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

1. Jun 14, 2006

### Greg Egan

Ralph Hartley wrote:

>Consider a polyhedron inscribed in a sphere of radius 1, centered at the
>origin. Let the surface of the polyhedron inherit the metric from R^3
>(which will be flat except at the vertexes).
>
>For any point p other than the origin, let p_1 be the intersection of
>the polyhedron with the ray from the origin through p. Let t(p) = |p|/|p_1|.
>
>The metric (on R^3-O) ds^2 = -dt(p)^2 + dp_1^2 is flat except at the
>rays from the origin through the vertexes, and any [space]like surface has
>total deficit 4Pi.

That's a really elegant construction, but (at least in the static case) I
think you can get rid of your Big Bang at t=0. Just take the Cartesian
product of the real line R with the polyhedron, with its inherited
metric, and put ds^2 = -dt^2 + dp_1^2, where now p_1 is the projection
from the Cartesian pair (t,p_1).

This would generalise to "polyhedra" formed by triangulating any compact
2-manifold and putting flat metrics on the triangles. The total deficit
in the general case will be 2pi*chi, where chi is the Euler
characteristic of the manifold.

2. Nov 4, 2006

### Ralph Hartley

Greg Egan wrote:
> Ralph Hartley wrote:
>
>> Consider a polyhedron inscribed in a sphere of radius 1, centered at the
>> origin. Let the surface of the polyhedron inherit the metric from R^3
>> (which will be flat except at the vertexes).
>>
>> For any point p other than the origin, let p_1 be the intersection of
>> the polyhedron with the ray from the origin through p. Let t(p) = |p|/|p_1|.
>>
>> The metric (on R^3-O) ds^2 = -dt(p)^2 + dp_1^2 is flat except at the
>> rays from the origin through the vertexes, and any [space]like surface has
>> total deficit 4Pi.

>
> That's a really elegant construction, but (at least in the static case) I
> think you can get rid of your Big Bang at t=0.

Yes, I realized, just after sending, that the area is constant, and the
space is just R x Polyhedron. The sphere was not needed either, any
convex polyhedron would do.

I'm pretty sure that the edges of the polyhedron are purely artifacts of
the construction as well. The space is flat except at the vertexes.

Given a reference frame on one face, you can extend it to the whole
space in many different ways, one for each way of cutting the polyhedron
up and spreading it out flat. How many there are depends on what you
count as a "different" cutting.

> This would generalise to "polyhedra" formed by triangulating any compact
> 2-manifold and putting flat metrics on the triangles. The total deficit
> in the general case will be 2pi*chi, where chi is the Euler
> characteristic of the manifold.

If you require the manifold to be orientable, and don't allow negative
masses, I think the only new case is a flat torus (which does have some
parameters).

If you don't require that it be compact (and allow some infinite
triangles), you get the cases with deficit between 0 and 2Pi.

Lets see, how many more static solutions are there? If you allow
timelike loops, quite a few more I think.

Ralph Hartley