Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 302 / 2

Table of contents
- a 2
  - Solution a
- b 2
  - Solution b
- c 2
  - Solution c
- d 2
  - Solution d

a

Solution

Differentiating

Λ^{T} η Λ = η

at the identity gives

X^{T} η + η X = 0

. The six matrices

(M^{μν})^{α}_{β} = η^{μα} δ_{β}^{ν} - η^{να} δ_{β}^{μ}, M^{μν} = - M^{νμ},

obey this condition and form a basis of the Lorentz algebra. With cyclic indices,

J_{1} = M^{23}, J_{2} = M^{31}, J_{3} = M^{12}, K_{i} = M^{0 i} .

The

J_{i}

mix the two spatial coordinates perpendicular to

x^{i}

and therefore generate rotations about that axis;

K_{i}

mixes

x^{0}

with

x^{i}

and generates a Lorentz boost in the

x^{i}

direction.

Solved by gpt-5.6-sol high.

b

Solution

Matrix multiplication gives

[J_{i}, J_{j}] = ϵ_{ijk} J_{k}, [J_{i}, K_{j}] = ϵ_{ijk} K_{k}, [K_{i}, K_{j}] = - ϵ_{ijk} J_{k} .

For example,

[J_{1}, J_{2}] = J_{3}

,

[K_{1}, K_{2}] = - J_{3}

, and

[J_{1}, K_{2}] = K_{3}

. Over

C

A_{i} = \frac{1}{2} (J_{i} + i K_{i}), B_{i} = \frac{1}{2} (J_{i} - i K_{i}) .

Then

[A_{i}, A_{j}] = ϵ_{ijk} A_{k}

,

[B_{i}, B_{j}] = ϵ_{ijk} B_{k}

, and

[A_{i}, B_{j}] = 0

. This proves the chiral decomposition of the complex Lorentz algebra. Its finite-dimensional irreducible representations are tensor products of irreducible representations of the two factors and are labelled by pairs

(j_{L}, j_{R})

of nonnegative half-integers.

Solved by gpt-5.6-sol high.

c

Solution

A coordinate transformation

x \mapsto P x

conjugates an infinitesimal generator, hence

M^{μν} \mapsto P M^{μν} P

. It fixes the rotation generators

J_{i}

and negates the boosts

K_{i}

. Therefore it exchanges

A_{i}

and

B_{i}

, and the parity action on a Lorentz representation is

(j_{L}, j_{R}) ⟼ (j_{R}, j_{L}) .

An irreducible representation is parity invariant precisely when

j_{L} = j_{R}

. If

j_{L} \neq = j_{R}

, parity invariance requires the reducible sum

(j_{L}, j_{R}) \oplus (j_{R}, j_{L})

. Thus the two Weyl-spinor representations

(1/2, 0)

and

(0, 1/2)

are exchanged, while their direct sum is the parity-invariant Dirac spinor.

Solved by gpt-5.6-sol high.

d

Solution

Parity preserves the Minkowski metric but reverses orientation. Hence ordinary tensor contraction makes

C_{1} = M_{μν} M^{μν}

parity even, while the Levi-Civita pseudotensor changes sign and makes

C_{2} = M_{μν} M^{μν}

parity odd. Equivalently,

C_{1}

is proportional to

J^{2} - K^{2}

and

C_{2}

to

J \cdot K

. The combinations

C_{1} \pm i C_{2}

are proportional to the two quadratic Lorentz Casimir invariants, and parity exchanges them exactly as it exchanges the two factors in the complexified algebra.

Solved by gpt-5.6-sol high.

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