# A Correction on a Proof of a Combinatorial Property of the Set of Minimal
Vectors in Root Lattices $\mathbb{A}_{n}$

Kong Xiaoran, Peking University

###### Abstract

In this papar, we point out some mistakes in a proof of an important
combinatorial property of $S(\mathbb{A}_{n})$, the set of all minimal vectors
of lattice $\mathbb{A}_{n}$, and correct them in the last section. This
property plays an essential role in classifying perfect lattices in euclidean
space.

##  1 Introduction.

As a member of the famous family of root lattices, lattice $\mathbb{A}_{n}$
has some interesting properties. In the traditional way to describe
$\mathbb{A}_{n}$, we usually embed $\mathbb{A}_{n}$ to the $(n+1)$-dimensional
Euclidean space $\mathbb{R}^{n+1}$. Choose
$\varepsilon_{0},\varepsilon_{1},\varepsilon_{2},\cdots,\varepsilon_{n}$ as
the orthonormal basis of $\mathbb{R}^{n+1}$, and let
$H\subset\mathbb{R}^{n+1}$ be the subspace orthogonal to the vector
$e=\varepsilon_{0}+\varepsilon_{1}+\cdots+\varepsilon_{n}$. We can define

$\mathbb{A}_{n}=\mathbb{Z}^{n+1}\cap
H=\left\\{\left.\sum_{k=0}^{n}a_{k}\varepsilon_{k}\right|\sum_{k=0}^{n}a_{k}=0,a_{i}\in\mathbb{Z}\right\\},$

which is a full-rank lattice in the $n$-dimensional Euclidean space $H$. We
can easily find out that $\\{e_{i}=\varepsilon_{0}-\varepsilon_{i}\mid
i=1,2,\cdots,n\\}$ form a basis of $\mathbb{A}_{n}$, which is usually called
Korkine-Zolotareff basis.

Since all components of a vector in $\mathbb{A}_{n}$ add up to 0, there does
not exist a vector of norm 1 in $\mathbb{A}_{n}$. Noticing that all $e_{i}$
are of norm111In this paper, the norm of a verctor means the square of its
euclidean length. 2, the shortest vectors in $\mathbb{A}_{n}$ should have norm
2. Hence, the set $S(\mathbb{A}_{n})$, consisting of all minimal vector of
$\mathbb{A}_{n}$, should be $\\{\varepsilon_{i}-\varepsilon_{j}\mid 0\leqslant
i\neq j\leqslant n\\}$, which could also be expressed as

$S(\mathbb{A})=\\{\pm e_{1},\pm e_{2},\cdots,\pm
e_{n}\\}\cup\\{\pm(e_{i}-e_{j})\mid 1\leqslant i<j\leqslant n\\}.$

The set $S(\mathbb{A}_{n})$ has a fantastic combinatorial property that every
collection of $n$ independent vectors in $S(\mathbb{A}_{n})$ spans the whole
lattice $\mathbb{A}_{n}$, which can be shown by reduction to absurdity (See
[1], Proposition 6.1.1). Surprisingly, the converse is also true! To classify
perfect lattices with maximal index 1, J. Martinet recorded the interesting
combinatorial property of $S(\mathbb{A}_{n})$ in his famous textbook on
perfect lattices in Euclidean Space [1]:

######  Theorem 1.1 ([1], Lemma 6.1.3).

Let $M$ be a free $\mathbb{Z}$-module of rank $n$ and let $\mathcal{S}$ be a
family of elements of $M$ which satisfy the following five properties:

1). $0\notin\mathcal{S}$;

2). $x\in\mathcal{S}\Longrightarrow-x\in\mathcal{S}$;

3). $\mathcal{S}$ is of rank $n$;

4). $|\mathcal{S}|\geqslant n(n+1)$;

5). $n$ arbitrary independent elements of $\mathcal{S}$ generate $M$.
Then $\mathcal{S}$ possesses exactly $\frac{n(n+1)}{2}$ pairs $\pm x$, from
which one can extract a basis $(e_{1},\cdots,e_{n})$ for $M$ such that

$\mathcal{S}=\\{\pm e_{i}\mid 1\leqslant i\leqslant
n\\}\cup\\{\pm(e_{i}-e_{j})\mid 1\leqslant i<j\leqslant n\\}.$

He also gave a proof on this proposition in Section 6.1 of [1]. Historically,
this proposition perhaps appeared first in the earlier paper [2] of A. Korkine
and G. Zolotareff, which is written in French and does not have English
version. They used this proposition to classify perfect quadratic forms, which
is equivalent to perfect lattices geometrically. However, both in [2] and [1],
the proofs of Theorem 1.1 include some mistakes.

