ScalingStacks

Proof. [02I9]

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Proof.

Given data as in the statement we constructed a 44–manifold MϵM_{\epsilon} and a 11–parameter family of closed definite triples 𝝎¯ϵ\bm{\underline{\omega}}_{\epsilon} which are approximately hyperkähler. For ϵ\epsilon sufficiently small we can apply Lemma 6.13 to find unique 𝒂¯ϵ∈Cδ1,α​(T∗​Mϵ)\bm{\underline{a}}_{\epsilon}\in C^{1,\alpha}_{\delta}(T^{\ast}M_{\epsilon}) for δ∈(−12,0)\delta\in(-\tfrac{1}{2},0) and 𝜻¯ϵ∈ℋϵ+\bm{\underline{\zeta}}_{\epsilon}\in\mathcal{H}^{+}_{\epsilon} such that ‖𝒂¯ϵ‖Cδ1,α+‖𝜻¯ϵ‖≤C​ϵ11−2​δ5\|\bm{\underline{a}}_{\epsilon}\|_{C^{1,\alpha}_{\delta}}+\|\bm{\underline{\zeta}}_{\epsilon}\|\leq C\epsilon^{\frac{11-2\delta}{5}} and 𝝎¯ϵ+d​𝒂¯ϵ+𝜻¯ϵ\bm{\underline{\omega}}_{\epsilon}+d\bm{\underline{a}}_{\epsilon}+\bm{\underline{\zeta}}_{\epsilon} is a hyperkähler structure on MϵM_{\epsilon}. In particular, since b1​(Mϵ)=0b_{1}(M_{\epsilon})=0 by Proposition 5.1, MϵM_{\epsilon} must be diffeomorphic to the K3 surface.

Away from the gluing regions 𝒂¯ϵ\bm{\underline{a}}_{\epsilon} solves the elliptic PDE d+​𝒂¯ϵ=ℱ⁡(d−​𝒂¯ϵ∗d−​𝒂¯ϵ)−𝜻¯ϵd^{+}\bm{\underline{a}}_{\epsilon}=\mathcal{F}(d^{-}\bm{\underline{a}}_{\epsilon}\ast d^{-}\bm{\underline{a}}_{\epsilon})-\bm{\underline{\zeta}}_{\epsilon}, d∗​𝒂¯ϵ=0d^{\ast}\bm{\underline{a}}_{\epsilon}=0. By elliptic regularity, for any k≥2k\geq 2 the Ck,αC^{k,\alpha}–norm of 𝒂¯ϵ\bm{\underline{a}}_{\epsilon} on compact sets of MϵghM^{\textup{gh}}_{\epsilon} and (after rescaling) on compact sets of the gravitational instantons MjM_{j} and NiN_{i} is controlled in terms of ‖𝒂¯ϵ‖Cδ1,α+‖𝜻¯ϵ‖\|\bm{\underline{a}}_{\epsilon}\|_{C^{1,\alpha}_{\delta}}+\|\bm{\underline{\zeta}}_{\epsilon}\|. In particular, on compact sets of MϵghM^{\textup{gh}}_{\epsilon} the hyperkähler metric induced by 𝝎¯ϵ+d​𝒂¯ϵ+𝜻¯ϵ\bm{\underline{\omega}}_{\epsilon}+d\bm{\underline{a}}_{\epsilon}+\bm{\underline{\zeta}}_{\epsilon} is Ck,αC^{k,\alpha}–close to g𝕋+ϵ2​θ2g_{\mathbb{T}}+\epsilon^{2}\theta^{2}. The statements (i), (ii) and (iii) about the limit ϵ→0\epsilon\rightarrow 0 now follow. ∎

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