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SU(8) gut models with fractionally charged color singlet fermions and low mass magnetic monopoles
Volume 119B, number 1,2,3
PHYSICS LETTERS
SU(8) GUT MODELS WITH FRACTIONALLY
16 December 1982
CHARGED COLOR SINGLET FERMIONS
AND LOW MASS MAGNETIC MONOPOLES
Fang-xiao DONG, Tung-sheng TU, Pei-you XUE and Xian-jian ZHOU Institute of High Energy Physics, Academia Sinica, Beijing, China Received 8 July 1982 Revised manuscript received 12 August 1982
Two SU (8) GUT models (one with fractionally charged color singlet fermions, another with low mass poles) are proposed. Both of them can accommodate three generations of ordinary light fermions.
1. Introduction The observation of fractionally charged objects has been reported several times [ 11. If these observations should be confirmed by further experiments, two possibilities may be considered: (1) Color SU(3) is broken, and liberated quarks [2], or diquarks [3], have been observed. (2) Color SU(3) is not broken, and fractionally charged objects are color singlets. Several authors [4] have discussed the second possibility in SU(7) GUT models. Unfortunately, these models can only accommodate two generations of ordinary fermions. So they are not compatible with the present experimental data. We have examined SU(9) GUT models [5]. Though they can accommodate massive fractional charged color singlets and low mass magnetic monopoles, there exist massless charged color singlets in the theories. In the standard assignment of SU(9) GUT models the asymptotic freedom is lost. These are contradictory with present observation. So, SU(8) GUT’s are worth trying. Here, the charge operator is Q = diag(-4 , - f , -$ , i,0,41,42,q3),butonlyanevennumberoftheq’scan be nonzero. We take q1 = --q2 = 4 f 0. In that case SU(8) reduces to the SU(7) subgroup. Otherwise, there would be charged massless particles in the models. In particular, if we assume 4 = 1, the SU(8) model is related to the SU(7) models discussed by us [6] and Kim [7]. 0 03 l-9 163/82/0000-0000/$02.75
0 1982 North-Holland
magneticmono-
On the other hand, although many GUT models can give a rather satisfactory explanation to the baryon asymmetry of the universe, the monopoles in these theories are very heavy (with mass - 1016 GeV). This is in conflict with present bounds of the energy density of the universe [8]. So, constructing GUT models with low mass magnetic monopoles is also of great interest. In this paper, we discuss two GUT models: one with fractionally charged color singlets, another with low mass magnetic monopoles. 2. Models with fractionally charged color singlets 2.1. Fermion assignment. If repetition of any representation of SU(8) is not allowed there is no solution at all in the SU(8) GUT which can accommodate three generations of ordinary fermions as welI as fractionally charged color singlets. So, we consider the models in which some representations are repeated. We found that the following representations 3 [8,11~+2[8,21~+[8,3I~ 3
(2.1)
are anomaly free. Where [N, m] denotes m rank totally anti-symmetric tensor of SU(N). Choosing the electric charge operator of SU(8) as e=diag(-~,-~,-;,l,O,q,-q,O),
(2.2)
We have the following decomposition of (2.1) with respect to SU(5): 121
Volume 119B, number 1,2, 3
PHYSICS LETTERS
318, 1]R + 218,2]L + [8,3]R 3 [5, 1] R(0) + 3 [5,0] R(A) + 2 [5,2] L(0) + 2 [5,0]L(B ) + 2[5, 1]L(A ) + [5,3]R(0) + [5,21R(A) + [5, llR(B) + [5,01R(0),
(2.3)
Where, [5, 0] L,R denote SU(5) singlets, A = (q, --q, 0), B = ( - q , q, 0) = - A , [5,n] (0) represents the standard charge assignment of SU(5), and [5,n] (A) denotes standard charge plus q, or - q or 0, respectively. Note that, [5,3] R = 10~ ~ 10L, eq. (2.3) becomes
16 December 1982
where a :# b 4=c, 4a + 3b + e = 0 and a, b, c, d ~ M 2 >~ 1015 GeV. The corresponding Higgs potential will get a minimum [9] at the condition of (2.6) and (2.7) which can break SU(8) down to SU(4) X SU(3) × U(1)'. The generator of U(1)' is Y' = diag ( - 3 , - 3 , - 3 , 4 , 4 , 4, - 3 , 0 ) .
