Proof. (i) For every $\Phi \in \operatorname {\mathrm {PSH}}_h(M'\times N')$ and every $(p-1,p-1)$ -form $\Omega $ with vertical support in $M"\times N$ we have

$$\begin{align*}|\langle{dd^c}\Phi,\Omega\rangle|=|\langle\Phi,{dd^c}\Omega\rangle|\le \|{dd^c}\Omega\|_{\mathcal C^l} \|\Phi\|_{-l} \lesssim \|\Omega\|_{\mathcal C^{l+2}} \|\Phi\|_{-l}, \end{align*}$$

where the implicit constant in the last inequality is independent of $\Omega $ , $\Phi $ , and l. The first inequality above follows by observing that the support of $dd^c \Omega $ is contained in the support of $\Omega $ , and hence in particular in $M"\times N$ .

(ii) For every compact subset K of $M"\times N$ , there exists a smooth function $\chi $ with vertical support in $M"\times N$ and such that $\mathbb 1_{K}\leq \chi \leq 1$ , where $\mathbb 1_K$ is the characteristic function of K. Since ${dd^c} \Phi \geq 0$ , we have

$$\begin{align*}\|{dd^c} \Phi\|_{K} = \int_K {dd^c} \Phi \wedge \omega^{p-1} \leq \langle {dd^c} \Phi, \chi \omega^{p-1} \rangle \leq \|{dd^c} \Phi\|_{-l-2} \|\chi \omega^{p-1}\|_{\mathcal C^{l+2}} \lesssim \|\Phi\|_{-l} \|\chi \omega^{ p-1}\|_{\mathcal C^{l+2}}, \end{align*}$$

where $\omega $ is the standard Kähler form of $\mathbb {C}^k$ and, by (i), the implicit constant in the last inequality is independent of l and $\Phi $ . As $\|\chi \omega ^{p-1}\|_{\mathcal C^{l+2}}$ can be bounded by a constant depending only on K and l, the assertion follows.

Proof. We can assume that $\|T\|_{\mathcal C^{l}}\leq 1$ . Since T is vertical in $M"\times N$ , it follows that, for every $\Phi _1, \Phi _2\in \operatorname {\mathrm {PSH}}_h (M'\times N')$ , we have

$$\begin{align*}|\langle T,\Phi_1-\Phi_2\rangle| \leq \|\Phi_1-\Phi_2\|_{- {l}} = \operatorname{\mathrm{dist}}_{- {l}} (\Phi_1,\Phi_2).\end{align*}$$

The assertion follows.

Proof. Fix a convex open set $M^\star $ with $M'\Subset M^\star \Subset M$ . The existence of a horizontal negative $L^1$ form $\Psi $ on $M^\star \times N$ satisfying ${dd^c} \Psi =\Omega $ and $ \|\Psi \|_{M^\star \times N'}\lesssim \|\Omega \|_{M\times N"}$ follows from [Reference Dinh, Nguyen and Sibony19, Theorem 2.7]. As $\Psi $ is defined by means of explicit kernels, and $M"\Subset M'\Subset M^\star $ , this also gives the second estimate in (3.1).

Proof. Fix a psh function u on D. By reducing D slightly and subtracting from u a constant, we can assume that $u\leq 0$ . For every $m\in \mathbb N$ , define $u_m:=\max (u,-m)$ . Observe that $u-u_m$ is negative, supported on $\{u\leq -m\}$ and equal to $u+m$ on this set. It follows that

(3.2) $$ \begin{align} |\langle \omega^k, u-u_m\rangle| =\int_{\{u\leq -m\}} |u-u_m| \lesssim e^{-\alpha_1 m} \int e^{\alpha_1 { |u|}} \lesssim e^{-\alpha_1 m}.\end{align} $$

for some positive constant $\alpha _1$ , where the integrals are taken with respect to the Lebesgue measure and last inequality follows from Skoda’s estimates [Reference Skoda35].

Fix a smooth positive closed horizontal form $\Omega _K$ on $M'\times N'$ such that $\Omega _K\geq \omega ^{k-p}$ on K, see for instance [Reference Bianchi, Dinh and Rakhimov9, Lemma 2.3]. Observe that we also have $\Omega _K \lesssim \omega ^{k-p}$ on $M"\times N'$ . Since u and $u_m$ are psh and $\Omega _K$ is closed, both $u\Omega _K$ and $u_m \Omega _{K}$ belong to $\operatorname {\mathrm {PSH}}_h (M'\times N')$ . Fix a $(p,p)$ -form $\Omega $ with vertical support in $M"\times N$ and such that $\|\Omega \|_{\mathcal C^l}\leq 1$ . As $\Omega _K \wedge \Omega \lesssim \omega ^k$ , it follows from (3.2) that $|\langle \Omega _K\wedge \Omega , u-u_m\rangle | \lesssim e^{-\alpha _1 m}$ , which implies that

$$ \begin{align*}\operatorname{\mathrm{dist}}_{-l} (u\Omega_K, u_m\Omega_K) \lesssim e^{-\alpha_1 m}.\end{align*} $$

The assumption on the super-potentials of T, together with the fact that T and $u_m-u$ are positive, implies that

$$ \begin{align*}0\leq \langle T\wedge \Omega_K, u_m- u \rangle =\langle T, u_m\Omega_K \rangle - \langle T, u\Omega_K\rangle \lesssim e^{-\alpha_2 m}\end{align*} $$

for some positive constant $\alpha _2$ . Hence, since $\Omega _K {\geq } \omega ^{k-p}$ , we have

$$ \begin{align*}0\leq \langle T_{|K}\wedge \omega^{k-p}, u_m-u\rangle \lesssim e^{-\alpha_2 m} \quad \text{ for every } m\in \mathbb N.\end{align*} $$

The assertion follows by choosing any $\beta <\alpha _2$ in Definition 3.7. Indeed, with this choice of $\beta $ and recalling that u is negative, we have

$$\begin{align*}\langle T_{|K}\wedge \omega^{k-p}, e^{\beta |u|}\rangle = \sum_{m\in \mathbb N} \int_{-m\leq u -\leq -m+1} e^{\beta |u|} T_{|K}\wedge \omega^{k-p} \lesssim \sum_{m\in \mathbb N} e^{\beta m } e^{-\alpha_2 m} \lesssim 1. \end{align*}$$

This completes the proof.

