Jul 4, 1973 - $E_{7}=[e_{1}-e_{2}, \cdots e_{5}-e_{6}, e_{5}+e_{6}, ..... Since -lXidentity is an element of Aut $(A_{n})$ , by Lemma 2-5,. (i) we can set ...
J. Math. Soc. Japan Vol. 28, No. 3, 1976
Note on the positive definite integral quadratic lattice By Michio OZEKI (Received July 4, 1973) (Revised Jan. 23, 1976)
\S 0. Introduction. By a quadratic lattice we understand a finitely generated module over , the ring of rational integers in which a metric is given in the sense of M. Eichler [2]. The bilinear form associated to the metric is denoted by $(x, y)$ where and are elements of . If for any in we have $(x, x)$ $L$ $>0$ , we shall say is positive definite, and if it holds that $(x, y)\in Z$ for any pair and in , we shall say is integral. Since we shall confine ourselves to the positive definite integral quadratic lattice only, we shall call such a lattice merely a lattice. Since for any element of a lattice $L(x, x)$ is a positive rational integer, we shall say is m-vector when $(x, x)$ is equal to a positive rational integer $m$ . It is known that the sublattice generated by 2vectors in a lattice plays an important role in the classification theory of positive definite integral quadratic lattices (E. Witt [8], M. Kneser [4], H.-V. $L$
$Z$
$x$
$x$
$y$
$L$
$y$
$L$
$x\neq 0$
$L$
$L$
$x$
$x$
Niemeier [5]). The first purpose of this paper is to show that when is an integer not a lattice has the largest smaller than 17 among all lattices of fixed rank number of 2-vectors if and only if contains or is equal to $B_{n}(D_{n}$ and are defined in \S 1). Roughly speaking, the set of 2-vectors and l-vectors in a lattice exibits the order of the subgroup generated by reflections in the group of units of that lattice. Our second purpose in this paper is to prove Theorem 2 which says that equals to the rank of , if the determinant of a lattice exceeds , where generated by 2-vectors is smaller than . then the rank of the sublattice of Though a fair part of the results in \S 2 is not new, we think it is not worthless to expose it with abbreviated proofs because some of the standpoints is not found in previous literatures as far as we know. $n$
$n$
$L$
$L$
$D_{n}$
$B_{n}$
$L$
$2^{n}$
$n$
$L$
$L$
$n$
\S 1. Some basic notations and definitions. We shall use in an Euclidean space $e_{1},$
$e_{2},$
$\cdots$
$R^{n}$
,
as orthonormal vectors of sufficiently large dimension $n(n=1,2, 3, )$ . We $e_{m},$
$f_{1},$
$f_{2},$
$f_{k}$
or
$g_{1},$
$\cdots$
,
$g_{p}$
422
M. OZEKI
shall order
$R^{n}’ s$
in a canonical manner; $ R^{1}\subset R^{2}\subset R^{3}\subset\ldots$
Let and be two lattices, we understand by an isomorphism a bijecsatisfying the following condition; tive Z-linear map from to $L_{1}$
$L_{2}$
$L_{2}$
$L_{1}$
$\sigma$
$(\sigma x, \sigma y)=(x, y)$
in . Since $Z$ is a principal ideal ring, any finitely holds for any pair generated torsion free Z-module is a free module of finite rank, then a lattice , has Z-basis. Let be a lattice of rank and let be its basis, then any element in can be written in the form; $x,$
$L_{1}$
$y$
$L$
$x$
$n$
$\xi_{i}’ s$
$\cdots$
$v_{n}$
$L$
$\xi_{i}\in Z$
$x=\sum_{i=1}^{n}\xi_{i}v_{i}$
If we think
$v_{1},$
as scalar
.
