ScalingStacks

Verified tagged author-source HTML · 1912.02360v1 · cited publication edition alignment unverified.

00RE

Lemma 4.3. Let Φ\Phi be any psh function on the open subset of {1<|ζi|<Λ,i=1,…n}⊂(ℂ∗)n\{1<|\zeta_{i}|<\Lambda,i=1,\ldots n\}\subset(\mathbb{C}^{*})^{n}. Then the TnT^{n}-invariant function

Φ¯​(log⁡|ζ1|,…,log⁡|ζn|)=1(2​π)n​∫TnΦ⁡(|ζ1|​ei​θ1,…​|ζn|​ei​θn)​d​θ1​…​d​θn\bar{\Phi}(\log|\zeta_{1}|,\ldots,\log|\zeta_{n}|)=\frac{1}{(2\pi)^{n}}\int_{T^{n}}\Phi(|\zeta_{1}|e^{i\theta_{1}},\ldots|\zeta_{n}|e^{i\theta_{n}})d\theta_{1}\ldots d\theta_{n}

is a convex function in the variables x1=log⁡|ζ1|,…,xn=log⁡|ζn|x_{1}=\log|\zeta_{1}|,\ldots,x_{n}=\log|\zeta_{n}|.

00RF

Proof. Since the TnT^{n}-action on (ℂ∗)n(\mathbb{C}^{*})^{n} is holomorphic, Φ⁡(ζ1​ei​θ1,…​ζn​ei​θn)\Phi(\zeta_{1}e^{i\theta_{1}},\ldots\zeta_{n}e^{i\theta_{n}}) is psh in ζ\zeta for any choice of θi\theta_{i}, so the average function Φ¯\bar{\Phi} is also psh. Any TnT^{n}-invariant psh function must be convex in the log coordinates, because of the formula

−1​∂∂¯​Φ¯=14​∑∂2Φ¯∂xi​∂xj​−1​d​log⁡ζi∧d​log⁡ζj¯≥0.\sqrt{-1}\partial\bar{\partial}\bar{\Phi}=\frac{1}{4}\sum\frac{\partial^{2}\bar{\Phi}}{\partial x_{i}\partial x_{j}}\sqrt{-1}d\log\zeta_{i}\wedge d\overline{\log\zeta_{j}}\geq 0.

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