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On the relation between covariant and canonicalQuantum Gravity [arXiv:gr-qc/1307.5885]
Antonia Zipfelwith Thomas Thiemann
Antonia Zipfel (FUW) ILQGS, 15.04.2014 1 / 30
Grant: 2012/05/E/ST2/03308
Motivation
Canonical LQG
Ts(A) = Tr[∏l
hjl(A)∏n
ιn]
Hkin =⊕γ∈ΣHkin,γ
Covariant LQG
v
m
ji
jk
jj
Z [κ] =∑c∏v
Av∏f
Af × B
H∂κ ≃Hkin,γ
Is it possible to merge covariant and canonical LQG?
Antonia Zipfel (FUW) ILQGS, 15.04.2014 2 / 30
The general idea
v
m
ji
jk
jj
⟨Ts ∣Z [κ]∣Ts′⟩
∂κ = γs ∪ γs′
Rovelli,Reisenberger,Barrett, Crane, Engle,
Pereira, Livine, Freidel, Krasnov,
Kaminski, Kisielowski, Lewandowski, . . .
Heuristic Idea
δ(H) = ∫ exp (itH)dt ↭∑κ
Z [κ]±
Feynmandiagrams
[Reisenberger, Rovelli]
“Physical Scalar Product”
η[Ts](Ts′) ∶= ∑κ∶γs′→γs
⟨Ts ∣Z [κ]∣Ts′⟩
Antonia Zipfel (FUW) ILQGS, 15.04.2014 3 / 30
Main Results
The extension of the [Engle,Pereira,Rovelli, Livine]-[Freidel, Krasnov]-model by[Kaminski, Kisielowski, Lewandowski] enables to investigate whether spin foams can beused to construct a rigging map for any of the presently definedHamiltonian constraint operators.We suggest a ‘rigging map’ closely following the ideas of [Reisenberger, Rovelli].
In the analysis of the resulting object we are able to identify an elementaryspin foam transfer matrix that allows to generate any finite foam as a finitepower of the transfer matrix.
It transpires that the postulated map does not define a projector on thephysical Hilbert space. However, it might be possible to construct a properrigging map in terms of a modified transfer matrix.
Antonia Zipfel (FUW) ILQGS, 15.04.2014 4 / 30
Outline
1 A spin foam rigging mapRigging maps and spin foams - a brief reviewProposal for a spin foam ‘rigging map’Properties
2 Why the would-be ‘rigging map’ is not a proper rigging mapTime orderingFactorizationWhy η is not a rigging map
3 Discussion and outlookWhat is going wrong?Open questions
Antonia Zipfel (FUW) ILQGS, 15.04.2014 5 / 30
A spin foam rigging map Rigging maps and spin foams - a brief review
Rigging maps
How to solve the constraints of canonical QG?
Generalized solution of C
.. is a state l ∈ D∗kin s.t [C∗l] (f ) ∶= l(C †f ) = 0 ∀f ∈ Dkin
Dkin ⊂Hkin dense domain, algebraic dual D∗kin
Construction principle? What is the scalar product?
Rigging mapGiven η ∶Hkin → D
∗kin s. t. ⟨η(f ), η(f ′)⟩phys = [η(f ′)](f )
Antonia Zipfel (FUW) ILQGS, 15.04.2014 6 / 30
A spin foam rigging map Rigging maps and spin foams - a brief review
Euclidean spin foam models
1 Action: SBF = ∫MTr[(B +
1β∗B) ∧ F ] + Simplicity constraint
M space-time, B bivector, F curvature of spin(4)-connection on M, β Barbero-Immirzi parameter
2 Discretize: T triangulation ofM with dual 2-complex κT ↝ SBF [κT]
3 Quantize ∫ DP eiSBF [κT] ↝ ABF [κT, ψin, ψout] = ⟨ψin,ZBF [κT ]ψout⟩
4 Implement simplicity constraint
Result
AEPRL−FK
[κT, ψin, ψout] = ⟨ψin,Z [κT]ψout⟩ =
∑j±f,ιEPRLe
∏f
Af ∏e∈κint
Qe ∏v∈κint
Av(j±f , ι
EPRLe ) B(ψin, ψout)
j± = ∣1±β∣2 j, ιEPRL intertwiner coupling j and (j+, j−)
Antonia Zipfel (FUW) ILQGS, 15.04.2014 7 / 30
A spin foam rigging map Proposal for a spin foam ‘rigging map’
Merging canonical and covariant LQG - Prerequisites
Can we construct η[Ts](Ts′) = ∑κ∶γs′→γs
⟨Ts ∣Z [κ]∣Ts′⟩ ?
