Given a holomorphic line bundle $L$ over complex manifold $X$, suppose $L$ is globally generated by sections $s_1,…,s_k\in H^0(X,L)$, i.e. For any $ x\in X$, not all $s_i(x) $ are $0$. Then there is a natural hermitian metric on $L$ locally defined by $h(t) =\frac{|\psi(t)|^2}{\sum |\psi(s_i(x) |^2} $ for any local trivialization $\psi$ and any $t\in L(x)$.
We can then define $h^*$ on $L^*$, by taking $h^*(\xi) =h(v_{\xi} ) $ where $v_{\xi} \in L$ is defined by $\xi(t) =h(t, v_{\xi} ) $. On the other hand we have natural inclusion $L^*\subset \mathcal O^{\oplus k} $ by $\xi\mapsto (x, \xi(s_1(x)),…,\xi(s_k(x)))$ where $\xi\in L^*(x)$. By taking the standard hermitian metric on the trivial bundle, denote the induced metric by $h'$. Specifically $h'(\xi) =\sum |\xi(s_i(x)) |^2$.
I wish to show that $h^*$ and $h'$ are in fact the same metric. So I have to prove that $$h'(\xi) =\sum |\xi(s_i(x)) |^2=h(v_{\xi}, v_{\xi} )=\xi(v_{\xi}) =h^*(\xi)$$
I speculated that $v_{\xi} =\sum\overline{\xi(s_i(x))} s_i(x)$, but I am not sure how to show this.
Best Answer
We just need to compare their hermitian matrix.
By abuse of language,we say that $h$ is given by $(\sum {|s_i|^2})^{-1}$.Then we have $h^*$ is given by $(\sum {|s_i|^2})$.Since $h'(\xi) =\sum |\xi(s_i(x)) |^2$,$h'$ is also given by $(\sum {|s_i|^2})$.Hence we get $h'=h^*$.