SUSY GUTs in 6D: Precise Gauge Coupling Unification

Abstract

In the previous chapter we argued that precise gauge coupling unification [PGCU], i.e. the absence of threshold corrections at the GUT scale, requires certain specific soft SUSY breaking masses at the weak scale. In particular, lighter gluinos than expected from unification of gaugino masses. In this chapter we argue that non-local GUT breaking will, in general, eliminate one-loop GUT threshold corrections. This is because the compactification scale is the scale of GUT symmetry breaking [342–344]. In this chapter (based on [345]) we present a 6D model with SU(6) gauge symmetry and N=2 supersymmetry. In terms of 4D language, such a 6D theory with N=2 SUSY contains one vector adjoint and three chiral adjoints. The model has gauge-Higgs unification with the Higgs doublets coming from one of the chiral adjoints. The group SU(6) is broken to SU(5) × U(1) _X via orbifold boundary conditions. Then SU(5) is broken to the Standard Model gauge group and, at the same time, Higgs doublet-triplet splitting is accomplished by a Wilson line. The two extra-dimensions are compactified on an orbifold that can be characterized as a sphere with a cross-cap (topologically equivalent to the manifold RP ²), as described in [342–344]. Quarks and leptons can be in the bulk (or localized at the fixed points), and their Yukawa couplings to the Higgs are localized at the orbifold fixed points which only retain an N=1 SUSY in 4D (see for example [313, 346] where this phenomenon has been discussed).

Appendices

Appendix 1: Kaluza-Klein Integrals

In order to compute the threshold corrections coming from an infinite tower of Kaluza-Klein states, we would like to evaluate the following integral (see Eq. (18.30)):

$$\displaystyle{ \sum _{(m,n)\in Z}\int _{\xi }^{\infty }\frac{dt} {t} e^{-\pi t\frac{M_{(m,n)}^{2}} {\mu ^{2}} }e^{-\pi \chi t} }$$

(18.41)

where, χ and ξ are IR and UV regulators, and the M _(m, n) are the masses of the (m, n)th KK mode. In the presence of Wilson lines they are given by:

$$\displaystyle{ M_{(m,n)}^{2} = \frac{(m +\rho _{1})^{2}} {R_{5}^{2}} + \frac{(n +\rho _{2})^{2}} {R_{6}^{2}} }$$

(18.42)

where in the scenario that we have, ρ ₁ can be either 0 or 2 and ρ ₂ is always 0. In general, we can solve the integral following Ghilencea [354] who evaluates the integral for the cases of one and two extra-dimensions. Again, in the current scenario that we have, we find that we only need to evaluate this integral in its one-dimensional limit. We follow Ghilencea and evaluate a 1-dimensional Kaluza-Klein integral of the form:

$$\displaystyle{ \mathcal{R}_{1}\left [\xi,\rho,\delta \right ] =\sum _{ m\in Z}^{^{{\prime}} }\int _{\xi }^{\infty }\frac{dt} {t} e^{-\pi t\left [(m+\rho )^{2}+\delta \right ] } }$$

(18.43)

where the prime over the summation in the second term represents that m ≠ 0, but runs over all other integer values.

We can make use of the Poisson re-summation formula:

$$\displaystyle{ \sum _{n\in Z}e^{-\pi A(n+\sigma )^{2} } = \frac{1} {\sqrt{A}}\sum _{\tilde{n}\in Z}e^{-\pi A^{-1}\tilde{n}^{2}+2i\pi \tilde{n}\sigma } }$$

(18.44)

to evaluate this integral. We have,

$$\displaystyle\begin{array}{rcl} \mathcal{R}_{1}\left [\xi,\rho,\delta \right ]& =& \int _{\xi }^{\infty }\frac{dt} {t} \left [-e^{-\pi t\rho ^{2} } +\sum _{m}e^{-\pi t(m+\rho )^{2} }\right ]e^{-\pi \delta t} \\ & =& \int _{\xi }^{\infty }\frac{dt} {t} \left [-e^{-\pi t\rho ^{2} } + \frac{1} {\sqrt{t}} + \frac{1} {\sqrt{t}}\sum _{m}^{^{{\prime}} }e^{-\pi m^{2}/t+2i\pi m\rho }\right ]e^{-\pi \delta t} \\ & =& -\varGamma \left [0,\pi \xi (\delta +\rho ^{2})\right ] + \frac{2e^{-\pi \delta \xi }} {\sqrt{\xi }} + 2\pi \sqrt{\delta }\text{Erf}\left [\sqrt{\pi \delta \xi }\right ] \\ & & -\text{log}\vert 2\ \text{sin}\pi (\rho +i\sqrt{\delta })\vert ^{2} {}\end{array}$$