Proof. Let $\rho $ be a smooth positive radial function compactly supported on the unit ball of $\mathbb C^k$ and such that $\int \rho \omega ^k =1$ . For every $\epsilon>0$ , set $\rho _\epsilon (\cdot ) := \epsilon ^{-2k} \rho (\epsilon ^{-1} \cdot )$ . For every $u\in \operatorname {\mathrm {Psh}}_1 (D)$ and every $\epsilon $ sufficiently small, the convolution $u_{\epsilon } := u * \rho _\epsilon $ is well-defined on $M'\times N'$ . One can check that we have

$$\begin{align*}u \leq u_{\epsilon} \quad \text{ and } \quad \|u_{\epsilon}-u\|_{L^1 { (M'\times N')} }\lesssim \epsilon |\log \epsilon|.\end{align*}$$

Since the $u_{\epsilon }$ ’s and u are psh, it follows that for every fixed positive closed smooth form $\Psi $ of bidimension $(p,p)$ and horizontal in $M'\times N'$ , all the products $u_{\epsilon }\Psi $ and $u \Psi $ belong to $\operatorname {\mathrm {PSH}}_h (M'\times N')$ . They also form a bounded family with respect to $\|\cdot \|_\bullet $ . Moreover, for every vertical $(p,p)$ -form $\Omega $ on $M"\times N$ , we have

$$\begin{align*}|\langle \Omega, (u_{\epsilon}-u)\Psi\rangle| = |\langle \Omega \wedge \Psi, (u_{\epsilon}-u)\rangle| \lesssim { \int_{M'\times N'} (u_{\epsilon}-u) \omega^k =} \|u_{\epsilon}-u\|_{L^1 { (M'\times N')} }, \end{align*}$$

which implies that $\operatorname {\mathrm {dist}}_{-l}(u_{\epsilon }\Psi , u\Psi )\lesssim \epsilon |\log \epsilon |$ . Since T has $(l,\alpha )$ -Hölder continuous super-potentials, we deduce that

$$ \begin{align*}\langle T,(u_{\epsilon}-u)\Psi\rangle \lesssim \|u_\epsilon-u\|_{L^1 { (M'\times N')}}^{\alpha}\lesssim ( \epsilon |\log \epsilon|)^{\alpha}\leq \epsilon^{\alpha'}\end{align*} $$

for every $\alpha '<\alpha $ . This and the inequality $u_{\epsilon }\geq u$ imply that

(3.3) $$ \begin{align} \operatorname{\mathrm{dist}}_{-l}(u_{\epsilon} T, u T) \lesssim \epsilon^{\alpha'}.\end{align} $$

Let now consider $u_1$ and $u_2$ as in the statement. Denote by $u_{1,\epsilon }=u_1 * \rho _\epsilon $ and $u_{2,\epsilon }= u_2 * \rho _\epsilon $ the regularizations by convolution of $u_1$ and $u_2$ respectively. It follows from the above and (3.3) that there exists $\alpha '>0$ such that

(3.4) $$ \begin{align} \operatorname{\mathrm{dist}}_{-l}(u_{1,\epsilon} T, u_1 T) \lesssim \epsilon^{\alpha'} \quad \text{ and } \quad \operatorname{\mathrm{dist}}_{-l}(u_{2,\epsilon} T, u_2 T) \lesssim \epsilon^{\alpha'}. \end{align} $$

We now get a similar estimate for $\operatorname {\mathrm {dist}}_{-l}(u_{1,\epsilon } T, u_{2,\epsilon }T)$ . By the definition of $u_{1,\epsilon }$ and $u_{2,\epsilon }$ , we have

$$ \begin{align*}\|u_{1,\epsilon} -u_{2,\epsilon}\|_{\mathcal C^l} = \| (u_1-u_2) * \rho_\epsilon \|_{\mathcal C^l} \lesssim \epsilon^{-2k-l} \|u_1-u_2\|_{L^1}.\end{align*} $$

Again by the fact that T has $(l,\alpha )$ -Hölder continuous super-potentials, we deduce that

(3.5) $$ \begin{align} \operatorname{\mathrm{dist}}_{-l}(u_{1,\epsilon} T, u_{2,\epsilon}T) \lesssim [\epsilon^{-2k-l} \|u_1-u_2\|_{L^1}]^{\alpha}.\end{align} $$

Together with (3.4), this gives

$$\begin{align*}\operatorname{\mathrm{dist}}_{-l}(u_{1} T, u_{2}T) \lesssim \epsilon^{\alpha'} + [\epsilon^{-2k-l} \|u_1-u_2\|_{L^1}]^{\alpha} \quad \text{ for every } \alpha'<\alpha.\end{align*}$$

Choosing $\epsilon := { c_0} \|u_1-u_2\|_{L^1}^{1/(1+2k+l)}$ for some sufficiently small constant $c_0$ , we see that

$$\begin{align*}\operatorname{\mathrm{dist}}_{-l}(u_{1} T, u_{2}T) \lesssim \|u_1-u_2\|_{L^1}^{\alpha\prime/(1+2k+l)}. \end{align*}$$

The assertion follows.

