Karlstads Universitet
Institutionen för Ingenjörsvetenskap, Fysik och Matematik
Examinator: Jürgen Fuchs
SU5 and its use in grand
unification theories
Arbete i Symmetrier, Grupper och Algebror
Fysikprogrammet
Jonas Björnsson
SU5 and its use in grand unification theories
I will begin with describing sl3 and the su3 and its use in describing flavor symmetry of the
quarks. Then I will explain a little about the standard model and its limitations and why the
SU5 group is interesting in grand unification.
Flavor symmetry, sl3 and su3
A basis of sl3 consists of 8 traceless matrices, which in Cartan-Weyl basis have the following
relations, i=1,2,θ, where θ denotes the highest root.
H
H
1
2

, E   2  
, E i  2   1,i   2,i   i , E i
i

  1,i   i , E i
2 ,i
One can construct new generators by linear combinations of elements in the Cartan
subalgebra of sl3.
1 0 0

1 1 1
I  H  0 1 0
2
2

0 0 0
1 0 0 

1 1
1
2
Y  H  2  H  0 1 0 
3
3

0 0  2


This basis fulfills.
 1 0 0 1 0 0 
1 0 0
1
 
 1 

tr ( I  Y )  tr  0  1 0    0 1 0    tr  0  1 0    0
 6 0 0 0 0 0  2 6 0 0 0
 


 

Because of sl3 is simple the trace of the matrices forms a scalar product between the
generators. This implies that the generators are orthogonal. The coordinates for the roots can
be calculated in the following way, set H  ( I , Y ) and calculate H , E i for i=1,2,θ.
1
1
1
 1

 (1) E 1  H , E 1   H 1 , E 1 , H 1 , E 1  2 H 2 , E 1     2,  2  2  (1)  E 1 
3
3
2
 2





 
 


 
 

 
 


 1,0 E 1
 ( 2 ) E 2  H , E 2   


1
1 1 2 1
 1

H , E  , H 1 , E 2  2 H 2 , E 2     (1),  (1)  2  (2)  E 2 
3
3
2
 2

  1 ,1 E 2
2
  E   H , E    
 
 1 ,1 E 
2

1 1  1
 1 1

H , E  , H 1 , E   2 H 2 , E     1, 1  2  1 E  
3
2
 2 3

To all of these roots there exists a corresponding negative root. If one draws all theses roots in
a two-dimensional root space one gets a hexagon. These vectors forms a basis for an 8
dimensional irreducible representation of sl3, with two distinct vectors pointing at the middle
of the hexagon which represent the I and Y generators. This is the adjoint representation of
sl3. The Lie algebra operation on the representation space can be seen as transforming the
base vectors to linear combinations of each other.
Y
α
θ
(2)
α(1)
I
There exist nine mesons and by associating I to the z-component of the isospin and Y to the
hypercharge of the particle one can then use the 8-dimensional representation of sl3. To every
base vector one associates a particle with the same values of these quantities, these particles
from a so-called meson octet, and with the ninth particle in a 1-dimensional irreducible
representation.
Y
K0
K+
πη
K-
π+ I
π0
K0
In the same way the light baryons from a baryon octet and the heavy baryons from a decuplet,
a 10-dim irreducible representation. A very important quantity, which is associated to the
isospin and the hypercharge generators, is the electric charge Q.
1
QI Y
2
The adjoint representation has not the lowest dimension of all irreducible ones. The nontrivial representation with the lowest dimension is the 3 dimensional, this one exists in two,
non-isomorphic, representations. From this the quark model arises. When you study this in
more detail the su3 is the one that explains this but there is a one to one correspondence
between the finite dimensional representation of the sl3 and su3. The quarks and anti-quarks
from 3 and 3 respectively.
Y
1/3
d
-1/2
-2/3
u
I
1/2
Y
s 2/3
-1/2
1/2
u -1/3
I
d
s
If one studies these pictures on can see that the up-quark corresponds to the Λ(1) the antistrange-quark corresponds to Λ(2). Where the scalar product between α(i) and Λ(j) is.

(i )

, ( j)   i j
By the quark model the mesons can be seen as bound states between one quark and one antiquark and a baryon as bound state between three quarks. The bound states are tensor products
between representations.
3  3  1 8
3  3  3  1  8  8  10
By investigating this one can see that one trivial and one 8-dimensional irreducible
representation construct the mesons and one trivial, two 8-dimensional and one 10dimensional construct the baryons. This was a description of flavor symmetry of quarks.
Relations between Lie algebras and Lie groups
The relation between Lie algebras and groups, which is interesting in this presentation, can be
seen like this. If one has a basis λa for a Lie algebra g, with finite dimension d, one can
describe the associated group H, where βi is any suitable choice of numbers.
 d

