I'll write $\mathbf{Z}^n$ for the integral quadratic form $x\_1^2 + \cdots + x\_n^2$. For which values of $n$ is $\mathbf{Z}^n$ unique in its genus, i.e. isolated in Kneser's graph? In particular can someone verify or shoot down the following guesses:


1. For $n \leq 8$, any lattice in the same genus as $\mathbf{Z}^n$ is isomorphic to $\mathbf{Z}^n$.
2. Any lattice in the same genus as $\mathbf{Z}^9$ is isomorphic to either $\mathbf{Z}^9$ or to $\mathbf{Z} \times \mathrm{E}\_8$.



What's in the genus of the cubic lattice?
[*edited* mostly to add information about $n > 9$]


**David Treumann**'s guess is correct: ${\bf Z}^n$ is unique in its genus
*iff* $n \leq 8$, and for $n = 9$ the genus consists of only ${\bf Z}^9$ and
${\bf Z} \oplus E\_8$.


The comments indicate two ways to prove this,
using $p$-neighbors (as implemented in MAGMA) or the mass formula.
Alternatively, one can use the fact that
the theta function $\theta\_L$ of any lattice $L$ in the genus of ${\bf Z}^n$
is a modular form of weight $n/2$ for an index-3
subgroup $\Gamma$ of the full modular group ${\rm PSL}\_2({\bf Z})$.


For $n < 8$, there is only one choice of $\theta\_L$
that has $q^0$ coefficient $1$, and we find that
$L$ must have $2n$ vectors of norm $1$, and is thus isomorphic with ${\bf Z}^n$.


For $n=8$, either there are $16$ vectors of norm $1$ or there is a
characteristic vector of norm zero. But in the latter case $L$ is
an even lattice, and thus not in the genus of ${\bf Z}^8$.
So we're back to having enough unit vectors to identify $L$ with ${\bf Z}^8$.


The case of $n=9$ requires a bit more work. We can always write
$L = {\bf Z}^m \oplus L\_0$ where $L\_0$ is a unimodular lattice of rank $n-m$
with no vectors of norm $1$. Using $\theta\_{L\_0}$ we soon find that
$L\_0$ is either $E\_8$ or the trivial lattice of rank zero. Hence
$L$ is either ${\bf Z} \oplus E\_8$ or ${\bf Z}^9$.


For the details, see for example my papers



> 
> A characterization of the ${\bf Z}^n$ lattice, *Math. Research Letters* **2** (1995), 321-326 (arXiv: [math.NT/9906019](http://arxiv.org/abs/math/9906019)).  
> 
>  Lattices and codes with long shadows, *Math. Research Letters* **2** (1995), 643-651 (arXiv: [math.NT/9906086](http://arxiv.org/abs/math/9906086)).
> 
> 
> 


Of course for every $n \geq 10$ the genus still contains at least the
two lattices ${\bf Z}^n$ and ${\bf Z}^{n-8} \oplus E\_8$, so $n=8$
is the last case where ${\bf Z}^n$ is unique in its genus.


*Added later*: The genus of ${\bf Z}^n$ is known at least for $n \leq 25$.
According to Table 2.2 on page 49 of SPLAG =



> 
> John Conway and Neil Sloane: *Sphere Packings, Lattices and Groups*, 3rd ed. New York: Springer 1999.
> 
> 
> 


the size of this genus for $9 \leq n \leq 25$ is as follows:
$$
\begin{array}{c|cccccccccccccc}
n & 9,10,11 & 12,13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 & 24 & 25
\cr\hline
\# & 2 & 3 & 4 & 5 & 6 & 9 & 13 & 16 & 28 & 40 & 68 & 117 & 273 & 665
\end{array}
$$
Look up "Lattice, unimodular" in the index (page 694) 
for pointers in the book for further information.


