Lorentz quotient and orientation $$ U \; = \;  
 \left(  \begin{array}{cc}
  0 & 1  \\\
   1  & 0  
\end{array} 
  \right)  ,
  $$
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.
 A: This particular question (``consistently orienting the reductions") and all those arising from your above link are interesting. 
However, the answer to this question seems `easy' -- and i'm wondering if i'm not missing something. 
First, let us fix an orientation on the total lorentz space $V':=V \oplus U$ and let $h$ denote our lorentz metric $h(v,u_1, u_2)=v^2+2u_1v_1$.
For a nonzero null vector $n \in V'$, your question is really to find a smooth co-orientation of the subspaces $n^\perp$, i.e. we want a smooth family of nonzero oriented transversals to $n^\perp$ (depending smoothly on the nonzero null vector $n$). 
Now for something that's almost silly. In analogy with almost-complex structures in symplectic geometry, consider those linear involutions $j$ of $V'$ (so $j^2=+id$) with the property that $h(x,jx)>0$ for all nonzero $x\in V'$ (we might say that $j$ `tames' $h$). Such involutions do exist, e.g. all  $j_\lambda:=\begin{pmatrix} \lambda & 0 \\ 0 & -\lambda^{-1} \end{pmatrix}$ for $\lambda \neq 0$ are involutions taming $U$ with respect to the standard null basis. 
The totality of such $j$'s (I wouldn't call them almost-complex structures ...) taming $h$ (or additionally acting as $h$-isometries) form a contractible symmetric space (by the same `fibration' argument that Gromov uses in his 1985 paper on pseudo-holomorphic curves).
As I see it, the existence of $j$'s taming your lorentz metric $h$ is exactly what is needed to co-orient the subspaces $n^\perp$, since indeed we have the splitting $V'=n^\perp \oplus \mathbb{R} jn$. 
This argument shows that we could not expect to co-orient the reductions $E(\mathbb{R}n)$ as they vary over the unoriented null lines $\mathbb{R}n$.
