Asymptotic Analysis of the EPRL Model with Timelike Tetrahedra

Asymptotic analysis of the EPRL model with timelike tetrahedra

Wojciech Kamiński∗1, Marcin Kisielowski†1,2, and Hanno Sahlmann‡2

1Instytut Fizyki Teoretycznej, Wydział Fizyki, Uniwersytet Warszawski, ul. Pasteura 5 PL-02093 Warszawa, Poland 2Institute for Quantum Gravity, Department of Physics, Friedrich-Alexander Universit¨at Erlangen-N¨urnberg (FAU), Staudtstr. 7 D-91058 Erlangen, Germany

June 12, 2019

We study the asymptotic behaviour of the vertex amplitude for the EPRL spin foam model extended to include timelike tetrahedra. We analyze both, tetrahedra of signature −−− (standard EPRL), as well as of signature +−− (Hnybida-Conrady extension), in a uniﬁed fashion. However, we assume all faces to be of signature −−. We ﬁnd that the critical points of the extended model are described again by 4-simplices and the phase of the amplitude is equal to the Regge action. Interestingly, in addition to the Lorentzian and Euclidean sectors there appear also split signature 4-simplices.

1. Introduction

Spin foam models have many origins. On the one hand they appear in the calculation of transition amplitudes in the (topological) quantum theory of 3d gravity (the Turaev Viro model [1] and, less formal, the Ponzano-Regge model [2, 3]). On the other hand spin foams can be regarded as Feynman diagrams of some auxiliary theory [4, 5, 6, 7, 8, 9, 10, 11], which can be chosen such that these diagrams correspond to triangulations of (pseudo-) manifolds.

arXiv:1705.02862v2 [gr-qc] 11 Jun 2019 The spin foam sum then becomes a sum over geometries, and the underlying theory, if deﬁned independently, gives a well deﬁned resummation. Prime examples for this viewpoint are matrix models. Both interpretations taken

∗[email protected] †[email protected] ‡[email protected]

1 together suggest that spin foam models are relevant also for quantum gravity also in dimension four, although it is not a topological theory anymore. Most of the modern 4d spin foam models are implementations of the path integral of (topological) BF theory, combined with the imposition of additional constraints on the B field [12, 13, 14, 15, 16, 17, 18, 19]. The resulting amplitudes can be interpreted as the transition amplitude between spin network states of loop quantum gravity (LQG) [20, 21]. For a careful analysis of the prospects and problems with linking spin foams and LQG see [22, 23]. States in LQG represent purely spatial geometry. Thus for the LQG interpretation of the spin foam amplitudes, at least the triangulations of the boundary must consist of spatial tetrahedra. In fact, in the standard EPRL- FK model [16, 17, 20], all tetrahedra are spacelike. On the other hand, the BF theory with group SU(1, 1) ≈ SL(2, R) (the Ponzano-Regge model in 3d in Lorentzian signature) contains sums over all (continuous and discrete series) representations. According to the coadjoint orbit method [24, 25] it should correspond to appearance of both timelike and spacelike edges and thus also mixed signature triangles (this is 3d analog of timelike tetrahedra). In fact it has been confirmed for specific cases [26, 27]. Moreover, the absence of timelike tetrahedra in LQG is in itself puzzling and attracted attention [28].1 Thus it is important to note that there is an extension of the standard EPRL model which incorporates tetrahedra with spacelike and with timelike normals [32]. In the present paper we will work with this extended EPRL model. The are now many spin foam models, but as there is no unique way of constructing them, it is not clear if any of them is a good candidate for quantum gravity. One important test is an asymptotic analysis of the models. The methodology for such analysis is now quite established (see for example [33]) based on many developments [34, 35, 36, 37, 38, 39] etc. The method of choice for 4d models, pioneered by [36], is an extended stationary phase approximation with Livine-Speziale coherent states as boundary states [40]. In this approach the amplitude is written as an integral with oscillatory integrant eS (see (2)) with S called an action. The key role of is played by critical points of S, i.e. stationary points for which the real part of the action vanishes. Each such point contributes to the asymptotic formula with an oscillatory term and if the action does not have any critical points, the amplitude is exponentially suppressed. The asymptotic behaviour of the vertex amplitudes is expected to be governed by the Regge action of a suitably constructed 4-simplex ∆ built out of the boundary data.2 Indeed, in the standard EPRL model, the asymptotic limit of the evaluation (see

1In fact, there is a way to reconcile LQG with BF theory by projective spin network technique [29] as well as canonical analysis that allow timelike signature [30, 31]. 2We note that the analysis of the asymptotic behaviour of the vertex amplitude is only the ﬁrst step in the process, because the proper behaviour of the vertex amplitude does not necessarily imply “right” asymptotic behaviour of the spin foam sum [41] (see however [42]).

2 [35, 36] and for an overview [43, 19]) is governed by the Regge action for discrete gravity without cosmological constant [44] X S∆ = Aijθij (1) i

3 Figure 1: Complete graph on ﬁve nodes 0,..., 4.

We expect (see 1.2) that in the case of timelike faces some nongeometric contributions will occur in the asymptotic behaviour. The other restriction is the exclusion of null tetrahedra and null faces. It is not known how to construct the vertex amplitude in this case. However, as we explain in section 1.3, we expect that one can still see contributions from those configurations in the EPRL model and it seems that they are responsible for problems in application of the stationary phase method. Our work is a first step towards obtaining asymptotic behaviour of the amplitude. In section 1.1 we will briefly review the setting and state our main result, Theorem 1. We describe remaining issues in the section 1.3. Since our result is not easy to state concisely, and since the proof is quite long, we will give a more thorough introduction in sections 1.1 and 1.2 and the outline of the proof in section 1.4.

1.1. Main results In the standard EPRL-FK model [16, 17, 20] the vertex amplitude assigns a complex number to each SU(2) spin network s = (Γ, , ι). The asymptotic limit has been studied in the case when Γ is a complete graph on ﬁve nodes 0,..., 4 (the unique graph connecting each pair of nodes with precisely one link) [35]. This means that spin networks with nodes labelled by the Livine- Speziale coherent intertwiners [38] were considered and the asymptotic limit when the spins  (all diﬀerent than 0) are uniformly rescaled by a large parameter Λ was studied. The technical tool used in the calculation is an extended stationary phase method [50]. The amplitude can be written in the

4 following form (see Section 2.7):   Z Z Y Y S(gi,zij ) A(s) = c dgiδ(g5)  Mij e , (2) 5 1 1 10 SL(2,C) i (CP ×CP ) i

1 1 10 where Mij is a measure on (CP × CP ) parametrized by spinors zij, zji ∈ 2 C \{0}, which is invariant under the scaling ij 7→ Λij, c is a factor that 20 scales approximately as c 7→ cΛ for large Λ and S(gi, zij) scales as S(gi, zij) 7→ ΛS(gi, zij). The number of critical points depends on the (coherent) spin- network s. Let us notice that inside the definition of a coherent state some arbitrary phase is hidden, so the total result will be influenced by an arbitrary phase that scales uniformly with the scaling of the labels. In the analysis of [35, 51] this phase was fixed. However the phase choice was done in a global way by considering the whole graph – not locally by choosing the phases separately for each node. This convention is useful only in considering vertices separately as the phases would be chosen differently for the same nodes glued to two different vertices (see however [41, 52, 42, 53])3. The choice of the phase also leads to additional phase terms in the action of [35] (π from thin wedges see Section 6.2 of [35]) that can be removed by a more judicious choice of overall coherent state phase. For these reasons we prefer to keep the phase unspecified. This leads to an additional arbitrary overall phase in the asymptotic limit. As a result we do not study the phase of each oscillatory term but a difference of phases between different terms. In the original formulation of the EPRL model only tetrahedra that are spacelike were considered. As a result s is an SU(2) spin network. Our goal is to study the asymptotics of the EPRL-FK vertex amplitude generalized to include tetrahedra that are timelike [32, 54]. We limit to the subcase where the faces are of signature −−. As a result [32] we will restrict to discrete series of the SU(1, 1) representations that are labelled with spins 4. We will consider a generalization of SU(2) spin networks, where the links are labelled with spins 5 but the nodes are labelled either with SU(2) intertwiners or SU(1, 1) intertwiners. In this paper we determine the critical points of the action S for the generalized vertex amplitude written in the form (2). This is a crucial step for determination of the asymptotic behaviour of the vertex amplitude. 4 To each coherent state we will associate vectors vij ∈ M in Minkowski can spacetime (see section 5.2.1) perpendicular to the standard normal Ni 3The sum of phases for 4-simplices glued together has a physical meaning. 4In the other subcase of faces of signature +− (see [32] for explanation of semiclassical origin of the notion) continuous series of SU(1,1) representations are used. We expect that that there may be some non-geometric critical points in the +− case and therefore we leave it for further research (see 1.2 for a comment). 5 In fact by representations (, ρ) of SL(2, C) satisfying simplicity constraint, that is in our case ρ = 2γ.

5 Figure 2: We consider a 4-simplex dual to the graph from ﬁgure 1.1. The nodes of the graph label tetrahedra of the simplex, links (sets of two nodes) label faces, edges between faces are labeled by sets of three nodes.

T T (e0 = (1, 0, 0, 0) or e3 = (0, 0, 0, 1) depending on the type of embedding) can such that bivectors ∗(vij ∧ Ni ) are spacelike. We will call them the boundary data. Following [35] we will say that (see 5.2.1) P • they satisfy closure condition if j vij = 0,

• they are non-degenerate if for every node i every 3 out of 4 vectors vij are independent. With such vectors for each i we can build, by the Minkowski theorem, [55, 56, 57] (for simplices in arbitrary signature, see section 6.3) a tetrahedron in can⊥ Ni , the faces of which have signature −−. The type of normal determines type of tetrahedra. For timelike normal e0 we have spacelike tetrahedra, for spacelike normal e3, timelike (in fact mixed signature) tetrahedra. We can determine lengths of the edges of each tetrahedron i. Let us consider a topological 4-simplex, the dual to the graph. The nodes of the graph label tetrahedra of the simplex, links (sets of two nodes) label faces, edges between faces are labeled by sets of three nodes (see figure 1.1. These nodes correspond to three tetrahedra sharing the edge. With every edge we can associate a length coming from application of Minkowski theorem to the boundary data in any node in the set. We will say that the lengths matching condition (definition 9) is satisfied if these lengths determined from all three tetrahedra are the same. If the lengths matching condition is satisfied then we can look for the Gram matrix constructed from such lengths (see section 6.7). We can reconstruct

6 Closure condition? No Yes Boundary data No critical points non-degenerate? No Yes Not addressed Length matching? in Theorem 1 Yes No At most 1 Orientation matching? critical point Yes No Reconstructed 4-simplex No critical non-degenerate? points Yes No 2 critical points 1 critical point

Figure 3: Classiﬁcation of the critical points of the action S according to Theorem 1.

a unique (up to reflections, rotations and shifts) 4 simplex from the Gram matrix. There are 5 cases for the signature of these simplices • Lorentzian (+ − −−), • Euclidean (− − −−), • split signature (+ + −−), • degenerate (+ − −), • degenerate (− − −). By the Minkowski theorem (for simplices in arbitrary signature, see section 6.3) we can also determine orientations of the reconstructed tetrahedra. If they match the orientations of tetrahedra of one of the reconstructed 4- simplices, we will say that orientations matching condition is satisfied (see Section 6.7 definition 10 for precise statement). Our main result is summarized by the following theorem (see also the diagram on figure 1):

Theorem 1. If the boundary data vij does not satisfy the closure condition, the action has no critical point. Let us assume that the boundary data satisﬁes the closure condition and is non-degenerate. If the lengths matching condition is not satisﬁed then there exists at most one critical point of the action S with the interpretation of a vector geometry [35]. The vector geometry can occur only if all normals are of the same type. If the lengths matching condition

7 is satisfied but orientations matching condition is not, then the action has no critical points. If both conditions are satisfied, then let us consider the reconstructed 4-simplex ∆ for the boundary data: • If the reconstructed 4-simplex ∆ is Lorentzian then there exist two critical points (gi, zij) and (˜gi, z˜ij). The difference

∆S = S(gi, zij) − S(˜gi, z˜ij)

is given by the Regge action S∆ without cosmological constant (1) for the ﬂat 4-simplex ∆

∆S = 2 i rS∆ mod 2π i, where r = ±1 is the Plebański orientation (definition 8). • If the reconstructed 4-simplex ∆ is Euclidean or of split signature then there exist two critical points (gi, zij) and (˜gi, z˜ij). The difference ∆S is given by the Regge action: 2 i r ∆S = S mod 2π i, γ ∆ where γ is the Barbero-Immirzi parameter. • If the reconstructed 4 simplex is degenerate then there exists a single critical point. Cases of signature − − −− (and + + −− respectively) can occur only if all can Ni are e0 (and e3 respectively). Degenerate cases can occur only if all normals are of the same type (either e0 or e3). can The case when all Ni are equal to e0 was proven before (standard EPRL [35, 58]) but the other cases are our new result. As one can see, the case with timelike tetrahedra is similar to the previously obtained case with spacelike tetrahedra. The main difference is that in addition to Euclidean and Lorentzian signature also split-signature case is present. Let us notice at the end that in our convention faces of the tetrahedra have 1 areas Aij = 2 ρij instead of ij in the convention of [35]. This is just a total rescaling of the action by the Barbero-Immirzi parameter γ. Our convention is compatible with LQG area operator spectrum [59, 60]6.

1.2. Type of faces As we mentioned earlier, our analysis is restricted to the case when all faces are of signature (−−) that is when the diagonal simplicity conditions of EPRL are [32]: ρ = 2γ. (3)

6 1 1 In LQG the area of the face would be Aij = 8πG ρij that would lead to the action S∆. We ~ 2 8πG~ are using units 8πG~ = 1.

8 In the case of signature (−+) the simplicity constraint is:

2 = γρ. (4)

This case is more complicated. Firstly, no EPRL like construction is known (see [48] for recent developments). Secondly, the coherent state proposition [32] in the line of Freidel-Krasnov model diﬀers from representation theoretic construction in the style of EPRL. We expect that the asymptotic analysis in this case may lead to some unphysical sectors due to non-extremality of the necessary embeddings7. We will give heuristic explanation of this fact. With the use of normal N we can identify bivector B in Minkowski spacetime M 4 with two vectors L,~ K~ perpendicular to N by formula

B = ∗(N ∧ L~ ) + N ∧ K.~ (5)

Regarding bivectors as generators of SO(1, 3) group we can write classical version of EPRL constraints imposed on the embedding. After some rear- rangements we obtain:

L~ · (L~ + γK~ ) = 0, (L~ + γK~ )2 = 0. (6)

These are quadratic constraints whereas simplicity would be:

L~ + γK~ = 0. (7)

Let us now describe case by case if from (6) follows (7). • In the standard EPRL case: N is timelike so L~ and K~ are spacelike so from (6) simplicity follows trivially as the only spacelike vector with norm zero is the zero vector. • In the extended EPRL case with spacelike face: N is spacelike but L~ is timelike. From (6) we know that L~ + γK~ is null. But there is no nontrivial null vector that is perpendicular to a nontrivial timelike vector, so L~ + γK~ = 0. • In the extended EPRL case with timelike face: N is spacelike and L~ is spacelike. In this situation it might happen that L~ +γK~ 6= 0 (simplicity fails). As the simplicity (7) is crucial for the reconstruction of 4 simplex we suspect that there might be some non-geometric conﬁgurations for timelike faces.

7This problem is probably absent in Barrett-Crane model [12].

9 1.3. Asymptotic analysis (conjecture) The vertex amplitude is given by an oscillatory integral over an infinite domain. We would like to apply version of stationary phase analysis as pre- sented for example in [50]. However in order to give a proper asymptotic result several technical conditions need to be satisfied. 1. finiteness. It is not known whether the evaluation of a spin network in the extended EPRL model is, in fact, finite. It was proven, but only for the standard EPRL setting (with all tetrahedra spacelike), that the integral defining the amplitude is absolutely convergent [61, 62]. The condition (well-definiteness of the definition) is necessary for any statement about the amplitude to make sense. 2. lack of boundary contributions. As the integral is over infinite domain with boundaries (or equivalently using compactification over domain with boundary) there might be some additional contributions to asymptotic expansion given by a version of Watson lemma. This is true as well in standard EPRL case (as the integration is over infinite domain), however in the extended set-up boundary is much more complicated. Preliminary check in the simpler Barrett-Crane model suggests that such contributions in fact appear but only for special boundary configurations that correspond to null (degenerate tetrahedra)8. In extended EPRL set-up one has additionally finite boundary of integrations. The integral at this boundary decays fast unless the spin of suitable representation is zero (degenerate or null face). This suggests that boundary contributions are related to null structures and we make the following conjecture: For nondegenerate boundary data (that excludes null faces and tetrahedra) there are no contributions from the boundary of integration. This issue was not addressed in any asymptotic analysis of the Lorentzian spin foam models so far. It is not a problem for the Euclidean models, as there is no boundary of the integration. 3. critical point non-degeneracy condition. An additional assumption is that, after suitably taking into account the symmetry of the action, the remaining matrix of second derivatives (Hessian) at the critical point is non-degenerate. This condition was never addressed in full generality in the analysis of the vertex amplitude. It was only checked for specific configurations in the standard EPRL model [35]. It is known that for the Barrett-Crane [12] spin foam models there are special configurations for which Hessian is degenerate [63]. However, this is probably related to the fact that the proper variables used in semiclassical description of this model are areas and conjugated angles [64, 65, 66] and there is some singularity in going from these variables to shapes. We conjecture

8In Barrett-Crane model only areas are speciﬁed but boundary contribution corresponds to a geometry of 4 simplex with null tetrahedra

10 that the Hessian is non-degenerate for the EPRL spin foam model for non-degenerate boundary data if the reconstructed 4 simplex is non- degenerate. The Hessian determinant is important for the asymptotic analysis as it determines asymptotic measure factor in the spin foam path integral. However, it is notoriously difficult to compute in terms of geometric quantities. The only known result for generic case is for the Barrett-Crane model [63]. If non-degeneracy fails, the scaling behaviour is different. These issues definitely deserve a separate treatment and we leave them for future research. Under these conditions our results would give a proper asymptotic expansion of the extended EPRL vertex amplitude (see [50] and for the form of asymptotic vertex amplitude [35]).

1.4. Outline of the proof In this paper we decided to give basically complete and detailed proof of the result. There are several reasons for that. First of all many of the subtle details of the classification of solutions are scattered in the literature and it is not easy to determined state of the affair. It is hard to find proof (extending to general case i.e. not using special properties of SU(2)) that there are at most two critical points or that there are no critical points if orientations matching condition is not satisfied. Moreover, the extended EPRL model has more complicated geometry than the standard one. For example relation between two different critical points is more complicated than in standard case (they are not merely related by conjugation). Because of that our paper is painfully long so for convenience of the reader we provide here an outline of the proof. We decided to work directly with the integral of EPRL Y -maps contracted using invariant bilinear form β. It is in 1 1 10 contrast to approach of [35] where β integrals over (CP ×CP ) are perform analytically and a critical point is described by SL(2, C) group elements gi. The reason is that in extended case there are many subcases and the result would lead to higher complexity of the problem. This however means that there are more variables and our critical points are characterized by SL(2, C) group elements gi and and in addition spinors zij. First task is to write all the cases in the unified manner allowing to treat them at once. This is done in Section 2 and Section 3. After achieving this we concentrate on describing critical point conditions. In order to do this we need to derive certain properties of the action Section 4 (detailed exposition in appendix D). Critical points are written in terms of group and spinor variables. We can then use traceless matrices instead of spinors (proved in Section 4) obtaining what we call SL(2, C) solution. Up to additional information of spin lifts this are equivalent to geometric SO(1, 3) solution introduced in Section 5. The latter consists of a bunch of SO(1, 3) group

11 elements that transform given boundary face bivectors such that they fit together. As we know from standard EPRL asymptotic analysis two degenerate critical points can correspond to one geometric solution but in different signature. In order to incorporate this we develop theory of geometric solutions in arbitrary signature in Section 6. We show relation of the 4 simplex to geometric solutions (classifications in terms of reconstructed 4 simplices) in general. We have now direct correspondence between critical points and 4 simplices. In Section 7 we show how to construct 4 simplex of different signature from two degenerate critical points (vector geometries) in SO(1, 3) signature. In order to finish the proof we need classification that there are only two nondegenerate SO(1, 3), two vector geometries or single critical point. This is done in Section 8 together with the proof that if lengths matching condition is satisfied, but not the orientations matching condition then there is no critical point. Parallelly to geometric description we compute difference of the phases between two critical points. As in the case of critical point description we lift the computation from original spinor description into the SO(1, 3) geometric description (or other signature computation). The problem with this approach is that in this way we loose some finite ambiguity of multiple of π that need to be regained by the topological deformation argument. Most of the phase computation is done in Section 9.