G  exp  i    i  i 
 i 1

GH
Standard model
The standard model combines the SU3c color gauge group for the strong interactions, this is
not the same as above described flavor SU3 group and the description is beyond the scope of
this presentation, and SU2xU1 gauge group that describes the weak and electromagnetic
interactions. This model has successfully explained, nearly, all of today’s experiments and
many think that it’s mathematically consistent. The standard model has three coupling
constants, g1 for the electromagnetic force, g2 for the weak force and g3 for the strong force.
This arises in the lagrangian for the description for each of the interactions. The model
includes 12 gauge bosons that are responsible for the interactions. These gauge particles are
the photon for the electromagnetic interaction, three weak bosons, W+/- and Z, for the weak
interactions and the eight gluons for the strong interactions.
The lagrangian for the electromagnetic interaction is.
1
L   F F   ( D  )( D   *)  m *
4
With Dμ and Fμν.
F    A   A
D     ig 1 A
Aμ is the field that describes the photon. This lagrangian is invariant under this gauge
transform and photon field satisfying.
  exp( i ( x))  
A ( x)  A ( x) 
1
   ( x)
g1
The lagrangian that describes the strong force is.
L
1 8
(a)
F F  ( a )  ( D  )( D   *)  m *

4 a 1
With Dμ and Fμν(a) a little different because of the non-commuting gauge fields.
F
(a)
   A
(a)
  A
8

D     ig 3  A
a 1
(a)
(a)
 a
i f
a
bc
A

(b )
, A
(c)


Aμ(a) is the fields for the gluons and λa and fabc the basis and structure constant for the Lie
algebra su3. This lagrangian is invariant under the this gauge transform, and Aμ(a) satisfying.

8


a 1

  exp   i    a ( x)  a   
A ( x)  UA U 1 
8
A   A
(a)
i
U  U 1
g3
 a
a 1
Where U is an element in the group SU3. The lagrangian for the weak interactions is, similar
as for the strong force but with a=1,2,3 and U is an element in the SU2 group instead and g2
instead of g3.
L
1 3
1 3
(a)
 ( a )


F
F

(
D

)(
D

*)

m

*

M ( a ) A A  




4 a 1
2 a 1
The last term arises because of that the weak bosons have mass, this is where the higgsfields
arises to make the lagrangian invariant.
A problem with the standard model is that it has many free parameters, at least 19 and as
many as 26 if the neutrino has mass. The free parameters arise from the coupling constants
and from the existence of massive particles. More problems are that it doesn’t explain why the
electric charge is quantised, in the u1 the eigenvalues aren’t quantised, and why the charge of
the quarks and leptons are related by a simple factor as three.
Grand Unification Theories, GUT’s
The simplest grand unification theory is the SU5 gauge group. The associated algebra is
simple and has a basis of 24 matrices that is traceless and hermitian. One of the reasons for
this algebra to be interesting is that its Cartan subalgebra has the same dimension as the one of
su3cxsu2xu1, four, and it doesn’t bring any new fermions into the theory. su5 have su3cxsu2xu1
as a subalgebra, which is necessary, and has only one coupling constant, g5. If one examines
the measured values of the coupling-constants g1, g2 and g3 at approximately 102 GeV one
concludes that they aren’t equal.
g 32
g12
g 22
1
1
1
1 

, 2 
 , 3 

4 100
4 30
4 10
But they aren’t constants, they vary as a function of energy. How they vary depends on what
groups one is examining, they depend of the number of particles in the theory, and for the SU5
case they vary as follows. α1 increasing slowly, α2 decreases slowly and α3 decreases a bit
more rapid with higher energy. With the known values in the mid-seventies one extrapolated
those and found that they met approximately at one single point at 1015 GeV. The value they
got were assumed to be the value of the coupling constant g5. Therefore it was interesting to
study the SU5 as a group for grand unification.
Imposing the following normalization on the elements of the Cartan subalgebra for the su5
algebra.
tr(i  j )  2   i , j
With this normalization one get the following, chosen, matrices as generators in the Cartan
subalgebra.
2

1

1







2


 1

 1

1 
1
0
1
2




 
, 
, 
2
0
2




3
15 
3
0 
0 







 3 
0 
0 



0



0


3


 
0


1



 1

One can now think of the upper 3x3 part of the 5x5 matrix as the basis for the su3c color
algebra, the lower 2x2 matrix as a basis for the su2 isospin algebra and λ0 as a generator for
the abelian u1 hypercharge algebra. By these attachments one can now write down the parts of
the representation for a part of the first family of fermions, where the transformation
properties for the su3cxsu2 algebra are described on the right hand side.
(e  , e ) R ; (1,2)
(e  , e ) L ; (1, 2 )
(d  ) R ; (3,1)
(d  ) L ; ( 3,1)
Where the L and the R is the handedness of the particle, R is labeling a right-handed particle
that interacts with left-handed anti-particle of the same kind.
If one now studies two of the simplest representations of the su5 algebra one get, the first is
the fundamental and the other the fundamental conjugated representation, With the su3cxsu2
part of the representation on the right hand side.
 i ; 5  (3,1)  (1,2)
 i ; 5  ( 3,1)  (1, 2)
In a more concrete form is.
d r 
 dr 
 g
 
d 
dg 
i


 i  db ,    d b 
 
 
 e 
e 
 
 
 e R
  e L
Where r, g and b label a basis for the 3-dimensional representation of the su3c algebra.
Charge quantisation
The su5 algebra has a simple explanation why charge is quantised because every simple Lie
algebra has quantised ladder operators for each of the diagonal generators, compare with E+/in sl2. The charge of each particle is given as a linear combination of the hypercharge and the
isospin.

