[proofplan] We prove the two implications separately. Positivity gives high tensor powers with enough holomorphic sections to separate both points and tangent directions, by the Kodaira jet separation theorem, and this produces a holomorphic embedding by the complete linear system. Conversely, a very ample tensor power embeds $X$ into projective space; pulling back the Fubini--Study metric gives a positive Hermitian metric on that tensor power, and taking the tensor-root metric divides curvature by the positive integer $m$. The projectivity assertion is then exactly the embedding obtained from the positive line bundle. [/proofplan]

[step:Use positivity to separate points and tangent vectors by high tensor powers]Assume that $L$ is positive. Thus there is a smooth Hermitian metric $h$ on $L$ such that the real $(1,1)$-form \begin{align*} \omega_h := \frac{i}{2\pi}\Theta_h(L) \end{align*} is positive on $X$. We use the Kodaira jet separation theorem as an external input. Applied to the positive line bundle $(L,h)$ over the compact complex manifold $X$, whose hypotheses are exactly compactness of $X$ and positivity of $(L,h)$, it gives an integer $m_0\geq 1$ such that for every integer $m\geq m_0$, the holomorphic sections of $L^{\otimes m}$ separate points and tangent vectors. For each point $x\in X$, let $T_xX$ denote the holomorphic tangent space of $X$ at $x$. Concretely, for such $m$: 1. for any distinct points $x,y\in X$, there exists a section $s\in H^0(X,L^{\otimes m})$ such that $s(x)=0$ and $s(y)\neq 0$; 2. for every point $x\in X$ and every nonzero tangent vector $v\in T_xX$, there exists a section $s\in H^0(X,L^{\otimes m})$ such that $s(x)=0$ and $d(s/e)_x(v)\neq 0$ in any local holomorphic frame $e$ for $L^{\otimes m}$ near $x$. The second condition is independent of the chosen local holomorphic frame. Indeed, if $e'=ge$ for a nowhere-vanishing [holomorphic function](/page/Holomorphic%20Function) $g$, and $s=fe=f'e'$, then $f'=f/g$. At a point where $s(x)=0$, equivalently $f(x)=0$, one has \begin{align*} d f'_x(v) = d(f/g)_x(v) = \frac{1}{g(x)}\,df_x(v), \end{align*} so nonvanishing of the derivative is unchanged.[/step]

[guided]Assume that $L$ is positive. By definition this means that we are given a smooth Hermitian metric $h$ on the holomorphic line bundle $\pi:L\to X$ whose Chern curvature form is positive after the standard normalization \begin{align*} \omega_h := \frac{i}{2\pi}\Theta_h(L). \end{align*} The form $\omega_h$ is a positive real $(1,1)$-form on every tangent space of $X$. The key analytic input is the Kodaira jet separation theorem, used here as an external result. Its hypotheses are satisfied here: $X$ is compact and complex, and $L$ is positive with respect to the metric $h$. Therefore there is an integer $m_0\geq 1$ such that, for every integer $m\geq m_0$, the finite-dimensional complex [vector space](/page/Vector%20Space) $H^0(X,L^{\otimes m})$ of global holomorphic sections separates both points and tangent vectors. For each point $x\in X$, let $T_xX$ denote the holomorphic tangent space of $X$ at $x$. Let us spell out the two separation properties because they are exactly what is needed for an embedding. First, if $x,y\in X$ are distinct points, there is a section $s\in H^0(X,L^{\otimes m})$ such that $s(x)=0$ and $s(y)\neq 0$. This will force the projective map defined by all sections to send $x$ and $y$ to different points. Second, if $x\in X$ and $v\in T_xX$ is nonzero, then there is a section $s\in H^0(X,L^{\otimes m})$ with $s(x)=0$ and with nonzero first derivative in the direction $v$ after trivializing the line bundle locally. We also verify that this tangent-vector condition is intrinsic. Let $e$ be a local holomorphic frame for $L^{\otimes m}$ near $x$, and write $s=fe$ for a holomorphic function $f$. If $e'=ge$ is another local holomorphic frame, where $g$ is nowhere vanishing and holomorphic, then $s=f'e'$ with $f'=f/g$. Since $s(x)=0$, we have $f(x)=0$, and therefore \begin{align*} d f'_x(v) = d(f/g)_x(v) = \frac{1}{g(x)}\,df_x(v). \end{align*} Because $g(x)\neq 0$, the condition $df_x(v)\neq 0$ is equivalent to $df'_x(v)\neq 0$. Thus tangent-vector separation is well defined.[/guided]