In this paper, we document the proof from [1] in Section 2, and illustrate the
mistakes of the proof in Section 3. In Section 4, we give our corrected proof
of Theorem 1.1.

##  2 Original Proof.

In this section, we document the original proof of Theorem 1.1 in Chapter 6 of
J. Martinet’s book [1]. We merely copy the proof literally here, with some
comments we make as footnotes. We do not correct any mistakes in the proof
while copying, which we postpone to the section 4. Both in the original
textbook [1] and our paper here, the components of an element with respect to
a given basis are written as a column-vector rather than a row-vector.

In [1], the proof of Theorem 1.1 goes as:

Before entering into the proof of Theorem 1.1, we introduce the notion of a
characteristic determinant:

######  Definition 2.1 ([1], Definition 6.1.4.).

Let $M$ be a free module of rank $n$ endowed with a basis
$\mathcal{B}=(e_{l},\cdots,e_{n})$, let $\mathcal{S}\supset\mathcal{B}$ be a
finite subset of $M$, and let $s=|\mathcal{S}|$. A characteristic determinant
for the pair $(\mathcal{B},\mathcal{S})$ is a determinant of order $r\leqslant
n$ extracted from the $n\times s$ matrix of the components in $\mathcal{B}$ of
the elements of $\mathcal{S}$.

######  Lemma 2.2 ([1], Lemma 6.1.5.).

With the notation of Definition 2.1, let $K$ be the maximum of the absolute
values of determinants $\det_{\mathcal{B}}(x_{1},\cdots,x_{n})$ for $x_{1}$,
$\cdots$, $x_{n}\in\mathcal{S}$. Then the inequality $|d|\leqslant K$ holds
for any characteristic determinant $d$ of $(\mathcal{B},\mathcal{S})$. In
particular, the absolute values of the components in $\mathcal{B}$ of any
element of $\mathcal{S}$ are bounded from above by $K$.

Proof. A characteristic determinant $d$ is the determinant of the matrix $D$
of $r\leqslant n$ elements $x_{1},\cdots,x_{r}$ of $S$ relatively to a system
of $r$ vectors $b_{1},\cdots,b_{r}$ of $\mathcal{B}$. After performing a
suitable permutation of the $e_{i}$, we may assume that $b_{i}=e_{i}$. We then
have

$\pm
d=\operatorname{det}_{(e_{1},\cdots,e_{r})}(x_{1},\cdots,x_{r})=\operatorname{det}_{(e_{1},\cdots,e_{n})}(x_{1},\cdots,x_{r},e_{r+1},\cdots,e_{n}),$

and the absolute value of the last determinant is bounded above by $K$.
$\square$

Proof of 1.1. We use induction on $n$, Theorem 1.1 being obvious for $n=1$,
and we first look for convenience at the cases when $n=2$ or $n=3$. We denote
by $(e_{1},e_{2},\cdots,e_{n})$ a basis for $M$ consisting of elements of
$\mathcal{S}$, and begin with a lemma that we shall use several times in the
proof:

######  Lemma 2.3 ([1], Lemma 6.1.6.).

Under the hypotheses of Theorem 1.1, we have:

1\. If the components of two elements of $\mathcal{S}$ on two elements of
$\mathcal{B}$ are nonzero, they are then equal or opposite.

2\. There does not exist in $\mathcal{S}$ any $3$-tuple
$(e_{i}-e_{j},e_{i}-e_{k},e_{j}+e_{k})$.

Proof. In the first case, after having changed if necessary the signs of
$e_{1}$ and $e_{2}$, we would have the characteristic determinant

$\operatorname{det}_{(e_{i},e_{j})}(e_{i}+e_{j},e_{i}-e_{j})=\begin{vmatrix}1&1\\\
1&-1\end{vmatrix}=-2;$

similarly, we would have in the second case a third-order determinant

$\operatorname{det}_{(e_{i},e_{j},e_{k})}(e_{i}-e_{j},e_{i}-e_{k},e_{j}+e_{k})=\begin{vmatrix}1&1&0\\\
-1&0&1\\\ 0&-1&1\end{vmatrix}=2.$

$\square$

If $n=2$, a half-system of elements of $\mathcal{S}$ is
$\\{e_{1}e_{2},e_{1}\pm e_{2}\\}$. By Lemma 2.3, $e_{1}+e_{2}$ and
$e_{1}-e_{2}$ cannot both belong to $\mathcal{S}$. Negating $e_{2}$ if
necessary, we obtain the half-system $\\{e_{1},e_{2},e_{1}-e_{2}\\}$ that we
need.

