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This is false. There exists a smooth projective surface $X$ with a strictly nef divisor $D$ (so $D\cdot C>0$ for all curves $C\subset X$) and yet $(D^2)=0$, so in particular $D$ is not ample, see e.g. …

Take $X$ any smooth projective variety of dimension at least $2$, let $H$ be an ample (i.e. positive) line bundle on $X$, let $\pi:Y\to X$ be the blow-up of a point in $X$ and let $L=\pi^*H$. Then $L$ …

It is a consequence of the Khovanskii-Teissier inequality for Kähler classes (which was proved by Gromov and Demailly on Kähler manifolds, the algebraic case is also in Lazarsfeld's book): $$\int_X \ …

This is theorem 7.9 in the book of Kobayashi-Nomizu "Foundations of Differential Geometry Vol.II". There the authors attribute it to Hawley and Igusa independently. These are probably the first papers …

The case when $n$ is even has been famously ruled out by Yau as a consequence of his proof of the Calabi Conjecture, see his original paper or these notes. In fact, thanks to results of Novikov, you …

As you hinted yourself, they were indeed discovered by W.V.D. Hodge. They appear explicitly in his 1941 book "The Theory and Applications of Harmonic Integrals", which you can find here, see e.g. the …

There are no constant scalar curvature Kähler metrics on $M$, the blowup of $\mathbb{CP}^n$ at one point in any Kähler class and for any $n>1$. This is because of the Lichnerowicz-Matsushima obstruc …

EDIT: this answer refers to a previous version of the question. Already for $n=3$ the answer is no. Indeed, $h^{3,3}=1$ so by your condition $h^{1,1}=h^{2,2}=0$ but a compact Kähler manifold has $h^{ …

This is not literally the answer that the OP wanted (a reference to the literature, which I am not aware of), but following the comments above let me write down the simple gluing argument. Let $\wide …

This result follows from Corollaire (p.212) of this paper by F. Campana, Coréduction algébrique d'un espace analytique faiblement kählérien compact, Invent. Math. 63 (1981), no. 2, 187–223. I had the …

If $M$ is compact this is obviously false: if we can write $\phi=u\sqrt{-1}\partial\overline{\partial}u$ then $\int_M \omega\wedge\phi\leq 0$ integrating by parts. So for every $\phi$ which does not s …

If I remember correctly, you can find this in the book "Complex Differential Geometry" by Fangyang Zheng in Chapter 7. The analogous result "with real coefficients" (i.e. for real vectors and with $S^ …

The answer is yes, as proved by Y. Ogawa in this paper, see Theorem 3.10. Apparently equality on functions and $1$-forms is enough to conclude that the metric is Kähler. There is also a related paper …

These manifolds have actually been studied to some extent by Campana and Peternell in their series of papers "Towards a Mori theory on compact Kähler threefolds I, II, III". In those papers they menti …

This problem was solved by Barlet and Varouchas in this paper. The base $X'$ is assumed to be reduced (surely this is OK for you), and the fibers pure dimensional (also certainly acceptable).