[step:Construct the complete linear system map and prove it is an embedding]Fix an integer $m\geq m_0$. Let \begin{align*} V_m := H^0(X,L^{\otimes m}) \end{align*} be the finite-dimensional complex vector space of global holomorphic sections. Since the sections of $L^{\otimes m}$ separate points, they have no common zero. If $X$ has at least two points, then for each $x\in X$ we choose $y\in X$ with $y\neq x$; point separation, with the ordered pair chosen appropriately, gives a section nonzero at $x$. If $X$ has one point, any nonzero element of the one-dimensional fiber is represented by a global section, so the same conclusion holds. For each $x\in X$, define the nonzero evaluation map \begin{align*} \operatorname{ev}_x:V_m&\to (L^{\otimes m})_x \\ s&\mapsto s(x). \end{align*} After choosing a nonzero element of the one-dimensional vector space $(L^{\otimes m})_x$, the map $\operatorname{ev}_x$ becomes a nonzero linear functional on $V_m$, well defined up to multiplication by a nonzero scalar. Hence it determines a point of $\mathbb{P}(V_m^*)$. Define the complete linear system map by \begin{align*} \Phi_m:X&\to \mathbb{P}(V_m^*) \\ x&\mapsto [\operatorname{ev}_x]. \end{align*} Equivalently, in a local holomorphic frame $e$ for $L^{\otimes m}$ with $s_j=f_j e$, by \begin{align*} \Phi_m(x)=[f_0(x):\cdots:f_N(x)] \end{align*} for any basis $s_0,\dots,s_N$ of $V_m$. This local expression is independent of the frame because changing $e$ multiplies all $f_j$ by the same nowhere-zero holomorphic function. Point separation implies that $\Phi_m$ is injective. Tangent-vector separation implies that $d(\Phi_m)_x:T_xX\to T_{\Phi_m(x)}\mathbb{P}(V_m^*)$ is injective for every $x\in X$. Hence $\Phi_m$ is a holomorphic injective immersion. Since $X$ is compact and $\mathbb{P}(V_m^*)$ is Hausdorff, $\Phi_m$ is a homeomorphism from $X$ onto its image. Therefore $\Phi_m$ is a holomorphic embedding. It follows that $L^{\otimes m}$ is very ample. Hence $L$ is ample.[/step]

[guided]Fix an integer $m\geq m_0$, and define \begin{align*} V_m := H^0(X,L^{\otimes m}). \end{align*} This is a finite-dimensional complex vector space because $X$ is compact and $L^{\otimes m}$ is a holomorphic line bundle over $X$. Before defining the projective map, we need the sections to have no common zero. Let $x\in X$. If $X$ has at least two points, choose $y\in X$ with $y\neq x$. Point separation gives a section distinguishing $x$ and $y$; applying the condition in the appropriate order gives a section that is nonzero at $x$. If $X$ has one point, any nonzero section of the one-dimensional fiber gives the same conclusion. Thus the evaluation map at every point is nonzero, so the complete linear system defines a map to projective space. Let \begin{align*} \Phi_m:X&\to \mathbb{P}(V_m^*) \end{align*} be the complete linear system map. In local coordinates this is easiest to read as follows. Choose a basis $s_0,\dots,s_N$ of $V_m$, and choose a local holomorphic frame $e$ for $L^{\otimes m}$ on an [open set](/page/Open%20Set) $U\subset X$. Write \begin{align*} s_j=f_j e,\qquad f_j:U\to \mathbb{C}\ \text{holomorphic},\qquad 0\leq j\leq N. \end{align*} Then \begin{align*} \Phi_m(x)=[f_0(x):\cdots:f_N(x)]. \end{align*} This formula is independent of the chosen frame because replacing $e$ by $e'=ge$, with $g:U\to\mathbb{C}$ nowhere zero and holomorphic, replaces every $f_j$ by $f_j/g$; projective coordinates are unchanged after multiplying every coordinate by the same nonzero scalar. Point separation now proves injectivity. If $x\neq y$, choose a section $s\in V_m$ such that $s(x)=0$ and $s(y)\neq 0$. Evaluation at $x$ and evaluation at $y$ are therefore not proportional linear functionals on $V_m$, so $\Phi_m(x)\neq \Phi_m(y)$. Tangent-vector separation proves that $\Phi_m$ is an immersion. Let $x\in X$ and let $v\in T_xX$ be nonzero. Choose $s\in V_m$ such that $s(x)=0$ and $d(s/e)_x(v)\neq 0$ in a local frame $e$. Extend $s$ to a basis $s_0,\dots,s_N$ of $V_m$ with $s=s_1$, and choose $s_0$ so that $s_0(x)\neq 0$. On the affine chart where the first projective coordinate is nonzero, the map $\Phi_m$ has local coordinates \begin{align*} x\mapsto \left(\frac{f_1(x)}{f_0(x)},\dots,\frac{f_N(x)}{f_0(x)}\right). \end{align*} Since $f_1(x)=0$ and $f_0(x)\neq 0$, the derivative of the first affine coordinate in the direction $v$ is \begin{align*} d\left(\frac{f_1}{f_0}\right)_x(v) = \frac{1}{f_0(x)}\,df_1{}_x(v), \end{align*} which is nonzero. Hence $d(\Phi_m)_x(v)\neq 0$. Since $v$ was arbitrary and nonzero, $d(\Phi_m)_x$ is injective. Thus $\Phi_m$ is a holomorphic injective immersion. Because $X$ is compact and projective space is Hausdorff, the continuous map $\Phi_m$ is a homeomorphism from $X$ onto its image. Therefore $\Phi_m$ is a holomorphic embedding. By definition, this says that $L^{\otimes m}$ is very ample. Hence $L$ is ample.[/guided]

