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The EPRL intertwiners andcorrect partition function.Wojciech Kaminski
Zakopane, 28.02.10
p.
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Spin-foams models
Is the EPRL map injective, doesnt it kill any SU(2)intertwiner?
p.
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Spin-foams models
Is the EPRL map injective, doesnt it kill any SU(2)intertwiner?
Is the EPRL map isometric, does it preserve the scalar
product between the SU(2) intertwiners?
p.
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Spin-foams models
Is the EPRL map injective, doesnt 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 complete, correct form of the partitionfunction written directly in terms of the SU(2)intertwiners, the preimages of the EPRL map?
p.
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Spin-foams
Spin-foam approach to LQG is an analog of the
Feynman path integral (Rovelli and Reisenberger)
p.
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Spin-foams
Spin-foam approach to LQG is an analog of the
Feynman path integral (Rovelli and Reisenberger)
Spin-foam is a history of spin-network.
p.
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Spin-foams
Spin-foam approach to LQG is an analog of the
Feynman path integral (Rovelli and Reisenberger)
Spin-foam is a history of spin-network.
p.
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Spin-foams
Spin-foam approach to LQG is an analog of the
Feynman path integral (Rovelli and Reisenberger)
Spin-foam is a history of spin-network.
Evolution without adding vertices.
p.
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Spin-foams
Spin-foam approach to LQG is an analog of the
Feynman path integral (Rovelli and Reisenberger)
Spin-foam is a history of spin-network.
Evolution by adding vertices.
p.
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Induced boundary spin-network
Spin network is obtained by transversal section of a
spin foam.
p.
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Induced boundary spin-network
Spin network is obtained by transversal section of a
spin foam.
Colouring are induced on the spin-network from the
spin foam.
p.
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Induced boundary spin-network
Spin network is obtained by transversal section of a
spin foam.
Colouring are induced on the spin-network from the
spin foam.
p.
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Euclidean 4d QG as spin-foam
4d QG is regarded as a BF theory with constraints.
Spin-networks consists of gauge invariant functions.
p.
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Euclidean 4d QG as spin-foam
In each step we add one vertex with evolution of BF
theory, obtaining a transition amplitude between spinnetworks.
p.
E lid 4d QG i f
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Euclidean 4d QG as spin-foam
In each step we add one vertex with evolution of BF
theory, obtaining a transition amplitude between spinnetworks.
Constraints are imposed as projections on edges(nodes of spin-network). p.
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Details
The simplicity constraints are imposed on the
elements of
H, locally at each vertex.
InvSimp(1 ...k k+1 ... N)
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Details
The simplicity constraints are imposed on the
elements of
H, locally at each vertex.
InvSimp(1 ...k k+1 ... N)In each subsequent spin-network intertwiners shouldbe in this subspace,
p.
D il
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Details
The simplicity constraints are imposed on the
elements of
H, locally at each vertex.
InvSimp(1 ...k k+1 ... N)In each subsequent spin-network intertwiners shouldbe in this subspace,
To sum with respect to the spin-network histories with
the amplitude as a weight, one fixes an orthonormal
basis in each space of simple intertwiners.
p.
Si l i t t i
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Simple intertwiners
There are three main proposals for the simple intertwiners:
1. that ofBarrett-Crane (BC) corresponding to the
Palatini action,
2. that ofEngle-Pereira-Rovelli-Livine (EPRL)
corresponding the Holst action with the value of the
Barbero-Immirzi parameter = 1,3. that ofFreidel-Krasnov (FK) also corresponding to
the Holst action with the value of the Barbero-Immirzi
parameter = 1,
p.
EPRL i t t i
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EPRL intertwiner
Given intertwiner Inv(k1 ...) with the spins kI
jI =1 2
kIkI = j
+ + j, if || < 1 and kI = |j+ j|, if || > 1. This follows from
adjusted/improved constraints.
p.
EPRL i t t i
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EPRL intertwiner
Given intertwiner Inv(k1 ...) with the spins kI
jI =1 2
kI
p.
EPRL i t t i
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EPRL intertwiner
Given intertwiner Inv(k1 ...) with the spins kI
jI =1 2
kI
p.
EPRL i t t i
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EPRL intertwiner
Given intertwiner Inv(k1 ...) with the spins kI
jI =1 2
kI
p.
EPRL intertwiner
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EPRL intertwiner
Given intertwiner Inv(k1 ...) with the spins kI
jI =1 2
kI
The map EPRL
EPRL() : = (P+ P)c1 ... cn.
P projections onto SU(2) invariants,
ci Clebsch-Gordon coefficients ki
j+i
ji
.
p.
Orthonormal basis
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Orthonormal basis
One can show that the map EPRL() isinjective, so one can labelled EPRL intertwiners by
the SU(2) intertwiners.
p.
Orthonormal basis
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Orthonormal basis
One can show that the map EPRL() isinjective, so one can labelled EPRL intertwiners by
the SU(2) intertwiners.The proof is based on the observation that in suitable
basis matrix of the EPRL is upper triangular andinjective on diagonal entries.
p.
