We analyze U(1)$_H$ as a horizontal symmetry and its possibilities to explain
the known elementary-fermion masses. We find that only two candidates, in the
context of SU(3)$_c\otimes$SU(2)$_L\otimes$U(1)$_Y\otimes$U(1)$_H$
nonsupersymmetric, are able to fit the experimental result m$_b<<$m$_t$.Comment: 10 pages, Accepted for publication in Z. Phys. 

Mira, Jesús M.

Ponce, William A.

Zepeda, Arnulfo

English

arXiv

We analyze U(1)_H as a horizontal symmetry and its possibilities to explain the known elementary-fermion masses. We find that only two candidates, in the context of SU(3)_c\otimesSU(2)_L\otimesU(1)_Y\otimesU(1)_H nonsupersymmetric, are able to fit the experimental result m_b<<m_t

Ponce, W A

Zepeda, A

Mira, J M

CERN Document Server

Po
st
Sc
rip
t〉  p
ro
ce
ss
ed
 b
y 
th
e 
SL
A
C/
D
ES
Y
 L
ib
ra
rie
s o
n 
28
 Ju
n 
19
95
.
H
EP
-P
H
-9
50
64
39
IS U(1)
H
A GOOD FAMILY SYMMETRY?
William A. Ponce
1;2
, Arnulfo Zepeda
2
and Jesus M. Mira
1
1-Departamento de Fsica, Universidad de Antioquia
A.A. 1226, Medelln, Colombia.
2-Departamento de Fsica, Centro de Investigacion y de Estudios Avanzados del IPN.
Apartado Postal 14-740, 07000 Mexico D.F., Mexico.
June 28, 1995
ABSTRACT
We analyze U(1)
H
as a horizontal symmetry and its possibilities to explain the known
elementary-fermion masses. We nd that only two candidates, in the context of
SU(3)
c

SU(2)
L

U(1)
Y

U(1)
H
nonsupersymmetric, are able to t the experimental
result m
b
<<m
t
.
Accepted for publication in Z. Phys. C
1
1 INTRODUCTION
The pattern of fermion masses, their mixing, and the family replication, remain
as the most outstanding problems of nowadays particle physics. The successful
standard model (SM) based on the local gauge group SU(3)
c

SU(2)
L

U(1)
Y
can
tolerate, but not explain the experimental results. Two main features that a con-
sistent family theory should provide are:
(i)-Within each charge sector, the masses increase with generation by large factors:
m
u
<< m
c
<< m
t
; m
d
<< m
s
<< m
b
; m
e
<< m

<< m

(ii)-Even if one restricts to the heaviest family, the masses are still quite dierent:
m

 m
b
<< m
t
:
The horizontal survival hypothesis[1] was invented in order to accommodate (i), un-
der the (wrong) assumption that m

 m
b
 m
t
. The idea of radiative symmetry
breaking in a supersymmetric extension of the SM[2] depends crucially on the ex-
istence of one quark with a mass comparable to the SM breaking scale, but it can
not explain why this was the top quark instead of the bottom quark. The modied
horizontal survival hypothesis[3] was introduced in order to explain the full extent
of (i) and (ii), but a dynamically realization of this hypothesis is still lacking. Of
course, these hypothesis and ideas rest on the assumption that all the dimension
four Yukawa couplings in a well behaved theory should be of order one.
Related to (i) and (ii) is the fact that the Cabibbo-Kobayashi-Maskawa quark
mixing matrix is near to the identity, but it is a common prejudice to assume that the
appropriate family symmetry may explain this fact as a consequence of (i) and (ii).
In what follows we will enlarge the SM gauge group with an extra U(1)
H
horizontal
local gauge symmetry (the simplest multi-generational continuous symmetry we can
think of). We then show that the structure SU(3)
c

SU(2)
L

U(1)
Y

U(1)
H
by itself
is able to explain (ii), and that the simplest supersymmetric (SUSY) extension of
this model without a -term can not cope with (ii).
2 SU(3)
c

