ScalingStacks

Remark 8.2 . [03JP]

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Remark 8.2.

If (X,g)(X,g) is a δ\delta-aymptotically Calabi space for n=2n=2, one can define a weight function

(8.4) ρδ,ν(k+α)​(𝒙)=eδ​z​(𝒙)⋅(z⁡(𝒙))ν+k+α2,\displaystyle\rho^{(k+\alpha)}_{\delta,\nu}(\bm{x})=e^{\delta z(\bm{x})}\cdot(z(\bm{x}))^{\frac{\nu+k+\alpha}{2}},

and defined the weighted space Cδ,νk,α​(X)C^{k,\alpha}_{\delta,\nu}(X) exactly as in Definition 8.1. On such a space the Laplace operator is a bounded linear mapping

(8.5) Δ:Cδ,νk,α​(X)→Cδ,ν+2k−2,α​(X).\displaystyle\Delta:C^{k,\alpha}_{\delta,\nu}(X)\rightarrow C^{k-2,\alpha}_{\delta,\nu+2}(X).

It is expected that this operator is Fredholm if δ\delta is sufficiently small and non-zero, and for arbitrary ν\nu. The analysis in Section 4 can likely be extended to prove this stronger result, but for the purposes of this paper we do not need this. Likewise, we expect that

(8.6) 𝒟:Cδ,νk,α​(Ω1)→Cδ,ν+1k−1,α​(Ω0⊕Ω+2)\displaystyle\mathscr{D}:C^{k,\alpha}_{\delta,\nu}(\Omega^{1})\rightarrow C^{k-1,\alpha}_{\delta,\nu+1}(\Omega^{0}\oplus\Omega^{2}_{+})

is Fredholm for δ\delta sufficiently small, but this would require a much more elaborate separation of variables argument.

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