(2.8)
+ [5L(0 ) + SR(0)] + [5L(q ) + 10R(q) ]
The Higgs representation at the first step does not depend on the value of q in eq. (2.2). But the Higgs representation at the second step depends on q. So, we first discuss the case o f q = - 1 / 3 . In that case the charge operator Q, the U(1) generator Y are as follows
+ [5L(--q) + 10R(--q)] + [SL(q) + 5R(q)l
Q= diag(--~,
+ [5L(--q) + 5R(--q) ] + 4 X 1R(0 ) + 2X 1L(0)
Y--diag(
+ 3 X I R ( q ) + 3 X I R ( - - q ) + 2 X 1L(q)+2X 1L(--q).
We choose mixed tensor Higgs ~mn}~. at this step, where the upper indices i/are anti-symmetric, the lower indices (lm n)are symmetric. The non.vanishing vev's of ¢ff(lmn)are 67 ~ 45 (2.11) <¢{888)) <~b{777))~M1< M 2 .
3 [5R(O) + lOL(O)] + [5L(O) + 1OR(O)]
(2.4) From eq. (2.4) we can see that there are three generations of ordinary fermions, one generation with standard charge and V+A weak coupling, two generations with non-standard charge and V+A weak coupling, a half generation with standard charge, two half generations with non-standard charge, and 16 SU(5) singlets with standard and non-standard charges. If we choose the spontaneous symmetry breaking mechanism appropriately to prevent fermions from getting superheavy mass, the number of flavour of low mass fermions is 15. Thus, the theory is asymptotically free. We shall see below, only in the two cases, q = - 1 / 3 and q = - 2 / 3 , we can obtain the right value of sin2Ow and accommodate fractionally charged color singlets at the same time. 2.2. Spontaneous symmetry breaking. We break SU(8) down to SU(3) C X U(1)em by three steps: SU(8) M2 SU(4) X SU(3) X U'(1) M1
SU(3)c × SU(2)w × U(1)
Me SU(3)C X U(1)e m .
(2.5)
At the first step, we use the adjoint ~} and a vector 4/with the non-vanishing vev's (~) = diag (a, a, a, b, b, b, a, c),
(2.6)
(¢i) = (0, O, O, O, O, O, O,d) T ,
(2.7)
122
1
3~
1 3, 1
3,
1 1 ½, 1, 0, - ~, ~, 0)
1 1 1
3,2,2,
1 1 0) "
3,3,
(2.9) (2.10)
It is easy to prove that, after the second step symmetry breaking we have the right residual symmetry. For the case q = - 2 / 3 2 0) , Q = d i a g ( - ~t, - ~1, - ~ , 1 1, 0 , - ~2, ~,
(2.12)
Y=diag(
(2.13)
x3,
t
3,
1
1
1
3,2,2,
2
2 O) "
3,3,
The Higgs representation is much more complicated. Of course we can still fred the appropriate Higgs representation to break SU(4) X SU(3) × U'(1), down to SU(3)c × SU(2) w × U(1). We shall not give it here. For the third step, we use another vector Higss with non-vanishing vev (~5) ~ M 0 ~, 102 GeV. This is the same as standard GUT models. In order to give fermion masses, other Higgs fields H are also needed. We shall not go into detail about H. In both masses ofq = - 1 / 3 and q = - 2 / 3 , U'(1) and U(1) are not orthogonal to each other. So, monopoles in these models obtain mass [ 10] of order of M2/ot. Where tx is the fme structure constant. Also, in these models, we have color singlets with charge -+1/3, +2/3, -+4/3, and 5/3. In order to prevent fermions from obtaining mas-
