ScalingStacks

7. Energy [01DB]

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7. Energy

In the complex case, the (Aubin-Mabuchi) energy functional is defined as follows. Fix a smooth semipositive reference metric ϕ0\phi_{0} and set

E⁡(ϕ):=1n+1​∑j=0n∫Xan(ϕ−ϕ0)​(d​dc​ϕ)j∧(d​dc​ϕ0)n−j.E(\phi):=\frac{1}{n+1}\sum_{j=0}^{n}\int_{{X^{\mathrm{an}}}}(\phi-\phi_{0})(dd^{c}\phi)^{j}\wedge(dd^{c}\phi_{0})^{n-j}. (7.1)

for any smooth metric ϕ\phi. Here (d​dc​ϕ)j∧(d​dc​ϕ0)n−j(dd^{c}\phi)^{j}\wedge(dd^{c}\phi_{0})^{n-j} is a mixed Monge-Ampère measure. It is a positive measure if ϕ\phi is semipositive.

In the non-Archimedean case, mixed Monge-Ampère measures can be defined using intersection theory when ϕ\phi and ϕ0\phi_{0} are model metrics, and the energy of ϕ\phi is then defined exactly as above.

For two smooth/model metrics ϕ\phi, ψ\psi we have

E⁡(ϕ)−E⁡(ψ)=1n+1​∑j=0n∫Xan(ϕ−ψ)​(d​dc​ϕ)j∧(d​dc​ψ)n−j.E(\phi)-E(\psi)=\frac{1}{n+1}\sum_{j=0}^{n}\int_{{X^{\mathrm{an}}}}(\phi-\psi)(dd^{c}\phi)^{j}\wedge(dd^{c}\psi)^{n-j}. (7.2)

This is proved using integration by parts in the complex case and follows from basic intersection theory in the non-Archimedean case.

We can draw two main conclusions from (7.2). First, the derivative of the energy functional is the Monge-Ampère operator, in the sense that

dd​t​E​(ϕ+t​f)|t=0=∫Xanf​MA⁡(ϕ)\frac{d}{dt}E(\phi+tf)\bigg|_{t=0}=\int_{{X^{\mathrm{an}}}}f\operatorname{MA}(\phi) (7.3)

for a smooth/model metric ϕ\phi on Lan{L^{\mathrm{an}}} and a smooth/model function ff on Xan{X^{\mathrm{an}}}.

Second, E⁡(ψ)≥E⁡(ϕ)E(\psi)\geq E(\phi) when ψ≥ϕ\psi\geq\phi are semipositive. It then makes sense to set

E(ϕ):=inf{E(ψ)∣ψ≥ϕ,ψ a semipositive smooth/model metric on Lan}.E(\phi):=\inf\{E(\psi)\mid\psi\geq\phi,\ \text{$\psi$ a semipositive smooth/model metric on ${L^{\mathrm{an}}}$}\}.

for any singular semipositive metric ϕ∈PSH⁡(Lan)\phi\in\operatorname{PSH}({L^{\mathrm{an}}}). The resulting functional

E:PSH(Lan)→[−∞,∞)E:\operatorname{PSH}({L^{\mathrm{an}}})\to[-\infty,\infty)

has many good properties: EE is concave, monotonous, and satisfies E⁡(ϕ+c)=E⁡(ϕ)+cE(\phi+c)=E(\phi)+c for c∈𝐑c\in{\mathbf{R}}. Further, EE is usc and continuous along decreasing nets.

The energy functional singles out a class ℰ1​(Lan){\mathcal{E}}^{1}({L^{\mathrm{an}}}) of metrics with finite energy, E⁡(ϕ)>−∞E(\phi)>-\infty. This class has good properties. In particular, one can (with some effort) define mixed Monge-Ampère measures (d​dc​ϕ)j∧(d​dc​ψ)n−j(dd^{c}\phi)^{j}\wedge(dd^{c}\psi)^{n-j} for ϕ,ψ∈ℰ1​(Lan)\phi,\psi\in{\mathcal{E}}^{1}({L^{\mathrm{an}}}), and (7.1) continues to hold.

Let us now go back to the variational approach to solving the Monge-Ampère equation. Fix a positive measure μ\mu on Xan{X^{\mathrm{an}}} of mass (Ln)(L^{n}). In the complex case we assume μ\mu is absolutely continuous with respect to Lebesgue measure, with density in LpL^{p} for some p>1p>1. In the non-Archimedean case we assume that μ\mu is supported on some dual complex. In both cases, one can show that the functional ϕ→∫(ϕ−ϕ0)​μ\phi\to\int(\phi-\phi_{0})\,\mu is (finite and) continuous on PSH⁡(Lan)\operatorname{PSH}({L^{\mathrm{an}}}), where ϕ0\phi_{0} is the same reference metric as in (7.1). Thus the functional Fμ:PSH(Lan)→[−∞,∞)F_{\mu}\colon\operatorname{PSH}({L^{\mathrm{an}}})\to[-\infty,\infty) defined by

Fμ​(ϕ):=E⁡(ϕ)−∫(ϕ−ϕ0)​μF_{\mu}(\phi):=E(\phi)-\int(\phi-\phi_{0})\mu

is upper semicontinuous. It follows from (7.2) that FμF_{\mu} does not depend on the choice of reference metric ϕ0\phi_{0}. We also have Fμ​(ϕ+c)=Fμ​(ϕ)F_{\mu}(\phi+c)=F_{\mu}(\phi) for ϕ∈PSH⁡(Lan)\phi\in\operatorname{PSH}({L^{\mathrm{an}}}), c∈𝐑c\in{\mathbf{R}}. Thus FμF_{\mu} descends to an usc functional on the quotient space PSH⁡(Lan)/𝐑\operatorname{PSH}({L^{\mathrm{an}}})/{\mathbf{R}}. By Theorem 6.1, the latter space is compact, so we can find ϕ∈PSH⁡(Lan)\phi\in\operatorname{PSH}({L^{\mathrm{an}}}) maximizing FμF_{\mu}. It is clear that ϕ∈ℰ1​(Lan)\phi\in{\mathcal{E}}^{1}({L^{\mathrm{an}}}), so the mixed Monge-Ampère measures of ϕ\phi and ϕ0\phi_{0} are well defined. However, equation (7.3) no longer makes sense, since there is no reason for the metric ϕ+t​f\phi+tf to be semipositive for t≠0t\neq 0. Therefore, it is not clear that MA⁡(ϕ)=μ\operatorname{MA}(\phi)=\mu, as desired. In the next section, we explain how to get around this problem.

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