Contents

1. Introduction1 1.1. Main results ...... 4 1.2. Type of faces ...... 8 1.3. Asymptotic analysis (conjecture) ...... 10 1.4. Outline of the proof ...... 11

2. Vertex amplitude in extended EPRL model 14 2.1. Notation ...... 15 2.2. Representations of SL(2, C) ...... 16 2.3. Subgroups preserving the normal N ...... 17 2.4. The Y map ...... 18 2.5. Coherent states ...... 18 2.6. Bilinear invariant form ...... 19 2.7. The vertex amplitude ...... 19

3. Stationary phase approximation 20 3.1. Scaled amplitude ...... 20 3.2. Action ...... 22 3.3. Gauge symmetries of the action ...... 22

12 3.4. Critical points ...... 23 3.5. Reality condition and holomorphic derivatives ...... 24 3.6. Holomorphic critical point conditions ...... 25

4. Traceless matrices, spinors and bivectors 25 4.1. Traceless matrices ...... 26

4.2. Variations δgi Sij and reality conditions ...... 26 4.3. Variations with respect to zij ...... 27 4.4. Critical point conditions and boundary data ...... 28 4.4.1. SL(2, C) solutions ...... 29 4.5. Determination of the phase ...... 31 4.5.1. Diﬀerence of the phase amplitude between two critical points ...... 31

5. The geometric solutions 33 5.1. Spin structures ...... 33 5.2. Bivectors and traceless matrices ...... 33 5.2.1. Characterization of the bivectors ...... 34 5.3. Geometric solutions ...... 34

6. Geometric reconstruction 36 6.1. Notation ...... 36 6.2. Geometric solution ...... 37 6.3. Geometric bivectors ...... 37 6.4. Reconstruction of normals from the bivectors ...... 38 6.5. Reconstruction of bivectors from the knowledge of ±Ni ... 38 6.6. Nondegenerate bivectors and 4-simplex ...... 40 6.7. Uniqueness of Gram matrix and reconstruction ...... 41 ∆ 6.8. Geometric rotations Gi ...... 44 ∆ 6.9. Relation of Gi to Gi ...... 45 6.10. Classiﬁcation of geometric solutions ...... 46

7. Other signature solutions 48 7.1. Vector geometries ...... 48 7.2. Other signature solutions ...... 49 7.3. Correspondence ...... 52 7.3.1. Orientations ...... 53

8. Classiﬁcation of solutions 54 8.1. Non-degenerate simplices ...... 54 8.2. Degenerate geometric solutions ...... 55 8.2.1. Possible signatures ...... 55 8.2.2. Possible degenerate points ...... 57 8.3. Degenerate SO(1, 3) geometric solutions ...... 59

13 8.4. Classiﬁcation of solutions ...... 60

9. Phase difference 60 9.1. The Regge term ∆S∆ ...... 60 9.2. Revisiting the phase difference ...... 62 9.3. Determination of rij and φij ...... 63 9.4. Geometric difference of the phase modulo π ...... 65 9.5. Deformation argument to fix the remaining ambiguity . . . . 67 9.5.1. Lorentzian signature case ...... 68 9.5.2. The case of other signatures ...... 69 9.6. Summary ...... 71

10.Computation of the Hessian 71

11.Outlook 72

Appendices 73 A. Notation summary ...... 73 B. Conventions ...... 74 C. Restriction of representations of SL(2, C) ...... 75 C.1. The embedding map in the case of St(N can) = SU(2) . 75 C.2. The embedding map in the case of St(N can) = SU(1, 1) 76 C.3. Coherent states ...... 76 D. Traceless matrices ...... 78 D.1. Spinors u and v ...... 80 D.2. Subalgebra of the subgroup preserving the normal N . 82 E. Characterization of the bivectors ...... 83 F. Geometric bivectors ...... 84 F.1. Nondegenerate case ...... 86 G. Generalized sine law ...... 88 H. Dihedral angles ...... 89

2. Vertex amplitude in extended EPRL model

In this section we will describe extended EPRL embeddings [16, 32] and the construction of the vertex amplitude for a simplicial graph Γ that was described in the introduction. The vertex amplitude in the spin foam models [67, 68, 21] is the evaluation of a certain spin network. This spin network consists of links labelled by irreducible unitary representations of SL(2, C) and nodes labelled by invariants in the tensor product of representations from links meeting in the node. These invariants are elements of a certain distribution space. Na¨ıve evaluation would be the contraction of invariants according to the prescription given by the spin network. This prescription, while valid for a compact group, gives inﬁnity for SL(2, C) and some gauge

14 fixing [69, 61, 12] is necessary (see 2.7). The spin network we consider is defined on a graph with five nodes 0,..., 4 connecting every node to every other one with precisely one link. We parametrize links by the two nodes (ij) that they connect. The invariants at the nodes are given by group averaging the EPRL Y-maps (see section 2.4), O O Inv can H → Inv D (8) St(Nn ) EPRL(,ρ) SL(2,C) l,ρl l3n l3n Z ψ → d[g] gψ, (9) can SL(2,C)/ St(Nn )

can where the group St(Nn ) ⊂ SL(2, C) is a subgroup stabilizing the normal can Nn , l are links meeting in the node n and HEPRL(,ρ) is certain representa- can can tion of St(Nn ). Invariants of the subgroup St(Nn ) are thus together with can labels and type of the subgroup (determined by Nn ) the boundary data for the vertex amplitude. We are interested in the asymptotic analysis for large labels. In such a case one needs to specify a family of boundary states with certain semiclassical limit. Examples of such states are given by coherent states (they scale nicely with the scaling of labels) integrated over the subgroup: Z ψ = dg g (⊗l3nΨl) . (10) can St(Nn ) We will now describe in uniﬁed manner the choice of coherent states and their image under the Y-map. We will present also the form of the vertex amplitude (using bilinear forms β used to contract invariants [35]).

2.1. Notation

Let us summarize notation about spinors and SL(2, C) → SO+(1, 3) double cover. We use signature + − −−. Let us introduce σµ as follows

σ0 = I, σi Pauli matrices (11)

We have the following isomorphism from Minkowski space M 4 into Hermitian 2 × 2 matrices

4 µ µ µ M 3 N → ηN = N σµ with NµN = det ηN (12)

The symplectic form is deﬁned as

0 1 ω = , ω¯ = ω, ω−1 = ωT = −ω, (13) −1 0

15 We will also use a notation for two spinors z and w T z0 w0 [z, w] = z ωw = z0w1 − z1w0, z = , w = (14) z1 w1

Let us also introduce T σ0 µ = 0 µ σˆµ = −ωσµ ω = , ηˆN = N σˆµ Hermitian (15) −σi µ = i

The following holds

σνσˆµ + σµσˆν = 2gµνI (16) σˆνσµ +σ ˆµσν = 2gµνI (17)

Together σµ and σˆµ form γµ matrix, but as we always work in either self-dual or anti-self-dual representation we prefer such a notation. From standard commutation relation we have 1 N µM = tr η ηˆ , (18) µ 2 N M

The isomorphism π from SL(2, C) to SO+(1, 3) is deﬁned by

µ µ † π(u)(N) σµ = u (N σµ) u (19)

Lie algebra of SO+(1, 3) can be identify with bivectors (2 forms). The action of bivectors on vectors is deﬁned by contraction with the use of the metric

B(v) = Byv (20)

The identiﬁcation of so(1, 3) with sl(2, C) is then given on the basic simple bivectors by 1 so(1, 3) 3 B = a ∧ b −→ = (η ηˆ − η ηˆ ) ∈ sl(2, ) (21) B 4 a b b a C With the right choice of the orientation the Hodge ∗ operation corresponds to multiplication by i of the traceless matrix.

2.2. Representations of SL(2, C)

Unitary irreducible representations (D,ρ) from the principal series (, ρ) are functions 2 z0 C \{0} 3 z → Ψ(z) ∈ C,, z = (22) z1 satisfying the condition

Ψ(eiφ+rz) = ei(2φ+ρr)−2rΨ(z), (23)

16 with the action of SL(2, C) deﬁned by

gΨ(z) = Ψ(gT z) (24)

We are using the convention of [35], as opposed to that of [70]. The latter is equivalent to the action gΨ(z) = Ψ(g−1z) (25) These two actions can be related due to

g−1 = ω−1gT ω (26)

The scalar product for two such functions Ψ1 and Ψ2 is deﬁned as follows: Let us introduce a form Ψ1(z)Ψ2(z)Ωz (27) where i z0 Ωz := (z0dz1 − z1dz0) ∧ (z0dz1 − z1dz0) , z = (28) 2 z1

The form (27) is invariant under the scaling transformation

∗ z → λz, λ ∈ C (29) and is annihilated by the generator of this transformation, thus it descends 1 to a form on CP Z hΨ1, Ψ2i = Ψ1(z)Ψ2(z)Ωz (30) 1 CP 2.3. Subgroups preserving the normal N

The subgroup of SO+(1, 3) that preserves the normal N is the image of the subgroup of SL(2, C) that preserves ηN as follows

† St(N) = {g ∈ SL(2, C): π(g)N = N} = {g ∈ SL(2, C): gηN g = ηN } (31) this is equivalent to gT preserving Hermitian form deﬁned by

† T T hz1, z2iN = z1ηN z2 = z2 ηN z1 (32)

It follows from the fact that

T T T T hg z1, g z2iN = z2 gηN g¯ z1 = hz1, z2iN (33) | {z } =ηN for g ∈ St(N).

17 2.4. The Y map Let us review the generalized EPRL construction. As explained in the introduction, we will work only with spacelike surfaces. Let us consider a normal vector N µ NµN = t, t ∈ {−1, 1} (34) where t = 1 for timelike, t = −1 for spacelike vector. Let us consider stabilizing group St(N). In the case of t = 1, St(N) =∼ SU(2) (35)

and it is exactly SU(2) for N = e0 = (1, 0, 0, 0). For t = −1 St(N) =∼ SU(1, 1) (36)

and it is exactly SU(1, 1) for N = e3 = (0, 0, 0, 1). We will consider only standard normals e0 = (1, 0, 0, 0) and e3 = (0, 0, 0, 1). We will also denote either of them by N can as we would like to work in a uniﬁed setup. We denote tcan = N can · N can. The EPRL Y map is an embedding of the unitary representation of the stabilizing group St(N) into unitary representation (, ρ = 2γ) of SL(2, C) that satisﬁes a certain extremality condition. Let us recall the choice of HEPRL(,ρ) made by [16, 32] can 1. Spin , D representation of SU(2) embedded into D,ρ for N = e0 ± 2. Discrete series D of spin  representations of SU(1, 1) embedded into can D,ρ for N = e3

2.5. Coherent states All three families of representations have certain common features. Let us consider the generator of rotations around the z axis. We can introduce bases of eigenfunctions. In all three cases there exists an extreme eigenvalue. The corresponding eigenfunctions are (Perelomov-)coherent states [71, 49, 70] (see Appendix C)9 can • For N = e0 r 2 + 1 i ρ −1− F (z) = hz|zi 2 hn |zi2 , (37)  2π N can 0 N can can • For N = e3

18 where n0 = (1, 0) and n1 = (0, 1) and θ is the Heaviside step function. All other coherent states are obtained by transforming these basic coherent states by group action of St(N can). In fact these states can be parametrized by spinors: r 2 + 1 i ρ −1− Ψn(z) = hz|zi 2 hn|zi2 , hn|nican = 1,N can = e (40) 2π N can N can N 0

can and for N = e3

± ± where hn |n iN can = ±1.

2.6. Bilinear invariant form

Let us recall the deﬁnition of the bilinear, SL(2, C)-invariant form [72] q ρ 2 2 +  Z ρ − i ρ + 0 2 0 −2 0 − i − 0 2 β(Φ, Φ ) := Ωz ∧ Ωz0 |[z, z ]| [z, z ] 2 [z, z ] 1 1 π CP ×CP | {z } β(z,z0) · Φ(z)Φ0(z0) (43)

0 where Φ, Φ are elements of the SL(2, C)-representation (, ρ), and Ωz is de- ﬁned in (28).

2.7. The vertex amplitude We can now consider the vertex amplitude in the spin foam model. Let us consider the pentagonal graph [35] with ﬁve nodes and links connecting each node with each other. can • for every node i we choose a canonical normal Ni that determines an embedded subgroup (either SU(2) or SU(1, 1)) • for every directed link ij starting in the node i we chose a type of + − embedded representation (in case of SU(1, 1) it can be Dij or Dij and

for SU(2) it is Dij )

• for every directed link ij starting in the node i we chose a spinor nij that determines a coherent state (in representation (ij, ρij) of SL(2, C)) that we will denote by Ψij(z)

19 This data is a boundary data for the vertex amplitude of the extended EPRL Spin Foam model [35, 32]. The vertex amplitude is given by an integral Z Y Y Av = dgiδ(g5) β(giΨij, gjΨji), 5 SL(2,C) i i

3. Stationary phase approximation

In this section we will write the amplitude from section 2.7 in a form suitable for the stationary phase method. We will consider the uniform scaling of the representations (we remind that ρ and  are related by the simplicity constraint) (ij, ρij) 7→ (Λij, Λρij), Λ ∈ N+ and we want to organize the integral in the form c(Λ) R feΛS. This will provide a deﬁnition of the action. The amplitude is written as integral of the product of many terms and we will focus on each term separately. The total action will be the sum of these partial contributions. We will write conditions for the critical points of this action in terms of holomorphic derivatives, since that is the form that we will use later. We will assume that (ij, ρij) 6= 0. Again, we expect that representations (ij, ρij) = (0, 0) are of special interest. They should correspond to null surfaces.

3.1. Scaled amplitude can We denote the normal stabilized by the embedded group by Ni , can can can can N · N = t , hn , n i can = s , s , t ∈ {−1, 1} (44) i i i ij ij Ni ij ij i The coherent states decompose under the scaling in the following way,

Λ Λ Ψij(zij) = cij(Λ)mij(z)pij(zij) where mij, pij independent of Λ −1 m (z ) = θ(s hz |z i can )(s hz |z i can ) (45) ij ij ij ij ij Ni ij ij ij Ni can 2 with notice that for N = e , s hz |z i can > 0 on the whole \{0}. i 0 ij ij ij Ni C Function pij is given explicitly (uniform description) by ρ  pij = pijpij (46)

20 where

ρij aux can ρ T T i 2  pij , ti = −1 p = s hg z , g z i can , p = (47) ij ij i ij i ij Ni ij aux can pij , ti = 1

with can T !ti 2ij s hg z , n i can aux ij i ij ij Ni pij = T T (48) (s hg z , g z i can )1/2 ij i ij i ij Ni

The factors cij(Λ) are independent from z and g  q 2Λij +1 can  2π ,Ni = e0 and D representation cij(Λ) = q (49) 2Λij −1 can ±  π ,Ni = e3 and D representation Similarly, we can decompose the integral kernel of β as

Λ β β β Λ βij(zij, zji) = cij(Λ)mij(zij, zji)pij(zij, zji) r ρ 2 ij +2 β 2 ij where cij(Λ) = Λ π ,

β −2 mij(zij, zji) = Ωzij ∧ Ωzji |[zij, zji]| (50) ρ ρij β ij − i +ij − i 2 −ij 2 pij(zij, zji) = [zij, zji] [zij, zji] (51) The diﬀerential form

β T T Mij := mij(zij, zji)mij(gi zij)mji(gj zji)

1 1 descends to CP × CP as a measure (smooth in the interior of its support). Similarly10, as the integrand form is invariant under rescaling of z also the part scaled uniformly with Λ need to be invariant. The amplitude

β T T Pij := pij(zij, zji)pij(gi zij)pji(gj zji)

is invariant under rescalings zij 7→ λijzij, zji 7→ λjizji and thus projects 1 1 down to a function on CP × CP . We will write the vertex amplitude as an integral of the form:    Λ Z Y Z Y Y Av(Λ, Λρ) = c(Λ) dgiδ(g5)  Mij  Pij , 5 1 1 10 SL(2,C) i (CP ×CP ) i

where c(Λ) ∼ Λ20 for large Λ,

Y β Y c(Λ) = cij(Λ) cij(Λ) (52) i

21 3.2. Action We observe, that pβ is of the form

ρ β α − i ij − p = , α = [z , z ] 2 ij , α ij ji

β so we conclude |p | = 1. Similarly we will prove that |pij| ≤ 1 (lemma 42). Now we deﬁne X S = ln Pij (53) i

X β S(g, z) = Sij(gi, zij) + Sji(gj, zji) + Sij(zij, zji), i

β Sij(zij, zji) = − i ρij ln |[zij, zji]| − ij ln[zij, zji] + ij ln [zij, zji].

The action Sij of the amplitude depending on gi and zij is given by

Sρ S eSij = e ij e ij (55) where ρ j Sij ρ Sij  e = pij, e = pij (56) aux Sij aux Similarly we can also deﬁne auxiliary e = pij . The integration is restricted to the domain where for all i 6= j

T T s hg z , g z i can > 0 (57) ij i ij i ij Ni

3.3. Gauge symmetries of the action ∗ The following transformations labeled by g ∈ SL(2, C), λij ∈ C

0 0 T −1 gi = ggi, zij = λij(g ) zij (58) preserve the non-gauge ﬁxed action. The g-part of this transformation is gauge ﬁxed by the δ(g5) term. We can still consider variations with respect to g5 of the action, they are just not independent of the others. 1 The subgroup of gauge transformations with g = 1 will be called CP -gauge transformations.

22 3.4. Critical points According to the stationary phase method11 [50], in the large Λ regime the amplitude is dominated by contributions from the critical points (manifolds) where the following conditions are satisﬁed: the reality condition

<(S) = 0

and the stationary point condition

δRS = 0.

Here δR denotes the standard variation, to be distinguished from the holomorphic variation δ that we will introduce later. We will use variational calculus (form) notation for derivatives: We denote 1 δRS(g, z) = lim (S(g(), z()) − S(g, z)) (59) →0 where the variations are of the form:

T δRgi δRg g()i = gie , z()ij = e ij zij (60)

with δRgi, δRgij taking values in the Lie algebra of SL(2, C). Let us notice that R R T δ zij = δ gijzij . (61) We will also use variations with respect to single variables that we will denote as R R δgi S, δzij S, (62)

that are variation with single δRgi (and respectively single δRzij) nonzero. can A critical point for the given boundary data (spinors nij, vectors Ni and the types of embedded representations) consists of a collection of

{gi, zij, i 6= j} (63)

satisfying the reality condition <(S) = 0 and the stationary point conditions (δRS = 0) i.e.

R R β ∀ijδzij Sij + δzij Sij = 0 (64) X R ∀i δgi Sij = 0. (65) j

There are gauge transformations acting on the critical points.

11Justiﬁcation of the applicability is left for future research, as discussed in the introduction.