If $n=3$, again by Lemma 2.3(1), there can be in $\mathcal{S}$ at most one
pair $\pm x$ with $x$ of the form $\pm e_{1}\pm e_{2}\pm e_{3}$. By
permutation and change of signs of $e_{1},e_{2},e_{3}$, we my assume that a
half-system of elements of $\mathcal{S}$ consists of the five elements
$e_{1},e_{2},e_{3},e_{4}=e_{1}-e_{2},e_{5}=e_{1}-e_{3}$ and a sixth one of the
form $e_{2}\pm e_{3}$ or $e_{1}\pm e_{2}\pm e_{3}$. Applying Lemma 2.2, we are
left with the two possibilities $e_{6}=e_{2}-e_{3}$ or
$e_{6}=e_{1}-e_{2}-e_{3}$. In the first one we are done, and the second case
reduces to the first case by taking $e_{2}^{\prime}=e_{1}-e_{2}$ instead of
$e_{2}$.

We now turn to the proof of Theorem 1.1 for $n\geqslant 4$. Let $M^{\prime}$
be the submodule of $M$ generated by $e_{2},e_{3},\cdots,e_{n}$, and let
$\mathcal{S}^{\prime}$ be the set of nonzero elements $x\in M^{\prime}$ such
that $x$ or $\pm e_{1}+x$ belongs to $\mathcal{S}$. Choose a half-system
$\overline{\mathcal{S}}$ in $\mathcal{S}$ in such a way that the component on
$e_{1}$ of any element of $\mathcal{S}$ be 0 or $+1$. One then obtains a map
of $\overline{\mathcal{S}}\setminus\\{e_{1}\\}$ onto a half-system
$\overline{\mathcal{S}^{\prime}}$ of $\mathcal{S}^{\prime}$ by suppressing the
component on $e_{1}$ of the elements of $\mathcal{S}$. The inverse image of an
element of $\overline{\mathcal{S}^{\prime}}$ possesses one or two elements,
this last possibility corresponding to twin systems $\\{x,e_{1}+x\\}$. To be
able to use induction from $\mathcal{S}^{\prime}$ to $\mathcal{S}$, we must
prove the inequality $|\mathcal{S}^{\prime}|\geqslant(n-1)n$. To this end, we
now prove the following lemma:

######  Lemma 2.4 ([1], Lemma 6.1.7.).

The number of twin systems $\\{x,e_{1}+x\\}$ in $\overline{\mathcal{S}}$ is at
most $n-1$.

Proof.  Lemma 2.3(1) shows that two elements of $\mathcal{B}$ belonging to two
twin systems may not have opposite nonzero components on one of the
$e_{i},i\geqslant 1$. Replacing $e_{i}$ by $-e_{i}$ when need be, we may
assume that all their components are non-negative on the $e_{i}$. Suppose that
there be at least $n$ twin systems, and let $x\in\mathcal{S}^{\prime}$ such
that $\\{x,e_{1}+x\\}$222In the textbook [1], it is writen as
$(e_{1},e_{1}+x)$ there. This may be just a slip of the pen and I correct it
in my paper here.  is a twin system and that $x$ has as many as possible
components $x_{i}$ equal to 1. After permuting the $e_{i}$ if need be, we may
assume that $x_{2}=x_{3}=1$. There do not exist elements
$y,z\in\mathcal{S}^{\prime}$ belonging to distinct twin systems for which the
components on $e_{2}$ and $e_{3}$ are respectively $(0,1)$ and $(1,0)$, for we
would have

$\operatorname{det}_{(e_{1},e_{2},e_{3})}(x,e_{1}+y,e_{1}+z)=\begin{vmatrix}0&1&1\\\
1&0&1\\\ 1&1&0\end{vmatrix}=2.$

