Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 18 / 4 / a

Solution

Let

v_{1, +} = e_{1} + i e_{2}

v_{1, -} = e_{1} - i e_{2}

v_{2, +} = e_{3} + i e_{4}

v_{2, -} = e_{3} - i e_{4}

and

v_{0} = e_{5}

. Direct multiplication by the printed rotation blocks gives their torus eigenvalues

e^{i x_{1}}, e^{- i x_{1}}, e^{i x_{2}}, e^{- i x_{2}}, 1.

Thus the differentiated weights of a representation are

i x_{1}, - i x_{1}, i x_{2}, - i x_{2}, 0

, with this sign convention fixed by the given matrix blocks.

A wedge of two distinct eigenvectors has the product eigenvalue, so its differentiated weight is the sum. The ten vectors in a basis of the exterior square split into three groups:

vectors v_{1, σ} \land v_{2, τ} v_{j, σ} \land v_{0} v_{1, +} \land v_{1, -}, v_{2, +} \land v_{2, -} weights i (σ x_{1} + τ x_{2}), σ, τ = \pm 1 iσ x_{j}, j = 1, 2, σ = \pm 1 0 number 442

The eight nonzero weights are distinct, so each has multiplicity one; the zero weight space has dimension two. The multiplicities add to

(2 5) = 10

, accounting for the entire representation.

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