Volume l19B, number 1,2,3
PHYSICS LETTERS
ses at the first and the second step of spontaneous symmetry breaking, we introduce the following discrete symmetry [ 11 ] ~/R-+--i~/R ,
~b~i~/L/,
~kiiRk-~--i~b~k,
¢i~¢i ,
¢: __>¢:,
(2.14) (2.15)
¢i~lrnn}-~¢i(lmn } ,
(2.16)
~oi ~ - - ~ ,
(2.17)
H~-H,
All gauge fields are unchanged. ~bi, ~ki/, ~iik denote the fermion representation [10,1], [10,2], [10.3] respectively. H represent all other Higgs fields introduced for developing fermion masses at the last step symmetry breaking. According to this discrete symmetry, fermions can only obtain mass at the last step of symmetry breaking. 2.3. Renormalization of sin2 0 w . Using the standard method of the renormalization equation of sin 2 O [12], we have the following coupled equations. sin20(M O) = a(Mo) + ~-~not(Mo)ln(M2/Mo) O~s(Mo)
(2.18)
16 December 1982
Table 1 M2
c~5(Mo) sin20(Mo) MI(-~)
MI(-~) MI(-I)
10Is 0.20 0.19 0.18 0.17
0.175 0.177 0.179 0.182
4.0X1011 2.9X106 2.2×103 2.7)< 1011 2.7× 106. 1.8× 103 1.8X1011 1.8×106 1.4×103 9.5× 10l° 1.2×106 1.0× 103
0.16 0.14 0.12
0.185 0.192 0.201
5.1x101° 8.5xlO s 7.6x102 1.2xlO I° 3.6xi0 s 4.5xI~02 1.9x I09 1.2xI0$ 1.6xI02
10x6 0.20 0.19 0.18 0.17 0.16 0.14 0.12
0.185 0.187 0.190 0.192 0.195 0.202 0.211
2.8× 101° 1.8×101° 9.9x 109 6.6x 109 3.6× 109 8.4×10 s 1.3x 108
1.2x106 9.2×10 s 6.4× 10s 5.0x lOs 3.5×105 1.5xlO s 4.9x 104
1.2x103 9.6×102 7.0× 102 5.7x 102 4.1×102 2 x102 -
1017
0.196 0.199 0.200 0.203 0.206 0.212 0.222
1.6X109 8.5 Xl0 s 6.9X 10 s 3.7X 10 a 2.0X 10 s 5.8X 107 7.5X 106
4.2X10 $ 2.9X 10 s 2.6X 10 s 1.8X 10 s 1.2X 10 s 5.9X 104 1.7X 104
5.8X102 4.2X 102" 3.8X 102 2.8X 102 2 X 102 -
0.20 0.19 0.18 0.17 0.16 0.14 0.12
8 $
1 + 4q 2 +
~ + 4 q 2 sin2O(Mo)=lla(Mo)ln(M1/Mo) ~5 + 4q 2 I33 _ 55(1+ 6q 2) - 2641] L
-5
j ln(M2/M1),
(2.19) where a(M0) and as(M0) are the f'me structure constant and the strong coupling constant at the energy scale M 0 respectively. TakingM 0 ~ 100 GeV, then [13] a(M0) = 1/128.5. Notice that eqs. (2.18) and (2.19), are obtained under l-loop approximation and neglecting Higgs boson contributions. Given M2, M 0 ~~s(M0), we can calculate sin 20(Mo) from eq. (2.18), t h e n M 1 from (2.19). Obviously, for q = - 1 / 3 and q = - 2 / 3 the values o f M 1 are different. We list all the parameters in table 1, where M l ( - 1 / 3 ) , M 1 ( - 2 / 3 ) denote the values ofM 1 for q = - 1 / 3 and - 2 / 3 respectively. And all masses are in the unit of GeV.