Need identification of boundary states ψin, ψout with states in Hkin
Canonical LQGContinuum theory
Quantize point separatingsub-algebra of Poisson algebra↝ all possible graphs in Σ
Hamiltonian ↝ 3-valent nodes
EPRL-FK Model (Eucl.)Discrete theory
κ dual to triangulation T ofM,↝ boundary graphs dual totriangulation of Σ
at least 4-valent nodes
Other issues: Path integral measure? Fate of observables?Implementation of simplicity constraint?
Antonia Zipfel (FUW) ILQGS, 15.04.2014 8 / 30
A spin foam rigging map Proposal for a spin foam ‘rigging map’
Abstract foams - The KKL-model
Spin foams - a tool to determine the physical scalar product ?
KKL-model[Kaminski, Kisielowski, Lewandowski]
Achievements:1 Z [κ] defined for arbitrary 2-complexes
allowing arbitrary boundary graphs ∂κ
2 For certain choice of β: H∂κ ≃Hkin,∂κ
In the following we will consider abstract foams
Abstract foam κ: p.l. homeomorphic to a 2-complex {f , e, v}whose boundary ∂κ is the disjoint union ofclosed graphs bordering κ
γ borders κ: exist 1-to-1 affine map γ × [0,1]→ κ
Antonia Zipfel (FUW) ILQGS, 15.04.2014 9 / 30
A spin foam rigging map Proposal for a spin foam ‘rigging map’
Proposed rigging-map
Abstract equivalence
Two embedded spin nets belong to the same abstract equivalence class[s]A if they are embeddings of the same abstract spin net sA.
A spin foam rigging map
η[Ts](Ts′) = ∑[s′]A∈NA
η[s]A,[s′]AL[s′]A with η[s]A,[s′]A = ∑κA(sA,s′A)
Z [κA(sA, s′A)]
and L[s′]A = η[s′]A ∑s∈[s′]A
⟨Ts , ⋅ ⟩
Z [κA(sA, s′A)] ∶= ∑
jf ,ιe
∏f
Af ∏e∈κint
Qe ∏v∈κint
Av(jf , ιe)∏l∈∂κ
δjl ,jfl ∏n∈∂κ
διn,ιen
[Kaminski, Kisielowski, Lewandowski], [Ding, Han, Rovelli],[Bahr, Hellmann, Kaminski, Kisielowski, Lewandowski]
Antonia Zipfel (FUW) ILQGS, 15.04.2014 10 / 30
A spin foam rigging map Properties
Properties of Z [κ]
Z [κA(sA, s′A)] ∶= ∑
jf ,ιe
∏f
Af ∏e∈κint
Qe ∏v∈κint
Av(jf , ιe)∏l∈∂κ
δjl ,jfl ∏n∈∂κ
διn,ιen
[Kaminski, Kisielowski, Lewandowski], [Ding, Han, Rovelli],[Bahr, Hellmann, Kaminski, Kisielowski, Lewandowski]
Can be defined s.t. Z [κA(sA, s′A)] is invariant if . . .
. . . faces labeled by trivial representation are added or removed
. . . internal edges splitting a face are added or removed
. . . 2-valent vertices are added or removed
Z [κA](sA, s′A) ∶Hkin,γ(s′
A) →Hkin,γ(sA) is cylindrically consistent!
