MATHEMATICS

# Lie Algebras in Particle Physics

# Errata and Notes

## Chapters 1

### 1.4 Irreducible representations

#### On equivalence

Note that \( S \) in Eq. (1.10) is in general not unitary. Therefore two equivalent representations do not just differ by *a trivial choice of basis* as claimed in the book. All representations of finite groups are equivalent to unitary representations, and we can always find a basis for any representation to have a unitary matrix form, although the basis will not be orthonormal in general. (N.B.: a non-unitary operator can have its matrix form being unitary by choosing a non-orthonormal basis. This does not make the operator itself unitary.)

#### On reducibility

The defintion of reducible representations should be those that have a *non-trivial* invariant subspace. Obviously \( \\{ 0 \\} \) and the original vector space are always invariant subspaces for any representations.

\( D(g) \) is always invertible for any representation \( D \) and \( g \in G \), because \( D(g) D(g^{-1}) = D(e) = 1 \). This is actually why Eq. (1.11) is valid as \( D(g) \) will never reduce the dimensionality of the invariant subspace.

#### On reducibility of equivalent representations

*If two representations \( D_1 \) and \( D_2 \) are equivalent, \( D_1 \) is reducible if and only if \( D_2 \) is reducible.* The proof is as follows.

If \( D_1 \) is reducible, there exists a non-trivial subspace \( X \) whose projector \( P \) satisfies Eq. (1.11)

\[ P D_1(g) P = D_1(g) P, \forall g \in G. \]

Since \( D_1 \) and \( D_2 \) are equivalent, there exists an invertible transformation \( S \) such that \( D_1(g) = S^{-1} D_2(g) S, \forall g \in G \), therefore

\[ P S^{-1} D_2(g) S P = S^{-1} D_2(g) S P, \]

or equivalently

\[ S P S^{-1} D_2(g) S P S^{-1} = D_2(g) S P S^{-1}, \]

where \( S P S^{-1} \) is the projector of subspace \( S X \), which has the same dimensionality of the original invariant subspace \( X \). Obviously \( S X \) is non-trivial since there exists \( v \not \in X \) and it is easy to show that \( S v \not \in S X \). This proves that \( D_2 \) is also reducible.

This fact is implicitly used in the proof of Theorem 1.2.

#### On reducibility of a representation and its Hermitian conjugate

*A representation \( D \) is reducible if and only if its Hermitian conjugate \( D^\dagger \) is reducible.* The proof is as follows.

If \( D \) is reducible, there is a non-trivial subspace \( S \) such that its projector \( P \) satisfies \( P D(g) P = D(g) P, \forall g \in G \).

Notice that the complementary subspace of \( S \) is also non-trivial, and \( I - P \) is the projector onto it. We have

\[ \begin{eqnarray} && (I - P) D^\dagger (g) (I - P) \\ &=& [(I - P) D(g) (I - P)]^\dagger \\ &=& [(I - P) D(g) - I D(g) P + P D(g) P]^\dagger \\ &=& [(I - P) D(g)]^\dagger \\ &=& D^\dagger (g) (I - P). \end{eqnarray} \]

This proves that \( D^\dagger \) is also reducible. The *if* part can be proved similarly.

#### On complete reducibility

Note that \( D_j(g) \) as in Eq. (1.13) should not be understood as an operator defined on the original space on which \( D(g) \) is. Instead, each \( D_j(g) \) is defined on a non-trivial subspace \( X_j \) of the original space \( X \), and \( X_i \) and \( X_j \) are orthogonal to each other for \( i \neq j \) and \( \cup_j X_j = X \).

One can easily prove that if two representations \( D_1 \) and \( D_2 \) are equivalent, \( D_1 \) is completely reducible if and only if \( D_2 \) is completely reducible.

This fact is implicitly used in the proof of Theorem 1.2.