variables, the form $(x, x)=,\sum_{t_{J}=1}^{n}(v_{i}, v_{j})\xi_{i}\xi_{j}$
becomes a quadratic form. We shall denote this quadratic form by $Q(L)$ , and up to integral equivalence. The deterthis is uniquely determined from $j=1,$ , is called the determinant of $Q(L)$ minant of the matrix or of . We denote it by $d(L)$ . We should remark that $Q(L)$ is always a positive definite integral quadratic form, because is integral and has a posiany quadratic form is expositive Conversely integral definite tive metric. pressed as $Q(L)$ for some lattice in , and to two integrally equivalent quadratic forms correspond isomorphic lattices. (As to the precise exposition see [2].) An isomorphism from onto itself is called an automorphism or a unit automorphisms . All the of of form a group and we denote it by Aut $(L)$ . “ Let be a lattice, then the dual of is defined by; $L$
$\Vert(v_{i}, v_{j})\Vert,$
$i,$
$\cdots$
$n$
$L$
$L$
$R^{n}$
$L$
$L$
$L$
$L$
$L$
$L$
$L^{\#}=\{y\in L\otimes_{Z}Q|(x, y)\in Z, \forall x\in L\}$
,
is the tensor product of and , the field of rational numbers, and is not necessarily integral but this is over . This lattice contains useful for our later consideration. We shall make a list of basic lattices namely;
where
$L$
$L\otimes_{Z}Q$
$Z$
$Q$
$L$
$A_{n}=[e_{1}-e_{2}, e_{2}-e_{3}, \cdots , e_{n}-e_{n+1}]_{Z}$
.
, $e_{n}-e_{n+1}$ This is a lattice in $R^{n+1}$ of rank with its generater $Z$ . In the following by we shall express similar meaning as above. $n$
$[$
$e_{1}-e_{2},$
$]_{Z}$
$B_{n}=[e_{1}, , e_{n}]_{Z}$
.
$\cdots$
over
423
Integral quadratic lattice
In the usual expression in the theory of Lie algebra, may be differently expressed, but both expressions are equivalent in the sense of quadratic lattice. $B_{n}$
$D_{n}=[e_{1}-e_{2}, \cdots , e_{n- 1}-e_{n}, e_{n-1}+e_{n}]_{Z}$
,
$n\geqq 4$
.
15 $E_{6}=[e_{1}-e_{2}, \cdots , e_{4}-e_{5}, e_{4}+e_{5’\overline{2}}(\sum_{i=1}e_{i}+\sqrt{3}e_{6})]_{Z}$
.
$E_{7}=[e_{1}-e_{2}, \cdots e_{5}-e_{6}, e_{5}+e_{6}, \frac{1}{2}(\sum_{i=1}^{6}e_{i}+\sqrt{2}e_{7})]_{Z}$
.
18 $E_{8}=[e_{1}-e_{2}, \cdots , e_{6}-e_{7}, e_{6}+e_{7’\overline{2}}\sum_{l=1}e_{i}]_{Z}$
.
We shall call these lattices basic lattices, these are all integral lattices but have different determinants. In the basic lattices the generators used in the above are also the basis of those lattices and we shall call them canonical basis. For each one of these basic lattices the structure of its automorphism group is known and [1] is a standard reference. A lattice is an orthogonal sum of sublattices and we write as if any in is and of expressed as $x=x_{1}+x_{2}$ with and such that the equation $=0$ holds for any and for any . A lattice is called irreducible if there is no non-trivial orthogonal decomposition. Otherwise is called reducible. $L$
$L$
$L_{2}$
$L_{1}$
$L=L_{1}\oplus L_{2}$
$x_{1}\in L_{1}$
$L$
$x$
$(x_{1}, x_{2})$
$x_{2}\in L_{2}$
$x_{1}\in L_{1}$
$L$
$x_{2}\in L_{2}$
$L$
\S 2. Some preliminary results.
If a lattice
LEMMA 2-1.
Pressed over .
has a l-vector , then is reducible and is exas $L=Zx\oplus L_{1}$ , where $Zx$ is a rank one sublattice of generated by $L$
$L$
$x$
$L$
$x$
$Z$
PROOF. Let ,
$\alpha_{1}-(\alpha_{1}, x)x,$
$\cdots$
$\alpha_{1},$
$\cdots$
,
$\alpha_{n}$
$L$
$\alpha_{n}-(\alpha_{n}, x)x$
$(\alpha_{i}-(\alpha_{i}, x)x,$
Set
be the basis of . Then and . Now we see that;
$x$
)
For the given lattice
is also generated by
$x$
$=(\alpha_{i}, x)-(\alpha_{i}, x)(x, x)=0$
,
.