3. Model with low mass magnetic monopoles. 3.1. Fermion assignment. The fermion assignment is almost the same as the models in section 2 except for the value o f q . Here, q = - 1 instead o f - l / 3 and -2/3.
3.2. Spontaneous symmetry breaking and low mass magnetic monopoles. We use the same pattern as described in eq. (2.5). The first and the third steps are exactly the same as in section 2. The second step is a bit different. When q = - 1 , the charge operator Q, the U'(1) generator Y' and the U(1) generator Y are as follows: Q = diag ( 3, 1
1 3,
½,1,0,-1,1,0),
(3.1)
Y' = diag ( - 3 , - 3 , - 3 , 4, 4, 4, - 3 , 0 ) ,
(3.2)
Y = d i a g ( - - ~ , - ~1, - ~i, ~ 1, ~ ,a - 1 , 1 , 0 ) .
(3.3)
From (3.2) and (3.3) we have tr (YY') = 0 .
(3.4)
This means that U(1) is orthogonal to U'(1). Thus, monopoles in this model will obtain mass [10] of order of M1/a. Because M 1 ~ 103 GeV (see section 3.3), the monopole mass ~ 105 GeV. This is compatible with the present bounds of the energy density of the universe. Now we come back to the second step of spontaneous symmetry breaking. Actually, it is easy to prove that three rank totally anti-symmetric Higgs ¢i/k with non-vanishing vev's 123
Volume 119B, number 1,2, 3
PHYSICS LETTERS
(~b675) ~ (¢456) ~.M1 will break SU(4) X SU(3) X U(1) down to SU(3)c X SU(2)w X U(1). In this model there are only color ringlet fermions with electric charges O, +1, +2, but no color singlet fermions with fractional charge. 3.3. Renormalization of sin20 w. Substituting q = - 1 into eq. (2.19), we obtain the following coupled equations ~20(MO)-
a(MO) + ~ a ( M o ) I n ( M 2 / M O ) , %(MO)
3 - 20sin2O(M O) =
(3.5)
1--~-7a(Mo)in(M1/Mo)
~ ot(Mo) 111(M2/M1) .
(3.6)
Using the same method illustrated in section 2.3, we calculate all the parameters and list them in table 1 too.
4. Conclusion We have investigated SU(8) GUT models systematically. We found that there is no SU(8) GUT model which can accommodate fractionally charged color singlets and low mass magnetic monopoles in the same time. In order to obtain the fight value of fin20 w we have to choose a spontaneous symmetry breaking m e c h ~ i s m with at least three steps, and to choose the charge operator Q carefully. For example, the value o f q, 1/2, 2/3, 1 does not give the right number o f sin20(M0).
