+}(\Cal Q_2)$;\hb
iii) the two dimensional vector space $V_2$ carrier of the definition 
action of $\glr2$ is the spinor space of the space time $\Cal Q_2$ and 
$\Cal Q_2=V_2\vee V_2$, where $\vee$ denotes the symmetric tensor product;\hb
iv) as E. Cartan taught us, there is a canonical bijective map $V_2\ni 
\pm v\mapsto \pm v\vee \pm v\in\partial_1\Cal C_+(\Cal Q_2)$ between the 
spinors, up to a sign, and the light vectors. This correspondence is a 
good computing tool when a basis as been chosen:
$$V_2\ni \pm v=\pm{\alpha\choose\beta}\mapsto v\vee v=\pmatrix{
\alpha^2&\alpha\beta\cr \alpha\beta&\beta^2\cr}. \eqno I5(11) $$

To draw a two dimensional picture (see figure I4-4b) we make a stereographic 
projection of the light cone on the plane $H$ of equation $\tr q=2$ of the 
space time $\Cal Q_2$:
$$\overline{\Cal C_+}(\Cal Q_2)\ni q\mapsto 2q/\tr q\in H\quad\Leftrightarrow 
\quad\xi={x\over t},\ \eta={y\over t};               \eqno I5(12) $$ 
more precisely, the cone $\overline{\Cal C_+}(\Cal Q_2)$ is mapped on the 
unit disc  $\xi^2+\eta^2\le1$ of $H$. Subspace  of the vector space 
$\Cal Q_2$ are mapped on their intersection with $H$. So $\glz2$ 
transforms straight line into straight line on the disc (beware that the 
middle of a line segment in not transformed into the middle of the
transformed segment!). To summarize, each point inside the disc 
represents a {\tt ray} $\lambda q$, $\lambda>0$ of positive quadratic forms.
\midinsert
\vskip 18truecm
\special{psfile=lat1-5f4.ps} 
Fig. I5-4. Stereographic projection of $\overline{\Cal C_+}(\Cal Q_2)$. \hb
{\er Each point inside of the circle represents a ray of positive quadratic 
forms $q$'s (i.e. defined up to a positive factor). The $q$'s belonging 
to the Bravais classes $pmm$, $cmm$, $p4m$, $p6m$ are represented 
respectively by the dotted lines, the full lines, the points $s$ 
(intersections of a dotted line and a full line: square lattices),  $h$ 
(intersections of three full line: hexagonal lattices). The dense  open set 
of the other points of the interior of the disc represent the generic $q$'s; 
they belong to the minimal Bravais class $p2$. See the text 
for the meaning of the pairs $p:q$ of relatively prime integers. Each 
triangle represents a fundamental domain in the action of $\glz2$ on 
$\overline{\Cal C_+}(\Cal Q_2)$.}
\medskip
\endinsert
The reflection $\sigma_3$ acts on $H$ as the symmetry through $LL'$ and 
the reflection $\sigma_1$ acts on $H$ as the symmetry through $KK'$. 
That explain the symmetry of the figure I4-4b\mge, which shows the strata of 
the actions of $\glz2$ on the disc. The computation of this figure 
yields another  determination of the five Bravais classes.

The dotted line $LL'$ represents the $q$'s invariant by $\pm\sigma_3$ and 
the full line $KK'$ those invariant by the $\pm\sigma_1$. The full and dotted 
lines form the $\glz2$ orbit of $LL'$ and $KK'$ respectively. They 
represent $q$'s belonging to the Bravais classe $pmm$ and $cmm$. The 
intersection of these lines must correspond to maximal Bravais classes.
Indeed $LL'$ and $KK'$ intersect at a point $s$ which represents $q$'s 
invariant by $\cv4$, the $\glz2$ subgroup generated by $\pm\sigma_3,
\pm\sigma_1$. All the intersections of a dotted line and a full line 
form the orbit $\glz2.s$ and represent the lattices of the Bravais class 
$p4m$, in plain words: the square lattices. The two points $h$ of $KK'$, 
represent two rays of $q$'s invariant under one of the contragredient $\cv6$ 
groups generated by  by the three pairs of 
reflections $s'=\pm\sigma_1$, and the solutions for $\pm s$ computed in 
\s5-3 iv\mgr). The orbit $\glz2.h$ contains all the intersections of three 
full lines; it represents the lattices of the Bravais class $p6m$, that 
is the hexagonal lattices. The points of the interior of the disc 
and outside the lines, represent the generic lattices, those of the 
minimal Bravais class $p2$.