$i=1,$
$\cdots$
$n$
.
Then we have . Q. E. D. by repeating use of Lemma 2-1 we get the follow-
$L^{\prime}=[\alpha_{1}-(\alpha_{1}, x)x, \cdots , \alpha_{n}-(\alpha_{n}, x)x]_{Z}$
$L$
$L$
$L=L^{\prime}\oplus Zx$
ing decomposition; $L=Zx_{1}\oplus\cdots\oplus Zx_{r}\oplus L^{\prime}$
,
are mutally orthogonal l-vectors and where is a rank one sublattice generated by of over $Z$ and does not contain any l-vector. Apparently is isomorphic to , so we can write without loss of generality as follows; $x_{i}’ s$
$L$
$Zx_{i}$
$L^{\prime}$
$x_{i}$
$Zx_{1}\oplus\cdots\oplus Zx_{r}$
$B_{r}$
$L=B_{r}\oplus L^{\prime}$
..........................
(1)
424
M. OZEKI
is l-vector free. As to 2-vectors, we must take precise care. After where the chain of lemmas we shall establish the following; is generated by 2-vectors over $Z$, then PROPOSITION 2-2. If a lattice has basis consisting of 2-vectors. To establish the above proposition we pose the following problem; (P) When is a basic lattice or an orthogonal sum of basic lattices and is a 2-vector with the condition that $(x, y)\in Z$ holds for any $y\in L$ (henceforth , how becomes the we shall write this condition symbolically as and over $Z$ ? lattice $L+Zx$ , generated by The answer to this problem is that $L+Zx$ is also isomorphic to a basic lattice or an orthogonal sum of basic lattices as the following consideration $(u, v)=0$ and is orthogonal shows. First we can set $x=u+v$ with (we shall write this condition as $L)=0$ ). For the basic lattice we have to the following; type and be a basic lattice other than be an LEMMA 2-3. (i) Let , then can be taken from , (ii) in element of such that such that the case of an element of can be satisfying certain condition specified in the prOOf. taken from type, then we can say that; PROOF. If be a basic lattice other than $L^{\prime}$
$L$
$L$
$L$
$x$
$(x, L)\subseteqq Z)$
$L$
$x$
$(u, L)\subseteqq Z,$
$L$
$v$
$(v,$
$L$
$A_{n}$
$(u, L)\subseteqq Z$
$R^{m}(\supseteqq L)$
$A_{n}(n\supseteqq 1)$
$u$
$L^{\#}$
$u$
$R^{m}(\supseteqq A_{n})$
$u$
$(u, A_{n})\subseteqq Z$
$R^{n+1}$
$L$
$A_{n}$
$L\otimes_{Z}R=R^{k}$
where
$k$
is the rank of
$u=\sum_{l=1}^{k}a_{i}w_{i}$
$L$
.
, where
The condition
$L$
. In this case and
$a_{i}\in R(1\leqq i\leqq k)$
$(u, L)\subseteqq Z$
(2) implies
$a_{i}\in Q(1\leqq i\leqq k)$
can be taken from and we write are the canonical basis of $R^{k}$
$u$
$w_{i}(1\leqq i\leqq k)$
implies that;
for
$(u, w_{i})\in Z$
In each case of basic lattice
,
$L$
$i=1,$
other than
and then
of
$B_{n+1}$
.
$a_{i}\in R(1\leqq i\leqq n+1)$
The given condition
,
.
...............
$\cdot$
where
$(u, A_{n})\subseteqq Z$
$(u, e_{i}-e_{i+1})=a_{i}-a_{i+1}\in Z$
$u$
$e_{1},$
(2)
type
$L^{\#}$
$u=\sum_{l=1}^{k}a_{i}w_{i}$
$A_{n},$
with
$k$
we can easily verify that belongs to . The part (i) can be embedded into from and we put , are the canonical basis
$A_{n}$
of the lemma is thus proved. In case of and without loss of generality we can take $u=\sum_{i=1}^{n+1}a_{i}e_{i}$
,
$\cdots$
$A_{n}$
$B_{n+1}$
$B_{n+1}\otimes_{Z}R=R^{n+1}$
$e_{2},$
$\cdots$
$e_{n+1}$
is equivalent to the condition; for
$i=1,$
$\cdots$
,
$n$
.
