Dr. Katz DEq Homework Solutions 47

Dr. Katz DEq Homework Solutions 47 - ~ Lie(D~o)k t(D(io)-...

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Algebraic Solutions of Differential Equations 47 Recall that (3.4.1.14) (V(D) p- V(DP))(Fr176176 '') (because V(D)(F~,(C~176176 ), and V(D) on gffvr ~176 is an inte- grable connection of p-curvature zero, given on Vio c~.-. n Vi, by D--+ Lie (D(i0))). F~o, (C), It follows from (3.4.1.13) Since a is congruent to z modulo ,+1 ,, that (3.4.1.15) (V(D) p- V(Dp))(a)---(V(D) p- V(D'))(z) modV~o +2. Thus the section O(D)(a) is represented by the component of bidegree (a+ 1, b- 1) in (V(DF- V(DP))(r). (3.4.2) Lemma. Let zeC"({Vi}, Y~x/s(logD)), and io<ia<. ..<i,+ 1. For each integer n > 1, the element (3.4.2.0) V(D)"(z)e ~ C"+'({V~}, ~xTs'(log D)) i>O has its components of bidegree (a, b) and (a+ 1, b- 1) given by (3.4.2.1) V(D)"(z)(io . ... , i,)=Lie(D(io))"(Z(io, . .., i~)), V(D)'(z)(io,. .. , i,) (3.4.2.2) ---(-1) b
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Unformatted text preview: ~ Lie(D~o)k t(D(io)- D(iO) Lie(D~f (z(i I . .... i,)). k+d~n--1 Proof The proof, by induction, is immediate; the case n = 1 is the definition. Q.E.D. Putting n = I', we find (V (D) p- V (DP))(z)(io, . .. , i~) (3.4.2.3) = [(Lie (D (i0))) P- Lie (D (io)P)] (z (i o . .... i,)) - 0, since (D(io)) p =- DP(io) (V(D) p- V(DP))(z)(io . .... ia+O (3.4.2.4) =(-1) b ~ Lie(D~o)kI(D(io)-D(iO)Lie(D,y(z(q .... ,i,+~)) k+d=p+l - ( -- 1 )b I (D (io) p - D (iO p) (z (il . .... i. + 1)). Thus (3.4.2.4) gives a formula for a cocycle representing ~h(D)(a) in R&quot;+lf. (3r D))), when we take for z a particular representing cochain for ~. (3.4.3) Combining all our explications (3.4.1.9), (3.4.1.10), and (3.4.2.4), we see that the commutativity of (3.4.1.6), and hence the truth of 3.0, is implied by the following assertion:...
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