The EPRL Intertwiners and Corrected Partition Function Wojciech Kamiski, Marcin Kisielowski, Jerzy Lewandowski

The EPRL intertwiners and corrected partition function Wojciech Kamiski, Marcin Kisielowski, Jerzy Lewandowski

To cite this version:

Wojciech Kamiski, Marcin Kisielowski, Jerzy Lewandowski. The EPRL intertwiners and corrected partition function. Classical and Quantum Gravity, IOP Publishing, 2010, 27 (16), pp.165020. 10.1088/0264-9381/27/16/165020. hal-00616260

HAL Id: hal-00616260 https://hal.archives-ouvertes.fr/hal-00616260 Submitted on 21 Aug 2011

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The EPRL intertwiners and corrected partition function

Wo jciech Kami´nski, Marcin Kisielowski, Jerzy Lewandowski

Instytut Fizyki Teoretycznej, Uniwersytet Warszawski, ul. Ho˙za 69, 00-681 Warszawa (Warsaw), Polska (Poland)

Abstract Do the SU(2) intertwiners parametrize the space of the EPRL solutions to the simplicity constraint? What is a complete form of the partition function written in terms of this parametrization? We prove that the EPRL map is injective in the general n-valent vertex case for the Barbero-Immirzi parameter less then 1. We ﬁnd, however, that the EPRL map is not isometric. In the consequence, a partition function can be deﬁned either using the EPRL intertwiners Hilbert product or the SU(2) intertwiners Hilbert product. We use the EPRL one and derive a new, complete formula for the partition function. Next, we view it in terms of the SU(2) intertwiners. The result, however, goes beyond the SU(2) spin-foam models framework and the original EPRL proposal.

PACS numbers: 04.60.Pp

I. INTRODUCTION

The main technical ingredient of the spin-foam models of 4-dimensional gravity is so called quantum simplicity constraint. Imposing suitably defined constraint on the domain of the (discrete) path integral turns the SU(2)×SU(2) (or SL(2,C)) BF theory into the spin-foam model of the Eu- clidean (respectively, Lorentzian) gravity [1]. The formulation of the simplicity constraint believed to be correct, or at least fitting gravity the best among the known approaches [5], is the one derived by Engle, Pereira, Rovelli, Livine (EPRL) [1] (and independently derived by Freidel and Krasnov [3]). The solutions to the EPRL simplicity constraint are EPRL SU(2)×SU(2) intertwiners. They are defined by the EPRL transformation, which maps each SU(2) intertwiner into a EPRL solution of the simplicity constraint. An attempt is made in the literature [1] to parametrize the space of the EPRL solutions by the SU(2) intertwiners. The vertex amplitude and the partition function of the EPRL model seem to written in in terms of that parametrization. The questions we raise and answer in this paper are: • Is the EPRL map injective, doesn’t it kill any SU(2) intertwiner? • Is the EPRL map isometric, does it preserve the scalar product between the SU(2) intertwiners? • If not, what is a form of a partition function derived from the SO(4) intertwiner Hilbert product written directly in terms of the SU(2) intertwiners, the preimages of the EPRL map? We prove the EPRL map is injective in the general, n valent vertex case and for the Barbero- Immirzi parameter |γ| < 1. The proof in the |γ| > 1hasalreadybeenprovidedin[4].Hence, there are as many SU(2)×SU(2) EPRL intertwiners as there are the SU(2) intertwiners. Owing to this result the SU(2) intertwiners indeed can be used to parametrize the space of the EPRL SU(2)×SU(2)intertwiners. However, we find the EPRL map is not isometric. In consequence, there are two inequivalent definitions of the partition function. One possibility is to use a basis in the EPRL intertwiners space orthonormal with respect to the SO(4) representations. And this is what we do in this paper. A second possibility, is to use the basis obtained as the image of an orthonormal basis of the SU(2) intertwiners under the EPRL map. The partition function derived in [1] corresponds to the second choice, whereas the first one is ignored therein. The goal of this part of our paper is pointing out the first possibility and deriving the corresponding partition function. After the derivation, we compare our partition function with that of EPRL on a possibly simple example. We conjecture, that the difference converges to zero for large spins. To make the paper intelligible we start presentation of the new results with the derivation of the partition function in Section II 4. The final formula for our proposal for the partition function 2

for the EPRL model is presented in Section II 5. The lack of the isometricity of the EPRL map is illustrated on speciﬁc examples in Section II 6. Finally, the proof of the injectivity of the EPRL map takes all the Section III B. This work is written in terms of the EPRL framework [1] combined with our previous paper [4] on the EPRL model.