(18.45)

where, it has been assumed that ξ ≪ 1 while evaluating the integral $\int _{\xi }^{\infty }\frac{dt} {t} e^{-\pi m^{2}/t-\pi \delta t }$ and Ghilencea [354] shows that the error by doing so vanishes when ξ is small.

We summarize the result of this integral in various useful limits:

• (m,n) = (0,0)

  $$\displaystyle\begin{array}{rcl} \int _{\xi }^{\infty }\frac{dt} {t} e^{-\pi t \frac{\rho _{1}^{2}} {R_{5}^{2}\mu ^{2}} }e^{-\pi \chi t}& =& \int _{\xi }^{\infty }\frac{dt} {t} e^{-\pi (\rho +\chi )t} \\ & =& \varGamma \left [0,\pi \xi (\chi +\rho )\right ] {}\end{array}$$

  (18.46)

  where $\rho = \frac{\rho _{1}^{2}} {R_{5}^{2}\mu ^{2}} + \frac{\rho _{2}^{2}} {R_{6}^{2}\mu ^{2}}$

• n=0, m ≠ 0

  $$\displaystyle\begin{array}{rcl} \sum _{m\in Z}^{^{{\prime}} }\int _{\xi }^{\infty }\frac{dt} {t} e^{-\pi t\frac{\frac{(m+\rho _{1})^{2}} {R_{5}^{2}} + \frac{\rho _{2}^{2}} {R_{6}^{2}} } {\mu ^{2}} }e^{-\pi \chi t}& =& \sum _{m\in Z}^{^{{\prime}} }\int _{\xi \nu _{1}}^{\infty }\frac{dt} {t} e^{-\pi t(m+\rho _{1})^{2} }e^{-\pi \frac{\delta _{1}} {\nu _{1}} t} \\ & =& \mathcal{R}_{1}\left [\xi \nu _{1},\rho _{1}, \frac{\delta _{1}} {\nu _{1}}\right ] {}\end{array}$$

  (18.47)

  where, $\nu _{1} = \frac{1} {\mu ^{2}R_{5}^{2}}$ and $\delta _{1} =\chi + \frac{\rho _{2}^{2}} {\mu ^{2}R_{6}^{2}}$.

• m=0, n ≠ 0

  Similar to the previous case with some parameters interchanged, we have:

  $$\displaystyle\begin{array}{rcl} \sum _{n\in Z}^{^{{\prime}} }\int _{\xi }^{\infty }\frac{dt} {t} e^{-\pi t\frac{ \frac{\rho _{1}^{2}} {R_{5}^{2}} + \frac{(n+\rho _{2})^{2}} {R_{6}^{2}} } {\mu ^{2}} }e^{-\pi \chi t}& =& \sum _{n\in Z}^{^{{\prime}} }\int _{\xi \nu _{2}}^{\infty }\frac{dt} {t} e^{-\pi t(n+\rho _{2})^{2} }e^{-\pi \frac{\delta _{2}} {\nu _{2}} t} \\ & =& \mathcal{R}_{1}\left [\xi \nu _{2},\rho _{2}, \frac{\delta _{2}} {\nu _{2}}\right ] {}\end{array}$$

  (18.48)

  where, $\nu _{2} = \frac{1} {\mu ^{2}R_{6}^{2}}$ and $\delta _{2} =\chi + \frac{\rho _{1}^{2}} {\mu ^{2}R_{5}^{2}}$

