Zee A. Quantum Field Theory in a Nutshell

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422 | VII. Grand Unification

Now iterate. Given the 2nγ matrices for SO(2n) we construct the (2n + 2) γ matrices

for SO(2n + 2) as follows

(n+1)

= γ

(n)

⊗ τ



(n)

0 −γ

(n)



, j = 1, 2,

...

,2n (2)

(n+1)

2n+1

= 1 ⊗ τ





(3)

(n+1)

2n+2

= 1 ⊗ τ



0 −i

i 0



(4)

(Throughout this book 1 denotes a unit matrix of the appropriate size.) The superscript in

parentheses is obviously for us to keep track of which set of γ matrices we are talking about.

Verify that if the γ

(n)

’s satisfy the Clifford algebra, the γ

(n+1)

’s do as well. For example,

{γ

(n+1)

, γ

(n+1)

2n+1

}=(γ

(n)

⊗ τ

)

(1 ⊗ τ

) + (1 ⊗τ

)

(γ

(n)

⊗ τ

)

= γ

(n)

⊗{τ

, τ

}=0

This iterative construction yields for SO(2n) the γ matrices

2k−1

= 1 ⊗ 1 ⊗

...

⊗ 1 ⊗ τ

⊗ τ

⊗

...

⊗ τ

(5)

and

= 1 ⊗ 1 ⊗

...

⊗ 1 ⊗ τ

⊗ τ

⊗

...

⊗ τ

(6)

with 1 appearing k − 1 times and τ

appearing n − k times. The γ ’s are evidently 2

by 2

matrices. When and if you feel confused at any point in this discussion you should work

things out explicitly for SO(4), SO(6), and so on.

In analogy with the Lorentz group, we define 2n(2n − 1)/2 = n(2n − 1) hermitean

matrices

≡

[γ

, γ

] (7)

Note that σ

is equal to iγ

for i = j and vanishes for i = j . The commutation of the σ ’s

with each other is thus easy to work out. For example,

[σ

, σ

] =−[γ

, γ

] =−γ

+ γ

=−[γ

, γ

] =2iσ

Roughly speaking, the γ

’s in σ

and σ

knock each other out. Thus, you see that the

’s satisfy the same commutation relations as the generators J

’s of SO(2n) (as given

in appendix B). The

’s represent the J

’s.

As 2

by 2

matrices, the σ ’s act on an object ψ with 2

components that we will call the

spinor ψ . Consider the unitary transformation ψ →e

iω

ψ with ω

=−ω

a set of real

numbers. Then

†

ψ → ψ

†

−iω

iω

ψ =ψ

†

ψ −iω

†

[σ

, γ

]ψ +

...

for ω

infinitesimal. Using the Clifford algebra we easily evaluate the commutator as

[σ

, γ

] =−2i(δ

− δ

). (If k is not equal to either i or j then γ

clearly commutes

VII.7. SO(10) Unification | 423

with σ

, and if k is equal to either i or j , then we use γ

=1.) We see that the set of objects

≡ ψ

†

ψ , k = 1,

...

,2n transforms as a vector in 2n-dimensional space, with 4ω

the

infinitesimal rotation angle in the ij plane:

→ v

− 2(ω

− ω

) = v

− 4ω

(8)

(in complete analogy to

ψγ

ψ transforming as a vector under the Lorentz group.) This

gives an alternative proof that

represents the generators of SO(2n).

We define the matrix γ

FIVE

=(−i)

...

, which in the basis we are using has the

explicit form

FIVE

= τ

⊗ τ

⊗

...

⊗ τ

(9)

with τ

appearing n times. By analogy with the Lorentz group we define the “left handed”

spinor ψ

≡

(1 −γ

FIVE

)ψ and the “right handed” spinor ψ

≡

(1 +γ

FIVE

)ψ , such that

FIVE

=−ψ

and γ

FIVE

= ψ

. Under ψ → e

iω

ψ , we have ψ

→ e

iω

and

→e

iω

since γ

FIVE

commutes with σ

. The projection into left and right handed

spinors cut the number of components into halves and thus we arrive at the important

conclusion that the two irreducible spinor representations of SO(2n) have dimension 2

n−1

(Convince yourself that the representation cannot be reduced further.) In particular, the

spinor representation of SO(10) is 2

10/2−1

= 2

= 16−dimensional. We will see that the

∗

and 10 of SU(5) can be fit into the 16 of SO(10).

