It turns out that each of Pete L. Clark's "euclidean" quadratic forms, as long as it has coefficients in the rational integers $\mathbb Z$ and is positive, is in a genus containing only one equivalence class of forms. In the language of (positive) integral lattices, the condition is that the covering radius be strictly smaller than $\sqrt 2.$ Please see these for background:

Intuition for the last step in Serre's proof of the three-squares theorem

Is the square of the covering radius of an integral lattice/quadratic form always rational?

Must a ring which admits a Euclidean quadratic form be Euclidean?

http://www.math.rwth-aachen.de/~nebe/pl.html

http://www.math.rwth-aachen.de/~nebe/papers/CR.pdf

I have been trying, for some months, to find an *a priori* proof that Euclidean implies class number one. I suspect, without much ability to check, that any such Euclidean form has a stronger property, if it represents any integral form (of the same dimension or lower) over the rationals $\mathbb Q$ then it also represents it over $\mathbb Z.$ This is the natural extension of Pete's ADC property to full dimension. Note that a form does rationally represent any form in its genus, with Siegel's additional restriction of "no essential denominator." If the ADC property holds in the same dimension, lots of complicated genus theory becomes irrelevant.

EDIT: Pete suggests people look at §4.4 of http://math.uga.edu/~pete/ADCFormsI.pdf .

EDIT 2: It is necessary to require Pete's strict inequality, otherwise the Leech lattice appears.

So that is my question, can anyone prove *a priori* that a positive Euclidean form over $\mathbb Z$ has class number one?

EDIT 3: I wrote to R. Borcherds who gave me a rough idea, based on taking the sum of a given lattice with a 2-dimensional Lorentzian lattice. From page 378 in SPLAG first edition, two lattices are in the same genus if and only if their sums with the same 2-dimensional Lorentzian lattice are integrally equivalent. I hope someone posts a fuller answer, otherwise I'll be spending the next six months trying to complete the sketch myself. The references: Lattices like the Leech lattice, J.Alg. Vol 130, No. 1, April 1990, p.219-234, then earlier The Leech lattice, Proc. Royal Soc. London A398 (1985) 365-376. Quoted:

But there is a good way to show that some genus of lattices L has class number 1 if its covering radius is small. What you do is look at the sum of L and a 2-dimensional Lorentzian lattice. Then other lattices in the genus correspond to some norm 0 vectors in the Lorentzian lattice. On the other hand, if the covering radius is small enough one can use this to show that a fundamental domain of the reflection group of the Lorentzian lattice has only one cusp, so all primitive norm 0 vectors are conjugate and therefore there is only 1 lattice in the genus of L. This works nicely when L is the E8 lattice for example. There are some variations of this. When the covering radius is exactly sqrt 2 the other lattices in the genus correspond to deep holes, as in the Leech lattice (Conway's theorem). The covering radius sqrt 2 condition corresponds to norm 2 reflections, and more generally one can consider reflections corresponding to vectors of other norms; see http://dx.doi.org/10.1016/0021-8693(90)90110-A for details

a priori. It would be great to have an argument which sheds light on the more general case. $\endgroup$ – Pete L. Clark Jul 4 '11 at 5:03