Furthermore: ∀ γ ∃ κ0γ s.t. Z [κ0
γ] ∶Hkin,γ →Hkin,γ with Z [κ0γ] = 1γ
Antonia Zipfel (FUW) ILQGS, 15.04.2014 11 / 30
A spin foam rigging map Properties
Minimal foams
NoteSince adding faces with jf = 0, splitting faces and edges leave theamplitude/spin net function invariant one should only sum over equivalenceclasses in the rigging map .
DefinitionAn abstract foam/graph is called minimal iff it cannot be obtained fromanother foam/graph by subdivisions.
RemarkMinimal representative is not unique . . .
. . . but amplitude does not depend on choice
Can always fix a representative for each class and only sum over those.
Antonia Zipfel (FUW) ILQGS, 15.04.2014 12 / 30
A spin foam rigging map Properties
Gluing
v1v2
v3
v1v2
v3
Suppose κ1 ∩ κ2 = ∂κ1 ∩ ∂κ2 = γ then
∑s(γ)
Z [κ1(s, s)]Z [κ2(s, s′)] = Z [κ1 ♯κ2(s, s
′)]
Antonia Zipfel (FUW) ILQGS, 15.04.2014 13 / 30
Why the would-be ‘rigging map’ is not a proper rigging map
Outline
1 A spin foam rigging mapRigging maps and spin foams - a brief reviewProposal for a spin foam ‘rigging map’Properties
2 Why the would-be ‘rigging map’ is not a proper rigging mapTime orderingFactorizationWhy η is not a rigging map
3 Discussion and outlookWhat is going wrong?Open questions
Antonia Zipfel (FUW) ILQGS, 15.04.2014 14 / 30
Why the would-be ‘rigging map’ is not a proper rigging map Time ordering
Spliting
v1v2
v3
v1v2
v3
Idea: Use this to split big complexes to get a better control on
η[Ts](Ts′) ∶= ∑κ∶γs′→γs
⟨Ts ∣Z [κ]∣Ts′⟩
Antonia Zipfel (FUW) ILQGS, 15.04.2014 15 / 30
Why the would-be ‘rigging map’ is not a proper rigging map Time ordering
Time ordering
v1v2
v3
Definitionv ∈ κint s.t. ∃n ∈ γi and ∃e ∈ κint withs(e) = n and t(e) = v
↝ vertex of first generation
Inductively: Vertex of nth generation
Only final graph ⇒ count backwards
∂κ = ∅ all v ∈ κ of first generation
TheoremA minimal foam κ can be uniquely split into minimal foams κi , called ‘onetime step’ foams, containing only vertices of i th generation with respect tothe original foam κ = κ1 ♯⋯ ♯κN where N is the maximal generation of κ.
Antonia Zipfel (FUW) ILQGS, 15.04.2014 16 / 30
Why the would-be ‘rigging map’ is not a proper rigging map Time ordering
Sketch of the Proof I
To prove the theorem we need the following observations:
LemmaIf e ∈ κint has a vertex of nth generation then the other vertex of e is eitherof generation n − 1,n or n + 1 or a vertex in a final graph.
Introduce additional vertices so that the second vertex of e is either ofgeneration n − 1,n or n + 1.
LemmaIf a face has an edge e joining a vertex of nth and (n + 1)th generation thenit contains at least one other edge connecting vertices of nth and (n + 1)th
generation (or v of generation m and v ′ in final graph).
Antonia Zipfel (FUW) ILQGS, 15.04.2014 17 / 30
Why the would-be ‘rigging map’ is not a proper rigging map Time ordering
Sketch of the Proof II
Proof of the theorem
2nd 2nd
1th
1th
2nd2nd
1th
1th
En,n+1 ∶= {e ∈ κint ∶joining vertices of nth and (n + 1)th generation}
Number of boundary edges e ∈ En,n+1 of f is even
Split faces until a face contains at most 2 edges inE1,2
Split all edges in E1,2 and join new vertices by edges
↝ κ = κ1 ♯κ′
Repeat the procedure until κ = κ1 ♯⋯ ♯κN
Antonia Zipfel (FUW) ILQGS, 15.04.2014 18 / 30
Why the would-be ‘rigging map’ is not a proper rigging map Factorization
Factorization of the rigging map I
If η is a proper rigging map then . . .