23 3.5. Reality condition and holomorphic derivatives We will consider action S as a function of holomorphic g and antiholomorphic g¯ variables12. We will prove later (see lemma 4) that

However this holds only when g¯ and g are complex conjugated (we will call this set real manifold). We will consider now complexified manifold where these group elements are independent. We denote holomorphic and antiholomorphic variations with respect to these group elements by δgS and δg¯S. Definition 1. A point satisfies the reality condition if it is in the real manifold and

~~Proof. The inequality X~~

~~is saturated only if ∀α~~

~~δgS = −δgS,¯ δg¯S = −δgS (70)~~

~~when in the second equality we took δg¯ = δg~~

~~Proof. The real variation of~~

~~R 0 = δ~~

~~that can be written as~~

~~δgS + δg¯S + δgS¯ + δg¯S¯ = 0 (72)~~

~~but we also have δgS = δg¯S,¯ δg¯S = δgS¯ (73) The equality is thus <(δgS + δgS¯) = 0, (74)~~

~~12 For our purpose of computing ﬁrst derivatives, we can assume that all variables are group elements in SL(2, C) considering gij instead of zij .~~

~~24 From arbitrariness of δg~~

~~δgS + δgS¯ = 0, δgS = −δgS¯ (75) We can also compute δg¯S = δgS¯ = −δgS (76) for δg¯ = δg.~~

~~Lemma 3. When the reality conditions are satisﬁed, then the stationary point conditions are equivalent to vanishing holomorphic derivatives. Proof. It follows from lemma 2 that~~

R δ S = δgS + δg¯S = δgS − δgS = i=δgS (77) where we took δg¯ = δg. As holomorphic variables can be multiplied by i we see that a vanishing holomorphic derivative is equivalent to a vanishing real variation.

3.6. Holomorphic critical point conditions Let us now rephrase the critical point conditions in holomorphic language: can A critical point for a given set of boundary data (spinors nij, vectors Ni and the type of embedded representation) consists of a collection of

~~{gi, zij} (78) (on the real manifold) satisfying the reality condition~~

~~∀ij~~

~~β ∀ijδzij Sij + δzij Sij = 0 (80) X ∀i δgi Sij = 0. (81) j There are gauge transformations acting on the critical points.~~

~~4. Traceless matrices, spinors and bivectors~~

In this section we will describe the connection between spinors and Lie algebra elements of SL(2, C) (traceless matrices). Using this connection, we will show that critical points can be described in terms of traceless matrices satisfying certain conditions (we will call it SL(2, C) solution). This will allow us later to translate it into geometric language of the Lorentz group and bivectors. We will also compute the (diﬀerence of the) phase between two critical points.

25 4.1. Traceless matrices We will now recall some properties of spinors that we will use to translate the critical point conditions into the language of traceless matrices. Proofs are provided for the convenience of the reader in Appendix D. Let us assume that δg is traceless then

~~ωδgT + δgω = 0 (82) for the symplectic form ω of (13), and for spinors u and v 1 [u, δgT v] = [v, δgT u] = tr(vuT + uvT )ωδgT (83) 2~~

~~T T as (vu −uv )ω = [u, v]I. Every traceless matrix with eigenvalues λ, −λ ∈ C can be written in the form~~

~~T T M = λ(vu + uv )ω, [u, v] = 1, (84) with u, v uniquely determined up to common rescaling (see lemma 39 in Appendix D). This property allows us to work purely in terms of traceless matrices.~~

4.2. Variations δgi Sij and reality conditions In order to avoid overburden the notation we will in this subsection suppress can indices ij and write Sij(gi, zij) as S(g, z). Let us consider normal N and normalized spinor n and ρ ∈ R, 2 ∈ Z

~~can can † T N · N = det ηN can = t, hn, niN can = n ηN can n = s, s, t ∈ {−1, 1} (85) the action part of the amplitude depending on g and z~~

~~ρ  eS = eS eS (86) where (the integration is restricted to the domain where~~

~~T T shg z, g ziN can > 0 (87) and~~

~~ρ ρ S T T i 2 e = shg z, g ziN can (88) Saux  e t = −1 eS = (89) eSaux t = 1 where T t 2 aux shg z, ni can eS = N (90) T T 1/2 (shg z, g ziN can )~~

~~26 In fact although there are many diﬀerent cases one can prove (see appendix D) that on real manifold~~

~~T ∃ξ∈C g z = ξn (92) Lemma 4. On real manifold~~

~~T δgi Sij = tr Bijδgi (95) where 1 ρ = i ij +  (u vT + v uT )ω (96) Bij 2 2 ij ij ij ij ij with spinors~~

~~u = s ωη can n , v = n (97) ij ij Ni ij ij ij and s = hn , n i can ij ij ij Ni~~

~~4.3. Variations with respect to zij The edge action can be divided into pieces~~

~~β Sij(gi, zij) + Sji(gj, zji) + Sij(zij, zji) (98)~~

~~We parametrize δzij by δgij traceless as follows~~

T δzij = δgij zij (99) all variations can be written this way but δgij is not unique. −1 Lemma 5. δzij Sij = δgi Sij, for δgi = gi δgijgi T Proof. Function Sij depends only on gi zij thus is invariant under variations T such that δ(gi zij) = 0. Examples of such variations are T T T T T −1 δzij = δgij zij and δgi = −gi δgij(gi ) (100)

~~In this case δzij Sij + δgi Sij = 0, so taking variation as in the thesis we get the result.~~

~~27 β β We will from now on abuse notation and regard Sji = Sij for i < j.~~

~~β Lemma 6. Let us introduce a traceless matrix Bij such that~~

~~β β T δzij Sij = tr Bijδgij (101) then 1 ρ β = i ij +  (uvT + vuT )ω (102) Bij 2 2 ij where 1 u = zij, v = zji, [u, v] = 1 (103) [zij, zji]~~

~~Spinors zij and zji are determined up to complex scaling by 1 ρ β z = i ij +  z (104) Bij ji 2 2 ij ji 1 ρ β z = − i ij +  z (105) Bij ij 2 2 ij ij~~

~~β β Moreover Bij = −Bji. Proof. We have~~

~~T T β ρij [δgijzij, zji] ρij [zij, δgijzji] δzij Sij = − i + ij = i + ij (106) 2 [zij, zji] 2 [zij, zji] and it can be written as~~

~~ ρij 1 T T i + ij tr zji zijωδgij (107) 2 [zij, zji]~~

~~Together with (418) it can be transformed into form from the lemma. Char- acterization of zij and zji follows from lemma 38.~~

4.4. Critical point conditions and boundary data can The boundary data (spinors nij, normal vectors Ni and types of embedded representations) can be summarized by can • Ni ∈ {e0, e3} normal vectors, can can can • ti = Ni · Ni ∈ {−1, 1},

~~• nij spinors,~~

~~• s = hn , n i can ∈ {−1, 1}, ij ij ij Ni A critical point for a given boundary data consists of a collection (real manifold) gi, zij (108)~~

~~28 satisfying (80)~~

~~β ∀ijδzij Sij + δzij Sij = 0 (109) X ∀i δgi Sij = 0 (110) j and reality conditions T gi zij = ξijnij, ξij ∈ C (111) Condition (109) is equivalent to (i 6= j)~~

β T −1 δgi Sij = −δzij Sij, δzij = δgijzij, δgi = gi δgijgi (112) There are gauge transformations acting on the critical points and they are ∗ parametrized by g ∈ SL(2, C) and λij ∈ C 0 0 T −1 gi = ggi, zij = λij(g ) zij (113)

~~Gauge ﬁxing condition g5 = 1 ﬁxes SL(2, C) gauge transformations.~~

~~4.4.1. SL(2, C) solutions We can now translate critical point conditions into language of traceless matrices~~

~~Deﬁnition 2. SL(2, C) solution for the boundary data consists of~~

~~{gi ∈ SL(2, C)} (114) β determining Bij, Bij traceless matrices by conditions 1 ρij T T 1. Bij = 2 i 2 + ij (uijvij + vijuij)ω where~~

~~u = s ωη can n¯ , v = n , s = hn , n i can ∈ {−1, 1} (115) ij ij Ni ij ij ij ij ij ij Ni~~

β T −1 T 2. Bij = −(gi ) Bijgi β β 3. Bij = −Bji P 4. j Bij = 0 The gauge transformations are parametrized by g ∈ SL(2, C) β T −1 β T gi = ggi, Bij = (g ) Bijg (116) Lemma 7. SL(2, C) solutions are in bijective correspondence with critical 1 points up to CP gauge transformations. An SL(2, C) solution is determined by group elements of the critical points and then T β T β tr Bijδgi = δgi Sij, tr Bijδgij = δgij Sij, (117) The SL(2, C) gauge transformations are acting the same way on both sides of the correspondence.

29 Proof. Critical point determines SL(2, C) geometric solution determining β traceless matrices from δzij Sij and δgi Sij. They satisﬁes all assumptions of SL(2, C) solution. In the opposite direction we need to ﬁnd zij such that traceless matrices β Bij and Bij are such that

~~T β T β tr Bijδgi = δgi Sij, tr Bijδgij = δgij Sij, (118)~~

This can be determined using lemma 38 and 6. In fact zij and zji need to β β be eigenvectors of Bij (or equivalently Bji) 1 ρ β z = i ij +  z (119) Bij ji 2 2 ij ji 1 ρ β z = − i ij +  z (120) Bij ij 2 2 ij ij

β β T −1 T Matrix Bij has exactly such eigenvalues because Bij = −(gi ) Bijgi and Bij 1 has such eigenvalues. This determines zij up to CP gauge transformations. β T −1 T From equality Bij = −(gi ) Bijgi we get

1 ρij 1 T T i + ij (zijzji + zjizij)ω = 2 2 [zij, zji] 1 ρij T −1 T T T − i + ij (gi ) (uijvij + vijuij) ωgi (121) 2 2 |{z} −1 =gi ω thus 1 T 1 T T T zij zji + zjizij ω = −(u˜ijv˜ij + v˜iju˜ij)ω (122) [zij, zji] [zij, zji] where T −1 T −1 v˜ij = (gi ) vij, u˜ij = (gi ) uij, (123) By lemma 39

~~T −1 T −1 zij = ξij(gi ) vij = ξij(gi ) nij, ξij ∈ C (124) and this is just reality condition. We see now that with our choice of zij~~

~~β T β T tr Bijδgij = δzij Sij, δzij = δgij zij (125) and from reality condition (by lemmas 4)~~

T tr Bijδgi = δgi Sij (126) The remaining conditions of critical point can be written in terms of matrices β Bij and Bij and are the same as conditions for SL(2, C) solution. 1 The zij are determined up to CP gauge transformations.

~~30 4.5. Determination of the phase Let us compute the value of the edge part of the action in the critical point We can choose a gauge for zij such that~~

~~T ∀ijgi zij = nij (127)~~

~~In such a case Sij = 0 (128) The only contribution to the action comes from the β amplitude~~

~~ ρ ρ β ij − i ij − S − i 2 +ij 2 ij e ij = [zij, zji] [zij, zji] (129)~~

~~From equality of traceless matrices from lemma 39 and (122) (taking into account gauge ﬁxing)~~

~~T −1 T −1 zij = (gi ) vij and zji = [zij, zji](gi ) uij (130) but similarly~~

~~T −1 T −1 zji = (gj ) vji and zij = [zji, zij](gj ) uji (131)~~

~~This gives equalities (where we introduced Λij = [zij, zji])~~

T −1 T −1 T T −1 (gj ) vji = Λij(gi ) uij =⇒ vji = Λij(gj )(gi ) uij (132) T −1 T −1 −1 T T −1 (gi ) vij = −Λij(gj ) uji =⇒ uji = −Λij (gj )(gi ) vij (133) T −1 T −1 T T −1 Λij = [(gi ) vij, (gj ) vji] = [(gj )(gi ) vij, vji] (134)

~~The phase amplitude is then~~

~~e−i(ρij ln |Λij |+2ij Arg Λij ) (135)~~

4.5.1. Difference of the phase amplitude between two critical points Changing phases of coherent states changes also the total phase. However difference of the phases of two different critical points is invariant under such transformation. Our goal is to determine this phase for two critical points

~~(gi, zij) and (˜gi, z˜ij) (136)~~

~~As uij and vij are determined by boundary data they are the same in both critical points thus~~

~~T T −1 −1 T T −1 vji = Λij(gj )(gi ) uij, uji = −Λij (gj )(gi ) vij (137) ˜ T T −1 ˜ −1 T T −1 vji = Λij(˜gj )(˜gi ) uij, uji = −Λij (˜gj )(˜gi ) vij (138)~~

~~31 ˜ −1 So we have (δij = ΛijΛij )~~

~~T T −1 T T −1−1 (˜gj )(˜gi ) (gj )(gi ) vji = δijvji (139) T T −1 T T −1−1 −1 (˜gj )(˜gi ) (gj )(gi ) uji = δij uji (140)~~

~~T −1 T −1 Using zji = (gj ) vji and zij ≈ (gj ) uji we can write~~

~~∆ij zji = δijzji (141) −1 ∆ij zij = δij zij (142) where we introduce~~

~~T −1 T T −1 T ∆ij = (gj ) (˜gj )(˜gi ) (gi ) (143)~~

~~We also have S˜β −Sβ e ij ij = ei(ρij ln |δij |+2ij Arg δij ) (144) We have for [u, v] = 1~~

~~T T T T eln |δ|(−i)i(uv +vu )ω+Arg δi(uv +vu )ωu = δ−1u (145) T T T T eln |δ|(−i)i(uv +vu )ω+Arg δi(uv +vu )ωv = δv (146)~~

~~Using fact that~~

~~2γ β ρij 1 T T Bij = i (zijzji + zjizij)ω (147) γ − i 2 [zij, zji] and the properties of ∆ij we see that~~

~~2 ˜ 2 ˜ Arg δij ρ Mij −i ln |δij | ρ Mij ∆ij = e ij ij (148)~~

˜ 2γ β where we introduced Mij = γ−i Bij. Lemma 8. The phase diﬀerence between critical points deﬁned by two SL(2, C) solutions {gi} and {g˜i} is equal X ∆S = S˜ − S = i ρijrij + 2ijφij mod 2πi (149) i

~~−1 −1 φij Mij −irij Mij gig˜i g˜jgj = e (150) where = 2 2γ ( β )T . Mij ρij γ−i Bij Proof. It is enough to transpose (148) and (143).~~

~~32 5. The geometric solutions~~

We will translate our SL(2, C) description into SO(1, 3) language. We will also extract necessary geometric boundary data from the boundary data given by spinors nij. This section is devoted to the relation between the two descriptions.

5.1. Spin structures Let us notice that EPRL construction makes sense only if certain integrability condition holds X ∀i ij is integer, (151) j6=i because only then the intertwiners space (deﬁned in the distributional sense) is nonempty. We assume this condition for every boundary tetrahedron. The vertex integral and the action (deﬁned up to 2πi) are then invariant under the following transformations

~~gi → sigi, si = ±1 (152)~~

~~The transformations of this kind will be called spin structure transformations.~~

5.2. Bivectors and traceless matrices 4 Generators of SO+(1, 3) are matrices in M that are antisymmetric after lowering an index. They can be identiﬁed with bivectors. Generators of SL(2, C) are traceless matrices. The isomorphism between these Lie algebras is given on simple bivectors by 1 π : so(1, 3) 3 B = a ∧ b −→ = (η ηˆ − η ηˆ ) ∈ sl(2, ). (153) B 4 a b b a C With the standard choice of the orientation, Hodge ∗ operation corresponds to multiplication by i of the traceless matrix. We also have for two bivectors B1 and B2 B1 · B2 = −2< tr B1B2. (154) The bivector is spacelike if B · B > 0 and mixed if B · B < 0 and ∗B · ∗B = −B · B. The image of traceless matrix with purely imaginary eigenvalues is spacelike. We assume that the SL(2, C) representations satisﬁes (spacelike faces conditions) ρij = 2γij. (155)

~~Then traceless matrices Bij have the property that 2γ 1 = i ρ (u vT + v uT )ω (156) γ − iBij 2 ij ij ij ij ij~~

~~33 5.2.1. Characterization of the bivectors~~

−1 2γ T We can characterize π − γ−i Bij (see lemma 46 in the Appendix) Lemma 9. We have 2γ π(B0 ) = − T (157) ij γ − iBij 0 can can where Bij = ∗(vij ∧ Ni ) = ∗(lij ∧ Ni ) and the null vector lij is given by

~~µ T µ lij = sijρijnijσ n¯ij (158)~~

~~It is future directed if sij = 1 and past directed if sij = −1. The vector vij is determined by~~

can can can lij · Ni lij = vij + cNi , vij ⊥ Ni , c = can can (159) Ni · Ni We can now characterize bivectors in terms of 3 dimensional geometry in can the space perpendicular to Ni . Deﬁnition 3. The geometric boundary data are sets of vectors

~~can vij ⊥ Ni (160)~~

~~can 2 with norm vij · vij = −ti ρij that are obtained form spinors nij as a projec- can tion onto space orthogonal to Ni of the null vectors lij deﬁned by~~

~~µ T µ lij = sijρijnijσ n¯ij (161)~~

Let us notice what follows from the previous subsection: can • For ti = 1 (sij = 1) vij is spacelike, can • For ti = −1 and sij = 1, vij is timelike future directed, can • For ti = −1 and sij = −1, vij is timelike past directed,

~~5.3. Geometric solutions The inversion I ∈ SO(1, 3) is deﬁned by~~

~~∀v Iv = −v (162)~~

~~It does not belong to SO+(1, 3). Let us introduce notion of geometric solution~~

~~Deﬁnition 4. The SO(1, 3) geometric solution is a collection~~

~~{Gi ∈ SO(1, 3)}i=0,...4 (163)~~

~~34 such that bivectors can {G} can Bij = ∗vij ∧ Ni ,Bij = Gi(∗vij ∧ Ni ) i 6= j (164)~~

with vij deﬁned by the boundary data (deﬁnition 3) satisfy {G} {G} ∀i6=jBij = −Bji , (165) X ∀i Bij = 0 (166) j6=i 0 Two geometric solutions {Gi}, {Gi} are gauge equivalent if there exists G ∈ SO(1, 3) and si ∈ {0, 1} such that

0 si ∀i Gi = GGiI (167) Gauge transformations 0 si ∀i Gi = GiI (168) are called inversion gauge transformation. Let us notice that necessary condition (166) for a critical point is in fact a condition for the boundary data. It is called closure condition [35] 13 X ∀i vij = 0 (169) j6=i We will always assume that boundary data satisﬁes the closure condition. Let us notice that from 5.2.1 we know that can Bij = ∗(vij ∧ Ni ) (170) is a spacelike bivector. Lemma 10. There exists bijection between SL(2, C) solutions up to spin structure transformations (152) for the given boundary data satisfying the closure condition and SO(1, 3) geometric solutions {Gi} up to inversion gauge transformations for the corresponding geometric boundary data. The map from SL(2, C) solutions is given by

~~Gi = π(gi) (171) and then also 2γ B = −π T (172) ij γ − iBij 2γ B{G} = π ( β )T (173) ij γ − i Bij~~

The SL(2, C) gauge transformations correspond to SO+(1, 3) gauge transformations. 13 can This is condition for non-decaying for Λ → ∞ of the invariant in the case of N = e0. Arguably it is can also condition in the case of N = e3 for non-decaying of the invariant deﬁned in the distributional sense. The issue deserve separate treatment and it will not be address in the present paper.

35 Proof. From any SL(2, C) solution we produce in this way SO(1, 3) geometric solution. The map identiﬁes only points that diﬀer by spin structure transformations. Two gauge equivalent SL(2, C) solutions are mapped into gauge equivalent geometric solutions. We need only to show that in every class up to gauge inversion transformations there exists an element that is the image of SL(2, C) solution. Let us choose an SO(1, 3) geometric solution. By inversion gauge transformation we can assume that all Gi ∈ SO+(1, 3). The preimages of such group elements constitute SL(2, C) solution. Comment: Transposition is a price one need to pay for following notation from [35].

