Algebres d'Operateurs by Pierre de la Harpe

April 3, 2017 | Science Mathematics | By admin | 0 Comments

By Pierre de la Harpe

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III-~ ~4o The modular equations. , ~ ( N = VJ(n)) be the representatives for Hn/G given by the matrices (i) with (a, b, d) = I. , N . These are the functions associated with N classes of Hn/G and, as HnG = Hn, js(Z) ~ Js(T(z)) is a permutation of Jl' "''' JN for any T in Go Hence, a symmetric function of Jl' "''' JN is invariant under z ~ T(z), T ~ G. N THEOREM 1. (a) Fn(t , j) - ~ - (t - js(Z)) is a polynomial of t and s =I j = j(z) ~rith integral rational coefficients, (b) i_~fn is not a square, the highest coefficient of j in Fn(J, j) is + l, (c) Fn(t , J) is an irreducible polynomial of t over the field C=(j), (d) Fn(t , j) = Fn(J, t), n > 1.

N) Thus ap(j(a) p, u, Ji(a), ~k(%, ~2)) TO rood p. Let us now put u = ~ ( ~ , =2 ) = ~ d ( ~ , =2 ). Then by (ii) and the definition of % ~ we get (12) (0(a)p - Sd(a_)) 7T (@d(~, ~2) - ~i(%, ~2)) -= 0 ~d p. Iv-9 Since ~divides (13) (p), we have afortiori (j(a)p - jd(z)) i7r ~ (~d (5, =2 ) " ~ i ( 5 ' ~2)) ----0 ~ d ~" By Theorem I, , a2) mod 2 and the right hand side generates the ideal ~12, which is prime to p. Therefore, (13) implies that J(a) p - Jd(a) --=0 rood p and this, in view of (9), is the congruence (8).

Let ~l' ~2 be a base of a and P , ~ H p be as _ Furthermore, let c,d < p+l be the indices such that -1 Iv-8 P~Mc, PGGH d. ~_) - j~(a_) - jd(~_) - Jd(%/~ 2) . We now apply Lemma 4 to the case where ~ i " Ji' ~i " ~ i (I <= i ~ p+l) and t - JP. This gi~s Gp(J(q)P,u,Ji(q) P ~k(q)) m (JP-Jp+l) 9 (u p- ~p) rood p which, together with (6), shows that Gp(J(q) p, u, Ji(q), ~k(q))=-0 rood p , G being considered as a power series in u and q. P principle (II ~6), we have then (lo) By the q-expansion N (jP,u,j) 9 o rood p P where (cf.

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