ScalingStacks

Proof. [041M]

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Proof.

The strategy is to improve the decay rate of d​ϕ′d\phi^{\prime} iteratively, until it becomes sufficiently fast. We rewrite the equation as a Poisson equation

12​(Δgℂ3​ϕ′)​ωℂ33=−3​(−1​∂∂¯​ϕ′)2∧ωℂ3−(−1​∂∂¯​ϕ′)3.\frac{1}{2}(\Delta_{g_{\mathbb{C}^{3}}}\phi^{\prime})\omega_{\mathbb{C}^{3}}^{3}=-3(\sqrt{-1}\partial\bar{\partial}\phi^{\prime})^{2}\wedge\omega_{\mathbb{C}^{3}}-(\sqrt{-1}\partial\bar{\partial}\phi^{\prime})^{3}.

Notice that −1​∂∂¯​ϕ′\sqrt{-1}\partial\bar{\partial}\phi^{\prime} lives in Cδ,τk,αC^{k,\alpha}_{\delta,\tau}, so its square lives in C2​δ,2​τk,αC^{k,\alpha}_{2\delta,2\tau}. As long as (δ,τ)(\delta,\tau) stays in the good range of weight exponents, Corollary 2.24 and the above vanishing lemma imply that the solution ϕ′\phi^{\prime} to this Poisson equation must satisfy ‖d​ϕ′‖C2​δ+1,2​τk+1,α​(ℂ3,Λ1)<∞\left\lVert d\phi^{\prime}\right\rVert_{C^{k+1,\alpha}_{2\delta+1,2\tau}(\mathbb{C}^{3},\Lambda^{1})}<\infty. This is an improved decay estimate because 2​δ+1<δ2\delta+1<\delta and 2​τ<τ2\tau<\tau. Since each iteration improves the decay rate by a definite amount, within a finite number of steps we can assume δ<−2\delta<-2 and τ<0\tau<0. Then ‖d​ϕ′‖Cδ+1,τk+1,α​(ℂ3,Λ1)<∞\left\lVert d\phi^{\prime}\right\rVert_{C^{k+1,\alpha}_{\delta+1,\tau}(\mathbb{C}^{3},\Lambda^{1})}<\infty implies that ‖ϕ′‖Cδ+2,τk+2,α​(ℂ3)<∞\left\lVert\phi^{\prime}\right\rVert_{C^{k+2,\alpha}_{\delta+2,\tau}(\mathbb{C}^{3})}<\infty after adjusting ϕ′\phi^{\prime} by a constant. Then we can use the standard integration by part argument for the complex Monge-Ampère equation to see

∫ℂ3|∇ϕ′|2​ϕ′p​ωℂ33=0,p≫1.\int_{\mathbb{C}^{3}}|\nabla\phi^{\prime}|^{2}\phi^{\prime p}\omega_{\mathbb{C}^{3}}^{3}=0,\quad p\gg 1.

Hence ϕ′\phi^{\prime} is a constant, and the metric is unique. ∎

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