3.2 The Spin(10) Theory

We now turn our attention to another grand unified theory. Physicists call it the `${\rm SO}(10)$ theory', but we shall call it the ${\rm Spin}(10)$ theory, because the Lie group involved is really ${\rm Spin}(10)$, the double cover of ${\rm SO}(10)$. This theory appeared in a 1974 paper by Georgi [10], shortly after his paper with Glashow on the ${\rm SU}(5)$ theory. However, Georgi has said that he conceived of the ${\rm Spin}(10)$ theory first. See Zee [40], Chapter VII.7, for a concise and readable account.

The ${\rm SU}(5)$ GUT has helped us explain the pattern of hypercharges in the Standard Model, and thanks to the use of the exterior algebra, $\Lambda {\mathbb{C}}^5$, we can interpret it in terms of a binary code. This binary code explains another curious fact about the Standard Model. Specifically, why is the number of fermions a power of 2? There are 16 fermions, and 16 antifermions, which makes the Standard Model rep have dimension

\begin{displaymath}\dim (F \oplus F^*) = 2^5 = 32. \end{displaymath}

With the binary code interpretation, it could not be any other way.

In actuality, however, the existence of a right-handed neutrino (or its antiparticle, the left-handed antineutrino) has been controversial. Because it transforms trivially in the Standard Model, it does not interact with anything except perhaps the Higgs.

The right-handed neutrino certainly improves the aesthetics of the ${\rm SU}(5)$ theory. When we include this particle (and its antiparticle), we obtain the rep

\begin{displaymath}\Lambda ^0 {\mathbb{C}}^5 \oplus \Lambda ^1 {\mathbb{C}}^5 \o...
...lus \Lambda ^4 {\mathbb{C}}^5 \oplus \Lambda ^5 {\mathbb{C}}^5 \end{displaymath}

which is all of $\Lambda {\mathbb{C}}^5$, whereas without this particle we would just have

\begin{displaymath}\Lambda ^1 {\mathbb{C}}^5 \oplus \Lambda ^2 {\mathbb{C}}^5 \oplus \Lambda ^3 {\mathbb{C}}^5 \oplus \Lambda ^4 {\mathbb{C}}^5 \end{displaymath}

which is much less appealing--it wants to be $\Lambda {\mathbb{C}}^5$, but it comes up short.

More importantly, there is increasing indirect evidence from experimental particle physics that right-handed neutrinos do exist. For details, see Pati [27]. If this is true, the number of fermions really could be 16, and we have a ready-made explanation for that number in the binary code.

However, this creates a new mystery. The ${\rm SU}(5)$ works nicely with the representation $\Lambda {\mathbb{C}}^5$, but ${\rm SU}(5)$ does not require this. It works just fine with the smaller rep

\begin{displaymath}\Lambda ^1 {\mathbb{C}}^5 \oplus \Lambda ^2 {\mathbb{C}}^5 \oplus \Lambda ^3 {\mathbb{C}}^5 \oplus \Lambda ^4 {\mathbb{C}}^5 .\end{displaymath}

It would be nicer to have a theory that required us to use all of $\Lambda {\mathbb{C}}^5$. Better yet, if our new GUT were an extension of ${\rm SU}(5)$, the beautiful explanation of hypercharges would live on in our new theory. With luck, we might even get away with using the same underlying vector space, $\Lambda {\mathbb{C}}^5$. Could it be that the ${\rm SU}(5)$ GUT is only the beginning of the story? Could unification go on, with a grand unified theory that extends ${\rm SU}(5)$ just as ${\rm SU}(5)$ extended the Standard Model?

Let us look for a group that extends ${\rm SU}(5)$ and has an irrep whose dimension is some power of 2. The dimension is a big clue. What representations have dimensions that are powers of 2? Spinors.

What are spinors? They are certain representations of ${\rm Spin}(n)$, the double cover of the rotation group in $n$ dimensions, which do not factor through the quotient ${\rm SO}(n)$. Their dimensions are always a power of two. We build them by exhibiting ${\rm Spin}(n)$ as a subgroup of a Clifford algebra. Recall that the Clifford algebra ${\rm Cliff}_n$ is the associative algebra freely generated by ${\mathbb{R}}^n$ with relations

\begin{displaymath}v w + w v = -2\langle v, w \rangle . \end{displaymath}

If we take products of pairs of unit vectors in ${\mathbb{R}}^n$ inside this algebra, these generate the group ${\rm Spin}(n)$: multiplication in this group coincides with multiplication in the Clifford algebra. Using this fact, we can get representations of ${\rm Spin}(2n)$ from modules of ${\rm Cliff}_n$.