Embedding unitary groups into orthogonal groups

The unitary group SU(5) can be naturally embedded into the orthogonal group SO(10).

In fact, I will now show you that embedding SU (n) into SO(2n) is as easy as z = x + iy.

Consider the 2n-dimensional real vectors x = (x

...

, x

, y

...

, y

) and



= (x



...

, x



, y



...

, y



). By definition, SO(2n) consists of linear transformations

on these two real vectors leaving their scalar product x



x =



j=1



+ y



) invariant.

Now out of these two real vectors we can construct two n-dimensional complex vectors

z = (x

+iy

...

, x

+iy

) and z



=(x



+iy



...

, x



+iy



). The group U (n) consists of

transformations on the two n-dimensional complex vectors z and z



leaving invariant their

scalar product



)

∗

z =



j=1



+ iy



)

∗

+ iy

)



j=1



+ y



) + i



j=1



− y



)

In other words, SO(2n) leaves



j=1



+ y



) invariant, but U (n) consists of the

subset of those transformations in SO(2n) that leave invariant not only



j=1



)

but also



j=1



− y



Now that we understand this natural embedding of U (n) into SO(2n), we see that the

defining or vector representation of SO(2n), which we will call simply 2n, decomposes

424 | VII. Grand Unification

upon restriction to U(n) into the two defining representations of U (n), n and n

∗

; thus

2n → n ⊕ n

∗

(10)

In other words, (x

...

, x

, y

...

, y

) can be written as (x

+ iy

...

, x

+ iy

) and

−iy

...

, x

−iy

) Note that this is the analog of (VII.5.4) indicating that the defining

representation of SU(5) decomposes into representations of SU(3) ⊗ SU(2) ⊗ U(1):

5 →(3

∗

,1,

) ⊕ (1, 2, −

). (11)

Given the decomposition law (10), we can now figure out how other representations of

SO(2n) decompose when restricted to the natural subgroup U(n). The tensor representa-

tions of SO(2n) are easy, since they are constructed out of the vector representation. [This is

precisely what we did in going from (VII.5.4) to (VII.5.7, 8, and 9).] For example, the adjoint

representation of SO(2n), which has dimension 2n(2n − 1)/2 = n(2n − 1), transforms as

an antisymmetric 2-index tensor 2n ⊗

2n and so decomposes into

2n ⊗

2n → (n ⊕ n

∗

) ⊗

(n ⊕ n

∗

) (12)

according to (10). The antisymmetric product ⊗

on the right hand side is, of course, to

be evaluated within U (n). For instance, n ⊗

n is the n(n − 1)/2 representation of U (n).

In this way, we see that

n(2n − 1) → n

− 1 (the adjoint)

⊕ 1 (the singlet)

⊕ n(n − 1)/2

⊕ (n(n − 1)/2)

∗

(13)

As a check, the total dimension of the representations of U (n) on the right hand side adds

up to (n

−1) + 1 + 2n(n − 1)/2 = n(2n − 1). In particular, for SO(10) ⊃ SU(5), we have

45 → 24 ⊕1 ⊕ 10 ⊕ 10

∗

and of course 24 + 1 + 10 + 10 = 45.

Decomposing the spinor

It is more difficult to figure out how the spinor representation of SO(2n) decompose

upon restriction to U(n). I give here a heuristic argument that satisfies most physicists,

but certainly not mathematicians. I will just do SO(10) ⊃ SU(5) and let you work out

the general case. The question is how the 16 falls apart. Just from numerology and from

knowing the dimensions of the smaller representations of SU(5) (1, 5, 10, 15) we see there

are only so many possibilities, some of them rather unlikely, for example, the 16 falling

apart into 16 1’s.