• Since the spectrum we are interested in has states that live either at odd or even integers, it is write down the result of this integral in these limit that the summation is over either even or odd integers:

  n = 0, m ≠ 0; m = even

  $$\displaystyle\begin{array}{rcl} \mathcal{R}_{1}^{E}\left [\xi \nu _{ 1},\rho _{1}, \frac{\delta _{1}} {\nu _{1}}\right ]& =& \sum _{m\in Z}^{^{{\prime}} }\int _{\xi }^{\infty }\frac{dt} {t} e^{-\pi t\frac{\frac{(2m+\rho _{1})^{2}} {R_{5}^{2}} + \frac{\rho _{2}^{2}} {R_{6}^{2}} } {\mu ^{2}} }e^{-\pi \chi t} \\ & =& \sum _{m\in Z}^{^{{\prime}} }\int _{\xi }^{\infty }\frac{dt} {t} e^{-4\pi t\frac{\frac{(m+ \frac{\rho _{1}} {2} )^{2}} {R_{5}^{2}} + \frac{\rho _{2}^{2}} {4R_{6}^{2}} } {\mu ^{2}} }e^{-\pi \chi t/4} \\ & =& \sum _{m\in Z}^{^{{\prime}} }\int _{4\xi \nu _{1}}^{\infty }\frac{dt} {t} e^{-\pi t(m+\rho _{1})^{2} }e^{-\pi \frac{\delta _{1}} {4\nu _{1}} t} \\ & =& \mathcal{R}_{1}\left [4\xi \nu _{1}, \frac{\rho _{1}} {2}, \frac{\delta _{1}} {4\nu _{1}}\right ] {}\end{array}$$

  (18.49)

  where, $\nu _{1} = \frac{1} {\mu ^{2}R_{5}^{2}}$ and $\delta _{1} =\chi + \frac{\rho _{2}^{2}} {\mu ^{2}R_{6}^{2}}$ is the same as previously defined.

  n = 0, m ≠ 0; m = odd

  $$\displaystyle\begin{array}{rcl} \mathcal{R}_{1}^{O}\left [\xi \nu _{ 1},\rho _{1}, \frac{\delta _{1}} {\nu _{1}}\right ]& =& \sum _{m\in Z}\int _{\xi }^{\infty }\frac{dt} {t} e^{-\pi t\frac{\frac{(2m-1+\rho _{1})^{2}} {R_{5}^{2}} + \frac{\rho _{2}^{2}} {R_{6}^{2}} } {\mu ^{2}} }e^{-\pi \chi t} \\ & =& \sum _{m\in Z}\int _{\xi }^{\infty }\frac{dt} {t} e^{-4\pi t\frac{\frac{(m+ \frac{\rho _{1}-1} {2} )^{2}} {R_{5}^{2}} + \frac{\rho _{2}^{2}} {4R_{6}^{2}} } {\mu ^{2}} }e^{-\pi \chi t/4} \\ & =& \int _{\xi \nu _{1}}^{\infty }\frac{dt} {t} e^{-\pi t\frac{(\rho _{1}-1)^{2}} {4} }e^{-\pi \frac{\delta _{1}} {4\nu _{1}} t} \\ & & +\sum _{m\in Z}^{^{{\prime}} }\int _{4\xi \nu _{1}}^{\infty }\frac{dt} {t} e^{-\pi t(m+\frac{\rho _{1}-1} {2} )^{2} }e^{-\pi \frac{\delta _{1}} {4\nu _{1}} t} \\ & =& \varGamma \left [0,\pi \xi (\nu _{1}(\rho _{1} - 1)^{2} +\delta _{ 1})\right ] + \mathcal{R}_{1}\left [4\xi \nu _{1}, \frac{\rho _{1} - 1} {2}, \frac{\delta _{1}} {4\nu _{1}}\right ] \\ & & {}\end{array}$$

  (18.50)

  In order to write the result of the integral in terms of the original $\mathcal{R}_{1}$, we separate the zeroth term from the rest in the summation. It is also useful to note that the function $\mathcal{R}_{1}$ is even in ρ and hence:

  $$\displaystyle{ \mathcal{R}_{1}\left [\xi,\rho,\delta \right ] = \mathcal{R}_{1}\left [\xi,-\rho,\delta \right ] }$$

  (18.51)