We can use this method to get a rep of ${\rm Spin}(10)$ on $\Lambda {\mathbb{C}}^5$ that extends the rep of ${\rm SU}(5)$ on this space. In fact, quite generally ${\rm Cliff}_{2n}$ acts on $\Lambda {\mathbb{C}}^n$. Then, because

\begin{displaymath}{\rm Spin}(2n) \hookrightarrow {\rm Cliff}_{2n}, \end{displaymath}

$\Lambda {\mathbb{C}}^n$ becomes a representation of ${\rm Spin}(2n)$, called the Dirac spinor representation.

To see this, we use operators on $\Lambda {\mathbb{C}}^n$ called `creation and annihilation operators'. Let $e_1, \ldots, e_n$ be the standard basis for ${\mathbb{C}}^n$. Each of these gives a creation operator:

a_j^* \colon \Lambda {\mathbb{C}}^n &\to...
...\mathbb{C}}^n \\
\psi &\mapsto& e_j \wedge \psi .

We use the notation $a_j^*$ because $\Lambda {\mathbb{C}}^n$ is a Hilbert space, so $a_j^*$ is the adjoint of some other operator

\begin{displaymath}a_j \colon \Lambda {\mathbb{C}}^n \to \Lambda {\mathbb{C}}^n , \end{displaymath}

which is called an annihilation operator.

In physics, we can think of the basis vectors $e_j$ as particles. For example, in the binary code approach to the ${\rm SU}(5)$ theory we imagine five particles from which the observed particles in the Standard Model are composed: up, down, red, green and blue. Taking the wedge product with $e_j$ `creates a particle' of type $j$, while the adjoint `annihilates a particle' of type $j$.

It may seem odd that creation is the adjoint of annihilation, rather than its inverse. One reason for this is that the creation operator, $a_j^*$, has no inverse. In some sense, its adjoint $a_j$ is the best substitute.

This adjoint does do what want, which is to delete any particle of type $j$. Explicitly, it deletes the `first' occurence of $e_j$ from any basis element, bringing out any minus signs we need to make this respect the antisymmetry of the wedge product:

\begin{displaymath}a_j ( e_{i_1} \wedge \cdots \wedge e_{i_p} ) =
( -1 )^{k+1} ...
...}} \wedge e_{i_{k+1}} \cdots \wedge e_{i_p}, \mbox{ if }j=i_k. \end{displaymath}

And if no particle of type $j$ appears, we get zero.

Now, whenever we have an inner product space like ${\mathbb{C}}^n$, we get an inner product on $\Lambda {\mathbb{C}}^n$. The fastest, if not most elegant, route to this inner product is to remember that, given an orthonormal basis $e_1, \ldots, e_n$ for ${\mathbb{C}}^n$, the induced basis, consisting of elements of the form $e_{i_1} \wedge
\cdots \wedge e_{i_p}$, should be orthonormal in $\Lambda {\mathbb{C}}^n$. But choosing an orthonormal basis defines an inner product, and in this case it defines an inner product on the whole exterior algebra, one that reduces to the usual one for the grade one elements, $\Lambda ^1 {\mathbb{C}}^n \cong {\mathbb{C}}^n$.

It is with respect to this inner product on $\Lambda {\mathbb{C}}^n$ that $a_j$ and $a_j^*$ are adjoint. That is, they satisfy

\begin{displaymath}\langle v, a_jw \rangle = \langle a_j^*v, w \rangle \end{displaymath}

for any elements $v,w \in \Lambda {\mathbb{C}}^n$. Showing this from the definitions we have given is a straightforward calculation, which we leave to the reader.

These operators satisfy the following relations:

\{ a_j, a_k \} & = & 0 \\
\{ a_j^*, a_k^* \} & = & 0 \\
\{ a_j, a_k^* \} & = & \delta_{jk}

where curly brackets denote the anticommutator of two linear operators, namely $\{a, b\} = ab + ba$.

