我们通过耦合一个N = 1,0 $$ \ mathcal {N} = \ left()来构建6D nonabelian N = 1,0 $$ \ mathcal {N} = \ left(1,\ 0 \ right)$$理论 1,\ 0 \ right)$$张量倍数到N = 1,0 $$ \ mathcal {N} = \ left(1,\ 0 \ right)$$超倍数。 虽然N = 1,0 $$ \ mathcal {N} = \ left(1,\ 0 \ right)$$张量多重数在量规组