~~6. Geometric reconstruction~~

In the previous section we determined that critical points modulo gauge transformations are in one to one correspondence with SO(1, 3) geometric solutions. In this section we will classify the latter14. Geometric SO(1, 3) solutions divide into non-degenerate and degenerate ones. With the ﬁrst class we can associate non-degenerate Lorentzian simplices. However with the pair of degenerate geometric solutions we will (in section 7) associate simplex in other than Lorentzian signature. For this reason we want to provide classiﬁcation in arbitrary signature.

6.1. Notation We will denote all operations in this arbitrary metric by underline i.e. (Hodge star ∗, scalar product · and contractions with use of the metric y and x). We can introduce reﬂections with respect to the normalized (to ±1 vector N) 2N µ N (R )µ = µ − ν ∈ O = O(p, q) (174) N ν Iν N·N 2 where we lowered index with use of the metric. Notice that RN = I. We can also introduce inversion

~~Iv = −v (175)~~

Depending on the signature inversion belongs (O(2, 2) and O(4)) or does not (O(1, 3)) to the connected component of identity. It is however always in special orthogonal subgroup in dimension 4. The reﬂections do not belong to special orthogonal subgroup thus neither to connected component of identity. We denote special orthogonal subgroup by SO and its connected component of identity by SO+.

~~14The classiﬁcation was done in [12, 69, 34], see also [36, 17], but in a bit diﬀerent set-up.~~

~~36 6.2. Geometric solution We will first define geometric version of the boundary data Definition 5. The ( SO geometric) boundary data is a collection~~

can 4 4 X can {Ni ∈ R , vij ∈ R : ∀i vij = 0, vij⊥Ni } (176) j6=i can can can can where Ni is a canonical normal, Ni ·Ni = ti ∈ {−1, 1}. We will say that the boundary data is non-degenerate if for every i, every 3 out of 4 vectors vij are linearly independent. Set of canonical normals is such that every normalized vector N, N·N = ±1 can be rotated by an element of SO to exactly one of canonical normals15. Definition 6. The geometric SO solution for the geometric boundary data is a collection {Gi ∈ SO}i=0,...4 (177) such that bivectors can {G} can Bij = ∗vij ∧ Ni ,Bij = Gi(∗vij ∧ Ni ) i 6= j (178) satisfy {G} {G} Bij = −Bji (179) 0 Two solutions {Gi}, {Gi}are gauge equivalent if there exists G ∈ SO and si ∈ {0, 1} such that 0 si ∀iGi = GI Gi. (180) {G} can We can associate with this data normals Ni = GiNi . Definition 7. Geometric solution is non-degenerate if every four out of five {G} 4 Ni span the whole R . It is degenerate if this condition is not satisfied.

6.3. Geometric bivectors For convenience of the reader we provide in Appendix F a construction of ∆ bivectors related to faces of the 4 simplex that we will denote by Bij and call geometric bivectors. The construction works also for degenerate cases and depends on the order of vertices. If the 4 simplex is nondegenerate we can introduce outer directed normal ∆ ∆ Ni (see 6.3)and certain numbers Wi related to volumes of tetrahedra (see Appendix F) such that

~~X ∆ ∆ ∆ Wi Ni = 0, −tiWi > 0. (181) i In fact,~~

15 For SO(1, 3) it can be chosen as {e0, e3}, for SO(4) as {e0}. For SO(2, 2) we can choose {e3, e1} if we assume that they have diﬀerent norms, but in the asymptotic analysis only e3 will play a role.

37 ∆ ∆ • Bij = −Bji , ∆ 1 ∆ ∆ ∆ ∆ • Bij = − V ol∆ Wi Wj ∗(Nj ∧ Ni ), ∆ ∆ • Bij yNi = 0. For a spacelike face its area is equal to 1 A∆ = |B∆|. (182) ij 2 ij and V ol∆ is volume of the simplex form (see Appendix F)16.

6.4. Reconstruction of normals from the bivectors {G} Suppose that we have bunch of bivectors Bij that come from some non- degenerate geometric solution. We can reconstruct from them Ni up to a sign. We assume that vij for the ﬁxed j span the whole space perpendicular can to Ni (for example the boundary data is non-degenerate).

Lemma 11. Let us assume that we have a geometric solution {Gi} for the non-degenerate geometric boundary data. Then the following are equivalent for the chosen i and vector N {G} •∀ j6=iBij yN = 0,N·N = ±1, {G} • N = ±Ni . Proof. We know that {G} {G} ∀j6=iBij yNi = 0 (183) If there is independent vector N satisfying the same equation, then {G} {G} ∃λij ,Bij = λij∗(N ∧ Ni ). (184) This contradicts non-degeneracy of the boundary data because every 3 out {G} of 4 Bij = GiBij should be independent. The vector N needs to be pro- {G} {G} portional to Ni and from normalization of Ni it follows that N = {G} ±Ni .

~~6.5. Reconstruction of bivectors from the knowledge of ±Ni~~

We will now reconstruct bivectors from normals Ni. In fact out theorem works in any dimension in arbitrary non-degenerate signature. n Theorem 2. Let us assume that in R we have normalized to ±1 vectors Ni, i = 0 . . . n, such that (nondegeneracy) any n out of n + 1 vectors Ni span n the whole R . Then there exist n − 2 vectors 0 Bij i 6= j, i = 0, . . . , n (185) such that 16It diﬀers by 4! from the volume of the simplex.

~~38 • for every i 0 X 0 BijyNi = 0, Bij = 0, (186) j6=i • for all i 6= j, 0 0 Bij = −Bji. (187) A solution to equations (186) and (187) is given by:~~

~~0 Bij = WiWj ∗Nj ∧ Ni, (188)~~

~~where constants Wi ∈ R are nonzero solutions of X WiNi = 0. (189) i~~

~~00 For any other solution Bij there exists a constant λ ∈ R such that~~

~~00 0 Bij = λBij (190)~~

The solution is independent of changing some of Ni by sign. Proof. Let us ﬁrst prove uniqueness of such solution up to scaling. Let us 0 assume such Bij are given. There is exactly one solution (up to a scaling by real constant) of the equation X WiNi = 0,Wi ∈ R (191) i as there are n + 1 vectors in n dimensional space and every n out of n + 1 are independent. The constant Wi are all nonzero (in the nontrivial solution) and in fact they will turn out to be proportional to signed volumes of the 17 tetrahedra of the 4 simplex with normal Ni . 0 Let us notice that Bij are simple n − 2 bivectors because they are annihilated by two independent normals

~~0 0 0 BijyNi = 0 BijyNj = −BjiyNj = 0 (192)~~

~~so there need to exists constant λij such that~~

~~0 Bij = λij∗Nj ∧ Ni, λij 6= 0 (193)~~

~~We have for every i~~

~~X 0 X 0 = Bij = ∗( λijNj) ∧ Ni (194) j6=i j6=i~~

~~17This is called Minkowski theorem [55] see also [56, 57], but in arbitrary signature and for simplices. We do not know any reference for general Minkowski theorem in other than euclidean signature.~~

~~39 so we have X λijNj = −λiNi, (195) j6=i for some λi ∈ R. This equation has a unique up to a constant solution so~~

~~Wj λij = λi (196) Wi From the symmetry~~

~~0 0 λij∗Nj ∧ Ni = Bij = −Bji = −λji∗Ni ∧ Nj = λji∗Nj ∧ Ni (197) so we have λij = λji and~~

~~Wj Wi 2 λi = λj ⇒ λi = λWi (198) Wi Wj for some constant λ and ﬁnally~~

~~0 Bij = λWiWj∗Nj ∧ Ni (199)~~

This shows uniqueness up to a scaling. To show existence it is enough to check that such constructed forms satisfy requirements (that is just reversing all arguments). 0 As changing Ni by sign also change Wi by the same sign, the Bij are independent of the choice of sign of normal vectors.

~~{G} {G} Bivectors Bij satisﬁes requirements for normals Ni .~~

~~6.6. Nondegenerate bivectors and 4-simplex We can now prove that non-degenerate geometric solution determines 4- simplex uniquely up to shifts and inversion.~~

~~Theorem 3. Given a non-degenerate geometric SO solution {Gi} there exist 4 exactly two 4-simplices (deﬁned up to shift R transformations) such that~~

~~∆ {G} Bij = rBij (200) where r = ±1. They are related by inversion transformation I and for both the sign r is the same.~~

Proof. We will ﬁrst prove that there exists such 4 simplex. We take any {G} 0 5 planes orthogonal to Ni . They cut out a 4 simplex ∆ . This simplex is unique up to shifts and scaling by a real number (changing the size and applying the inverse).

~~40 ∆0 The bivectors of any of these 4-simplices Bij satisfy from reconstruction from normals (see Theorem 2):~~

~~{G} ∆0 ∗ Bij = cBij , c ∈ R (201)~~

Under scaling transformation (by real number λ, so also under inverse) the bivector changes by λ2. There exist exactly two scalings ±p|c| that brings the bivectors to {G} ∆ Bij = rBij r = ±1 (202) The sign cannot be changed but it depends on the choice of orientation. {G} Uniqueness: From bivectors we can reconstruct ±Ni (sign ambiguity). For any choice of the signs we reobtain the same 4 simplices.

~~The sign r seems to be an additional data in the reconstruction.~~

~~Deﬁnition 8. We call the constant r = ±1 from the reconstruction for geometric solution, geometric Plebański orientation.~~

4 Constant r relates chosen orientation of R with the orientation deﬁned by order of tetrahedra. We have 1 B{G} = − rW ∆W ∆∗(N ∆ ∧ N ∆) (203) ij V ol∆ i j j i where V ol∆ is 4! volume of the 4 simplex (see 6.3).

~~6.7. Uniqueness of Gram matrix and reconstruction~~

For the non-degenerate geometric solution {Gi} the edge lengths of the tetra- ∆ {G} hedron i in the 4 simplex ∆ can be reconstructed from bivectors Bij = rBij , j 6= i. Let us now consider single i-th tetrahedron. As Gi is a rotation the {G} can shape of the tetrahedron with bivectors Bij = Gi∗(vij ∧ Ni ) is the same as the shape of the tetrahedron with bivectors

~~can Bij = ∗(vij ∧ Ni ) , (204) which are determined by the geometric boundary data.~~

Lemma 12. If the geometric boundary data is non-degenerate then for every i there exists a unique up to inversion and translations tetrahedron with face bivectors can Bij = ri∗(vij ∧ Ni ) . (205) can⊥ in the subspace Ni with ri = ±1.

41 Proof. Let us ﬁx i. We cut a tetrahedron with planes perpendicular to vij in can⊥ 0 can Ni in generic position. Its bivectors Bij are proportional to ∗(vij ∧ Ni ) thus 0 0 can Bij = λij∗(vij ∧ Ni ) (206) From the closure condition we know   X 0 X 0 can X 0 can Bij = λij∗(vij ∧ Ni ) = ∗  λijvij ∧ Ni (207) j6=i j6=i j6=i

~~can As vij are perpendicular to Ni from non-degeneracy~~

~~X 0 0 λijvij = 0 ⇒ ∃λ : λij = λ (208) j6=i By rescaling of the tetrahedron we can get λ = ±1.~~

~~This determines edge lengths uniquely as functions of vij. Let us denote the signed square lengths of the edge~~

i 2 ljk between faces (ij) and (ik) of the tetrahedron i. (209) This numbers are defined for i, j, k pairwise different and are symmetric in j, k. Definition 9. The geometric boundary data satisfies lengths matching con- k 2 dition if lij is symmetric in all its indices. The lengths matching condition is necessary for existence of non-degenerate geometric solution. If lengths matching condition is satisfied we define signed square lengths

2 k 2 lml = lij , for m, l the remaining missing indices diﬀerent from i, j, k (210) These lengths determines 4 simplex unique up to orthogonal transformation and shifts (see [73, 74] for description of the matching conditions). 2 Theorem 4. For the signed square lengths lij from non-degenerate geometric boundary data satisfying lengths matching condition we introduce lengths Gram matrix of the 4 simplex  0 1 1 ··· 1   1 0 l2 ··· l2   01 04  l  1 l2 0 ··· l2  G =  10 14  (211)  ......   . . . . .  2 2 1 l40 l41 ··· 0 Let us denote the signature of Gl by (p + 1, q + 1, n ) | {z } | {z } |{z} + − 0

42 4 • If n = 0 then there exists a unique up to O(p, q) o R transformations non-degenerate 4 simplex in the spacetime with signature (p, q) with 4 these lengths. There are two inequivalent 4-simplices up to SO(p, q)oR transformations p+q • If n > 0 then there exists a unique up to O(p, q) o R transformations degenerate 4 simplex in the signature (p, q) with these lengths. Proof. It is more convenient to work with the matrix with elements 1 M = (l2 + l2 − l2 ), i, j ∈ {1,... 4} (212) ij 2 i0 j0 ij that should correspond to the matrix of suitable scalar products. By a change of basis we can transform Gl to the block diagonal form where one block is 0 1 and the second block is M. The signature of M is thus (p, q, n). 1 0 There exist matrix R with 4 columns and p + q rows with full range such that M = RT ηR, η = diag(1, ··· , 1, −1, ··· , −1) (213) | {z } | {z } p q because M is symmetric with given signature. The 4 simplex can be constructed as follows. Choose arbitrary x0 and define xi for i 6= 0 by µ µ µ Ri = xi − x0 (214) This 4 simplex has a prescribed lengths. Let us now compare two different 0 0 4 simplices with vertices xi and xi. Both R and R need to satisfy (213). Since R and R0 have the same kernel (kernel of M) and full range there exists G ∈ GL(p + q) such that R0 = GR (215) and it satisfies GT ηG = η =⇒ G ∈ O(p, q) (216) We have thus 0 0 xi = Gxi + v, where v = Ox0 − x0 (217) as v and G ∈ O(p, q) are arbitrary we obtained uniqueness of the solution up to desired transformation.

Suppose that p0 ≥ p and q0 ≥ q then we can aﬃnely embed spacetime of signature (p, q) into spacetime of signature (p0, q0). Even if n > 0 we 0 0 4 can reconstruct degenerate 4-simplex in 4 dimensions up to O(p , q ) o R transformations. However signature (p0, q0) is not unique. If we introduce normals to tetrahedra of two such 4 simplices {Ni} and 0 {Ni } then there exists G ∈ O and si ∈ {0, 1} such that

~~0 si si Ni = (−1) GNi = GI Ni (218) that are exactly gauge transformations of the SO geometric solution.~~

43 ∆ 6.8. Geometric rotations Gi Suppose that we have a non-degenerate boundary data and the Gram matrix (211) is non-degenerate. From the lengths Gram matrix we can reconstruct 4 geometric non-degenerate 4 simplex (up to O nR transformations). Let us ∆ choose one of the simplices and compute geometric bivectors Bij and normals ∆ Ni . ∆ For normals Ni we will introduce vectors j 6= i

~~∆ ∆ ∆ ∆ ! ∆ 1 ∆ ∆ ∆ Wi Wj Nj ·Ni ∆ ∆ ∆ vij = − ∆ Wi Wj Nj − ∆ ∆ Ni , vij ⊥Ni (219) V ol Ni ·Ni~~

~~that are normals to the faces of i-th tetrahedron recovered from geometric bivectors. We have ∆ ∆ ∆ Bij = ∗(vij ∧ Ni ) (220) Lemma 13. If the lengths matching condition is satisﬁed then~~

~~∆ ∆ vij ·vik = vij·vik (221)~~

~~Proof. This is equivalent to~~

~~∆ ∆ Bij ·Bik = Bij·Bik (222)~~

Both bivectors are bivectors of either reconstructed ith tetrahedron in 4 simplex or reconstructed boundary tetrahedron in the space perpendicular can to Ni . Similarly as for 4 simplex we can prove that the two tetrahedra diﬀers by rotation G ∈ O. The scalar products are thus preserved18.

~~∆ We can introduce group elements Gi for any i by conditions~~

~~∆ can ∆ ∆ ∆ Gi Ni = Ni , ∀j6=i Gi vij = vij (223)~~

~~There are 5 conditions but only 4 are independent (closure conditions are the same for v∆ and v vectors).~~

∆ Lemma 14. Elements Gi ∈ O. ∆ can Proof. Both Ni and Ni are of the same type normalized, and perpendicular to the rest of the vectors. The shapes of tetrahedra are also the same so scalar product between vectors are preserved. This is however the deﬁnition of being orthogonal.

~~18These scalar product can be computed by a version of Cayley-Menger determinant [75].~~

~~44 ∆ 6.9. Relation of Gi to Gi~~

We would like to compare Gi from the deﬁnition of geometric solution with ∆ ∆ s {G} s Gi obtained from Bij = (−1) Bij (where r = (−1) for s ∈ {0, 1}). We know that there exist si ∈ {0, 1} such that

~~{G} si ∆ Ni = (−1) Ni , (224) and~~

~~{G} {G} s ∆ ∗(Givij ∧ Ni ) = Bij = (−1) Bij =~~

~~s ∆ ∆ s+si ∆ {G} = ∗((−1) vij ∧ Ni ) = ∗((−1) vij ∧ Ni )~~

~~∆ {G} so because both vij and Givij are orthogonal to Ni we have~~

s+si ∆ can si ∆ Givij = (−1) vij ,GiNi = (−1) Ni (225) thus ∆ s s G = G I i (IR can ) (226) i i Ni Lemma 15. Geometric SO solutions are in 1-1 correspondence with reconstructed from lengths 4-simplices and choices s ∈ {0, 1} and si ∈ {0, 1} such that ∆ s ∀i det Gi = (−1) . (227) The relation is given by

~~∆ s s G = G I i (IR can ) . (228) i i Ni~~

~~Proof. The condition for {Gi} to be a geometric solution is det Gi = 1. This is equivalent by (226) to~~

~~∆ s s s ∆ 1 = det G I i (IR can ) = (−1) det G (229) i Ni i~~

~~Let us notice that as there is only one reconstructed 4 simplex up to rotation from O thus two geometric rotations are always related by~~

~~∆0 ∆ Gi = GGi ,G ∈ O (230) thus ∆0 det Gi ∀i ∆ = det G (231) det Gi and we can introduce a deﬁnition~~

45 Definition 10. Suppose that the geometric boundary data satisfies the lengths matching condition. We say that it satisfies orientations matching condition if for any (and thus for all) reconstructed 4 simplices

∆ ∃r∈{−1,1} ∀i det Gi = r (232) Let us notice that after we choose reconstructed 4 simplex (O transformation), the choice of si is arbitrary and corresponds to involution gauge transformations. The value of s is ﬁxed by

r = (−1)s (233) and it is Plebański orientation. Theorem 5. The non-degenerate SO geometric solution exists if and only if the geometric boundary data satisﬁes the lengths and orientations matching conditions. If we have one gauge equivalence class of geometric solutions {Gi} then the representative of the second one is given as follows

G˜ = R G R can (234) i eα i Ni where eα is any normalized to ±1 vector. The second class corresponds to reflected 4 simplex. Any non-degenerate solution is gauge equivalent to one of these two. Proof. The identification comes from lemma 15 and the description of the gauge transformations. There are two reconstructed 4-simplices up to SO rotations. The choice of si corresponds to involution gauge transformations. Given such simplex from one class the representative of the second class can be obtained by applying a reflection

~~{G˜} {G} 0 Bij = Reα (Bij ), s = s + 1. (235) Thus for geometric SO solutions~~

~~G˜ = R G (IR can ), (236) i eα i Ni where eα is normalized. The equation (234) follows by applying an inversion gauge transformation.~~

6.10. Classiﬁcation of geometric solutions We will consider only the case of non-degenerate boundary data that is that for every i, every 3 out of 4 vectors vij are independent. General classiﬁcation was done in [34].

