PomeLemma 12.8.1.47843166E777 Suppose $T_ j\to T$ weakly in $\mathscr G_\omega $ and $\mathfrak D_ F(T_ j)\to M(F)$. Then $T\in \mathscr M(F)$ and $\Gamma _ F(T_ j)\rightharpoonup \Gamma _ F(T)$. On $\mathscr M(F)$ the map $T\mapsto \Gamma _ F(T)$ is affine and continuous.
12.8 Supporting probabilities represented by maximizing currents7C2BB0FDDDF5
7C2BB0FDDDF5
By theorem 12.7.3, $B_0=-n\log M$, and $M(F)$ is the maximum of the functionals $T\mapsto \mathfrak D_ F(T)=\int _ X\mathrm{e}^{F/n}\, \mathrm{d}\Gamma _0(T)$, each of which is linear in $\mathrm{e}^{F/n}$. The supporting probabilities of $B_0$ should therefore be the normalized determinant measures of maximizing currents, and we show that they are exactly these. The point that needs proof is the continuity of the determinant measures: on the set of maximizers, upper semicontinuity of the total determinant integral becomes equality. Recall from section 11.34 that the set $\mathcal A_0(F)$ of supporting probabilities of $B_0$ at $F\in C(X)$ consists of the Borel probability measures $\nu $ with $B_0(G)\leq B_0(F)-\int _ X(G-F)\, \mathrm{d}\nu $ for all $G\in C(X)$; by theorem 11.34.2 it is nonempty, convex and weakly compact.
For $T\in \mathscr M(F)$ put
\[ \nu _{T,F}=\Gamma _ F(T)/M(F). \]
This is a probability measure.
Proof. Upper semicontinuity gives $\mathfrak D_ F(T)\geq M(F)$, hence equality. The measures $\Gamma _ F(T_ j)$ have bounded mass; let $\Gamma '$ be a weak cluster point, whose mass is $M(F)$. For a continuous real function $h$, choose $\varepsilon >0$ with $1\pm \varepsilon h>0$. Since $\Gamma _{F+n\log (1\pm \varepsilon h)}=(1\pm \varepsilon h)\Gamma _ F$, upper semicontinuity of $\mathfrak D$ for these two continuous logarithmic densities gives
\[ \int _ X(1\pm \varepsilon h)\, \mathrm{d}\Gamma ' \leq \int _ X(1\pm \varepsilon h)\, \mathrm{d}\Gamma _ F(T). \]
Since both measures have mass $M(F)$, the two signs give equality of the integrals of $h$. Thus every cluster point equals $\Gamma _ F(T)$, which proves the convergence.
For affineness, take $T_1,T_2\in \mathscr G_\omega $ and $t\in [0,1]$, and express the coefficient measures of both currents through densities with respect to the sum of their trace measures. Concavity and homogeneity of the $n$th root of the determinant, applied to these matrix densities, give
\[ \Gamma _ F(tT_1+(1-t)T_2) \geq t\Gamma _ F(T_1)+(1-t)\Gamma _ F(T_2),\qquad 0\leq t\leq 1. \]
If $T_1$ and $T_2$ maximize, so does the current on the left, by concavity of $\mathfrak D_ F$. Both measures in the inequality then have mass $M(F)$, so equality holds. This proves affineness, and continuity is the special case of the first assertion in which every $T_ j$ is a maximizer. $\square$
Theorem 12.8.2.E9E119AAA269 For every $F\in C(X,\mathbb R)$,
\[ \mathcal A_0(F)=\{ \nu _{T,F}:T\in \mathscr M(F)\} . \]
Consequently the determinant measures $\Gamma _ F(T)$ of all maximizing currents are absolutely continuous with respect to $\mathrm{d}V$. (This concerns the measures $\Gamma _ F(T)$, not the currents $T$ themselves.)
Proof. For a maximizing $T$ and any $G\in C(X)$,
\[ M(G)\geq \mathfrak D_ G(T) =M(F)\int _ X\mathrm{e}^{(G-F)/n}\, \mathrm{d}\nu _{T,F}. \]
Since $B_0=-n\log M$, taking $-n$ times the logarithm and applying Jensen's inequality gives
\[ B_0(G)\leq B_0(F)-n\log \int _ X\mathrm{e}^{(G-F)/n}\, \mathrm{d}\nu _{T,F} \leq B_0(F)-\int _ X(G-F)\, \mathrm{d}\nu _{T,F}. \]
Hence every probability $\nu _{T,F}$ with $T\in \mathscr M(F)$ belongs to $\mathcal A_0(F)$.
Denote the set of these probabilities by $\mathscr N(F)$; by the preceding lemma it is compact and convex. To prove the reverse inclusion we compute the right derivative $(B_0)'_+(F;h)=\lim _{t\downarrow 0}(B_0(F+th)-B_0(F))/t$ for $h\in C(X)$. Since $\mathfrak D_{F+th}(T)=\int _ X\mathrm{e}^{th/n}\, \mathrm{d}\Gamma _ F(T)$ and the masses of the measures $\Gamma _ F(T)$ are bounded on $\mathscr G_\omega $, a Taylor expansion of the exponential gives, as $t\downarrow 0$,
\[ \mathfrak D_{F+th}(T) =\mathfrak D_ F(T)+\frac tn\int _ Xh\, \mathrm{d}\Gamma _ F(T)+O(t^2), \]
with the error bounded uniformly in $T$. Testing maximizers at $F$ gives
\[ \liminf _{t\downarrow 0}\frac{M(F+th)-M(F)}t \geq \frac{M(F)}n\max _{\nu \in \mathscr N(F)}\int _ Xh\, \mathrm{d}\nu . \]
For the reverse inequality, choose a maximizer $T_ t$ at $F+th$. The multiplicative bounds in the previous proof imply $\mathfrak D_ F(T_ t)\to M(F)$. Every sequence $t\downarrow 0$ has a subsequence along which $T_ t$ converges, and the preceding lemma identifies the limit of $\Gamma _ F(T_ t)$ with the determinant measure of a maximizer at $F$. Since $\mathfrak D_ F(T_ t)\leq M(F)$, the Taylor expansion gives the matching upper bound for the difference quotients of $M$. Since $B_0=-n\log M$, therefore
\[ (B_0)'_+(F;h) =-\max _{\nu \in \mathscr N(F)}\int _ Xh\, \mathrm{d}\nu . \]
If $\nu \in \mathcal A_0(F)$, its supporting inequality in direction $th$ implies
\[ \int _ Xh\, \mathrm{d}\nu \leq \max _{\widetilde\nu \in \mathscr N(F)} \int _ Xh\, \mathrm{d}\widetilde\nu \quad \text{for every }h\in C(X). \]
If $\nu $ were not in the weakly compact convex set $\mathscr N(F)$, the Hahn–Banach separation theorem in the dual of $C(X)$ would give some $h\in C(X)$ violating this inequality. Hence $\nu \in \mathscr N(F)$, and the two sets are equal. The absolute continuity of their members was proved in theorem 11.34.2. $\square$
For continuous $F$, the supporting probability is therefore unique exactly when all maximizing currents have the same normalized determinant measure, which is weaker than uniqueness of the maximizing current. Whether a maximizing current can be written as a conformal multiple of a power $(\omega +\mathrm{dd}^{\mathrm{c}}u)^{n-1}$ of a weak solution form is a further question, discussed in section 25.17; the construction above works directly with the coefficient measures of the currents.
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.