The Minkowksi theorem in Minkowski space

4³ǫ_αβµνp^α[T1]p^β[T2]p^µ[T3]p^ν[T4], (3.105) withTi for i= 1, . . . ,5 labelling the five tetrahedra bounding the four-simplex. With the conservation law of the four-momenta (3.101) fulfilled, the four-volume (3.105) would be obviously the same for whatever quadruple of tetrahedra we take to calculate it. In this sense the additional torsional constraint (3.101) has the same intention as the original volume constraint, and guarantees the volume of a four-simplex is the same from whatever side we look at it.

The torsional equations (3.99), (3.100) and (3.101) have an important geometrical interpretation provided by Minkowski’s theorem [156]. The Minkowski theorem holds in any dimensionN, irrespective of the metric signature^*. It states that given a number of covectorsv¹, . . . v^M,M > N that close to zero, there exists a uniqueN-dimensional convex polytope in R^N, bounded by N −1-dimensional facets normal to v¹, . . . v^M, with their volume given by the magnitude of v¹, . . . v^M. The role of the Minkowski theorem for the three-dimensional geometry is well explored, [51, 52, 55, 150, 180–183].

The idea is, that the conservation law (3.101) provides the geometry of the spinfoam vertices, just as the Gauß law (3.100) uncovered the geometry at the nodes of the spin network functions.

3.5.2 The Minkowksi theorem in Minkowski space

In this section we will prove that Minkowski’s theorem holds irrespective of the metric signature. To this goal, let us first recall the Minkowski theorem in R⁴. We choose Cartesian coordinates(X⁰, X¹, X², X³), and introduce the Euclidean metric:

ds² =δ_µνdX^µdX^ν = (dX⁰)²+ (dX¹)²+ (dX²)²+ (dX³)². (3.106) Consider now a set of N positive numbers V(1), . . . , V(N) and covectors n_µ(1), . . . , nµ(N) normalised to one: δ^µνnµ(i)nν(i) = 1(δ^µν is the inverse metric). Suppose that (i) the normals span all ofR⁴, and (ii) close to zero if we weight them byV(i):

span

n^µ(i) _i=1,...,N =R⁴, and:

XN

i=1

V(i)n_µ(i) = 0. (3.107)

*The metric plays actually little role in the Minkowski theorem, a point that has so far been largely ignored to my knowledge. In the next section, we will in fact prove that the Minkowski theorem also holds in Minkowski space.

The Minkowski theorem states that (i) and (ii) are both sufficient and necessary to reconstruct a four-dimensional convex polytope P ⊂R⁴ out of this data. Its bound-ary ∂P splits into N three-dimensional polytopes Ti, with their respective three-volumina given by{V(i)}i=1,...,N while their outwardly oriented four-normals are given by {n^µ(i)}i=1,...,N (with the Euclidean metric (3.106) and its inverse moving the in-dices). The resulting polytope is unique up to rigid translations in R⁴. We can remove the translational symmetry by demanding that the center of mass lies at the origin:

Z of the three-dimensional polytopes T_i. We call them the facets of P.

We will now show that the metric (3.106) plays little role in the reconstruction of P from the volumes and normals of the bounding facets. In fact the pseudo-normals

V_µ(i) =V(i)n_µ(i), (3.109) together with the four-volume element d⁴X are the only ingredients needed to recon-struct the polytope.

To understand why this is true, let us first define the following covectors attached to each bounding polytope: Notice that no metric structure enters the definition of these covectors, the only ingredi-ent is the four-dimensional volume elemingredi-ent (which is in one-to-one correspondence with theǫ-tensorǫ₀₁₂₃ = 1). It is immediate to see that any such covectorV_µ[T_i]annihilates the tangent space ofT_i: for ifZ^µbe tangent toT_i: V_µ[T_i]Z^µ= 0, henceV_µ[T_i]∝n_µ(i).

The proportionality is given by the volume, and therefore Vµ[Ti] = V(i)nµ(i). This follows^* from the determinant formula:

ǫ_α1α2α3α4ǫ_β1β2β3β4 = X

π∈S4

sign(π)δ_α1βπ(1)δ_α2βπ(2)δ_α3βπ(3)δ_α4βπ(4), (3.111) where S4 is the group of permutations of four elements. In other words, the volume V(i) of T_i determines the magnitude ofV_µ[T_i]according to

V_µ[T_i] =V_µ(i) =V(i)n_µ(i), (3.112) whereV(i)equals the Euclidean volume of the three-dimensional bounding polytopes.

Let us make the dependence of (3.112) on the metric tensor more explicit, we thus write:

*The proof is simple: Multiply Vµ[Ti] by the normal n^µ(i) and write the resulting integral as Vµ[Ti]n^µ(i) =R coordi-nates in Ti. If we then employ (3.111) we get the volume as the integral of the square root of the determinant of the induced metric onTi.

where we have introduced (positively oriented) coordinates {x¹, x², x³} on T_i, and δ(∂_i, ∂_j) = δ_µν^∂X_∂xi^µ∂X^ν

∂x^j . We can now also see that n_µ(i) is indeed the metrical nor-mal vector ofTi:

n_µ(i) =n_µ[T_i, δ] = V_µ[T_i]

Vol[T_i, δ]. (3.114) So far, we have done nothing new. Now we should ask the crucial question: How does the reconstruction of the polytope out of normals and volumes depend on the metric tensor—given two metric tensors δ_αβ and δ˜_αβ would we still get the same polytope?

The answer is yes provided the two metrics induce the same four-dimensional volume element, i.e. detδ= det ˜δ= 1. This can be seen as follows.

