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A line bundle is ample if some power of it is very ample. A line bundle is positive if the chern class in $H^2(X,\mathbb{Z})$ is represented by a Kahler metric in $H^{1,1}(X,\mathbb{Z})$.(Regarded as elements in $H^{1,1}(X,\mathbb{C})$ through the map $H^2(X,\mathbb{Z})\to H^2(X,\mathbb{C})$)

Are ample and positive the same concept for projective algebraic manifolds?

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Yes. For ample implies positive, use the fact that $c_1(O(1))$ on projective space is the Kähler form of the Fubini-Study metric, and then restrict to $X$. For the converse, you need the Kodaira embedding theorem (in fact, this is more or less the content of that theorem).

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Well more precisely ample may be defined in more general situation than positive, no? For example projective nonsingular algebraic varieties can be defined in any characteristic, and in fact nonsingularity is not essential, etc –  მამუკა ჯიბლაძე May 18 at 7:19
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Yes, sure. Ampleness and positivity can both be defined in situations where the other can't. But they coincide for line bundles on a smooth complex projective variety. I assume that's what the OP meant. –  Donu Arapura May 18 at 11:45

A line bundle $L$ on a complex manifold is positive if there is an Hermitian metric $h$ on $L$ such that $\frac{1}{2\pi}\Theta(L,h)$ is a Kahler form. By Kodaira's embedding theorem if $L$ is a positive line bundle on a compact Kahler manifold the $L^{\otimes m}$ induces an embedding $f:X\rightarrow\mathbb{P}^n$ for some $m>0$ and $L^{\otimes m} = f^*\mathcal{O}_{\mathbb{P}^n}(1)$. Therefore $L$ is ample. Conversely if such an embedding exists the pullback of the Fubini-Study metric on $*\mathcal{O}_{\mathbb{P}^n}(1)$ determine a positive metric on $L^{\otimes m}$ and hence on $L$.

Similarly, if $L$ has an Hermitian metric such that $c_1(L) = \frac{i}{2\pi}\Theta(L,h)$ is non-negative then $L$ is nef. On the other hand it could happen that a nef bundle does not have such a metric, see for instance Example $1.7$ in this paper http://www-fourier.ujf-grenoble.fr/~demailly/manuscripts/dps1.pdf.

If $\omega$ is a Kahler form on $X$, then $L$ is nef if and only if for any $\epsilon >0%$ there is an Hermitian metric $h_\epsilon$ on $L$ such that $\frac{i}{2\pi}\Theta(L,h_\epsilon)>-\epsilon \omega$. Note that a compact Kahler manifold may not contain any curve. However, in this way one can define nefness on any compact Kahler manifold.

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