II. OUR PROPOSAL FOR A PARTITION FUNCTION OF THE EPRL MODEL

1. Partition functions for the spin-foam models of 4-gravity: deﬁnition

Consider an oriented 2-complex κ whose faces (2-cells) are labeled by ρ with the irreducible representations of G =SU(2)×SU(2),

κ(2) 3 f 7→ ρ(f), and denote by H(f)thecorrespondingHilbertspace.Foreveryedge(1-cell)e we have the set/set of incoming/outgoing faces, that is the faces which contain e and whose orientation agrees/disagrees with the orientation of e.WeusethemtodeﬁnetheHilbertspace

H(e)= O H(f) ⊗ O H(f 0)∗. (2.1) f incoming f 0 outgoing

The extra data we use, is a subspace

H(e)SIMPLE ⊂H(e)(2.2) deﬁned by some constraints called the quantum simplicity constraints. In this paper, starting from section below, we will be considering the subspace proposed by Engle-Pereira-Rovelli-Livine. For the time being H(e)SIMPLE is any subspace of the space of invariants of the representation 0 ∗ Nf incoming ρ(f) ⊗ Nf 0 outgoing ρ(f ) ,

  SIMPLE 0 ∗ H(e) ⊂ Inv × O H(f) ⊗ O H(f ) . (2.3) SU(2) SU(2)   f incoming f 0 outgoing

SIMPLE 0 (The subspace He may be trivial for generic representations ρ(f)andρ(f ). Typically the simplicity constraints constrain also the representations themselves.) To every edge we assign the operator of the orthogonal projection onto H(e)SIMPLE,

SIMPLE 0 ∗ 0 ∗ Pe : O H(f) ⊗ O H(f ) → O H(f) ⊗ O H(f ) (2.4) f incoming f 0 outgoing f incoming f 0 outgoing

Our index notation is as follows (we drop ‘SIMPLE’ for simplicity)

A... A...B0... A0... (Pev) B... = PeA0...B...v B0... (2.5)

0 ∗ where the upper/lower indices of any vector v ∈ Nf incoming H(f) ⊗ Nf 0 outgoing H(f ) correspond to incoming/outgoing faces. In the operator Pe,foreachfacecontaininge,therearetwoindices,an upper and a lower one corresponding to the Hilbert space H(f). If f is incoming (outgoing), then we assign the corresponding lower/upper index of Pe to the beginning/end (end/beginning) of the edge. That rule is illustrated on Fig. 1. Now, for every pair of edges e and e0,whichbelongtoa same face f,andshareavertexv,thereisdeﬁnedthenaturalcontractionatv of the corresponding vertex of Pe with the corresponding vertex of Pe0 .Thecontractiondeﬁnesthefollowingtrace,

  O P SIMPLE 7→ Tr O P SIMPLE . (2.6) e  e  e∈κ(1) e∈κ(1) 3

FIG. 1: According to this rule, given an edge e ( e0 )containedinincoming(outgoing)facef,theindices of Pe ( Pe0 )correspondingtoH(f)areassignedtothebeginningand,respectively,totheendoftheedge. The oriented arc only marks the orientation of the polygonal face f.

Deﬁne partition function Z(κ, ρ)tobethefollowingnumber:

  Z(κ, ρ):= Y d(f)Tr O P SIMPLE A(boundary),d(f):=dimH(f)(2.7)  e  f∈κ(2) e∈κ(1) where A(boundary) is a factor that depends only on the boundary of (κ, ρ), and we derive it elsewhere.

2. Partition functions for the spin-foam models of 4-gravity: the amplitude form

The partition function is usually rewritten in the spin-foam amplitude form [6–8]. For that purpose one needs an orthonormal basis in each Hilbert space H(e)SIMPLE;denoteitselementsby SIMPLE ιe,α ∈H(e) , α =1, 2,...,n(e). Then

ne SIMPLE † Pe = X ιe,α ⊗ ιe,α (2.8) α=1 where by ‘†0,foreveryHilbertspaceH we denote the duality map

H3v 7→ v† ∈H∗ deﬁned by the Hilbert scalar product. In the Dirac notation

† ιe,α = |e, αi, and ιe,α = he, α|.

SIMPLE Substituting the right hand side of (2.8) for each Pe in (2.7), one writes the partition function in terms of the vertex amplitudes in the following way: • For each edge of κ choose an element of the corresponding orthonormal basis; denote this assignment by

ι : e → ιe,αe (2.9)

• At each vertex v ∈ κ(0): – take ι ,...,ι where e ,...,e are the incoming edges e1,αe1 em,αem 1 m † † 0 0 – take ι 0 ,...,ι 0 where e1,...,e 0 are the outgoing edges e1,αe0 e 0 ,αe0 m 1 m m0 4

– deﬁne the vertex amplitude ⊗ ⊗ ⊗ † ⊗ ⊗ † Av(ι):=Tr ιe1 ,α1 ... ιem,αm ιe0 ,α0 ... ιe0 ,α0 (2.10) 1 1 m0 m0 where ‘Tr’ stands for the contraction (2.6) and can be deﬁned by the evaluation of the spin-networks corresponding to the vertices (see [4]). • to each face f assign the face amplitude d(f). With this data, with the vertex amplitudes and face amplitudes, the partition function takes the famous form

Z(κ, ρ)= Y d(f) X Y Av(ι) A(boundary), (2.11) f∈κ(2) ι v∈κ(0) SIMPLE The result is independent of the choice of the orthonormal basis of each He .