~~Lemma 16. For a geometric solution {Gi} for a non-degenerate geometric boundary data for any diﬀerent i, j, k one of the following holds~~

46 {G} {G} {G} {G} a) Ni = ±Nk and Nj = ±Nk {G} {G} b) Ni 6= ±Nj {G} Proof. The conditions are exclusive. Suppose neither holds, then Ni = {G} {G} ±Nj and their are not parallel to Nk . In this case

~~Gkvki ≈ Gkvkj, (237) but this contradicts non-degeneracy of the boundary data. Thus exactly one of the two conditions must be satisﬁed.~~

Lemma 17. There are two exclusive possibilities for solution {Gi} for a non-degenerate geometric boundary data: {G} a) All Ni are parallel. b) Geometric solution is non-degenerate. {G} Proof. From lemma 16 we can conclude that either all Ni are parallel or they are pairwise independent

{G} {G} ∀i6=jNi 6= ±Nj (238) Let us consider the second case. We will prove that there exists only one (up to a scaling) solution Wi of X {G} WiNi = 0, (239) i and for nontrivial solution all Wi 6= 0. As there are 5 vectors in 4 dimensional space at least one solution exists. We know that

~~{G} {G} {G} Givij ∧ Ni = λijNj ∧ Ni , λij 6= 0 (240)~~

~~{G} {G} as Ni is independent of Nj . For any i   X {G} {G} X Wj {G} 0 = WjNj ∧ Ni =  Givij ∧ Ni (241) λij j j6=i~~

~~{G} as Givij are perpendicular to Ni~~

X Wj Wj Wk 0 = G v =⇒ = (242) λ i ij λ λ j ij ij ik P from non-degeneracy of vij as for any i j6=i Givij = 0. The ratio of Wj to Wk is ﬁxed and nonzero for j, k 6= i. However choice of i is arbitrary so the solution is unique up to a constant and with all Wi 6= 0. This is equivalent to the solution being non-degenerate.

~~47 7. Other signature solutions~~

can Let us notice that from Lemma 17 it follows that in the case when Ni are of different types and the boundary data is non-degenerate then degenerate can geometric solutions cannot occur. The case of all Ni timelike was describe in [35, 51]. In this case, if lengths and orientations matching conditions are satisfied then either the Gram matrix is degenerate and geometric solution corresponds to degenerate 4-simplex or critical points occur in pairs (there are exactly two geometric solutions) and one can associate with them an Euclidean 4-simplex. The difference of the phases is proportional to Regge action again, but the proportionality constant is different. can Our goal is to provide uniform treatment of both cases N = e0 and can N = e3 where can can ∀i Ni = N (243)

~~7.1. Vector geometries Let us denote V = {v ∈ M 4 : v ⊥ N can} (244) We can consider subgroup of SO(1, 3) that preserves N can~~

~~SO(V ) = {G ∈ SO(1, 3): GN can = N can} (245)~~

~~Let us recall a notion of vector geometry introduced in [35].~~

~~Deﬁnition 11. For the geometric SO(1, 3) non-degenerate boundary data vij (satisfying closure condition) we call a vector geometry a collection~~

~~{Gi ∈ SO(V )} (246) such that {G} {G} vij = −vji (247) {G} where vij = Givij, modulo gauge transformations~~

~~Gi → GGi,G ∈ SO(V ) (248)~~

~~We have~~

~~Lemma 18. There is 1-1 correspondence between SO(1, 3) geometric degenerate solutions and vector geometries (up to gauge equivalence on both sides).~~

{G} {G} Proof. As all Ni are parallel there exists G ∈ SO(1, 3) such that GNi = s can (−1) i N , where si ∈ {0, 1}. We can apply rotation and inversion gauge can can {G} transformations to get GiN = N . We can thus assume that Ni = N can. The remaining gauge freedom is as in vector geometries.

~~48 can can In this situation GiN = N so Gi ∈ SO(V )~~

~~{G} can can Bij = ∗Gi(vij ∧ N ) = ∗(Givij ∧ N ) (249)~~

~~{G} and we can deﬁne vij = Givij ∈ V . One can check that all conditions for geometric solutions are equivalent to conditions for vector geometries.~~

7.2. Other signature solutions In this section we will relate pairs of degenerate solutions with non-degenerate simplex but of different than Lorentzian signature. We describe it in unified can can language applicable to both N = e0 and N = e3. Let us introduce auxiliary space M 40 that differs from Minkowski space M 4 by flipping the norm of N can N canN can g0 = g − 2 µ ν (250) µν µν N can · N can where we used gµν for lowering indices. We will use prime to distinguish operations related to this metric (Hodge star ∗0, scalar product ·0, contraction 0 with use of the metric y ). We introduce

~~0 tcan = N can ·0 N can = −N can · N can ∈ {−1, 1}. (251)~~

Let us notice that restricted to V both scalar products coincide, thus V can be regarded as subspace of both M 4 as well M 40. For vectors in V we can use exchangeably both scalar products. The Hodge ∗0 operation satisﬁes in M 40 ∗02 = 1 (252) and inversion I ∈ SO(M 40). Let us introduce

± 2 40 ± can0 0 0 can Φ :Λ M → V, Φ (B) = t (±B + ∗ B)y N (253) where we regard V as a subspace of M 40. Let us notice also that for v ∈ V ⊂ M 40 Φ±(∗0v ∧ N can) = v (254) The map Φ˜ = (Φ+, Φ−):Λ2M 40 → V ⊕ V is an isomorphism 1 Φ˜ −1(v+, v−) = [(v+ − v−) ∧ N can + ∗0(v+ + v−) ∧ N can] (255) 2 We can use this isomorphism to transform the action of SO(M 40) from bivectors into V ⊕ V . Let us notice that as SO(M 40) preserves the decomposition into self-dual and anti-self-dual forms, the action is diagonal and we can deﬁne 0 SO(M 4 ) 3 G → (Φ+(G),Φ−(G)) (256)

~~49 Lemma 19. Φ±(G) ∈ O(V ) Proof. The action on Λ2M 40 preserves the scalar product~~

~~B ·0 B = ∗0(B ∧ ∗0B) (257)~~

~~It can be translated into V ⊕ V as follows. We need to compute~~

Φ˜ −1(v+, v−) · Φ˜ −1(v+, v−) (258) but this is by (255) after cancellations 1 ∗0[v+ ∧ N can ∧ ∗0(v+ ∧ N can) + v− ∧ N can ∧ ∗0(v− ∧ N can)] = 2 1 0 tcan (v+ · v+ + v− · v−) (259) 2 Thus 1 0 Φ˜ −1(v+, v−) · Φ˜ −1(v+, v−) = tcan (v+ · v+ + v− · v−) (260) 2 So Φ±(G) preserves the scalar product on V .

~~Let us notice that the condition for simplicity of a bivector B~~

~~1 0 0 = ∗0(B ∧ B) = tcan (v+ · v+ − v− · v−) (261) 2 is equivalent to |v+|2 = |v−|2. Lemma 20. The following exact sequence holds~~

~~40 Φ˜ ˜sgn 0 → {1,I} → SO(M ) → SO(V ) × SO(V ) → Z2 (262) where we deﬁned~~

Φ˜(G) = (Φ+(G),Φ−(G)), (263) ˜sgn(G+,G−) = sgn(G+) sgn(G−), (264) and  can 1 N = e0 ±  can ± sgn(G ) = 1 N = e3 and G is time direction preserving, can ±  −1 N = e3 and G is time direction reversing. (265)

can Proof. Let us consider N = e3. The only group elements preserving all bivectors are 1 and I. We need to ﬁnd now the image of SO(M 40). The image 40 of a connected component SO+(M ) is SO+(V ) × SO+(V ) (dimensions of groups match). We need to determine the image of the group element

~~G = Re1 Re3 (266)~~

50 40 40 as it is the generator of SO(M )/ SO+(M ). We have G(e3) = −e3 and Φ±(G) are equal and ± ± ± Φ (G)(e0) = −e0,Φ (G)(e1) = e1,Φ (G)(e2) = −e2, (267) so Φ±(G) ∈ SO(V ), sgn(Φ±(G)) = −1 and the image is in the kernel of ˜sgn as sgn(Φ+(G)) sgn(Φ−(G)) = 1. can In the case of N = e0 0 ˜ 0 → {1,I} → SO(M 4 ) →Φ SO(V ) × SO(V ) → 0 (268) by similar reasoning as before.

Lemma 21. Let us suppose that we have two vector geometries {Gi} and 0 {Gi} then 0 sgn Gi sgn Gi (269) is equal for every i. can can Proof. It is a trivial statement for N = e0. Let us consider N = e3. It is enough to show that 0 0 0 s = sgn Gi sgn Gj is equal to s = sgn Gi sgn Gj (270) We know that {G} {G} vij = −vji , (271) so if one is future directed then the second is past directed. Group elements changing future into past directed vectors have sgn = −1 thus 1 v and v has opposite time direction s = ij ji (272) −1 vij and vji has the same time direction for s0 it is the same conditions thus s = s0. Lemma 22. The two statement are equivalent for G ∈ SO(M 40) • GN can = (−1)sN can, s ∈ {0, 1} • Φ+(G) = Φ−(G) and then for GIs regarded as an element of SO(V ) Φ±(G) = GIs (273) Proof. In one direction: follows from the deﬁnition of Φ± and (254) that Φ±(G)(v) = Φ±(G)Φ±(∗0(v ∧ N can) = Φ±(∗0(Gv ∧ GN can )) = (−1)sGv | {z } =(−1)sN can (274) so Φ+(G) = Φ−(G). If Φ±(G) = G˜ ∈ SO(V ) then 0 G = GI˜ s ∈ SO(M 4 ) (275) where s ∈ {0, 1}.

51 7.3. Correspondence can can Let us notice that SO(1, 3) geometric boundary data with all Ni = N can be regarded as SO(M 40) geometric boundary data. We will call it ﬂipped geometric boundary data. Theorem 6. There is a 1-1 correspondence between • ordered pairs of two non-gauge equivalent vector geometries, • geometric SO(M 40) non-degenerate solutions (up to the inversion gauge transformations) for ﬂipped geometric boundary data. 40 ± From the SO(M ) solution {Gi} the two vector geometries {Gi } are obtained as ± ± Gi = Φ (Gi) (276) ± ± {G} {G } and Φ (Bij ) = vij . Proof. Let us consider geometric SO(M 40) solution. We have

~~± ± {G} ± Φ (Bij) = vij, Φ (Bij ) = Gi vij (277) so the following conditions are equivalent~~

{G} {G} {G+} {G+} {G−} {G−} Bij = −Bji ⇐⇒ vij = −vji and vij = −vji (278) + − Let us now suppose that we have two vector geometries Gi and Gi . From + − lemma 21, s = sgn Gi sgn Gi is the same for all i. By gauge transforming − Gi by G such that sgn G = s we can obtain situation when + − ∀i sgn Gi sgn Gi = 1 (279) s 40 so there exist (unique up to I i ) elements Gi ∈ SO(M ) that constitute an SO(M 40) solution. Gauge equivalent SO(M 40) geometric solutions are obtained from gauge equivalent pairs of vector geometries. The only thing that is left is to prove that the SO(M 40) geometric solution is non-degenerate exactly when two vector geometries are not gauge equivalent. Let us assume that the geometric solution is degenerate. As boundary data {G} is non-degenerate, the geometric solution is degenerate when all Ni are parallel. By gauge transformation (including inversion gauge transformation) {G} can we can assume that Ni = N then + − Gi = Gi (280) so vector geometries were gauge equivalent. Other way around, if two vector geometries are equivalent then by gauge + − {G} can si can transformation we can assume Gi = Gi and Ni = GiN = (−1) N by lemma 22.

52 Lemma 23. Suppose that we have a non-degenerate geometric boundary data satisfying lengths and orientations matching conditions. There is a 1-1 correspondence between gauge equivalent classes of geometric non-degenerate SO(M 40) solutions and reconstructed 4 simplices in M 40 (up to shifts and SO(M 40) rotations). Gauge equivalent classes of non-degenerate SO(M 40) geometric solutions occur in pairs related to reﬂected 4-simplices and their representatives are related by

~~0 Gi = RN can GiRN can (281) Let us notice the following identity for G ∈ SO(M 40)~~

~~± ∓ Φ (RN can GRN can ) = Φ (G) (282)~~

~~Two non-equivalent geometric solutions thus satisfy~~

~~± 0 ∓ Φ (Gi) = Φ (Gi) (283)~~

~~7.3.1. Orientations We will now describe orientations matching condition in terms of self-dual and anti-self-dual forms. Let us introduce~~

~~∆± ± ∆ vij = Φ (Bij ) (284)~~

~~∆ From simplicity of Bij and (259) and (261)~~

~~∆± ∆± vij · vik = vij · vik (285)~~

~~∆± We can introduce Gi ∈ O(V ) by conditions~~

~~∆± ∆± ∀j6=i Gi vij = vij (286)~~

~~∆ ∆± Lemma 24. det Gi = det Gi~~

~~∆ ˜ si si ∆ ˜ Proof. We can write Gi = Gi(IRN can ) where (−1) = det Gi and Gi ∈ SO(M 40). Let us notice that~~

~~˜−1 ∆ can Gi Ni = N (287) We have~~

± ˜−1 ∆± ± ˜−1 ∆± Φ (Gi )Gi vij = Φ (G )vij = can0 ˜−1 0 ∆ ∆ ˜−1 ∆ ∆ 0 can = t (±Gi ∗ (vij ∧ Ni ) + Gi (vij ∧ Ni ))y N = can0 0 ˜−1 ∆ can ˜−1 ∆ can 0 can = t (±∗ (Gi vij ∧ N ) + Gi vij ∧ N )y N = ˜−1 ∆ ˜−1 ∆ = Gi vij = Gi Gi vij

53 and as also ˜−1 ∆ can can Gi Gi N = N (288) we see that ∆± ± ˜−1 ∆± ˜−1 ∆ ∆ det Gi = det Φ (Gi )Gi = det G Gi = det Gi (289) ˜ 40 ˜± because Gi ∈ SO(M ) and Gi ∈ SO(V ).

~~8. Classiﬁcation of solutions~~

In this section we will classify solutions and determine under which circum- stances which geometric SO(1, 3) solution can occur. We assume that the boundary data is non-degenerate, that is, that for every i, every 3 out of the 4 vectors vij are independent. Using the correspondence between the geometric SO(1, 3) solutions and critical points, this gives, in fact, a classiﬁcation of the latter.

8.1. Non-degenerate simplices Let us recall that non-degenerate SO(1, 3) geometric solution can occur only if the boundary data is non-degenerate. Theorem 7. Let us assume that the geometric boundary data is non-degenerate. Then the following are equivalent: • lengths and orientations matching conditions are satisﬁed and the reconstructed 4-simplex is non-degenerate Lorentzian, • there exists a non-degenerate SO(1, 3) geometric solution, • there exist exactly two classes of gauge equivalent non-degenerate SO(1, 3) geometric solutions. We can choose their representatives {Gi} and {G˜i} to be related by

~~G˜ = R G R can , e normalized. (290) i eα i Ni α Proof. Follows from theorem 5.~~

~~If we have one SO(1, 3) geometric solution with additional choice of the gauge such that Gi ∈ SO+(1, 3), then the representative of the second gauge equivalence class is given as follows:~~

~~r G˜ = R G R can I i ∈ SO (1, 3), (291) i e0 i Ni + where we added can 0 when Ni = e0, ri = can (292) 1 when Ni = e3, such that also the G˜i are in the connected component of the identity.~~

54 Theorem 8. Let us assume that the geometric boundary data is non-degenerate. Then the following are equivalent: 1. lengths and orientations matching conditions are satisfied, and the reconstructed 4-simplex is non-degenerate of split or Euclidean signature, 2. there exists a non-degenerate SO(M 40) geometric solution for flipped geometric boundary data, 3. there exist exactly two gauge equivalence classes of non-degenerate SO(M 40) geometric solutions for flipped geometric boundary data. We can choose their representatives {Gi} and {G˜i} to be related by

~~˜ can Gi = Reα GiRN , eα normalized, (293)~~

4. there exist at least two gauge equivalence classes of vector geometries + − {Gi } and {Gi }, 5. there exist exactly two gauge equivalence classes of vector geometries. + − 40 The vector geometries {Gi } and {Gi } are obtained from geometric SO(M ) solution {Gi} by ± ± Gi = Φ (Gi) (294) Proof. The equivalences 1. ⇔ 2. ⇔ 3. follow from lemma 23. 3. ⇒ 5.: By theorem 6 from an ordered pair of non-equivalent vector geometries one obtains one non-degenerate geometric solution. If there existed a third class of vector geometries, then taking all 6 possible ordered pairs, one would obtain 6 diﬀerent geometric solutions. This contradicts 3. 5 ⇒ 4 is trivial. 4 ⇒ 2 is just an application of theorem 6.

~~8.2. Degenerate geometric solutions In this section we will prove several results about vector geometries under assumption that the lengths matching condition is satisﬁed.~~

8.2.1. Possible signatures Let us now suppose that lengths matching condition is satisﬁed and we have a vector geometry {Gi ∈ SO(V )}. We would like to determine possible signatures of 4 simplices reconstructed from the Gram matrix. Let us denote reconstructed spacetime by M 4. For that let us choose an edge 34 of the reconstructed 4-simplex. As all faces are spacelike, the edge needs also to be spacelike. Let us denote its ∆ ∆ vector by e34. The space perpendicular to e34 can be identiﬁed with the subspace W of bivectors of the form

~~∆ ∆ 1 2 1 2 W = {B : ∃n ⊥ e34,B = n ∧ e34},B ·B = −n ·n (295)~~

~~55 This space is spanned by geometric bivectors of the faces (kl), k, l ∈ {0, 1, 2} ∆ sharing the edge e34~~

~~∆ W = span{Bkl, k < l, k, l ∈ {0, 1, 2}} (296)~~

~~As every two such faces share a tetrahedron we can compute the scalar product between their bivectors without actual knowledge about the reconstruction, for example (as all faces are spacelike)~~

~~∆ ∆ ∆ ∆ − nkl·nkl0 = Bkl·Bkl0 = Bkl·Bkl0 . (297)~~

~~This can be also computed in the Lorentzian signature~~

∆ ∆ can can ∆ ∆ can {G} {G} Bkl·Bkl0 = Bkl ·Bkl0 = −∗Bkl ·∗Bkl0 = −N ·N vkl ·vkl0 = −t vkl ·vkl0 (298) where tcan = N can · N can. ∆ Let us now notice that vectors e3i can be expressed as a linear combinations ∆ ∆ of nkl for k, l∈ / {3, 4} and e34. Therefore the vectors

~~∆ ∆ nkl, (kl) ∈ {(01), (12), (20)}, e34 (299)~~

~~∆ span the whole space of the 4 simplex as vectors {e3i : i ∈ {0, 1, 2, 4}} do. ∆ ∆ Let us now consider the matrix of scalar products of e34 and vectors nkl, (kl) ∈ {(01), (12), (20)}:~~

~~ ∆ ∆ 0 nkl·nk0l0 0 G = ∆ ∆ . (300) 0 e34·e34~~

∆ ∆ can {G} {G} The submatrix nkl·nk0l0 has the same signature as t vkl · vk0l0 . Lemma 25. If lengths matching condition is satisﬁed, the reconstructed 4 simplex is non-degenerate and there exists a vector geometry then the signature of the reconstructed 4-simplex is either + + −− or − − −−.

{G} {G} Proof. The matrix vkl · vk0l0 ((kl) ∈ {(01), (02), (12)} has rank 3 thus is non-degenerate and has the same signature as a matrix of scalar products in V . Let us consider two cases: can can 0 • N = e0. Since t = 1 the signature of G is (− − −) plus (−) can can 0 • N = e3. Since t = −1 the signature of G is (− + +) plus (−).