Appendix 2: Useful Limits of Relevant Functions

The result of the Kaluza-Klein integrals were evaluated in the previous section, in terms of the two functions, Γ[0, π ξ χ] and $\mathcal{R}_{1}\left [\xi,\rho,\delta \right ]$. χ and ξ are the regulators and in the limit that they are zero, we can replace them with the relevant mass scales.

$$\displaystyle{ Q^{2} \equiv \pi e^{\gamma }\chi \mu ^{2}\Big\vert _{\chi \rightarrow 0}\qquad \varLambda ^{2} \equiv \frac{\mu ^{2}} {\xi } \Big\vert _{\xi \rightarrow 0} }$$

(18.52)

As evaluated in the previous section:

$$\displaystyle\begin{array}{rcl} \mathcal{R}_{1}\left [\xi,\rho,\delta \right ]& =& -\varGamma \left [0,\pi \xi (\delta +\rho ^{2})\right ] + \frac{2e^{-\pi \delta \xi }} {\sqrt{\xi }} + 2\pi \sqrt{\delta }\text{Erf}\left [\sqrt{\pi \delta \xi }\right ] {}\\ & & -\text{log}\vert 2\ \text{sin}\pi (\rho +i\sqrt{\delta })\vert ^{2} {}\\ \end{array}$$

We use the following expansions:

$$\displaystyle\begin{array}{rcl} -\varGamma \left [0,z\right ]& =& \gamma +\text{ln}\ z +\sum _{k\geq 1}\frac{(-z)^{k}} {k!\ k} \qquad z > 0{}\end{array}$$

(18.53)

$$\displaystyle\begin{array}{rcl} \text{Erf}\left [x\right ]& =& \frac{2x} {\sqrt{\pi }} -\frac{2x^{3}} {3\sqrt{\pi }} + \mathcal{O}(x^{5})\qquad x \ll 1{}\end{array}$$

(18.54)

Then,

$$\displaystyle\begin{array}{rcl} \varGamma \left [0,\pi \xi \chi \right ]& =& -\gamma -\text{ln}\ \pi \xi \chi \\ & =& -\text{ln}\ \pi \frac{e^{\gamma }\mu ^{2}Q^{2}} {\varLambda ^{2}\pi e^{\gamma }\mu ^{2}} \\ & =& -\text{ln}\ \frac{Q^{2}} {\varLambda ^{2}}{}\end{array}$$

(18.55)

With these approximations, $\mathcal{R}_{1}\left [\xi,\rho,\delta \right ]$ simplifies to:

$$\displaystyle\begin{array}{rcl} \mathcal{R}_{1}\left [\xi,\rho,\delta \right ]& =& -\text{ln}\left [4\pi e^{-\gamma }\frac{1} {\xi } e^{-2/\sqrt{\xi }}\right ] -\text{ln}\left \vert \frac{\text{sin}(\rho +i\sqrt{\delta })} {\pi (\rho +i\sqrt{\delta })} \right \vert ^{2}{}\end{array}$$

(18.56)

We summarize, the various terms that come up in the calculation of threshold corrections in Sect. 18.5. In the expressions below, we have also introduced the compactifications scales $M_{5} = \sqrt{\pi e^{\gamma }}/R_{5}$ and $M_{6} = \sqrt{\pi e^{\gamma }}/R_{6}$.

