46 M. L. RACINE

With respect to the new bas e HA becomes D HAD. We must show that

D HAD c C . Since HA e ^ it suffices t o show that the non-diagonal

n n

entries of the first row of HA € $, i . e . d a + c a

€

? , 1 j n. If j = 2

l l j l

£)

from (4') c a

6

©. But v(c ) = 1 an d a € & . Since v(j& ) = 22£,

-1 2

a € C. Also c a € C implies a

€

$ . But d e ? , s o

d-a_ _ + c_a__ € *p. If j 2 from (4") c , a . , , c. , a,, or d.a., « 0 , Since

1 12 1 22 J j l J- l j l J j l

v (c ) = 0, s 1 and v(d.) = 0, i 2r, we have a c ©. Similarly a * ©.

Hence a . . = a., c C, a^. = a._ € 0 an d d a , . + c , a

0

. € ? . Hence

lj Jl 2j j2 1 l j 1 2j

M C ^nE(LD) . Now D" 1 HAD = D ^ H D " 1 DtAD = H'E^AD. Let a eJ0 n C ,

—t — 2

D a H e il D = pa il pe i l € ^ e l l " B u t H ' ( ^

0

n r ) e i l C E ( L D ) n ^ in particular

pe e E(LD) n p. So M C J n E(LD) and the exception i s indeed not maximal.

/ d i

° \

LEMMA 14. Let H = , d. e & ; v(d ) v(d ),

\0 d

2

/

l

r = v(d

2

) - vfd^ , M1 = { H J A I A = H(®2, fiQ), H ^ * ©

2

} . Then

/ - v ( d l

M'

-v(d )

P (

d

2

( j 5

0

n 1 } ) ) , + , p

-v(d )

(d,(fi n ? ))' denotes the o-subalgebra of 0 generated by

1 -v(d )

d.(*

0

n ? ).