We may thus assume that the system of components $(1,0)$ does not occur. Then
replacing $e_{2}$ by $e_{2}+e_{3}$333Note: this element may not in
$\mathcal{S}^{\prime}$; fortunately, we do not need this condition, since we
just need a basis of $M^{\prime}$ to label the twin systems in some sense.
reduces the number of components equal to $1$ for $x$, and does not increase
it for the members of the other twin systems. If the component of $x$ on
$e_{4}$ is nonzero, we can in the same way consider the new basis obtained by
replacing $e_{2}$ by $e_{2}+e_{4}$, until each member of a twin system will
possess a single nonzero component on the new basis. This contradicts the
existence of at least $n$ twin systems, since $\mathcal{S}^{\prime}$444Here
shall be ${M}^{\prime}$ rather than $\mathcal{S}^{\prime}$, for the new basis
we adopt may not and need not belong to $\mathcal{S}^{\prime}$ as we explained
in the former footnote.  contains only $n-1$ elements of this type. $\square$

End of the proof of 1.1. The induction hypothesis first shows that
$\mathcal{S}^{\prime}$ contains exactly $(n-1)n$ pairs $\pm x$. Since one
obtains $\mathcal{S}$ by adjoining to $\mathcal{S}^{\prime}$ the elements $\pm
e_{1}$ and at most $n-1$ pairs $\pm(e_{1}+x)$ with $x\in\mathcal{S}^{\prime}$,
we see that $\mathcal{S}$ contains exactly $\frac{n(n+1)}{2}$ pairs $\pm x$,
namely those of $\mathcal{S}^{\prime}$, the pair $\pm e_{1}$, and $n-1$ pairs
$\pm(e_{1}+x)$ with $x\in\mathcal{S}^{\prime}$.555 The first mistake in the
proof.

There remains to prove that $\mathcal{S}$ contains a basis for $M$ of the type
we want. The induction hypothesis implies that up to a permutation of the
elements of $\mathcal{S}^{\prime}$, we may assume that $\mathcal{S}^{\prime}$
contains a half-system of the form

$\overline{\mathcal{S}^{\prime}}=\\{\pm e_{i}\mid 2\leqslant i\leqslant
n\\}\cup\\{\pm(e_{i}-e_{j})\mid 2\leqslant i<j\leqslant n\\};$

this implies that $\mathcal{S}$ contains the half-system
$\overline{\mathcal{S}}$ which one obtains from
$\overline{\mathcal{S}^{\prime}}$ by adjoining $e_{1}$ and of $n-1$ elements
of the form $e_{1}\pm e_{i}$ or $e_{1}\pm(e_{i}-e_{j})$.

Let $r$ be the number of elements of $\mathcal{S}$ of the first kind. We shall
show that we can restrict ourselves by a suitable permutation of the elements
of $\mathcal{S}^{\prime}$ to the case where $r=n-1$.

We remark that the following transformations stabilize $\mathcal{S}^{\prime}$
666 The second mistake in the proof.  and thus allow a partition of
$\mathcal{S}^{\prime}$ into two half-systems, one of which is of the desired
form:

  * •

The simultaneous change of signs of $e_{2},\cdots,e_{n}$.

  * •

The permutations of $e_{2},\cdots,e_{n}$.

  * •

The transformations $p_{i,j}(i,j\geqslant 2,i\neq j)$ which map $e_{j}$ onto
$e_{i}-e_{j}$ and fix $e_{k}$ for $k\neq j$.

Since $p_{i,j}$ transforms $e_{1}\pm(e_{i}-e_{j})$ into $e_{1}\pm e_{j}$, we
may assume that $r\geqslant 1$. By a permutation of $e_{2},\cdots,e_{n}$, we
restrict ourselves to the case where the $r$ elements of the first kind are
$e_{1}\pm e_{2},\cdots,e_{1}\pm e_{r+1}$ for a suitable choice of the signs.
If there exists in $\mathcal{S}$ an element $e_{1}\pm(e_{i}-e_{j})$ with
$i,j>r+1$, applying $p_{i,j}$ allows us to increase $r$ to $r+1$.

From now on, we suppose that such an element does not exist. Negating
$e_{2},\cdots,e_{n}$ if necessary, we may assume that there is a minus sign in
front of $e_{2}$. Lemma 2.3 then shows that these $r$ elements are
$e_{1}-e_{2},\cdots,e_{1}-e_{r+1}$.

If $r<n-1$, let $x=e_{1}\pm(e_{i}-e_{j})$ be another element of $\mathcal{S}$.
We may assume that $i\leqslant r+1$. Lemma 2.3 shows that there must be a
minus sign in front of $e_{i}$, hence that $x=e_{1}-(e_{i}-e_{j})$. Since
there is a plus sign in front of $e_{j}$, we have $j>r+1$.