[step:Pull back the Fubini--Study metric from a very ample tensor power]Conversely, assume that $L$ is ample. Then there exists an integer $m\geq 1$ such that $L^{\otimes m}$ is very ample. Let $N\in \mathbb{N}\cup\{0\}$ be the projective dimension determined by a basis of $H^0(X,L^{\otimes m})$, and let \begin{align*} \Phi:X&\to \mathbb{P}^N \end{align*} be the associated holomorphic embedding. By very ampleness, there is a holomorphic line bundle isomorphism \begin{align*} \alpha:L^{\otimes m}&\to \Phi^*\mathcal{O}_{\mathbb{P}^N}(1). \end{align*} Let $h_{\mathrm{FS}}$ denote the Fubini--Study Hermitian metric on $\mathcal{O}_{\mathbb{P}^N}(1)$. Its curvature form satisfies \begin{align*} \frac{i}{2\pi}\Theta_{h_{\mathrm{FS}}}(\mathcal{O}_{\mathbb{P}^N}(1))=\omega_{\mathrm{FS}}, \end{align*} where $\omega_{\mathrm{FS}}$ is the positive Fubini--Study $(1,1)$-form on $\mathbb{P}^N$. Pull back this metric and transport it through $\alpha$ to obtain a smooth Hermitian metric $h_m$ on $L^{\otimes m}$. Its curvature satisfies \begin{align*} \frac{i}{2\pi}\Theta_{h_m}(L^{\otimes m}) = \Phi^*\omega_{\mathrm{FS}}. \end{align*} Since $\Phi$ is a holomorphic embedding and $\omega_{\mathrm{FS}}$ is positive, $\Phi^*\omega_{\mathrm{FS}}$ is a positive real $(1,1)$-form on $X$.[/step]

[guided]Assume conversely that $L$ is ample. By definition, there is an integer $m\geq 1$ such that the tensor power $L^{\otimes m}$ is very ample. Very ampleness gives a holomorphic embedding \begin{align*} \Phi:X&\to \mathbb{P}^N \end{align*} for some integer $N\geq 0$, constructed from a basis of the section space $H^0(X,L^{\otimes m})$. It also gives the natural holomorphic line bundle isomorphism \begin{align*} \alpha:L^{\otimes m}&\to \Phi^*\mathcal{O}_{\mathbb{P}^N}(1). \end{align*} This is the precise meaning of saying that the embedding is given by the complete linear system of $L^{\otimes m}$. Now equip $\mathcal{O}_{\mathbb{P}^N}(1)$ with its Fubini--Study Hermitian metric $h_{\mathrm{FS}}$. The defining curvature property of this metric is \begin{align*} \frac{i}{2\pi}\Theta_{h_{\mathrm{FS}}}(\mathcal{O}_{\mathbb{P}^N}(1))=\omega_{\mathrm{FS}}, \end{align*} where $\omega_{\mathrm{FS}}$ is the positive Fubini--Study $(1,1)$-form on projective space. Pull back $h_{\mathrm{FS}}$ by $\Phi$ to a Hermitian metric on $\Phi^*\mathcal{O}_{\mathbb{P}^N}(1)$, and then use the isomorphism $\alpha$ to transport this metric to $L^{\otimes m}$. Denote the resulting smooth Hermitian metric on $L^{\otimes m}$ by $h_m$. Curvature is functorial under holomorphic pullback, so \begin{align*} \frac{i}{2\pi}\Theta_{h_m}(L^{\otimes m}) = \Phi^*\omega_{\mathrm{FS}}. \end{align*} Because $\Phi$ is an embedding, $d\Phi_x:T_xX\to T_{\Phi(x)}\mathbb{P}^N$ is injective for every $x\in X$. Since $\omega_{\mathrm{FS}}$ is positive on projective space, its pullback is positive on every nonzero tangent vector of $X$. Thus $h_m$ is a positive Hermitian metric on $L^{\otimes m}$.[/guided]