3. The EPRL map

Now we turn to the EPRL intertwiners. For every edge e ∈ κ(1) H(e)SIMPLE = H(e)EPRL,P(e)SIMPLE = P (e)EPRL. (2.12) The definition of H(e)EPRL uses a fixed number γ ∈ R called Barbero-Immirzi parameter. The Hilbert space H(e)EPRL can be non-empty only if the 2-complex κ is labeled by EPRL represen- ± 1 − + × ∈ tations. A representation ρ =(ρj ,ρj )ofSU(2) SU(2), where j 2 N define the SU(2) ∈ 1 representations in the usual way, is an EPRL representation, provided there is k 2 N such that |1 ± γ| j± = k. (2.13) 2 Therefore, we will be considering here labellings of the faces of the 2-complex κ with EPRL representations

f 7→ ρ(f)=(ρj−(f),ρj+(f)), (2.14) each of which is deﬁned in the Hilbert space

H(f)=Hj−(f) ⊗Hj+(f). Each labeling deﬁnes also a labeling with SU(2) representations given by (2.13),

f 7→ ρk(f), (2.15) deﬁned in the Hilbert space Hk(f).Givenanedgee and the corresponding Hilbert space ∗ ∗ H(e)= O Hj−(f) ⊗Hj+(f) ⊗ O Hj−(f 0) ⊗Hj+(f 0), (2.16) f incoming f 0 outgoing the natural isometric embeddings

C : Hk →Hj− ⊗Hj+ (2.17) and the orthogonal projection operator P : H(e) →H(e)(2.18) onto the subspace InvSU(2)×SU(2) (H(e)) deﬁnes the natural map, the EPRL map:

EPRL ∗ ι :InvSU(2) O Hk(f) ⊗ O Hk(f 0) f incoming f 0 outgoing ∗ ∗ → O Hj−(f) ⊗Hj+(f) ⊗ O Hj−(f 0) ⊗Hj+(f 0) f incoming f 0 outgoing Its image is promoted to the Hilbert space (2.2),    EPRL EPRL ∗ H := ι Inv O H ⊗ O H 0 . (2.19) e  SU(2)  k(f) k(f ) f incoming f 0 outgoing 5

4. The problem with the EPRL intertwines

All the EPRL intertwiners can be constructed from the SU(2) intertwiners by using the EPRL map. The point is, that one has to be more careful while doing that. First, one has to make sure that the map ιEPRL is injective. If not, then Hilbert space of the SU(2)×SU(2) EPRL intertwiners is smaller then the corresponding space of the SU(2) intertwiners and we should know how big it is. For γ ≥ 1, the injectivity was proved in [4]. In the next section we present a proof of the injectivity for |γ| < 1. Secondly, one should check whether or EPRL not the map ι is isometric. Given an orthonormal basis Ie,1,...,Ie,ne of the Hilbert space ∗ InvSU(2) Nf incoming Hk(f) ⊗ Nf 0 outgoing Hk(f 0) ,wehaveacorrespondingbasis EPRL EPRL EPRL ι (Ie,1),...,ι (Ie,ne )ofthecorrespondingHilbertspaceHe .Thequestionis,whether or not the latter basis is also orthonormal. We show in Section II 6, this is not the case. The direct procedure would be to orthonormalize the basis. We propose, however, a simpler solution.

5. A solution

An intelligent way, is to go back to the formula (2.7) for the partition function and repeat the SIMPLE EPRL step leading to (2.11) with each projection Pe = Pe written in terms of the corresponding EPRL EPRL basis ι (Ie,1),...,ι (Ie,ne ). The suitable formula reads

ne EPRL a¯b EPRL EPRL † Pe = X he ι (Ie,a) ⊗ ι (Ie,b) (2.20) a,b=1

a¯b where he , a, b =1, ..., ne deﬁne the inverse matrix to the matrix

EPRL EPRL he,¯ba := (ι (Ie,b)|ι (Ie,a)) (2.21)

EPRL given by the Hilbert product (·|·)intheHilbertspace(2.16).(IntheDiracnotation,Pe = a¯b he |e, aihe, b|.) Now, we are in a position to write the resulting spin-foam amplitude formula for the partition function. It is assigned to a ﬁxed 2-complex κ and a ﬁxed labeling of the faces by the EPRL representations

f 7→ ρ(f)=(ρ − ,ρ + ). (2.22) jf jf The labeling is accompanied by the corresponding labeling with the SU(2) i

f 7→ ρkf , (2.23)