Proof. It follows from Definition 3.3 that the intersection $T_v \wedge T_h$ is well-defined. It is a positive measure since both $T_v$ and $T_h$ are positive. By [Reference Dinh and Nguyen18], in order to show that this measure is a Monge-Ampère as in the statement, it is enough to show that $u\mapsto \langle T_v\wedge T_h, u\rangle $ is Hölder continuous on $\operatorname {\mathrm {Psh}}_1 (D)$ with respect to the distance induced by the norm $L^1$ . By Lemma 3.9 and Definition 3.3, for every $u_1, u_2\in \operatorname {\mathrm {Psh}}_1 (D)$ we have

$$\begin{align*}|\langle T_v\wedge T_h, u_1 - u_2 \rangle| =|\langle T_h, u_1 T_v - u_2 T_v\rangle| \lesssim \|u_1 T_v - u_2 T_v\|_{-l}^{\alpha} \lesssim \|u_1-u_2\|^{\alpha \gamma}, \end{align*}$$

for some positive constant $\gamma $ as in Lemma 3.9. The first assertion follows. The last property is true for the Monge-Ampère measure associated with any Hölder continuous psh function, see for instance [Reference Sibony34, Théorème 1.7.3].

Proof. We can assume that $l_0< q$ (as otherwise the statement is clear) and it is enough to consider the case where $\alpha $ is of the form $\alpha = \beta \wedge \Omega $ for some positive $(\ell ,\ell )$ -form $\beta $ with $\ell \le l_0$ and $(2l_0-2\ell +1,0)$ -form $\Omega $ , and show that $\langle S \wedge (\pi _D)^* \alpha , \Phi \rangle =0$ for every compactly supported smooth $(q+\ell -2l_0-1, q-\ell )$ -form $\Phi $ of the form $\Phi = \gamma \wedge \Theta $ , for some positive $(q+\ell -2l_0-1,q+\ell -2l_0-1)$ -form $\gamma $ and $(0,2l_0-2\ell +1)$ -form $\Theta $ .

Applying Cauchy-Schwarz inequality, we have

$$\begin{align*}\begin{aligned} |\langle S\wedge (\pi_D)^* \alpha, \Phi\rangle|^2 & = |\langle S \wedge (\pi_D)^* \beta \wedge \gamma, (\pi_D)^* \Omega \wedge \Theta\rangle|^2\\ & \leq |\langle S \wedge (\pi_D)^* \beta \wedge \gamma, (\pi_D)^* (\Omega \wedge \overline \Omega)\rangle| \cdot |\langle S \wedge (\pi_D)^* \beta \wedge \gamma, \Theta\wedge \overline \Theta\rangle|=0. \end{aligned}\end{align*}$$

Since $(\pi _D)^*(\beta \wedge \Omega \wedge \overline \Omega )$ is a smooth $(2l_0-\ell +1,2l_0-\ell +1)$ -form and $\ell \le l_0$ , the first factor in the last term vanishes by the assumption (ii) in Definition 5.4.

Proof. Since S is a positive current on $D\times {\mathbb {G}}({q},k)$ , by (5.3) $\underline {S}$ is a positive current of bidimension $(l_0,l_0)$ on D. Since $S\wedge (\pi _D)^*({\omega _D^{l_0+1}})=0$ we have

$$ \begin{align*} \|\underline{S}\|_{{\tilde{D}}}=\int_{{\tilde{D}}} \underline{S}\wedge \omega_D^{l_0} &=\int_{{\tilde{D}}\times {\mathbb{G}}({q},k)} S\wedge (\pi_{{\mathbb{G}}})^*\omega_{{\mathbb{G}}}^{{q}-l_0}\wedge (\pi_D)^*\omega_D^{l_0}\\ &=\int_{{\tilde{D}}\times {\mathbb{G}}({q},k)} S\wedge (\pi_D)^*\omega_D^{l_0}\wedge \left((\pi_{{\mathbb{G}}})^*\omega_{{\mathbb{G}}}+ (\pi_D)^*\omega_D\right)^{{q}-l_0} \\ &=\|S\wedge (\pi_D)^*\omega_D^{l_0}\|_{{\tilde{D}}\times {\mathbb{G}}({q},k)}. \end{align*} $$

Hence, (5.4) holds.

Let us now prove the last assertion. Since the problem is local, we can assume that U is a small ball compactly supported in D. Assume first that we have $l_0=0$ . In this case, since $\underline {S}$ is a measure on D, it is clear that $\underline {S}$ is closed. Assume now that we have $l_0>0$ and let $\theta $ be a smooth $(2l_0-1)$ -form compactly supported on U. By using Stokes’ formula twice and the fact that S is a closed current of bidimension $({q},{q})$ on $U\times {\mathbb {G}}({q},k)$ , we have

$$ \begin{align*} \langle d\underline{S},\theta\rangle&=\langle \underline{S},d\theta\rangle =\langle {S},(\pi_{{\mathbb{G}}})^*\omega_{{\mathbb{G}}}^{{q}-l_0}\wedge\pi_D^*(d\theta)\rangle\\ &=\langle {S}\wedge (\pi_{{\mathbb{G}}})^*\omega_{{\mathbb{G}}}^{{q}-l_0},d(\pi_D^*(\theta))\rangle=\langle d{(S\wedge (\pi_{{\mathbb{G}}})^*\omega_{{\mathbb{G}}}^{{q}-l_0})},\pi_D^*(\theta)\rangle\\ &=\langle dS\wedge (\pi_{{\mathbb{G}}})^*\omega_{{\mathbb{G}}}^{{q}-l_0},\pi_D^*(\theta)\rangle =0. \end{align*} $$

Hence, $\underline {S}$ is closed on U and the proof is complete.

Proof of Theorem 5.9.

