Pome

Theorem 17.2.53E698B84D63D

Theorem 17.2.5.3E698B84D63D Let $a,b$ be bounded quasi-psh differences on $X$, and let $T_1,\ldots ,T_{n-1}$ be closed positive $(1,1)$-currents on $X$. Then

93ACF4FD4967 17.2.6
\begin{equation} \int _ Xa\, \mathrm{dd}^{\mathrm{c}}b\wedge \langle T_1\wedge \cdots \wedge T_{n-1}\rangle =\int _ Xb\, \mathrm{dd}^{\mathrm{c}}a\wedge \langle T_1\wedge \cdots \wedge T_{n-1}\rangle , \end{equation}

where the Hessian factors are the signed measures defined in (17.2.1).

Proof. Step 1: a repeated factor and a common background. Let $(\varphi _1,\varphi _2)$ and $(\psi _1,\psi _2)$ be presentations of $a$ and $b$, and suppose that these four functions and a further potential $\eta $ all belong to $\mathrm{PSH}(X,\theta )$ for one smooth closed real $(1,1)$-form $\theta $. For $0\leq t\leq 1$ the functions $\psi _ j^ t=(1-t)\eta +t\psi _ j$ are $\theta $-psh, and $\psi _1^ t-\psi _2^ t=tb$ is bounded. Lemma 17.2.3, applied to $\varphi _1,\varphi _2,\psi _1^ t,\psi _2^ t$, gives

\[ \int _ Xa(\mu _{\psi _1^ t}-\mu _{\psi _2^ t})=t\int _ Xb(S_1^ t-S_2^ t), \qquad S_ j^ t=\sum _{k=0}^{n-1}\left\langle \theta _{\varphi _ j}\wedge \theta _{\psi _1^ t}^ k\wedge \theta _{\psi _2^ t}^{n-1-k}\right\rangle . \]

Since $\theta _{\psi _ j^ t}=(1-t)\theta _\eta +t\theta _{\psi _ j}$, multilinearity expands every measure here as a polynomial in $t$ with finite measures as coefficients, so both sides are polynomials in $t$. We compare their coefficients of $t$. On the left this coefficient is

\[ n\int _ Xa\left(\langle \theta _{\psi _1}\wedge \theta _\eta ^{n-1}\rangle -\langle \theta _{\psi _2}\wedge \theta _\eta ^{n-1}\rangle \right) =n\int _ Xa\, \mathrm{dd}^{\mathrm{c}}b\wedge \langle \theta _\eta ^{n-1}\rangle , \]

where we used that (17.2.1) may be computed in any smooth background with respect to which the presentation is psh. On the right it is $\int _ Xb(S_1^0-S_2^0)$, and $S_ j^0=n\langle \theta _{\varphi _ j}\wedge \theta _\eta ^{n-1}\rangle $, so it equals $n\int _ Xb\, \mathrm{dd}^{\mathrm{c}}a\wedge \langle \theta _\eta ^{n-1}\rangle $. This proves (17.2.6) when $T_1=\cdots =T_{n-1}=\theta _\eta $ and all potentials share the background $\theta $. For $n=1$ there are no currents $T_ j$, and (17.2.6) is (17.2.4) for a Kähler background with respect to which the four functions are psh.

Step 2: arbitrary currents. Let $n>1$. By theorem 10.2.15, write $T_ j=\theta _ j+\mathrm{dd}^{\mathrm{c}}\eta _ j$ with $\theta _ j$ smooth closed and $\eta _ j\in \mathrm{PSH}(X,\theta _ j)$. For positive numbers $r_1,\ldots ,r_{n-1}$ with $\sum _ jr_ j=1$ put $\theta _ r=\sum _ jr_ j\theta _ j$ and $\eta _ r=\sum _ jr_ j\eta _ j$, so that $\sum _ jr_ jT_ j=\theta _ r+\mathrm{dd}^{\mathrm{c}}\eta _ r$. The potential $\eta _ r$ is $(\theta _ r+s\omega )$-psh for every $s\geq 0$, since $\theta _ r+s\omega +\mathrm{dd}^{\mathrm{c}}\eta _ r=\sum _ jr_ jT_ j+s\omega \geq 0$. Let $\kappa $ be a multiple of $\omega $ such that the presentations of $a$ and $b$ are $\kappa $-psh. Since the forms $\theta _ r$ are convex combinations of the finitely many smooth forms $\theta _ j$, there is $s_0$ with $\theta _ r+s\omega \geq \kappa $ for all $r$ and all $s\geq s_0$, and then the presentations are $(\theta _ r+s\omega )$-psh as well. Step 1 applied to the current $\sum _ jr_ jT_ j+s\omega =(\theta _ r+s\omega )+\mathrm{dd}^{\mathrm{c}}\eta _ r$ gives

\[ \int _ Xa\, \mathrm{dd}^{\mathrm{c}}b\wedge \Bigl \langle \Bigl (\sum _ jr_ jT_ j+s\omega \Bigr )^{n-1}\Bigr \rangle =\int _ Xb\, \mathrm{dd}^{\mathrm{c}}a\wedge \Bigl \langle \Bigl (\sum _ jr_ jT_ j+s\omega \Bigr )^{n-1}\Bigr \rangle , \]

where, by the independence of (17.2.1) from the background, the Hessian factors may be computed in one fixed Kähler background. Expanding the repeated current by multilinearity, both sides become polynomials in $s$, and comparing constant coefficients gives

\[ \int _ Xa\, \mathrm{dd}^{\mathrm{c}}b\wedge \Bigl \langle \Bigl (\sum _ jr_ jT_ j\Bigr )^{n-1}\Bigr \rangle =\int _ Xb\, \mathrm{dd}^{\mathrm{c}}a\wedge \Bigl \langle \Bigl (\sum _ jr_ jT_ j\Bigr )^{n-1}\Bigr \rangle . \]

Both sides are homogeneous polynomials of degree $n-1$ in $(r_1,\ldots ,r_{n-1})$. Since they agree on the open simplex, homogeneity makes them agree on the open positive cone of $\mathbb R^{n-1}$. Hence their coefficients agree: a polynomial that vanishes on a nonempty open set is zero, because all its partial derivatives vanish at a point $r_0$ of that set and it equals its finite Taylor expansion at $r_0$. The coefficient of $r_1\cdots r_{n-1}$ is $(n-1)!$ times the corresponding side of (17.2.6). $\square$

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