We rewrite this condition as; $a_{1}\equiv a_{2}\equiv\cdots\equiv a_{n+1}$
The part (ii) of the lemma is proved.
$(mod 1)$
. ................. $\cdot$
(3)
Q. E. D.
Integral quadratic lattice
425
be an orthogonal sum of basic lattices and $x=u+v$ be $(u, v)=0$ and $(v, L)=0$ , then a 2-vector in the problem (P) such that we can write as $u=u_{1}+\cdots+u_{t}$ with the following conditions;
Let
$L=L_{1}\oplus\cdots\oplus L_{t}$
$(u, L)\subseteqq Z,$
$u$
$(i\neq j)$
$(u_{i}, u_{j})=0$
$(u_{i}, L_{j})=0$
$(i\neq j)$
$(u_{i}, L_{i})\subseteqq Z$
$i=1,$
,
..................
and $\cdots$
,
(4)
.
$t$
We shall justify the above settings. If type, then is a basic other than , where type, then we is we know is the rank of . If $A_{n}\otimes_{Z}R=R^{n}$ $[e_{1}-e_{2}, \cdots , e_{n}-e_{n+1}]_{Z}$ can not say that as far as we adopt as the , namely; . model of But there is another model of $A_{n}$
$L_{i}$
$L_{i}\otimes_{Z}R=R^{n_{i}}$
$L_{i}$
$L_{i}$
$n_{i}$
$A_{n}$
$A_{n}$
$A_{n}$
$\tilde{A}_{n}=[\sqrt{2}e_{i},$
$-\sqrt{\frac{1}{2}}e_{1},$
... ,
$+\sqrt{\frac{3}{2}}e_{2},$
$\cdots$
$-\sqrt{\frac{r-1}{r}}e_{r- 1}+\sqrt{\frac{r+1}{r}}e_{r},$
$-\sqrt{\frac{7l-1}{n}}e_{n- 1}+\sqrt{\frac{n+1}{n}}e_{n}]_{Z}$
$\cdots$
.
It is easily seen that; $[e_{1}-e_{2}, ’ e_{n}-e_{n+1}]_{Z}\cong\tilde{A}_{n}$
.
This time we can say that; $\tilde{A}_{n}\otimes_{Z}R=R^{n}$
and rank
$A_{n}=n$
.
If we can use type, then the settings (4) can be easily as the model of $L_{j}(i\neq j)$ justified because and are separated by the ambient spaces and ). Though ) is the rank of (resp. (resp. , where is theoretically simpler model than $A_{n}=[e_{1}-e_{2}, \cdots , e_{n}-e_{n+1}]_{Z}$ , the calculations attached to . Let be a and we shall not use are more complicated than those of type, then it is known that each element of basic lattice other than Aut $(L)$ is naturally extended to an orthogonal transformation of $L\otimes_{Z}R=R^{k}$ , case Aut is generated by the reflections where is the rank of . In $(-1)$ xidentity, and each element of and with respect to 2-vectors in orthogonal of $B_{n+1}\otimes_{Z}R=R^{n+1}$ . This to an is transformation extended Aut remark will be used later. be a lattice and $x=u+v$ be a 2-vector in the problem (P) (the deLet composition of is like as above). Since our metric is positive, $(x, x)=2$ im$\tilde{A}_{n}$
$A_{n}$
$R^{n_{i}}$
$L_{i}$
$R^{n_{j}}$
$n_{i}$
$L_{i}$
$n_{j}$
$A_{n}$
$L_{j}$
$\tilde{A}_{n}$
$A_{n}$
$A_{n}$
$L$
$A_{n}$
$L$
$k$
$\sigma$
$(A_{n})$
$A_{n}$
$A_{n}$
$(A_{n})$
$L$
$x$
plies; $(u, u)\leqq 2$
Besides $A_{n}$
type
$A_{n}$
we
type the vector must take precise
.