Let $\Sigma $ be a horizontal submanifold of $M\times N'$ of pure dimension p and such that $d^{-n}(f^n)_*[\Sigma ]$ converge to $T^-$ . Let $v_n$ and $\hat {v}_n$ be the volume of $\Sigma _n:=f^{n}(\Sigma )$ on $M"\times N$ and of $\hat {\Sigma }_n$ on $M"\times N \times {\mathbb {G}}(p,k)$ , respectively, where $\hat {\Sigma }_n$ is the submanifold of $D\times {\mathbb {G}}(p,k)$ defined by means of (5.2). Note that, since $d^{-n}(f^n)_*[\Sigma ]$ converge to $T^-$ we have

(5.5) $$ \begin{align} d^n\lesssim v_n\lesssim d^n. \end{align} $$

By definition of $\hat {\Sigma }_n$ and F we can see that $\hat {\Sigma }_n=F^n(\hat {\Sigma })$ . Hence, we have

$$\begin{align*}\hat v_n=\|(F^n)_*[\hat{\Sigma}]\|_{M"\times N\times{\mathbb{G}}(p,k)}.\end{align*}$$

We deduce from Proposition 5.10 that $\hat {v}_n=O(d^n)$ as $n\to \infty $ . It follows from (5.5) that we have $\hat {v}_n=O(d^n)=O(v_n)$ . Hence, thanks to Theorem 5.2, we conclude that $(T^-)_{|M"\times N}$ is woven on D (or, equivalently, that $T^-$ is woven as a current on $M"\times N$ ). In order to conclude the proof of the first assertion, we need to deduce that $T^-$ is woven on all of D.

Recall that $\mathcal {B}_p$ denotes the set of p-dimensional connected complex (not necessarily closed) submanifolds of D with locally finite $2p$ -dimensional volume. Denote similarly by $\mathcal {B}^{\prime \prime }_p$ the set of p-dimensional connected complex submanifolds of $M"\times N$ with locally finite $2p$ -dimensional volume. As $(T^-)_{|M"\times N}$ is woven, we can write it as

(5.6) $$ \begin{align} (T^-)_{|M"\times N}=\int_{Z\in\mathcal{B}^{\prime\prime}_p }[Z]d\nu(Z), \end{align} $$

where $\nu $ is a $\mathcal {B}^{\prime \prime }_p$ -finite measure on $\mathcal {B}^{\prime \prime }_p$ . Define a map

$$\begin{align*}\mathcal{F}:\mathcal{B}^{\prime\prime}_p \to \mathcal{B}_p \quad \text{ as }\quad Z\mapsto f(Z\cap D_{v,1}).\end{align*}$$

The map $\mathcal F$ is well-defined by (4.1). Let us also define a measure $\tilde {\nu }$ on $\mathcal {B}_p$ as

$$\begin{align*}\tilde \nu := d^{-1} \mathcal F_* \nu. \end{align*}$$

Since $\nu $ is $\mathcal {B}^{\prime \prime }_p$ -finite on $\mathcal {B}^{\prime \prime }_p$ , $\tilde {\nu }$ is $\mathcal {B}_p$ -finite on $\mathcal {B}_p$ . Indeed, let K be any compact subset of D. Thanks to (4.1) there exists a convex open set $M^\star \Subset M"$ such that $f^{-1}(K)\subset M^\star \times N$ . Then, we have

$$ \begin{align*} \int_{Z\in\mathcal{B}_{p}}\langle[Z],\omega^{p}|_K\rangle d\tilde{\nu}(Z) &= d^{-1}\int_{Z\in{\mathcal{B}_{p}"}}\langle[f(Z)],\omega^{p}|_K\rangle d{\nu}(Z)\\ &=d^{-1}\int_{Z\in{\mathcal{B}_{ p}"}}\langle[Z], f^*(\omega^{p}|_K)\rangle d{\nu}(Z) \\ &\lesssim \int_{Z\in{\mathcal{B}_p"}}\langle[Z],\omega^{p}|_{M^\star\times N'}\rangle d{\nu}(Z), \end{align*} $$

where in the last inequality we used [Reference Bianchi, Dinh and Rakhimov9, Lemma 3.3] and the fact that we have $Z\subset M"\times N'$ for $\nu $ -almost every Z. Since $\nu $ is $\mathcal {B}^{\prime \prime }_p$ -finite, we deduce from the last inequality that $\tilde {\nu }$ is $\mathcal {B}_p$ -finite on $\mathcal {B}_p$ .

Again by (4.1), we also have $d\cdot T^-=f_*\left ((T^-)_{|M"\times N}\right )$ on D. Hence, by (5.6), we have

$$\begin{align*}T^- =d^{-1}f_*\left((T^-)_{|M"\times N}\right) =d^{-1}\int_{Z\in\mathcal{B}^{\prime\prime}_p}f_*[Z]d\nu(Z) =\int_{Z\in \mathcal{B}_p}[Z]d\tilde{\nu}(Z) \end{align*}$$

on D. Hence, $T^-$ is woven on D.

In order to conclude the proof of Theorem 5.9, we need to show that $T^-$ is laminar assuming that $p=k-1$ (observe that this implies $d^{-}_{k-p-1}=d^-_0=1<d$ ). Recall that the condition $p=k-1$ implies that $T^-$ is of bidegree $(1,1)$ . By Theorem 4.1, $T^-$ has Hölder continuous local potentials. So $T^-\wedge T^-$ is a well-defined horizontal positive closed $(2,2)$ -currentFootnote ¹ . By Proposition 5.3, it is enough to show that $T^-\wedge T^-=0$ .

Assume by contradiction that $T^-\wedge T^-\ne 0$ . By the f-invariance of $T^-$ and the fact that f is invertible, we have $f_* (T^-\wedge T^-)=d^2(T^-\wedge T^-)$ . By Definition 2.2, this implies that $d^2\le d^+_{k-2}$ . Since by assumption we have $d^+_{k-2}<d=d^+_{k-1}$ , this gives $d^2<d$ , which implies that $d<1$ . Since we always have $d\geq 1$ , this gives the desired contradiction and shows that $T^-\wedge T^-=0$ . The proof is complete.

