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Example 16.3.22E548BFB3BF04

Example 16.3.22.E548BFB3BF04 We continue our favorite example example 5.20.6. Let $X=\mathbb {P}^1$ and $\omega $ be the Fubini–Study metric. We define $\varphi \in \mathrm{PSH}(X,\omega )$ as follows: For $z\in \mathbb {C}$, we let

\[ \varphi (z)= \begin{cases} -\log \left(|z|^2+1\right)+\Bigl (-\log \left(-\log |z|^2\right)\Bigr ) \lor \Bigl (2+\log |z|^2\Bigr ),& \textup{if }|z|<1/\sqrt{2},\\ 2+\log \frac{|z|^2}{|z|^2+1},& \textup{otherwise}, \end{cases} \]

while $\varphi (\infty )=2$. Note that $\varphi $ is $\omega $-psh on $X=\mathbb {P}^1$ by the gluing lemma 5.14.16. The singularity of $\varphi $ only occurs at $z=0$, close to which, $\varphi $ is approximately $-\log \left(-\log |z|^2\right)$. This type of singularity is therefore called the log-log type singularity.

We claim that

B2DFFDD6A0F3 16.3.23
\begin{equation} P_{\omega }[\varphi ]=0. \end{equation}

In particular, we find that $\varphi $ and $P_{\omega }[\varphi ]$ have different singularity types.

By theorem 16.3.15, in order to verify (16.3.23), it suffices to verify that

C825DA5CEC8C 16.3.24
\begin{equation} \int _ X \omega _{\varphi }^1=1. \end{equation}

Here $\omega _{\varphi }^1$ is taken in the non-pluripolar sense. Since $\{ 0,\infty \} \subseteq \mathbb {P}^1$ is pluripolar, this reduces to showing that

\[ \int _{\mathbb {C}^*} \mathrm{dd}^{\mathrm{c}}\psi =\frac{1}{4\pi }\int _{\mathbb {C}^*} (\Delta \psi )\, \mathrm{d}\mu =1, \]

where $\psi (z)=\varphi (z)+\log (|z|^2+1)$ and $\mu $ is the standard Lebesgue measure on $\mathbb {C}$.

Note that the Laplacian vanishes outside $\overline{B(0,0.7)}$ since $\psi (z)=2+\log |z|^2$ there, which is harmonic. Therefore,

\[ \int _{\mathbb {C}^*} \mathrm{dd}^{\mathrm{c}}\psi =\frac{1}{4\pi }\int _{|z|<1/\sqrt{2}} (\Delta \psi )(z)\, \mathrm{d}\mu . \]

It is an elementary exercise to see that the right-hand side is exactly equal to $1$. If you are familiar with toric geometry, this is more or less trivial since

\[ \nabla _ r \Bigl ( \left(-\log (-r)\right)\lor (2+r) \Bigr )\bigl ((-\infty ,-\log 2)\bigr )= (0,1]. \]

Otherwise, just try to evaluate the integral using Green's identities. Therefore, (16.3.24) is proved and our assertion (16.3.23) follows.

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