Understand this hermitian vector bundle

algebraic-geometrycomplex-analysiscomplex-geometryhodge-theory

I am reading Huybrechts's complex geometry. In page 166, he gives the definition of hermitian structure.

Let $E$ be a complex vector bundle over a real manifold $M$. An hermitian structure $h$ on $E\to M$ is an hermitian scalar product $h_{x}$ on each fibre $E(x)$ which depend smoothly on $x$. This pair $(E,h)$ is called an hermitian vector bundle.

This definition is very natural. However, then he gives an example which make me confused.

Let $L$ be a holomorphic line bundle and let $s_{1},s_{2},…,s_{k}$ be global sections generating $L$ every where. Picking a trivialization $\phi: L_{U}\cong U\times \mathbb{C}$ around a point $x$. Then one defines an hermitian structure on $L$ by

$$h(t)=\frac{|\phi(t)|^{2}}{\sum |\phi(s_{i})|^2}.$$

where $t$ is a point in the fiber $L(x)$.

I am very confused. We know $L$ is a line bundle, so $L(x)$ is an complex one dimensional vector space. So if we want to find a hermitian metric on $L(x)$, it is enough to find a positive number for it. Thus we should construct a funtion $h:=h(x)$ which is a smooth function on $U$. I do not know why he uses $h(t)$. $h(t)$ is a function on the fiber $L(x)$, not the function on the manifold.

One explantion is he may want to refer $h(x)=\frac{|\phi(t)|^{2}}{\sum |\phi(s_{i})|^2},$ where $t$ is a point in $L(x)$. This is also strange, since then this hermitian metric will depond on the choice of $t$ and hence it is not natural.

Could you tell me how to explain this?

Best Answer

Let $t$ is an arbitrary section of $L$ representing the germ. Now you use the local trivialization $\phi$ to define $h(t)\colon U\to\mathbb{R}$ by $$ \frac{\lvert\phi_2(t(\cdot))\rvert^2}{\sum_i\lvert\phi_2(s_i(\cdot))\rvert^2} $$ where $\phi_2$ is the $\mathbb{C}$-coordinate of $\phi\colon L\vert_U\to U\times\mathbb{C}$ (at the appropriate point of $U$).

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