 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-2/10/c/solution

Put

Y_{p} = D (Y)

Y_{p}^{2} = (10 p - p 1) = I_{2}

for every real

p

. Since

D (X) = - I_{2}

commutes with all matrices and

D (Z) = - Y_{p}

, the remaining relations also hold:

D (X)^{2} = D (Z)^{2} = I_{2}, D (X) D (Y) = D (Z), D (X) D (Y) D (Z) = I_{2} .

All matrices are invertible, and the relations reproduce the multiplication table of the Klein four-group. Therefore

p \in R is arbitrary .

A representation need not be an orthogonal representation; imposing orthogonality on the displayed matrices would be an extra condition.

The characters are their traces, independent of

p

(χ (I), χ (X), χ (Y), χ (Z)) = (2, - 2, 0, 0) .

In terms of part (b),

χ = χ_{-+} + χ_{--}

, so the representation is reducible. For an explicit change of basis, take

S = (10 - p /2 1) .

Then

S^{- 1} D (Y) S = diag (- 1, 1)

and

S^{- 1} D (Z) S = diag (1, - 1)

, while

D (X)

remains

- I_{2}

. These are the equivalent sheared representations of the Klein four-group; the parameter changes the basis, not the two irreducible constituents.

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