In answer to Pete L. Clark's question [Must a ring which admits a Euclidean quadratic form be Euclidean?](https://mathoverflow.net/questions/39510/) on Euclidean quadratic forms, I gave an example in seven variables, repeated below. Pete's Euclidean property is simply that for any point $\vec x \in \mathbf Q^7$ but $\vec x \notin \mathbf Z^7,$ we require that there be at least one $\vec y \in \mathbf Z^7$ such that $$ q(\vec x - \vec y) < 1. $$ 


[Edit: This is the definition for positive definite integral quadratic forms. --PLC]


I think my answer works (and the easier 6 variable one), based on extensive computer calculations, and Pete has been too polite to express much doubt.


Could someone please try to prove that this example works (and the 6 variable one)? It seems likely that this lies in the field <http://en.wikipedia.org/wiki/Geometry_of_numbers> but who can say? 


$$ q( \vec x) = x\_1^2+ x\_1 x\_2 + x\_2^2 + x\_2 x\_3 + x\_3^2 + x\_3 x\_4 + x\_4^2 + x\_4 x\_5 + x\_5^2 + x\_5 x\_6 + x\_6^2 + x\_6 x\_7 + x\_7^2 + x\_7 x\_1. $$ This has the Euclidean property, its worst behavior is either when all $x\_i = \frac{1}{4}$ or when all $x\_i = \frac{3}{4},$ with ``Euclidean minimum'' equal to $\frac{7}{8}.$
Notice that with $\vec x = \left( \frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4} \right),$ the integer lattice points $\vec y$ such that $ q( \vec x - \vec y)=\frac{7}{8} $ include $\vec y = \left( 0,0,0,0,0,0,0\right)$ and all seven cyclic permutations (including the identity) of
$\vec y = \left( 0,1,0,1,0,1,0\right),$ another seven for
$\vec y = \left( 1,0,0,0,0,0,0\right),$ another seven for
$\vec y = \left( 1,0,1,0,0,0,0\right),$ finally seven for
$\vec y = \left( 1,0,0,1,0,0,0\right),$ a total of 29 lattice points on the ellipsoid, of 128 in the standard unit 7-cube. The Gram matrix for the form is
$$ Q \; \; = \; \; 
\left( \begin{array}{ccccccc}
 1 & \frac{1}{2} & 0 & 0 & 0 & 0 & \frac{1}{2}\\\
 \frac{1}{2} & 1 & \frac{1}{2} & 0 & 0 & 0 & 0 \\\
 0 & \frac{1}{2} & 1 & \frac{1}{2} & 0 & 0 & 0 \\\
 0 & 0 & \frac{1}{2} & 1 & \frac{1}{2} & 0 & 0 \\\
 0 & 0 & 0 & \frac{1}{2} & 1 & \frac{1}{2} & 0 \\\
 0 & 0 & 0 & 0 & \frac{1}{2} & 1 & \frac{1}{2} \\\
 \frac{1}{2} & 0 & 0 & 0 & 0 & \frac{1}{2} & 1
\end{array} 
 \right) , $$
which has determinant $\frac{1}{32}$ and characteristic polynomial 
$$ \left( \frac{1}{64} \right) \left(x - 2 \right) \left(8 x^3 - 20 x^2 + 12 x - 1 \right)^2. $$ So the ellipsoids described are not oblate spheroids, there is less symmetry than that.


EDIT. I think it wise to describe what I am completely certain about and what is unclear. What I did is make a cubic grid, where each variable takes on values $\frac{i}{M}$ for $0 \leq i < M.$ So that makes a grid with $M^7$ points. For each point $\vec r$ in the grid, I find the 128 different values of $q(\vec r - \vec y)$ for
$\vec y \in \mathbf Z^7$ and all coordinates of $\vec y$ are either 0 or 1. For that point $\vec r,$ I take the smallest of the 128 values. Now, for every $M$ I have tried, and for every $\vec r$ in the grid, this best value out of 128 has never been larger than $\frac{7}{8}.$ Now, using the fact that for any $\vec x \in \mathbf Q^7$ that is not in the grid, there is some point $\vec r$ such that $ | \vec r -\vec x | \leq \frac{1}{2 M \sqrt 7},$
I get that I can always find a $\vec y \in \mathbf Z^7$ such that
$$ g(x-y) \leq \frac{7}{8} + \frac{1}{7 M} + \frac{1}{14 M^2}, $$ using Cauchy-Schwarz and the maximum eigenvalue of $Q$ being 2. Anyway, this does not show that the Euclidean minimum is really $\frac{7}{8},$ although I believe it is. What it does show is that the Euclidean minimum is less than 1, as soon as $M \geq 2.$ 



Verifying an example in the Geometry of Numbers and Quadratic Forms
Consider the form
$$ Q(x) = 2q(x) = (x\_1+x\_2)^2 + (x\_2+x\_3)^2 + \ldots + (x\_7+x\_1)^2.$$
You have to show that it has Euclidean minimum $\frac74$ attained at $X\_1 = x\_1+x\_2 = \frac12$, ..., $X\_7 = x\_7 + x\_1 = \frac12$, but unfortunately not over the lattice ${\mathbb Z}^7$, where it would be trivial, but over a lattice of index $2$ defined by the condition $y\_1 + y\_2 + \ldots + y\_7 \equiv 0 \bmod 2$. It is clear that $(\frac12, \ldots, \frac12)$ has Euclidean minimum $\frac 74$, and it remains to show that all other points in the fundamental domain of the lattice have a minimum at most $\frac74$.


This is not difficult to see: assume you have the point $(\frac12 + \delta, \frac12 + \varepsilon, ...)$ for small $\delta, \varepsilon \ge 0$. Subtracting the point $(1,1,0,\ldots, 0)$ you will get a point with coordinates $(-\frac12 + \delta, -\frac12 + \varepsilon, ...)$. Repeating this procedure you will eventually reach a point with
$|X\_2|, \ldots, |X\_7| \le \frac12$. If $|X\_1| \le\frac12$, we are done. If not, you can
make $|X\_1| < \frac12$ at the cost of making another coordinate $> \frac12$ in absolute values. If you think about this for a minute you will see that we can always reach a point of the form 
$$ X = \Big(\frac12 + \delta\_1, \frac12 - \delta\_2, . . . , \frac12 - \delta\_7\Big)$$
with $0 \le \delta\_j \le \frac12$ and $\delta\_1 \le \delta\_i$ for all $i > 1$. It remains to show that $q(X) \le \frac74$, which is equivalent to 
$$ \delta\_1 - \delta\_2 - \ldots - \delta\_7 + \delta\_1^2 + \ldots + \delta\_7^2 \le 0. $$
Does this inequality hold?