Since the transformation $p_{i,j}$ preserve the elements $e_{1}-e_{k}$ for
$2\leqslant k\leqslant r+1$ and transforms $e_{1}-(e_{i}-e_{j})$ into
$e_{1}-e_{j}$, it can be used to increase again $r$ to $r+1$. Iterating the
process finally gives $r$ the value $n-1$. $\square$

##  3 Some Mistakes and a Counter Example.

There are several mistakes in the proof given above. One of them is easy to
fix, while two of others are hard.

###  3.1 A Mistake Easy to Fix.

At the beginning of the proof of cases $n\geqslant 4$, the author constructed
a set $\overline{\mathcal{S}^{\prime}}$ by choosing a half-system of
$\mathcal{S}$ and suppressing the component on $e_{1}$ of elements in
$\overline{\mathcal{S}}\setminus\\{e_{1}\\}$. This may cause that
$\overline{\mathcal{S}^{\prime}}$ is not a half-system. In fact, if $\\{\pm
e_{1},\pm(e_{1}-e_{2})\\}$ is a twin system, one happens to choose
$e_{1}-e_{2}$ and $e_{2}$ when he determines $\overline{\mathcal{S}}$.
Suppressing $\overline{\mathcal{S}}$, one may obtain that both $-e_{2}$ and
$e_{2}$ belong to $\overline{\mathcal{S}^{\prime}}$.

This phenomena is easy to overcome. We can exchange the order of constructing
a half-system and suppressing the component on $e_{1}$. Namely, we can
construct the set $\mathcal{S}^{\prime}=\\{x\in M^{\prime}\mid x\in
M\;\text{or}\;x\pm e_{1}\in M,x\neq 0\\}$ first, and next let
$\overline{\mathcal{S}^{\prime}}$ be an arbitrary half-system of
$\mathcal{S}^{\prime}$. This guarantees that $\mathcal{S}^{\prime}$ will
satisfy the induction conditions and $\overline{\mathcal{S}^{\prime}}$ be a
true half-system of $\mathcal{S}^{\prime}$.

###  3.2 Mistakes not trivial.

The first mistake in the original proof, which is marked by footnote 5 in page
5, is recovering the set $\mathcal{S}$ by adding $n-1$ pairs of $\pm(e_{1}+x)$
to $\mathcal{S}^{\prime}$.

Although there are exactly $n-1$ twin systems in $\mathcal{S}$, which
contribute $2(n-1)$ pairs of vectors to $\mathcal{S}$, the inverse images of
other elements of $\mathcal{S}^{\prime}$ may not be themselves. In fact,
according to the definition of $\mathcal{S}^{\prime}$, the inverse image of
$x\in\mathcal{S}^{\prime}$ could be $x$, $x+e_{1}$ or $x-e_{1}$. Therefore, we
should recover $\mathcal{S}$ by not only adding $n-1$ twin system to
$\mathcal{S}^{\prime}$, but also modifying each vector
$x\in\mathcal{S}^{\prime}$ not in a twin system with adding $e_{1}$, $-e_{1}$
or 0 accordingly. In the following subsection 3.3, we can see that, for any
integer $r$ less than $n$, one can easily construct a choice of initial basis
$e_{1},e_{2},\cdots,e_{n}$ such that there are $(r-1)(n-r)$ pairs of elements
of $\mathcal{S}^{\prime}$ whose inverse images lie out of
$M^{\prime}=\operatorname{span}_{\mathbb{Z}}\\{e_{2},\cdots,e_{n}\\}$.

The second mistake, marked by footnote 6 in page 6, is that the
transformations $p_{i,j}$ do not stabilize the forms of elements in
$\mathcal{S}$. For example, it maps the vector $e_{\ell}-e_{j}\,(\ell\neq i)$,
if it belongs to $\mathcal{S}$, to $e_{\ell}-(e_{i}-e_{j})$, which has 3 non-
zero components. As a consequence, we eliminate a vector $e_{1}-(e_{i}-e_{j})$
with three non-zero components by $p_{i,j}$, but introduce a series new
vectors $e_{\ell}-(e_{i}-e_{j})\,(\ell\neq i)$ of this form.

Hence, we cannot conclude at the end that we obtain a basis of $M$ with
respect to which $\mathcal{S}$ has the desire form. We can only say that under
the basis we obtain, all of $\pm e_{i}(1\leqslant i\leqslant n)$ and
$\pm(e_{1}-e_{j})\,(2\leqslant j\leqslant n)$ belong to $\mathcal{S}$.
Luckily, we can complete the proof based on the conclusion, although there is
a more straightforward way to show this as presented in the proof of Lemma
4.1.