SU(2)
L

U(1)
Y

U(1)
H
as an anomaly-free model
Our attempt is to keep the number of assumptions and parameters down to the
minimum possible, and to try to construct a model which explains both features
(i) and (ii) at the lowest possible energy scale. We therefore demand cancellation
2
of the triangular (chiral) anomalies[4] by the power counting method, including the
mixed gravitational(grav) anomaly[5]. The alternative of cancelling the anomalies
by a Green-Schwarz mechanism[6] has been already considered in Refs.[7], and corre-
sponds to the construction of a model string-motivated which demands the inclusion
of physics near the Plank scale.
SU(3)
c

SU(2)
L

U(1)
Y

U(1)
H
as a continuous gauge group, with U(1)
H
as a
family symmetry, was introduced long ago in Ref.[8], (revived recently in the context
of SUSY string-motivated models in Refs.[7, 9, 10]). There are two dierent versions
of the model, corresponding to two dierent ways of cancelling the chiral anomalies.
One is demanding cancellation of the anomalies for each generation and the other
one is cancelling the anomalies between generations.
2.1 Cancellation of anomalies in each generation
Assuming there are no right-handed neutrinos, using the U(1)
Y
and U(1)
H
charges
displayed in Table (1), and demanding freedom from chiral anomalies for
SU(3)
c

SU(2)
L

U(1)
Y

U(1)
H
, we get:
[SU(2)
L
]
2
U(1)
H
: Y
 
i
+ 3Y

i
= 0 (1)
[SU(3)
c
]
2
U(1)
H
: 2Y

i
+ Y
U
i
+ Y
D
i
= 0 (2)
[U(1)
Y
]
2
U(1)
H
: 2Y
 
i
+ 4Y
E
i
+
2
3
Y

i
+
16
3
Y
U
i
+
4
3
Y
D
i
= 0 (3)
U(1)
Y
[U(1)
H
]
2
:  Y
2
 
i
+ Y
2
E
i
+ Y
2

i
  2Y
2
U
i
+ Y
2
D
i
= 0 (4)
[grav]
2
U(1)
H
: 2Y
 
i
+ Y
E
i
= 0 (5)
[U(1)
H
]
3
: 2Y
3
 
i
+ Y
3
E
i
+ 6Y
3

i
+ 3Y
3
U
i
+ 3Y
3
D
i
= 0: (6)
The solution to Eqs.(1)-(6) is [8, 11]
Y
i
H

= 
i
Y
SM

;
where 
i
is an arbitrary number dierent for each generation, and Y
SM

is the U(1)
Y
charge for the  multiplet.
These U(1)
H
charges cannot explain the feature (ii) which demands that at tree
level only the top quark acquires a mass, and therefore that the Higgs eld with
U(1)
H
charge Y
H
satises:
Y

3
+ Y
U
3
= Y
H
3
Y
3
+ Y
D
3
6=  Y
H
= Y
H
?
:
But once the rst of these equations is satised, Eq. (2) above implies Y

3
+ Y
D
3
=
 Y
H
. Therefore, if a top quark mass arises at tree level (Y
H
= 
3
), a bottom mass
arises as well at the same level.
Adding right-handed neutrinos N
c
i;
L
to our set of fundamental elds does not
change this conclusion since Eq.(2) stays valid [the only changes are in Eqs. (5) and
(6) which are now replaced by
[grav]
2
U(1)
H
: 2Y
 
i
+ Y
E
i
+ Y
N
i
= 0 (7)
[U(1)
H
]
3
: 2Y
3
 
i
+ Y
3
E
i
+ 6Y
3

i
+ 3Y
3
U
i
+ 3Y
3
D
i
+ Y
3
N
i
= 0]:
2.2 Cancellation of anomalies between generations
If the U(1)
H
anomalies are cancelled by an interplay among generations, Eqs. (1) 
(6) should be understood with a summ over i = 1; 2; 3. Eq. (4) then reads
X
i
( Y
2
 
i
+ Y
2
E
i
+ Y
2

i
  2Y
2
U
i
+ Y
2
D
i
) = 0: (8)
Obviously a solution to the new anomaly constraint equations which are linear or
cubic in the Y

i
is
3
X
i=1
Y
i

= 0
for each . We will limit ourselves to this type of solutions and within this set we
will consider only those for which the  
i
and U
i
H-hypercharges are xed to satisfy
either
Y
 
