# Download Analysis, Manifolds and Physics [Part II] (rev.) [math] by Y. Choquet-Bruhat, et. al., PDF

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11) holds in the cases 1 < i < Ni — l,j = 1. Proof. Let fm(h,... IV0(1 + *i, • • •, 1 + tN), so that fm(h,. • •, tN) is identically zero if m < Ni(Nx - 1). 15) with j = 1 it suffices to show that E «i^5 = 0. 17, (1) Since ((i — l)&)rfc = 0 when Tj > 0, the summation is over the partitions with n = 0. 16) gives fm(tl, • • • ,ti-i,0, Since nAr0+i

When we set x = 1 in Eq. 17), we have in a limit q —> 1 H^9' ~ / dn exp [ n log x + - n2 log q - J * \ 2 / dt logq J1 1 . 24) where we have set w = qn. With a solution of Eq. 24), we have ~(s) ~ exp ( -—*— L(l - w)) , V logq) which gives the effective central charge as c(g) = \L{1- W). , Ref. 27a) L(x) + L(y) = L(xy) + L{^—^-) + L(^—^-). 25) as a function of the statistical interaction g in Figure 1. 8 0 0 1 5 I 10 I < 15 20 9 Figure 1. Effective central charge c(g) is plotted for the statistical interaction g.

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