Dr. Katz DEq Homework Solutions 82

Dr. Katz DEq Homework Solutions 82 - 82 N.M. Katz: t hat...

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Unformatted text preview: 82 N.M. Katz: t hat whenever h orizontal s ections m w , o ne for each exponent-system 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 w-v 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 IVI--IWI,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 ~s-module 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.

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