Let $k$ be an algebraically closed field of char($k$)=p>0, $X$ a smooth projective variety over $k$, $F:X\rightarrow X^{(1)}$ the relative Frobenius morphism. Let $E$ be an ample vector bundle on $X$. Then Frobenius direct image $F\_\*(E)$ is also an ample vector bundle on $X^{(1)}$?



Ample bundle under Frobenius morphism
Definitely no. Take $X=\mathbb{P}^1$, $E=\mathcal{O}(1)$. Write 
$$F\_\*(E)= \bigoplus\_{i=1}^p \mathcal{O}(a\_i)$$
If $a\_i\ge 0$, for all $i$, then 
$$2=h^0(E)=h^0(F\_\*(E)) \ge \sum (a\_i+1) \ge p$$
so it's not even semipositive when $p>2$.


Note this works for any finite map $F:\mathbb{P}^1\to \mathbb{P}^1$,
so in particular for the $p$th power map in characteristic $0$.


