Clebsch-Gordan decomposition for SU2

For

m, n \geq 0

V_{m} \otimes V_{n} ≅ ⨁_{j = 0}^{m i n (m, n)} V_{m + n - 2 j} .

The identity follows by multiplying the weight characters and comparing their nested weight strings.

Flip parity in the SU2 tensor square

On the multiplicity-one summand

V_{2 n - 2 j} \subset V_{n} \otimes V_{n}

, interchange of tensor factors acts by

(- 1)^{j}

. A highest-weight vector exhibiting the sign is

\sum_{k = 0}^{j} (- 1)^{k} (k j) x^{n - k} y^{k} \otimes x^{n - j + k} y^{j - k} .

The odd-parity summands give

⋀^{2} V_{n} ≅ ⨁_{k = 0}^{⌊(n - 1) /2 ⌋} V_{2 n - 4 k - 2} .

The alternating summands in the tensor square give

⋀^{2} V_{4} ≅ V_{6} \oplus V_{2}

. Since

V_{4}

is self-dual and has trivial determinant,

⋀^{3} V_{4} ≅ (⋀^{2} V_{4})^{*} \otimes det V_{4} ≅ V_{6} \oplus V_{2} .

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