Proof. Clearly if \(\gamma\Delta+f:L^2(M,g)\rightarrow L^2(M,g)\) has entirely positive spectrum then \(\int_Mf\mathop{\mathrm{dvol}}_g>0\) as \[\int_Mf\mathop{\mathrm{dvol}}_g=\langle(\gamma\Delta+f)1,1\rangle_{L^2(M,g)}>0.\] Conversely let us assume that \(\int_Mf\mathop{\mathrm{dvol}}_g>0\). Without loss of generality we can assume that \(f\) is smooth, as for every \(\epsilon>0\) there exists \(f_{\epsilon}\in C^{\infty}(M,\mathbb{R})\) such that \(|f(x)-f_{\epsilon}(x)|<\epsilon.\) It is clear that for a sufficiently small \(\delta>0\) we have \[P\leq \frac{\int_M \delta f \mathop{\mathrm{dvol}}_g}{\|\delta f\|_{L^{\infty}(M)}(4\mathop{\mathrm{vol}}_g(M)\|\delta f\|_{L^{\infty}(M)}+\int_M \delta f\mathop{\mathrm{dvol}}_g)}\] with \(P\) denoting the Poincaré constant of \((M,g)\), that is the inverse of the first non-zero eigenvalue of \(\Delta\), see [13]. Then by [13] we can conclude that \(\Delta+\delta f:L^2(M,g)\rightarrow L^2(M,g)\) has entirely positive spectrum and thus also \(\frac{1}{\delta}\Delta+ f:L^2(M,g)\rightarrow L^2(M,g)\) has entirely positive spectrum. ◻

Proof. For simplicity, we will always denote the Kähler form of a Hermitian metric \(g\) by \(\omega_g\). Because \(E\) is mean curvature \(k\)-positive in the total sense, then there exists a Hermitian metric \(h\) on \(E\) and a Gauduchon metric \(g\) on \(X\) such that \[\label{equa-condition-positivity} \int_X \frak{r}_{k,g}(E,H)\cdot\frac{\omega_g^m}{m!}>0.\tag{5}\] Let \(s\) be a holomorphic section of \(\Lambda^{p}E^{-1}\otimes L^{-1}\). Since \(L\) is pseudo-effective, it admits a singular semi-positive metric \(h_L\), i.e., \(\mathop{\mathrm{im}}\partial\bar{\partial}\log\Theta_{h_L}\geq0\). One can choose a finite covering \(\{U_\alpha\}\) consists of coordinates neighborhoods such that on each \(U_\alpha\), \[s=\eta_\alpha\otimes e_\alpha^{-1}\] for some \(\eta_\alpha\in\Lambda^{p}E^{-1}|_{U_\alpha}\) and a generator \(e_\alpha\) of \(L|_{U_\alpha}\) with \(|e_\alpha|_{h_L}^2=e^{-\varphi_\alpha}\) for some \(\varphi_\alpha\in\mathrm{PSH}(U_\alpha)\). There is a induced singular Hermitian metric \(H:=\Lambda^{p}H^{-1}\otimes h_L^{-1}\) on \(\Lambda^{p}E^{-1}\otimes L^{-1}\) and \[\begin{align} |s|_H^2=|s|^2_{\Lambda^{p}H^{-1}\otimes h_L^{-1}}=e^{\varphi_\alpha}|\eta_\alpha|_{\Lambda^{p}H^{-1}}^2. \end{align}\] We argue by the contradiction. Suppose that \(s\neq 0\) and then \(\eta_\alpha\neq 0\). Recall that we have the following Bochner-type formula in the sense of currents (see e.g. [14]), \[\label{equa-inequality} \begin{align} \mathop{\mathrm{im}}\partial\bar{\partial}\log|s|^2_H\wedge \frac{\omega_g^{m-1}}{(m-1)!}&\geq \mathop{\mathrm{im}}\partial\bar{\partial}\varphi_\alpha\wedge \frac{\omega_g^{m-1}}{(m-1)!}-\frac{\langle \mathop{\mathrm{tr}}_gR^{\Lambda^{p}H^{-1}}\eta_\alpha,\eta_\alpha\rangle_H}{|\eta_\alpha|^2_H}\cdot\frac{\omega_g^m}{m!}\\ &\geq\frak{r}_{p,g}(E,H)\cdot\frac{\omega_g^m}{m!}, \end{align}\tag{6}\] where the last inequality follows from that the smallest eigenvalue of \(\mathop{\mathrm{tr}}_gR^{\Lambda^{p}H^{-1}}\) equals to \(-\frak{r}_{k,g}(E,H)\). Integrate 6 over \(X\), one has \[\int_M \frak{r}_{p,g}(E,H)\cdot\frac{\omega_g^m}{m!}\leq 0,\] which contradicts to 5 . The proof is completed. ◻

Proof. Vanishing results directly follows from Proposition 2. Let \(f:X\dashrightarrow Y\) be a MRC fibration of \(X\) and denote the indeterminacy locus of \(f\) by \(Z\). We argue by contradiction. If \(\mathop{\mathrm{rd}}(X)\leq k-1\), then \(p:=\dim Y\geq n-k+1\). By [15], any rational curves in \(Y\) can be lifted into \(X\) and thus \(Y\) is not uniruled by the above (2). Then it follows from Ou’s recent result [16] that \(K_Y\) is pseudo-effective. Applying Grauert-Remmert’s extension theorem [17], we conclude that \(f|_{X\setminus Z}^*K_Y\) extends to a pseudo-effective invertible subsheaf \(\mathcal{F}\subset \Lambda^{p,0}X\) since \(\mathrm{codim}_XZ\geq 2\), which corresponds to a holomorphic section of \(\Lambda^{p,0}X\otimes L^{-1}\) and we derive a contradiction. ◻

Proof. If \(\gamma\Delta+r_1:L^2(M,h)\rightarrow L^2(M,h)\) has entirely positive spectrum for some \(\gamma\in [0,\frac{4}{2m-1})\) then \(M\) is compact, see [1]. Now, by Lemma 1, we know that \(\int_Mr_1\mathop{\mathrm{dvol}}_g>0\) and thus we can conclude that \(M\) is rationally connected by Prop. 1. Analogously, if \(\gamma\Delta+r_1:L^2(M,h)\rightarrow L^2(M,h)\) has entirely positive spectrum for some \(\gamma\in [0, +\infty)\) then \(\int_Mr_1\mathop{\mathrm{dvol}}_g>0\) and the conclusion follows again by Prop. 1. ◻

Proof. It is well known that all the above properties follow from the fact that \(M\) is rationally connected. More precisely we know that \(H^{p,0}_{\overline{\partial}}(M)=\{0\}\) for each \(p>0\), see e.g. Prop. 1. Since \(M\) is compact and Kähler with \(h^{p,0}(M)=0\) we know that \(M\) is projective, see [18]. Finally, given that \(M\) is projective and rationally connected we know that it is simply connected, see [12]. ◻

