!q (z) Phu;qi (z) dz?K;
i 2
@x Phu;qi(z) d;
see eq. (17). It is easy to see that@ !q= 0, if@ di erentiates only the variables (; z ). Set (; u; z; y )= !1
; (z1? y1;1 )^ !; (y1;1? y1;2)^:::^ !;K (y1;K?1? y1;K )^
:::::::::::::::::::::::::::::::::^! N; (zN? yN;1)^ ! N; (yN;1? yN;2)^:::^ ! N;KN (yN;KN?1? yN;KN ) h j n (e n;::: e n;Kn ) n (43)11 12 1 1 1 1 2 1 ( )
with n;i=
Kn P i0=i
n;i0
and (; u; z; y )=X X
K 2SK
(?1)j j K; (; u; z; y):
We still have to dress de ne the form1
with the duj di erentials. The convenient way to do this is toX
= e? S dw1^:::^ dwr^ h; i+ 2 1i 16
r
j=1
dw1^:::d:::^ dwr@uj h; ib
j
(44)
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