Proof. Set

$$ \begin{align*}L:=\{\tilde l:\limsup_{n\to \infty} C_{\tilde l}(n)=\infty\}=:\{l^{\prime}_1,l^{\prime}_2,...,l^{\prime}_{m'}\},\end{align*} $$

where $1\le m'\le p$ and $l^{\prime }_1< \dots <l^{\prime }_{m'}$ . By assumption, we have $p\notin L$ and so $l^{\prime }_{m'}<p$ . Moreover, by the definition of L it follows that if $\tilde l\notin L$ then $\{C_{\tilde l}(n)\}_{n\in \mathbb N}$ is bounded. Set

$$ \begin{align*}a_n:=\max\{C_{\tilde l}(n): {\tilde l}\in L\}.\end{align*} $$

Since by assumption we have $\limsup _{n\to \infty } a_n=\infty $ , there exists a subsequence $a_{\tilde n_j}$ such that $a_n< a_{\tilde n_j}$ for every j and $n< {\tilde n_j}$ . We will only use this sequence $\{\tilde n_j\}$ or a subsequence of it. For every j, define

$$ \begin{align*}L_j:=\{\tilde l\in L: C_{\tilde l}({\tilde n_j})=a_{{ \tilde n_j}}\}\end{align*} $$

As L is finite, up to replacing the sequence $\{{ \tilde n_j}\}$ by a subsequence, we can assume that $L_j$ does not depend on j, i.e., that we can write

$$\begin{align*}\mathcal L := L_j =: \{l_1, \dots, l_m\} \quad \text{ for every } j. \end{align*}$$

It is clear that the pair $(l_1,\{{\tilde n_j}\}_{j\in \mathbb N})$ satisfies (i) for $\beta =1$ . We now describe a procedure that, by possibly modifying the pair and $\beta $ , will also lead to (ii).

We will play the following simple game between $l_1$ and the other indexes $l_r$ , for $1<r\leq m$ . If there exists $s\in \mathbb N$ such that

(5.7) $$ \begin{align} \limsup\limits_{j\to \infty} \frac{C_{{l}_r}( \tilde n_j-s)}{C_{{l_1}}( \tilde n_j)}>0 \end{align} $$

then $l_r$ is the winner. Otherwise, $l_1$ is the winner.

We start playing the game above between $l_1$ and $l_m$ . If the winner is $l_m$ , we stop playing. Otherwise, we play the game between $l_1$ and $l_{m-1}$ . Again, if the winner is $l_{m-1}$ we stop playing, otherwise we continue the game between $l_1$ and $l_{m-2}$ . After repeating the procedure at most $m-1$ times, we have the final winner. If the final winner is $l_1$ then it is clear that the pair $(l_1,\{{\tilde n_j}\}_{j=1}^\infty )$ satisfies both (i) and (ii). In this case, the proof is complete. We can then assume that the final winner is $l_{r_0}$ for some $1<r_0\le m$ . Observe that this means that $l_{r_0}$ was the winner between $l_1$ and $l_{r_0}$ (i.e., (5.7) holds with $r_0$ instead of r, for some given choice $s_0$ of s), but also that $l_1$ was the winner against all $l_{r'}$ , for all $r'> r_0$ .

After again choosing a subsequence, if necessary, we can assume that the limsup in (5.7) (with $r=r_0$ and $s=s_0$ ) is actually a limit, i.e., that we have

(5.8) $$ \begin{align} \lim \limits_{j\to \infty} \frac{C_{{l}_{r_0}}({\tilde n_j}-s_0)}{C_{{l_1}}({\tilde n_j})} =: A>0. \end{align} $$

We can also assume that $\tilde {n}_1>s_0$ . Take $(l, \{n_j\}_{j\in \mathbb N})$ as a new pair, where ${n}_j:={\tilde n_j}-s_0$ and $l:=l_{r_0}$ . We now show that this pair satisfies both the requests (i) and (ii).

By (5.8), we have $C_{l_1}({\tilde n_j})\le \beta \cdot C_{l}(n_j)$ for every j, for some positive constant $\beta $ independent of j. By construction, for any $n\le {\tilde n_j}$ and $0\le \tilde {l}\le p$ , we also have $C_{\tilde {l}}(n)\le C_{l_1}({\tilde n_j})$ . As a consequence, we have

$$ \begin{align*}C_{\tilde{l}}(n) \le C_{l_1}({\tilde n_j}) \le {\beta}\cdot C_{l}({{n}_j}) \quad \text{ for all } n\le n_j \text{ and }0\le {\tilde{l}}\le p.\end{align*} $$

So, $(l,\{{n}_j\}_{j\in \mathbb N})$ satisfies (i). For the sake of contradiction, assume that $(l,{n}_j)$ does not satisfy (ii). In this case, there must exist $l_{r_1}$ with $r_0<r_1\le m$ and $s_1$ such that

(5.9) $$ \begin{align} \limsup\limits_{j\to \infty} \frac{C_{{l}_{r_1}}({n}_j-s_1)}{C_{l}({n}_j)}>0. \end{align} $$

Again, after choosing a subsequence if necessary, we can assume that the above $\limsup $ is a limit. Since $l_1$ was the winner when playing the game between $l_1$ and $l_{r_1}$ , we have

$$ \begin{align*}0=\lim\limits_{j\to \infty} \frac{C_{{l}_{r_1}}({\tilde{n}_j}-s_0-s_1)}{C_{{l_1}}({\tilde n_j})} = \lim\limits_{j\to \infty}\frac{C_{l}({\tilde n_j}-s_0)}{C_{{l_1}}({\tilde n_j})}\cdot \frac{C_{{l}_{r_1}}({n_j}-s_1)}{C_{{l}}({n_j})}>0.\end{align*} $$

The last inequality follows from (5.8) and (5.9) and the assumption that the $\limsup $ in (5.9) is actually a limit. This gives a contradiction. Hence, the pair $(l,\{{n}_j\}_{j\in \mathbb N})$ satisfies (ii) and the proof is complete.

