On the Flavor Problem in Strongly Coupled Theories - THEP Mainz

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52 Chapter 2. The Randall Sundrum Model and its Holographic Interpretation F G H UV brane IR brane F (ND) G (DD) I (NN) H (DN) Figure 2.6: Illustration of the breaking by BCs at the UV and IR brane. In the dual theory, the subgroup F is gauged by elementary fermions and the group H leaves the composite sector invariant after confinement. The right hand side shows a Venn diagram of the groups. The BCs of the fields living in the corresponding complements are also given (For example, fields in G may have any combination of BCs, but fields in G\(F ∪ H) must have Dirichlet BCs on both branes). for all other combinations. We will consider the consequences of Goldstones theorem in a situation in which breaking on both branes can appear, and its implications for the dual theory. In general, one has a bulk gauge group G, which is broken down to the subgroup F on the UV brane by explicit breaking through weakly gauging this subgroup in the elementary sector, see Figure 2.6. The dual theory is given by the Lagrangian (2.41), where a denotes the indices of the generators of F . On the IR brane, G is broken down to the subgroup H, which is the subgroup that still leaves the composite sector invariant after confinement sets in. We will call the intersection of these groups I = F ∩ H. One therefore expects dim G/H = dim G − dim H Goldstone bosons from the spontaneous symmetry breaking. However, dim F/I of those will be eaten, because there exist elementary gauge fields corresponding to the generators of this spontaneously broken subgroup. In the end, one thus has dim I massless gauge bosons in the elementary sector and NPNGB = dim G/H −dim F/I pseudo-Nambu Goldstone bosons, since they correspond to explicitly broken generators. The general setup is depicted in Figure 2.6, where the corresponding BCs for the vector part of the bulk gauge fields have been given. Since the breaking takes place via BCs, there are no scalar degrees of freedom on the brane. The longitudinal degrees of freedom of the gauge bosons are in this setup provided by the A5 components of the residual dim G/(F ∪ H), which must have Neumann BCs on both branes and thus develop zero modes, compare Section 4 of [62]. Both approaches, symmetry breaking via a brane field or via BCs, give the same number of degrees of freedoms. One can further infer, that the A5 mode of a bulk gauge field is the holographic dual of a composite pseudo Nambu Goldstone boson. Higgsless TC is therefore described by an RS model without a brane scalar and the following symmetries realized in the bulk and on the branes respectively (here, and in the following scenarios, only the electroweak sector plays a role and the not noted
SU(3)C is considered to be unbroken on both branes ) G = SU(2)L × U(1)B−L F = SU(2)L × U(1)B−L H = U(1)Y ⇒ dim G/H = 3 dim F/I = 3 NPNGB = 0 2.2. AdS/CFT 53 . (2.48) In this scenario, the U(1) charges are chosen in a way that ensures that the single massless gauge boson, I = U(1)Q, corresponds to the photon. The remaining three gauge bosons correspond to W ± and the Z and are massive only due to BCs on the IR brane, which will put their masses in the right range (compare the upper left panel of Figure 2.8 and the discussion in Section 2.3). The extension to a model with a custodial global symmetry is straightforward. Consider the setup [63], G = SU(2)L × SU(2)R × U(1)B−L F = SU(2)L × U(1)Y H = SU(2)V × U(1)B−L ⇒ dim G/H = 3 dim F/I = 3 NPNGB = 0 . (2.49) Here, the spontaneous breaking preserves a custodial SU(2)V symmetry, just like in the SM. The group F tells us for which composite Goldstone bosons there are elementary gauge bosons. This is again a W ± and the Z (there is not much room for choosing F ). The intersection I will again be the electromagnetic U(1)Q. But now there are three gauge bosons, corresponding to the bulk SU(2)R, which do not have a zero mode due to Dirichlet BCs on both branes. They will lead to a tower of massive W ±′ s and Z ′ s, which will couple with the same coupling strength as the KK modes of the electroweak gauge bosons and therefore cancel their contributions to the T parameter, see Section 3.1 for details. In the holographic dual of the composite Higgs scenario, which realizes the Higgs as a Nambu Goldstone boson, we expect to have a fully unbroken electroweak gauge group in the theory and the confining phase will only give us 4 Goldstone Bosons in the correct representation to form a composite Higgs. We will demonstrate this by means of a scenario which also implements custodial symmetry, see [37, Sec. 2], G = SO(5) × U(1)X F = SU(2)L × U(1)Y H = SO(4) × U(1)X ⇒ dim G/H = 4 dim F/I = 0 NPNGB = 4 . (2.50) Here, the U(1) charges are chosen in a way that results in I = SU(2)L × U(1)Y . Obviously, this choice of groups will leave all four electroweak gauge bosons massless, preserves the SU(2)V ∼ SO(4) custodial symmetry in the composite sector, and four PNGBs remain in the theory. One can construct the holographic dual of models which implement the collective breaking mechanism in an analogous way. For example, the corresponding scenario
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On the Flavor Problem in Strongly C
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UNIVERSITY OF MAINZ ABSTRACT THEORE
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Contents Acknowledgements vii Prefa
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Acknowledgements First and foremost
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x One of the most fascinating insig
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102 Chapter 3. Solving the Flavor P
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146 Chapter 4. The Asymmetry in Top
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176 Chapter 5. Conclusions describe
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178 Appendix A. Generating Paramete
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180 Appendix A. Generating Paramete
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182 Appendix B. Neutral Meson Mixin
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184 Appendix B. Neutral Meson Mixin
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C Input Parameters This appendix is
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Parameter Value ± Error Reference
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192 Appendix D. Higgs Potential for
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194 Bibliography [13] S. Dimopoulos
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196 Bibliography [43] G. Nordström
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198 Bibliography [72] K. S. Babu,
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200 Bibliography [104] R. Contino a
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202 Bibliography [133] M. Baak, M.
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204 Bibliography [160] J. Charles,
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206 Bibliography [185] C. Csaki, G.
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208 Bibliography [213] E. Thomson e
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210 Bibliography [239] V. Ahrens, A
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