The Category of H-modules over a Spectrum 13
We have
a/(l+d) e a/(l-d) ;
t(l-d)/a _ t/e _ t(l+d)/a;
(t-d)/a t/e (t+d)/a.
We also have e a/(l-d); e-ed a; and e a+ed a+d. If t _ a, then
(xx[t-d,t+d]) n (Xxi) c L , and
(x,t/e) e (xx[(t-d)/a,(t+d)/a]) n (Xxi) ch(L). Thus
F(x,t/e) = p (F,G,e)(x,t) e U. If t a, then (x,t) e Kn (xx[a-d,a+d] ) ,
and x x [a-d,a+d] c L . Then we have
(a-d)/a (t-d)/a t/e _ 1 (a+d)/a, and so
(x,t/e) e (xx[(a-d)/a,(a+d)/a]) n (Xxi) c h(L ). Again
F(x,t/e) = p (F,G,e)(x,t) eU.
Suppose now that e ^ t. If t _ a, then (x,t) e Kn (Xx[a-d,a+d]);
x x [a-d,a+d] e L ; and x x [0,d/(l-a)] c k(L ). It is readily checked
2
that (a-d-a )/(l-d-a) a-d e, and it follows that (a-e)/(l-e) d/.(l-a) .
We have 0 _ (t-e)/(l-e) _ (a-e)/(l-e) d/(l-a), and so
(x,(t-e)/(l-e)) ek(L2). Therefore p3(F,G,e)(x,t) = G(x,(t-e)/(1-e)) eU.
Next suppose that e _ t and that a _ t _ 1. Then
e (a+d-a )/(l+d-a) _ (a+d-at)/(1+d-t), and it follows that
(t-d-a)/(l-a) (t-e)/(1-e). The function r(t) = (a-d-at)/(1-d-t) is
decreasing in the interval (-°°,l-d). Therefore, for a _ t 1-d,
(a-d-at)/(1-d-t) _ r (a) = (a-d-a2)/(1-d-a) e. If a _ t 1-d, then
xx [t-d,t+d] c L ; (xx[(t-d-a)/(l-a),(t+d-a)/(1-a)]) n (Xxi) ck(L2); and
hence (x, (t-e)/(1-e) ) ek(L). Thus p (F,G,e)(x,t) eU. If 1-d _ t _ 1,
then x x [t-d,l] c L ; x x [(t-d-a)/(1-a),1] c k(L ); and hence
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