WORLD RELIGIONS III
WORLD RELIGIONS III Rel Std 5
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This 14 page Class Notes was uploaded by Amya Torphy on Saturday September 12, 2015. The Class Notes belongs to Rel Std 5 at University of California - Irvine taught by Staff in Fall. Since its upload, it has received 36 views. For similar materials see /class/201941/rel-std-5-university-of-california-irvine in Religious Studies at University of California - Irvine.
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Date Created: 09/12/15
susvoe Irvine SSM from Z G Alex Love amp David Bailin Department of Physics amp Astronomy University of Sussex dbailinsussexacuk Based on hepthO603172 119 University of Sussex susvoe Irvine SSM from Zg Outline 0 Background 0 Z6 orientifold Honeckeramp Ott Z6 orientifold Summary amp conclusions SUSY06 Irvine SSM from Z2 Background Open strings that begin and end on a stack a of NC D6branes wrapping a 3cycle of T6 T12 8 T22 8 T32 give the massless gauge bosons of Um 11 sum At the intersections of two stacks a and b there is chiral matter in the bifundamental Na Nb representation of UNC UNb where NC and Nb respectively have charges Qa 1 and Qb 1 with respect to U1C and U1b To get the standard model we start with Na 3 and Nb 2 and then arrange that there are a H b 3 intersections This will give 3 quark doublets QL 3 However 33 has Qb 9 The only other chiral doublets in the standard model are the 3 lepton doublets L 1 2 If we arrange that there are b H C 3 such intersections there would remain Qb 6 units of doublet charge to be found from other sources to cancel Qb overall SUSY06 Irvine SSM from Zg This requires additional vectorlike nonstandard model matter If instead we wrap an orientifold T6Q where Q is the worldsheet parity operator then at the intersections of a and the orientifold image b of b the chiral matter is in the Na Nb 3 2 representation Then 23 l 13 2 has Qb 3 which can be cancelled by 3 lepton doublets To getjust the standard model we require that a H b a 9 lb hez Marchesano amp Rabad n D6branes wrapping 3cycles of T6Q generally give a nonsupersymmetric spectrum This requires a low TeVscale unificationstring scale to avoid the hierarchy problem Such lowscale models have unacceptable levels of flavour changing neutral currents induced by worldsheet instantons Abel Lebedev amp Santiago susvoe Irvine SSM from Zg o Nonsusy theories generally have uncancelled NSNS tadpoles which reflect the instability of the complex structure moduli of T6 We can stabilise some of these moduli using an orbifold T6P We shall be concerned only with the point groups P Z6 and Z6 We shall study orientifolds in which the orbifold is quotiented with the worldsheet parity operator 9 o If the embedding of P is supersymmetric then RR tadpole cancellation also ensures NSNS tadpole cancellation too Cvetic Shiu amp Uranga SUSY06 Irvine SSM from Z2 l Z6 orientifold We may use a complex coordinate zk k 1 2 3 in each torus T13 The point group generator acts on these as sz e27mkzk For the Z6 orbifold V 1 1 2 The action of 8 must be an automorphism of the lattice so we may use an SU3 root lattice in each of the tori T13 The embedding R of 2 acts as Rzk 2k and for this too to be an automorphism each lattice must be in either the A or B configuration Then there are 6 essentially different lattices to consider Since b3T6Z6 2 there are 2 independent invariant untwisted bulk 3cycles p12 The only nonzero intersection is even p1 p2 2 To get odd intersection numbers you have to use fractional branes 1 a 2 where ngceptional consists of a collapsed 2 cycle stuck at a Z2 fixed point bulk exceptional Ha Ha l SUSY06 Irvine SSM from Zg in T12 8 T22 times a 1cycle in T32 They occur only in the 83 twisted sector There are 10 independent exceptional cycles 67 6 1 26 with nonzero intersection numbers 67 O Q 267j The pairs of fixed points to be used are determined by the wrapping numbers in T122 of the bulk part On the Z6 orientifold there is just one R invariant combination for each lattice and all supersymmetric stacks are automatically Rinvariant ln orientifolds there is also chiral matter on the branes as well as gauge particles The matter is in the symmetric representation Sa Na gtlt NaSymm and the antisymmetric representation Aa Na gtlt Naantisymm of UNa Now Sa 6 for Na 3 and Sb 3 for Nb 2 and both are excluded phenomenologically susvoe Irvine SSM from Zg o Orientifolding induces topological defects O6planes which are sources of RR charge If a is supersymmetric then HBUIk oc H06 and a H H06 0 Consequently Sa Aa a H 0 and likewise for b To avoid symmetric and hence also antisymmetric representations of the nonabelian gauge groups we require that a H 0 0 b H b Honecker amp Ott have shown that it is then impossible for a H b a 9 susvoe Irvine SSM from Z2 2g orientifold 0 Can we get the required intersection numbers using the Z6 orientifold The twist vector is V 12 3 On T122 the point group can again be realised using an SU3 root lattice P acts as a reflection on T32 so the complex structure U3 is unconstrained There are still two Rinvariant orientations of T32 Re U3 0 in A and Re U3 in B but Im U3 is arbitrary o b3T6Zg 4 so there are 4 independent invariant untwisted bulk 3cycles p1234 The intersection numbers p7 pj are even so again we must use fractional branes to get the standard model 0 Supersymmetry constrains H211 On the AABlattice for example 2A3 A6 A6 2A42 Im U3 4A1 2143 2144 A6 gt Im U3 SUSY06 Irvine SSM from Zg There are two independent Rinvariant combinations for each lattice On the AABlattice 2A3A6 0A6 2A4 Both are supersymmetric provided m U3 has the correct sign But in this case there are supersymmetric combinations that are not R invariant Exceptional cycles occur in the 62 4 and 63 twisted sectors The latter generates 8 independent cycles 67 Q 1 4 5 6 with 67 Ej 26 j Only this sector looks rich enough to generate the required intersection numbers We must exclude symmetric matter but not necessarily antisymmetric matter Aa 3 X 3antisymm 3 N qz could be quark singlet states and Ab 2 X 2antisymm 1 N 60 could be lepton singlet states However we do not want more than three rightchiral quarks qz or lepton singlets 2 with the same hypercharge So we also require that Aa g 3 and Ab g 3 SUSY06 Irvine SSM from Z25 1 1 0 We find that we can find supersymmetric stacks a and b satisfying a H b a 9 and the other conditions In all cases there is neither symmetric nor antisymmetric matter on one of the stacks Whether or not there is any antisymmetric matter on the other stack depends upon the lattice We have examples of both cases 0 Supersymmetry fixes the complex structure Im U3 on T32 SUSY06 Irvine SSM from zg Summary amp conclusions SUSY06 Irvine SSM from Zg It can be done using the Z8 orientifold By adding further branes C d with Nady 1 it may be possible to construct a model with just the spectrum of the supersymmetric standard model consistent with tadpole cancellation Different lattices give different solutions and numbers of solutions Some have q or 5 states as antisymmetric matter on one of the stacks SUSY06 Irvine SSM from Z25
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