Picture the spinor 16 of SO(10) breaking up into a bunch of representations of SU(5).

By definition, the 45 generators of SO(10 ) scramble all these representations together. Let

us ask what the various pieces of 45, namely 24 ⊕1 ⊕10 ⊕10

∗

, do to these representations.

The 24 transform each of the representations of SU(5) into itself, of course, because they

are the 24 generators of SU(5) and that is what generators were born to do. The generator

VII.7. SO(10) Unification | 425

1 can only multiply each of these representations by a real number. (In other words, the

corresponding group element multiplies each of the representations by a phase factor.)

What does the 10, which as you recall from chapter VII.5 is represented as an antisym-

metric tensor with two upper indices and hence also known as [2], do to these represen-

tations? Suppose the bunch of representations that S breaks up into contains the singlet

[0] =1ofSU(5). The 10 = [2] acting on [0] gives the [2] =10. (Almost too obvious for words!

An antisymmetric tensor of two indices combined with a tensor with no indices is an anti-

symmetric tensor of two indices.) What about 10 = [2] acting on [2]? The result is a tensor

with four upper indices. It certainly contains the [4], which is equivalent to [1]

∗

= 5

∗

.But

look, 1 ⊕10 ⊕ 5

∗

already add up to 16. Thus, we have accounted for everybody. There can’t

be more. So we conclude

→ [0] ⊕[2] ⊕[4] = 1 ⊕ 10 ⊕ 5

∗

(14)

The 5

∗

and the 10 of SU(5) fit inside the 16

of SO(10)!

We will learn later that the two spinor representations of SO(10) are conjugate to each

other. Indeed, you may have noticed that I snuck a superscript plus on the letter S. The

conjugate spinor S

−

breaks up into the conjugate of the representations in (14):

−

→ [1] ⊕[3] ⊕[5] = 5 ⊕ 10

∗

⊕ 1

∗

(15)

The long lost antineutrino

The fit would be perfect if we introduce one more field transforming as a 1, that is, a singlet

under SU(5) and hence a fortiori a singlet under SU(3) ⊗ SU(2) ⊗ U(1). In other words,

this field does not participate in the strong, weak, and electromagnetic interactions, or in

plain English, it describes a lepton with no electric charge and is not involved in the known

weak interaction. Thus, this field can be identified as the “long lost” antineutrino field ν

This guy does not listen to any of the known gauge bosons.

Recall that we are using a convention in which all fermion fields are left handed, and

hence we have written ν

. By a conjugate transformation, as explained earlier, this is

equivalent to the right handed neutrino field ν

Since ν

is an SU(5) singlet, we can give it a Majorana mass M without breaking SU(5).

Hence we expect M to be larger than or of the same order of magnitude as the mass scale

at which SU(5) is broken, which as we saw in chapter VII.5 is much higher than the mass

scales that have been explored experimentally. This explains why ν

has not been seen.

On the other hand, with the presence of ν

we can have a Dirac mass term m( ¯ν

h.c.). Since this term breaks SU(2) ⊗ U(1) just like the mass terms for the quarks and

leptons we know, we expect m to be of the same order of magnitude

as the known quark

and lepton masses (which for reasons unknown span an enormous range).

Explicitly, with ν

now available we can add to the SU(2) ⊗U(1) theory of chapter VII.2 the term f



˜ϕ ¯ν

where ˜ϕ ≡ τ

†

. In the absence of any indication to the contrary, we might suppose that f



is of the same order

of magntiude as the coupling f that leads to the electron mass.

426 | VII. Grand Unification

Thus, in the space spanned by (ν, ν

) we have the (Majorana) mass matrix

M =



0 m



(16)

with M m. Since the trace and determinant of M are M and −m

, respectively, M has a

large eigenvalue ∼ M and a small eigenvalue ∼ m

/M. A tiny mass ∼ m

/M, suppressed

relative to the usual quark and lepton masses by the factor m/M , is naturally generated

for the (observed) left handed neutrino. This rather attractive scenario, known as the

seesaw mechanism for obvious reason, was discovered independently by Minkowski and

by Glashow, and somewhat later by Yanagida and by Gell-Mann, Ramond, and Slansky.