[step:Take the tensor-root metric and divide the curvature by $m$]Choose a smooth Hermitian metric $h$ on $L$ such that its induced tensor-power metric on $L^{\otimes m}$ is $h_m$. Locally, if $e$ is a nowhere-vanishing holomorphic frame for $L$ and $h_m(e^{\otimes m},e^{\otimes m})=\rho^m$ for a positive smooth function $\rho$, define \begin{align*} h(e,e):=\rho. \end{align*} This is independent of the frame because transition functions transform both sides by the same absolute-square factor. For Hermitian line bundles, the Chern curvature of a tensor power satisfies \begin{align*} \Theta_{h_m}(L^{\otimes m})=m\,\Theta_h(L). \end{align*} Therefore \begin{align*} \frac{i}{2\pi}\Theta_h(L) = \frac{1}{m}\,\frac{i}{2\pi}\Theta_{h_m}(L^{\otimes m}) = \frac{1}{m}\Phi^*\omega_{\mathrm{FS}}. \end{align*} Since $m\geq 1$ and $\Phi^*\omega_{\mathrm{FS}}$ is positive, $\frac{i}{2\pi}\Theta_h(L)$ is positive. Hence $L$ is positive.[/step]

[guided]We have produced a positive metric $h_m$ on the tensor power $L^{\otimes m}$. To get a metric on $L$ itself, we take the fiberwise $m$-th root. Let $e$ be a nowhere-vanishing local holomorphic frame for $L$ on an open set $U\subset X$. Then $e^{\otimes m}$ is a nowhere-vanishing local holomorphic frame for $L^{\otimes m}$ on $U$. Since $h_m$ is a Hermitian metric, there is a positive smooth function $\rho:U\to (0,\infty)$ such that \begin{align*} h_m(e^{\otimes m},e^{\otimes m})=\rho^m. \end{align*} Define the local metric $h$ on $L$ by \begin{align*} h(e,e):=\rho. \end{align*} On overlaps, if $e'=ge$ for a nowhere-vanishing holomorphic transition function $g$, then \begin{align*} (e')^{\otimes m}=g^m e^{\otimes m}. \end{align*} The metric $h_m$ transforms by multiplication with $|g|^{2m}$, so its positive $m$-th root transforms by multiplication with $|g|^2$. This is exactly the transformation rule for a Hermitian metric on $L$. Hence these local definitions glue to a smooth Hermitian metric $h$ on $L$, and the induced metric on $L^{\otimes m}$ is $h_m$. Now we compare curvatures. For a Hermitian line bundle, tensor powers add Chern curvature, so the metric induced by $h$ on $L^{\otimes m}$ satisfies \begin{align*} \Theta_{h_m}(L^{\otimes m})=m\,\Theta_h(L). \end{align*} Equivalently, \begin{align*} \frac{i}{2\pi}\Theta_h(L) = \frac{1}{m}\,\frac{i}{2\pi}\Theta_{h_m}(L^{\otimes m}) = \frac{1}{m}\Phi^*\omega_{\mathrm{FS}}. \end{align*} The form $\Phi^*\omega_{\mathrm{FS}}$ is positive, and multiplying a positive $(1,1)$-form by the positive scalar $1/m$ preserves positivity. Therefore $\frac{i}{2\pi}\Theta_h(L)$ is positive, so $L$ is positive.[/guided]

[step:Conclude projectivity from the embedding] If $X$ carries a positive holomorphic line bundle $L$, the first implication gives an integer $m\geq 1$ such that $L^{\otimes m}$ is very ample. The complete linear system of $L^{\otimes m}$ then gives a holomorphic embedding \begin{align*} \Phi_m:X&\to \mathbb{P}(H^0(X,L^{\otimes m})^*). \end{align*} Thus $X$ is biholomorphic to a closed complex submanifold of projective space. Hence $X$ is projective. This completes the proof. [/step]