$$ U \; = \; \left( \begin{array}{cc} 0 & 1 \\\ 1 & 0 \end{array} \right) , $$

Given a real oriented vector space $V$ with inner product, form Lorentzian $L = V \oplus U.$ Elements are of the form $ (v; x,y). $ The norm on $L$ is given by $$ (v; m,n)^2 = v^2 + 2 xy.$$ We infer the inner product $$ (v_1; x_1,y_1) \cdot (v_2; x_2,y_2) = v_1 \cdot v_2 + x_1 y_2 + x_2 y_1.$$

Given any $ l \in L,$ a null vector, we have $l \cdot l = 0,$ and so $l \in l^\perp.$ Furthermore, if $k \in l^\perp$ as well, then $ (k + l)^2 = k^2.$ As a result, we may form the one-dimensional space $\langle l \rangle$ spanned by $l$ itself, then form another space with norm, $$ E(l) = l^\perp / \langle l \rangle.$$ We have that $E\left((0;0,1) \right) = V,$ and $E(l)$ is positive definite.

**Question:** is there a reasonably consistent way to assign an orientation to all the $E(l)?$

Motivation: the Leech lattice is chiral. That is, there is no automorph of the Leech lattice with negative determinant. All the even lattices that Pete Clark and I found are achiral, they all possess improper automorphs. So, not only are they in genera of class number one, they are in genera of proper class number one. I am trying, with a good deal of frustration, to decide whether Conway's argument, as I report in A Priori proof that Covering Radius strictly less than $\sqrt 2$ implies class number one really imples proper class number one, or is it just luck? After some email with Daniel Allcock, this question is a beginning. Daniel emphasizes that orientation of a lattice is an orientation of the surrounding real vector space. Conway's proof is that (in my case) every (primitive) null vector is equivalent by a sequence of Lorentz reflections in roots, which ought certainly to be said to reverse orientation on $L.$ But the $E(l)$ construction does not seem to care about that, it does not know how we got $l.$ At the end, and I may need several questions to get through this, does Conway's argument actually show proper class number one?

Very confused.