Fragmentos de una conversación con Julio Maruri:

M⊗ N 

P⊗ Q

= (M P) ⊗ (N Q). (8.28)

For example we can construct(4 × 4) matrices in this way, such as σ1⊗ σ3=

and the like. Let us apply this construction to Clifford algebras.

• Start with the (2 × 2) matrices

which is a diagonal matrix in this representation. Note that in this construction the Clifford matrices are all hermitian. We take their bilinear product to build the S O(4) generators commute with one another and form two SU(2) Lie algebras. In block form

L⁺_k =

The S O(4) algebra is semi-simple, the sum of two SU(2)s. We can also define the operators

P_± ≡ 1

2(1 ± ₅), (8.36)

which satisfy the projection operator relations

P_±P_±= P_±, P_±P_∓= 0, (8.37)

so that they split the four-dimensional space into two parts. Since they clearly commute with the S O(4) generators, the four-dimensional spinor space on which the generators act is reducible.

However, the full four-dimensional space is needed to realize the S O(5) generators, since we need to add

X₄₅= −i

2 _{4 5}, X_{i 5}= −i

2 _{i 5}, (8.38)

showing that S O(5) has a four-dimensional irreducible representation.

• We repeat the procedure to build (8 × 8) Clifford matrices, by multiplying the five previous matrices byσ1, and adding one alongσ2

₁= σ₁⊗ σ₁⊗ σ₁, ₂= σ₁⊗ σ₁⊗ σ₂, 3= σ1⊗ σ1⊗ σ3, 4= σ1⊗ σ2⊗ 1,

₅= σ₁⊗ σ₃⊗ 1, ₆= σ₂⊗ 1 ⊗ 1, (8.39) and one more by taking their product

7= (−i) _{1 2 3 4 5 6}= σ₃⊗ 1 ⊗ 1. (8.40) This means that we can represent both S O(6) and SO(7) algebras in this eight-dimensional space, using the projection operators

P_± ≡ 1

2(1 ± ₇), (8.41)

8.5 Spinor representations 161 to separate the eight-dimensional spinor space into two four-dimensional spaces. Indeed, the fifteen S O(6) generators

X_{i j} = −i

2 _{i j}, i = j = 1, 2, . . . , 6, (8.42) are 8× 8 matrices which split (in this representation) into two 4 × 4 block-diagonal matrices, since 7is diagonal. Thus, S O(6) acts independently on two four-component spinors, and because S O(6) has the same algebra as SU(4), they must correspond to its four-dimensional complex representation and its conjugate. Write the infinitesimal changes in the form usual 4× 4 Gell-Mann matrices. The hatted matrices are closely related, as it is only rotations containing 6that act onψ and χ with different sign. For instance the diagonal generator corresponding toω3is actually the same on both spinors

1

However, the generator corresponding toλ8is actually different

⎧⎪ where the tilde matrix is the representation matrix on the 4 representation, and the(4×4) charge conjugation matrix is Is there a direct way to see how conjugation arises? Our Clifford matrices are man-ifestly hermitian, but not symmetric nor real. Note that the Clifford algebra is equally satisfied by the transposed and complex conjugated matrices

±

Similarity transformations on Clifford matrices

i → S iS⁻¹ (8.49)

preserve the Clifford algebra, and they can be used to change representations; in our case we have opted for a representation where 2n+1is diagonal. This also means the existence of a similarity transformation which transposes or takes the complex conjugate of the matrix.

Conjugation of spinors is tricky. We have seen that the S O(3) spinor representation is pseudoreal in the sense that it has a charge conjugation operation

C= σ₂, (8.50)

which is a member of the algebra, so that the 2 and 2 are the same representation.

These examples suggest a recursive construction of Clifford algebras in higher dimensions. Given the 2n−1 Clifford (2ⁿ⁻¹×2ⁿ⁻¹) matrices, ˜ i, we build(2ⁿ×2ⁿ) Clifford matrices as follows

i = σ1⊗ ˜ i, i = 1, 2, . . . , 2n − 1.

2n = σ2⊗ 1,

2n+1= (−i)ⁿ 1 2. . . 2n = σ3⊗ 1. (8.51) These(2ⁿ× 2ⁿ) matrices, a, satisfy the(2n + 1)-dimensional Clifford algebra

{ a, b} = 2 δab, a, b = 1, 2, . . . , 2n + 1. (8.52) The S O(2n + 1) generators can be written as (2ⁿ× 2ⁿ) matrices

X_ab = −i

2 a b, a = b. (8.53)

By forming the projection operators out of 2n+1, we split the 2ⁿ spinor into two 2ⁿ⁻¹spinors that transform according to S O(2n). Hence we see in this way that there are two spinor representations in even dimensions and only one in odd dimensions.

The product of two S O(3) spinors yields a triplet which describes a vector in three-space, and a singlet. Mathematically, the outer product of two two-component spinors ψ, χ is a (2 × 2) matrix, which can be written as a linear combination of the three Pauli matrices and the unit matrix

ψ χ^T = 1 v0+ σivi, (8.54)

where the coefficients are obtained by taking the appropriate traces; for example Tr(ψ χ^Tσj) = Tr(σiσj) vi = 2 vj.

8.5 Spinor representations 163 This is called the Fierz expansion of spinor products. Let us apply it to the spinor representations of S O(6), where

4× 4 = 10 + 6, 4× 4 = 15 + 1, 4× 4 = 10 + 6. (8.55) Representations appearing on the right-hand side are easily identified as S O(6) tensors: 6 is the vector, 15 is the second-rank antisymmetric tensor. The third-rank antisymmetric tensor Ti j k has twenty real components, while only 10 and 10 are in the product. The real combination is of course their sum which has twenty compo-nents. The way to understand this is to note that in S O(6), there is the totally antisymmetric Levi–Civita tensor on six indices, i j klmn, which can be used to generate self- and antiself-dual third rank tensors

T_{i j k}^± = 1 Similarly, we can use the Levi–Civita symbol to rewrite a four-component anti-symmetric tensor in terms of a second rank tensor.

The numbers check since we can express the 8· 8 = 2⁶components break up as 2⁶= (1 + 1)⁶=

each being interpreted as rank p antisymmetric tensors (called p-forms by mathe-maticians: a scalar is a zero-form, a vector is a one-form, etc.). Their construction is straightforward in terms of products of different anticommuting Clifford matrices

i₁i₂...in = i₁ i₂ · · · i_n, i1= i2= · · · = in. (8.57) Their antisymmetry is obvious since all the matrices anticommute with one another. They are also traceless

Tr

 i1i2...in

= 0, (8.58)

as one can verify from our explicit construction. The Fierz expansion expresses the outer product of two spinors as a linear combination of all these matrices

 ^T = 6

j=1

i1i2...ij vi1i2...ij. (8.59)

The coefficients are obtained by taking the appropriate traces. For example

^T i j = 8 vi j, (8.60)

where we have used Tr

 i j kl

= 8 (δi kδjl − δj kδil). (8.61)

Cn ∼ Sp(2n)

Symplectic algebras are similar to the orthogonal ones except that symmetriza-tion and antisymmetrizasymmetriza-tion are interchanged. The fundamental representasymmetriza-tions move along the Dynkin diagram; all, including the nth node, are represented by symmetric tensors.