ScalingStacks

Example 2.9 . [0336]

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Example 2.9.

The capacity C​a​pω​(⋅)Cap_{\omega}(\cdot) does not distinguish between ”big sets”. Assume indeed there exists an ample divisor DD such that [D]∼k​ω[D]\sim k\omega, k∈ℕk\in\mathbb{N}. Then there exists φ∈P​S​H​(X,ω)\varphi\in PSH(X,\omega) such that d​dc​φ=k−1​[D]−ωdd^{c}\varphi=k^{-1}[D]-\omega. Note that φ∈𝒞∞​(X∖D)\varphi\in{\mathcal{C}}^{\infty}(X\setminus D), eφ∈𝒞0​(X)e^{\varphi}\in{\mathcal{C}}^{0}(X) and {φ=−∞}=D\{\varphi=-\infty\}=D. Replacing φ\varphi by φ−supXφ\varphi-\sup_{X}\varphi if necessary, we may assume supXφ=0\sup_{X}\varphi=0. Consider φc=max⁡(φ,−c)∈P​S​H​(X,ω)∩𝒞0​(X)\varphi_{c}=\max(\varphi,-c)\in PSH(X,\omega)\cap{\mathcal{C}}^{0}(X). Then φc≡φ\varphi_{c}\equiv\varphi outside some neighborhood Vc={φ<−c}V_{c}=\{\varphi<-c\} of DD. Since 0≤1+φ1≤10\leq 1+\varphi_{1}\leq 1 and ω1+φ1=ωφ=0\omega_{1+\varphi_{1}}=\omega_{\varphi}=0 in X∖V1X\setminus V_{1}, we get

C​a​pω​(X)=∫X(ω1+φ1)n=∫V1(ω1+φ1)n≤C​a​pω​(V1),Cap_{\omega}(X)=\int_{X}(\omega_{1+\varphi_{1}})^{n}=\int_{V_{1}}(\omega_{1+\varphi_{1}})^{n}\leq Cap_{\omega}(V_{1}),

hence C​a​pω​(V1)=C​a​pω​(X)Cap_{\omega}(V_{1})=Cap_{\omega}(X).

As a concrete example take X=ℂ​ℙnX=\mathbb{C}\mathbb{P}^{n} and ω=ωF​S\omega=\omega_{FS}, DD being some hyperplane H∞H_{\infty} ”at infinity” (k=1k=1). Set φ[z:t]=log|t|−12log[||z||2+|t|2]\varphi[z:t]=\log|t|-\frac{1}{2}\log[||z||^{2}+|t|^{2}] where zz denotes the euclidean coordinates in ℂn=ℂ​ℙn∖H∞\mathbb{C}^{n}=\mathbb{C}\mathbb{P}^{n}\setminus H_{\infty} and H∞=(t=0)H_{\infty}=(t=0). Observe that supℂ​ℙnφ=0\sup_{\mathbb{C}\mathbb{P}^{n}}\varphi=0. One then computes

ℂℙn∖V1={z∈ℂn/|z|≤e2−1}.\mathbb{C}\mathbb{P}^{n}\setminus V_{1}=\left\{z\in\mathbb{C}^{n}\,/\,|z|\leq\sqrt{e^{2}-1}\right\}.

Thus the capacity of the complement of any euclidean ball of radius smaller than e2−1\sqrt{e^{2}-1} equals 11.

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