By the theorem: Let $X$ be a compact Hodge manifold. Then $X$ is a projective algebraic manifold, it follows that any compact complex manifold $X$ is projective algebraic iff it admits a positive line bundle $L\to X$.
I can see if it admits a positive line bundle $L\to X$, then $X$ is projective algebraic. However, I can't see the converse.
Here is what I tried: If $X$ is projective algebraic, (embedded into $P^n$), then recall there is a hyperplane section bundle $\pi:H\to P^n$, which is a positive line bundle. I am thinking we can use the pullback bundle regarding to the embedding $i:X\to P^n$ and $\pi:H\to P^n$ to show there is a positive line bundle over $X$. However, the fiber product of the fiber bundle is not necessary to be a manifold. I wonder if this way is correct now. Or are there other methods to prove the claim?
Best Answer
You're making this too difficult. The pullback of a bundle on $\Bbb P^n$ by an embedding $X\hookrightarrow\Bbb P^n$ is the same as the restriction of the bundle to $X$, and this is indeed a line bundle over $X$. (This is true, in fact, for an immersion, too.) [If you have an open cover $\{U_\alpha\}$ of $\Bbb P^n$ so that the bundle is trivial on each open set, then $\{U_\alpha\cap X\}$ gives an open cover of $X$ so that the bundle is trivial on each open set. Obviously, transition functions match up.]