Let us introduce for any pairwise diﬀerent i, j, k, ∈ {0, 1 ... 4} a matrix   vij · vij −vji · vjk −vij · vik ˜ Gijk =  −vjk · vji vjk · vjk −vkj · vki  . (301) −vik · vij −vki · vkj vki · vki

~~56 Lemma 26. If lengths matching condition is satisﬁed and there exists a vector geometry then the following are equivalent • the reconstructed 4-simplex is degenerate~~

• for any pairwise diﬀerent i, j, k, ∈ {0, 1 ... 4} the matrix G˜ijk from (301) satisﬁes det G˜ijk = 0 (302) Proof. The Matrix G0 from (300) is degenerate exactly when the matrix of the {G} {G} scalar products vkl · vk0l0 , (kl) ∈ {(01), (12), (20)} is degenerate. The latter is equal to G˜012. As choice of indices is arbitrary we obtain equivalence.

~~8.2.2. Possible degenerate points We will now prove a version of Plebański classiﬁcation (see [76, 77, 78, 79, 67]).~~

Lemma 27. Let us assume that we have non-degenerate geometric boundary can can data for Ni = N . There exist at most two (up to an overall rotation by G ∈ O(V )) sets of vectors {wij}i6=j,i,j∈{0,...4} in V satisfying

~~1. ∀i6=j wij = −wji P 2. ∀i j6=i wij = 0~~

~~3. ∀i,j,kwij · wik = vij · vik If for any pairwise diﬀerent i, j, k, ∈ {0, 1 ... 4} the matrix from (301) satisﬁes~~

~~det G˜ijk = 0 (303) then there exists at most one solution.~~

Proof. We will ﬁrst prove that there are at most two matrices Gij,kl (indices are pairs i 6= j and k 6= l) satisfying X Gij,ik = vij ·vik,Gij,kl = −Gji,kl = −Gij,lk = Gkl,ij, Gij,kl = 0 (304) j6=i

Matrix Gij,kl should be the matrix of scalar product wij · wkl. If there are at most two such matrices there will be at most two sets of vectors {wij} up to rotations from O(V ). The only undetermined entries are Gij,kl where all indices are diﬀerent. Let us assume that i, j, k, l are diﬀerent and m is the lacking index. We know that

~~expressed by vab·vac X z X }| { 0 = Gin,kl = Gin,kl +Gij,kl + Gim,kl. (305) n6=i n∈{k,l}~~

57 So Gij,kl can be expressed by Gim,kl and vab ·vac. Similarly we can replace by m any other index. As all permutations are generated by transpositions (mi), (mj), (mk), (ml), we see that we can express any Gij,kl with diﬀerent indices as a linear combination of G01,23 and vab · vac. Let us denote G01,23 = α. As wij are vectors in V , every determinant of a 4 by 4 minor of Gij,kl need to vanish. Let us take the determinant of G G 0i,0j 0i,23 , i, j ∈ {1, 2, 3} (306) G23,0j G23,23

The only unknown entry is α = G01,23 = G23,01 and it appears twice in the matrix. The coeﬃcients in the contribution of α2 to the determinant is 2 −(G02,03 −G02,02G03,03) 6= 0 as v02 and v03 are not parallel. The determinant is thus quadratic polynomial and there are at most two solutions in α to the equation G G det 0i,0j 0i,23 = 0. (307) G23,0j G23,23

If Gij,kl is a matrix of scalar product then its range of is 3 dimensional (non- degeneracy of boundary data) and the signature is the same as signature of µ V . There exist vectors Mij ∈ V such that

~~µ ν Gij,kl = MijηµνMkl, ηµν scalar product in V. (308)~~

0 µ If there exists another set of vectors Mij then there exists G ∈ O(V ) such that 0 Mij = GMij. (309) This proves that there are at most two solutions up to rotations. If det G˜012 = 0 then w01, w12, w20 are not independent (if they were the signature of G˜012 would be the same as V ). Thus     G01,12 G01,20 G01,23 −v10 · v12 −v01 · v02 G01,23 0 = det  G12,12 G12,20 G12,23  = det  v21 · v21 −v21 · v20 −v21 · v23  . G20,12 G20,20 G20,23 −v20 · v21 v20 · v20 v20 · v23 (310)

As v21 and v20 are independent (boundary non-degeneracy condition) this 2 gives linear equation for G01,23 as (v21 · v21)(v20 · v20) − (v20 · v21) 6= 0. So there exists at most one matrix of scalar products.

Lemma 28. There are at most two vector geometries up to gauge transformations. If for any pairwise diﬀerent i, j, k, ∈ {0, 1 ... 4} the matrix G˜ijk from (301) satisﬁes det G˜ijk = 0 (311) then there exists at most one.

~~58 Proof. For the sets of vectors {wij} satisfying assumptions of lemma 27 we {w} can introduce Gi deﬁned by (for non-degenerate boundary data)~~

~~{w} Gi vij = wij. (312)~~

{w} We have Gi ∈ O(V ). Vectors {wij} are deﬁned by vector geometry if and {w} only if det Gi = 1 for all i. There exist at most two up to an overall O(V ) transformation sets of vectors from lemma 27. From two sets diﬀering by {w} rotations not from SO(V ) at most one has ∀i det Gi = 1, so there are at most two vector geometries up to gauge transformations. If det G˜ijk = 0 then there exists at most one set of possible vectors {wij} {w} up to rotations from O(V ). As before only one can have ∀i det Gi = 1.

8.3. Degenerate SO(1, 3) geometric solutions Lemma 29. Let us assume that the lengths matching condition is satisfied, the geometric boundary data is non-degenerate and there exists a vector geometry {G˜i} and if in addition • the reconstructed 4-simplex is non-degenerate in M 40 (flipped with respect to N can) then orientations matching condition is satisfied and there exist exactly two gauge inequivalent classes of vector geometries, • the reconstructed 4-simplex is degenerate then orientations matching condition is satisfied and there exists exactly one gauge equivalent class of vector geometries Proof. If lengths matching condition is satisfied, the reconstructed 4 simplex is non-degenerate and there is one vector geometry then we can consider 3 sets of vectors satisfying assumptions of lemma 27

~~∆± G˜ vij , vij (313)~~

∆+ ∆− 19 and vij and vij from equation (284) are not related by a rotation . From lemma 27 we know that there exists s ∈ {+, −} and G ∈ O(V ) such that ∆ s G˜ vij = Gvij (314) thus by lemma 24 ∆ s ˜ ∆ ˜ Gi = GGi, det Gi = det GGi = r, (315) where we introduced r = det G. As a result, the orientations matching condition is satisﬁed and by theorem 8 there exist two gauge inequivalent

19 ∆+ ∆− ∆+ ∆− −1 ∆+ ∆− If vij = Gvij then G = Gi (Gi ) ∈ SO(V ) (det Gi = det Gi ) for non-degenerate boundary data. In such case all bivectors of the reconstructed 4-simplex are annihilated by Φ−(G)N can, which contradicts its non-degeneracy.

59 vector geometries. By lemma 28 there exists no other gauge class of vector geometry. If the reconstructed 4 simplex is degenerate then by lemma 26 det G˜012 = 0 and by lemma 27 there exists at most one solution wij (up to rotation from G˜ O(V )). We know one solution vij and the second follows from the geometric ∆+ construction of vij from (284) ( it follows that they diﬀer by G ∈ O(V ))

~~˜ ∆+ ∆ r = det G = det GGi = det Gi = det Gi (316)~~

~~Thus orientations matching condition is satisﬁed. By lemma 28 there exists no other vector geometry.~~

8.4. Classification of solutions Degenerate and non-degenerate solutions cannot occur at the same time. If lengths matching condition is satisfied, but orientations matching condition is not satisfied then we cannot have any solution. When lengths matching condition is not satisfied we can have neither non- degenerate SO(1, 3) geometric solution (reconstructed + − −− simplex satisfies lengths matching condition) nor two degenerate solutions (reconstructed + + −− or − − −− simplex need to satisfy lengths matching condition). There still might exist a single vector geometry in this case. By lemma 10 this classification applies to real critical points of the action.

~~9. Phase diﬀerence~~

In this section we will give an interpretation in terms of Regge geometries of the difference of the phases between two critical points. The overall phase can be changed by adjusting the phases of the coherent states. Therefore, in the study of the asymptotic behaviour of the vertex amplitude the phase difference ∆S is of main interest. It is defined up to 2πi because it appears in the exponent. In Section 9.4 we will show that it agrees with the expected Regge term ∆S∆ up to π i. We will improve this result in Section 9.5 by using certain deformation argument and we will show that this agreement holds exactly, i.e. up to 2π i:

~~∆S = ∆S∆ mod 2π i .~~

9.1. The Regge term ∆S∆ Regge calculus [44] was devised as an extension of gravitational action to certain distributional metrics that are ﬂat everywhere except on the faces of the simplicial decomposition. On these faces, the geometry is not smooth

~~60 anymore, and deﬁcit angles appear. Faces are assumed to have ﬂat inner simplicial geometry. The action is a sum of contributions from single simplices given by X S∆ = Aijθij, i~~

~~can0 ∆ 0 ∆ can0 cos θij = t Ni · Nj , t θij ∈ (0, π)~~

~~∆ ∆ can can • Lorentzian signature (+ − −−) and Ni · Nj > 0 and Ni = Nj . The dihedral angle θij > 0 is the unique angle such that~~

∆ ∆ Ni · Nj = cosh θij This case contains thick wedge for both normals timelike and thin wedge for both normals spacelike. ∆ ∆ can can • Lorentzian signature (+ − −−) and Ni · Nj < 0 and Ni = Nj . The dihedral angle θij < 0 is the unique angle such that

∆ ∆ Ni · Nj = − cosh θij This case contains thick wedge for both normals spacelike and thin wedge for both normals timelike. can can • Lorentzian signature (+ − −−) and Ni 6= Nj . The dihedral angle θij is the unique angle such that

~~∆ ∆ Ni · Nj = sinh θij.~~

~~61 thick wedges thin wedges~~

~~Figure 4: Thick and thin wedges.~~

~~The area of the triangle (ij) is equal to 1 A = |B∆|. ij 2 ij By the reconstruction theorem (Theorem 3) we know that~~

~~∆ G Bij = ±Bij .~~

~~G 2 2 Since |Bij | = ρij, it immediately follows that the Regge action for the geometries reconstructed from the geometric solution {Gi} becomes~~

~~1 X S = ρ θ . ∆ 2 ij ij i~~

~~We deﬁne the geometric diﬀerence of the phase ∆S∆ to be equal to~~

~~ 1 reconstructed simplex is Lorentzian, ∆S∆ := 2irS (318) ∆ γ−1 reconstructed simplex in other signature,~~

~~where r is the Plebański orientation (deﬁnition 8).~~

9.2. Revisiting the phase diﬀerence Let us recall that the diﬀerence of the phases between two critical points is given in terms of its SL(2, C) solutions {gi} and {g˜i} by lemma 8 as X ∆S = S˜ − S = i ρijrij + 2ijφij (319) i

~~20Our sign is opposite to [46] and we subtract π in the case of Euclidean and split signatures. It also diﬀers from [81]. Our dihedral angles are always real opposed to [82].~~

~~62 where −1 −1 φij Mij −irij Mij gig˜i g˜jgj = e , (320) where = 2 2γ ( β )T . Taking the image in SO (1, 3) (see lemma 10) Mij ρij γ−i Bij + we obtain~~

1 {G} 1 {G} 2φij B −2rij ∗B ˜−1 ˜ −1 ρij ij ρij ij GiGi GjGj = e , (321) as π( ) = 2 B{G}. Mij ρij ij Let us notice that B{G} · B{G} = ρ2 , and that 1 B{G} generates rotation ij ij ij ρij ij with period 2π. The diﬀerence of the contribution to the phase from the link action is i(ρijrij + 2ijφij). (322)

~~9.3. Determination of rij and φij~~

~~In this section we will determine rij and φij in a geometric way. The price we pay is an ambiguity of π in the phase that will be ﬁxed by a deformation argument in section 9.5.~~

~~Lemma 30. The contributions φij and rij to the diﬀerence in phase for two critical points corresponding to two non-degenerate SO(1, 3) solutions {Gi} and {G˜i} satisfy~~

~~1 {G} 1 {G} −2rij ∗B +2φij B s ρij ij ρij ij I R {G} R {G} = e , (323) Ni Nj where s = 0 if both canonical normals are of the same type, and s = 1 if they are of diﬀerent types.~~

~~Proof. We can compute~~

~~−1 −1 r −1 r −1 G G˜ G˜ G = G I i R can G R R G R can I j G = i i j j i Ni i e0 e0 j Nj j ri+rj = I R {G} R {G} , (324) Ni Nj where ri ∈ {0, 1} are such that~~

~~r G˜ = R G R can I i ∈ SO (1, 3). (325) i e0 i Ni +~~

~~We see that s = ri + rj = 0 modulo 2 if and only if both canonical normals are the same.~~

~~We will now obtain a similar result for two degenerate SO(1, 3) geometric solutions (vector geometries).~~

~~63 40 Lemma 31. Let Gi,Gj ∈ SO(M ) be such that~~

~~± ± ± ± Φ (Gi) = Gi ,Φ (Gj) = Gj . (326)~~

~~{G} can {G} can Let Ni = GiN and Nj = GjN . Then~~

~~± ± ∓ −1 ∓ ± −1 Φ (R {G} R {G} ) = (Gi )(Gi ) (Gj )(Gj ) . (327) Ni Nj Proof. From the equality (282) we have~~

~~± −1 −1 Φ (GiRN can Gi GjRN can Gj ) = ± −1 −1 = Φ (Gi(RN can Gi RN can )(RN can GjRN can )Gj ) = ± ∓ −1 ∓ ± −1 = (Gi )(Gi ) (Gj )(Gj ) . (328) That prove the equality.~~

~~Lemma 32. Suppose that we have a bivector B in Λ2M 40 then~~

~~± Φ±(eB) = eB , (329) where B± ∈ so(V ) ≈ Λ2V are determined by the equality~~

~~Φ±(B±) = Φ±(B) ∈ V, (330) and we embed bivectors from V into M 40.~~

~~Proof. We can decompose B into its self-dual/anti-self-dual part, B = B+ + B−. Group elements generated by them commute, thus~~

~~Φ±(eB) = Φ±(eB+ )Φ±(eB− ) = Φ±(eB± ). (331)~~

+ − As B± are independent and B is determined by B+ (B is determined ± B by B−), it is enough to check what the image of Φ (e ± ) is in terms of ± ± can Φ (B±) = Φ (B). Let us assume that B preserves N , then by lemma 22

~~± eB = eB ∈ SO(V ), (332) where we identify subgroups SO(V ) from Minkowski and from M 40. In this special case~~

~~can0 ± 0 ± 0 can ± ± t (±B + ∗ B )y N = Φ (B) = Φ (B±), (333)~~

~~± can as B preserves N . Let us notice that as B+ = B− is arbitrary in this case so we have~~

~~can0 ± 0 ± 0 can ± ± t (±B + ∗ B )y N = Φ (B±) = Φ (B) (334) for arbitrary B.~~

~~64 We have~~

Lemma 33. The contributions φij and rij for the difference in phase between + − two critical points corresponding to two vector geometries {Gi } and {Gi } 40 satisfies an identity written in terms of SO(M ) geometric solutions Gi and G˜i for flipped geometric boundary data, where ± ± ˜ Φ (Gi) = Gi , Gi = RN can GiRN can . (335) Namely 1 0 {G} 2φij ∗ B ρij ij R {G} R {G} = e , rij = 0 (336) Ni Nj

~~{G} 40 where Bij is the bivector from the SO(M ) geometric solution. Proof. From lemma 31 and 8 we know that (regarding SO(V ) as subgroup of SO(1, 3))~~

± ± 2 {G } 2 {G } ∓rij ∗B ±φij B ± ± ∓ −1 ∓ ± −1 ρij ij ρij ij Φ (R {G} R {G} ) = (Gi )(Gi ) (Gj )(Gj ) = e , Ni Nj (337) as + counts diﬀerence from G+ to G− and − in opposite direction. If rij 6= 0 then right hand side would not be in SO(V ) ⊂ SO(1, 3). However ± every element Gi belongs to this subgroup. Thus rij = 0. ± {G } 40 Regarding Bij as bivectors in M , we have with use of theorem 6

~~± 0 {G} ± {G} {G±} ± {G±} Φ (∗ Bij ) = ±Φ (Bij ) = ±vij = ±Φ (Bij ). (338) Thus by lemma 32 we have~~

~~± 1 0 {G} 1 {G } 2φij ∗ B ±2φij B Φ± e ρij ij = e ρij ij . (339)~~

~~Comparing the images of two group elements under Φ˜ we obtain their equality up to I 1 0 {G} 2φij ∗ B s ρij ij I R {G} R {G} = e . (340) Ni Nj~~

~~0 {G} As ∗ Bij is simple, we see that s = 0.~~

9.4. Geometric diﬀerence of the phase modulo π We know that s s I R {G} R {G} = I RN ∆ RN ∆ (341) Ni Nj i j This is a group element that appears in lemmas 30 and 33. ∆ ∆ Let us recall that ∗(Ni ∧ Nj ) is spacelike. We have (see appendix H):

~~65 can can • For Ni = Nj~~

~~N∆∧N∆ can can0 i j 2t t θij j i |∗N∆∧N∆| i j R {G} R {G} = e . (342) Ni Nj~~

~~can can • For Ni 6= Nj , (in Lorentzian signature)~~

N∆∧N∆ N∆∧N∆ can can i j i j 2t t θij +π∗ j i |∗N∆∧N∆| |∗N∆∧N∆| i j i j IR {G} R {G} = e . (343) Ni Nj Moreover {G} 2 2 |Bij | = ρij, (344) so the rotation generated by either

1 {G} 1 0 {G} Bij , or ∗ Bij in ﬂipped signature, (345) ρij ρij has period 2π. From the geometric reconstruction theorem we know that 1 B{G} = − rW ∆W ∆∗(N ∆ ∧ N ∆), (346) ij V ol∆ i j j i and the sine law shows that 1 rW ∆W ∆ |∗N ∆ ∧ N ∆| = ρ . So V ol∆ i j j i ij

~~1 {G} 1 ∆ ∆ Bij = σij ∆ ∆ ∗(Nj ∧ Ni ), (347) ρij |∗Ni ∧ Nj | where ∆ ∆ σij = −r sign Wi Wj . (348) Let us notice (see (505)) that~~

can can can can −r Ni = Nj σij = −rti tj = can can (349) r Ni 6= Nj By comparing (342), (343) with (323) and (336) we get the following • For Lorentzian signature (non-degenerate solutions) and normals of the same type 2φij = 0 mod 2π, 2rij = 2rθij (350) • For Lorentzian signature (non-degenerate solutions) and normals of different types 2φij = π mod 2π, 2rij = 2rθij (351) • For other signature solutions (degenerate solutions)

~~2φij = 2rθij mod 2π, 2rij = 0 (352)~~

~~66 π We will now consider contributions of 2 appearing in the action from (351). Lemma 34. The following holds for given boundary data: X π X π s ∈ π , s 2 ∈ π , (353) ij 2 Z ij ij 2 Z i~~

~~π As we determined S modulo π we can skip the 2 terms coming from (351) by lemma 34 and write ∆S = ∆S∆ mod πi, (356) where P ∆ i~~

9.5. Deformation argument to ﬁx the remaining ambiguity We only need to determine the remaining ambiguity of π in the action. This however depends on the choice of SL(2, C) lifts, and it needs to be done consistently for the whole 4-simplex. Taking into account that the only source of ambiguity are the contributions φij for half-integer spins, we have ( X φij (non-deg. Lorentzian solutions) ∆S − ∆S∆ = i φij − rθij (two degenerate solutions) i