Proof. Let us consider the Dolbeault complex \[0\rightarrow C^{\infty}(M)\stackrel{\overline{\partial}_{0,0}}{\rightarrow}\Omega^{0,1}(M)\stackrel{\overline{\partial}_{0,1}}{\rightarrow}...\stackrel{\overline{\partial}_{0,m-1}}{\rightarrow}\Omega^{0,m}(M)\rightarrow 0\] Then, by Cor. 1, we have \(H^{0,q}_{\overline{\partial}}(M)=\{0\}\) and consequently \[L_{\overline{\partial}}(f):=\sum_{q=0}^m(-1)^q\mathrm{Tr}\left(f^*: H^{0,q}_{\overline{\partial}}(M)\rightarrow H^{0,q}_{\overline{\partial}}(M)\right)=1.\] The conclusion now follows from the holomorphic Lefschetz fixed point theorem, see e.g. [19]. ◻

Proof. In order to prove this theorem it is enough to prove that if \(\pi_1(M)\) is infinite then \(\chi(M,\mathcal{O}_M)=0\). So, let us assume that \(\pi_1(M)\) is infinite. Let \(\pi:\tilde{M}\rightarrow M\) be the universal covering of \(M\) and let us consider the operator \[\gamma\tilde{\Delta}+\tilde{\frak{r}}_p:C^{\infty}_c(\tilde{M})\rightarrow C_c^{\infty}(\tilde{M})\] with \(\tilde{\frak{r}}_p:=\frak{r}_p\circ\pi\) and \(p=1,...,m\). Since \(\tilde{\Delta}:C^{\infty}_c(\tilde{M})\rightarrow C_c^{\infty}(\tilde{M})\) is essentially self-adjoint when viewed as an unbounded and densely defined operator acting on \(L^2(\tilde{M},\tilde{h})\) and given that \(\tilde{\frak{r}}_p\) is a bounded real-valued function on \(\tilde{M}\), we can conclude that \(\gamma\tilde{\Delta}+\tilde{\frak{r}}_p:C^{\infty}_c(\tilde{M})\rightarrow C_c^{\infty}(\tilde{M})\) is also essentially self-adjoint in \(L^2(\tilde{M},\tilde{h})\) for each \(1\leq p\leq m\). With a little abuse of notation let us denote with \[\label{lift} \gamma\tilde{\Delta}+\tilde{\frak{r}}_p:L^2(\tilde{M},\tilde{h})\rightarrow L^2(\tilde{M},\tilde{h})\tag{7}\] the \(L^2\)-closure of \(\gamma\tilde{\Delta}+\tilde{\frak{r}}_p:C^{\infty}_c(\tilde{M})\rightarrow C_c^{\infty}(\tilde{M})\). Since \(\tilde{\frak{r}}_p\) is bounded it is also easy to deduce that the domain of 7 equals the domain of the \(L^2\)-closure of the Laplace-Beltrami operator \(\tilde{\Delta}\) and that the corresponding graph norms are equivalent. Note that using the pushing-down technique, see [20], or alternatively using [21] or [22], we can conclude that 7 has still non-negative spectrum. Let us now denote with \(\tilde{\Delta}_{\overline{\partial},p,q}:\Omega^{p,q}(\tilde{M})\rightarrow \Omega^{p,q}(\tilde{M})\) the Hodge-Kodaira Laplacian on \(\tilde{M}\) and let \[\label{L2closure} \tilde{\Delta}_{\overline{\partial},p,q}:L^2\Omega^{p,q}(\tilde{M},\tilde{h})\rightarrow L^2\Omega^{p,q}(\tilde{M},\tilde{h})\tag{8}\] be the \(L^2\)-closure of \(\tilde{\Delta}_{\overline{\partial},p,q}:\Omega_c^{p,q}(\tilde{M})\rightarrow L^2\Omega^{p,q}(\tilde{M},\tilde{h})\). We denote with \(\mathcal{H}^{p,q}_{2,\overline{\partial}}(\tilde{M},\tilde{h})\) the kernel of 8 . Let now \(\psi\in \mathcal{H}^{p,0}_{2,\overline{\partial}}(\tilde{M},\tilde{h})\) and let \(W^{1,2}(\tilde{M},\Lambda^{p,0}(\tilde{M}))\) be the Sobolev space of \(L^2\) \((p,0)\)-forms \(\eta\) such that \(\nabla \eta\in L^2(\tilde{M},T^*\tilde{M}\otimes \Lambda^{p,0}(\tilde{M}), \tilde{h})\). Since \(M\) is compact and \((\tilde{M},\tilde{h})\) is the universal covering of \(M\) endowed with the pull-back metric, we know that \(\psi\in W^{1,2}(\tilde{M},\Lambda^{p,0}(\tilde{M}))\), see [23]. Consequently \(|\psi|_h\in W^{1,2}(\tilde{M},\tilde{h})\), see e.g. [24] where \(W^{1,2}(\tilde{M},\tilde{h})\) denotes the Sobolev space of \(L^2\) functions on \((\tilde{M},\tilde{h})\) with \(L^2\) gradient. Hence, using the refined Kato inequality [25], we obtain \[\begin{align} 0&=\langle 2\tilde{\Delta}_{\overline{\partial},p,0}\psi,\psi\rangle_{L^2\Omega^{p,0}(\tilde{M},\tilde{h})}\\ &=\int_{\tilde{M}}\left(|\tilde{\nabla} \psi|^2_{\tilde{h}}+\tilde{h}(\tilde{\mathrm{ric}}^{p,0}\psi,\psi)\right)\mathop{\mathrm{dvol}}_{\tilde{h}}\\ &\geq \int_{\tilde{M}}\left(2|\tilde{d}_0|\psi|_{\tilde{h}}|^2_{\tilde{h}}+\tilde{\frak{r}}_p|\psi|^2_{\tilde{h}}\right)\mathop{\mathrm{dvol}}_{\tilde{h}}\\ &\geq \int_{\tilde{M}}\left((2-\gamma)|\tilde{d}_0|\psi|_{\tilde{h}}|^2_{\tilde{h}}+\gamma|\tilde{d}_0|\psi|_{\tilde{h}}|^2_{\tilde{h}}+\tilde{\frak{r}}_p|\psi|^2_{\tilde{h}}\right)\mathop{\mathrm{dvol}}_{\tilde{h}}\\ &\geq \int_{\tilde{M}}(2-\gamma)|\tilde{d}_0|\psi|_{\tilde{h}}|^2_{\tilde{h}}\mathop{\mathrm{dvol}}_{\tilde{h}}\\ &\geq 0. \end{align}\] We can thus conclude that \(|\psi|_{\tilde{h}}\) is a constant function on \(\tilde{M}\) and since \(|\psi|_{\tilde{h}}\in L^2(\tilde{M},\tilde{h})\) and \((\tilde{M},\tilde{h})\) has infinite volume, we can deduce that \(\psi\equiv 0\). This shows that \(\mathcal{H}^{p,0}_{2,\overline{\partial}}(\tilde{M},\tilde{h})=\{0\}\) for each \(p=1,...,m\). Moreover, we also know that \(\mathcal{H}^{0,0}_{2,\overline{\partial}}(\tilde{M},\tilde{h})=\{0\}\), since \(\mathcal{H}^{0,0}_{2,\overline{\partial}}(\tilde{M},\tilde{h})\) coincides with the space of \(L^2\)-harmonic functions on \((\tilde{M},\tilde{h})\), which is trivial given that \((\tilde{M},\tilde{h})\) is complete and has infinite volume. Now, using the conjugation and the Hodge star operator, we obtain that \[\label{cucu} \mathcal{H}^{0,q}_{2,\overline{\partial}}(\tilde{M},\tilde{h})=\{0\}\tag{9}\] for each \(q=0,...,m\). According to the celebrated Atiyah’s \(L^2\)-index theorem, see [26], we know that \[\label{L2-A} \chi(M,\mathcal{O}_M)=\sum_{k=0}^m(-1)^k\dim_{\pi_1(M)}\left(\mathcal{H}^{0,q}_{2,\overline{\partial}}(\tilde{M},\tilde{h})\right).\tag{10}\] The above expression, that is \(\dim_{\pi_1(M)}\left(\mathcal{H}^{0,q}_{2,\overline{\partial}}(\tilde{M},\tilde{h})\right)\), denotes the Von Neumann dimension of \(\mathcal{H}^{0,q}_{2,\overline{\partial}}(\tilde{M},\tilde{h})\) with respect to \(\pi_1(M)\). For the purpose of this proof it is not necessary to recall the definition of \(\dim_{\pi_1(M)}\left(\mathcal{H}^{0,q}_{2,\overline{\partial}}(\tilde{M},\tilde{h})\right)\) for which we refer to [26]. It is only enough to keep in mind that \(\dim_{\pi_1(M)}\left(\mathcal{H}^{0,q}_{2,\overline{\partial}}(\tilde{M},\tilde{h})\right)\) is a non-negative real number such that \[\label{L2v} \dim_{\pi_1(M)}\left(\mathcal{H}^{0,q}_{2,\overline{\partial}}(\tilde{M},\tilde{h})\right)=0\quad \mathrm{if\;and\;only\;if}\quad \mathcal{H}^{0,q}_{2,\overline{\partial}}(\tilde{M},\tilde{h})=\{0\}.\tag{11}\] Finally, by 9 , 10 and 11 , we can conclude that \(\chi(M,\mathcal{O}_M)=0\), as required. ◻

