 Solution

ID: past-exam-of-the-groups-and-representations-course-of-the-university-of-oxford/2018/1/e/solution

The stated coordinate substitution splits the polynomial basis into two three-cycles:

x^{2} ⟼ y^{2} ⟼ z^{2} ⟼ x^{2}, x y ⟼ yz ⟼ x z ⟼ x y .

The identity fixes all six basis vectors; neither nonidentity element fixes any. A permutation representation has character equal to the number of fixed basis vectors, hence

χ_{R} (0) = 6, χ_{R} (1) = χ_{R} (2) = 0.

m_{k} = ⟨ χ_{k}, χ_{R} ⟩ = \frac{1}{3} \cdot 6 = 2, R ≅ 2 ρ_{0} \oplus 2 ρ_{1} \oplus 2 ρ_{2} .

An explicit eigenbasis confirms the result. For

k = 0, 1, 2

, put

f_{k} = x^{2} + α^{- k} y^{2} + α^{- 2 k} z^{2}, h_{k} = x y + α^{- k} yz + α^{- 2 k} x z .

Then

R (1) f_{k} = α^{k} f_{k}

and

R (1) h_{k} = α^{k} h_{k}

. Thus each three-cycle carries one copy of every irreducible representation. These signs use precisely the source's substitution action; taking an inverse substitution would interchange

ρ_{1}

and

ρ_{2}

, leaving the total decomposition unchanged.

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