Notice that {\sl in the interior of the disc}, every dotted line 
contains only one point $s$ and each full line one $s$ and two $h$'s.
Each triangle (whose sides are carried by two full lines and one dotted 
line, whose vertices are $s,h$ and one point of the boundary circle)
represent a fundamental domain of the $\glz2$ action. The Lagrange 
foundamental domain given in I5(1\mge) corresponds to the triangle of 
vertices $L',\xi=-1,\eta=0$, $s,\xi=0,\eta=0$, $h,\xi=0,\eta=-\frac12$.

To build these figure we have use the action of $\glz2$ on the (real) spinor 
space $V_2$. As shown in I5(11\mge) to the pair $\pm v$ of spinors 
correspond in $\Cal Q_2$ to a light vector $v\vee v$; to a ray of light 
vectors correspond a point of the circle, boundary of the disc. We have 
have shown (I\s1-6d\mgr) that the stabilizer $\glr n_v$, $v\in V_n$ is $\aff 
n^\#$; we obtain that of $\glz n$ by taking the intersection with that 
group. It is as easy to compute directly this stabilizer of $v$:
$$v={\alpha\choose0},\quad\glz2_v =\{\pm\pmatrix{1&b\cr 0&1\cr},\ 
\pm\pmatrix{1&b'\cr 0&-1\cr},\ b,b'\in Z\}=\glz2_L. \eqno I4(52) $$
It is isomorphic to $d_\infty\times Z_2$ where $d_\infty$ is the infinite 
dihedral group of generators and relations $s_1^2=s_2^2=1$.

The spinors with integral components form a real lattice whose automorphism
group is $\glz2$. Taking $v={1\choose0}$ one obtains the orbit $\glz2.v$ 
of visible spinors. It is an enumerable set, so its faithful image on the 
circle must have accumulation points. There are four of them, 
$L,L',K,K'$. We have plotted the image of visible spinors on the circle, 
giving their components in the form $\alpha:\beta$. By the reflections 
$-\sigma_3$ and $\sigma_1$ they are respectively tranformed into 
$-\alpha:\beta$ and $\beta:\alpha$.\hb
It is easy to obtain the orbit of the dotted lines: the matrix 
$\pmatrix{a&b\cr c&d\cr}$ tranforms $L=1:0$ and $L'=0:1$ into the 
representative points of the visible spinors $a:c$ and $b:d$ with $ad-bc=\pm1$. 
Similarly we obtain that the full lines join the representative points of the 
pairs of spinors $a+b:c+d$ and $a-b:c-d$.

\proclaim I5-5. $\Cal L^{od}_2$ as fundamental domain of the modular group.

We are interested only in the 2-dimensional manifold $\Cal L^{od}$. So 
we can for instance normalize $N(\vec s)=1$. Another method is to 
represent the vectors of $E_2$ as points of the complex plane:
for instance the basis $\vec s,\vec b$ chosen in I\s4-4b),
$\vec s\mapsto \sigma\in\CC$, $\vec b\mapsto \beta\in\CC$. Then 
$\zeta=\beta/\sigma$ represents the vector $\vec b$ when $\vec s$ is 
rotated on the positive real axis and its norm is normalized to 1.
The action of $\glz2$ on $\zeta\in\CC$ is defined by
$$g=\pmatrix{a&b\cr c&d\cr}\in\glz2;\quad g.\zeta=
 \pmatrix{a\zeta&b\cr c\zeta&d\cr}\in\glz2.  \eqno I5(14)     $$
We verify easily that 
$$-I_2.\zeta=\zeta,\qquad\Im m\, g.\zeta=\Im m\,\zeta(\det g). \eqno 
I5(15) $$ 
A fundamental domain of this action represents $\Cal L^{od}$. The last 
equation shows that we can choose the fundemental domain in the upper 
half plane, it is the fundamental domain of $\slz2$ acting on the upper 
half plane
\fnote{The center $C(\slz2)\sim Z_2$ acts trivially; so the action on the upper 
half plane is effective only for the quotient $\slz2/C(\slz2)=PSL_2(Z)$, 
the special projective group.}.
A classical choice for this domain is drawn in fig. I5-5; its 
construction can be find in number theory or ``arithmetic'' books; e.g.
{\er[SER70] chap.7}.