$\cdot$
.........................
(5)
by Lemma 2-3. As to can be taken from later. care and we shall discuss the case of
$u$
$L$
$A_{n}$
426
M. OZEKI
of We shall call an element (we shall abbreviate as $m$ . ) of
$L^{\#}$
$u$
$r.$
( $L$
is a lattice) is a minimal representative modulo if satisfies the following con$L$
$L^{\#}$
$u$
dition; $(u, u)\leqq(u+y, u+y)$
$\forall y\in L$
.
When is a basic lattice, the structure of $L^{\#}/L$ is known and Niemeier [5] remarked at pages 148-150 the following (the notations are a little different from his); modulo LEMMA 2-4. (i) $A_{n}^{\#}/A_{n}\cong Z/(n+1)Z$ and complete $m$ . . of $L$
are given by and calculation we have; $0$
$u_{r}=\frac{r}{n+1}\sum_{l-1}^{n-r+1}e_{i}-\frac{n-r+1}{n+1}\sum_{i=n-r+2}^{n+1}e_{i}$
for
$(u_{r}, u_{r})=\frac{r(n+1-r)}{n+1}$
(ii)
are
$u_{0},$
and complete
$D_{2n+1}^{\#}/D_{2n+1}\cong Z/4Z(n\geqq 2)$
$u_{1}=_{2}^{1^{2n}}--z_{-,-1}^{+1}e_{i},$
$u_{2}=e_{2n+1}$
and
(iii)
are
$D_{2n}^{\#}/D_{2n}\cong Z/2Z\times Z/2Z(n\geqq 2)$
$u_{0}=0,$
$u_{1}=\frac{1}{2}\sum_{i=1}^{2n}e_{i},$
$u_{2}=e_{2n}$
and
(v)
$E_{8}^{*}=E_{8}$
(vi)
$E_{7}^{\#}/E_{7}\cong Z/2Z$
$e_{6}+\frac{e_{7}}{\sqrt{2}}$
,
and
$B_{n}^{\#}=B_{n}$
$e_{5}-\frac{e_{6}}{\sqrt{3}}$
$r$
and
$r=1,$
$\cdots$
,
$n$
and by
,
modulo
$D_{2n+1}^{\#}$
$D_{2n+1}$
and we have;
$(u_{2}, u_{2})=1$
$m$
,
. . of $r$
modulo
$D_{2n}^{\#}$
$D_{2n}$
and we haue;
$u_{3}=\frac{1}{2}\sum_{i=1}^{2n}e_{i}-e_{2n}$
$(u_{2}, u_{2})=1$
,
,
and compleie
$m$
.
$r$
. of
$E_{7}$
modulo
are
$E_{7}$
$0$
and
and we have; $(e_{6}+\frac{e_{7}}{\sqrt{2}},$
(vii)
. . of
and complete
$(u_{1}, u_{1})=(u_{3}, u_{3})=\frac{2n}{4}$
(iv)
$m$
$u_{3}=\frac{1}{2}\sum_{i=1}^{2n+1}e_{i}-e_{2n+1}$
$(u_{1}, u_{1})=(u_{3}, u_{3})=\frac{2n+1}{4}$
for
$1\leqq r\leqq n$
$A_{n}$
$A_{n}^{\#}$
$r$
$E_{6}^{\#}/E_{6}\cong Z/3Z$
$e_{6}+\frac{e_{7}}{\sqrt{2}})=\frac{3}{2}$
and complete
$m$
. . of $r$
$E_{6}^{\#}$
,
modulo
$E_{6}$
are
$0,$
$e_{5}+\frac{e_{6}}{\sqrt{3}}$
and
and we have; $(e_{5}+\frac{e_{6}}{\sqrt{3}},$
$e_{5}+\frac{e_{6}}{\sqrt{3}})=(e_{5}-\frac{e_{6}}{\sqrt{3}},$
$e_{5}-\frac{e_{6}}{\sqrt{3}})=\frac{4}{3}$
.