(1) EPRL according to (2.13). For every edge e ∈ κ ,inadditiontotheHilbertspaceHe ⊂ InvSU(2)×SU(2) (He), we also have its preimage, the Hilbert space

  ∗ Inv O H(k ) ⊗ O H . (2.24)  f kf0  f incoming f 0 outgoing

Therein, we ﬁx an orthonormal basis

Ie,a,a= 1, 2,...,ne. (2.25) To deﬁne the partition function we proceed as follows: • assign to every edge of κ apairofelementsofthebasis,

II : e 7→ (I , I† ), (2.26) e,ae e,be

more speciﬁcally, I is assigned to the end point and I† to the beginning point of e,and e,ae e,be we denote the assignment by the double symbol II; 6

• deﬁne for every edge an edge amplitude to be

bea¯e he(II):=h

• to every vertex v of κ assign the vertex amplitude with the trace deﬁned by Fig.1, (2.6) and (2.20)

  EPRL EPRL † Av(II):=Tr O ι (Ie,a ) ⊗ O ι (Ie0,b 0 )  e e  e incoming e0 outgoing

• to every face f assign the amplitude df Finally, the spin-foam amplitude formula for the partition function reads

Z(κ, ρ)=Y df X Y he(II) Y Av(II)A(boundary). (2.27) f II e v

The matrix (2.21) can be written in terms of the EPRL fusion coeﬃcients,

EPRL c−c+ ι (Ie,a)=:fea ιe,c− ⊗ ιe,c+ , (2.28) deﬁned by the decomposition into an orthonormal basis ιe,c− ⊗ ιe,c+ in ∗ ∗ InvSU(2)×SU(2) Nf incoming Hj−(f) ⊗Hj+(f) ⊗ Nf 0 outgoing Hj−(f 0) ⊗Hj+(f 0) . Then, we have

c−c+ c−c+ he,ab¯ = X fea feb . (2.29) c+,c−

We are in a position now, to compare our partition function with that of [1]. The partition function of [1] is given by replacing the matrix he,ab¯ in (2.27) with δab.Theexamplebelowgives quantitative idea about the diﬀerence between the two possible deﬁnitions of partition function for the EPRL model.

6. Example of the edge amplitude h¯ba showing that the EPRL map is not isometry

We will show in this section, that the EPRL map is not isometric. We calculate the edge amplitude deﬁned in the previous section in a simple example, and see that its matrix is not proportional to the identity matrix, or even not diagonal. Consider an edge at which exactly four faces meet. Assume the orientation of each face is opposite to the orientation of the edge.

We have an intertwiner I∈InvSU(2) (Hk1 ⊗Hk2 ⊗Hk3 ⊗Hk4 )assignedtotheendpointofthis edge. We choose a basis |kimii (the eigenvector of the third component of angular momentum operator with eigenvalue mi)ineachspaceHki , i ∈{1,...,4}.Wechoosearealbasisofthespace

InvSU(2) (Hk1 ⊗Hk2 ⊗Hk3 ⊗Hk4 )inthefollowingform[9]:

√ a a a+m k1 k2 a k3 k4 a 0 (Ia)k1m1k2m2k3m3k4m4 = 2a +1 X X (−1) δm,−m 0 m1 m2 m m3 m4 m m=−a m0=−a

j1 j2 j3 where is the Wigner 3j-Symbol, δm,m0 is the Kronecker Delta. m1 m2 m3

Let ιa+ ⊗ ιa− be the basis of the space Inv H + ⊗ ...⊗H + ⊗ Inv H − ⊗ ...⊗H − .The j1 j4 j1 j4 EPRL intertwiner ι (Ia)expressedinthisbasistakesthefollowingform(weskipinthissectionthe subscript e indicating the dependence on edge):

EPRL a+a− ι (Ia)=fa ιa+ ⊗ ιa− 7

a+a− where fa are real and are known as fusion coeﬃcients [1]. The tensor hab¯ could be expressed in terms of them:

EPRL EPRL a+a− a+a− hab¯ =(ι (Ia)|ι (Ib)) = X fa fb a+a−

1 As an example we give the result of the calculation of the ha¯b matrix for γ = 2 ,j1 =2, j2 =4, j3 = 4,j1 =2;a,b∈{2,...,6}):