Again, the tight fit of the 5

∗

and the 10 of SU(5) inside the 16

of SO(10) has convinced

many physicists that it is surely right.

A binary code for the world

Given the product form of the γ matrices in (5) and (6), and hence of σ

, we can write the

states of the spinor representations as

|ε

...

 (17)

where each of the ε’s takes on the values ±1. For example, for n = 1, τ

|+=|−and

|−=|+, while τ

|+=i |−and τ

|−=−i |+. From (9) we see that

FIVE

|ε

...

=(

j=1

) |ε

...

 (18)

The right handed spinor S

consists of those states |ε

...

 with (

j=1

) =+1, and

the left handed spinor S

−

those states with (

j=1

) =−1. Indeed, the spinor represen-

tations have dimension 2

n−1

Thus, in SO(10) unification the fundamental quarks and leptons are described by a five-

bit binary code, with states such as |++−−+and |−+−−−. Personally, I find

this a rather pleasing picture of the world.

Let us work out the states explicitly. This also gives me a chance to make sure that

you understand the group theory presented in this chapter. Start with the much simpler

case of SO(4). The spinor S

consists of |++and |−−while the spinor S

−

consists

of |+−and |−+. As discussed in chapter II.3, SO(4) contains two distinct SU(2)

subgroups. Removing a few factors of i from the discussion in chapter II.3 we see that

the third generator of SU(2), call it σ

, can be taken to be either σ

− σ

or σ

+ σ

The two choices correspond to the two distinct SU(2) subgroups. We choose (arbitrarily)

(σ

−σ

). From (5) and (6) we have σ

=iγ

=i(τ

⊗τ

)(τ

⊗τ

) =−τ

⊗1and

=−1 ⊗ τ

, and so σ

(−τ

⊗ 1 + 1 ⊗ τ

). To figure out how the four states |++,

|−−, |+−, and |−+transform under our chosen SU(2), let us act on them with σ

For example,

|++=

(−τ

⊗ 1 + 1 ⊗ τ

) |++=

(−1 + 1) |++=0

VII.7. SO(10) Unification | 427

and

|−+=

(−τ

⊗ 1 + 1 ⊗ τ

) |−+=

(1 + 1) |−+=|−+

Aha, under SU(2) |++and |−−are two singlets while |+−and |−+make up a

doublet.

Note that this is consistent with the generalization of (14) and (15), namely that upon

the restriction of SO(2n) to U(n) the spinors decompose as

→ [0] ⊕[2] ⊕

...

(19)

and

−

→ [1] ⊕[3] ⊕

...

(20)

I have not indicated the end of the two sequences: A moment’s reflection indicates that it

depends on whether n is even or odd. In our example, n =2,and thus 2

→[0] ⊕[2] =1 ⊕ 1

and 2

−

→ [1] = 2. Similarly, for n = 3, upon the restriction of SO(6) to U(3),4

→

[0] ⊕ [2] = 1 ⊕ 3

∗

and 4

−

→ [1] ⊕ [3] = 3 ⊕ 1. (Our choice of which triplet representation

of U(3) to call 3 or 3

∗

is made to conform to common usage, as we will see presently.)

We are now ready to figure out the identity of each of the 16 states such as |++−−+

in SO(10) unification. First of all, (18) tells us that under the subgroup SO(4) ⊗ SO(6) of

SO(10) the spinor 16

decomposes as (since 

j=1

=+1 implies ε

= ε

)

→ (2

) ⊕ (2

−

) (21)

We identify the natural SU(2) subgroup of SO(4) as the SU(2) of the electroweak interac-

tion and the natural SU(3) subgroup of SO(6) as the color SU(3) of the strong interaction.