Proof. Let \(\psi\) be a holomorphic \(p\)-form, with \(1\leq p\leq m\). Arguing as in the proof of Th. 6 we know that \(\psi\in W^{1,2}(M,\Lambda^{p,0}(M))\) and consequently \(|\psi|_h\in W^{1,2}(M,h)\). Applying the refined Kato inequality [25], we obtain \[\begin{align} 0&=\langle 2\Delta_{\overline{\partial},p,0}\psi,\psi\rangle_{L^2\Omega^{p,0}(M,h)}\\ &=\int_{M}\left(|\nabla \psi|^2_{h}+h(\mathrm{ric}^{p,0}\psi,\psi)\right)\mathop{\mathrm{dvol}}_{h}\\ &\geq \int_M\left(2|d_0|\psi|_{h}|^2_{h}+\frak{r}_p|\psi|^2_{h}\right)\mathop{\mathrm{dvol}}_{h}\\ &\geq \int_{M}\left((2-\gamma)|d_0|\psi|_{h}|^2_{h}+\gamma|d_0|\psi|_{h}|^2_{h}+\frak{r}_p|\psi|^2_{h}\right)\mathop{\mathrm{dvol}}_{h}\\ &\geq \int_{M}(2-\gamma)|d_0|\psi|_{h}|^2_{h}\mathop{\mathrm{dvol}}_{h}\\ &\geq 0. \end{align}\] We can thus conclude that \(|\psi|_{h}\) is a constant function on \(M\). Finally let \(\psi_1,...,\psi_{\ell}\) be \(\ell\)-holomorphic and linearly independent \(p\)-forms. If \(\ell>\frac{m!}{p!(m-p)!}\) then there exists a point \(x\in M\) and \(\mu_1,...,\mu_{\ell}\in \mathbb{C}\) such that \(\mu_1\psi_1(x)+...+\mu_{\ell}\psi_{\ell}(x)=0\) and \(\mu_j\neq 0\) for some \(j=1,...,\ell\). Let \(\psi:=\mu_1\psi_1+...+\mu_{\ell}\psi_{\ell}\). Since \(\psi\) is holomorphic we know that \(|\psi|_h\) is constant. Thus \(\psi(x)=0\) implies that \(\psi\equiv 0\). This contradicts the fact that \(\psi_1,...,\psi_{\ell}\) are linearly independent. We can thus conclude that \(\ell\leq \frac{m!}{p!(m-p)!}\), as required. ◻

Proof. Note that \(\gamma\Delta+mr_1:L^2(M,h)\rightarrow L^2(M,h)\) has entirely non-negative spectrum if and only if \(\frac{\gamma}{m}\Delta+r_1:L^2(M,h)\rightarrow L^2(M,h)\) has entirely non-negative spectrum. Since \(\frac{\gamma}{p}\geq \frac{\gamma}{m}\) we can deduce that if \(\frac{\gamma}{m}\Delta+r_1:L^2(M,h)\rightarrow L^2(M,h)\) has entirely non-negative spectrum then also the spectrum of \(\frac{\gamma}{p}\Delta+r_1:L^2(M,h)\rightarrow L^2(M,h)\) is entirely non-negative. This latter property amounts to saying that \(\gamma\Delta+pr_1:L^2(M,h)\rightarrow L^2(M,h)\) has entirely non-negative spectrum and finally we can conclude that also the spectrum of \(\gamma\Delta+\frak{r}_p:L^2(M,h)\rightarrow L^2(M,h)\) is entirely non-negative given that \(\frak{r}_p\geq pr_1\). The conclusion follows now from Th. 6. ◻