√ √ √ 2265 5  53723 − 7 5093 5 − 3 55  175616√ 50176 1053696 25088√ √0 2265 5 45 11 3 13  − 7 117853 − 12805√ 7 − 7   50176 501760 301056 7 7168 8960   √ √ √  5093 5 − 12805√ 741949 − 781 11 5 13  1053696 3512320 752640 5376   √ 301056√ 7 √   45 11   − 3 55 7 − 781 11 583 0   25088 7168√ 7√52640 2560   3 13  − 7 5 13 13 0 8960 5376 0 40 We used the analytic expression for the fusion coeﬃcient presented in [9]. Clearly this matrix is ba¯ nondiagonal. It shows that the EPRL map is not isometric. The edge amplitude he is given by the inverse matrix: √ 31728718 7  46976713 5 − 75194882√ − 3865813√ 13066606√  14064543 70322715 257849955 5 70322715 55 773549865 65 √ √ √ √ 31728718 7 318127222 7 7212044 7  5 7682388364 67078172 7 11 13   −   70322715 1758067875√ 586022625 1758067875 1758067875   − 75194882√ 67078172 7 112636131412 1305090458√ − 12462294236√   257849955 5 586022625√ 23636245875 6446248875 11 70908737625 13   318127222 7   − 3865813√ − 11 1305090458√ 85031744497 − 192524374√   1758067875 19338746625   70322715 55 √ 6446248875 11 19338746625 143  7212044 7  13066606√ 13 − 12462294236√ − 192524374√ 8510451083428  773549865 65 1758067875 70908737625 13 19338746625 143 2765440767375

III. INJECTIVITY OF THE MAP I7→ιEPRL(I)

This part of the paper is devoted to the injectivity of EPRL intertwiner. More explicitly, we will prove the result stated in 1.

A. Statement of the result

We assume that γ ∈ R and |γ| < 1. Suppose that 1 (k ,...,k ) ∈ N 1 n 2 1 ± γ 1 ∀ j± = k ∈ N. i i 2 i 2 We consider the EPRL map

EPRL − − ι :Inv(Hk1 ⊗···⊗Hkn ) → Inv H + ⊗···⊗H + ⊗ Inv H ⊗···⊗H (3.1) j1 jn j1 jn + + − − EPRL k1C1 knCn j1 D1...jn Dn j1 E1...jn En ι (I) + + − − = I C + − ···C + − P + + P − − j A1...jn Anj B1...jn Bn k1C1...knCn 1 1 j1 D1j1 E1 jn Dnjn En j1 A1...jn An j1 B1...jn Bn with P standing for the orthogonal projections onto the subspaces of the SU(2) invariants of the Hilbert spaces H + ⊗···⊗H + ,andrespectively,H − ⊗···⊗H − . j1 jn j1 jn Now we can state our result. 8

∈ 1 H ⊗···⊗H Theorem 1. For any sequence (k1,...,kn) 2 N such that Inv ( k1 kn ) is nontrivial the map

EPRL + + − − ι :Inv(Hk1 ⊗···⊗Hkn ) → Inv H ⊗···⊗H ⊗ Inv H ⊗···⊗H j1 jn j1 jn is injective.

B. Proof of the theorem

In order to make the proof transparent, we divide it into subsections. In subsection III B 1 some auxiliary deﬁnitions are introduced. We state also an inductive hypothesis, that will be proved in subsection III B 5. The injectivity of EPRL map follows from that result. The main technical tool of the proof is placed in subsection III B 3, where the map 7 is deﬁned.

1. Auxiliary deﬁnitions

Let us introduce some notations Deﬁnition 2. For x ∈ R we deﬁne • − 1 1 [x]+ as the only half integer number in the interval x 4 ,x+ 4 • − 1 1 [x]− as the only half integer number in the interval x 4 ,x+ 4 and

Deﬁnition 3. Asequenceofhalfnaturalnumbers(k1,...,kn) satisﬁes triangle inequality if

∀i ki ≤ X kj . j=6 i One can deﬁne map ι under condition ± ± Con n: Sequences of half natural numbers (k1,...,kn)and(j1 ,...,jn )aresuchthat

• (k1,...,kn)satisﬁestriangleinequality, + − • ji + ji = ki for i =1, ...,n, • ± 1±γ 6 ji = 2 ki for i =1and

+ 1+γ − 1 − γ j1 = k1 ,j1 = k1 2 + 2 − or

+ 1+γ − 1 − γ j1 = k1 ,j1 = k1 . 2 − 2 + Let us deﬁne

+ + − − ιk1...kn :Inv(Hk1 ⊗···⊗Hkn ) → Inv H ⊗···⊗H ⊗ Inv H ⊗···⊗H j1 jn j1 jn + + − − k1C1 knCn j1 D1...jn Dn j1 E1...jn En ι (I) + + − − = I C + − ···C + − P + + P − − ... j A1...jn Anj B1...jn Bn k1C1...knCn 1 1 j1 D1j1 E1 jn Dnjn En j1 A1...jn An j1 B1...jn Bn with P standing for projections onto invariant subspaces. We will use the letter ιk1...kn for all ± ± sequences (k1,...,kn), (j1 ,...,jn )ifitdonotcauseanymisunderstanding. We will base our prove on the following inductive hypothesis: ± ± Hyp n: Suppose that (k1,...,kn)and(j1 ,...,jn )satisfyconditionCon n and that

