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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 $L'$?

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    $\begingroup$ 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 Dec 3 '19 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 Dec 3 '19 at 18:10
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    $\begingroup$ @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... :) $\endgroup$ – Sándor Kovács Dec 4 '19 at 3:02
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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 $Y$.

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  • $\begingroup$ It is a cute reasoning, I love it. Thanks Mohan! $\endgroup$ – aglearner Dec 4 '19 at 8:09

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