Dr. Katz DEq Homework Solutions 54

Dr. Katz DEq Homework Solutions 54 - g; [gl = k, S(g) = a'-...

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54 N.M. Katz: Returning to the expansion of (3.5.4.3), we find Lie (P)R( l-I a(v)) = ~ (1~}) l-I Lie(P)e~(~r(v)) v*i ~'; [lI=k, iCs(O x " /v*i (3.5.4.9) r * a = {~ ,. .-, b}- {i1 e; lel = k, StO = a \t,/ "111 Lie(P)e~(tr(v)) H atv). vezl vC~d u{i} Substituting (3.5.4.9) into (3.5.4.2), we are faced with proving that dr(~ Obs (log D~)) contains ,,_1 UL, E (-- 1) i+1 ~ (- 1)kPP-l-k(f(i)) ~, sgn(A, i) ~,z~i e; Iel=k.s(e)=~\ f ] (3.5.4.10) i k=I 1--[ Lie(P)e~(a(v)) [I a(v). YeA vCdu{i} The key point now is to reindex the expression (3.5.4.10) by the sub- sets A'=A w {i} having at least two elements; then (3.5.4.10) becomes p-1 ~ (- 1) i+1 sgn(A'-i, i) ~ (- 1)kPP-l-~(f(i)) d' i~d' k~l (3.5.4.11) g; [gl = k, S(g) = a'-
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Unformatted text preview: g; [gl = k, S(g) = a'- {i} v~a'- {i} yea' We now calculate the sign(-1)l+lsgn(A'-i,i) as a function of the position of i in A' and of the position of A' in {1 . ... , b}. (_ 1)i+lsgn(A,_i,i)=sg n (i, {1 . ... 1, . .., b , b}-{i}]] sgn \(A'-i'{1 .... {1,, b}- {i}""b}-A') (i,{1 . ... ,b}-{i})sgn(i,A'-i,{1,. ..,b}-A'] =sgn 1,. ..,b i,{i . .... b}-i / (3.5.4.12) =sgn (i, zl'- {i}, {1 . ... , b} -A'] 1 . .... b ] /i,A'-{i},{1 . ... ,b}-d'\ (A', {1 .... ,b}-A'] =sgn ~ d',{1 . ... ,b}--d' )sgn 1 .... ,b / ' {1 . ... ,b}-A"~ Because each tr(v) is a closed form, in order to show that (3.5.4.11) lies in ~C2 dF(V, f2~/s (log D~)), it suffices to show that, for each subset zl { 1,. .., b}...
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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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