Let $X$ be a smooth complex projective variety and $p:Y\to X$ be a locally trivial in analytic topology $\mathbb CP^k$-bundle. Suppose we have a line bundle $L$ on $Y$, restricting to $\mathcal O(1)$ on $\mathbb CP^k$-fibres.

Question. Is it true that there is a line bundle $L'$ on $X$ such that $p^*L'\otimes L$ is very ample on $Y$?

I think one could be able to prove this using Serre's (+ Kodaira?) vanishing, but I can't prove it so far.

Idea. Here is an idea of how one could try to solve this. So, first all, one can take $L''$ such that $p^*L''\otimes L$ is ample on $Y$. Next, one can try to use Kodaira vanishing, it says that $K_Y\otimes p^*L''\otimes L$ has zero higher cohomology. Now, we could take the line bundle $K_Y\otimes (p^*L''\otimes L)^{k+2}$, and this bundle will restrict to each fibre as $\mathcal O(1)$. By taking $L''$ positive enough, it should be possible to make $K_Y\otimes (p^*L''\otimes L)^{k+2}$ ample. I think that since this bundle is ample and its higher cohomology vanish, by Grothendiek-Riemann-Roch it will have a lot of sections (especially if $L''$ is very ample and has a lot of sections). This seems to be not far from proving that $K_Y\otimes (p^*L''\otimes L)^{k+2}$ is very ample... And I guess this bundle is $p^*L'\otimes L$ for some $L'$?

  • 1
    $\begingroup$ If I understand you correctly, you are assuming $Y$ is algebraic and $p$ is a morphism in the category of algebraic varieties. Then, it is fairly easy to reduce to the case when $Y=\mathbb{}^k\times X$ and then the assertion is trivial. $\endgroup$
    – Mohan
    Commented Dec 3, 2019 at 18:02
  • $\begingroup$ Dear Mohan, thanks for this comment! You understand me correctly. But I am not familiar with the notation $Y=^k\times X$. Do you think you could write a complete answer? If you do, would you be so kind, please, to write an answer as an answer and not as a comment? $\endgroup$
    – aglearner
    Commented Dec 3, 2019 at 18:10
  • 2
    $\begingroup$ @aglearner: I suppose Mohan meant to write $\mathbb CP^k\times X$, but possibly used his own macro for $\mathbb CP$ which the system did not recognize, and hence ignored... :) $\endgroup$ Commented Dec 4, 2019 at 3:02

1 Answer 1


Since $R^ip_*L=0$ for $i>0$, by semi-continuity theorem, you see that $p_*L$ is a vector bundle of rank $k+1$. You can twist by a sufficiently ample bundle $L'$ on $X$ to make it globally generated. Thus, you have $O_X^m\to p_*L\otimes L'$ surjective and thus you get an embedding $Y\subset X\times \mathbb{P}^{m-1}$. Further, the $O(1)$ of the second factor restricts to $L\otimes p^*L'$ on $Y$. Now, taking a very ample line bundle $M$ on $X$, $M\otimes O(1)$ is very ample on $X\times\mathbb{P}^{m-1}$ and then its restriction $L\otimes p^*(L'\otimes M)$ is very ample on $Y$.

  • $\begingroup$ It is a cute reasoning, I love it. Thanks Mohan! $\endgroup$
    – aglearner
    Commented Dec 4, 2019 at 8:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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