η[Tsf ](HTsi )!= 0 ∀Tsi ∈Hkin
Ô⇒ η[T∅](HTsi ) = (∑n(ηc[T∅](T∅))n)ηc[T∅](HTsi )
!= 0
where ∑κ in ηc is restricted to connected foams
Thus to test the rigging map it suffices to only consider connected foams!!
But when splitting a connected κ into ‘one time step’ foamsκ1 ♯⋯ ♯κn = κ then κi are not necessarily connected
However, if T∅ is excluded and either γ(sf ) and or γ(si) areconnected then any foam κ1 ♯⋯ ♯κn is connected
Test wether ηc[Tsf ] annihilates H for any sf with connected graph γ(sf )
Antonia Zipfel (FUW) ILQGS, 15.04.2014 19 / 30
Why the would-be ‘rigging map’ is not a proper rigging map Factorization
Factorization of the rigging map II
Spin foam transfer matrix
Kγ,γ′ set of ‘one time step’ foam and Pγ ∶Hkin →Hkin,γ
Z ∶= ∑γ,γ′
Pγ′ [ ∑κ∈Kγ,γ′
Z(κ)] Pγ
We then obtain for γ(sf ) connected . . .
ηc[Tsf ](Tsi ) = ∑κ∈Kγ(si ),γ(sf )
⟨Tsf , Z(κ1 ♯⋯ ♯κN) Tsi ⟩ =∞∑N=0
⟨Tsf , ZNTsi ⟩
RemarksAbove factorization would lead to ordering ambiguities fordisconnected foamsZ is symmetric
Antonia Zipfel (FUW) ILQGS, 15.04.2014 20 / 30
Why the would-be ‘rigging map’ is not a proper rigging map Why η is not a rigging map
A naive argument
If η would be a proper rigging map then
ηc[Tsf ](HTsi ) = ⟨Tsf ,∞∑N=0
ZNHTsi ⟩!= 0
But . . .
A ∶=∞∑N=0
ZN⇒ A = 1 + ZA
leads to a contradiction:
0 = AH = (1 + ZA)H = H
Of course, this expression is only formal as A is very likely diverging andtherefore requires a regularization
Antonia Zipfel (FUW) ILQGS, 15.04.2014 21 / 30
Why the would-be ‘rigging map’ is not a proper rigging map Why η is not a rigging map
Regularization of Z
Cut-Off: Kγ,γ′ is infinite, therefore introduce cut-off (J,Nf ,Ne)
Weight: Suppose there exist a weight ω(κ ♯κ′) = ω(κ)ω(κ′) s.t.Z ′
∶= ∑
κ∈Kγ,γ′ω(κ)Z [κ] has finite norm
Even if Z ′ is now densely defined (Z ′)n is not necessarily densely defined.
Recall: Z is symmetric and so is Z ′
Suppose: Z ′ can be extended to an self-adjoint operator withprojection valued measure E and let
Then: Z ′ acts on the states ψq ∶= ∫q−q dE(λ) ψ, ψ ∈Hkin, by
multiplication with λ ∈ [−q,q] where 0 < q < 1
But: Aψq = ∫q−q dE(λ) (1 − λ)−1 ψ ↝ A = 1 + ZA ☇
Antonia Zipfel (FUW) ILQGS, 15.04.2014 22 / 30
Discussion and outlook
Outline
1 A spin foam rigging mapRigging maps and spin foams - a brief reviewProposal for a spin foam ‘rigging map’Properties
2 Why the would-be ‘rigging map’ is not a proper rigging mapTime orderingFactorizationWhy η is not a rigging map
3 Discussion and outlookWhat is going wrong?Open questions
Antonia Zipfel (FUW) ILQGS, 15.04.2014 23 / 30
Discussion and outlook What is going wrong?