Proof. First part. We divided it in four steps.
First step. \(M\) is a real homology sphere.
Since we assumed that ?? has entirely positive spectrum we know that \(\lambda_{k,1}\), the smallest eigenvalue of ?? , is positive for each \(k=1,...,m-1\). Furthermore, we have already observed that the domain of ?? equals the domain of the \(L^2\)-closure of the Laplace-Beltrami operator \(\Delta\) and that the corresponding graph norms are equivalent. Therefore, from the min-max characterization of the eigenvalues we can deduce that \[\frac{m}{m-1}\|d_0\phi\|^2_{L^2\Omega^1(M,g)}+\langle w_{k,1}\phi,\phi\rangle_{L^2(M,g)}\geq \lambda_{k,1}\|\phi\|^2_{L^2(M,g)}\] for each \(\phi\in W^{1,2}(M,g)\), with \(W^{1,2}(M,g)\) denoting the Sobolev space of \(L^2\) functions on \((M,g)\) with \(L^2\) gradient. Let us now consider an arbitrarily fixed harmonic \(k\)-form \(\eta\) for some \(k\in \{1,...,m-1\}\). We know that \(|\eta|_g\in W^{1,2}(M,g)\), see e.g. [24]. In this way, keeping in mind the refined Kato inequality [29] or [25] and the min-max characterization of the eigenvalues, we have: \[\begin{align} 0&=\langle \Delta_k\eta,\eta\rangle_{L^2\Omega^k(M,g)}\\ &=\int_{M}\left(|\nabla \eta|^2_g+g(W_k\eta,\eta)\right)\mathop{\mathrm{dvol}}_{g}\\ &\geq \int_{M}\left(\frac{m}{m-1}|d_0|\eta|_g|^2_g+w_{k,1}|\eta|^2_g\right)\mathop{\mathrm{dvol}}_{g}\\ &\geq \lambda_{k,1}\|\eta\|^2_{L^2(M,g)}. \end{align}\] Since \(\lambda_{k,1}>0\) we can conclude that \(\eta\equiv0\) and thus \(H^k(M,\mathbb{R})=\{0\}\) for each \(k=1,...,m-1\). Moreover \(H^m(M,\mathbb{R})\cong H^1(M,\mathbb{R})\cong \mathbb{R}\) as \(M\) is compact and oriented. This shows that \(M\) is a real homology sphere and, using the fact that it is even dimensional, we can also conclude that \(\chi(M)=2\).
Second step: \(\mathcal{H}^{k}_{2}(\tilde{M},\tilde{g})=\{0\}\) for each \(k=1,...,m-1\).
Let \(\pi:\tilde{M}\rightarrow M\) be the universal covering of \(M\). Let us consider the operator \[\label{fvfv} \frac{m}{m-1}\tilde{\Delta}+\tilde{w}_{k,1}:C^{\infty}_c(\tilde{M})\rightarrow C_c^{\infty}(\tilde{M})\tag{15}\] with \(\tilde{w}_{k,1}:=w_{k,1}\circ \pi\), \(\tilde{\Delta}:C^{\infty}_c(\tilde{M})\rightarrow C_c^{\infty}(\tilde{M})\) the Laplace-Beltrami operator of \((\tilde{M},\tilde{g})\) and \(\tilde{g}=\pi^*g\). Arguing as in the proof of Th. 6 we know that 15 is essentially self-adjoint in \(L^2(\tilde{M},\tilde{g})\). Let us denote with \[\label{liftz} \frac{m}{m-1}\tilde{\Delta}+\tilde{w}_{k,1}:L^2(\tilde{M},\tilde{g})\rightarrow L^2(\tilde{M},\tilde{g})\tag{16}\] its \(L^2\)-closure. Since \(\tilde{w}_{k,1}\) is bounded, it is easy to deduce that the domain of 16 equals the domain of the \(L^2\)-closure of the Laplace-Beltrami operator \(\tilde{\Delta}\) and that the corresponding graph norms are equivalent. Furthermore using again the pushing-down technique, see [20], or alternatively using [21] or [22], we can conclude that 16 has entirely positive spectrum. In particular there exists \(\tilde{\lambda}_{k,1}\geq \lambda_{k,1}>0\) such that \[\frac{m}{m-1}\|\tilde{d}_0\phi\|^2_{L^2\Omega^1(\tilde{M},\tilde{g})}+\langle\tilde{w}_{k,1}\phi,\phi\rangle_{L^2(\tilde{M},\tilde{g})}\geq \tilde{\lambda}_{k,1}\|\phi\|^2_{L^2(\tilde{M},\tilde{g})}\] for each \(\phi\in W^{1,2}(\tilde{M},\tilde{g})\). Let us now denote with \(\tilde{\Delta}_{k}:\Omega^{k}(\tilde{M})\rightarrow \Omega^{k}(\tilde{M})\) the Hodge Laplacian on \(\tilde{M}\) and let \[\label{L2closureH} \tilde{\Delta}_{k}:L^2\Omega^k(\tilde{M},\tilde{g})\rightarrow L^2\Omega^{k}(\tilde{M},\tilde{g})\tag{17}\] be the \(L^2\)-closure of \(\tilde{\Delta}_{k}:\Omega_c^{k}(\tilde{M})\rightarrow L^2\Omega^{k}(\tilde{M},\tilde{g})\). We label with \(\mathcal{H}^{k}_{2}(\tilde{M},\tilde{g})\) the kernel of 17 . Let now \(\psi\in \mathcal{H}^{k}_{2}(\tilde{M},\tilde{g})\) with \(k\in \{1,...,m-1\}\). By arguing again as in the proof of Th. 6 we know that \(\psi\in W^{1,2}(\tilde{M},\Lambda^k(\tilde{M}))\) and \(|\psi|_{\tilde{g}}\in W^{1,2}(\tilde{M},\tilde{g})\). In this way, using again the Weitzenböck formula and the refined Kato inequality, we obtain \[\begin{align} 0&=\langle \tilde{\Delta}_k\psi,\psi\rangle_{L^2\Omega^{k}(\tilde{M},\tilde{g})}\\ &\geq \int_{M}\left(\frac{m}{m-1}|\tilde{d}_0|\psi|_{\tilde{g}}|^2_{\tilde{g}}+\tilde{w}_{k,1}|\psi|^2_{\tilde{g}}\right)\mathop{\mathrm{dvol}}_{\tilde{g}}\\ &\geq \tilde{\lambda}_{k,1}\|\psi\|^2_{L^2(\tilde{M},\tilde{g})}. \end{align}\] We can thus conclude that \(\mathcal{H}^{k}_{2}(\tilde{M},\tilde{g})=\{0\}\) for each \(k=1,...,m-1\).
Third step: \(\pi_1(M)\) is finite.
We prove this step by contradiction. Let us assume that \(\pi_1(M)\) is infinite and, as above, let \(\pi:\tilde{M}\rightarrow M\) be the universal covering of \(M\). We proved in the second step that \(\mathcal{H}^{k}_{2}(\tilde{M},\tilde{g})=\{0\}\) for each \(k=1,...,m-1\). Since \(\pi_1(M)\) is infinite we also know that \((\tilde{M},\tilde{g})\) is a complete Riemannian manifold of infinite volume. Therefore \(\mathcal{H}^{0}_{2}(\tilde{M},\tilde{g})=\{0\}\) and since \(\tilde{M}\) is oriented we can use the Hodge star operator to obtain \(\mathcal{H}^{m}_{2}(\tilde{M},\tilde{g})=\{0\}\). Summarizing \[\mathcal{H}^{k}_{2}(\tilde{M},\tilde{g})=\{0\}\] for each \(k=0,...,m\). Applying now Atiyah’s \(L^2\)-index theorem again, we have \[\label{L2} \chi(M)=\sum_{k=0}^m(-1)^k\dim_{\pi_1(M)}\left(\mathcal{H}^{k}_{2}(\tilde{M},\tilde{g})\right)=0\tag{18}\] and clearly this is in contradiction with the conclusion of the first step above, namely that \(\chi(M)=2\). We can thus conclude that \(\pi_1(M)\) is finite.
Forth step \(M\) is simply connected.
Once again let \(\pi:\tilde{M}\rightarrow M\) be the universal covering of \(M\). Then \((\tilde{M},\tilde{g})\) is a compact and oriented Riemannian manifold with \(\mathcal{H}^{k}_2(\tilde{M},\tilde{g})=\{0\}\) for each \(k=1,...,m-1\) and thus we have \(\chi(\tilde{M})=2\). If we now denote with \(\ell\) the cardinality of \(\pi_1(M)\) we obtain by means of the Chern-Gauss-Bonnet theorem: \[2=\chi(\tilde{M})=\int_{\tilde{M}}\mathrm{e}(\tilde{M})=\ell\int_{M}\mathrm{e}(M)=2\ell.\] We can thus conclude that \(\ell=1\) and so \(M\) is simply connected, as required. This completes the first part of the proof.
Second part.
Since \(M\) is not orientable there exists a connected, oriented double covering \(\pi:\hat{M}\rightarrow M\). Once again let us consider the operator \[\label{fvfvx} \frac{m}{m-1}\hat{\Delta}+\hat{w}_{k,1}:C^{\infty}(\hat{M})\rightarrow C^{\infty}(\hat{M})\tag{19}\] with \(\hat{w}_{k,1}:=w_{k,1}\circ \pi\), \(\hat{\Delta}:C^{\infty}_c(\hat{M})\rightarrow C_c^{\infty}(\hat{M})\) the Laplace-Beltrami operator of \((\hat{M},\hat{g})\) and \(\hat{g}=\pi^*g\). Arguing as in the proof of Th. 6 we know that 19 is essentially self-adjoint in \(L^2(\hat{M},\hat{g})\) and its unique \(L^2\)-closed extension, denoted here with \[\label{liftza} \frac{m+1}{m}\hat{\Delta}+\hat{w}_{k,1}:L^2(\hat{M},\hat{g})\rightarrow L^2(\hat{M},\hat{g})\tag{20}\] has entirely positive spectrum. We are therefore in the position to apply the first part of this theorem to conclude that \((\hat{M},\hat{g})\) is a simply connected real homology sphere. This in turn implies that \(\pi:\hat{M}\rightarrow M\) is the universal covering of \(M\) and so we can conclude that \(\pi_1(M)\cong \mathbb{Z}_2\). Let now \(\eta \in \Omega^k(M)\), \(k=\{1,...,m-1\}\) be a closed form. Then there exists \(\beta\in \Omega^{k-1}(\hat{M})\) such that \(d_{k-1}\beta=\pi^*\eta\). Let \(\gamma:\hat{M}\rightarrow \hat{M}\) be the non-trivial lift of the identity \(\mathrm{Id}:M\rightarrow M\). Then \(d_{k-1}(\gamma^*\beta)=\pi^*\eta\) and thus \(\frac{1}{2}(d_{k-1}(\beta+\gamma^*\beta))=\pi^*\eta\). Since \(\beta+\gamma^*\beta\) is invariant with respect to the action of the Deck transformations group of \(\hat{M}\), we know that there exists \(\alpha\in \Omega^{k-1}(M)\) such that \(\frac{1}{2}(\beta+\gamma^*\beta)=\pi^*\alpha\). We can thus conclude that \(\eta=d_{k-1}\alpha\), as required.
Third part. By repeating the same argument used in the first step of the first part we can conclude that \(H^k(M,\mathbb{R})=\{0\}\) for \(k=1,...,m-1\). On the other hand, given that \(m\) is odd, we know that \(\chi(M)=0\). This implies that \(H^m(M,\mathbb{R})\cong \mathbb{R}\) and so we can conclude that \(M\) is a real homology sphere. Finally \(M\) is orientable as \(H^m(M,\mathbb{R})\neq \{0\}\). ◻