$$\displaystyle\begin{array}{rcl} \varGamma \left [0,\pi \xi \chi \right ]& =& \text{ln} \frac{\varLambda ^{2}} {Q^{2}} \\ \varGamma \left [0,\pi \xi \nu _{1}\right ]& =& -\gamma -\text{ln} \frac{\pi } {\varLambda ^{2}R_{5}^{2}} \\ & =& -\text{ln}\left [\frac{M_{5}^{2}} {\varLambda ^{2}} \right ] \\ \varGamma \left [0,\pi \xi \nu _{2}\right ]& =& -\gamma -\text{ln} \frac{\pi } {\varLambda ^{2}R_{6}^{2}} \\ & =& -\text{ln}\left [\frac{M_{6}^{2}} {\varLambda ^{2}} \right ] \\ \mathcal{R}_{1}\left [4\xi \nu _{1},0, \frac{\chi } {4\nu _{1}}\right ]& =& -\text{ln}\left [\pi e^{-\gamma -\varLambda R_{5} }(\varLambda R_{5})^{2}\right ] \\ \mathcal{R}_{1}\left [4\xi \nu _{1}, \frac{1} {2}, \frac{\chi } {4\nu _{1}}\right ]& =& -\text{ln}\left [\pi e^{-\gamma -\varLambda R_{5} }(\varLambda R_{5})^{2}\right ] -\text{ln}\left [\frac{2} {\pi } \right ]^{2} \\ \mathcal{R}_{1}\left [4\xi \nu _{2},0, \frac{\chi } {4\nu _{2}}\right ]& =& -\text{ln}\left [\pi e^{-\gamma -\varLambda R_{6} }(\varLambda R_{6})^{2}\right ] \\ \mathcal{R}_{1}\left [4\xi \nu _{2}, \frac{1} {2}, \frac{\chi } {4\nu _{2}}\right ]& =& -\text{ln}\left [\pi e^{-\gamma -\varLambda R_{6} }(\varLambda R_{6})^{2}\right ] -\text{ln}\left [\frac{2} {\pi } \right ]^{2} \\ \varGamma \left [0,\pi \xi \left (\frac{\nu _{1}} {4} + \frac{\nu _{2}} {4}\right )\right ]& =& -\text{ln}\left [\frac{M_{5}^{2} + M_{6}^{2}} {4\varLambda ^{2}} \right ] {}\end{array}$$

(18.57)

Appendix 3: 6D → 4D Matching

We calculated the corrections to the gauge couplings coming from the KK states of the 6D orbifold model that was constructed. At the lowest compactification scale (largest compactification radius), we said that the couplings from 4D MSSM and 6D orbifold model should match. In this section, we will compare the two sets of equations, from the two theories:

$$\displaystyle\begin{array}{rcl} \alpha _{i}^{-1}(Q)& =& \alpha ^{-1}(\varLambda ) +\sum _{\rho }\varOmega _{ i,\rho }(Q) {}\\ \alpha _{i}^{-1}(Q)& =& \alpha _{ GUT}^{-1} + \frac{b_{i}} {2\pi } log\frac{M_{GUT}} {Q} -\alpha _{GUT}^{-1} \frac{\epsilon _{3}} {(1 +\epsilon _{3})}\delta _{i3} {}\\ \end{array}$$

and solve for the three scales of the orbifold model, Λ, M ₅, and M ₆ as well as coupling constant, α at the cut-off scale.

Since the two expressions have to match at all scales below the smallest compactification scale of the orbifold model, we can rewrite the above two equations as:

$$\displaystyle\begin{array}{rcl} & & \alpha _{GUT}^{-1} + \frac{b_{i}} {4\pi } \text{ln}\frac{M_{GUT}^{2}} {Q^{2}} -\alpha _{GUT}^{-1} \frac{\epsilon _{3}} {(1 +\epsilon _{3})}\delta _{i3} \\ & & \quad =\alpha ^{-1}(\varLambda ) + \frac{b_{i}^{++}(I_{\rho } = 0)} {4\pi } \text{ln} \frac{\varLambda ^{2}} {Q^{2}} + \left (\frac{b_{i}^{+-}(I_{\rho } = 0) + b_{i}^{-+}(I_{\rho } = 0)} {4\pi } \right )\text{ln}\left [ \frac{\pi \varLambda } {2M_{5}}\right ]^{2} \\ & & \qquad + \left (\frac{b_{i}^{+-}(I_{\rho } = 0) + b_{i}^{--}(I_{\rho } = 0)} {4\pi } \right )\text{ln}\left [ \frac{\pi \varLambda } {2M_{6}}\right ]^{2} + \frac{b_{i}^{+-}(I_{\rho } = 2)} {4\pi } \text{ln}\left [ \frac{4\varLambda ^{2}} {M_{5}^{2} + M_{6}^{2}}\right ]{}\end{array}$$

(18.58)

where we have used the complete expression we estimated for the corrections to couplings in (18.38).