###  3.3 A Counter Example to Illustrate the First Fault.

Here is a simple example which explains why we cannot obtain $\mathcal{S}$
just by adding $n-1$ new pairs of vectors to $\mathcal{S}^{\prime}$.

Let $f_{1},\cdots,f_{n}$ be the $n$ independent vectors such that
$\mathcal{S}=\\{\pm f_{i}\mid 1\leqslant i\leqslant
n\\}\cup\\{\pm(f_{i}-f_{j})\mid 1\leqslant i<j\leqslant n\\}$. If we happen to
choose the initial basis as

$e_{1}=f_{1},\cdots,e_{r}=f_{r},\quad
e_{r+1}=f_{1}-f_{r+1},\cdots,e_{n}=f_{1}-f_{n}.$

Then, the vectors of $\mathcal{S}$ can be expressed under the basis we choose
by

$\begin{matrix}\pm e_{1}&\pm(e_{1}-e_{2})&\cdots&\pm(e_{1}-e_{r})&\pm
e_{r+1}&\cdots&\pm e_{n}\\\ &\pm
e_{2}&\cdots&\pm(e_{2}-e_{r})&\pm(e_{1}-e_{2}-e_{r+1})&\cdots&\pm(e_{1}-e_{2}-e_{n})\\\
&&\ddots&\vdots&\vdots&&\vdots\\\ &&&\pm
e_{r}&\pm(e_{1}-e_{r}-e_{r+1})&\cdots&\pm(e_{1}-e_{r}-e_{n})\\\
&&&&\pm(e_{1}-e_{r+1})&\cdots&\pm(e_{r+1}-e_{n})\\\ &&&&&\ddots&\vdots\\\
&&&&&&\pm(e_{1}-e_{n})\\\ \end{matrix}.$

In above table, the elements in main diagonal are $\pm f_{i}$, and the element
in position $(i,j)$ is the difference $f_{i}-f_{j}$.

Hence, as was done in Section 2, the submodule
$M^{\prime}=\operatorname{span}_{\mathbb{Z}}\\{e_{2},e_{3}\cdots,e_{n}\\}$
contains

$\mathcal{S}_{1}=\\{\pm e_{i}\mid 2\leqslant i\leqslant
n\\}\cup\\{\pm(e_{i}-e_{j})\mid 2\leqslant i<j\leqslant
r\;\text{or}\;r+1\leqslant i<j\leqslant n\\},$

which are also contained in $\mathcal{S}$. To obtain the compressed set
$\mathcal{S}^{\prime}$, we need to combine $\mathcal{S}_{1}$ with the set

$\mathcal{S}_{2}=\\{\pm(e_{i}-e_{j})\mid 2\leqslant i\leqslant
r,\;r+1\leqslant j\leqslant n\\},$

the elements of which are not in $\mathcal{S}$. Then
$\mathcal{S}^{\prime}=\mathcal{S}_{1}\cup\mathcal{S}_{2}$.

In this way, the $n-1$ twin systems are $\\{-e_{i},e_{1}-e_{i}\\}_{2\leqslant
i\leqslant n}$. But we cannot recover $\mathcal{S}$ by just adding these $n-1$
pairs of vectors to $\mathcal{S}^{\prime}$. In fact, we should modify each
element in $\mathcal{S}_{2}$ by adding or subtracting $e_{1}$ accordingly to
pull it back to $\mathcal{S}$. We need to modify
$|\mathcal{S}_{2}|/2=(r-1)(n-r)$ pairs of vectors in total.

##  4 The Corrected Proof.

Since the main mistakes in the original proof are the construction of
$\mathcal{S}$ from a basis of $\mathcal{S}^{\prime}$ and the usage of
transformations $p_{i,j}$, which have nothing to do with Lemma 2.3, Lemma 2.4
and the assertion that $|\mathcal{S}|=(n+1)n$, we can give our corrected proof
based on these propositions.

Actually, despite of the irregular form of the components of some element, we
can conclude at the end of the original proof that:

######  Lemma 4.1.

There exist $n$ independent elements $e_{1},e_{2},\cdots,e_{n}$ of
$\mathcal{S}$ such that every $e_{1}-e_{i}(2\leqslant i\leqslant n)$ belongs
to $\mathcal{S}$.