$m$ . REMARK 1. The values $(u_{r}, u_{r})(1\leqq r\leqq n)$ of . of modulo are necessary for our later argument and we give some as the table. $u_{r},$
$r$
$A_{n}^{\#}$
$A_{n}$
427
Integral quadratic lattice
The following lemma may simplify later arguments. LEMMA 2-5. Let $L=L_{1}\oplus\cdots\oplus L_{t}(t\geqq 1)$ be an orthogonal sum of basic latbe a 2-vectors in the problem(P), where $u=$ tices and $x=u+v$ and $u_{1}+\cdots+u_{t}$ and and in the be the decompOsitiOns of $(v, L)=(v^{\prime}, L)=0$ manner of (4) and and . (i) It is clear that Aut $(L)$ contains the direct prOduct Aut ... $\times Aut(L_{t})$ as the subgroup. If there exist ( $i=1,$ , t) such that $(i=1, \cdots , t)$ , then $L+Zx$ is isomorphic to . $i=1,$ (ii) If there exist ( , t) such that and $i=1,$ isomorphic $L+Zx$ , , to is then . for (iii) If the equation holds for any element of and for $i=1,$ , , then $L+Zx$ is isomorphic to . $y=y_{1}+\cdots+y_{t}$ (i). general element of be the Let PROOF OF with defined by; and be the mapping from $L+Zx$ to $x^{\prime}=u^{\prime}+v^{\prime}$
$u^{\prime}$
$u$
$u^{\prime}=u_{1}^{\prime}+\cdots+u_{t^{\prime}}$
$(v, u^{\prime})=(v^{\prime}, u)=0$
$(L_{1})\times Aut(L_{2})\times$
$\sigma_{i}\in Aut(L_{i})$
$\cdots$
$L+Zx^{\prime}$
$u_{i^{\prime}}=\sigma_{i}u_{i}$
$w_{i}\in L_{i}$
$(u_{i}, u_{i})$
$\cdots$
$u_{i^{\prime}}=u_{i}+w_{i}$
$\cdots$
$L+Zx^{\prime}$
$t$
$(u_{i}, y_{i})=(u_{i^{\prime}}, y_{i})$
$\cdots$
$(u_{i^{\prime}}, u_{i^{\prime}})=$
$L_{i}$
$y_{i}$
$L+Zx^{\prime}$
$t$
$L$
$(1\leqq i\leqq t)$
$y_{i}\in L_{i}$
$L+Zx^{\prime}$
$\varphi$
$\varphi(x)=x^{\prime}$
,
$\varphi(y)=\sum_{i=1}^{t}\sigma_{i}y_{i}$
and $\varphi(y+kx)=\varphi(y)+k\varphi(x)$
for
$k\in Z$
,
. is an isomorphism from $L+Zx$ to then we can easily verify that $L+Zx=L+Zx^{\prime\prime}$ , where PROOF OF (ii). By the given condition we can say $L+Zx^{\prime}$
$\varphi$
$x^{\prime\prime}=\sum_{i=1}^{t}u_{i^{\prime}}+v$
. Let
$\varphi$
be the mapping from
$L+Zx^{\prime\prime}$
to
$L+Zx^{\prime}$
defined by;
428
M. OZEKI $\varphi(y)=y$
for
$\forall y\in L$
,
$\varphi(x^{\prime})=x^{\prime}$
and
for
$\varphi(y+kx^{\prime\prime})=\varphi(y)+k\varphi(x^{\prime\prime})$
$\forall k\in Z$
,
then we can say that is an isomorphism from to PROOF OF (iii). Let be the mapping from $L+Zx$ to $L+Zx^{\prime\prime}$
$\varphi$
$\varphi$
$\varphi(y)=y$
$L+Zx^{\prime}$
$L+Zx^{\prime}$
for
$\forall y\in L$
.
dePned by;
,
$\varphi(x)=x^{\prime}$
and
for
$\varphi(y+kx)=\varphi(y)+k\varphi(x)$
then we can say that is an isomorphism. We shall treat part of 2-vector in
$\forall k\in Z$
$\varphi$
$u$
more
precisely.
and (3).
case
we can set
By Lemma 2-3
$a_{i}\in R(1\leqq i\leqq n+1)$
$A_{n}$
$x$
Moreover
$u$
,
Q. E. D. (described in Lemma 2-3 (ii))
with the conditions
$u=\sum_{i=1}^{n+1}a_{i}e_{i}$
satisPes the inequality (5), so we have;
$(u, u)=\sum_{i=1}^{n+1}a_{i}^{2}\leqq 2$
.