124
16 December 1982
References [1] G.S. La Rue, J.D. Phillips and W.M. Fairbank, Phys. Rev. Lett. 46 (1981) 967, and references therein. [2] A. de Rujula, R.C. Giles and R.L. Jaffe, Phys. Rev. D17 (1978) 285. [3] R. Slansky, T. Goldman and G.L. Shaw, to be pubfished. [4] L.F. Li and F. Wilczek, Phys. Lett. 107B (1981) 64; H. Goldberg, T.W. Kephart and M.T. Vaughn, Phys. Rev. Lett. 47 (1981) 1429. [5] F.S. Dong, T.S. Tu, P.Y. Xue and X.J. Zhou, unpublished. [6] Z. Q. Ma, T.S. Tu, P.Y. Xue, and Z.W. Yue, BIHEP-TH-1 (May 1980); SeA.Sinica 24 (1981) 1538; 1536; Z.Q. Ma, T.S. Tu, P.Y. Xue and X.J. Zhou, BIHEP-TH-3 (May, 1980), Physiea Energiae Forts at Physica Nuel. 5 (1981) 664. [7] J.E. Kim, Phys. Rev. Lett. 45 (1980) 1916; Phys. Rev. D23 (1981) 2706. [8] A.H. Guth, SLAC-PUB-2576 (1980) (T); A.H. Guth, and E.J. Weinberg, CU-TP-183 (1980); Y.B. Zeldovich and M.Y. Khlopov, Phys. Lett. 79B (1978) 239; J.P. Preskill, Phys. Rev. Lett. 43 (1979) 1365. [9] F. BucceUa, H. Ruegg and C.A. Savoy, Nucl. Phys. B169 (1980) 68. [101 D.M. Scott, Nucl. Phys. B171 (1980) 95; P.Y. Xue, Phys. Lett. 105B (1981) 147. [11] N.P. Chang, P.Y. Xue and X.J. Zhou, A new method for flavor unifications, Sci. Sinica (1982), to be published. [12] H. Georgi, H. Quinn and S. Weinberg, Phys. Rev. Lett. 33 (1974) 451; S. Dawson and H. Georgi, Phys. Rev. Lett. 43 (1979) 821; I. Umemura and K. Yamamoto, Prog. Theor. Phys. 64 (1980) 278. [131 W.J. Marciano, Phys. Rev. D20 (1979) 274.
PHYSICS LETTERS
SU(8) GUT MODELS WITH FRACTIONALLY
16 December 1982
CHARGED COLOR SINGLET FERMIONS
AND LOW MASS MAGNETIC MONOPOLES
Fang-xiao DONG, Tung-sheng TU, Pei-you XUE and Xian-jian ZHOU Institute of High Energy Physics, Academia Sinica, Beijing, China Received 8 July 1982 Revised manuscript received 12 August 1982
Two SU (8) GUT models (one with fractionally charged color singlet fermions, another with low mass poles) are proposed. Both of them can accommodate three generations of ordinary light fermions.
1. Introduction The observation of fractionally charged objects has been reported several times [ 11. If these observations should be confirmed by further experiments, two possibilities may be considered: (1) Color SU(3) is broken, and liberated quarks [2], or diquarks [3], have been observed. (2) Color SU(3) is not broken, and fractionally charged objects are color singlets. Several authors [4] have discussed the second possibility in SU(7) GUT models. Unfortunately, these models can only accommodate two generations of ordinary fermions. So they are not compatible with the present experimental data. We have examined SU(9) GUT models [5]. Though they can accommodate massive fractional charged color singlets and low mass magnetic monopoles, there exist massless charged color singlets in the theories. In the standard assignment of SU(9) GUT models the asymptotic freedom is lost. These are contradictory with present observation. So, SU(8) GUT’s are worth trying. Here, the charge operator is Q = diag(-4 , - f , -$ , i,0,41,42,q3),butonlyanevennumberoftheq’scan be nonzero. We take q1 = --q2 = 4 f 0. In that case SU(8) reduces to the SU(7) subgroup. Otherwise, there would be charged massless particles in the models. In particular, if we assume 4 = 1, the SU(8) model is related to the SU(7) models discussed by us [6] and Kim [7]. 0 03 l-9 163/82/0000-0000/$02.75
0 1982 North-Holland
magneticmono-
