 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ii/paper-4/18g/c/solution

Codex  0 Created 2026-09-24 Updated 2026-10-05

V_{n} \otimes V_{n}

, interchange of the two factors commutes with SU2. On the highest-weight vector

D^{r} x_{1}^{n - r} x_{2}^{n - r}

it acts by

(- 1)^{r}

. Each irreducible summand in the Clebsch-Gordan decomposition has multiplicity one, so this sign is its scalar action on the entire summand. The symmetric square is the

+ 1

- 1

eigenspace. Therefore

S^{2} V_{n} ≅ j = 0 ⨁ ⌊ n /2 ⌋ V_{2 n - 4 j}, Λ^{2} V_{n} ≅ j = 0 ⨁ ⌊(n - 1) /2 ⌋ V_{2 n - 2 - 4 j} .

For instance,

S^{2} V_{1} = V_{2}

and

Λ^{2} V_{1} = V_{0}

. The dimensions agree respectively with

(n + 1) (n + 2) /2

and

n (n + 1) /2

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