Proof. We can show this by using the transformations $p_{i,j}$ to increase $r$
repeatedly until $r=n-1$ as was done in the original proof. However, there is
a more straightforward way to the conclusion.

By the induction hypothesis and Lemma 2.4, we have $|\mathcal{S}|=n(n+1)$,
which implies that there are exactly $n-1$ pairs of twin systems
$\\{x_{1},e_{1}+x_{1}\\},\cdots,\\{x_{n-1},e_{1}+x_{n-1}\\}$. Let
$e_{i}=-x_{i-1}(2\leqslant i\leqslant n)$, and we will obtain that
$e_{1},e_{2},\cdots,e_{n}$ belong to $\mathcal{S}$ and $e_{1}-e_{i}(2\leqslant
i\leqslant n)$ also belong to $\mathcal{S}$. As for the independency of
$e_{2},\cdots,e_{n}$, we have seen in the proof of Lemma 2.4(also see footnote
3 in Page 3) that we can choose a suitable basis of $M^{\prime}$ such that
every $x_{i}$ has exactly one non-zero component under this basis. Hence,
$x_{1},\cdots,x_{n-1}$ form a basis of $M^{\prime}$ and of course are
independent. So are $e_{2},\cdots,e_{n}$. $\square$

Choosing the $n$ independent elements $e_{1},e_{2},\cdots,e_{n}$ obtained in
the above lemma as a basis of $M$, we can characterize the remaining elements
in $\mathcal{S}$ as following:

######  Lemma 4.2.

Beside $\pm e_{1},\pm e_{2},\cdots,\pm e_{n}$ and $\pm(e_{1}-e_{i}),2\leqslant
i\leqslant n$ belonging to $\mathcal{S}$, other elements in $\mathcal{S}$ can
only have two non-zero components with the form $\pm(e_{i}-e_{j}),2\leqslant
i<j\leqslant n$ or three non-zero components with the form
$\pm(e_{1}-e_{i}-e_{j}),2\leqslant i<j\leqslant n$. Moreover, for each index
pair $(i,j),2\leqslant i<j\leqslant n$, one and only one pair of
$\pm(e_{i}-e_{j})$ or $\pm(e_{1}-e_{i}-e_{j})$ belong to $\mathcal{S}$.

Proof. Let $y\in\mathcal{S}$ be a vector not listed in Lemma 4.1. If the
component of $y$ with respect to $e_{1}$ is 0, arbitrary two components of $y$
must have opposite signs. Otherwise one of $\pm y$ has component 1 on some
$e_{i}$ and $e_{j}$ both, then we will have

$\operatorname{det}_{(e_{1},e_{i},e_{j})}(y,e_{1}-e_{i},e_{1}-e_{j})=\begin{vmatrix}0&1&1\\\
1&-1&0\\\ 1&0&-1\end{vmatrix}=2,$

which is impossible. Therefore $y$ can only have two non-zero components,
since there must exist two integers with the same sign among three or more
non-zero integers.

If the component of $y$ with respect to $e_{1}$ is non-zero, using $-y$ to
substitute $y$ when necessary, we can assume the component is 1. In this case,
the other non-zero components of $y$ have to be $-1$. Otherwise the
characteristic determinant

$\operatorname{det}_{(e_{1},e_{i})}(y,e_{1}-e_{i})=\begin{vmatrix}1&1\\\
1&-1\end{vmatrix}=-2$

would appear. Moreover, the number of non-zero components of $y$ could not be
four or more, in which case we would have a forth-order characteristic
determinant

$\operatorname{det}_{(e_{1},e_{i},e_{j},e_{k})}(y,e_{1}-e_{i},e_{1}-e_{j},e_{1}-e_{k})=\begin{vmatrix}1&1&1&1\\\
-1&-1&0&0\\\ -1&0&-1&0\\\ -1&0&0&-1\end{vmatrix}=2.$

Hence, $y$ can only have the form $e_{1}-e_{i}-e_{j}$.