1
= 
1
 ; Y
 
2
= 
2
=  ; Y
 
3
= 
3
= 0;
Y
U
1
= 
0
1
 
0
; Y
U
2
= 
0
2
=  
0
; Y
U
3
= 
0
3
= 0;
or any set of relations obtained from the former equations by a permutation of the
indices i = 1; 2; 3. The solutions can then be divided onto four classes according to
the way the cancellations occur in Eq.(8).
CLASS A
Y
E
i
= Y
 
i
= 
i
and Y
D
i
= Y

i
= Y
U
i
= 
0
i
; i = 1; 2; 3. A model with a tree-level
4
top quark mass arises if Y
H

= Y

i
+ Y
U
j
for some i and j. There are ve dierent
models in this class characterized by Y
H

= 2
0
;
0
and 0 respectively. Any of
this ve models becomes nonviable if it gives rise to a tree-level bottom mass. That
is if there exists a k and a l for wich Y

k
+ Y
D
l
=  Y
H

. For example, if Y
H

= 2
0
then i = j = 1 and k = l = 2 satisfy the previous equations; this is signaled in
Table (2) by the entry (1,1)
U
; (2,2)
D
in the Class A column and the 2
0
row. The
fact that in Table (2) there is at least one D-type entry for every U-type one for
all the ve models of Class A, means that none of them is viable. This fact can
be easily understood by noticing that Y

i
+ Y
U
j
changes sign under the interchange
1$ 2 in the i; j indices, and that in Class A Y
D
i
= Y
U
i
. Therefore, for a xed Y
H

,
Y

i
+ Y
U
j
=  (Y

k
+ Y
D
l
).
CLASS B
Y

i
= Y
 
i
= 
i
and Y
D
i
= Y
E
i
= Y
U
i
= 
0
i
; i = 1; 2; 3. There are nine dierent models
in this class characterized by Y
H

= (  
0
); (  
0
);;
0
and 0 respectively.
Since again Y
D
i
= Y
U
i
none of those models is viable.
CLASS C
Y
D
i
= Y
 
i
= 
i
and Y

i
= Y
E
i
= Y
U
i
= 
0
i
; i = 1; 2; 3. There are now eleven dierent
models in this class characterized by Y
H

= 2
0
; (  
0
); (  
0
);;
0
and 0.
As can be seen from Table (2) for  6= 
0
;2
0
;3
0
and 
0
6= 0, there are two
models in which only one U-type mass and none D-type one develops at tree-level.
These models are:
Mark I. For a Higgs eld with (U(1)
Y
,U(1)
H
) hypercharges ( 1; 2
0
).
Mark II. For a Higgs eld with (U(1)
Y
,U(1)
H
) hypercharges ( 1; 2
0
).
The rest of the models in this classd are non-viable because a tree-level bottom mass
arises in them.
CLASS D
This is a special class which is a particular case of Classes A,B, and C, for which
 = 
0
, which in turn implies Y
E
i
= Y
 
I
= Y
D
i
= Y

i
= Y
U
i
. As far as the quark
mass spectrum is concerned this class is equivalent to class A.
Two comments: rst, in Ref.[8], the class of solutions A, B, and C were all lumped
together in class D, which in turn forbids the two models classied as Mark I and
Mark II above. Second, adding right-handed neutrino elds does not change our
analysis at all, either by setting Y
N
1
=  Y
N
2
= ; Y
N
3
= 0 (or permutations of the
indices 1,2,3); or by imposing Y
N
i
= 0; i = 1; 2; 3 in order to implement the seesaw
mechanism[12].
5
3 SU(3)
c

SU(2)
L

U(1)
Y

U(1)
H
SUSY
For the supersymmetric extension of the standard model (SSM) new particles of
spin 1/2 are introduced which are the supersymmetric partners of the Higgs elds
and gauge bosons. These higgsinos and gauginos do not contribute to the U(1)
Y
anomaly because they are chosen vector-like with respect to the quantum numbers
of the SM, in such a way that the SM relationship Q
EM
=T
3L
+Y/2 holds, which in
turn implies the U(1)
Y
charges in Table (3). As in the minimal SSM two dierent
supermultiplets of Higgs elds 
D
and 
U
are introduced.
Now, the simplest way to have a gauge symmetry U(1)
H
anomaly-free when the
new spin 1/2 members of the supermultipletes are included, is to demand that these
new fermions are vector-like with respect to U(1)
H
. That is, to impose Y

s
= Y
SM

,
with  an arbitrary constant. For  = 
3
, 
U
and 
D
will produce tree level masses
for the top and bottom quark respectively. This particular solution is not consistent
with (ii).
But is there other solution to the anomaly constraint equations which is consistent
with (ii)? Let us see:
If all the electrically neutral gauginos are allowed to have Majorana masses,
then Y

