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I saw a statement used in the paper that E of rank 2 with $c_1(E) = 0$ is self-dual. I was wondering, how does one prove this statement? If it makes a difference, let the underlying variety be algebraic and rationally connected.

I thought of taking a sufficiently ample bundle bundle, to take a map $E \to L$ which would then give an exact sequence $$ 0 \to L^* \to E \to L \to 0 $$ To dualize it and to compare the classes in $Ext^1$, but I couldn't find anything on dualizing extensions.

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    $\begingroup$ This is true only if $c_1(E)$ is taken in the Chow group, that is in the Picard group. Then $\det(E)$ is trivial, and the isomorphism $\wedge^2E\overset{\sim}{\rightarrow}\mathcal{O}$ defines a non-degenerate skew-symmetric form on $E$, hence an isomorphism $E\overset{\sim}{\rightarrow}E^*$. $\endgroup$
    – abx
    Commented Dec 20, 2018 at 8:03
  • $\begingroup$ @abx I do not know much algebraic geometry.. only recently I started reading about characteristic classes... Does the same holds in differential geometry set up? I can ask a separate question but if it is too trivial it would be of no interest to users here... $\endgroup$
    – Praphulla Koushik
    Commented Dec 20, 2018 at 9:32
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    $\begingroup$ @Praphulla Koushik: Yes, the same argument works in differential geometry; the point is that $c_1(E)=0$ implies that the line bundle $\det(E)$ is trivial. $\endgroup$
    – abx
    Commented Dec 20, 2018 at 9:36
  • $\begingroup$ @abx sorry for second question. I am reading characteristic classes from Kobayashi and Nomizu's Foundations of Differential geometry.. I could not find that result.. Can you give a reference. Thanks. $\endgroup$
    – Praphulla Koushik
    Commented Dec 20, 2018 at 9:52
  • $\begingroup$ A reference to what? That complex line bundles are classified by $c_1$? $\endgroup$
    – abx
    Commented Dec 20, 2018 at 10:02

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For a rank 2 vector bundle you can show that $E^* \cong E \otimes \mathcal{O}(-\det(E))$ (exercise!), and that $\mathcal{O}(c_1(E)) = \det(E)$, which gives the result.

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