Dr. Katz DEq Homework Solutions 51

# Dr. Katz DEq Homework Solutions 51 - make use of it through...

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Algebraic Solutions of Differential Equations 51 and substituting via (3.5.1) and especially via (3.5.1.3), (3.5.2.1) becomes (3.5.2.2) Lie(P)P-lI(P-Q)(a(1))-h(1)= -g(1) p. A final substitution via (3.5.1) gives the assertion (3.5.2.3) PP-1 (f(1))- h(1)= -g(1) p. We now "decode" (3.5.2.3) by returning to the definitions (3.5.0.7-9) of f, h, g; (3.5.2.3) becomes the assertion (3.5.2.4) if 1~, PP-I(P(xOI\ xl ! PP(XO_x, (~0) p, (3.5.2.5) if 1r PP-1(x~-lp(xl))-x(-IPV(xO=-(P(Xl)) p. Both (3.5.2.4) and (3.5.2.5) are in fact true, and follow ((3.5.2.5) directly, (3.5.2.4) after dividing by x 0 from Hochschild's identity (cf. [20]), according to which, if P is any derivation of any commutative ring A of characteristic p>0, then for any element xeA, (3.5.2.6) Pp-l(xo-~ p(x))-xp-l Pp(x)= -(P(x)) p. This concludes the proof of (3.5.0.3) in case b = 1. In the following, we will make use of it, through the identity (3.5.2.3).
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Unformatted text preview: make use of it, through the identity (3.5.2.3). (3.5.3) Because our proof of (3.5.0.3) in the general case is so unen- lightening we first present the proof in the case b = 2, which is somewhat more intelligible. The assertion to be verified is Lie (p)k I (P -- Q) Lie (Q)e (~ (r (l) ^ z (2))) k+g'=p- 1 (3.5.3.0) - I (P p- Qp)(o~ (z (1) ^ z (2))) - o~ (I (P - Q) (, (1) ^ z (2))) modulo dF(V, (gv). Substituting via (3,5.1.1) and noting that by (3.5.1.3) the terms under the summation sign vanish for (~0, (3.5.3.0) becomes Lie (P)P-x (I (P - Q))(~r (1) ^ a (2)) - I(Pp - QO)(a (1) ^ a (2)) (3.5.3.1) - ~----~(I(P- Q)(,(I) ^ ,(2))) modulo dF(V, (9~,). Expanding the interior products and substituting via (3.5.1), (3.5.3.1) becomes Lie (P)P-1 (f(1) a (2) - f(2) a (1)) - (h (1) a (2) - h (2) a (1)) (3.5.3.2) -- ~(g(1)z(2)-g(2)z(1))modulodr(V, Cv). 4*...
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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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