“…3), it is sufficient to combine A, B, and C. In this regard, we invoke Eq. (E 22),. and then we obtain:√ a ς,l,k det −bz z z + a −ς (g −1 ⋄ z z z) • (g −1 ⋄ z z z) k Y l−2k,m g −1 ⋄ z z z = l ′′′ k ′′′ m ′′′ U (ς,0) l,k,m;l ′′′ ,k ′′′ ,m ′′′ (g) √ a ς,l ′′′ ,k ′′′ (z z z • z z z) k ′′′ Y l ′′′ −2k ′′′ ,m ′′′ (z z z) ,,m;l ′′′ ,k ′′′ ,m ′′′ (g) are the matrix elements of the Sp(4, R) scalar representations U (ς,0) such that:U (ς,0) l,k,m;l ′′′ ,k ′′′ ,m ′′′ (g) = ′ L ′′ m ′′ L ′ −L ′′ =even L √ a ς,l,k a (ς,k),l,k a ′ k,l,k √ a ς,l ′′′ ,k ′′′ U L,k,m;L ′ ,L ′′ ,m ′′ (g) × δ l ′′′ ,l+l+L ′ δ 2k ′′′ ,l ′′′ −L ′′′ det a −ς−k+1 det b k (w • w) k Y l−2k,m (w) (u • u) k Y l−2k,m (u) × (−1) m (2l − 4k + 1)(2l − 4k + 1)(2L + 1) 4π 1/2 l − 2k l − 2k L m m −m l − 2k l − 2k L 0 0 (−1) m ′′′ (2L ′′ + 1)(2L + 1)(2L ′′′ + 1) 4π 1/2 L L ′′ L ′′′ m m ′′ −m ′′′ L L ′′ L ′′′ 0…”