Proof. By the definitions of $\hat R$ and F, we have

(5.11) $$ \begin{align} (\pi_D)_*( (F^n)_*(\hat{R}))= (f^n)_*(R). \end{align} $$

Indeed, if $\theta $ is a smooth $(p,p)$ -form compactly supported on D, we have

$$ \begin{align*} \langle (\pi_D)_*( (F^n)_*(\hat{R})), \theta \rangle_D&=\langle (F^n)_*(\hat{R}), (\pi_D)^*(\theta) \rangle_{D \times {\mathbb{G}}(p, k)}\\ &=\int_{F^n(\hat \Sigma)}(\pi_D)^*(\theta)=\int_{f^n( \Sigma)}\theta=\langle (f^n)_*({R})), \theta \rangle_D, \end{align*} $$

which proves (5.11). By using this identity, we have

$$ \begin{align*} I_p(n) & =\int_{D\times{\mathbb{G}}(p,k)} (\pi_D)^*(\chi) \cdot (F^n)_*(\hat{R})\wedge (\pi_D)^*\omega_D^p\\ & =\int_D (\pi_D)_*\left((\pi_D)^*\chi \cdot (F^n)_*(\hat{R})\wedge (\pi_D)^*\omega_D^p\right)\\ & =\int_D \chi \cdot (f^n)_*(R)\wedge \omega_D^p\le \int_{M'\times N} (f^n)_*(R)\wedge \omega_D^p\lesssim d^n. \end{align*} $$

The last equality is due to (2.2). Similarly, since the dynamical degrees do not depend on the choice of $M'$ and $N'$ , by using (2.2) again, we also have

$$ \begin{align*} I_p(n) & =\int_D \chi \cdot (f^n)_*(R)\wedge \omega_D^p\ge \int_{M"\times N} (f^n)_*(R)\wedge \omega_D^p\gtrsim d^n. \end{align*} $$

This completes the proof.

Proof. By a direct computation, for all h as in the statement we have

$$ \begin{align*} \|S_j^{(s)}\wedge (\pi_D)^*\omega_D^{h}\|_{D\times {\mathbb{G}}(p,k)}= &\int_{D\times {\mathbb{G}}(p,k)} S_j^{(s)} \wedge (\pi_D)^*\omega_D^{h}\wedge ((\pi_D)^*(\omega_D)+(\pi_{{\mathbb{G}}})^*\omega_{\mathbb{G}})^{p-h}\\ =&\sum_{m=0}^{p-h}{p-h \choose m}\int_{D\times {\mathbb{G}}(p,k)} {S_j^{(s)}\wedge (\pi_D)^*(\omega_D^{m+h})\wedge(\pi_{\mathbb{G}})^*(\omega_{\mathbb{G}}^{p-h-m}) }\\ =& \sum_{m=0}^{p-h}{p-h \choose m}\frac{I_{m+h}(n_j-s)}{C_l(n_j)d^{n_j-s}}=\sum_{m=0}^{p-h}{p-h \choose m}\frac{C_{m+h}(n_j-s)}{C_l(n_j)}, \end{align*} $$

where in the last step we used (5.12). The assertion follows.

Proof. It is immediate to see that (i) follows from (5.13). Since f is horizontal-like, R has horizontal support in $M\times N'$ , and $\chi $ has vertical support in $M'\times N$ it follows that $S_j^{(s)}$ is supported on $D'\times {\mathbb {G}}(p,k)$ for every $s\in \mathbb N$ . Moreover, since by (4.1) the current $f_* (R)$ is supported on $M\times N"$ , for every $j\geq 1$ the current $S_j^{(s)}$ is supported on $\left (\mathrm {supp}(\chi )\cap (M\times N" ) \right )\times {\mathbb {G}} (p,k) \Subset {D'}\times {\mathbb {G}}(p,k) $ . Hence, $S^{(s)}_\infty $ has compact support in $D'\times {\mathbb {G}}(p,k)$ . Since by definition $S_j^{(s)}$ is closed in a neighbourhood of $\overline {M"}\times N\times {\mathbb {G}}(p,k)$ (as $\chi $ is constantly equal to 1 on an open neighbourhood of such set), the currents $S_\infty ^{(s)}$ are closed in $M"\times N\times {\mathbb {G}}(p,k)$ . So, (ii) follows.

In order to prove (iii), as $S_\infty ^{(s)}$ is positive, it is enough to show that

(5.14) $$ \begin{align} S_\infty^{(s)}\wedge (\pi_D)^*\omega_D^{l+1}=0. \end{align} $$

Since $S_j^{(s)}\wedge (\pi _D)^*\omega _D^{l+1}$ is positive, it suffices to show that its mass tends to zero as $j\to \infty $ . By taking $h=l+1$ in Lemma 5.13 (recall that $l\leq p-1$ ) and the fact that l was given by Lemma 5.11 (and in particular the quantities $C_{\tilde l}$ as in (5.12) satisfy Lemma 5.11 (ii)) we have

$$ \begin{align*} \lim_{j\to \infty}\|S_j^{(s)}\wedge (\pi_D)^*\omega_D^{l+1}\|& = \lim_{j\to \infty} \sum_{m=0}^{p-l-1} {p-l-1 \choose m}\frac{C_{l+1+m}(n_j-s)}{C_l(n_j)} =0. \end{align*} $$

This gives (5.14). The assertion (iii) follows.