Let us start again with a set ofN covectorsVµ(i)i= 1, . . . , N that span the algebraic dual ofR⁴ and close to zero. Using the metric δ_µν we can separate V_µ(i) =V(i)n_µ(i) into its magnitude V(i) and its normal direction n_µ(i): δ^µνn_µ(i)n_ν(i) = 1. We can now use this data to reconstruct a polytope P bounded by facets Ti and centered at the origin.

Suppose now that we would have used another Euclidean metricδ˜_µν =δ_αβΛ^α_µΛ^β_ν with det Λ > 0 without loss of generality. We would then write Vµ(i) = ˜V(i)˜nµ(i), and use this splitting (together with the metricδ˜_αβ) to reconstruct the corresponding polytope P˜ now bounded by three-dimensional facets that we call T˜_i. Their normals aren˜µ(i) : ˜δ^µνn˜µ(i)˜nν(i) = 1, while V˜(i) gives the volume of T˜i as measured by δ˜αβ. Again we assume bothP andP˜ to be centered at the origin.

Consider now the determinant formula (3.111) which is now modified only by an overall factor proportional to the determinant ofΛ:

ǫα1α2α3α4ǫ_β1β2β3β4 = 1 det Λ²

X

π∈S4

sign(π)˜δ_α1βπ(1)δ˜_α2βπ(2)δ˜_α3βπ(3)˜δ_α4βπ(4). (3.115) We can thus repeat the argument that has led us to (3.112) in order to find:

V_µ[T_i] =V_µ(i) = det ΛV_µ[ ˜T_i]. (3.116) Looking back at the definition of the volume and the normal (i.e. equations (3.113) and (3.114)) we see immediately that:

Vol[T_i, δ] = det Λ Vol[ ˜T_i, δ], and: n_µ[ ˜T_i, δ] =n_µ[T_i, δ]. (3.117) We have thus found two convex polytopesP andP˜(both centered at the origin). With respect to the original δ_αβ-metric any two bounding facets T_i and T˜_i have identical normals, while the three-volumina coincide only if det Λ = 1. In this case the two data are the same, and the uniqueness of the reconstruction guarantees that the two polytopes are the same, henceP = ˜P provided det Λ = 1.

Let us make an intermediate summary: Taking N covectors Vµ(i) that span all of R^4∗ and close to zero we can reconstruct a unique convex polytope inR⁴ centered at the origin. We then introduce an auxiliary metric δ_αβ with detδ = 1 to facilitate the Minkowski-reconstruction of the polytope from these covectors. The result of the construction is independent of the metric chosen: We could have picked any other Euclidean metric δ˜_αβ that satisfies det ˜δ = 1 and would have found yet the same polytope.

This tell us a lot for the Minkowski theorem in Minkowski space(R⁴, η_αβ). Again we look at a set ofN covectorsV_µ(i)that span all ofR^4∗ (i.e. the dual ofR⁴) and close to zero. We can now pick a future oriented timelike normal and construct the Euclidean metric δ_µν = 2T_µT_ν +η_µν. The next step is to use this Euclidean metric to run the reconstruction algorithm. We end up with a unique convex^* polytope centered at the origin. This polytope is bounded by three-dimensional polytopes Ti. We compute the conormals V_µ[T_i](as defined by (3.110)) and find again V_µ(i) = V_µ[T_i]. The resulting polytope P is independent of the metric chosen: If we chose another future oriented time-normalT˜µwe would use another metric ˜δµν = 2 ˜TµT˜ν+ηµν. The two normals are related by a Lorentz transformation: Λ^α_β ∈SO(1,3) : ˜T_α = Λ^β_αT_β, and so are the two metrics: δ˜_αβ =δ_µνΛ^µ_αΛ^ν_β. But det Λ = 1 implies P = ˜P, the two polyhedra are the same and thus independent of the metric chosen.

We are now only left to understand the metrical interpretation of these conormals.

This is again uncovered by the determinant formula: In fact, equation (3.111) also holds in Minkowski space, it just picks a minus sign:

ǫα1α2α3α4ǫ_β1β2β3β4 =− X

π∈S4

sign(π)η_α1βπ(1)η_α2βπ(2)η_α3βπ(3)η_α4βπ(4). (3.118) This equation implies that the magnitude of the pseudo-normals measures theLorentzian three-volume ofT_i:

η^µνV_µ(i)V_ν(i) =ε_iVol[T_i, η]², (3.119) where η^µν is the inverse Minkowski metric, Vol[T_i, η] measures the Lorentzian three-volume ofT_i, andε_i =±1for ifT_i is a spacelike (timelike) three-surface. In the case of T_i being null V_µ(i)is a null-vector, henceVol[T_i, η] = 0, and we thus have for all three cases:

Vol[Ti, η] = Z

Ti

dx¹dx²dx³ q

|detη(∂i, ∂j)|=qη^µνVµ[Ti]Vν[Ti], (3.120) withη(∂_i, ∂_j) =η_αβ^∂X_∂x^αi ∂X^β

∂x^j , and{x¹, x², x³} positively oriented coordinates inT_i. Let us summarise this result: Given a set of N covectorsV_µ(i) ∈ R^4∗ that close to zero and span all ofR^4∗, we can construct a convex polytopeP ⊂R⁴ unique up to rigid translations. This polytope is bounded byN three-dimensional polytopesT_i such that V_µ(i) = _3!¹ R

Tiǫ_µναβdX^µ∧· · ·∧X^β. These covectors acquire a geometrical interpretation only if we introduce a metric (say a Lorentz metric η_αβ): In this case the magnitude η^µνV_µ(i)V_ν(i)measures the metrical volume ofT_iwhile the vectorV^µ(i)∝n_µ(i)points into the direction perpendicular to T_i.