1st Chern class is invariant under choice of section? 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.
 A: 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.
A: 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.
