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Unformatted text preview: 82 N.M. Katz: t hat whenever h orizontal s ections m w , o ne for each exponentsystem W
a s above verify ~ XW mw=O (6.0.3.1) in M W t hen all m w= O. IWl=~ W~eZ, ; t his has a meaning
and put D W=[I
i
W~!
b ecause all Wi__<p 1. For any two exponent V, W as above, we have
L et (6.0.3.2) D v(Xw)=o
DV (x v) = 1 u nless W~> V~ g for all i ~ ~
Dv ( x w) = ( ~ I ( WVi ) ) X wv
i i f W~>V~ for all i. S uppose that (6.0.3.1) holds, but that not all m w=O. A mong exponent
s ystems W having m w 9 O, l et V have a maximal weight IV]. Let's apply
V(D v) t o (6.0.3.1): ~ V (Dv)(x w mw) = O. (6.0.3.3) W B ecause all the mw a re horizontal, V(Dv)(x w row)= Dv (Xw) 9row, w hence
(6.0.3.3) becomes ~ Dv (Xw) 9 (6.0.3.4) =0. W B y construction, IWI>IV[ implies row=0, while [Wi<[V! implies
IVIIWI,v . w i mplies DV(xW)=O, as follows from
(6.0.3.2). Thus (6.0.3.4) reduces to D v(Xw)=o, a nd D V( X v) my = 0
w hence m v = 0, a contradiction. Q.E.D. (6.0.4) Let T be any scheme, and S a scheme 6tale over A~ via a section
X o f (3s. Let (M, V) be a free ~smodule of rank n with integrable Tc onnection, such that there exists a basis eo ..... e,_l of M in terms of
w hich the connection takes the form [7 (d~)(el)=ei+ 1
(6.0.4.0) n V f or O _ < i _ n  2
l (e,,_O = ~aiei;
i=0 aier(S,(gs). ...
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This note was uploaded on 12/21/2011 for the course MAP 4341 taught by Professor Normankatz during the Fall '11 term at UNF.
 Fall '11
 NormanKatz

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