I∈Inv (Hk1 ⊗···⊗Hkn ). Then, there exists

φ ∈ Inv H + ⊗···⊗H + ⊗ Inv H − ⊗···⊗H − j1 jn j1 jn

such that hιk1 ...kn (I),φi6=0. This in fact proves the injectivity. 9

2. Useful inequalities

∈ ∈ 1 Both [x]± are increasing functions and satisfy (x, y R, j 2 N)

a) [x + j]=[x]+j and [j]± = j,

b) if x>ythen [x]− ≥ [y]+ and if x ≥ y then [x]+ ≥ [y]−

c) if x + y ∈ Z then [x]+ +[y]− = x + y

d) if x + y ≥ j then [x]+ +[y]− ≥ j − 1 ≥ − 1 In order to prove the last point, we notice that [x]+ >x 4 and [y]− y 4 so [x]+ +[y]− > − 1 ≥ − 1 ≥ x + y 2 j 2 but as j is an half integer number [x]+ +[y]− j. 1+γ ∈ 1 Lemma 4. Suppose that (k, l, j) satisﬁes triangle inequality and that 2 k 2 N then

• 1+γ 1+γ 1+γ both triples 2 k, 2 l± , 2 j± satisfy triangle inequalities if k + l = j or k + j = l

• 1+γ 1+γ 1+γ both triples 2 k, 2 l± , 2 j∓ satisfy triangle inequalities if k + l>jand k + j>l.

1+γ 1+γ 1+γ Proof. In the ﬁrst case suppose that k + l = j holds, then 2 k + 2 l± = 2 j± that proves triangle inequality. 1+γ 1+γ 1+γ In the second case, we restrict our attention to 2 k, 2 l+ , 2 j− .Wehave

• 1+γ 1+γ ≥ 1+γ 1+γ 1+γ 1+γ 2 k + 2 j− 2 l+ because 2 k + 2 l> 2 j, • 1+γ k + 1+γ l ≥ 1+γ j because 1+γ k + 1+γ j ≥ 1+γ l, 2 2 + 2 − 2 2 2 • 1+γ ≤ 1+γ 1+γ 2 k 2 l+ + 2 j− from the property d) listed above.

1+γ 1+γ 1+γ The case of 2 k, 2 l− , 2 j+ is analogous.

1+γ ∈ 1 Lemma 5. Suppose that (k1,...,kn) satisﬁes triangle inequality and that 2 ki 2 N for i = 1+γ 1+γ 1+γ 2,...,n, then 2 k1± , 2 k2,..., 2 kn also satisfy triangle inequalities.

Proof. That follows from the monotonicity of functions [x]± and the fact that in the inequality 1+γ 1+γ k ≤ X k 2 i 2 j j=6 i all terms but one are half-integer.

Lemma 6. Suppose that (k1,...,kn) satisﬁes triangle inequality then

Inv (Hk1 ⊗···⊗Hkn ) is nontrivial.

Proof. We will ﬁnd an kα such that both (kα,k1,k2)and(kα,k3,...,kn)satisfytriangleinequalities. By induction there would be

0 =6 φ ∈ Inv (Hkα ⊗Hk3 ⊗···Hkn ) , and then

0 =6 CkαA φ k1A1k2A2 kαAk3A3...knAn 10

proves nontriviality. Now we extract conditions on kα from triangle inequalities (we assume for simplicity that k1 ≥ k2)

k1 + k2 ≥ kα ≥ k1 − k2

X ki ≥ kα ≥ ki − X kj ,i≥ 3 i≥3 j=6 i, j≥3

For the existence of such kα we need only to show that

k1 + k2 ≥ ki − X kj ,i≥ 3 j=6 i, j≥3

X ki ≥ k1 − k2 i≥3 but these are exactly conditions for (k1,...,kn)tosatisfytriangleinequality.

3. Important maps

Every I∈Inv (Hk1 ⊗···⊗Hkn )maybeuniquelywrittenas

I = X CkαAα Ikα , k1A1k2A2...knAn k1A1k2A2 kαAαk3A3...knAn kα

kα where I ∈ Inv (Hkα ⊗Hk3 ⊗···⊗Hkn ). Summation is taken over such kα that (kα,k1,k2)and (kα,k3,...,kn)satisfytriangleinequality.