$\cdot$
.....................
Suppose that one of is an integer, then by (3) each integer for $i=1,$ , $n+1$ . By the inequality (6) at most two of (I)
$a_{i}’ s$
$a_{i}$
is also
are not
$a_{i}’ s$
$\cdots$
(6)
zero. ), then and When exactly two of are not zero (say $|a_{i_{1}}|=|a_{i_{2}}|=1$ . Since we aim the and we can say that solution of the problem (P) and the lattice $L+Zx$ is identical to $L+Z(-x)$ , we . Since . and must be one of the forms can assume that ) of (resp. there exists an element such that $w+e_{i_{1}}-e_{i_{2}}=e_{n}-e_{n+1}$ (resp. $w^{\prime}+e_{i1}+e_{i_{2}}=e_{n}+e_{n+1})$ $(e_{i_{1}}-e_{i_{2}}, e_{i_{1}}-e_{i_{2}})=(e_{n}-e_{n+1}, e_{n}-e_{n+1})$ (resp. , and $e_{i_{1}}+e_{i_{2}})=(e_{n}+e_{n+1}, e_{n}+e_{n+1}))$ , by Lemma 2-5, (ii) we can set $u=e_{n}-e_{n+1}$ (resp. $e_{n}+e_{n+1})$ . When $x=u=e_{n}-e_{n+1}$ , then $A_{n}+Zx=A_{n}$ and we shall neglect this case henceforth. When $x=u=e_{n}+e_{n+1}$ , then $A_{n}+Zx=D_{n+1}(n\geqq 3),$ $A_{2}+Z(e_{2}+e_{3})$ and $A_{1}+Z(e_{1}+e_{2})\cong A_{1}\oplus A_{1}$ . We shall call this vector $e_{n}+e_{n+1}$ singular . vector of first kind for $(I)-(ii)$ is not zero, then we have $|a_{io}|=1$ When only one , by Lemma 2-5, . Since -lXidentity is an element of Aut and (i) we can set , by Lemma . Since and (ii) 2-5, we can set $u=-e_{n+1}$ . We shall call this vector singular vector of . second kind for (II) Suppose that one of is not is not integer, then by (3) each $(I)-(i)$
$a_{i}’ s$
$a_{i_{2}},$
$a_{i_{1}}$
$i_{1}\neq i_{2}$
$x=u=a_{i_{1}}e_{i_{1}}+a_{i_{2}}e_{i_{2}}$
$a_{i_{1}}=1$
$u$
$w$
$e_{i_{1}}-e_{i_{2}}$
$w^{\prime}$
$e_{t_{1}}+e_{i2}$
$A_{n}$
$(e_{i_{1}}+e_{i_{2}}$
$\cong A_{3}$
$A_{n}$
$a_{i_{0}}(1\leqq i_{0}\leqq n+1)$
$(A_{n})$
$u=\pm e_{i_{0}}$
$u=e_{i_{0}}$
$e_{i_{0}}-e_{n+1}\in A_{n}$
$(e_{i_{0}}, e_{i_{0}})=(e_{n+1}, e_{n+1})$
$A_{n}$
$a_{i}’ s$
$a_{i}$
Integral quadratic lattice
429
integer for $i=1,$ , $n+1$ . In this case we can take 2-5 (ii). By (3) we can set as; $\cdots$
$a_{1}$
as positive by Lemma
$a_{i}$
$(2\leqq i\leqq n+1)$
$a_{i}=a_{1}+k_{i}$
,
and . By the inequality (6) at most one of has the absolute value larger than one and less than two. $(II)-(i)$ When each $|a_{i}|(1\leqq i\leqq n+1)$ takes the value between zero and one, we can set as;
where
$k_{i}\in Z$
$a_{1}\not\in Z$
$a_{i}’ s$
$0
