# Re: This Week's Finds in Mathematical Physics (Week 232)

1. Nov 4, 2006

### Greg Egan

I wrote:

>One way to see why the products of some rotations will be boosts is as
>follows.
>
>Suppose that for a pair of linearly independent vectors u and v in R^3 we
>can find an element of O(2,1) with the following properties:
>
> * it preserves both u and v
> * its determinant is -1
> * its square is the identity
>
>If we can find such an element, we'll call it ref(u,v).
>
>Now, if we pick three linearly independent vectors u, v and w so that
>ref(u,v), ref(u,w) and ref(v,w) all exist, we can construct the following
>elements of SO(2,1):
>
> g1 = ref(u,v) ref(v,w) which preserves v
> g2 = ref(v,w) ref(u,w) which preserves w
> g3 = ref(u,v) ref(u,w) which preserves u
>
>and we have g3 = g1.g2 because ref(v,w)^2 = I.
>
>So if we pick a spacelike u and a timelike v and w, we'll have a boost
>that's equal to a product of rotations.
>
>The one question that remains is, when can we find ref(u,v)? This is
>trivial in O(3), but O(2,1) is trickier. Generically we can construct a
>normal n to the plane spanned by u and v:
>
> n^a = g^{ab} eps_{bcd} u^c v^d
>
>and if it's not a null vector we can construct ref(u,v) as the projector
>into the plane minus the projector onto n. But if n is null, I don't
>think ref(u,v) exists.

I think the only way the normal to the plane can be null is if the plane
is tangent to the light cone, in which case it will contain a single null
ray (which is itself normal to the plane), and all the other vectors in
it will be spacelike.

So if (but not only if) u and v are timelike, ref(u,v) will exist.

What's the significance of this construction failing if two of the
vectors lie on a tangent plane to the light cone? Well, if u and v are
*not* on such a plane, then you can choose w to be almost anywhere:
anywhere such that (u,w) and (v,w) also don't lie on tangent planes to
the light cone. In other words, given such generic u and v, you can find
SO(2,1) elements g_u and g_v that preserve them, *and* such that the
eigenvector of g_u g_v lies *almost* anywhere in R^3.

However, if u and v lie on a tangent plane to the light cone, then I
think the eigenvector of (g_u g_v) will always lie on that same tangent
plane, for all g_u and g_v that preserve u and v respectively.

The upshot of this for collisions in 2+1 gravity would be that certain
tachyon-tachyon and tachyon-luxon collisions (which yield either tachyons
or luxons as the outgoing particle, never tardyons) involve *coplanar*
vector-valued momenta.

What I'm claiming amounts to the statement that for each null ray n there
is a subgroup H(n) of SO(2,1), consisting of those elements whose
eigenvectors are normal to n.

So the simplest way to check this would be to look for a subalgebra of
so(2,1) consisting of elements whose null space lies in the plane normal
to n.

If we take the example of n being the null ray generated by (1,1,0), then
so(2,1) elements of the form:

| 0 a b |
| a 0 b |
| b -b 0 |

have their null space generated by (b,b,-a), which is always normal to
(1,1,0). And this set turns out to be closed under the Lie bracket, with
[e_a,e_b]=e_b (where by e_a I mean the element above with a=1, b=0, and
by e_b vice versa).

So it looks like those subgroups of H(n) do exist. But they won't be
Abelian! So those weird coplanar collisions in 2+1 gravity still won't
obey the old vector addition law.