67 9.5.1. Lorentzian signature case Lemma 35. Suppose that we have non-degenerate geometric boundary data 0 can0 0 vij and Ni with spins ij with non-degenerate SO(1, 3) geometric solutions 0 {Gi }. Let us assume that there exists a continuous path

~~can can0 Gi(t), vij(t),Ni (t) = Ni , (359) such that: • For all i 6= j 0 0 Gi = Gi(0), vij = vij(0), (360)~~

~~• for all t ∈ [0, 1], {Gi(t)} is a SO(1, 3) geometric solution for vij(t) boundary data,~~

~~• for all t 6= 1 the boundary data vij(t) is non-degenerate,~~

~~• for all i 6= j vij(1) 6= 0,~~

~~• for all t 6= 1 solution {Gi(t)} is non-degenerate, and~~

~~• for t = 1 solution {G (t)} and {G˜ (t) = R G (t)R can } are gauge i i eα i Ni equivalent. Then ∆S0 = ∆S∆0 mod 2πi. (361)~~

~~Proof. The function X f(t) = φij(t) mod 2π (362) 0 i~~

~~Thus from Mij 6= 0 it follows that φij(1) = (si + sj)π mod 2π. We have~~

~~0 ∆0 X ∆S − ∆S = i (si + sj)π mod 2πi (365) 0 i~~

~~68 We can consider an Euler cycle in the subgraph consisting of edges with odd spin. For such a cycle X X (si + sj)π = 2siπ = 0 mod 2π. (366) i~~

~~As the subgraph of half-integer spins has even-valent nodes, we can decompose it into Euler cycles thus ∆S0 = ∆S∆0 mod 2πi.~~

Let us deform our boundary data by deforming solution as follows: We choose a spacelike plane described by a simple, normalized bivector V in generic position i.e. {G} ∀i6=jV ∧ ∗Bij 6= 0. (367) We now contract directions in ∗V (perpendicular to directions in V ). All {G(t)} the time Bij 6= 0 and the solution {Gi(t)} obtained by reconstruction from the bivectors is non-degenerate. Let us now consider the limit where we shrink ∗V to zero. We will denote this time as t = 1. Due to (367), {G(t)} the limits limt→1 Bij exist and are nonzero. The shrinking has a dual action on the geometric normals (their directions in ∗V expands but we ∆ need to apply normalization). As the geometric normal vectors Ni (t) do not lie in the V plane (see (367)), they also have a limit and it is equal to their normalized components lying in the plane ∗V . By suitable deﬁnition ∆ of vij(t) we can assume that the limits Gi(1) = limt→1 Gi (t) exist. The 4-simplex is now highly degenerate, contained in a 2d plane. All bivectors are proportional to V . So we have for any non-degenerate boundary data by lemma 35

~~∆S = ∆S∆ mod 2πi. (368)~~

~~9.5.2. The case of other signatures The difference ∆S is well-defined if we fix between which two degenerate points we need to compute the difference of the phase.~~

Lemma 36. Let us consider a set of non-degenerate geometric boundary 0 can0 can data, with spins ij and all canonical normals Ni = N . Let us suppose that there are two non-equivalent vector geometries for this boundary data. 40 0 We associate with them a non-degenerate SO(M ) geometric solution {Gi } 0 with boundary data vij. Let us suppose that there exists a continuous path

~~can can Gi(t), vij(t),Ni (t) = N , (369) such that • one has 0 0 Gi = Gi(0), vij = vij(0), k ∈ {0, 1}, (370)~~

~~69 40 • for all t ∈ [0, 1], {Gi(t)} is a SO(M ) geometric solution for vij(t) boundary data,~~

• for all t ∈ [0, 1] the boundary data vij(t) is non-degenerate, 40 • for t ∈ [0, 1) the SO(M ) geometric solution {Gi(t)} is non-degenerate, 40 •{ Gi(1)} is a degenerate SO(M ) geometric solution. Then ∆S0 = ∆S∆0 mod 2πi. (371) Proof. The function X f(t) = φij(t) − rθij(t) mod 2π (372) 0 i

~~+ − {G(1)} ri can Gi (1) = Gi (1) ∈ SO+(1, 3),Ni = (−1) N , ri ∈ {0, 1}. (373)~~

~~+ si − We have for the lifts gi (1) = (−1) gi (1). Similar considerations using Euler cycles as for Lorentzian signature show that X φij(1) = 0 mod 2π. (374) 0 i~~

Let us deform the boundary data as follows: We choose N of the same type as N can, in generic position, i.e., such that {G} ∀ij N ∧ Bij 6= 0. (378) We contract in the direction of N in M 40. During contraction we have a continuous path of non-degenerate SO(M 40) geometric solutions (with non- degenerate boundary data). At the end we obtain a degenerate 4-simplex for non-degenerate boundary data. By lemma 36 we have ∆S = ∆S∆ mod 2πi. (379)

70 9.6. Summary ∆ Let us now choose all outer pointing normals Ni . We obtained the following formula for the diﬀerence of the phase between two critical points X 1 reconstructed simplex is Lorentzian, ∆S = ir ρ θ ij ij γ−1 reconstructed simplex in other signature, i

~~2tcantcanθ j i ij N ∆∧N ∆ |∗N∆∧N∆| i j R ∆ R ∆ = Oe j i (381) Ni Nj where ∆ ∆ can can inversion in the plane Ni ∧ Nj for Ni 6= Nj , O = can can (382) identity for Ni = Nj .~~

~~10. Computation of the Hessian~~

In this chapter we will ﬁnish our analysis by computing the scaling property of the measure factor in the formal application of stationary phase approximation. We assume that after taking into account gauge transformations the remaining Hessian is non-degenerate. The contribution to the stationary phase approximation of the amplitude A from the Hessian is √ −1/2 det H . (383) The integrand depends on the following variables

• zij for i 6= j, giving 80 real variables, • gi ∈ SL(2, C), giving 30 real variables. But these variables are subject to gauge transformations, which reduces the eﬀective number of nontrivial variables:

• zij → λijzij, giving 40 real gauge parameters, • SL(2, C) gauge gives 6 real gauge parameters. This gives 64 nontrivial variables, and the scaling √ −1/2 det H = CΛ−32. (384) Together with the scaling Λ20 (see (52)) of c(Λ) this gives the scaling of the measure factor, Λ−32Λ20 = Λ−12. (385) Let us notice that H does not depend on the choice of the phase of the coherent states (phase of nij). Its determinant can be expressed by geometric {G} {G} quantities like vij and Ni , Bij in a covariant way. It is thus a geometric quantity depending on lengths and orientations.

~~71 11. Outlook~~

In this paper we performed the complete analysis of the critical points in the extended EPRL setting. Besides the important open questions discussed in section 1.3, there are several points which merit further analysis: • Extension of our results to the case of surfaces of mixed signature. A spin foam model for surfaces of this kind was introduced in [54] based on coherent state techniques. There is no known EPRL-like construction. We suspect that there might be some nongeometric contributions to the asymptotic formula in this case. • The determinant of the Hessian is a complex number, thus it contributes to the phase. It is an important part of the asymptotics and can be described in geometric terms if the boundary state is geometric [2, 45, 63]. The spread of the coherent states has no obvious geometric meaning but it inﬂuences the measure factors. In the Euclidean EPRL model the values of the Hessian at two diﬀerent critical points are equal thus, their phase can be cancelled by proper choice of the phase of coherent states. It is tempting to conjecture that similar statement is true for the Lorentzian models. However nothing of this kind was proven even in the standard EPRL setup. • A determination of the absolute value of the determinant of the Hessian is interesting for the following reason. In order to evaluate the spin foam amplitudes of more complicated triangulations, one needs to take a product over many vertex amplitudes. In the semiclassical limit one can hope that one can replace the vertex amplitudes in the product by their asymptotic forms. In such a situation, the phase is exactly the Regge action [44] of discrete gravity (with some problems regarding orientations [84]) and the measure of the path integral is obtained from the absolute value of the determinant of the Hessian. However, it seems that in the case of current spin foam models the amplitude of the whole foam cannot be obtained by this approximation [41, 52] (but see also [42] for possible resolutions). • Another open problem is the extension of our result to the case of non-vanishing cosmological constant. The asymptotic analysis of a corresponding version of the EPRL model was given in [85, 86]. How- ever, moving away from Euclidean signature even the formulation of the model becomes very formal. There is no satisfactory proposal for the intertwiners representing timelike tetrahedra for the situation with cosmological constant (see however [87, 88] for possible, alternative rig- orous deformations).

~~72 Acknowledgements~~

We would like to thank Bianca Dittrich, Sebastian Steinhaus and Hal Hag- gard for interactions at the early stage of this work. We would like to express our gratitude for Jeff Hnybida for fruitful discussions about the topic. We also thank the anonymous referees for their valuable comments on a previous version of this work. This article is partially based upon the work from COST Action MP1405 QSPACE, supported by COST (European Cooperation in Science and Technology). WK thanks the Institute for Quantum Gravity at the Friedrich-Alexander UniversitätErlangen-Nürnberg (FAU) for hospital- ity. This work was also partially supported by the Polish National Science Centre grant No. 2011/02/A/ST2/00300.

~~Appendices~~

~~A. Notation summary Notation summary • i, j, k nodes (tetrahedra) numbers (indices), • spacetime indices µ, ν,~~

~~• v, u, r, z spinors, zij spinors labelled by pairs of indices (tetrahedra), nij boundary data spinors,~~

• B, M traceless matrices sl(2, C), Bij traceless matrices in sl(2, C) labelled by two tetrahedra, µ µ • ηN = N σµ, ηˆN = N σˆµ, † T •h u, viN = u ηN v Hermitian scalar product deﬁned by normal N, • [u, v] = uT ωv, • ω symplectic form for spinors, ∆ • N normal vectors, Ni outer pointing normal vector to i-th tetrahedron, can Ni canonical normal vector either e0 = (1, 0, 0, 0) or e3 = (0, 0, 0, 1), {G} Ni normal vector to i-th tetrahedron obtained from geometric solution,

~~• v, l vectors (l null vector), vij boundary data vectors, •· scalar product,~~

• RN reﬂection with respect to normal N, • I inversion, • g ∈ SL(2, C) group elements, {G} ∆ • Bij bivectors, Bij bivectors from the geometric solution, Bij geometric bivectors from the simplex,

73 • M 4 Minkowski spacetime, • M 40 spacetime with ﬂipped N can (split or euclidean signature), • r Plebanski orientation (relating orientation of M 4 or M 40 to the com- binatoric orientation of the simplex), • G ∈ O(M 4) or O(M 40) (SO(M 4), SO(M 40)), 4 4 • we denote SO+(M ) = SO+(1, 3) connected component of SO(M ) = SO(1, 3), • V space perpendicular to N can in Minkowski (also embeddable in M 40), • Φ± maps from bivectors in M 40 into V (self-dual and anti-self-dual forms), Φ± corresponding maps from SO(M 40) into SO(V ), • Hodge star ∗, • if we are working explicitly in the ﬂipped spacetime then we use ∗0, ·0 0 0 and for contraction with use of the metric x and y , • if we are working in arbitrary signature spacetime then we use ∗, ·, x and y, • representation labels (, ρ) for SL(2, C) group, ij and ρij representation labels for edge connecting tetrahedron i with j, • λ number, ∗ ∗ • C invertible complex numbers, R invertible real numbers • Λ integer scaling of spins,

~~• θij dihedral angle between tetrahedron i and j,~~

~~• Aij area of the face between tetrahedron i and j, β • S action, Sij, Sij parts of the action.~~

B. Conventions We are using abstract deﬁnition of Grassmann algebra V X and ∧ by its uni- ∗ versal properties. For b ∈ X we deﬁne left (x) and right (y) antiderivatives of order −1 ^ bxw, wyb, w ∈ X (386) by its action on X ⊂ V X

~~bxx = xyb = (b, x), x ∈ X (387) Suppose that we have a metric g . The norm of k vectors is given by µν~~

~~k X sgn σ Y a1 ∧ ... ∧ ak·b1 ∧ ... ∧ bk = (−1) ai·bσ(i) (388)~~

~~σ∈Sk i=1~~

~~74 The Hodge dual is deﬁned by~~

~~∗(a1 ∧ ... ∧ ak) = akxak−1x ··· a1xΩ (389) where Ω is the normalized volume n-vector (choice of the orientation) and x is the contraction with dualized vector by the scalar product.~~

C. Restriction of representations of SL(2, C) In this appendix we will collect results of [70] and [32] about restriction of can the irreducible unitary representations of SL(2, C) to the subgroup St(N ). We introduce spinors

~~ 1 0 n = , n = (390) 0 0 1 1~~

Let us consider generator of rotation around z axis. We can introduce bases of eigenfunction in the representations of SU(1, 1) and SU(2). On the groups St(N can) we consider right regular representations.

C.1. The embedding map in the case of St(N can) = SU(2) In this case following [16] we consider embedding of the spin  representation. We can realize this representation as functions on SU(2) given by matrix elements of the representation in Lz eigenbasis. The embedding of the functions p  Ψm(u) = 2 + 1Dm(u), u ∈ SU(2) into the representation space of unitary irreducible representation (, ρ = 2γ) is given by 1 i ρ −1 F (z) = √ hz|zi 2 Ψ (u(z)), m π N can m can where N = e0 and

~~1 z z z u(z) = 0 1 , z = 0 p −z z z hz|ziN can 1 0 1 and~~

hu|viN can = u0v0 + u1v1. Using the explicit expression for the representation matrices of the SU(2) group we write F explicitly: s Γ(2 + 2) i ρ −1− F (z) = hz|zi 2 hn |zi+m hn |zi−m , m 2πΓ( + m + 1)Γ( − m + 1) N can 0 N can 1 N can

75 C.2. The embedding map in the case of St(N can) = SU(1, 1) Following Hnybida and Conrady [32, 49] we consider embeddings of the basis functions of the ± discrete series unitary irreducible representations of spin  of the SU(1, 1) group. Again we can realize them as a functions on the group

~~+ p +, Ψm(v) = 2 − 1Dm (v), m ≥  (391) − p −, Ψm(v) = 2 − 1D−m(v), m ≤ − (392)~~

~~Their image in the representation space of unitary irreducible representation (, ρ = 2γ) is given by~~

~~ρ τ 1 i 2 −1 τ τ F (z) = √ θ(τ hz|zi can ) hz|zi Ψ (v (z)), jm π N N can m~~

~~can where τ = ±1, N = e3,~~

~~1 z z z v+(z) = 0 1 , z = 0 (393) p z1 z0 z1 hz|ziN can 1 z z v−(z) = 1 0 , (394) p z z − hz|ziN can 0 1 and~~

hu|viN can := u0v0 − u1v1 is the SU(1, 1)-invariant scalar product. Using the explicit expression for the representation matrices of the SU(1, 1) group [70] we write the distributions F ± explicitly: s τ Γ(τm + ) F (z) = θ(τ hz|zi can ) m πΓ(2 − 1)Γ(τm −  + 1) N ρ i 2 −1+ −−m −+m hz|ziN can hz|n0iN can (τ hz|n1iN can ) , (395)

~~C.3. Coherent states~~

~~All three families of representations have certain common feature. In the Lz eigenbasis~~

Ψm, m ∈ {−, − + 1, . . . } for SU(2) + Ψm, m ∈ {,  + 1,...} for SU(1, 1) (396) − Ψm, m ∈ {..., − − 1, −} for SU(1, 1) we can consider extremal eigenfunctions. These are basic coherent states [71]. They usefulness for asymptotic analysis comes from their simple form (see [70, 49])

~~76 can • For N = e0 r 2 + 1 i ρ −1− F (z) = hz|zi 2 hn |zi2 (397)  2π N can 0 N can can • For N = e3~~

r ρ + 2 − 1 i 2 −1+ −2 F (z) = θ(hz|zi can ) hz|zi hz|n i , (398)  π N N can 0 N can r − 2 − 1 i ρ −1+ −2 F (z) = θ(− hz|zi can )(− hz|zi can ) 2 (− hz|n i can ) − π N N 1 N (399) 1 0 where n = and n = . 0 0 1 1 All other coherent states are obtained through transformation of these basic coherent states by group action of St(N can). Let us notice that the group action preserves h·|·iN can and we can move group elements from z into ni obtaining n T T −1 Ψ (z) = F(u z), n = (u ) n0, (400) +,n+ + T + T −1 Ψ (z) = F (u z), n = (u ) n0, (401) −,n− − T − T −1 Ψ (z) = F−(u z), n = (u ) n1 (402) We have the following classiﬁcation: can • Every spinor n+ such that hn+, n+iN can = 1 with N = e3 is obtained T −1 as n+ = (u ) n0 for u ∈ SU(1, 1) can • Every spinor n− such that hn−, n−iN can = −1 with N = e3 is ob- T −1 tained as n− = (u ) n1 for u ∈ SU(1, 1) can • Every spinor n such that hn, niN can = 1 with N = e0 is obtained as T −1 n = (u ) n0 for u ∈ SU(2) This allow us to write coherent states as follows

r ρ + +,n 2 − 1 i 2 −1+ + −2 Ψ (z) = θ(hz|zi can ) hz|zi z|n , (403) N π N can N can r −,n− 2 − 1 i ρ −1+ − −2j Ψ (z) = θ(− hz|zi can ) (− hz|zi can ) 2 (− z|n ) . N π N N can (404) can with N = e3 and arbitrary + + − − hn , n iN can = 1, hn , n iN can = −1 (405) and for SU(2) embedding r 2 + 1 i ρ −1− Ψn(z) = hz|zi 2 hn|zi2 . 2π N can N can can with N = e0 and arbitrary hn, niN can = 1.

~~77 D. Traceless matrices In this section we will describe relation between traceless matrices in two complex dimensions and spinors. Let us assume that δg is traceless then~~

~~ωδgT + δgω = 0 (406) thus for spinors u and v~~

~~[u, δgT v] = uT ωδgT v = −vT δgωu = vT ωδgT u = [v, δgT u] (407) where we transposed the whole formula in the middle equality.~~

~~Lemma 37. Let us assume that [u, v] = 1 then~~

~~T T (vu − uv )ω = I (408)~~

~~Proof. We apply left hand side to v and u~~

(vuT − uvT )ωv = v uT ωv −u vT ωv = v (409) | {z } | {z } =1 =0 (vuT − uvT )ωu = v uT ωu −u vT ωu = u (410) | {z } | {z } =0 =−1 so as u and v are independent ([u, v] 6= 0) and span the whole space of spinors we show that it is identity operator.

~~We can write (δg traceless [u, v] = 1)~~

~~ 1 1 [u, δgT v] = tr vuT ωδgT = tr vuT ω − δgT = tr(vuT + uvT )ωδgT 2I 2 (411) 1 T T In the last expression 2 (vu + uv )ω is traceless. Lemma 38. Let us assume that [u, v] = 1 then~~

~~T T (vu − uv )ω = I (412)~~

~~Furthermore the matrix~~

~~T T M = (vu + uv )ω (413) is traceless and Mv = v, Mu = −u (414) are two eigenvectors. Every traceless matrix with eigenvalues ±1 is of this form.~~

78 Proof. We can compute tr(vuT + uvT )ω = [u, v] + [v, u] = 0 (415) and (vuT + uvT )ωv = v uT ωv +u vT ωv = v (416) | {z } | {z } =1 =0 (vuT + uvT )ωu = v uT ωu +u vT ωu = −u (417) | {z } | {z } =0 =−1 Matrix M with eigenvectors u and v such that Mu = −u and Mv = v is of this form.

