ScalingStacks

Proof. [044P]

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Proof.

We focus on γ1\gamma_{1}. The 2-form −1​d​η2∧(d​η¯2−d​η¯1)\sqrt{-1}d\eta_{2}\wedge(d\bar{\eta}_{2}-d\bar{\eta}_{1}) on SS is exponentially small along the 𝔇2,𝔇3\mathfrak{D}_{2},\mathfrak{D}_{3} ends, so the only divergence problem happens at infinity along the 𝔇1\mathfrak{D}_{1} end.

Applying Lemma 4.1 allows us to replace γ\gamma by the much simpler function

−14​π​𝔸​|(y1−y1′,y2−y2′,μ)|a′−1.-\frac{1}{4\pi\sqrt{\mathbb{A}}}|(y_{1}-y_{1}^{\prime},y_{2}-y_{2}^{\prime},\mu)|_{a}^{\prime-1}.

The integral

∫S∩{y2′<Λ}γ(η1−η1′,η2−η2′,μ)−1dη2′∧(dη¯2′−dη¯1′)\int_{S\cap\{y_{2}^{\prime}<\Lambda\}}\gamma(\eta_{1}-\eta_{1}^{\prime},\eta_{2}-\eta_{2}^{\prime},\mu)\sqrt{-1}d\eta_{2}^{\prime}\wedge(d\bar{\eta}_{2}^{\prime}-d\bar{\eta}_{1}^{\prime})

has the same divergence behaviour as

∫Λ−12​π​𝔸|(y1,y2−y2′,μ)|a′−1dy2′∼−12​π​A​a2​2¯logΛ,\int^{\Lambda}-\frac{1}{2\pi\sqrt{\mathbb{A}}}|(y_{1},y_{2}-y_{2}^{\prime},\mu)|_{a}^{\prime-1}dy_{2}^{\prime}\sim-\frac{1}{2\pi\sqrt{Aa_{2\bar{2}}}}\log\Lambda,

which is cancelled by the log term we put in the limit. ∎

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