Thus, according to the preceding discussion, (2

) are the SU(2) singlets of the stan-

dard U(1) ⊗ SU(2) ⊗ SU(3) model, while (2

−

) are the SU(2) doublets. Here is the

lineup (all fields being left handed as usual):

SU(2) doublets:

ν =|−+−−−

−

=|+−−−−

u =|−+++−, |−++−+, and |−+−++

d =|+−++−, |+−+−+, and |+−−++

SU(2) singlets:

=|+++++

=|−−+++

=|+++−−, |++−+−, and |++−−+

=|−−+−−, |−−−+−, and |−−−−+.

I assure you that this is a lot of fun to work out and I urge you to reconstruct this

table without looking at it. Here are a few hints if you need help. From our discussion

428 | VII. Grand Unification

of SU(2) I know that ν =|−+ε

 and e

−

=|+−ε

, but how do I know that

= ε

=−1? First, I know that ε

=−1. I also know that 4

−

→ 3 ⊕ 1 upon

restricting SO(6) to color SU(3). Well, of the four states |−−−, |++−, |+−+,

and |−++the “odd man out” is clearly |−−−. By the same heuristic argument,

among the 16 possible states |+++++is the “odd man out” and so must be ν

There are lots of consistency checks. For example, once I identify ν =|−+−−−,

−

=|+−−−−, and ν

=|+++++, I can figure out the electric charge Q,

which, since it transforms as a singlet under color SU(3), must have the value Q =

aε

+bε

+c(ε

+ε

) when acting on the state |ε

. The constants a , b, and

c can be determined from the three equations Q(ν) =−a + b − 3c =0, Q(e

−

) =−1, and

Q(ν

) = 0. Thus, Q =−

(ε

+ ε

Living in the computer age, I find it intriguing that the fundamental constituents

of matter are coded by five bits. You can tell your condensed matter colleagues that

their beloved electron is composed of the binary strings +−−−−and −−+++.An

intriguing possibility

suggests itself, that quarks and leptons may be composed of five

different species of fundamental fermionic objects. We construct composites, writing a

+ if that species is present, and a − if it is absent. For example, from the expression for

Q given above, we see that species 1 carries electric charge −

, species 2 is neutral, and

species 3, 4, and 5 carry charge

. A more or less concrete model can even be imagined by

binding these fundamental fermionic objects to a magnetic monopole.

I emphasize that particles transforming in 16

−

, such as |+−+++, have not been

observed experimentally.

A speculation on the origin of families

One of the great unsolved puzzles in particle physics is the family problem. Why do quarks

and leptons come in three generations {ν

, e, u, d}, {ν

, μ, c, s}, and {ν

, τ , t , b}? The way

we incorporate this experimental fact into our present day theory can only be described

as pathetic: We repeat the fermionic sector of the Lagrangian three times without any

understanding whatsoever. Three generations living together gives rise to a nagging family

problem.

Our binary code view of the world suggests a wildly speculative (perhaps too speculative

to mention in a textbook?) approach to the family problem: We add more bits. To me,

a reasonable possibility is to “hyperunify” into an SO(18) theory, putting all fermions

into a single spinorial representation S

= 256

, which upon the breaking of SO(18) to

SO(10) ⊗ SO(8) decomposes as

256

→ (16

) ⊕ (16

−

) (22)

We have a lot of 16

’s. Unhappily, we see that group theory [see also (21)] dictates that we

also get a bunch of unwanted 16

−

’s. One suggestion is that Nature might repeat the trick

For further details, see F. Wilczek and A. Zee, Phys. Rev. D25: 553, 1982, Section IV.

VII.7. SO(10) Unification | 429

She uses with color SU(3), whose strong force confines fields that are not color singlets

(chapter VII.3). Interestingly, we can exploit a striking feature of SO(8), which some people

regard as the most beautiful of all groups. In particular, the two spinorial representations

have the same dimension as the vectorial representation 8

(the equation 2

n−1

= 2n

has the unique solution n =4). There is a transformation that cyclically rotates these three

representations 8

−

, and 8

into each other (in the jargon, the group SO(8) admits an

outer automorphism). Thus, there exists a subgroup SO(5) of SO(8) such that when we

break SO(8) into that SO(5) 8

behaves like 8

while 8

−

behaves like a spinor, namely

→ 5 ⊕1 ⊕ 1 ⊕1 and 8

−

→ 4 ⊕ 4

∗

(23)

If we call this

SO(5) hypercolor and assumes that the strong force associated with it con-

fines all fields that are not hypercolor singlets, then only three 16

’s remain! Unfortunately,

as the relevant physics occurs in the energy regime above grand unification, our knowl-

edge of the dynamics of symmetry breaking is far too paltry for us to make any further

statements.