By taking $h=l$ in Lemma 5.13 and using Lemma 5.11 (ii) again we have

$$ \begin{align*} \lim_{j\to \infty}\|S_j^{(0)}\wedge (\pi_D)^*\omega_D^{l}\|& = 1+ \lim_{j\to \infty} \sum_{m=1}^{p-l} {p-l \choose m}\frac{C_{l+m}(n_j)}{C_l(n_j)} =1. \end{align*} $$

By taking $h=l$ in Lemma 5.13 and using Lemma 5.11 (i) we have

$$ \begin{align*} \|S_j^{(s)}\wedge (\pi_D)^*\omega_D^{l}\|& =\frac{C_{l}(n_j-s)}{C_l(n_j)} + \sum_{m=1}^{p-l} {p-l \choose h}\frac{C_{l+m}(n_j-s)}{C_l(n_j)} \le c \end{align*} $$

for some constant c independent of s and j. On the other hand, since $(\pi _D)^*((f^s)_*{\omega _D^{l}})$ is strictly positive on the support of $S^{(0)}_\infty $ we have

$$ \begin{align*} 0 < \|S^{(0)}_\infty\wedge (\pi_D)^*((f^s)_*{\omega_D^{l}})\| &= \|S^{(0)}_\infty\wedge (F^s)_*(\pi_D)^*({\omega_D^{l}})\| \\& \le \|(F^s)_*(S^{(s)}_\infty)\wedge (F^s)_*(\pi_D)^*({\omega_D^{l}})\| = \|(F^s)_*(S^{(s)}_\infty\wedge (\pi_D)^*({\omega_D^{l}}))\|. \end{align*} $$

This gives (iv) and completes the proof.

Proof. Let us first show that the equality

(5.15) $$ \begin{align} (\pi_D)^*{\chi} {\partial} S_\infty^{(s)}=(\pi_D)^*({ \partial}\chi)\wedge S_\infty^{(s)} \end{align} $$

holds for every $s\geq 0$ . By the definition of $S_j^{(s)}$ and the fact that $ (F^{n_j-s})_*(\hat {R})$ is closed we have

$$ \begin{align*}{ \partial} S_{j}^{(s)} =\frac{(\pi_D)^*( {\partial} \chi){\wedge} (F^{n_j-s})_*(\hat{R})}{C_l(n_j)d^{n_j-s}}\end{align*} $$

for every $s\geq 0$ and every j such that $n_j>s$ . Multiplying $(\pi _D)^*\chi $ to both sides we obtain

$$ \begin{align*}(\pi_D)^*(\chi) { \partial} S_{j}^{(s)} =(\pi_D)^*( { \partial} \chi)\wedge \frac{ (\pi_D)^*\chi\cdot (F^{n_j-s})_*(\hat{R})}{C_l(n_j)d^{n_j-s}}=(\pi_D)^*( { \partial} \chi)\wedge S_{j}^{(s)},\end{align*} $$

where we used the fact that the support of $ \partial \chi $ is contained in the support of $\chi $ . (5.15) follows taking $j\to \infty $ .

It follows from (5.15) that, for any $\theta $ as in the statement and every $s\geq 0$ , we have

(5.16) $$ \begin{align} (\pi_D)^*{\chi} {\partial}\left( S_\infty^{(s)} \wedge (\pi_D)^*((f^s)^*\theta)\right) =S_\infty^{(s)}\wedge (\pi_D)^*(\Theta), \end{align} $$

where

$$\begin{align*}\Theta := { \partial} \chi \wedge(f^s)^*\theta + \chi\cdot (f^s)^*( { \partial} \theta) \end{align*}$$

is a smooth vertical $(l+1,{l})$ -form in D. Since the D-dimension of $S_\infty ^{(s)}$ is equal to l, by Lemma 5.5 the right hand side of (5.16) is equal to $0$ . As $\mathrm {Supp} \,S_\infty ^{(s)}\subseteq \mathrm {Supp} \left ((\pi _D)^*\chi \right )$ , this implies ${ \partial } \left ( S_\infty ^{(s)} \wedge (\pi _D)^*((f^s)^*\theta )\right )=0$ . So, we have

$$ \begin{align*}{ \partial} \left( (F^s)_*(S_\infty^{(s)}) \wedge (\pi_D)^*\theta\right)=(F^s)_* { \partial}\left( S_\infty^{(s)} \wedge (\pi_D)^*((f^s)^*\theta)\right)=0.\end{align*} $$

The proof is complete.

Proof. Assertion (i) follows from Proposition 5.7 (applied with $D'$ instead of $\tilde D$ ) and Lemma 5.14 (iv). Assertion (ii) follows from Lemma 5.14 (ii). Hence, we only have to prove the assertion (iii). Fix $s\geq 0$ and let $\theta $ be a positive $(l,l)$ -form compactly supported in D. We are going to show that

(5.17) $$ \begin{align} \langle \underline{S}^{(0)}_\infty,\theta \rangle \leq \langle (f^s)_*(\underline{S}_\infty^{(s)}),\theta \rangle. \end{align} $$

Recall that F can also be seen as a map from $D_{v,1} \times \mathbb {P}^{L(p, k)}$ to $D_{h,1} \times \mathbb {P}^{L(p, k)}$ , and that $\omega _{{\mathbb {G}}}$ is the form induced on ${\mathbb {G}}(p,k)$ by the Fubini-Study form on $\mathbb P^L= \mathbb P^{L(p,k)}$ . For every $z_0\in D_{v,1}$ we can identify the second coordinate of $F^s(z_0,\cdot )$ as an automorphism of ${\mathbb P}^{L}$ . By the invariance of the class of $\omega _{{\mathbb P}^{L}}$ under the actions of such automorphisms, we have

$$\begin{align*}(F^s)_*( (\pi_{{\mathbb P}^L})^* { \omega_{{\mathbb P}^L}})=(\pi_{{\mathbb P}^L})^*{\omega_{{\mathbb P}^L}} + dd^c {u_s}+ v_s,\end{align*}$$

for some smooth $2$ -form $v_s$ which vanishes on the fibres of $\pi _{D}$ and smooth function $u_s$ , on the common domain of definition of $(F^s)_*( (\pi _{{\mathbb P}^L})^*\omega _{{\mathbb P}^L})$ and $(\pi _{{\mathbb P}^L})^*\omega _{{\mathbb P}^L}$ . It follows that, on the same set, we have