On the other hand, although many GUT models can give a rather satisfactory explanation to the baryon asymmetry of the universe, the monopoles in these theories are very heavy (with mass - 1016 GeV). This is in conflict with present bounds of the energy density of the universe [8]. So, constructing GUT models with low mass magnetic monopoles is also of great interest. In this paper, we discuss two GUT models: one with fractionally charged color singlets, another with low mass magnetic monopoles. 2. Models with fractionally charged color singlets 2.1. Fermion assignment. If repetition of any representation of SU(8) is not allowed there is no solution at all in the SU(8) GUT which can accommodate three generations of ordinary fermions as welI as fractionally charged color singlets. So, we consider the models in which some representations are repeated. We found that the following representations 3 [8,11~+2[8,21~+[8,3I~ 3
(2.1)
are anomaly free. Where [N, m] denotes m rank totally anti-symmetric tensor of SU(N). Choosing the electric charge operator of SU(8) as e=diag(-~,-~,-;,l,O,q,-q,O),
(2.2)
We have the following decomposition of (2.1) with respect to SU(5): 121
Volume 119B, number 1,2, 3
PHYSICS LETTERS
318, 1]R + 218,2]L + [8,3]R 3 [5, 1] R(0) + 3 [5,0] R(A) + 2 [5,2] L(0) + 2 [5,0]L(B ) + 2[5, 1]L(A ) + [5,3]R(0) + [5,21R(A) + [5, llR(B) + [5,01R(0),
(2.3)
Where, [5, 0] L,R denote SU(5) singlets, A = (q, --q, 0), B = ( - q , q, 0) = - A , [5,n] (0) represents the standard charge assignment of SU(5), and [5,n] (A) denotes standard charge plus q, or - q or 0, respectively. Note that, [5,3] R = 10~ ~ 10L, eq. (2.3) becomes
16 December 1982
where a :# b 4=c, 4a + 3b + e = 0 and a, b, c, d ~ M 2 >~ 1015 GeV. The corresponding Higgs potential will get a minimum [9] at the condition of (2.6) and (2.7) which can break SU(8) down to SU(4) X SU(3) × U(1)'. The generator of U(1)' is Y' = diag ( - 3 , - 3 , - 3 , 4 , 4 , 4, - 3 , 0 ) .
(2.8)
+ [5L(0 ) + SR(0)] + [5L(q ) + 10R(q) ]
The Higgs representation at the first step does not depend on the value of q in eq. (2.2). But the Higgs representation at the second step depends on q. So, we first discuss the case o f q = - 1 / 3 . In that case the charge operator Q, the U(1) generator Y are as follows
+ [5L(--q) + 10R(--q)] + [SL(q) + 5R(q)l
Q= diag(--~,
+ [5L(--q) + 5R(--q) ] + 4 X 1R(0 ) + 2X 1L(0)
Y--diag(
+ 3 X I R ( q ) + 3 X I R ( - - q ) + 2 X 1L(q)+2X 1L(--q).
We choose mixed tensor Higgs ~mn}~. at this step, where the upper indices i/are anti-symmetric, the lower indices (lm n)are symmetric. The non.vanishing vev's of ¢ff(lmn)are 67 ~ 45 (2.11) <¢{888)) <~b{777))~M1< M 2 .
3 [5R(O) + lOL(O)] + [5L(O) + 1OR(O)]
(2.4) From eq. (2.4) we can see that there are three generations of ordinary fermions, one generation with standard charge and V+A weak coupling, two generations with non-standard charge and V+A weak coupling, a half generation with standard charge, two half generations with non-standard charge, and 16 SU(5) singlets with standard and non-standard charges. If we choose the spontaneous symmetry breaking mechanism appropriately to prevent fermions from getting superheavy mass, the number of flavour of low mass fermions is 15. Thus, the theory is asymptotically free. We shall see below, only in the two cases, q = - 1 / 3 and q = - 2 / 3 , we can obtain the right value of sin2Ow and accommodate fractionally charged color singlets at the same time. 2.2. Spontaneous symmetry breaking. We break SU(8) down to SU(3) C X U(1)em by three steps: SU(8) M2 SU(4) X SU(3) X U'(1) M1
SU(3)c × SU(2)w × U(1)
Me SU(3)C X U(1)e m .
