Let $X$ be a compact Kaehler manifold.
What is a good, possibly algebraic-geometric, way to think to Nakano semipositivity of holomorphic vector bundles on $X$?


Is the trivial line bundle $\mathcal{O}\_X$ Nakano semi-positive as a vector bundle?



Nakano semipositivity
I would say more properly that nowadays it is not known any satisfactory algebraic description or characterization of the concept of Nakano's positivity for a hermitian vector bundle.


I would like also to add some precisions to Sándor's answer.


First, the positivity in the sense of Nakano is a good notion to obtain vanishing theorems for vector bundles. For instance, we have the following result due to Nakano


**Theorem (Nakano, 1955).** Let $X$ be a compact connected Kähler manifold of dimension $n$ and $E\to X$ a hermitian vector bundle. Then


* if $E\ge\_{\text{Nak}}0$, strictly in one point, $H^{n,q}(X,E)=0$, for $q\ge 1$;
* if $E\le\_{\text{Nak}}0$, strictly in one point, $H^{p,0}(X,E)=0$, for $p<n$.


On the other hand, Nakano's positivity is not well-behaved with respect to taking duals: we have that $E$ is Griffiths positive if and only if $E^\*$ is Griffiths negative, but if we take $H\to\mathbb P^n$ to be the vector bundle of rank $n$ defined by
$$
0\to\mathcal O(-1)\to\underline{\mathbb C}^{n+1}\to H\to 0,
$$
where $\underline{\mathbb C}^{n+1}$ si the trivial vector bundle over $\mathbb P^n$ with fiber $\mathbb C^{n+1}$, then $H$ is Griffiths (semi)positive and $H^\*$ is Nakano (semi)negative but $H$ is neither Nakano (semi)positive nor Nakano (semi)negative.


If we look to short exact sequences of vector bundles
$$
0\to S\to E\to Q\to 0,
$$  

then the Nakano's negativity of $E$ implies the Nakano's negativity of $S$, but nothing can be said about the Nakano's positivity of $Q$ when $E$ is Nakano positive (the desired property holds if we look instead to Griffiths' positivity).


Of course Nakano's positivity implies the Griffiths' one; the following "partial converse" is due to Demailly and Skoda:


**Theorem (Demailly-Skoda, 1979).** For any hermitian vector bundle $E$, $E>\_{\text{Grif}}0$ implies $E\otimes\det E>\_{\text{Nak}}0$.


