By A. Church

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However, the fifth line determines φ0i |ρ=0 by the (n/2+1) requirement that Ri∞ = O(ρ(n−1)/2 ). The third line then determines φij |ρ=0 by the requirement that Rij = O(ρ(n+1)/2 ). 13), we already know that the first two lines hold, and the third (n/2+1) = O(ρ(n−1)/2 ). Thus the φIJ |ρ=0 have been line tells us that R∞∞ (n/2+1) uniquely determined and we have RIJ = O(ρ(n+1)/2 ) for I, J = ∞ (n/2+1) and RI∞ = O(ρ(n−1)/2 ). 11). By the observation ((n+1)/2) noted above that (ρ1−n/2 Ri∞ )|ρ=0 is a constant multiple of hij , j , we deduce that φ0i |ρ=0 is a constant multiple of hij , j , and in particular φ0i |ρ=0 vanishes if and only if hij , j = 0.

We obtain a = 2ρ and so u = −2ρ. By the definition of χ, we see that u = r is the coordinate in the second factor of M × [0, ∞). 3) on M × [0, ∞). Clearly g+ is an even asymptotically hyperbolic metric with conformal infinity (M, [g]) in normal form relative to g. √ In the general case, we have g+ = u−2 (h + du2 ) with u = −a and h = h(x, ρ, dx, dρ). Since r is given by r = 2|ρ| and a vanishes exactly to first order at ρ = 0, we can write u = rb(x, r2 ) for a positive smooth function b.

The value of g ij ∂ρn gij is determined by the third equation, and all higher derivatives are then determined by the first equation. If the initial metric is Einstein, one can identify explicitly the solution gij (x, ρ). 17). In general it is feasible to carry out the first few iterations by hand. 18) where Wijkl is the Weyl tensor, Cijk = Pij ,k −Pik ,j is the Cotton tensor, and Bij = Cijk , k − P kl Wkijl is the Bach tensor. The traces are given by ′′ = 2Pij P ij g ij gij ′′′ (n − 4)g ij gij = −8Pij B ij , n = 4.