As an algebra, ${\rm Cliff}_{2n}$ is generated by the standard basis vectors of ${\mathbb{R}}^{2n}$. Let us call the elements of ${\rm Cliff}_{2n}$ corresponding to these basis vectors $\gamma_1, \ldots, \gamma_{2n}$. From the definition of the Clifford algebra, is easy to check that

\begin{displaymath}\{ \gamma_k , \gamma_\ell \} = -2\delta_{k\ell} .\end{displaymath}

In other words, the elements $\gamma_k$ are anticommuting square roots of $-1$. So, we can turn $\Lambda {\mathbb{C}}^n$ into a ${\rm Cliff}_{2n}$-module by finding $2n$ linear operators on $\Lambda {\mathbb{C}}^n$ that anticommute and square to $-1$. We build these from the raw material provided by $a_j$ and $a_j^*$. Indeed, it is easy to see that

\phi_j & = & i(a_j + a_j^*) \\
\pi_j & = & a_j - a_j^*

do the trick. Now we can map $\gamma_1, \ldots, \gamma_{2n}$ to these operators, in any order, and $\Lambda {\mathbb{C}}^n$ becomes a ${\rm Cliff}_{2n}$-module as promised.

Now for $n > 1$ we may define ${\rm Spin}(2n)$ to be the universal cover of ${\rm SO}(2n)$, with group structure making the covering map

{\rm Spin}(2n) \ar[d]^p \\
{\rm SO}(2n)

into a homomorphism. This universal cover is a double cover, because the fundamental group of ${\rm SO}(2n)$ is ${\mathbb{Z}}_2$ for $n$ in this range.

This construction of ${\rm Spin}(2n)$ is fairly abstract. Luckily, we can realize ${\rm Spin}(2n)$ as the multiplicative group in ${\rm Cliff}_{2n}$ generated by products of pairs of unit vectors. This gives us the inclusion

\begin{displaymath}{\rm Spin}(2n) \hookrightarrow {\rm Cliff}_{2n} \end{displaymath}

we need to make $\Lambda {\mathbb{C}}^n$ into a representation of ${\rm Spin}(2n)$. From this, one can show that the Lie algebra ${\mathfrak{so}}(2n)$ is generated by the commutators of the $\gamma_j$. Because we know how to map each $\gamma_j$ to an operator on $\Lambda {\mathbb{C}}^n$, this gives us an explicit formula for the action of ${\mathfrak{so}}(2n)$ on $\Lambda {\mathbb{C}}^n$. Each $\gamma_j$ changes the parity of the grades, and their commutators do this twice, restoring grade parity. Thus, ${\mathfrak{so}}(2n)$ preserves the parity of the grading on $\Lambda {\mathbb{C}}^n$, and ${\rm Spin}(2n)$ does the same. This breaks $\Lambda {\mathbb{C}}^n$ into two subrepresentations:

\begin{displaymath}\Lambda {\mathbb{C}}^n = \Lambda^{\rm ev}{\mathbb{C}}^n \oplus \Lambda^{\rm odd}{\mathbb{C}}^n \end{displaymath}

where $\Lambda^{\rm ev}{\mathbb{C}}^n$ is the direct sum of the even-graded parts:

\begin{displaymath}\Lambda^{\rm ev}{\mathbb{C}}^n = \Lambda ^0 {\mathbb{C}}^n \oplus \Lambda ^2 {\mathbb{C}}^n \oplus \cdots \end{displaymath}

while $\Lambda^{\rm odd}{\mathbb{C}}^n$ is the sum of the odd-graded parts:

\begin{displaymath}\Lambda^{\rm odd}{\mathbb{C}}^n = \Lambda ^1 {\mathbb{C}}^n \oplus \Lambda ^3 {\mathbb{C}}^n \oplus \cdots. \end{displaymath}

In fact, both these representations of ${\rm Spin}(2n)$ are irreducible, and ${\rm Spin}(2n)$ acts faithfully on their direct sum $\Lambda {\mathbb{C}}^n$. Elements of these two irreps of ${\rm Spin}(2n)$ are called left- and right-handed Weyl spinors, respectively, while elements of $\Lambda {\mathbb{C}}^n$ are called Dirac spinors.

All this works for any $n$, but we are especially interested in the case $n=5$. The big question is: does the Dirac spinor representation of ${\rm Spin}(10)$ extend the obvious representation of ${\rm SU}(5)$ on $\Lambda {\mathbb{C}}^5$? Or, more generally, does the Dirac spinor representation of ${\rm Spin}(2n)$ extend the representation of ${\rm SU}(n)$ on $\Lambda {\mathbb{C}}^n$?

