If we insist on equal n s which is not strictly

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If we insist on equal N ’s, which is not strictly necessary, we are limited to the models 2.a) and 2.b) with representations given by (5.16), (5.17), Dirac operators and chirality assignments by : D 2 .a = 0 0 0 0 0 0 0 A + 0 0 0 B + 0 A B 0 ; χ 2 .a = χ 1 N 0 0 0 0 1 N 0 0 0 0 1 N 0 0 0 0 - 1 N . (5.25) D 2 .b = 0 B + A + 0 B 0 0 0 A 0 0 0 0 0 0 0 ; χ 2 .b = χ - 1 N 0 0 0 0 1 N 0 0 0 0 1 N 0 0 0 0 1 N . (5.26) 5.3 The product The product of T 1 with T 2 yields the total triple T = {A , H , D ,χ, J } with algebra A = A 1 ⊗ A 2 . Since T 1 is S 0 -real, the product is also S 0 -real with hermitian involution Σ 0 = σ 0 1 2 . The total Hilbert space H = H 1 ⊗ H 2 is decomposed as H = H (+) ⊕ H ( - ) , (5.27) where H ( ± ) = H 1( ± ) ⊗ H 2 . Elements of H ( ± ) are represented by column matrices of the form Ψ ( ± ) ( x ) = ψ ( ± ) aa ( x ) ψ ( ± ) ab ( x ) ψ ( ± ) ba ( x ) ψ ( ± ) bb ( x ) , (5.28) where each ψ ( ± ) αβ ( x ) is a Pensov spinor of H 1( ± ) with N ”internal” indices (not explicitely written down). The total Dirac operator is D = D 1 1 2 + χ 1 ⊗ D 2 , (5.29) 39
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and the chirality is given by χ = χ 1 χ 2 . (5.30) The continuum spectral triple T 1 of 5.1 is of dimension two and its real structure J 1 obeys J 1 2 = 1 1 1 ; J 1 D 1 = 1 D 1 J 1 ; J 1 χ 1 = 1 χ 1 J 1 , where 1 = - 1 was fixed but 1 and 1 were independent free ± 1 factors. On the other hand the discrete triple T 2 of 5.2 is of zero dimension and J 2 obeys J 2 2 = 2 1 2 ; J 2 D 2 = 2 D 2 J 2 ; J 2 χ 2 = 2 χ 2 J 2 , with 2 = 2 = 2 = +1. The real structure J of the product triple should obey J 2 = 1 ; J D = D J ; J χ = χ J . (5.31) If we require that Connes’ sign table be satisfied, i.e. for T 1 , n 1 = 2 and 1 = - 1 , 1 = +1 , 1 = - 1, for T 2 , n 2 = 0 and 2 = 2 = 2 = +1 for n 2 = 0, for the product T , n = 2 and = - 1 , = +1 , = - 1, it is seen that, with J = J 1 J 2 , the sign table for the product is not obeyed, since such a J implies the consistency conditions = 1 2 , = 1 = 1 2 , = 1 2 , (5.32) and the second condition is not satisfied. If we keep the same Dirac operator (5.29), it is the definition of J that should be changed 19 to J = J 1 ( J 2 χ 2 ) , (5.33) 19 A general examination of the sign table and the product of two real spectral triples is done in [19]. 40
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and with this J the consistency conditions become = 1 2 2 , = 1 = - 1 2 , = 1 2 , (5.34) and these are satisfied. In the rest of this section, we shall assume that these choices are made. Also, in order to simplify the forthcoming formulae, we take α = - 1 so that C 1 = 0 1 - 1 0 and a (+ s ) = a ( - s ) = +1 which imply that J 1( - ) = - J 1(+) + = J 1(+) and D 1( ± ) = D ( ± s ) . The change of J 2 to J 2 χ 2 will not change the representation π o 2 since π 2 is even with respect to χ 2 . The S 0 -real structure implies that π and π o , D and χ are block diagonal in the decomposition (5.27) of H and we obtain the representations : let f ∈ C ( S 2 , C C ), then π ( f ) = π (+) ( f ) π ( - ) ( f ) ; π o ( f ) = π o (+) ( f ) π o ( - ) ( f ) , (5.35) π ( ± ) ( f ( x )) = f a ( x ) 1 N 0 0 0 0 f a ( x ) 1 N 0 0 0 0 f b ( x ) 1 N 0 0 0 0 f b ( x ) 1 N , π o ( ± ) ( f ( x )) = f a ( x ) 1 N 0 0 0 0 f b ( x ) 1 N 0 0 0 0 f a ( x ) 1 N 0 0 0 0 f b ( x ) 1 N .
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