We can write (for δg traceless and [u, v] = 1) 1 [u, δgT v] = tr(vuT + uvT )ωδgT (418) 2 Lemma 39. Suppose that spinors u, v, u˜, v˜ satisfy [u, v] = 1, [u˜, v˜] = 1 (419) and • for all traceless matrices δg tr(vuT + uvT )ωδgT = tr(v˜u˜T + u˜v˜T )ωδgT (420)

~~∗ then there exists λ ∈ C such that u˜ = λu v˜ = λ−1v (421)~~

~~• and for all traceless matrices δg tr(vuT + uvT )ωδgT = − tr(v˜u˜T + u˜v˜T )ωδgT (422)~~

~~∗ then there exists λ ∈ C such that u˜ = λv v˜ = −λ−1u (423)~~

Proof. From lemma 38 matrices 1 1 = (vuT + uvT )ω, ˜ = (v˜u˜T + u˜v˜T )ω (424) M 2 M 2 are traceless thus M = M˜ . The result now follows from uniqueness of eigenvectors up to scaling and lemma 38. As for the second part we can write 1 1 − tr(v˜u˜T + u˜v˜T )ω = tr(vˆuˆT + uˆvˆT )ω (425) 2 2 where uˆ = v˜ and vˆ = −u˜. Notice that [uˆ, vˆ] = 1 and we can apply previous part.

~~79 D.1. Spinors u and v We introduce spinors u = sωηN can n¯, v = n (426) that satisﬁes † T T [u, v] = sn ηN can ω ωn = shn, niN can = 1 (427) Lemma 40. For any spinor r holds~~

~~hn, riN can = s[u, r], [n, r] = −sthu, riN can (428) Proof. The ﬁrst equality follows from~~

~~T T T s[u, r] = n¯ ηN can ω ωr = hn, riN can (429) T The second by ηN can ωηN can = tω T T T † T [n, r] = n ωr = tn ηN can ωηN can r = −stu ηN can r = −sthu, riN can (430)~~

~~† † where we used ηN can = ηN can and ω = −ω. Lemma 41. We have equality~~

~~T T † † T ηN can = ηN can (snn + stuu )ηN can (431) Proof. We will prove that~~

~~† † T I = (snn + stuu )ηN can (432) It is enough to check that action of both sides coincides on spinors v = n and u. We have by lemma 40~~

~~† † T (snn + stuu )ηN can n = sn hn, niN can +stu hu, niN can = n (433) | {z } | {z } =s =−st[n,n]=0 and similarly~~

~~† † T (snn + stuu )ηN can u = sn hn, uiN can +stu hu, uiN can = u (434) | {z } | {z } =s[u,u]=0 =−st[v,u]=st~~

~~Consider Sρ,S,Saux deﬁned in section 4.2. Lemma 42. On real manifold~~

~~T ∃ξ∈C g z = ξn (436)~~

~~80 Proof. From construction~~

T † T T T † T † T T T † T † T T ts(g z) ηN can (g z) = t(g z) ηN can nn ηN can (g z) + (g z) ηN can uu ηN can (g z) T † T † T T ≥ t(g z) ηN can nn ηN can (g z) (437) This is equivalent to

~~T T 0 = hu, g ziN can = st[n, g z] (440) that is gT z = ξn.~~

~~Lemma 43. On real manifold~~

 aux δgS = −tδgS (444) The latter can be computed T T T aux hg z, δg g ziN can T δgS = −t T T = −t[u, δg v] (445) hg z, g ziN can thus the total variation ρ δ S = δ Sρ + δ S = i +  [u, δgT v] (446) g g g 2 By (418) it can be written in the given form.

81 D.2. Subalgebra of the subgroup preserving the normal N We will use description of St(N) from 2.3. Diﬀerentiating (31) we obtain that the Lie subalgebra of SL(2, C) consists of these traceless matrices B that satisﬁes † † −1 BηN + ηN B = 0, B = −ηN B ηN (447) T Lemma 44. Suppose that M is a traceless matrix such that M belongs to the Lie algebra of the group preserving normal N, then the eigenvalues of M are either real or purely imaginary. Proof. We can always write

~~T T M = iλ(vu + uv )ω, [u, v] = 1 (448) with v being iλ and u being −iλ eigenvectors with λ ∈ C. The condition for M is −1 T † T M = −(ηN ) M ηN (449)~~

~~† means that M has the same eigenvalues as −M (±iλ) so either λ ∈ R or λ ∈ iR.~~

~~We will call M spacelike if its eigenvalues are purely imaginary (see section 5.2).~~

T Lemma 45. Suppose that M is a traceless matrix such that M belongs to the Lie algebra of the group preserving normal N and is spacelike then there exists λ > 0, and n spinor hn, niN = s ∈ {−1, 1} (450) such that T T M = iλ(vu + uv )ω, [u, v] = 1 (451) where u = sωηN n¯, v = n, (452) Proof. We can always write

~~T T M = iλ(vu + uv )ω, [u, v] = 1 (453) with v being iλ and u being −iλ eigenvectors with λ ≥ 0. The condition for M is~~

~~−1 T † T † M = −(ηN ) M ηN = tωηN (−ω)M ηN (454)~~

~~−1 T where t = det ηN = N · N from the identity t(ηN ) = ωηN (−ω). It is equivalent to~~

~~T T 2 T T 2 iλ(vu + uv )ω = itλωηN (−ω) (v¯u¯ + u¯v¯ )ηN ω (455)~~

~~82 We can write it in the form~~

~~(vuT + uvT )ω = −(v˜u˜T + u˜v˜T )ω (456) where we introduced u˜ = ωηN u¯, v˜ = tωηN v¯ (457) Let us notice that~~

~~T T T [u˜, v˜] = tu¯ ηN ωηN v¯ = u¯ ωv¯ = [u, v] = 1 (458) | {z } =tω~~

~~From lemma 39 there exists ξ ∈ C −1 u = ξωηN v¯, v = −ξ tωηN u¯ (459)~~

~~We have also the equality~~

~~† T T [u, v] = ξv ηn ω ωv = ξhv, viN (460) Normalizing v we can introduce spinor n such that~~

~~u = sωηN n¯, v = n (461)~~

~~Spinor n is unique up to a phase.~~

E. Characterization of the bivectors −1 2γ T µ We will now characterize π − γ−i Bij . Let us introduce a vector lij by an identity T µ † lijσˆµ = 2sijρij nijnij (462) It is null because the matrix on the right-hand side has rank one. It is future directed if sij = 1 and past directed if sij = −1. Contracting with σν and taking the trace we obtain

~~µ T µ lij = sijρijnijσ n¯ij (463)~~

~~Let us consider decomposition (introducing vector vij) can can can lij · Ni lij = vij + cNi , vij ⊥ Ni , c = can can (464) Ni · Ni We have~~

~~can T l · N = s ρ n η can n¯ = s ρ hn |n i can = ρ (465) ij i ij ij ij Ni ij ij ij ij ij Ni ij thus can 2 (lij · Ni ) 2 vij · vij = lij · lij − can can = −tiρij (466) Ni · Ni~~

~~83 Lemma 46. We have 2γ π(B0 ) = − T (467) ij γ − iBij 0 can can where Bij = ∗(vij ∧ Ni ) = ∗(lij ∧ Ni ). 0 0 Proof. The traceless matrix Bij = π(Bij) satisﬁes~~

0 1 1 can = i (ηl ηˆN can − ηN can ηˆl ) = −i (ηN can ηˆl − lij · N ) (468) Bij 4 ij i i ij 2 i ij i I For any traceless matrix K 2γ tr ij = iρijsijhnij, nijiN can (469) γ − iB K K i but also

0T † T can tr = − tr(iρijsijnijn η can − lij · N ) = −iρijsijhnij, nijiN can B ijK ij Ni i I K K i (470) 0 2γ T This equality shows that the traceless matrices Bij and − γ−i Bij are equal.

F. Geometric bivectors We will perform the following construction of the k form corresponding to n the simplex spanned on the points x0, . . . xk in R . n+1 Let us introduce auxiliary space R and in this space vectors  0    xi 0  .   .  y =  .  , and a =  .  (471) i  n     xi   0  1 1 and covector A = (0,... 0, 1). (472) Let us introduce k + 1 vectors ˜ Vα0···αk = yα0 ∧ · · · ∧ yαk (473)

~~n n+1 k-vectors in R can be identiﬁed with k-vectors Ω in R that satisfy~~

~~AxΩ = 0 (474) ˜ n From Vα0···αk we can produce a k-vector in R ˜ Vα0···αk = AxVα0···αk (475) ˜ as AxAxVα0···αk = 0.~~

~~84 We can check that~~

~~Axyα0 ∧ · · · ∧ yαk =~~

~~= Axyα0 ∧ (yα1 − yα0 ) · · · ∧ (yαk − yα0 ) = (yα1 − yα0 ) · · · ∧ (yαk − yα0 ) (476)~~

~~as the only nonzero last component is in yα0 . This gives (after restriction to n R )~~

Vα0···αk = (xα1 − xα0 ) · · · ∧ (xαk − xα0 ) (477) that is the volume k-vector of the k-simplex multiplied by k!. Let us notice that this k-vector depends on the order of α0, . . . , αk. As ˜ sgn σ Vα0···αk change by (−1) under permutation of points (even permutations preserves it) the same is true for Vα0···αk . Suppose that we have a simplex determined by points with indices 0, . . . n. We can deﬁne codimension 1 and 2 simplices by indicating which points we skip. Let us introduce i Vi = (−1) V0···ˆi···n (478) where ˆ· means omission,

( i+j+1 (−1) V0···î···ˆj···n i < j Bij = i+j (479) (−1) V0···ˆj···î···n i > j and similarly V˜i and B˜ij. With this definition Bij = −Bji. Theorem 9. The following holds X X Vi = 0, ∀i Bij = 0, (480) i j6=i

~~Proof. Let us consider ˜ ˜ B = AxVi = AxV0···ˆi···n (481) It can be written as~~

~~X sj X sj (−1) y0 ∧ · · · ∧ (Axyj) ∧ · · · ∧ yn = (−1) y0 ∧ · · · ∧ yˆj ∧ · · · ∧ yn (482) j6=i j6=i where sj is the number of site on which we contract.~~

~~ j + 1 j < i s = (483) j j j > i~~

~~The sum can be written as i X (−1) B˜ij (484)~~

~~85 However this n − 1 vector contracted with A is zero~~

~~˜ i X 0 = AxAxVi = AxB = (−1) Bij (485)~~

~~This ﬁnishes the proof of the second equality in (480). We prove the ﬁrst equality in similar way:~~

~~˜ X i ˜ X 0 = AxAxV0···n = Ax (−1) V0···ˆi···n = Vi. (486) i i~~

Let us restrict to the case of n = 4, Deﬁnition 12. The geometric bivectors of the 4 simplex determined by ver- 4 tices x0, . . . x4 ∈ R are ∆ Bij = Bij. (487) Let us consider a scaling transformation

~~∗ xi → λxi, λ ∈ R . (488) Under this transformation bivectors changes~~

~~∆ 2 ∆ Bij → λ Bij . (489) Let us notice that in particular inversion transformation λ = −1 preserves ∆ Bij .~~

~~F.1. Nondegenerate case~~

~~Let us now assume that xi do not lay in the hyperplane that is yi are linearly independent. We introduce a dual basis yˆi deﬁned by~~

~~yˆiyyj = δij (490)~~

~~Let us also introduce y˜i deﬁned by~~

~~yˆi =y ˜i + µiA, y˜iya = 0 (491)~~

~~n These covectors can be regarded as belonging to R . We have ˜ Vi = AxyˆixV0···n = −yˆixV0···n. (492)~~

~~n This can be written with the use of y˜i in terms of R~~

~~Vi = −y˜ixV0···n (493) and similarly Bij =y ˆjxyˆixV0···n =y ˜jxy˜ixV0···n. (494)~~

~~86 Lemma 47. We have n X yˆi = A (495) i=0 Proof. It is enough to check that both sides of equality give the same value contracted with the elements of the basis yj~~

~~n n X X yˆixyj = δij = 1 = Axyj (496) i=0 i=0~~

~~Thus we have n X y˜i = 0 (497) i=0~~

~~Covectors y˜i are conormal to subsimplices V0···ˆi···n. n Let us now add metric tensor on R . It deﬁnes also a scalar product on k-vectors by~~

~~k X sgn σ Y a1 ∧ · · · ak·b1 ∧ · · · bk := (−1) ai·bσi (498) σ∈Sk i=1~~

~~We can also introduce normalization of V0···n by~~

~~d ∆ 2 ∆ V0···n·V0···n = (−1) (V ol ) , V ol > 0 (499)~~

~~By the deﬁnition of Hodge star (see Appendix B)~~

~~∆ ∆ Bij = −V ol ∗(˜yj ∧ y˜i),Vi = −V ol ∗y˜i (500)~~

If codimension 1 subsimplices are not null we can introduce normal vectors Ni and positive numbers Wi and ti ∈ {−1, 1} such that 1 y˜ = W N ,N ·N = t (501) i V ol∆ i i i i i then we have 1 X B = − W W ∗(N ∧ N ), W N = 0 (502) ij V ol∆ i j j i i i i

Let us consider an altitude for our simplex from point xi. Its base we will denote by hi. Let us notice that as it lays inside the hyperplane of remaining points X X ∃αk,k6=i : αk = 1, αkxk = hk (503) k6=i k6=i

87 We have (identifying vectors and covectors using scalar product) X y˜i·(xi − hi) =y ˆix(yi − αkyk) = 1 ≥ 0 (504) k6=i Thus vectors ∆ Ni = −tiNi (505) ∆ are outer directed and introducing Wi = −tiWi (Wi > 0) we get X ∆ ∆ Wi Ni = 0 (506) i Let us notice that ∆ ∆ • Bij = −Bji ∆ 1 ∆ ∆ ∆ ∆ • Bij = − V ol∆ Wi Wj ∗(Nj ∧ Ni ) ∆ ∆ • Bij yNi = 0. Let us notice that for a spacelike face its area is equal to 1 A∆ = |B∆|. (507) ij 2 ij

~~• If Ni and Nj are of diﬀerent types and signature in the plane Ni ∧ Nj is mixed then ∆ ∆ 2 2 ∆ ∆ |Nj ∧ Ni | = − cosh θij,Ni ·Nj = sinh θij. (512)~~

• If signature in the plane Ni ∧ Nj is ++ then ∆ ∆ 2 2 ∆ ∆ |Nj ∧ Ni | = sin θij,Ni ·Nj = cos θij. (513) ∆ ∆ • If signature in the plane Ni ∧ Nj is −− then ∆ ∆ 2 2 ∆ ∆ |Nj ∧ Ni | = sin θij,Ni ·Nj = − cos θij. (514)

~~88 H. Dihedral angles~~

~~We will now describe R ∆ R ∆ in terms of dihedral angles. Let us notice Ni Nj ∆ ∆ that ∗Ni ∧ Nj is spacelike. We have~~

~~∆ ∆ q ∆ ∆ ∆ ∆ |∗Ni ∧ Nj | = (∗Ni ∧ Nj ) · (∗Ni ∧ Nj ). (515)~~

Let us introduce can can 0 Ni = Nj dij = can can , 1 Ni 6= Nj and also sin θ if the plane spanned by normals is euclidean s (θ) = (516) ij sinh θ if the plane spanned by normals is mixed cos θ if the plane spanned by normals is euclidean c (θ) = . (517) ij cosh θ if the plane spanned by normals is mixed

~~We will ﬁrst prove~~

~~Lemma 48. The following identities hold~~

~~dij ∆ ∆ ∆ ∆ sij(2θij) = (−1) 2(Ni ·Nj )|∗Ni ∧ Nj | (518) dij can can ∆ ∆ 2 cij(2θij) = (−1) (2ti tj (Ni ·Nj ) − 1) (519)~~

~~Proof. Let us consider cases. Using results of appendix G can can ∆ ∆ • ti = tj and Ni ·Nj > 0 in Lorentzian signature dij = 0~~

~~∆ ∆ ∆ ∆ 2(Ni ·Nj )|∗Ni ∧ Nj | = 2 cosh θij| sinh θij| = sinh(2θij), (520) can can ∆ ∆ 2 2 2ti tj (Ni ·Nj ) − 1 = 2 cosh θij − 1 = cosh(2θij), (521)~~

~~because θij > 0. can can ∆ ∆ • ti = tj and Ni ·Nj < 0 in Lorentzian signature dij = 0~~

~~∆ ∆ ∆ ∆ 2(Ni ·Nj )|∗Ni ∧ Nj | = −2 cosh θij| sinh θij| = sinh(2θij), (522) can can ∆ ∆ 2 2 2ti tj (Ni ·Nj ) − 1 = 2 cosh θij − 1 = cosh(2θij) (523)~~

~~because θij < 0. can can • ti 6= tj in Lorentzian signature dij = 1~~

~~∆ ∆ ∆ ∆ 2(−1)(Ni ·Nj )|∗Ni ∧ Nj | = 2 sinh θij cosh θij = sinh(2θij), (524) can can ∆ ∆ 2 2 (−1)(2ti tj (Ni ·Nj ) − 1) = 2 sinh θij + 1 = cosh(2θij). (525)~~

~~89 can can • ti = tj in Euclidean or split signature dij = 0~~

~~∆ ∆ ∆ ∆ can 2(Ni ·Nj )|∗Ni ∧ Nj | = 2t cos θij| sin θij| = sin(2θij), (526) can can ∆ ∆ 2 2 2ti tj (Ni ·Nj ) − 1 = 2 cos θij − 1 = cos(2θij), (527)~~

~~can because t θij ∈ (0, π).~~

~~∆ ∆ Let us introduce the inversion in the plane spanned by Ni ,Nj :~~

~~∗N∆∧N∆ π i j |∗N∆∧N∆| Oij = Ie i j . (528)~~

~~Lemma 49. The following holds for geometric normals:~~

~~N∆∧N∆ can can i j 2t t θij dij i j |∗N∆∧N∆| R ∆ R ∆ = O e i j , Ni Nj ij where θij is dihedral angle.~~

∆ Proof. We can restrict ourselves to two dimensional plane spanned by Ni ∆ and Nj . The connected component of the group of Lorentz transforma- ∆ ∆ 0 tions is generated by Ni ∧ Nj . Moreover for two elements G and G from connected component

~~G = G0 ⇐⇒ G − G−1 = G0 − G0−1 and tr G = tr G0 (529)~~

~~Let us notice that~~

~~dij −1 dij G = O R ∆ R ∆ ,G = O R ∆ R ∆ . (530) ij Ni Nj ij Nj Ni is in connected component and for any vector v in the plane R ∆ R ∆ v is Ni Nj equal to~~

~~can ∆ ∆ can ∆ ∆ can can ∆ ∆ ∆ ∆ v−2ti (Ni ·v)Ni −2tj (Nj ·v)Nj +4ti tj (Ni ·Nj )(Nj ·v)Ni , (531) thus~~

~~−1 dij can can ∆ ∆ ∆ ∆ G − G = 4(−1) ti tj (Ni ·Nj )(Ni ∧ Nj ) (532) dij can can ∆ ∆ 2 tr G = (−1) (4ti tj (Ni ·Nj ) − 2). (533)~~

~~Similarly expanding in the Taylor series~~

~~N∆∧N∆ can can i j ∆ ∆ 2t t θij i j |∗N∆∧N∆| Ni ∧ Nj 0 i j can can G = e = cij(2θij) + sij(2θij)ti tj ∆ ∆ (534) |∗Ni ∧ Nj |~~

~~90 and can can 0 0−1 2sij(2θij)ti tj ∆ ∆ G − G = ∆ ∆ Ni ∧ Nj , (535) |∗Ni ∧ Nj | 0 tr G = 2cij(2θij). (536) The equality now follows from lemma 48~~

~~dij ∆ ∆ ∆ ∆ sij(2θij) = (−1) 2(Ni ·Nj )|∗Ni ∧ Nj |, (537) dij can can ∆ ∆ 2 cij(2θij) = (−1) (2ti tj (Ni ·Nj ) − 1). (538) which holds for the dihedral angles.~~

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