 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2003/paper-2/6/b/solution

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

T_{ik}^{'} = Q_{ia} Q_{kb} T_{ab}

. In every complete tensor contraction, summation over a transformed index contracts a pair of

Q

factors to a Kronecker delta. In detail,

α^{'} = Q_{ia} Q_{ib} T_{ab} = δ_{ab} T_{ab} = α,

β^{'} = Q_{ia} Q_{kb} Q_{k c} Q_{i d} T_{ab} T_{c d} = δ_{b c} δ_{a d} T_{ab} T_{c d} = T_{ab} T_{ba} = β,

γ^{'} = Q_{ia} Q_{kb} Q_{k c} Q_{m d} Q_{m e} Q_{i f} T_{ab} T_{c d} T_{e f} = δ_{b c} δ_{d e} δ_{a f} T_{ab} T_{c d} T_{e f} = T_{ab} T_{b d} T_{d a} = γ .

Thus all three are invariant under rotations. They are

tr T

tr (T^{2})

and

tr (T^{3})

α = \sum_{i} λ_{i}, β = \sum_{i} λ_{i}^{2}, γ = \sum_{i} λ_{i}^{3} .

Expanding

\prod_{i} (λ - λ_{i})

λ^{3} - α λ^{2} + \frac{α ^{2} - β}{2} λ - \frac{α ^{3} - 3 α β + 2 γ}{6} = 0.

Comparison identifies

c_{2} = - 1, c_{1} = \frac{1}{2}, c_{0} = - \frac{1}{6}

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