Proof. Let \[L(f)=\sum_{k=0}^m(-1)^k\mathrm{Tr}\left(f^*: H^{k}_{dR}(M)\rightarrow H^{k}_{dR}(M)\right).\] According to the Lefschetz fixed point theorem if \(L(f)\neq 0\) then \(f\) must have some fixed point. In our setting the above formula boils down to \[L(f)=1\] if \(M\) is not orientable and to \[L(f)=1+\mathrm{deg}(f)\neq 0\] in the other case. We thus can conclude that \(\{x\in M : f(x)=x\}\neq \emptyset\), as desired ◻

Proof. It is enough to prove the case where both \(M\) and \(N\) are not orientable. So let \(\hat{\pi}_M:\hat{M}\rightarrow M\) and \(\hat{\pi}_N:\hat{N}\rightarrow N\) be the corresponding oriented double coverings. By the fact that \(\hat{M}\times \hat{N}\) is not a real homology sphere, namely \(H^m(\hat{M}\times \hat{N},\mathbb{R})\neq \{0\}\) and \(H^n(\hat{M}\times \hat{N},\mathbb{R})\neq \{0\}\), we can conclude that \(\hat{M}\times \hat{N}\) carries no Riemannian metric \(\hat{g}\) such that \[\frac{m+n}{m+n-1}\hat{\Delta}+w_{k,1}:L^2(\hat{M}\times \hat{N},\hat{g})\rightarrow L^2(\hat{M}\times \hat{N},\hat{g})\] has entirely positive spectrum for each \(k=1,...,m-1\). Finally, using the pushing-down technique, we can conclude that also \(M\times N\) carries no Riemannian metric \(g\) such that \(\frac{m+n}{m+n-1}\Delta+w_{k,1}:L^2(M\times N,g)\rightarrow L^2(M\times N,g)\) has entirely positive spectrum for each \(k=1,...,m-1\). ◻

