Let $E \to X$ be a homomorphic vector bundle over a projective variety $X$. Does $\mathbb{P}(E)$ always have holomorphic sections? If not what is the obstruction?

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    $\begingroup$ Are you asking about section of $\mathbb{P}(E)$ (as in the title) or about section of $E$ (as in the question)? $\endgroup$
    – Sasha
    Nov 5, 2019 at 11:49

1 Answer 1


I'll assume you're asking about sections of $PE$. The bundle $E$ has Chern classes $c_i(E)\in H^{2i}(X)$ and thus a Chern polynomial $f_E(t)=\sum_{i=0}^nc_{n-i}(E)t^i$, where $n=\dim(E)$. A section of $PE$ corresponds to a line bundle $L\leq E$ and thus a factorisation $f_E(t)=f_L(t)f_{E/L}(t)$, and thus a root $-c_1(L)\in H^2(X)$ of $f_E(t)$. Thus, if $f_E(t)$ has no roots in $H^2(X)$ then $PE$ does not even have a topological section. One can give similar arguments in the topological category using any complex oriented multiplicative cohomology theory such as $MU^*(X)$ or $KU^*(X)$. I believe that you can give similar arguments in the algebraic category using the Chow ring or algebraic $K$-theory, but I will not swear to the technical details of that.


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