(5.18) $$ \begin{align} (F^s)_*( (\pi_{{\mathbb P}^L})^* \omega_{{\mathbb P}^L}^{p-l})=(\pi_{{\mathbb P}^L})^* \omega_{{\mathbb P}^L}^{p-l} + dd^c U_s + V_s, \end{align} $$

for some form $U_s$ and a form $V_s$ which vanishes on the fibres of $\pi _{D}$ . Hence, $V_s$ can be written as a linear combination of forms of the form $(\pi _{D})^*\alpha \wedge V^{\prime }_s$ , where $\alpha $ is a $1$ -form on D.

By using (5.3), Lemma 5.14 (i), and (5.18), we have

$$ \begin{align*} \langle \underline{S}^{(0)}_\infty,\theta \rangle & =\int\limits_{D\times{\mathbb{G}}(p,k)}S_\infty^{(0)}\wedge (\pi_{{\mathbb{G}}})^*\omega_{{\mathbb{G}}}^{p-l}\wedge (\pi_D)^*\theta \\ & \le d^{-s}\int\limits_{D\times{\mathbb P}^L} (F^s)_*(S_\infty^{(s)}) \wedge (\pi_{\mathbb{P}^L})^*\omega_{{\mathbb P}^L}^{p-l}\wedge (\pi_D)^*\theta\\ & = d^{-s}\int\limits_{D\times{\mathbb P}^L} (F^s)_*(S_\infty^{(s)}) \wedge (F^s)_* ((\pi_{\mathbb{P}^L})^*\omega_{{\mathbb P}^L}^{p-l}) \wedge (\pi_D)^*\theta \\ & \quad - d^{-s}\int\limits_{D\times{\mathbb P}^L} (F^s)_*(S_\infty^{(s)}) \wedge dd^c U_s \wedge (\pi_D)^*\theta\\ & \quad - d^{-s}\int\limits_{D\times{\mathbb P}^L} (F^s)_*(S_\infty^{(s)}) \wedge {V_s} \wedge (\pi_D)^*\theta. \end{align*} $$

The second integral in the last term vanishes by Stokes’ formula and Lemma 5.15. We claim that the third integral also vanishes. Indeed, by the above description of $V_s$ , it is enough to show this claim for $V_s$ of the form $(\pi _D)^*\alpha \wedge V^{\prime }_s$ , where $\alpha $ is a $1$ -form on D. This implies that $(\pi _{D})^*(\alpha \wedge \theta )$ is $(2l+1)$ -form and thus Lemma 5.5 implies that

$$ \begin{align*}(F^s)_*(S_\infty^{(s)}) \wedge (\pi_{D})^*(\alpha\wedge\theta)=(F^s)_*(S_\infty^{(s)} \wedge (\pi_{D})^*((f^s)^*(\alpha\wedge\theta)))=0.\end{align*} $$

As $F^s$ is invertible, this implies

$$ \begin{align*} \langle \underline{S}^{(0)}_\infty,\theta \rangle & \leq d^{-s}\int\limits_{D\times{\mathbb P}^L} (F^s)_*\left(S_\infty^{(s)}\wedge (\pi_{\mathbb{P}^L})^*\omega_{{\mathbb P}^L}^{p-l}\right)\wedge (\pi_D)^*\theta \\ &=d^{-s}\int\limits_{D\times{\mathbb{G}}(p,k)} S_\infty^{(s)}\wedge (\pi_{{\mathbb{G}}})^*\omega_{{\mathbb{G}}}^{p-l}\wedge (F^s)^*( (\pi_D)^*\theta) \\ &=d^{-s} \langle \underline{S}_\infty^{(s)},(f^s)^*(\theta) \rangle=d^{-s} \langle (f^s)_*(\underline{S}_\infty^{(s)}),\theta \rangle. \end{align*} $$

This shows (5.17) and completes the proof.

End of the proof of Proposition 5.10.

We continue to use the notations introduced above. By Lemma 5.16 (i) and (ii), we have $\|\underline {S}^{(0)}_\infty \|_{M'\times N}=1$ . By Lemma 5.16 (iii), this gives

$$ \begin{align*}d^s=d^s\|\underline{S}_\infty^{(0)}\|_{M'\times N}\le\|(f^s)_*(\underline{S}^{(s)}_\infty\|_{M'\times N}\end{align*} $$

for every $s\geq 0$ . Since $\underline {S}^{(s)}_\infty $ is a closed horizontal current on $M"\times N$ , by (4.1) we see that ${f_*(\underline {S}^{(s)}_\infty )}$ is a closed horizontal current on $M'\times N$ . Moreover, since the mass of ${\underline {S}^{(s)}_\infty }$ is uniformly bounded from above by a constant independent of s, the mass of ${f_*(\underline {S}^{(s)}_\infty })$ is also bounded by a constant independent of s. By using the Definition 2.2 of the dynamical degrees, it follows that

$$ \begin{align*}d\le \limsup_{s\to \infty}\|(f^{s})_*(f_*(\underline{S}^{(s+1)}_\infty))\|_{M'\times N}^{1/s}\le d^+_l.\end{align*} $$

On the other hand, since $l<p$ , by [Reference Bianchi, Dinh and Rakhimov9, Theorem 1.1] and the assumption $d^+_{p-1}< d$ we have $d_l^+\le d^+_{p-1}< d$ . This gives the desired contradiction. So, we have $I(n)=O(d^n)$ and the proof is complete.