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 ? ).
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