What is going wrong?
Option I: Z is itself a projector, e.g. BF-theory
Option II: Vertex amplitude too local
Option III: Necessity to restrict to one Plebanski sector?
Option IV: Wrong assumption on the weight!
Antonia Zipfel (FUW) ILQGS, 15.04.2014 24 / 30
Discussion and outlook What is going wrong?
What is going wrong?
Option I: Z is itself a projector, e.g. BF-theory
1 It can be shown that for β = 1 the operator ∑κ∈KMelon
Z [κ] annihilates the
Euclidean constraint. Here, κ ∈ KMelon are foams with just one internal
vertex and boundary graphs of the form: and
[Alesci, Thiemann, A.Z.]
2 Z can be interpreted as an object obtained from some sort of coarsegraining. In [Dittrich, Hellmann, Kaminski] a similar object was derived forholonomy spin foam models via coarse graining. For BF-theory thisalready incorporates the dynamics.
Antonia Zipfel (FUW) ILQGS, 15.04.2014 25 / 30
Discussion and outlook What is going wrong?
What is going wrong?
Option II: Vertex amplitude too local
The would-be rigging map can be factorized since foams can besplit, that is, κ = κ1 ♯⋯ ♯κN
Suppose AvQe(v , v′)Av ′ /∝ AvQe(v ,m)AmQe(m, v
′)Av ′
then splitting is no longer possible.
In this sense: vertex amplitude too local
Other hints into this direction: E.g. [Dittrich], [Hellmann, Kaminski],. . .
Antonia Zipfel (FUW) ILQGS, 15.04.2014 26 / 30
Discussion and outlook What is going wrong?
What is going wrong?
Option III: Necessity to restrict to one Plebanski sector?
2πδ(C) = limT→∞∫
T
−Te itC = lim
T→∞∫T
0[e itC + e−itC ]
= limT→∞
limn→∞
n
∑k=1
T
n[e iCT /n
]k
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶Uk
+ [e−iCT /n]k
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶(U†)k
But: Z is symmetric ↝ Z = U +U†
⇒ ∑k∈N
Z k= ∑
k∈N(U +U†
)k≠ δ(H)
A similar problem occurs in the asymptotic expansion.Can it be solved by the proper vertex proposal [Engle]?
Antonia Zipfel (FUW) ILQGS, 15.04.2014 27 / 30
Discussion and outlook What is going wrong?
What is going wrong?
Option IV: Wrong assumption on the weight!
We assumed that Z can be regularized by introducing a weight ω thatobeys ω(κ1)ω(κ2) = ω(κ1 ♯κ2). Is that sensible?
1 Regularization might require symmetry factors that destroy splitting:
δ(x) = ∫ dk eikx = ∫ dk(∑n
in
n!(kx)n)
For example in standard GFT λ∣γ(0)∣
∣sym(γ)∣2 Spin foam amplitudes suffer from bubble divergencies. However, if
foams with bubbles are excluded then gluing might no longer bepossible, since even if κ1, κ2 contain no bubbles, κ1 ♯κ2 might havebubbles.
Antonia Zipfel (FUW) ILQGS, 15.04.2014 28 / 30
Discussion and outlook Open questions
Open questions
RegularizationExplore regularization methods of generalized GFT [Oriti]
Explore latest results on regularization of the EPRL-model[Bonzom,Carrozza,Oriti, Puchta, Riello, Rovelli, Smerlak,. . . ]
Connection to coarse graining [Dittrich, . . . ]?
Proper vertexGeneralize proper vertex proposal [Engle] to n-valent vertices and test wetherthe modified rigging map still has the same problems.
Technical aspectsSingle time step ↝ better control on the sumFormalism to design single time step foams;[Kisielowski, Lewandowski, Puchta]
Antonia Zipfel (FUW) ILQGS, 15.04.2014 29 / 30