Remember, we can think of a unitary representation as a group homomorphism

\begin{displaymath}G \to {\rm U}(V) \end{displaymath}

where $V$ is the Hilbert space on which $G$ acts as unitary operators. Here we are concerned with two representations. One of them is the familiar representation of ${\rm SU}(n)$ on $\Lambda {\mathbb{C}}^n$:

\begin{displaymath}\rho \colon {\rm SU}(n) \to {\rm U}(\Lambda {\mathbb{C}}^n), \end{displaymath}

which acts as the fundamental rep on $\Lambda ^1 {\mathbb{C}}^n \cong {\mathbb{C}}^n$ and respects wedge products. The other is the representation of ${\rm Spin}(2n)$ on the Dirac spinors, which happen to form the same vector space $\Lambda {\mathbb{C}}^n$:

\begin{displaymath}\rho' \colon {\rm Spin}(2n) \to {\rm U}(\Lambda {\mathbb{C}}^n). \end{displaymath}

Our big question is answered affirmatively by this theorem, which can be found in a classic paper by Atiyah, Bott and Shapiro [2]:

Theorem 2   . There exists a Lie group homomorphism $\psi$ that makes this triangle commute:

{\rm SU}(n) \ar[r]^{\psi} \ar[dr]_\rho & {\rm Spin}(2n) \ar[d]^{\rho'} \\
& {\rm U}(\Lambda {\mathbb C}^n)

Proof. The complex vector space ${\mathbb{C}}^n$ has an underlying real vector space of dimension $2n$, and the real part of the usual inner product on ${\mathbb{C}}^n$ gives an inner product on this underlying real vector space, so we have an inclusion ${\rm U}(n)
\hookrightarrow \O (2n)$. The connected component of the identity in $\O (2n)$ is ${\rm SO}(2n)$, and ${\rm U}(n)$ is connected, so this gives an inclusion ${\rm U}(n) \hookrightarrow {\rm SO}(2n)$ and thus ${\rm SU}(n) \hookrightarrow {\rm SO}(2n)$. Passing to Lie algebras, we obtain an inclusion ${\mathfrak{su}}(n) \hookrightarrow {\mathfrak{so}}(2n)$. A homomorphism of Lie algebras gives a homomorphism of the corresponding simply-connected Lie groups, so this in turn gives the desired map $\psi \colon {\rm SU}(n) \to {\rm Spin}(2n)$.

Next we must check that $\psi$ makes the above triangle commute. Since all the groups involved are connected, it suffices to check that this diagram

{\mathfrak{su}}(n) \ar[r]^{d\psi} \ar[dr]_{d\rho}...
...k{so}}(2n) \ar[d]^{d\rho'} \\
& \u (\Lambda {\mathbb C}^n)

commutes. Since the Dirac representation $d\rho'$ is defined in terms of creation and annihilation operators, we should try to express $d\rho$ this way. To do so, we will need a good basis for ${\mathfrak{su}}(n)$. Remember,

\begin{displaymath}{\mathfrak{su}}(n) =
\{n \times n \mbox{ traceless skew-adjoint matrices over } {\mathbb{C}}\}. \end{displaymath}

If $E_{jk}$ denotes the matrix with 1 in the $jk$th entry and 0 everywhere else, then the traceless skew-adjoint matrices have this basis:

E_{jk} - E_{kj} & j>k \\
i(E_{jk} + E_{kj...
...k \\
i(E_{jj} - E_{j+1,j+1}) & j=1, \ldots, n-1.
\end{array} \end{displaymath}

For example, ${\mathfrak{su}}(2)$ has the basis

0 & -1 \\
1 & 0
i & 0 \\
0 & -i

and our basis for ${\mathfrak{su}}(n)$ simply generalizes this.

