MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $X^{(1)}$?

share|cite|improve this question
up vote 6 down vote accepted

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 $0$.

share|cite|improve this answer
Thank you very much Prof. Arapura – Universe Apr 18 '11 at 14:54
You're welcome. – Donu Arapura Apr 18 '11 at 15:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.