(2.5)
At the first step, we use the adjoint ~} and a vector 4/with the non-vanishing vev's (~) = diag (a, a, a, b, b, b, a, c),
(2.6)
(¢i) = (0, O, O, O, O, O, O,d) T ,
(2.7)
122
1
3~
1 3, 1
3,
1 1 ½, 1, 0, - ~, ~, 0)
1 1 1
3,2,2,
1 1 0) "
3,3,
(2.9) (2.10)
It is easy to prove that, after the second step symmetry breaking we have the right residual symmetry. For the case q = - 2 / 3 2 0) , Q = d i a g ( - ~t, - ~1, - ~ , 1 1, 0 , - ~2, ~,
(2.12)
Y=diag(
(2.13)
x3,
t
3,
1
1
1
3,2,2,
2
2 O) "
3,3,
The Higgs representation is much more complicated. Of course we can still fred the appropriate Higgs representation to break SU(4) X SU(3) × U'(1), down to SU(3)c × SU(2) w × U(1). We shall not give it here. For the third step, we use another vector Higss with non-vanishing vev (~5) ~ M 0 ~, 102 GeV. This is the same as standard GUT models. In order to give fermion masses, other Higgs fields H are also needed. We shall not go into detail about H. In both masses ofq = - 1 / 3 and q = - 2 / 3 , U'(1) and U(1) are not orthogonal to each other. So, monopoles in these models obtain mass [ 10] of order of M2/ot. Where tx is the fme structure constant. Also, in these models, we have color singlets with charge -+1/3, +2/3, -+4/3, and 5/3. In order to prevent fermions from obtaining mas-
Volume l19B, number 1,2,3
PHYSICS LETTERS
ses at the first and the second step of spontaneous symmetry breaking, we introduce the following discrete symmetry [ 11 ] ~/R-+--i~/R ,
~b~i~/L/,
~kiiRk-~--i~b~k,
¢i~¢i ,
¢: __>¢:,
(2.14) (2.15)
¢i~lrnn}-~¢i(lmn } ,
(2.16)
~oi ~ - - ~ ,
(2.17)
H~-H,
All gauge fields are unchanged. ~bi, ~ki/, ~iik denote the fermion representation [10,1], [10,2], [10.3] respectively. H represent all other Higgs fields introduced for developing fermion masses at the last step symmetry breaking. According to this discrete symmetry, fermions can only obtain mass at the last step of symmetry breaking. 2.3. Renormalization of sin2 0 w . Using the standard method of the renormalization equation of sin 2 O [12], we have the following coupled equations. sin20(M O) = a(Mo) + ~-~not(Mo)ln(M2/Mo) O~s(Mo)
(2.18)
16 December 1982
Table 1 M2
c~5(Mo) sin20(Mo) MI(-~)
MI(-~) MI(-I)
10Is 0.20 0.19 0.18 0.17
0.175 0.177 0.179 0.182
4.0X1011 2.9X106 2.2×103 2.7)< 1011 2.7× 106. 1.8× 103 1.8X1011 1.8×106 1.4×103 9.5× 10l° 1.2×106 1.0× 103
0.16 0.14 0.12
0.185 0.192 0.201
5.1x101° 8.5xlO s 7.6x102 1.2xlO I° 3.6xi0 s 4.5xI~02 1.9x I09 1.2xI0$ 1.6xI02
10x6 0.20 0.19 0.18 0.17 0.16 0.14 0.12
0.185 0.187 0.190 0.192 0.195 0.202 0.211
2.8× 101° 1.8×101° 9.9x 109 6.6x 109 3.6× 109 8.4×10 s 1.3x 108
1.2x106 9.2×10 s 6.4× 10s 5.0x lOs 3.5×105 1.5xlO s 4.9x 104
1.2x103 9.6×102 7.0× 102 5.7x 102 4.1×102 2 x102 -
1017
0.196 0.199 0.200 0.203 0.206 0.212 0.222
1.6X109 8.5 Xl0 s 6.9X 10 s 3.7X 10 a 2.0X 10 s 5.8X 107 7.5X 106
4.2X10 $ 2.9X 10 s 2.6X 10 s 1.8X 10 s 1.2X 10 s 5.9X 104 1.7X 104
5.8X102 4.2X 102" 3.8X 102 2.8X 102 2 X 102 -
0.20 0.19 0.18 0.17 0.16 0.14 0.12
8 $
1 + 4q 2 +
~ + 4 q 2 sin2O(Mo)=lla(Mo)ln(M1/Mo) ~5 + 4q 2 I33 _ 55(1+ 6q 2) - 2641] L
