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How do I see that the 1st Chern class is invariant under choice of section? I know metric invariance follows from how two metrics on line bundle have to be conformally equivalent, but how do we show invariance under choice of section? Thanks in advance.

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    $\begingroup$ What definition/construction of first Chern class are you using? $\endgroup$ Nov 7, 2015 at 23:18

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There are many definitions of the $1$-Chern class of a complex line bundle $L\to M$, $M$ compact $CW$-complex. The topological one goes as follows. The line bundle $L$ is an oriented rank $2$ real vector bundle over $M$. As such it has a Thom class $\tau_L\in H^2(D(L), S(L))$, where $D(L)$ and $S(L)$ are the disk and respectively the unit sphere bundle of $L$. If $\zeta:M\to L$ is the zero section, then the first Chern class of $L$ coincides with the Euler class of $L$ defined as the pullback

$$ e(L):=\zeta^*\tau_L\in H^2(M). $$

If additionally $M$ is an oriented $m$-dimensional manifold, then the Poincare dual is a homology class $e(L)^*\in H_{m-2}(M)$. This class is also represented by the zero set of a generic section of $L$; see Chapter 4 of these notes for details and proofs.

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Let $s$ be a local holomorphic section of a hermitian line bundle $(L, h)$.

In local coordinates set $h(z)=h(1,1)$:

$$ \partial\bar\partial \log h(s(z), s(z))=\partial\bar\partial \log (h(z)s(z)\bar s(z))=\partial\bar\partial\log h(z) + \partial\bar\partial\log s(z) + \partial\bar\partial\log\bar s(z). $$ Since $s(z)$ is holomorphic, $\bar\partial s(z)=\partial \bar s(z)=0$, so the last two terms in the formula above vanish and the curvature term (representing $c_1(L)$) does not depend on the choice of a section.

Note, that it is important that the section $s(z)$ is holomorphic.

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  • $\begingroup$ However, the OP does not even mention complex manifolds. On a complex manifold, one gets more information by interpreting $c_1(L)$ as a divisor class. And then the equivalence relation on divisors takes care that $c_1(L)$ is independent of the section. $\endgroup$ Nov 9, 2015 at 19:34
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    $\begingroup$ Agree, I've hastened a bit. Probably, because this is the definition where you essentially have to check both invariance under the changes of section and metric. However, this analytic approach has its own advantages, as you need only local holomorphic sections to define the curvature form $\Theta(L)$, as opposed to a global meromorphic form to define divisor $[L]$. $\endgroup$ Nov 9, 2015 at 20:32

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