We use the following redefinitions:

$$\displaystyle\begin{array}{rcl} \frac{b_{i}^{MSSM}} {4\pi } = \frac{b_{i}^{++}(I_{\rho } = 0)} {4\pi } & =& \beta _{i} \\ \frac{b_{i}^{+-}(I_{\rho } = 0) + b_{i}^{-+}(I_{\rho } = 0)} {4\pi } & =& -A_{i} \\ \frac{b_{i}^{+-}(I_{\rho } = 0) + b_{i}^{--}(I_{\rho } = 0)} {4\pi } & =& -B_{i} \\ \frac{b_{i}^{+-}(I_{\rho } = 2)} {4\pi } & =& -C_{i}{}\end{array}$$

(18.59)

and

$$\displaystyle{ (A_{i} + B_{i})\text{ln}\left [ \frac{\pi } {2}\right ]^{2} + C_{ i}\text{ln}\left [4\right ] = D_{i} }$$

(18.60)

and hence end up with a set of three equations that can be simply written as:

$$\displaystyle\begin{array}{rcl} & & \alpha _{GUT}^{-1} -\alpha ^{-1}(\varLambda ) -\alpha _{ GUT}^{-1} \frac{\epsilon _{3}} {(1 +\epsilon _{3})}\delta _{i3} -\beta _{i}\text{ln} \frac{\varLambda ^{2}} {M_{GUT}^{2}} \\ & & \quad + A_{i}\text{ln} \frac{\varLambda ^{2}} {M_{5}^{2}} + B_{i}\text{ln} \frac{\varLambda ^{2}} {M_{6}^{2}} + C_{i}\text{ln} \frac{\varLambda ^{2}} {M_{5}^{2} + M_{6}^{2}} + D_{i} = 0{}\end{array}$$

(18.61)

where, A _i  = A _1i + A _2i and i = 1, 2, 3. We look at the equations corresponding (i) (i = 1) - (i =2) (ii) i = 2 (iii) i = 3 and solve for Λ, M ₅, and M ₆. It is usually considered that the 4D unification scale is around 3. 0 × 10¹⁶ GeV and the couplings at this scale are unified at α _GUT ⁻¹ = 24. In standard scenarios of MSSM with gaugino mass unification, ε ₃ = −3%. Depending of the spectrum of low energy SUSY, these quantities are subject to change. The first equation we get by simplifying Eq. (18.61) for (i = 1) - (i =2) is:

$$\displaystyle\begin{array}{rcl} & & -(\beta _{1} -\beta _{2})\text{ln} \frac{\varLambda ^{2}} {M_{GUT}^{2}} + (A_{1} - A_{2})\text{ln} \frac{\varLambda ^{2}} {M_{5}^{2}} + (B_{1} - B_{2})\text{ln} \frac{\varLambda ^{2}} {M_{6}^{2}} \\ & & \quad + (C_{1} - C_{2})\text{ln} \frac{\varLambda ^{2}} {M_{5}^{2} + M_{6}^{2}} + (D_{1} - D_{2}) = 0 {}\end{array}$$

(18.62)

Defining, $\frac{\varLambda ^{2}} {M_{5}^{2}} = X$ and $\frac{\varLambda ^{2}} {M_{6}^{2}} = Y$, we get:

$$\displaystyle\begin{array}{rcl} \text{ln} \frac{\varLambda ^{2}} {M_{GUT}^{2}}& =& \left (\frac{A_{1} - A_{2}} {\beta _{1} -\beta _{2}} \right )\text{ln}X + \left (\frac{B_{1} - B_{2}} {\beta _{1} -\beta _{2}} \right )\text{ln}Y \\ & & -\left (\frac{C_{1} - C_{2}} {\beta _{1} -\beta _{2}} \right )\text{ln}\left ( \frac{1} {X} + \frac{1} {Y }\right ) + \left (\frac{D_{1} - D_{2}} {\beta _{1} -\beta _{2}} \right ){}\end{array}$$

(18.63)

Next, we look at Eq. (18.61) when i = 2:

$$\displaystyle\begin{array}{rcl} & & \alpha _{GUT}^{-1} -\alpha ^{-1}(\varLambda ) -\beta _{ 2}\text{ln} \frac{\varLambda ^{2}} {M_{GUT}^{2}} \\ & & \quad + A_{2}\text{ln}X + B_{2}\text{ln}Y - C_{2}\text{ln}\left ( \frac{1} {X} + \frac{1} {Y }\right ) + D_{2} = 0{}\end{array}$$