Proof. We will be brief as the proof follows by adapting to this setting the proof of Th. 6. Let us first deal with the case when \(M\) is oriented. By contradiction let us assume that \(\pi_1(M)\) is infinite. Let \(\pi:\tilde{M}\rightarrow M\) be the universal covering of \(M\). As in the previous proof let us consider the operator \[\gamma\tilde{\Delta}+\tilde{w}_{k,1}:C^{\infty}_c(\tilde{M})\rightarrow C_c^{\infty}(\tilde{M})\] and, with a little abuse of notation, let \[\label{lift3} \gamma\tilde{\Delta}+\tilde{w}_{k,1}:L^2(\tilde{M},\tilde{g})\rightarrow L^2(\tilde{M},\tilde{g})\tag{21}\] its unique \(L^2\)-closed extension. Then by the pushing-down technique we know that 21 has still non-negative spectrum. Let now \(\psi\in \mathcal{H}^{k}_{2}(\tilde{M},\tilde{g})\), \(k\in \{1,...,m-1\}\). Since \(\psi\in W^{1,2}(\tilde{M}, \Lambda^{p,0}(\tilde{M}))\) and \(|\psi|_{\tilde{g}}\in W^{1,2}(\tilde{M},\tilde{h})\) we can use the Weitzenböck formula and the refined Kato inequality to obtain \[\begin{align} 0&=\langle \tilde{\Delta}_k\psi,\psi\rangle_{L^2\Omega^{k}(\tilde{M},\tilde{g})}\\ &\geq \int_{\tilde{M}}\left(\frac{m}{m-1}|\tilde{d}_0|\psi|_{\tilde{g}}|^2_{\tilde{g}}+\tilde{w}_{k,1}|\psi|^2_{\tilde{g}}\right)\mathop{\mathrm{dvol}}_{\tilde{g}}\\ &\geq \int_{\tilde{M}}\left(\frac{m}{m-1}-\gamma\right)|\tilde{d}_0|\psi|_{\tilde{g}}|^2_{\tilde{g}}\mathop{\mathrm{dvol}}_{\tilde{g}}\geq 0. \end{align}\] We can thus conclude that \(|\psi|_{\tilde{g}}\) is a constant function on \(\tilde{M}\) and since \(|\psi|_{\tilde{g}}\in L^2(\tilde{M},\tilde{g})\) and \((\tilde{M},\tilde{g})\) has infinite volume, we can deduce that \(\psi\equiv 0\). This shows that \(\mathcal{H}^{k}_{2}(\tilde{M},\tilde{g})=\{0\}\) for each \(k=1,...,m-1\). The vanishing of \(\mathcal{H}^0_2(\tilde{M},\tilde{g})\) follows by the fact that \((\tilde{M},\tilde{g})\) is complete with infinite volume while the vanishing of \(\mathcal{H}^m_2(\tilde{M},\tilde{g})\) follows because the Hodge star operator \(*\) induces an isomorphism between \(\mathcal{H}^0_2(\tilde{M},\tilde{g})\) and \(\mathcal{H}^m_2(\tilde{M},\tilde{g})\). Now by using Atiyah’s \(L^2\)-index theorem as in the proof of Th. 7, we deduce that \(\chi(M)=0\) which contradicts the assumption \(\chi(M)\neq 0\). We can thus conclude that \(\pi_1(M)\) is finite, as required. If \(M\) is not oriented let us consider a connected oriented double covering \(\pi:\hat{M}\rightarrow M\). We know that \(\chi(\hat{M})\neq 0\) and by the pushing-down technique we also know that \[\gamma\hat{\Delta}+\hat{w}_{k,1}:L^2(M,g)\rightarrow L^2(M,g)\] has entirely non-negative spectrum. We can argue now as in the first part of this proof to conclude that \(\pi_1(\hat{M})\) is finite and consequently we can conclude that also \(\pi_1(M)\) is finite. ◻

Proof. This follows immediately by Th. 8. ◻

Proof. Let us start by showing that if \(m\geq 4\) then \(M\) is compact. As in the previous section let us denote with \(r_1:M\rightarrow \mathbb{R}\) the function that to each \(x\in M\) assigns the smallest eigenvalue of \(\mathrm{ric}_{g,x}:T_xM\rightarrow T_xM\), with \(\mathrm{ric}_g\) the tangent bundle endomorphism such that \(\mathrm{Ric}_g(\cdot,\cdot)=g(\mathrm{ric}_g\cdot,\cdot)\). Thanks to [27] we have \[\begin{align} \left\langle \frac{4}{m-1}\Delta\phi+r_{1}\phi,\phi\right\rangle_{L^2(M,g)}&\geq \left\langle \frac{4}{m-1}\Delta\phi+(m-1)\sigma_1\phi,\phi\right\rangle_{L^2(M,g)}\\ &=(m-1)\left\langle \frac{4}{(m-1)^2}\Delta\phi+\sigma_1\phi,\phi\right\rangle_{L^2(M,g)} \end{align}\] Since ?? has entirely positive spectrum and \(\frac{1}{(m-1)^2}>\frac{1}{m^2-m}\) for each positive integer \(m\), we can conclude that \[\frac{4}{(m-1)^{2}}\Delta+\sigma_1:L^2(M,g)\rightarrow L^2(M,g)\] has entirely positive spectrum and thus that also \[\frac{4}{m-1}\Delta\phi+r_{1}:L^2(M,g)\rightarrow L^2(M,g)\] has entirely positive spectrum. We can thus conclude that \(M\) is compact thanks to [1]. Let us now continue by denoting with \(\mu\) the smallest eigenvalue of ?? . Clearly \(\mu>0\) as we assumed that ?? has entirely positive spectrum. We want to show that \[\label{entpos} \frac{m}{m-1}\Delta+w_{k,1}:L^2(M,g)\rightarrow L^2(M,g)\tag{22}\] has entirely positive spectrum for each \(k=1,...,m-1\). In order to accomplish this goal we prove that \[\left\langle \frac{m}{m-1}\Delta\phi+w_{k,1}\phi,\phi\right\rangle_{L^2(M,g)}\geq k(m-k)\mu\|\phi\|^2_{L^2(M,g)}\] for each \(\phi\in C^{\infty}(M)\) and \(k=1,...,m-1\). Indeed, thanks to [27], we have \[\begin{align} \left\langle \frac{m}{m-1}\Delta\phi+w_{k,1}\phi,\phi\right\rangle_{L^2(M,g)}&\geq \left\langle \frac{m}{m-1}\Delta\phi+k(m-k)\sigma_1\phi,\phi\right\rangle_{L^2(M,g)}\\ &=k(m-k)\left\langle \frac{m}{k(m-k)(m-1)}\Delta\phi+\sigma_1\phi,\phi\right\rangle_{L^2(M,g)}\\ & \geq k(m-k)\left\langle \frac{4}{m^2-m}\Delta\phi+\sigma_1\phi,\phi\right\rangle_{L^2(M,g)}\\ &\geq k(m-k)\mu\|\phi\|^2_{L^2(M,g)}. \end{align}\] Now that we established that the operator 22 has entirely positive spectrum for each \(k=1,...,m-1\), we can conclude by applying Th. 7. ◻

