By Eisenhart L.P.
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Additional info for Coordinate geometry
We indicate the details for R00 ; the cases R0j and R0∞ are similar. In all cases, one knows ahead of time that the powers of t work out correctly. 29) gives t2 R00 = tg IK ∂K EI0 − tg IK g P Q ΓIKP EQ0 − tg IK g P Q Γ0KP EIQ . It follows using EI0 ∈ t1−#(I) ρn/2+1 A and t2−#(IK) g IK ∈ A that when the first term on the right-hand side is expanded, all terms are in ρn/2+1 A except possibly those with K = ∞. Since E00 = 0, only the terms with K = ∞ and I = 0 need be considered. But t2−#(I) g I∞ ∈ ρA for I = 0, so the first term is in ρn/2+1 A.
The components Ri∞ and √ ((n+1)/2) are given by formal power series in ρ. 17) that the ρ3/2 term in gij ((n+1)/2) generates a ρ−1/2 term in R∞∞ . 13) holds for m ∈ 21 Z √ and for the g that we are considering with expansions in ρ, still under the same hypotheses: RIJ = O(ρm−1 ) for I, J = ∞ and RI∞ = O(ρm−2 ). 9, now using R0I = 0 and ((n+1)/2) ((n+1)/2) ((n+1)/2) Rij = O(ρn/2 ), we find Ri∞ , R∞∞ = O(ρn/2−1 ). We make 33 FORMAL THEORY ((n+1)/2) ((n+1)/2) ; namely, (ρ1−n/2 Ri∞ )|ρ=0 is one further observation about Ri∞ j a constant multiple of hij , .
11) does not vanish, so one can uniquely choose φ00 and φ0i at ρ = 0 to (m) (m) make R00 and R0i be O(ρm ). The map φij → (m − n/2)φij − 21 g kl φkl gij is bijective on symmetric 2-tensors unless m = n/2 or m = n, so except for (m) these m one can similarly make Rij = O(ρm ). Thus the induction proceeds up to m < n/2 for n even and up to m < n for n odd. Consider the next value of m in each case. For n even and m = n/2, one can uniquely determine φ00 and φ0i at ρ = 0 to make R00 , R0i = O(ρn/2 ) just as before.
Coordinate geometry by Eisenhart L.P.