This gives us decomposition of Inv (Hk1 ⊗···⊗Hkn )intoorthogonalsubspaces

⊕αHα,

where each Hα is isomorphic to Inv (Hkα ⊗Hk3 ⊗···⊗Hkn ). Let us deﬁne maps which assign these partial isometries

kα Qkα :Inv(Hk1 ⊗···⊗Hkn ) → Inv (Hkα ⊗Hk3 ⊗···⊗Hkn ) ,Qkα I = I

Adjoints to them are embeddings Q∗ . kα

∗ Q Ikα = CkαAα Ikα . kα k1A1k2A2...knAn k1A1k2A2 kαAαk3A3...knAn

These maps are also well deﬁned in a case that α does not occur in the decomposition ⊕Hα but

(kα,k1,k2)satisﬁestriangleinequalities.ThenthespaceInv(Hkα ⊗Hk3 ⊗···⊗Hkn )istrivial and the maps Q and Q∗ too. kα kα ± ± + − Let us ﬁx (k1,...,kn)and(j1 ,...,jn )satisfyingtriangleinequalitiesandsuchthatji +ji = ki. ± + − ± ± ± Lemma 7. Suppose kα,jα are such that jα + jα = kα and (kα,k1,k2) and (jα ,j1 ,j2 ) satisfy triangle inequalities. Then there exists an operator

G + − :InvH + ⊗H + ⊗···⊗H + ⊗ InvH − ⊗H − ···⊗H − → kαjα jα jα j3 jn jα j3 jn

→ InvH + ⊗···⊗H + ⊗ InvH − ⊗···⊗H − j1 jn j1 jn

+ + + such that for all I∈Inv Hkβ ⊗Hk3 ⊗···⊗Hkn and φ ∈ Inv H ⊗H ⊗···⊗H ⊗ jα j3 jn

Inv H − ⊗H − ···⊗H − jα j3 jn

 hιkαk3...kn I,φi,kβ = kα ∗ hι Q I,G + − φi =  0,k>k k1...kn kβ k j j β α α α α ∗  ,kβ

Proof. We deﬁne G ± as kαjα

+ − jα Aα jα Bα G ± (φ) + + − − = βC + + C − − φ + + − − , kαjα j A1...jn Anj B1...jn ...Bn jα Aαj A3...jα Bαj B3... 1 1 j1 A1j2 A2 j1 A1j2 A2 3 3 with β nonzero constant to be deﬁned later. Let us compute

∗ hι Q I,G ± φi k1 ...kn kβ kαjα

In the deﬁnition of ι one can skip projection because both φ and G ± φ are invariants. Let us kαjα ∗ write explicitly hι Q I,G ± φi.Wehave k1...kn kβ kαjα

+ + − − + + − − k A j B1j B2 j C1j C2 β k1A1 knAn 1 2 1 2 jα Bj3 B3...jα Cj3 C3... I C C + − ···C + − βC + C − φ = kβ A...knAn k1A1k2A2 j1 B1j1 C1 jn Bnjn Cn jα B jα C + + − − kβ A k1A1 k2A2 j1 B1j2 B2 j1 C1j2 C2 = βC C − C − C C − k1A1k2A2 + + + j1 B1j1 C1 j2 B2j2 C2 jα B jα C + + − − k3A3 knAn jα Bj3 B3...jα Cj3 C3... Ikβ A...knAn C + − ···C + − φ . j3 B3j1 C3 jn Bnjn Cn We need only to show that

kβ A + − k A j+B j+B j− C j−C ( C ,k = j + j β k1A1 k2A2 1 1 2 2 1 1 2 2 j+Bj−C β α α βC C + − C + − C + C − = α α k1 A1k2A2 j B1j C1 j B2j C2 jα B jα C + − 1 1 2 2 0 kβ >jα + jα .

+ − The second equality is obvious because there exists no intertwiner if kβ >jα + jα .Theﬁrstwill be proved in the next subsection III B 4 1.

4. Relation among intertwiners

+ + − − kβ A k1A1 k2A2 j1 B1j2 B2 j1 C1j2 C2 kβ A We know that C C + − C + − C + C − is proportional to C + − . k1A1k2A2 j1 B1j1 C1 j2 B2j2 C2 jα B jα C jα Bjα C In order to prove that the factor of proportionality is nonzero we will show that

+ + − − + − kαA k1A1 k2A2 j1 B1j2 B2 j1 C1j2 C2 jα Bjα C C C + − C + − C + C − C =06 k1A1k2A2 kαA j1 B1j1 C1 j2 B2j2 C2 jα B jα C and that would be β−1.Infactitisenoughtoshowthattheintertwiner

+ + − − + − k1A1 k2A2 j1 B1j2 B2 j1 C1j2 C2 jα Bjα C C + − C + − C + C − C kαA j1 B1j1 C1 j2 B2j2 C2 jα B jα C is not equal zero or equivalently the same for

+ − + − + − k1A1 j2 B2j2 C2 j1 B1 j1 C1 jα Bjα C C + − C C + + C − − C k2A2 kαA j1 B1j1 C1 j2 B2jα B j2 C2jα C We only sketch the proof. First of all, we remind some facts about intertwiners and diagrammatic notation. k H⊗2k H Let P stands for projection onto symmetric subspace in 1/2 equivalent to k (k is a half natural number).

k H⊗2k →H⊗2k P : 1/2 1/2 .