Proof. Since \(\gamma\Delta+\sigma_1:L^2(M,g)\rightarrow L^2(M,g)\) has entirely non-negative spectrum for some \(\gamma\in [0,\frac{4}{m^2-m})\) we also know that \[\gamma\tilde{\Delta}+\tilde{\sigma}_1:L^2(\tilde{M},\tilde{g})\rightarrow L^2(\tilde{M},\tilde{g})\] has entirely non-negative spectrum for the same \(\gamma\in [0,\frac{4}{m^2-m})\). Let now \(\psi\in \mathcal{H}_2^k(\tilde{M},\tilde{g})\) with \(1\leq k\leq m-1\). We have \[\begin{align} 0&=\langle \tilde{\Delta}_k\psi,\psi\rangle_{L^2\Omega^{k}(\tilde{M},\tilde{g})}\\ &\geq \int_{M}\left(\frac{m}{m-1}|\tilde{d}_0|\psi|_{\tilde{g}}|^2_{\tilde{g}}+\tilde{w}_{k,1}|\psi|^2_{\tilde{g}}\right)\mathop{\mathrm{dvol}}_{\tilde{g}}\\ &\geq k(m-k)\int_{M}\left(\frac{4}{m^2-m}|\tilde{d}_0|\psi|_{\tilde{g}}|^2_{\tilde{g}}+\tilde{\sigma}_1|\psi|^2_{\tilde{g}}\right)\mathop{\mathrm{dvol}}_{\tilde{g}}\\ &\geq k(m-k)\int_{M}\left(\frac{4}{m^2-m}-\gamma\right)|\tilde{d}_0|\psi|_{\tilde{g}}|^2_{\tilde{g}}\mathop{\mathrm{dvol}}_{\tilde{g}}\\ &\geq 0 \end{align}\] and so we can conclude that \(|\psi|_{\tilde{g}}\) is a constant function on \(\tilde{M}\). We can now repeat word by word the proof of Th. 8 to achieve the desired conclusion. ◻

Proof of Theorem 12. Building on structure theorems established in [14], either \(\mathop{\mathrm{rd}}(X)\geq m-k\), or the universal cover \((X_{\mathrm{univ}},g_{\mathrm{univ}})\) of \((X,g)\) is the product of a rationally connected manifold \((F,g_F)\) with semi-positive holomorphic sectional curvature and the universal cover \(Y_{\mathrm{univ}}\) of a compact Kähler manifold \((Y,g_Y)\) with semi-positive Ricci curvature satisfies the following diagram: \[\begin{tikzcd} F_{\mathrm{univ}} & X_{\mathrm{univ}}=Y_{\mathrm{univ}}\times F_{\mathrm{univ}} \arrow[l,"\mathrm{pr}_2"'] \arrow[r] \arrow[rd,"\mathrm{pr}_1"'] \arrow[rr,bend left=10,"\mu"] & Y_{\mathrm{univ}}\times F \arrow[r] \arrow[d] & (Y_{\mathrm{univ}}\times F)/\pi_1(Y)\cong X \arrow[d,"f"] \\ & & Y_{\mathrm{univ}} \arrow[r,"\pi"] & Y. \end{tikzcd}\] Let \(\eta\) be a holomorphic \((p,0)\)-form with \(p\geq k\). In the first case, any holomorphic \(p\)-form with \(p\geq k\) are identically zero by Corollary 8. In the second case, the pull-back \(\widetilde{\eta}\) of \(\eta\) on \(\widetilde{X}\) must be holomorphic forms on \(\widetilde{Y}\) because any holomorphic forms on \(F\) are identically zero. In particular, \(\eta\) is the pull-back of some holomorphic form on \(Y\), and so is parallel because \(\mathrm{Ric}(g_Y)\geq 0\). The proof is complete. ◻

Proof. Let \(\beta\) be some fixed big movable class. For any quotient torsion-free sheaf \(\mathcal{Q}\) of \(\Lambda^p(E)\), \[\mu_{\alpha+\epsilon \beta}(\mathcal{Q})=\mu_{\alpha}(\mathcal{Q})+\epsilon\mu_\beta(\mathcal{Q})\geq \mu_{\alpha,\min}(\mathcal{Q})+\epsilon \mu_{\beta,\min}(\mathcal{Q})>0,\] so \(\mu_{\alpha+\epsilon\beta,\min}(\Lambda^{p}E)>0\) for sufficiently small \(\epsilon\) and we may assume that \(\alpha\) is big.

Let \(g\) be a Gauduchon metric such that \(\alpha=[\omega_g^{m-1}]\) and \(H(t)\) be the solution of the Hermitian-Yang-Mills flow on \(E\) with respect to \(g\), then \(\Lambda^{p}H(t)\) is the solution of the Hermitian-Yang-Mills flow on \(\Lambda^{p}E\). It follows from [39] and [32] that the smallest \(p\) eigenvalue of \(\mathop{\mathrm{tr}}_gR^{\Lambda^{p}H(t)}\) increases to \(\mu_{\alpha,\min}(\Lambda^{p} E)\). Thus, there exists some Hermitian metric \(H\) on \(E\) such that \(\mathop{\mathrm{tr}}_gR^{\Lambda^{p}H}\) is positive definite and so \(\mathop{\mathrm{tr}}_g R^H\) is \((p-1)\)-positive. ◻