\midinsert
\medskip
{\vbox{\hsize= 8 truecm
Figure I5-5. The gray part of the complex plane represents the orbit 
space  $\Cal L_2^{od}=\Cal C_+(\Cal Q_2)|\glz2\times R_+^\times$. \hb
{\er The strata correspond to the five Bravais classes. The points $q$ and 
$h$ represents respectively the squa\-re and the hexagonal lattices (each 
stratum has a u\-ni\-que orbit). They are the two maximal Bravais classes; 
in crystallography thet are denoted respectively by $p4m$ and $p6m$.
The open segment $qy$ represents the rectangular Bravais class $pmm$ 
(also called $p$-orthorhombic) while the disjoint union 
of the open arc $h'q\cup qh$ and the open segment $hy'\cup h'y''$ represents 
the lattices of the rhombohedral Bravais class $cmm$ (they have a basis in 
$S$); those two Bravais classes form a unique Bravais cristallographic system  
corresponding to the conjugacy class in $O_2$ of the group $\sim Z_2^2$ 
generated by the symmetry through two orthogonal axes. The inside of the 
domain, image of the generic stratum, corresponds to the 
stabilizer $C(\glz2)$ $\sim$ $Z_2$; this minimal Bravais class is denoted by 
$2\ p$ in crystallography}.\vskip 1truecm}}
\special{psfile=lat1-5f5.ps} 
\endinsert
\medskip
\proclaim I\s5-6. The Vorono{\"\i} method and the Delone symbols in 2 dimensions.

To conclude this subsection, we explain here a third method for 
classifying the 2-dimensional lattices. It is due to Vorono{\"\i} 
who wrote it for two dimensions ({\er[VOR09]} p.157) and extended it for 
arbitrary dimension for the study of what he called type I lattices.

We have shown in I\s5-1 that we can always choose a basis of 
a lattice such that the coefficient of the associated quadratic form 
$q_{11}x^2+2q_{12}xy+q_{22}y^2$ satisfy I5(1) \mge
$0\le-2q_{12}\le q_{11}\le q_{22}$, $q_{11}>0$.
With the variable $\lambda=q_{11}+q_{12}$, $\mu=q_{22}+q_{12}$, $\nu=-q_{12}$,
the quadratic form becomes a sum of squares:
$$\lambda x^2+\mu y^2+\nu(x-y)^2,\quad \lambda\ge0,\mu\ge0,\nu\ge0. 
\quad\det(q_{ij})=D=\lambda\mu+\mu\nu+\nu\lambda>0. \eqno I5(16)    $$
As the value of the determinant shows, the quadratic form is positive if 
no more than one of the three parameter vanishes. I5(16) and the next 
equation given the norm of some vectors

$$N{1\choose0}=\lambda+\nu;\quad N{0\choose1}=\mu+\nu;\quad
N{1\choose1}=\lambda+\mu;   \eqno I5(16)   $$
show a complete syntactic symmetry among the parameters $\lambda,\mu,\nu$.
The domain in $\cqf$ of 
quadratic forms defined by I4(55) is invariant by the group of permutation 
$\Cal S_3$ of the three parameters $\lambda,\mu,\nu$. Indeed it 
corresponds to the triangle $L'K'L$ of fig I5-4 and $\Cal S_3$ permutes 
the six fundamental domains contained in the domain of I5(16).

It is straightforward to describe the five Bravais strata by studying 
the strata in the parameter space (which is always a simplex in $\cqf$ 
in any dimension). They can be labelled by an elegant symbol invented by 
Delone: the three parameters are represented by the three sides of a 
triangle.\hb
{\bf first case}: $\lambda\mu\nu\ne0$:

\noindent{\sl generic Bravais class} $p2$: represented by the Delone 
symbol \hfil $\btu$

\noindent When two parameters are equal, an order 2 symmetry appears: 
the invariance by $\sigma_1$ in I4(10) (for instance $\lambda=\mu$); it
exchanges the two equal sides of the triangle; it corresponds to the 
Bravais class $cmm$: \hfil $\btuc$  

\noindent When the three parameters are equal: we have the full symmetry 
$S_3$ of the triangle; with the inversion through the origin (=rotation 
by $\pi$), one describes the hexagonal Bravais class $p6m$:\hfil$\btuh$

\noindent{\bf second case}: one of the three parameters vanishes
\fnote{If we choose $\nu=0$ the quadratic form is diagonal (= invariant 
by $\sigma_3$ of I4(10)). The cases $\lambda=0$ and $\mu=0$ are obtained 
from the preceeding one by transforming the quadratic form by the $\slz2$ 
matrices $\pmatrix{1&0\cr -1&1\cr}$ and $\pmatrix{1&-1\cr 0&1\cr}$ 
respectively.}:

\noindent the two other parameters are different: Bravais class $pmm$ 
\hfil$\btup$
  
\noindent the two other parameters are equal: Bravais class $p4m$ of the 
square \hfil $\btus$
\bye