Proof 1
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Proof 1
Suitable basis is given by an intermediate spin k12
p.1
Proof 1
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Proof 1
Suitable basis is given by an intermediate spin k12 and j12.
Instead of projecting we consider contractions with the el-
ements ofInvSU(2)
SU(2).
p.1
Proof 1
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Proof 1
Suitable basis is given by an intermediate spin k12 and j12.
+ Instead of projecting we consider contractions with the el-
ements ofInvSU(2)SU(2).
p.1
Proof 2
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Proof 2
Such a basis is graded by k12 and j+12 + j
12 respectively,
+
p.1
Proof 2
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Proof 2
Such a basis is graded by k12 andj+12 +j
12 respectively,
+ then
(k12), j+12 j12 = 0, k12 > j+12 +j12.
p.1
Proof 2
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Proof 2
Such a basis is graded by k12 andj+12 +j
12 respectively,
+ then
(k12), j+12 j12 = 0, k12 > j+12 +j12.We can restrict our attention to k12 = j
+12 +j
12
p.1
Proof 3
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Proof 3
The contraction of
+ k12 = j
+12 +j
12
p.1
Proof 3
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Proof 3
...is equivalent to contraction of
+
p.1
Proof 3
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Proof 3
...is equivalent to contraction of
+ Imposing additional constraints on j12 allows forinductive procedure.
p.1
Proof 3
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Proof 3
...is equivalent to contraction of
+ Imposing additional constraints on j12 allows forinductive procedure.
There are some technical details...
p.1
Additional factor in the amplitude
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p
However, the map is not isometric for || = 1.
p.1
Additional factor in the amplitude
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p
However, the map is not isometric for || = 1. In orderto cure the lack of unitarity one should replace the former
spin -foam amplitude
Z[] =
jf ,e
f face
dim j+f , jf
v vertex
Av({(e)})({(bdr)}),
p.1
Additional factor in the amplitude
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p
However, the map is not isometric for || = 1. In orderto cure the lack of unitarity one should replace the former
spin -foam amplitude by
Z[] =
jf ,in/oute
f face
dim j+f , jf
e edge
A(ine , oute )
v vertex
Av({(e)})({(bdr)}),
p.1
Additional factor in the amplitude
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p
However, the map is not isometric for || = 1. In orderto cure the lack of unitarity one should replace the former
spin -foam amplitude by
Z[] =
jf ,in/oute
f face
dim j+f , jf
e edge
A(ine , oute )
v vertex
Av({(e)})({(bdr)}),
A(1, 2) is the inverse of the matrix (1), (2) so
PEPRL =
in/out
Ain,out|EPRL(out)EPRL(in)| p.1
Asymptotic behavior ofEPRL
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Even if the map EPRL is not unitary the asymptoticbehavior ofA may occure to be rather simple.
p.1
Asymptotic behavior ofEPRL
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Even if the map EPRL is not unitary the asymptoticbehavior ofA may occure to be rather simple.
=1
2 , j1 = 2, j2 = 4, j3 = 4, j1 = 2; a, b {2, . . . , 6}:
53723
175616
2265
57
50176
50935
1053696
355
250880
22655750176
117853
501760 12805
3010567
45 1177168
31378960
50935
1053696 12805
3010567
741949
3512320781
11
752640
513
5376
35525088
45 1177168
78111752640
5832560
0
0
3
137
8960
513
53760 13
40
p.1
Asymptotic behavior ofEPRL
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Even if the map EPRL is not unitary the asymptoticbehavior ofA may occure to be rather simple.
=1
2 , j1 = 2, j2 = 4, j3 = 4, j1 = 2; a, b {2, . . . , 6}:
0.3 0 0 0 0
0 0.23 0 0 00 0 0.21 0 0
0 0 0 0.22 0
0 0 0 0 0.32
p.1
Asymptotic behavior ofEPRL
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Even if the map EPRL is not unitary the asymptoticbehavior ofA may occure to be rather simple.
Examples of low j indicate that matrix A isaproximately diagonal (conjecture).
p.1
Asymptotic behavior ofEPRL
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Even if the map EPRL is not unitary the asymptoticbehavior ofA may occure to be rather simple.
Examples of low j indicate that matrix A isaproximately diagonal (conjecture).
If this is true A does not change (spoil) asymptoticbehavior of the spin foam amplitude.
p.1
Summary
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The map
EPRL()
is injective for all ,
p.1
Summary
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The map
EPRL()
is injective for all ,
Simple examples shows that it is not unitary for
||
= 1.
p.1
Summary
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The map
EPRL()
is injective for all ,
Simple examples shows that it is not unitary for
||
= 1.
The basis labelled by SU(2) intertwiners is notorthonormal and we should introduce an additional
factor A in the spin-foam amplitude
1|A12 = EPRL(1)|EPRL(2)
p.1