= Y

0
= Y
g
= Y
(W;B)
=0. Now, the higgsinos


U
and


D
do not carry a
generational index, but if they are to produce masses at least for the third generation,
then their charges have to be related to the charges of the third family (see the third
paper in Ref.[7]). If this is the case then the anomaly cancellation equations are
Eqs. (1)-(6) for i = 1; 2, but for i = 3 they are:
[SU(2)
L
]
2
U(1)
H
: Y
 
3
+ 3Y

3
+ Y

U
+ Y

D
= 0 (9)
[SU(3)
c
]
2
U(1)
H
: 2Y

3
+ Y
U
3
+ Y
D
3
= 0 (10)
[U(1)
Y
]
2
U(1)
H
: 4Y
E
3
+
2
3
Y

3
+
16
3
Y
U
3
+
4
3
Y
D
3
+ 2(Y
 
3
+ Y

U
+ Y

D
) = 0(11)
U(1)
Y
[U(1)
H
]
2
:  Y
2
 
3
+ Y
2
E
3
+ Y
2

3
  2Y
2
U
3
+ Y
2
D
3
+ Y
2

U
  Y
2

D
= 0 (12)
[grav]
2
U(1)
H
: 2Y
 
3
+ Y
E
3
+ 2Y

U
+ 2Y

D
= 0 (13)
[U(1)
H
]
3
: 2Y
3
 
3
+ Y
3
E
3
+ 6Y
3

3
+ 3Y
3
U
3
+ 3Y
3
D
3
+ 2Y
3

U
+ 2Y
3

D
= 0 (14)
There are two solutions to these equations. The rst one is Y

3
= Y
E
3
=6 =  Y
U
3
=4 =
Y
D
3
=2 =  Y

D
=3 =  Y
 
3
=3 and Y

U
=  Y

D
. The second one is Y

3
= Y
E
3
=6 =
 Y
U
3
=4 = Y
D
3
=2 =  Y

D
=3 and Y
 
3
=  Y

U
. For the rst solution masses for
the Up and Down sector are generated simultaneously, and for the second solution
6
YU
3
+ Y

3
6=  Y

U
, failing to give a mass for the Up sector.
So for this extension of the SSM, the U(1)
H
anomalies can not vanish simultane-
ously with the generation of only a tree-level mass for the top quark. The alternative
is to go to higher mass scales and cancell the U(1)
H
anomalies by a Green-Schwarz
mechanism[6]. Then, for SUSY to work there must be a -term, meaning that the
hypercharges of the Higgs elds can be changed[10]. But this analysis has allready
been carried through in the literature[7, 9, 10].
4 ACKNOWLEDGMENTS
This work was partially supported by CONACyT in Mexico and COLCIENCIAS in
Colombia.
7
Table (1)
U(1)
Y
and U(1)
H
charges for the known fermions. i=1,2,3 is a avor index re-
lated to the rst, second and third generations. The Y
SM
values stated are family
independent.
 
i;L
=(N
i
,E
i
)
L
E
c
i;L

i;L
= (U
i
,D
i
)
L
U
c
i;L
D
c
i;L
N
c
i;L
Y
SM
 1 2 1/3  4=3 2/3 0
Y
i
H

Y
 
i
Y
E
i
Y

i
Y
U
i
Y
D
i
Y
N
i
Table(2)
Summary of three-level mass term for all the possible models for the local gauge
group SU(3)
c

SU(2)
L

U(1)
Y

U(1)
H
. A Higgs eld with a hypercharge Y
H

dif-
ferent to the ones in the rst column does not produce a mass term in the quark
sector.
Y
H

CLASS A CLASS B CLASS C
2
0
(1,1)
U
;(2,2)
D
(1,1)
U
 2
0
(2,2)
U
;(1,1)
D
(2,2)
U
0 (1,2)
U
;(2,1)
U
;(3,3)
U
;(1,2)
D
;(2,1)
D
;(3,3)
D
(3,3)
U
;(3,3)
D
(1,2)
U
;(2,1)
U
;(3,3)
U
;(3,3)
D

