ScalingStacks

Proof. [045L]

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Proof.

(Sketch) The method is the same as in Proposition 4.16, so we only mention the key points. The long distance contribution to the integral has improved decay, so there is no need for the log factor. The short distance contribution appeals to Lemma 4.15. In the first derivative estimates, notice the mean curvature vector vanishes because SS an algebraic curve.

For second derivative estimates, notice for R≲A−1/2R\lesssim A^{-1/2}, the magnitudes of tensors are inhomogeneous:

|dμ|gNUT≤A−1/4R1/2,|dξ1|gNUT≤A−1/4R1/2,|dξ2|gNUT≤CA−1/2.|d\mu|_{g_{\text{NUT}}}\leq A^{-1/4}R^{1/2},\quad|d\xi_{1}|_{g_{\text{NUT}}}\leq A^{-1/4}R^{1/2},\quad|d\xi_{2}|_{g_{\text{NUT}}}\leq CA^{-1/2}.

These RR factors make the gNUTg_{\text{NUT}}-magnitudes bounded even though the gag_{a}-magnitudes can be unbounded. Inside the tensor Ψ∗​(wp​q¯​d​ηp⊗d​η¯q)\Psi^{*}(w^{p\bar{q}}d\eta_{p}\otimes d\bar{\eta}_{q}), the coeffient of d​ξ1⊗d​ξ¯2d\xi_{1}\otimes d\bar{\xi}_{2} and d​ξ2⊗d​ξ¯1d\xi_{2}\otimes d\bar{\xi}_{1} correspond to imposing f⁡(0)=0f(0)=0, and the coeffient of d​ξ2⊗d​ξ¯2d\xi_{2}\otimes d\bar{\xi}_{2} correspond to imposing f⁡(0)=0f(0)=0 and d​f​(0)=0df(0)=0. ∎

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