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 also gave an example in six or fewer variables, repeated below. Pete's Euclidean property (in the case of positive definite integral quadratic forms) is simply that for any point $\vec x \in \mathbf Q^n$ but $\vec x \notin \mathbf Z^n,$ we require that there be at least one $\vec y \in \mathbf Z^n$ such that $$ q(\vec x - \vec y) < 1. $$


The question on the example with 7 variables was a big success, see [Verifying an example in the Geometry of Numbers and Quadratic Forms](https://mathoverflow.net/questions/40195/)
Could some kind soul please verify the example(s) below. Note how very symmetric this one is, I have little doubt that the "worst" point(s) must occur on the main diagonal $x\_1 = x\_2 = \cdots x\_n.$ Indeed, I think that for any point the orthogonal projection onto the main diagonal has worse "Euclidean" value.


For six or fewer variables we can use one of the easiest constructions, include all mixed terms so that the Gram matrix becomes
$$ P\_6 \; \; = \; \; 
\left( \begin{array}{cccccc}
 1 & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2}\\\
 \frac{1}{2} & 1 & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\\
 \frac{1}{2} & \frac{1}{2} & 1 & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\\
 \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & 1 & \frac{1}{2} & \frac{1}{2} \\\
 \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & 1 & \frac{1}{2} \\\
 \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & 1 
\end{array} 
 \right) . $$
Then the worst $\vec x$ is either
$$ \vec x = \left( \frac{3}{7}, \frac{3}{7}, \frac{3}{7} , \frac{3}{7}, \frac{3}{7}, \frac{3}{7} \right) $$
or 
$$ \vec x = \left( \frac{4}{7}, \frac{4}{7}, \frac{4}{7} , \frac{4}{7}, \frac{4}{7}, \frac{4}{7} \right) $$
with ``Euclidean minimum'' $\frac{6}{7}.$


This construction is much easier to figure out. In dimension $ n$ we have determinant $\frac{n +1}{2^n}$ and characteristic polynomial
 $$ \left( \frac{1}{2^n} \right) \left(2 x - (n+1) \right) \left(2 x - 1 \right)^{n-1}. $$ For even $n $ the worst $\vec x$ has either all entries $\frac{n}{2(n + 1)}$ or $\frac{n + 2}{2(n + 1)}$ with a Euclidean minimum of $\frac{n^2 + 2 n}{8 (n+1)}.$ 
For odd $n $ the worst $\vec x$ has all entries $\frac{1}{2}$ with a Euclidean minimum of $\frac{n+1}{8}.$ 


The formulas $\frac{n^2 + 2 n}{8 (n+1)}$ and $\frac{n+1}{8}$ show that this recipe fails (just barely) for $n=7$ and more obviously for larger $n.$ I don't believe there are any examples with $n \geq 9$ and I have my doubts that there can be any for $n=8.$ I did try half of Gosset's root lattice for $E\_8,$ see 
<http://en.wikipedia.org/wiki/E8_lattice> 
 but it does not seem to work to have any of the squared terms with a coefficient other than $1,$ in all likelihood as soon as $n \geq 4.$ 



Verifying my other example in the Geometry of Numbers and Quadratic Forms
It's strange that I didn't see this question before, since Will and I have started thinking about these issues off-site. Anyway, recently I found (in the sense of located, not discovered) the answer to this and a bit more.


First, the MAGMA computational system has a built in command to compute the Euclidean minimum of (the lattice associated to) a positive definite integral quadratic form, namely ${ \tt CoveringRadius }$. In the examples I tried, this computation was almost instantaneous up until about $6$ variables, relatively quick for $7$ and $8$ variables, and didn't terminate when run on my local server starting in $9$ variables. Here is how it goes for Will's six dimensional lattice:



> 
> $>$ F61 := MatrixRing(IntegerRing(), 6) ! [2,1,1,1,1,1, 1,2,1,1,1,1, 1,1,2,1,1,1\
>  , 1,1,1,2,1,1, 1,1,1,1,2,1, 1,1,1,1,1,2];  
> 
>  $>$ L61 := LatticeWithGram(F61);  
> 
>  $>$ L61;  
> 
>  Standard Lattice of rank 6 and degree 6  
> 
>  Determinant: 7  
> 
>  Factored Determinant: 7  
> 
>  Inner Product Matrix:  
> 
>  [2 1 1 1 1 1]  
> 
>  [1 2 1 1 1 1]  
> 
>  [1 1 2 1 1 1]  
> 
>  [1 1 1 2 1 1]  
> 
>  [1 1 1 1 2 1]  
> 
>  [1 1 1 1 1 2]  
> 
>  $>$ (1/2)\*CoveringRadius(L61);  
> 
>  6/7  
> 
>  $>$ #GenusRepresentatives(L61);  
> 
>  1  
> 
>  $>$ IsIsomorphic(L61,Lattice("A",6));  
> 
>  true 
> 
> 
> 


Some comments: you can see that the Gram matrix is twice the one Will gives. This is because MAGMA's lattice package likes matrices with integral entries. Thus we compute half the covering radius rather than the covering radius to take care of this, and the answer is indeed what Will said it would be: $\frac{6}{7}$. (In particular, this *is* a Euclidean quadratic form.)


The second calculation shows that this lattice has class number one in the sense of quadratic forms, i.e., its genus has only one class. Will, Jon Hanke and I believe that any Euclidean quadratic form should have class number one, but we haven't (yet!) been able to prove this, so this was an interesting data point for that. (I computed nearly $60$ other examples late last week.) 


The last calculation shows that Will's lattice (when rescaled by multiplication by $2$) is isomorphic to 
the $A\_6$ root lattice. The covering radii for all the root lattices are well known to the experts (though not to me): for instance, the covering radius of $A\_n$ is given on p. 109 of Conway and Sloane's *Sphere Packings, Lattices and Groups* and the formula agrees with the one given in the question above: in particular it is asymptotic to $\frac{n}{8}$. (Of course, in their book they explain what is going on geometrically, which is a much more desirable answer than just running a software package.)