0
(1,3)
U
;(3,1)
U
;(2,3)
D
;(3,2)
D
(3,1)
U
;(3,2)
D
(1,3)
U
;(3,1)
U
;(2,3)
D
 
0
(2,3)
U
;(3,2)
U
;(1,3)
D
;(3,1)
D
(3,2)
U
;(3,1)
D
(2,3)
U
;(3,2)
U
;(1,3)
D
 + 
0
(1,1)
U
;(2,2)
D
(2,2)
D
  + 
0
(2,1)
U
;(1,2)
D
(2,1)
D
   
0
(1,2)
U
;(2,1)
D
(1,2)
D
    
0
(2,2)
U
;(1,1)
D
(1,1)
D
 (1,3)
U
;(2,3)
D
(3,1)
D
  (2,3)
U
;(1,3)
D
(3,2)
D
Table (3)
U(1)
Y
and U(1)
H
charges for the sparticles of spin 1/2.


and


0
are the gauginos
related to U(1)
Y
and U(1)
H
respectively, ~g stand for the eight gluinos, etc.


U


D




0
~g (
~
W,
~
B)
Y
SM
1  1 0 0 0 0
Y

s
Y

U
Y

D
Y

Y

0
Y
g
Y
(W;B)
8
References
[1] R.Barbieri and D.V.Nanopoulos, Phys. Lett. 91B, 369 (1980);
95B, 43 (1980).
[2] L.Iba~nez and G.G.Ross, Phys. Lett. 110B, 215 (1982);
L.Alvarez Gaume, J.Polchinki and M.Wise, Nucl. Phys. B221,
495 (1983).
[3] W.A.Ponce, A.Zepeda, A.H.Galeana and R.Martnez, Phys.
Rev D44, 2166 (1991).
[4] S.L.Adler, Phys. Rev. 177, 2426 (1969); J.S.Bell and R.Jackiw,
Nuovo Cimento 51, 47 (1969).
[5] A.Salam and R.Delburgo, Phys. Lett. 40B, 381 (1972);
L.Alvarez Gaume and E.Witten, Nucl. Phys. B234, 269 (1983)
[6] M.Green and J.Schwarz, Phys. Lett. 149B, 117, (1984).
[7] L.Iba~nez and G.G.Ross, Phys. Lett 332B, 100 (1994);
E.Papageorgiu, Z.Physik C64, 509 (1994); P.Binetruy and
P.Ramond: \Yukawa Textures and Anomalies", LPTHE-
ORSAY 94/115, UFIFT-HEP-94-19 preprint, unpublished.
[8] A.Davidson, M.Koca and K.C.Wali, Phys. Rev. Lett. 43, 92
(1979); Phys. Rev. D20, 1195 (1979); D21, 787 (1980).
[9] V.Jain and R.Shrock: \Models of Fermion Mass Matrices based
on Flavor and Generation-Dependent U(1) gauge symmetry",
ITP-SB-94-55 preprint, unpublished.
[10] Yosef Nir: \Gauge Unication, Yukawa hierarchy and the 
problem", hep-ph/9504312, Wis-95/18/Apr-PH preprint, un-
published.
[11] W.A.Ponce, Phys. Rev. D36, 962 (1987).
[12] M.Gell-Mann, P.Ramond, and R.Slansky, in \Supergravity",
proceedings of the workshop, Stony Brook, N.Y. (1979), edited
by P.van Nieuwenhuizen and D.Z.Freedman (North-Holland,
Amsterdam 1979). p.315; T.Yanahida, in \Proceedings of the
9
workshop on Unied Theories and the Baryon Number in the
Universe", edited by A.Sawada and A. Sugamoto (KEK report
No. 79-18, Tsukuba-Gun, Ibaraki-Ken, Japan, 1979).
10


Is U(1) $_H$ a Good Family Symmetry?

Abstract

Similar works

Full text

Available Versions

CERN Document Server

Is U(1)H_HH​ a Good Family Symmetry?

Abstract

Similar works

Full text

Available Versions

CERN Document Server

Is U(1) $_H$ a Good Family Symmetry?