-5
j ln(M2/M1),
(2.19) where a(M0) and as(M0) are the f'me structure constant and the strong coupling constant at the energy scale M 0 respectively. TakingM 0 ~ 100 GeV, then [13] a(M0) = 1/128.5. Notice that eqs. (2.18) and (2.19), are obtained under l-loop approximation and neglecting Higgs boson contributions. Given M2, M 0 ~~s(M0), we can calculate sin 20(Mo) from eq. (2.18), t h e n M 1 from (2.19). Obviously, for q = - 1 / 3 and q = - 2 / 3 the values o f M 1 are different. We list all the parameters in table 1, where M l ( - 1 / 3 ) , M 1 ( - 2 / 3 ) denote the values ofM 1 for q = - 1 / 3 and - 2 / 3 respectively. And all masses are in the unit of GeV.
3. Model with low mass magnetic monopoles. 3.1. Fermion assignment. The fermion assignment is almost the same as the models in section 2 except for the value o f q . Here, q = - 1 instead o f - l / 3 and -2/3.
3.2. Spontaneous symmetry breaking and low mass magnetic monopoles. We use the same pattern as described in eq. (2.5). The first and the third steps are exactly the same as in section 2. The second step is a bit different. When q = - 1 , the charge operator Q, the U'(1) generator Y' and the U(1) generator Y are as follows: Q = diag ( 3, 1
1 3,
½,1,0,-1,1,0),
(3.1)
Y' = diag ( - 3 , - 3 , - 3 , 4, 4, 4, - 3 , 0 ) ,
(3.2)
Y = d i a g ( - - ~ , - ~1, - ~i, ~ 1, ~ ,a - 1 , 1 , 0 ) .
(3.3)
From (3.2) and (3.3) we have tr (YY') = 0 .
(3.4)
This means that U(1) is orthogonal to U'(1). Thus, monopoles in this model will obtain mass [10] of order of M1/a. Because M 1 ~ 103 GeV (see section 3.3), the monopole mass ~ 105 GeV. This is compatible with the present bounds of the energy density of the universe. Now we come back to the second step of spontaneous symmetry breaking. Actually, it is easy to prove that three rank totally anti-symmetric Higgs ¢i/k with non-vanishing vev's 123
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PHYSICS LETTERS
(~b675) ~ (¢456) ~.M1 will break SU(4) X SU(3) X U(1) down to SU(3)c X SU(2)w X U(1). In this model there are only color ringlet fermions with electric charges O, +1, +2, but no color singlet fermions with fractional charge. 3.3. Renormalization of sin20 w. Substituting q = - 1 into eq. (2.19), we obtain the following coupled equations ~20(MO)-
a(MO) + ~ a ( M o ) I n ( M 2 / M O ) , %(MO)
3 - 20sin2O(M O) =
(3.5)
1--~-7a(Mo)in(M1/Mo)
~ ot(Mo) 111(M2/M1) .
(3.6)
Using the same method illustrated in section 2.3, we calculate all the parameters and list them in table 1 too.
4. Conclusion We have investigated SU(8) GUT models systematically. We found that there is no SU(8) GUT model which can accommodate fractionally charged color singlets and low mass magnetic monopoles in the same time. In order to obtain the fight value of fin20 w we have to choose a spontaneous symmetry breaking m e c h ~ i s m with at least three steps, and to choose the charge operator Q carefully. For example, the value o f q, 1/2, 2/3, 1 does not give the right number o f sin20(M0).
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