(18.64)

Then, using the expression we just derived in Eq. (18.63), we get an expression for α ⁻¹(Λ):

$$\displaystyle\begin{array}{rcl} \alpha ^{-1}(\varLambda )& =& \alpha _{ GUT}^{-1} + \left [A_{ 2} -\beta _{2}\left (\frac{A_{1} - A_{2}} {\beta _{1} -\beta _{2}} \right )\right ]\text{ln}X + \left [B_{2} -\beta _{2}\left (\frac{B_{1} - B_{2}} {\beta _{1} -\beta _{2}} \right )\right ]\text{ln}Y \\ & & -\left [C_{2} -\beta _{2}\left (\frac{C_{1} - C_{2}} {\beta _{1} -\beta _{2}} \right )\right ]\text{ln}\left ( \frac{1} {X} + \frac{1} {Y }\right ) + \left [D_{2} -\beta _{2}\left (\frac{D_{1} - D_{2}} {\beta _{1} -\beta _{2}} \right )\right ]{}\end{array}$$

(18.65)

Finally, we look at Eq. (18.61) when i = 3, and simplify it using the relations obtained in Eqs. (18.63) and (18.65) and we get a final expression:

$$\displaystyle\begin{array}{rcl} & & \left [A_{3} - A_{2} + \left (\frac{A_{1} - A_{2}} {\beta _{1} -\beta _{2}} \right )(\beta _{2} -\beta _{3})\right ]\text{ln}X + \left [B_{3} - B_{2} + \left (\frac{B_{1} - B_{2}} {\beta _{1} -\beta _{2}} \right )(\beta _{2} -\beta _{3})\right ]\text{ln}Y \\ & & \quad -\left [C_{3} - C_{2} + \left (\frac{C_{1} - C_{2}} {\beta _{1} -\beta _{2}} \right )(\beta _{2} -\beta _{3})\right ]\text{ln}\left ( \frac{1} {X} + \frac{1} {Y }\right ) \\ & & \quad + \left [D_{3} - D_{2} + \left (\frac{D_{1} - D_{2}} {\beta _{1} -\beta _{2}} \right )(\beta _{2} -\beta _{3})\right ] -\alpha _{GUT}^{-1} \frac{\epsilon _{3}} {1 +\epsilon _{3}} = 0 {}\end{array}$$

(18.66)

The above three equations can be rewritten in a simple manner as (in the order Eqs. (18.66), (18.63), (18.65)):

$$\displaystyle\begin{array}{rcl} \mathcal{A}\text{ln}X + \mathcal{B}\text{ln}Y -\mathcal{C}\text{ln}\left ( \frac{1} {X} + \frac{1} {Y }\right ) + \mathcal{D}& =& 0 \\ \mathcal{F}\text{ln}X + \mathcal{G}\text{ln}Y -\mathcal{H}\text{ln}\left ( \frac{1} {X} + \frac{1} {Y }\right ) + \mathcal{I}& =& \text{ln} \frac{\varLambda ^{2}} {M_{GUT}^{2}} \\ \mathcal{K}\text{ln}X + \mathcal{L}\text{ln}Y -\mathcal{M}\text{ln}\left ( \frac{1} {X} + \frac{1} {Y }\right ) + \mathcal{N}& =& \alpha ^{-1}(\varLambda ) {}\end{array}$$

(18.67)

