"Regular polytopes" in Minkowski spacetime

Question

Is there an analogue of regular "polytopes" (hyperbolic honeycombs?) in the 4D Minkowski spacetime of special relativity, just as there are six regular polytopes in Euclidean 4D space?

If so, what is their classification?

Answer 1

I don't think that's possible.

First of all one have of course know how you generalize some concepts. The concepts of flatness must be the same, even if you defines flats in terms of distance you will end up with a linear equation anyway. This means that convex polytopes are the same as for euclidean metric (since convexity is defined in terms of line segments between points inside the polytope).

In addition we must define the concept of angles and that could be defined by the bilinear form that defines the metric $d(x,y)^2 = \langle x-y, x-y\rangle$ as if it were a inner product.

Now we can see that regular polygons only can exist in planes that has only space like lines (or time like if you have more than one time dimension). This is because if for example there's both time and space like lines we can find a coordinate system for the space where $\langle u, v\rangle = u_xv_x-u_yv_y$ and if we consider the vectors for the edges of $u_n = r_n(\cos\varphi_n, \sin\varphi_n)$, $r_n>0$ we have that without loss of generality:

$$\langle u_n, u_n \rangle=r_n^2 \cos^2\varphi_n - r_n^2\sin^2\varphi_n = r_n^2\cos(2\varphi_n) = 1$$ $$\langle u_n, u_{n+1}\rangle = r_nr_{n+1}\cos(\varphi_n + \varphi_{n+1}) = C$$

in addition due to convexity $\varphi_n$ must be strictly monotone and $r_n$ and $\varphi_n$ must repeat the same values after one turn. In addition if the polygon is to be closed it must have $\varphi$ both in the intervals $(-\pi/4,\pi/4)$ and $(3\pi/4, 5\pi/4)$. Now that means we have one in the first interval followed by one in the second and we can assume these are $\varphi_0$ and $\varphi_1$, this makes $C<0$ and $\varphi_n$ must alter between the intervals and since this makes $\varphi_3$ is in the second after more than one turn it must have repeated and $\varphi_3=\varphi_1$, that is we have only two edges.

The case where $\cos(2\varphi_n)=0$ has to be handled separately. There's also the case where the plane contains light like lines.

For regular polyhedra we can in a similar way see that it's required that they are in a hyperplanes with only space like lines. For example if the hyperplane contains perpendicular spacelike lines and timelike line (ie we can form a coordinate system with two space axes and one time axes). Since it's faces will be in space like planes their normal has to be time like, if we place the coordinate system around the center (in some sense) of the polygon we can see that it has to have faces with normals to the positive and negative time axis and two of these has to be adjecent which means that all adjecent faces must have time like normals pointing in opposite directions. This rules out more than two faces.

Same reasoning will rule out regular polychora entirely since the space contains both time and space like lines.

Answer 2

Yes, and there are exactly two regular Minkowski polytope for each regular hyperbolic polytope in one less dimension.

The important part of a regular polytope is that it has a certain kind of symmetry. Normally in Euclidean space you use compositions of reflections, rotations and translations because they are the distance preserving transformations. So in Minkowksi space the symmetries of the polytopes would be compositions of reflections, translations and lorentz boosts since they are the distance preserving transformations (together these make up the Lorentz group).

If we disallow translations, the vertices of the polytope must all have the same distance from the origin, so they live on a "sphere". In Minkowski space they will be on a hyperboloid.

One sheet of a two sheeted hyperboloid is an isometric embedding of hyperbolic space into Minkowksi space (meaning distances are preserved), so any polytope in n hyperbolic dimensions can be converted to a polytope on the hyperboloid in 1 time and n space dimensions and vice versa. So the polytopes on this hyperboloid are exactly the polytopes in hyperbolic space. There are infinitely many honeycomb polychora in 3 hyperbolic dimensions, so there are the same polychora on this hyperboloid.

The other possible "spheres" are the cone of lightlike points and the one sheeted hyperboloid of spacelike points. These are also connected to hyperbolic space, but we first need to add some points.

An ideal hyperbolic point is the intersection of two lines that converge at infinity. An ultraideal point is the intersection of two lines that diverge. A line of ordinary hyperbolic points can be converted to plane in Minkowski space by just taking the plane through the line and the origin. Using this, for any point, we can find two hyperbolic lines that intersect at it, turn them into planes, and find their intersection (which will be a line). With this we can map any hyperbolic point to a line through the origin in Minkowski space, and we can also reverse it so it is a bijection. This mapping turns ordinary hyperbolic points into timelike lines, ideal points into lightlike lines and ultraideal points into spacelike lines.

The next step is to identify a n-polytope with its n mirrors, R0 through R(n-1), and its initial point. The initial point must be on all the mirrors except for R0. Using this point and the mirrors we can generate the whole polytope.

Converting the initial point between hyperbolic and Minkowski space is simple, we just use the mapping above. The mirrors are more complicated because we are converting from hyperbolic points into lines through the origin. Lines through the origin can be reflected through the origin without changing them, so when we convert a hyperbolic mirror to a Minkowski mirror we actually have two options: one mirror we get directly from the mapping and that mirror's orthgonal complement (which only differ by a reflection through the origin). This actually lets us convert from a Minkowski mirror to a hyperbolic mirror, since either it or its orthogonal complement will contain timelike points which we can convert to ordinary points using the mapping.

So any Minkowski polytope can be converted to a hyperbolic polytope by converting its initial point and mirrors. Going the other way is more complicated, since each hyperbolic mirror corresponds to two Minkowski mirrors. However, R1 through R(n-1) must contain the initial vertex so they will be forced into one option. R0 will have two options, which will differ by a central inversion.

So for every hyperbolic polytope, there are exactly 2 Minkowski polytopes in one more dimension.

Because 4 dimensions is hard to visualise, here are some polyhedra in 3D Minkowski space. The first two are on one sheet of the two sheeted hyperbola.

{∞, 3}

{5,4}

This one is on the cone, and corresponds to a polyhedra with ideal points.

{3, ∞}