Noticing that $e_{i}-e_{j}$ and $e_{1}-e_{i}-e_{j}$ cannot belong to
$\mathcal{S}$ both due to

$\operatorname{det}_{(e_{i},e_{j})}(e_{i}-e_{j},e_{1}-e_{i}-e_{j})=\begin{vmatrix}1&-1\\\
-1&-1\end{vmatrix}=-2,$

and that there are $\frac{(n-1)(n-2)}{2}$ index pairs $(i,j),2\leqslant
i<j\leqslant n$, the cardinality of $\mathcal{S}$ is bound above by

$\frac{1}{2}|\mathcal{S}|\leqslant
n+(n-1)+\frac{(n-1)(n-2)}{2}=\frac{n(n+1)}{2}.$

But $|\mathcal{S}|=n(n+1)$, which implies that one of $e_{i}-e_{j}$ or
$e_{1}-e_{i}-e_{j}$ belongs to $\mathcal{S}$ exactly for each index pair
$(i,j),2\leqslant i<j\leqslant n$. $\square$

With the help of above lemma, we can arrange the elements of $\mathcal{S}$ in
a table shown following:

$\begin{matrix}\pm
e_{1}&\pm(e_{1}-e_{2})&\pm(e_{1}-e_{3})&\cdots&\pm(e_{1}-e_{n-1})&\pm(e_{1}-e_{n})\\\
&\pm e_{2}&\pm y_{2,3}&\cdots&\pm y_{2,n-1}&\pm y_{2,n}\\\ &&\pm
e_{3}&\cdots&\pm y_{3,n-1}&\pm y_{3,n}\\\ &&&\ddots&\vdots&\vdots\\\ &&&&\pm
e_{n-1}&\pm y_{n-1,n}\\\ &&&&&\pm e_{n}\\\ \end{matrix}$ (4.1)

Each $y_{i,j}(2\leqslant i<j\leqslant n)$ has the form $e_{i}-e_{j}$ or
$e_{1}-e_{i}-e_{j}$.

The Proof of Theorem 1.1.  Let us consider the last column in the table from
$y_{2,n}$ to $y_{n-1,n}$. If $y_{j,n}=e_{j}-e_{n}$, we are done and consider
the next $y_{j+1,n}$. If $y_{j,n}=e_{1}-e_{j}-e_{n}$, we adopt a new basis in
$\mathcal{S}$ as

$e_{k}^{\prime}=e_{k}\,(k\neq j),\qquad e_{j}^{\prime}=e_{1}-e_{j}.$

Then, we still have all of

$e_{1}^{\prime}-e_{k}^{\prime}=e_{1}-e_{k}\,(k\neq j),\qquad
e_{1}^{\prime}-e_{j}^{\prime}=e_{j}$

belong to $\mathcal{S}$. Therefore the new basis also satisfy the desired
condition in Lemma 4.1, and the elements in the first row of table (4.2) do
not change their forms in the new basis. Moreover, the element
$y_{j,n}=e_{1}-e_{j}-e_{n}$ has the form $e_{j}^{\prime}-e_{n}^{\prime}$.

When we have done suitable substitutes of basis for all $j=2,3,\cdots,n-1$, we
will eventually obtain a basis $e_{1},\cdots,e_{n}$ with the elements in last
column having the form $y_{j,n}=e_{j}-e_{n}$.

Now, elements who may have three non-zero coordinates can only be located at
$(i,j)$-place in the table, $2\leqslant i<j\leqslant n-1$. Fortunately, for
each $y_{i,j}$, on the one hand, as was shown before, it can only have the
form $e_{i}-e_{j}$ or $e_{1}-e_{i}-e_{j}$ since $\left\\{e_{1}-e_{i}\mid
2\leqslant i\leqslant n\right\\}\subseteq\mathcal{S}$. On the other hand,
$\left\\{e_{j}-e_{n}\mid 1\leqslant j\leqslant
n-1\right\\}\subseteq\mathcal{S}$ forces $y_{i,j}$ can only be $e_{i}-e_{j}$
or $\pm(e_{n}-e_{i}-e_{j})$. Therefore, $y_{i,j}$ must be $e_{i}-e_{j}$. In
this way, we obtain $n$ independent elements $e_{1},e_{2},\cdots,e_{n}$ of
$\mathcal{S}$ such that

$\mathcal{S}=\left\\{\pm e_{i}\mid 1\leqslant i\leqslant
n\right\\}\cup\left\\{\pm(e_{i}-e_{j})\mid 1\leqslant i<j\leqslant
n\right\\}.$

## References

  * [1] Jacques Martinet, Perfect Lattices in Euclidean Spaces, Grundlehren der mathematischen Wissenschaften 327, Springer-Verlag, Berlin Heidelberg(2003).
  * [2] A. Korkine, G. Zolotareff, Sur les formes quadratiques positives, Math.Ann. 11(1877), 242-292.