Charge conjugation

The product ⊗notation we use here allows us to construct the conjugation matrix C explic-

itly. By definition C

−1

∗

C =−σ

(so that C changes e

iθ

into its complex conjugate.)

From (2), (3), and (4) we see that we can construct

(n+1)

(n)

⊗ τ

if n odd

(n)

⊗ τ

if n even

(24)

You can check that this gives C

−1

∗

C = (−1)

and hence the desired result.

Explicitly, C is a direct product of an alternating sequence of τ

and τ

and so we deduce

an important property. Acting on |ε

...

, C flips the sign of all the ε’s. Thus C

changes the sign of (

j=1

) for n odd, and does not for n even. For n odd, the two

spinor representations S

and S

−

are conjugates of each other, while for n even, they

are conjugates of themselves, or in other words, they are real. This can also be seen

directly from C

−1

FIVE

C = (−1)

FIVE

. You can check this with all the explicit examples

we have encountered: SO(2), SO(4), SO(6), SO(8), SO(10), and SO(18). See also

exercise VII.7.3.

Anomalies

What about anomalies in SO(2n) grand unification? According to the discussion in chap-

ter VII.5 we have to evaluate A

ij klmn

≡tr(J

, J

}) over the fermion representation.

The reader savvy with group theory would recognize that SO(5) is isomorphic with the symplectic group

Sp(4) and that the Dynkin diagram of SO(8) is the most symmetric of all.

430 | VII. Grand Unification

Applying an SO(2n) transformation J

→O

O we see easily that A

ij klmn

is an invari-

ant tensor. Can we construct an invariant 6-index tensor with the appropriate symmetry

properties (e.g., A

ij klmn

=−A

jiklmn

) in SO(2n)? We can’t, except in SO(6), for which we

have ε

ij klmn

. Thus, A

ij klmn

vanishes except in SO(6), where it is proportional to ε

ij klmn

An elegant one line proof that any grand unified theory based on SO(2n) for n = 3 is free

from anomaly!

The cancellation of the anomaly between 5

∗

and 10 at the end of chapter VII.5 doesn’t

seem so miraculous any more. Miracles tend to fade away as we gain deeper understanding.

Amusingly, by discussing a physics question, namely whether a gauge theory is renor-

malizable or not, we have discovered a mathematical fact. What is so special about SO(6)?

See exercise VII.7.5.

Exercises

VII.7.1 Work out the Clifford algebra in d-dimensional space for d odd.

VII.7.2 Work out the Clifford algebra in d-dimensional Minkowski space.

VII.7.3 Show that the Clifford algebra for d = 4k and for d = 4k + 2 have somewhat different properties. (If you

need help with this and the two preceding exercises, look up F. Wilczek and A. Zee, Phys. Rev. D25: 553,

1982.)

VII.7.4 Discuss the Higgs sector of the SO(10). What do you need to give mass to the quarks and leptons?

VII.7.5 The group SO(6) has 6(6 − 1)/2 = 15 generators. Notice that the group SU(4) also has 4

− 1 = 15

generators. Substantiate your suspicion that SO(6) and SU(4) are isomorphic. Identify some low

dimensional representations.

VII.7.6 Show that (unfortunately) the number of families we get in SO(18) depends on which subgroup of SO(8)

we take to be hypercolor.

VII.7.7 If you want to grow up to be a string theorist, you need to be familiar with the Dirac equation in various

dimensions but especially in 10. As a warm up, study the Dirac equation in 2-dimensional spacetime.

Then proceed to study the Dirac equation in 10-dimensional spacetime.

Part VIII Gravity and Beyond