with,

$$\displaystyle\begin{array}{rcl} \mathcal{A}& =& A_{3} - A_{2} + (A_{1} - A_{2})\left (\frac{\beta _{2} -\beta _{3}} {\beta _{1} -\beta _{2}}\right ), {}\\ \mathcal{B}& =& B_{3} - B_{2} + (B_{1} - B_{2})\left (\frac{\beta _{2} -\beta _{3}} {\beta _{1} -\beta _{2}}\right ), {}\\ \mathcal{C}& =& C_{3} - C_{2} + (C_{1} - C_{2})\left (\frac{\beta _{2} -\beta _{3}} {\beta _{1} -\beta _{2}}\right ), {}\\ \mathcal{D}& =& D_{3} - D_{2} + (D_{1} - D_{2})\left (\frac{\beta _{2} -\beta _{3}} {\beta _{1} -\beta _{2}}\right ) -\alpha _{GUT}^{-1} \frac{\epsilon _{3}} {(1 +\epsilon _{3})} {}\\ \mathcal{F}& =& \frac{A_{1} - A_{2}} {\beta _{1} -\beta _{2}}, {}\\ \mathcal{G}& =& \frac{B_{1} - B_{2}} {\beta _{1} -\beta _{2}}, {}\\ \mathcal{H}& =& \frac{C_{1} - C_{2}} {\beta _{1} -\beta _{2}}, {}\\ \mathcal{I}& =& \frac{D_{1} - D_{2}} {\beta _{1} -\beta _{2}}, {}\\ \mathcal{K}& =& A_{2} -\beta _{2}\left (\frac{A_{1} - A_{2}} {\beta _{1} -\beta _{2}} \right ), {}\\ \mathcal{L}& =& B_{2} -\beta _{2}\left (\frac{B_{1} - B_{2}} {\beta _{1} -\beta _{2}} \right ), {}\\ \mathcal{M}& =& C_{2} -\beta _{2}\left (\frac{C_{1} - C_{2}} {\beta _{1} -\beta _{2}} \right ), {}\\ \mathcal{N}& =& D_{2} -\beta _{2}\left (\frac{D_{1} - D_{2}} {\beta _{1} -\beta _{2}} \right ) +\alpha _{ GUT}^{-1}, {}\\ \end{array}$$

These quantities can be calculated using the beta-function coefficients given in Table 18.3. The numerical values of all the above coefficients are summarized in Table 18.5.

With these coefficients, we get a simple quadratic equation in terms in of the variables X and Y:

$$\displaystyle{ \left ( \frac{Y } {X}\right )^{2} + \frac{Y } {X} -\text{Exp}\ \left (-\frac{7\pi \mathcal{D}(\epsilon _{3})} {3} \right ) = 0 }$$

(18.68)

Recall that $X = \frac{\varLambda ^{2}} {M_{5}^{2}}$ and $Y = \frac{\varLambda ^{2}} {M_{6}^{2}}$, which implies that the above equation turns into a quadratic equation in $\left (\frac{M_{5}} {M_{6}} \right )^{2}$ with the solution.

$$\displaystyle{ M_{5}^{2} = \frac{-1 \pm \sqrt{1 + 4\text{Exp} \ \left (-\frac{7\pi \mathcal{D}(\epsilon _{3 } )} {3} \right )}} {2} M_{6}^{2} }$$

(18.69)

which we write as $M_{5} = \sqrt{m(\epsilon _{3 } )}M_{6}$. The slope m, is the positive solution from the above expression and is shown in Fig. 18.3. The other two equations then yield us M ₅ and M ₆ uniquely and one expression relating α ⁻¹(Λ) and Λ.

$$\displaystyle\begin{array}{rcl} M_{5}& =& \left (m(\epsilon _{3})^{(\mathcal{G}-\mathcal{H})/2}(m(\epsilon _{ 3}) + 1)^{\mathcal{H}/2}e^{\mathcal{I}/2}\right )M_{ GUT} \\ M_{6}& =& \left (m(\epsilon _{3})^{(\mathcal{G}-\mathcal{H}-1)/2}(m(\epsilon _{ 3}) + 1)^{\mathcal{H}/2}e^{\mathcal{I}/2}\right )M_{ GUT} \\ \alpha ^{-1}(\varLambda )& =& -\frac{3} {\pi } \text{ln} \frac{\varLambda ^{2}} {M_{GUT}^{2}} + \frac{3} {\pi } \text{ln}\left (m(\epsilon _{3})^{(\mathcal{G}-\mathcal{H})}(m(\epsilon _{ 3}) + 1)^{\mathcal{H}}e^{\mathcal{I}}\right ) \\ & & +\text{ln}\left (m(\epsilon _{3})^{(\mathcal{L}-\mathcal{M})}(m(\epsilon _{ 3}) + 1)^{\mathcal{M}}e^{\mathcal{N}}\right ) {}\end{array}$$

(18.70)