V_{4} \otimes V_{3} ≅ V_{7} \oplus V_{5} \oplus V_{3} \oplus V_{1},

and

V_{3} \otimes V_{3} ≅ V_{6} \oplus V_{4} \oplus V_{2} \oplus V_{0} .

On the summand

V_{6 - 2 j}

, interchange of the two tensor factors acts by

(- 1)^{j}

j

summands make the symmetric square, so

S^{2} V_{3} ≅ V_{6} \oplus V_{2} .

ii

Solution

Repeatedly using

V_{r} \otimes V_{1} ≅ V_{r + 1} \oplus V_{r - 1}

shows that the multiplicity of

V_{n - 2 j}

in the nth tensor power satisfies Pascal's recursion. Consequently

V_{1}^{\otimes n} ≅ j = 0 ⨁ ⌊ n /2 ⌋ ((j n) - (j - 1 n)) V_{n - 2 j},

where

(- 1 n) = 0

n - 2 j

and

n - 2 j + 2

The weights of

V_{2}

are

z^{2}, 1, z^{- 2}

. Degree-three monomials in these weights show that the character of the symmetric power

S^{3} V_{2}

z^{6} + z^{4} + 2 z^{2} + 2 + 2 z^{- 2} + z^{- 4} + z^{- 6} = χ_{V_{6}} + χ_{V_{2}} .

Characters determine finite-dimensional

S U (2)

representations, hence

S^{3} V_{2} ≅ V_{6} \oplus V_{2} .

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