[proofplan] Nakano positivity is positivity of the curvature Hermitian form on the full [tensor product](/page/Tensor%20Product) $T_x^{1,0}X \otimes E_x$, while Griffiths positivity tests the same form only on decomposable tensors $\xi \otimes v$. Since every decomposable tensor is a special case of a tensor, Nakano positivity immediately implies Griffiths positivity. When $E=L$ is a line bundle, every tensor in $T_x^{1,0}X \otimes L_x$ is decomposable, so the two tests become identical. [/proofplan]

[step:Write the two curvature tests in a local frame] Fix a point $x \in X$. Let $(U,z)$ be a holomorphic coordinate chart around $x$, with coordinates $z=(z_1,\dots,z_n)$, and let $(e_1,\dots,e_r)$ be a local holomorphic frame for $E$ over $U$. Write the Chern curvature tensor at $x$ in components as \begin{align*} \Theta_h(E)_x = \sum_{i,j=1}^n \sum_{\alpha,\beta=1}^r R_{i\bar j\alpha\bar\beta}(x)\, dz_i \wedge d\bar z_j \otimes e_\alpha^* \otimes e_\beta . \end{align*} For a tensor \begin{align*} \tau \in T_x^{1,0}X \otimes E_x, \qquad \tau = \sum_{i=1}^n \sum_{\alpha=1}^r \tau_i^\alpha \frac{\partial}{\partial z_i}\bigg|_x \otimes e_\alpha(x), \end{align*} the Nakano curvature form is \begin{align*} \mathcal N_x(\tau,\tau) = \sum_{i,j=1}^n \sum_{\alpha,\beta=1}^r R_{i\bar j\alpha\bar\beta}(x)\, \tau_i^\alpha \overline{\tau_j^\beta}. \end{align*} Nakano positivity means that $\mathcal N_x(\tau,\tau)>0$ for every $x \in X$ and every nonzero $\tau \in T_x^{1,0}X \otimes E_x$. For vectors \begin{align*} \xi = \sum_{i=1}^n \xi_i \frac{\partial}{\partial z_i}\bigg|_x \in T_x^{1,0}X, \qquad v = \sum_{\alpha=1}^r v_\alpha e_\alpha(x) \in E_x, \end{align*} the Griffiths curvature test is \begin{align*} \mathcal G_x(\xi,v) = \sum_{i,j=1}^n \sum_{\alpha,\beta=1}^r R_{i\bar j\alpha\bar\beta}(x)\, \xi_i \overline{\xi_j}\, v_\alpha \overline{v_\beta}. \end{align*} Griffiths positivity means that $\mathcal G_x(\xi,v)>0$ for every $x \in X$, every nonzero $\xi \in T_x^{1,0}X$, and every nonzero $v \in E_x$. [/step]

[step:Specialize Nakano positivity to decomposable tensors] Assume $(E,h)$ is Nakano positive. Fix $x \in X$, a nonzero vector $\xi \in T_x^{1,0}X$, and a nonzero vector $v \in E_x$. Define the decomposable tensor \begin{align*} \tau := \xi \otimes v \in T_x^{1,0}X \otimes E_x. \end{align*} Since $\xi \ne 0$ and $v \ne 0$, the tensor $\tau$ is nonzero. In the coordinate chart and frame from the previous step, $\tau$ has components \begin{align*} \tau_i^\alpha = \xi_i v_\alpha. \end{align*} Therefore \begin{align*} \mathcal N_x(\tau,\tau) &= \sum_{i,j=1}^n \sum_{\alpha,\beta=1}^r R_{i\bar j\alpha\bar\beta}(x)\, \tau_i^\alpha \overline{\tau_j^\beta} \\ &= \sum_{i,j=1}^n \sum_{\alpha,\beta=1}^r R_{i\bar j\alpha\bar\beta}(x)\, \xi_i v_\alpha \overline{\xi_j v_\beta} \\ &= \sum_{i,j=1}^n \sum_{\alpha,\beta=1}^r R_{i\bar j\alpha\bar\beta}(x)\, \xi_i \overline{\xi_j}\, v_\alpha \overline{v_\beta} \\ &= \mathcal G_x(\xi,v). \end{align*} Nakano positivity gives $\mathcal N_x(\tau,\tau)>0$, hence $\mathcal G_x(\xi,v)>0$. Since $x$, $\xi$, and $v$ were arbitrary, $(E,h)$ is Griffiths positive. [/step]

[step:Use the one-dimensional fiber to prove the converse for line bundles] Now suppose $E=L$ is a Hermitian holomorphic line bundle. The implication from Nakano positivity to Griffiths positivity has already been proved, so it remains to prove the converse. Assume $L$ is Griffiths positive. Fix $x \in X$, and let \begin{align*} 0 \ne \tau \in T_x^{1,0}X \otimes L_x. \end{align*} Choose a nonzero vector $e \in L_x$. Since $L_x$ is one-dimensional, there is a unique vector $\xi \in T_x^{1,0}X$ such that \begin{align*} \tau = \xi \otimes e. \end{align*} Because $\tau \ne 0$ and $e \ne 0$, we have $\xi \ne 0$. Applying Griffiths positivity to the nonzero pair $(\xi,e)$ gives \begin{align*} \mathcal G_x(\xi,e)>0. \end{align*} As in the decomposable-tensor computation above, \begin{align*} \mathcal N_x(\tau,\tau) = \mathcal N_x(\xi \otimes e,\xi \otimes e) = \mathcal G_x(\xi,e) > 0. \end{align*} Thus $\mathcal N_x(\tau,\tau)>0$ for every nonzero $\tau \in T_x^{1,0}X \otimes L_x$ and every $x \in X$. Hence $L$ is Nakano positive. Combining this converse with the first implication, Nakano positivity and Griffiths positivity agree for Hermitian holomorphic line bundles. [/step]