### 1.9 Useful theorems

#### On theorem 1.1

\( U \) is implicitly assumed to be unitary in Eq. (1.28) and this guarantees that \( X \) is hermitian in Eq. (1.31). The proof that a unitary \( U \) always exists to diagonalize a hermitian matrix can be found in most QM books. In fact \( U \) can be simply made of orthonormal eigenvectors of \( S \) juxtaposed next to each other.

### 1.10 Subgroups

The statement *Every element of \( G \) must belong to one and only one coset* is made without proof. The *one* part is rather obvious since \( e \) must be in the subgroup. Below is the proof of the *only one* part.

Assuming that \( x \in H g_1 \) and \( x \in H g_2 \), we want to prove that \( H g_1 = H g_2 \) by showing that \( H g_1 \subset H g_2 \) and \( H g_2 \subset H g_1 \).

There must exist \( x_1 \in H \) and \( x_2 \in H \) such that \( x = x_1 g_1 = x_2 g_2 \). This implies that \( g_1 = {x_1}^{-1} x_2 g_2 \).

For any \( y \in H g_1 \), there exists \( y_1 \in H \) such that \( y = y_1 g_1 = y_1 {x_1}^{-1} x_2 g_2 \). Note that \( y_1 \), \( {x_1}^{-1} \) and \( x_2 \) are all elements of \( H \) and therefore \( y_1 {x_1}^{-1} x_2 \in H \) and \( y \in H g_2 \). This proves \( H g_1 \subset H g_2 \).

\( H g_2 \subset H g_1 \) can be proved in similarly.

### 1.11 Schur's lemma

#### On theorem 1.3

In the proof it is rather hand-waving to simply state *A similar argument shows that \( A \) vanishes if there is a \( \langle v| \) which annihilates \( A \)*, because it actually requires the irreducibility of \( D_1^\dagger \) rather than that of \( D_1 \) itself.

Fortunately it can be proved that for any \( D \), \( D \) is reducible if and onlly if \( D^\dagger \) is reducible. The proof is given here. Note that this proof does not require \( D \) to be a representation of a finite \( G \). It is equally valid even if \( G \) is infinite.

#### On the discussion immediately following Schur's lemma

Note that we first *choose* a unitary representation \( D \) and then an orthonormal basis such that \( D \) has block diagonal form. If we choose a non-unitary represenation in the first place, we will never manage to make it have block diagonal form in an orthonormal basis.

Suppose \( D \) has the following matrix form

\[ \begin{bmatrix} D_1 & 0 \\ 0 & D_2 \end{bmatrix}, \]

where \( D_1 \) and \( D_2 \) are actually equivalent.

We can always redefine the orthonormal basis so that \( D_1 \) and \( D_2 \) have exactly the same matrix form. Now the matrix form of \( D \) becomes

\[ \begin{bmatrix} D_1 & 0 \\ 0 & D_1 \end{bmatrix}. \]

Since there is only one non-equivalent irreducible sub-representaton, \( a \) can take only one value. But this sub-representaton appears twice and therefore \( x \) can be 1 or 2. Suppose \( D_1 \) is a 3-by-3 matrix. We will label the basis as \( | 1, 1, 1 \rangle \), \( | 1, 2, 1 \rangle \), \( | 1, 3, 1 \rangle \), \( | 1, 1, 2 \rangle \), \( | 1, 2, 2 \rangle \), and \( | 1, 3, 2 \rangle \). Note \( x \) is used to differentiate different basis vectors that share the same symmetry properties. The statement *\( x \) represents whatever other physical parameters there are* is rather misleading.

### 1.12 Orthogonality relations

\( A^{a b}_{j l} \) is an operator on the subspace consisting of the tensorial product of all invariant subspaces corresponding to \( D_b \). \( l \) corresponds to the combination of \( k \) and \( y \) as in Eq. (157), and runs from 1 to the product of the number of the invariant subspaces and the dimensionality of \( D_b \).

It is implicitly assume that \( G \) is a finite group in the derivation from Eq. (161) to Eq. (162).