20.2.29 Relation to the classical restricted volumesA437169E35B2

Proof. We start with the first equality of (20.2.33). Since $Y\not\subseteq \operatorname{nK}(\alpha )$, $V_{\theta }|_ Y \not\equiv -\infty $ as a consequence of 13.13.8, hence also $\nu (V_{\theta },Y)=0$.

Take a quasi-equisingular approximation $(\varphi _ j)_{j>0}$ of $V_{\theta }$ with $\varphi _ j\in \mathrm{PSH}(X,\theta +\epsilon _ j\omega )$ for a decreasing sequence $(\epsilon _ j)_{j>0}$ converging to $0$. By the monotonicity theorem theorem 13.13.11, we have for each $j>0$,

for every $\chi \in \mathrm{PSH}(X,\theta )$ with $\chi |_ Y\not\equiv -\infty $. Since $\chi $ is arbitrary, we conclude that

For any fixed $\epsilon >0$, we have

Letting $\epsilon \to 0+$ and applying example 20.2.17, we conclude that the $\leq $ direction in the first equality of (20.2.33) holds. Moreover, $Y\not\subseteq \operatorname{nK}(\alpha )$ implies $\operatorname{vol}_{X|Y}(\alpha )>0$ by (20.2.30). Hence the limiting mass in example 20.2.17 is positive, and $\operatorname{Tr}_ Y^\theta (V_\theta )$ is indeed defined in the present situation.

For the reverse direction, by definition, for any fixed $\epsilon >0$, we have

for all large enough $j$. Therefore, for large $j$, by the monotonicity theorem theorem 13.13.11, we have

Letting $j\to \infty $ and $\epsilon \searrow 0$, using the continuity of $\operatorname{vol}_{X|Y}$ ([Corollary 4.11, Mat13]) together with example 20.2.17, we conclude the first equality of (20.2.33).

Now we address the second equality. Due to theorem 13.13.11, the defining formula (20.2.30), and the definition of $V_\theta $, we obtain that

The reverse inequality now follows from the first equality of (20.2.33), theorem 13.13.11 and the fact that $V_\theta |_ Y \preceq _ P \operatorname{Tr}_ Y^\theta (V_\theta )$ as proved in proposition 20.2.12. $\square$

Proof. Since $\nu (V_\theta , Y)=0$, we have $Y\not\subseteq \operatorname{nK}(\alpha +\epsilon \{ \omega \} )$ for all $\epsilon >0$. Therefore, by (20.2.31) and (20.2.33), only the last equality of (20.2.35) needs to be argued.

Thanks to 16.9.50, we have

as $\epsilon \to 0+$.

Therefore, using proposition 20.2.22, we find

Set

By example 20.2.17, if $\operatorname{Tr}_ Y^{\theta }(V_{\theta })$ is defined, then

whereas if $\operatorname{Tr}_ Y^{\theta }(V_{\theta })$ is not defined, then $M=0$. Thus $M$ is the last term of (20.2.35), with the convention made in the statement.

Thanks to theorem 16.9.30, for any $\epsilon _0>0$, we have

By the definition of $M$, this implies the existence of $(\epsilon _ j)_{j>0}$ such that $0<\epsilon _ j<1/j$ and

Moreover, for each $j>0$,

Indeed, the first inequality follows from the multi-linearity of the non-pluripolar products proposition 13.11.4 6. For the second one, if $M=0$ there is nothing to prove; if $M>0$, then example 20.2.17 shows that $\operatorname{Tr}_ Y^\theta (V_\theta )$ is defined, and the inequality follows from the multi-linearity of the non-pluripolar products proposition 13.11.4 6 and the monotonicity theorem proposition 16.8.22. Putting (20.2.36) and (20.2.37) together, it follows that

Finally, since $\int _ Y \left(\theta |_ Y + \epsilon \omega |_ Y + \mathrm{dd}^{\mathrm{c}}\operatorname{Tr}_ Y^{\theta + \epsilon \omega }(V_{\theta + \epsilon \omega }) \right)^ m$ depends monotonically on $\epsilon >0$, we conclude that

This completes the proof. $\square$