1
1
Q  I  Y  3  c   0
2
2

The constant c is determined by assuming that the hypercharge is known for the particles in
ψi.
 2 3

2





2 3
2




c 



Y ( i ) 

2 3
2



15 
1 
3






1
 3 


1
c   15   5
3
3
One than determine the values for the charge of the particles in units of the elementary charge.
1 3



1 3




Q i  
1 3


1 


0 

Qd   1
3

 Qe   1
 Q 0
 e
By using that all generators are traceless in su5 one then get a simple expression between the
charges of the d-quark and the positron.
3Qd  Qe   0
This is also a prediction of the SU5 group.
Gauge particles
As earlier described the standard model has 12 gauge particles, 8 gluons, 3 weak bosons and
the photon. The number of gauge particles is the same as the dimension of the algebra that
describes the interaction. The dimension of the Lie algebra is equal to how many basis
elements that span the algebra. From this one concludes that the su5 has 24 dimensions. When
one studies the gauge particles one uses the adjoint representation and how they looks like in
su3cxsu2 algebra can be seen below.
24  (8,1)  (1,3)  (1,1)  (3,2)  ( 3,2)
From this one sees that (8,1) corresponds to the 8 gluons, (1,3) to the 3 weak bosons and (1,1)
to the photon, but the 12 extra gauge particles don’t exist the standard model. These gauge
particles can, in the same time, change both color and flavor, within its family, of the
particles. These gauge particles causes, for example, proton decay. The lagrangian that is
invariant under a su5 gauge transform is.
1 24
1 24
(a)
 ( a )

L    F F
 ( D  )( D  *)  m *   M ( a ) A A  
4 a 1
2 a 1
Where not all M(a) is zero and with Dμ, and Fμν(a).
F
(a)
   A
(a)
  A
24

D     ig 5  A
a 1
(a)
(a)

 i  f a bc A , A
 a
(b)
(c)


Aμ(a) is the field for the gauge particle and where λa and fabc is the basis and the structure
constant for the Lie algebra su5.
This lagrangian is invariant under the following gauge transform and Aμ(a) satisfying.

24


a 1

  exp   i    a ( x)  a   
A ( x)  UAU 1 
24
A   A
(a)
i
U U 1
g5
 a
a 1
Where U is an element in the group SU5.
Proton decay
As previously described their exist new gauge particles in this grand unification theory. We
are now going to deal with a special case. This is about proton decay which is forbidden in the
standard model were baryon number is a conserved quantity. B=1/3 for all quarks and –1/3
for all anti-quarks and 0 for all other elementary particles.
d  u  u  d  X  d  d  e     e
This reaction does not only violate the baryon number conservation but also violate the
conservation of the lepton number as well. L=1 for all leptons, –1 for all anti-leptons and 0 for
all others. Those two conservation laws are exchanged in this new theory to conservation of
B-L. This is a reaction that hasn’t been detected in any laboratory, so one can now doubt the
trueness of this theory. But the characteristic decay time of such a reaction is proportional to
the unifying energy and the decay time is about 1031 years and the consequence of this is that
the gauge particles is very heavy. This implies more, and more complicated, higgsfields in
this theory than in the standard model. But the decay time is small enough to be able to detect
any proton decay to see if the theory is right or not. Many experiments were done in the
eighties in large ponds of water in mines. But the results of these experiments were that the
theory was not correct but there exists theories that predict longer decay times that are still
valid. Another thing that dismisses the SU5 as a unification group is that with better
measurements of the couplings constants the lines don’t intersect at a single point.
SU5 theory is interesting because many other grand unifying theories have the same basic
idée.
Things that aren’t nice about GUT theories
1. They reduce the number of coupling constants but with expense of more and more
complicated higgsfields and we haven’t found the first one yet.
2. They do not include gravity and in that sense do not reach full unification.
3. They introduce new gauge particles and with these new interactions with fermions.
4. They give no answer to the question about multiple family degenerations of the
particles, and why each family looks and behaves like each other but with different
masses.
5. They say that nothing special happens between 102 and 1015 GeV, which is rather
strange.
6. They introduce new fermions, except the SU5 group.
Literature
Mathematical methods for Physicists, Arfken and Weber
Cheng-Li kap 14
Rolnick kap 17.2
Grand unification theories and proton decay, Langacker
Symmetries, Lie algebras and representations, Fuchs and Schweigert