Let $E$, $X$ be irreducible smooth algebraic varieties over the complex numbers and let $p \colon E \to X$ be a morphism which is locally trivial with respect to the Zariski topology. Since $p$ is a vector bundle and the varieties are smooth, we get a natural isomorphism $p^\ast \colon\textrm{Pic}(X) \to \textrm{Pic}(E)$ by the pull-back of divisors, see e.g. Chp. 3 in [Fu84].

Is it true that $p^\ast \omega_X = \omega_E$ where $\omega_X$ and $\omega_E$ denote the canonical divisors of $X$ and $E$ respectively?

Any proof, counter-example or textbook reference would be perfect.

[Fu84] W. Fulton. Intersection theory, Springer-Verlag, Berlin, 1984.

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    $\begingroup$ Take a vector bundle $E$ of rank r+1 on $Y$. We have $K_X=\mathscr O_X(-r-1)\otimes \pi^\ast \det E^\vee\otimes \pi^\ast K_Y$ where $\pi:X=\mathbb P(E)\to Y$ Here is your answer with details :math.stackexchange.com/questions/1559927/… $\endgroup$
    – user21574
    Commented Oct 17, 2017 at 7:58
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    $\begingroup$ Philosophically, the condition would mean that the fibers have trivial canonical bundle. That can be turned into a computation like Jolany describes to show that unless the fibers have trivial canonical bundle, this can't happen $\endgroup$
    – meh
    Commented Oct 17, 2017 at 16:26
  • $\begingroup$ @ aginensky , I gave a related answer, to canonical divisor of projective bundle of $E$ . Th OP asking for canonical divisor of $E$. $\endgroup$
    – user21574
    Commented Oct 17, 2017 at 17:19
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    $\begingroup$ I think the confusion between the commenters comes about because the question is written in a very confusing way. First you say "let $p: E \rightarrow X$ be a morphism which is locally trivial with respect to the Zariski topology". OK, so now I am thinking about a projective bundle, or maybe an isotrivial family of elliptic curves, or... But then in the next sentence you say "Since $p$ is a vector bundle..." Where did that come from? If you really just want to ask about vector bundles, please make that clear. $\endgroup$
    – Pooter
    Commented Oct 18, 2017 at 8:34
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    $\begingroup$ I'm very sorry, yes, I wrote the question in a confusing way. I meant that $p \colon E \to X$ is a vector bundle. $\endgroup$
    – Anonymous
    Commented Oct 18, 2017 at 8:42

1 Answer 1


I'm sorry for asking the question. There is an easy counter-example. Let $\mathbb{F}_1$ be the first Hirzebruch surface and let $\pi \colon \mathbb{F}_1 \to \mathbb{P}^1$ be a projection which makes $\mathbb{F}_1$ a $\mathbb{P}^1$-bundle over $\mathbb{P}^1$. Let $S \subset \mathbb{F}_1$ be the unique section of $\pi$ which has self-intersection $-1$. Then $\pi$ restricts to a vector bundle $p \colon E \to \mathbb{P}^1$ where $E = \mathbb{F}_1 \setminus S$. The canonical divisor of $\mathbb{P}^1$ is $-2 P$ where $P$ is some point on $\mathbb{P}^1$. On the other hand, $\mathbb{F}^1$ is the blow-up of $\mathbb{P}^2$ in one point and thus the canonical divisor of $E$ is given by $-3 F$ where $F$ is a fiber of $p$. But the pull-back of $-2P$ on $E$ is $-2 F$.

  • $\begingroup$ Here is an even easier counter-example: Any such $p:E\to X$ where the canonical bundle of the fibers is not trivial. $\endgroup$ Commented Oct 17, 2017 at 20:21
  • $\begingroup$ @Sándor Kovács Interesting comment $\endgroup$
    – user21574
    Commented Oct 18, 2017 at 4:42

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