Solution

Write

g = (a - \overset{ˉ}{b} b \overset{a}{ˉ}) \in S U_{2}, ∣ a ∣^{2} + ∣ b ∣^{2} = 1.

The homogeneous polynomial representation of SU2 is the symmetric-power action determined by

g \cdot x = a x - \overset{ˉ}{b} y, g \cdot y = b x + \overset{a}{ˉ} y,

and extended multiplicatively to

V_{n}

In the ordered basis

(x^{3}, x^{2} y, x y^{2}, y^{3})

, the matrix for

g

a^{3} - 3 a^{2} \overset{ˉ}{b} 3 a \overset{ˉ}{b}^{2} - \overset{ˉ}{b}^{3} a^{2} b a^{2} \overset{a}{ˉ} - 2 ab \overset{ˉ}{b} b \overset{ˉ}{b}^{2} - 2 a \overset{a}{ˉ} \overset{ˉ}{b} \overset{a}{ˉ} \overset{ˉ}{b}^{2} a b^{2} 2 ab \overset{a}{ˉ} - b^{2} \overset{ˉ}{b} a \overset{a}{ˉ}^{2} - 2 b \overset{a}{ˉ} \overset{ˉ}{b} - \overset{ˉ}{b} \overset{a}{ˉ}^{2} b^{3} 3 b^{2} \overset{a}{ˉ} 3 b \overset{a}{ˉ}^{2} \overset{a}{ˉ}^{3} .

The character of a representation

(V, ρ)

χ_{V} (g) = tr ρ (g)

. Every

g \in S U_{2}

is conjugate to

diag (z, z^{- 1})

with

∣ z ∣ = 1

χ_{n} (g) = z^{n} + z^{n - 2} + \dots + z^{- n} .

For

z = e^{i θ}

this is

sin ((n + 1) θ) / sin θ

, interpreted by continuity when

sin θ = 0

Solved by gpt-5.6-sol high.

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