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let $X$ be a compact Riemann surface and $D$ a divisor on $X$. I'm looking for a argument for the statement $c_1(\mathcal{O}_X(D)) = \deg(D)$, where $\mathcal{O}_X(D)$ is the associated line bundle to $D$.

Of course, one can consider the exponential sequence, i.e. $H^1(X, \mathcal{O}_X^*) = Pic(X) \stackrel{c_1}{\to} \underbrace{H^2(X,\mathbb{Z})}_{\cong \mathbb{Z}} \to 0$ and examine the connecting homomorphism $c_1$.

I'm looking for another, nicer way to see the above statement. Is there one?

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  • $\begingroup$ So, can you show that this result holds for the tautological line bundle $\mathcal O_{\Bbb P^1}(-1)$ on $\Bbb P^1$? $\endgroup$ Commented Dec 22, 2014 at 14:43
  • $\begingroup$ @TedShifrin As we defined the Chern classes in an axiomatic way (like in en.wikipedia.org/wiki/…), cf. my comment below, we immediately have that $c_1(\mathcal{O}_{X}(1)) = 1$ when $X = \mathbb{P}_1$. The result for the tautological bundle follows. Or, alternatively, we know that $\mathbb{Z} \to Pic(X), d \mapsto \mathcal{O}_{X}(d)$ is an isomorphism which maps the tautological bundle to $-1$. But this isomorphism is exactly the Chern class, if you make the identification $H^2(X,\mathbb{Z}) \cong \mathbb{Z}$. $\endgroup$
    – Slash_
    Commented Dec 22, 2014 at 21:23

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It is quite immediate if you know the definition of Chern classes via smooth sections: You start with your meromorphic section $\sigma$ and near the poles you replace it with a smooth section $s$ transverse to the zero section ${\bf 0}$, using the fact that on the unit circle $z^{-1}=\bar{z}$. Now, just compute the oriented intersection number of the image of $s$ and of the zero section. The result, by the construction, equals the degree of your line bundle.

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  • $\begingroup$ Thanks for your answer. Unfortunately, we defined the Chern classes in an axiomatic way through their properties (for instance, Chern classes and pullback of de Rham cohomology commutes). Could you recommend a good reference to me where I can have a closer look at the things you told me? $\endgroup$
    – Slash_
    Commented Dec 20, 2014 at 14:46

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