j 8 If AIS5 with I S disjoint then 5C I5 dimS dimS dimIdimS C If L is a vector

# J 8 if ais5 with i s disjoint then 5c i5 dims dims

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j (8) If A(I)S'=5, with I, S disjoint, then 5'C. I+5, dimS~& dimS'~& dimI+dimS. (C) If L is a vector not in 5, then A({I. ))S'=5 if and only if either S'=S+{L) or there exist numbers Ni I„ with S'= {Li+NiL, Ls+NsL, , L, +N, L), where S= {I i ~ I r). (These two alternatives represent the possibilities dimS'= dimS+1 and dimS'= dimS, respectively. ) (D) If I and 5 are disjoint subspaces then 5'+5+I if and only if there is some subspace 5"( 5 such that A(I)S'=S". LProof: The subspace 5" is just {Li' L, '). g (E) If S, Si, Ss are disjoint subsPaces, and A(Si)S'= S, then A(Ss)5"=5' if and only if A(St+5&)5"=S. t Proof: We can write S'={Li+Li', , L„+L, ') where 5= {Li. . L, ), Li'. . L„'eSi. Then A(Ss)5"=5' means that S"= {Li+Li'+Li", , L, +L, '+L, "}, where Li" . L„"eSs. Also, A(Si+Ss)5"=5 means that 5"=Li+Li"', , L„+L„"' where Li"' L„"'eSi+Ss. The most general L;"'eSi+Ss may be written L, "'= L +L;", so these statements are equivalent. J ACKNOWLEDGMENTS It is a pleasure to thank Professor N. M. Kroll, Professor T. D. Lee, Professor S. B. Treiman, and Professor A. S. Wightman for discussions on this work and related matters. I particularly wish to thank Pro- fessor Wightman for his many valuable suggestions, and for enabling this paper to satisfy Salam's criterion. bIote to be added iN proof For the sake of ma. t— hematical rigor, the de6nition in Equation (III-1) of the class A requires a slight modification. The coeKcients P should be taken as functions of the individual vectors L&, 12, ~ and not only of the subspaces {L&, le, . }. The proof in Sec. IV that if fed„then freA„sis then correct, with no changes (except minor notational ones) required.

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