H H⊗2k In this subsection we regard k as this subspace of 1/2 .Letusalsodenotethecanonicalmap : C 7→ H1/2 ⊗H1/2.

1 Although it seems to be standard, we include it for a sake of completeness. 12

k1 P

k12 k13 k23

P P

k2 k3

k1 FIG. 2: An intertwiner proportional to C , k12 = k1 + k2 − k3 and etc. k2k3

k1 k2 = k1 + k2

2k1 2k2 2k1+2k2 FIG. 3: An equality between H1/2 ⊗H1/2 and H1/2 .

k1k2

P k1k2k3 k 3 = P k1k2

P k1k2k3

k1k2k3

FIG. 4: An equality P k1+k2+k3 ◦ P k1+k2 ⊗ I = P k1+k2+k3 .

k1k2k3 k1 k2 The intertwiner C : C →Hk1 ⊗Hk2 ⊗Hk3 is proportional to P ⊗ P ⊗ P k3 ⊗2k1+2k2+2k3 .

In the diagrammatic language this can be depicted as on figure 2. We skip the index k in P k on H⊗2k the diagrams for notations’ brevity. The line with symbol k denotes 1/2 . We have to notice important properties,that in diagrammatic language is shown on figures 3 and 4. Our intertwiner can be written as shown on the figure 5. + − Now using properties mentioned earlier we see that one can merge jij with jij into kij and that intertwiner is equal to intertwiner shown on the figure 2 and is nonzero.

5. Inductive steps

± 0 Induction starts with n =1.Inthiscasek1 =0andsoalsoj1 =0.Themapι0 = C00 : C → C ⊗ C is the identity. Suppose now, that we have just proved Hyp n − 1.

In the decomposition of given I∈Inv (Hk1 ⊗···⊗Hkn )intosubspacesHα we choose minimal kα such that Qkα I is nonzero. We know, by lemmas 4 and 5 that for either

+ − 1+γ 1 − γ ! (jα ,jα )= kα , kα 2 + 2 − 13

+ − j1 j1

P P − j12 + + + j12 j2α j1α − j1α j− P P 2α P P j+ 2 j+ − α − j2 jα

P P

k2 kα

j+B j−C j+B j−C + − FIG. 5: Intertwiner proportional to Ck1A1 C 2 2 2 2 C 1 1 C 1 1 Cjα Bjα C . + − k2A2 + + − − kαA j1 B1j1 C1 j2 B2jα B j2 C2jα C or for

+ − 1+γ 1 − γ ! (jα ,jα )= kα , kα 2 − 2 + all necessary assumptions of lemma 7 are satisﬁed. From the Hyp n − 1forthesequences ± ± ± (kα,k3,...kn), (jα ,j3 ,...jn )weknowthatthereexists

φ ∈ Inv H + ⊗···⊗H + ⊗ Inv H − ⊗···⊗H − jα jn jα jn such that

hιkαk3...kn Qkα I,φi =06 . We have

∗ Dι I,G ± φE = X Dι Q Q I,G ± φE = hι Q I,φi =06 . k1...kn kαjα k1...kn kβ kβ kαjα kαk3...kn kα kβ ≥kα This ﬁnishes inductive step and the proof.

IV. SHORT DISCUSSION

We studied in this paper properties of the solutions to the EPRL simplicity constraints which were derived in [1]. We also pointed out two different possibilities of defining the partition function out of them. Our definition is (2.7). It uses only the subspace of the SO(4) intertwiners which solve the EPRL simplicity constraint. The comparison and contrast between our definition and that of [1] is provided by (2.27) and the comments which follow that equality. The difference follows from the fact proven in Section II 6 above, that the EPRL map that is not isometric. The example considered in that section gives also quantitative idea of the difference. The question of which definition of the partition function is correct can not be answered at this stage. Finally, we studied the “size of the space of the EPRL solutions”. We have shown that the EPRL map does not kill any SU(2) intertwiner. The proof is presented in detail in Section III.

Acknowledgments We would like to thank John Barrett, Jonathan Engle for coming to Warsaw and delivering lectures on the SFM. JL acknowledges the conversations with Laurent Freidel, and 14

exchange of e-mails with John Baez, Carlo Rovelli, Roberto Pereira and Etera Livine. MK would like to thank Jonathan Engle from Albert Einstein Institute in Potsdam and Carlo Rovelli, Matteo Smerlak from Centre de Physique Theorique de Luminy for discussions and hospitality during his visits at their institutes. The work was partially supported by the Polish Ministerstwo Nauki i Szkolnictwa Wyzszego grants 182/NQGG/ 2008/0, 2007-2010 research project N202 n081 32/1844, the National Science Foundation (NSF) grant PHY-0456913, by the Foundation for Polish Science grant Master and a Travel Grant from the QG research networking programme of the European Science Foundation.

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