 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2008/paper-2/10/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

Let

J = (0 - 1 10)

. Since

J^{2} = - I_{2}

and

J^{4} = I_{2}

ρ (1) = I_{2}, ρ (- 1) = - I_{2}, ρ (- i) = - J .

For the quaternion group, denote the three displayed complex matrices by

A, B, C

. Direct multiplication gives

A^{2} = B^{2} = C^{2} = - I_{2}

A B = C

B A = - C

, and

A BC = - I_{2}

. Together with the images of

\pm 1

, these are exactly the quaternion multiplication relations, proving that the assignment defines a group homomorphism into

GL (2, C)

M = X + iY

R (M) = (X Y - Y X)

. It acts on grouped real and imaginary coordinates and satisfies

R (MN) = R (M) R (N)

. The explicit generators are

D (i) = 00010010 0 - 1 00 - 1 000, D (j) = 0010 000 - 1 - 1 000 0100, D (k) = 0 - 1 00 1000 000 - 1 0010 .

D (1) = I_{4}

and

D (- q) = - D (q)

to specify all eight elements.

For an invertible real

S

D (q) = S D (q) S^{- 1}

obeys

D (q_{1}) D (q_{2}) = S D (q_{1} q_{2}) S^{- 1} = D (q_{1} q_{2})

. The trace is invariant under similarity, so their characters agree. Explicitly,

χ (1) = 4, χ (- 1) = - 4, χ (\pm i) = χ (\pm j) = χ (\pm k) = 0.

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