Now, it is easy to guess a formula for $d\rho$ in terms of creation and annihilation operators. After all, the elementary matrix $E_{jk}$ satisfies

E_{jk} (e_\ell) =
\left\{ \begin{array}{cl}
e_j & \textrm{i...
...ll = k \\
0 & \textrm{if} \; \ell \ne k
\end{array} \right.

and $a^*_j a_k$ acts the same way on $\Lambda ^1 {\mathbb{C}}^n \subseteq \Lambda {\mathbb{C}}^n$. So, we certainly have

d\rho \left(E_{jk} - E_{kj}\right) & = & ...
...\right) & = &
i(a^*_j a_j - a^*_{j+1} a_{j+1}) \\

on the subspace $\Lambda ^1 {\mathbb{C}}^n$. But do these operators agree on the rest of $\Lambda {\mathbb{C}}^n$? Remember, $\rho$ preserves wedge products:

\begin{displaymath}\rho(x)(v \wedge w) = \rho(x)v \wedge \rho(x)w \end{displaymath}

for all $x \in {\rm SU}(n)$. Differentiating this condition, we see that ${\mathfrak{su}}(n)$ must act as derivations:

\begin{displaymath}d\rho(X)(v \wedge w) = d\rho(X)v \wedge w + v \wedge d\rho(X)w\end{displaymath}

for all $X \in {\mathfrak{su}}(n)$. Derivations of $\Lambda {\mathbb{C}}^n$ are determined by their action on $\Lambda ^1 {\mathbb{C}}^n$. So, $d\rho$ will be given on all $\Lambda {\mathbb{C}}^n$ by the above formulas if we can show that

a^*_j a_k - a^*_k a_j ,
i(a^*_j a_k + a^*_k a_j),
\textrm{and} \quad
i(a^*_j a_j - a^*_{j+1} a_{j+1})

are derivations.

Now, the annihilation operators are a lot like derivations: they are antiderivations. That is, if $v \in \Lambda ^p {\mathbb{C}}^n$ and $w \in \Lambda ^q {\mathbb{C}}^n$, then

\begin{displaymath}a_j(v \wedge w) = a_j v \wedge w + (-1)^p v \wedge a_j w .\end{displaymath}

However, the creation operators are nothing like derivations. They satisfy

\begin{displaymath}a^*_j (v \wedge w) = a^*_j v \wedge w = (-1)^p v \wedge a^*_j w, \end{displaymath}

because $a^*_j$ acts by wedging with $e_j$, and moving this through $v$ introduces $p$ minus signs. Luckily, this relation combines with the previous one to make the composites $a^*_j a_k$ into derivations for every $j$ and $k$. We leave this for the reader to check.

So, $d\rho$ can really be expressed in terms of annihilation and creation operators as above. Checking that

{\mathfrak{su}}(n) \ar[r]^{d\psi} \ar[dr]_{d\rho}...
...k{so}}(2n) \ar[d]^{d\rho'} \\
& \u (\Lambda {\mathbb C}^n)

commutes is now a straightforward but somewhat tedious job, which we leave to the dedicated reader. $\sqcap$ $\sqcup$

This theorem had a counterpart for the ${\rm SU}(5)$ GUT--namely, Theorem 1. There we saw a homomorphism $\phi$ that showed us how to extend the Standard Model group ${G_{\mbox{\rm SM}}}$ to ${\rm SU}(5)$, and made this square commute:

{G_{\mbox{\rm SM}}}\ar[r]^\phi \ar[d] & {\rm SU...
...F^*) \ar[r]^-{{\rm U}(f)} & {\rm U}(\Lambda {\mathbb C}^5)

Now $\psi$ says how to extend ${\rm SU}(5)$ further to ${\rm Spin}(10)$, and makes this square commute:

{\rm SU}(5) \ar[r]^{\psi} \ar[d]_\rho & {\rm Spin...
...bda {\mathbb C}^5) \ar[r]^1 & {\rm U}(\Lambda {\mathbb C}^5)

We can put these squares together to get this commutative diagram:

{G_{\mbox{\rm SM}}}\ar[r]^{\psi \phi} \ar[d] & {\...
... F^*) \ar[r]^-{{\rm U}(f)} & {\rm U}(\Lambda {\mathbb C}^5)

This diagram simply says that ${\rm Spin}(10)$ is a GUT: it extends the Standard Model group ${G_{\mbox{\rm SM}}}$ in a way that is compatible with the Standard Model representation, $F \oplus F^*$. In Section 3.1, all the hard work lay in showing the representations $F \oplus F^*$ and $\Lambda {\mathbb{C}}^5$ of ${G_{\mbox{\rm SM}}}$ were the same. Here, we do not have to do that. We just showed that ${\rm Spin}(10)$ extends ${\rm SU}(5)$. Since ${\rm SU}(5)$ already extended ${G_{\mbox{\rm SM}}}$, ${\rm Spin}(10)$ extends that, too.