Proof. Since \[T_{kt}:=f_*\big([F_k]\cdot [F_t]\cdot [\omega_X]^{d-1}\big)\] is a closed \((1,1)\)-current supported in \(E_k\cap E_t\), it vanishes if \(E_k\ne E_t\). Hence \(M_{kt}=0\) unless \(E_k=E_t\). Therefore, to show the negativity of \((M_{kt})_{k,t\in K}\), it suffices to consider the case where \(f(F_k)=E_k=E\) is the same \(E\), for all \(k\in K\). There exist the index \(J\) and some \(a_j>0\) (\(j\in J\)) such that \(f^*E=\sum\limits_{j\in J} a_j F_j\) and \(K\subset J\). Set \[M_{ij}:=[F_i]\cdot [F_j]\cdot[\omega_X]^{d-1}\cdot [f^*\omega_Z]^{m-d-1}.\] Note that \[\begin{align} \sum\limits_{j\in J}a_j M_{ij}&=[F_i]\cdot [f^*E]\cdot[\omega_X]^{d-1}\cdot[f^*\omega_Z]^{m-d-1}\\ &=\int_{F_{i,\text{reg}}}(\omega_X|_{F_{i,reg}})^{d-1}\wedge (f|_{F_{i,\text{reg}}})^*(c_1(\mathcal{O}(E),h)\wedge \omega_Z^{m-d-1})\\ &=0 \end{align}\] for some Hermitian metric \(h\) of \(\mathcal{O}(E)\) because \((f|_{F_{k,\text{reg}}})^*(c_1(\mathcal{O}(E),h)\omega_Z^{m-d-1})=0\). Furthermore, for any \(i\neq j\), \([F_i]\cdot[F_j]=m_p[C_{ijp}]\) for some \(m_p\in\mathbb{N}^*\), where \(C_{ijp}\) is the irrducible component of \(F_i\cap F_j\), thus \(M_{ij}\geq 0\) for any \(i\neq j\) and \(M_{ij=0}\) if and only if \(\dim f(F_i\cap F_j)\leq m-d-2\). Then for any \(\{x_j\}\in \mathbb{R}^{|J|}\), we have \[-\sum\limits_{i,j\in J} x_i x_j M_{ij}=\sum\limits_{i\in J}a_i (\frac{x_i}{a_i})^2(\sum\limits_{j\in J}a_i M_{ij})-\sum\limits_{i,j\in J} x_i x_j M_{ij}=\frac{1}{2}\sum\limits_{i,j\in J, i\neq j}M_{ij}a_ia_j (\frac{x_i}{a_i}-\frac{x_j}{a_j})^2\geq 0\] and so \((M_{ij})_{i,j\in J}\) is a semi-negative matrix. We claim that the graph \(\Gamma_J\), whose vertices are \(J\) and whose edges are given by \(M_{ij}>0\), is connected, which shall complete the proof of (1). It follows that \[-\sum_{i,j\in J}x_ix_jM_{ij} = \frac{1}{2}\sum_{i\ne j}M_{ij}a_ia_j \left(\frac{x_i}{a_i}-\frac{x_j}{a_j}\right)^2\ge 0.\] Since \(\Gamma_J\) is connected, equality holds only when \(\frac{x_i}{a_i}\) is constant. Thus the kernel of \((M_{ij})_{i,j\in J}\) is generated by \((a_i)_{i\in J}\). Since \(K\subsetneq J\), the principal submatrix \((M_{kt})_{k,t\in K}\) is negative definite. Now let us focus on showing the claim. We argue by contradiction. Otherwise, there exist nonempty sets \(J_1\) and \(J_2\) such that \(J=J_1\sqcup J_2\) and \(M_{j_1j_2}=0\) for any \(j_1\in J_1,j_2\in J_2\). Let \(B=\bigcup\limits_{j_1\in J_1,j_2\in J_2}f(F_i\cap F_i)\). \(\dim B\leq m-d-2\) implies \(B\subsetneq E\). Fix \(z\in E\setminus B\), \(\left(F_{j_1}\cap X_z\right)\bigcap\left(F_{j_2}\cap X_z\right)= \emptyset\) for any \(j_1\in J_1,j_2\in J_2\), where \(X_z=f^{-1}(z)\). Then \[X_z= \left(\bigcup\limits_{j_1\in J_1}F_{j_1}\cap X_z\right)\bigsqcup \left(\bigcup_{j_2\in J_2}F_{j_2}\cap X_z\right),\] which contradicts the assumption that \(X_z\) is connected. The proof of (1) is complete.

Now let us show (2). We may assume that \(f_*([\omega_X]^d)=1\). Let \[\beta'=[\omega_X]^{d}\cdot f^*\beta+\sum\limits_{k\in K}c_k[F_k]\cdot [\omega_X]^{d-1}\cdot[f^*\omega_Z]^{m-d-1},\] where \(c_k\in \mathbb{R}\) are to be determined. One obviously has: \(f_*\beta'=\beta\) and \(\beta'\cdot [F_s]=0\) for any \(s\in S\setminus K\). Building on the statement (1), \((M_{kt})_{k,t\in K}\) is invertible and every coefficient of its inverse matrix must be positive. Then, we obtain the existence and non-negativity of the solutions \(c_k,k\in K\), of the equations: \[-(\sum\limits c_k[F_k]\cdot[\omega_X]^{d-1}\cdot[f^*\omega_Z]^{m-d-1})\cdot[F_t]=[\omega_X]^d\cdot[f^*\beta]\cdot[F_t], \forall t\in K.\] One can see that \(\beta'\cdot[F_k]=0\) for all \(k\in K\) by the construction, and \(F_s, s\in S\setminus K\) is \(f\)-exceptional divisor implies \(\beta'\cdot[F_s]=0\). It remains to check that \(\beta'\) is movable. For any pseudo-effective class \(\alpha\), there exists a unique divisorial Zariski decomposition \[\alpha=P(\alpha)+[N(\alpha)]\] such that \(P(\alpha)\) is modified nef and \(N(\alpha)=\sum\limits_{\text{prime divisor }D}v(\alpha,D)D\) (cf. [40]). For any effective prime divisor \(D\), we have \(\beta'\cdot D\ge 0\). Indeed, if \(D=F_s\) for some \(s\in S\), this follows from the construction above. If \(D\) is different from all \(F_k\), then every correction term has non-negative intersection with \(D\), while \([\omega_X]^d\cdot f^*\beta\cdot D\ge0\). Thus \(\beta'\cdot [N(\alpha)]\ge0\). For any \(\epsilon>0\), we can take a positive current \(T\) in \(P(\alpha)\) with analytic singularities in codimension at least \(2\) such that \(T\geq -\epsilon\omega_X\) (c.f. [40]). Then one can see that \[[F_k]\cdot[\omega_X]^{d-1}\cdot[f^*\omega_Z]^{m-d-1}\cdot P(\alpha)\geq -\epsilon\int_{F_{k,\text{reg}}} \omega_X^{d}\wedge f^*\omega_Z^{m-d-1}\] for every \(\epsilon>0\). We conclude that \[[F_k]\cdot[\omega_X]^{d-1}\cdot[f^*\omega_Z]^{m-d-1}\cdot\alpha=[F_k]\cdot[\omega_X]^{d-1}\cdot[f^*\omega_Z]^{m-d-1}\cdot P(\alpha)+[F_k]\cdot[\omega_X]^{d-1}\cdot[f^*\omega_Z]^{m-d-1}\cdot [N(\alpha)]\geq 0.\] Note that \([\omega_X]^d\cdot[f^*\beta]\) is movable, one has that \([\omega_X]^d\cdot[f^*\beta]\cdot\alpha\geq 0\) and thus \(\beta'\cdot\alpha\geq 0\). The proof is complete. ◻