Prove the holomorphic line bundle $\lambda(p+q)$ is the dual of the natural projective bundle

Let $M=\mathbb{C}P^1$ be the complex projective space, $U_0=\{[z_0,z_1]:z_0\ne 0\}$, $U_1=\{[z_0,z_1]:z_1\ne 0\}$ be the coordinate charts and define $E=\{([z_0,z_1],x):x\in\operatorname{span}_{\mathbb{C}}[z_0,z_1]\}$ the complex line bundle over $M$ with the natural projection $\pi([z_0,z_1],x)=[z_0,z_1]$ and the bi-holomorphic maps:

$$\phi_0:\pi^{-1}(U_0)\rightarrow U_0\times\mathbb{C},\ ([z_0,z_1],a(1,z_1/z_0))\mapsto ([z_0,z_1],a)$$ $$\phi_1:\pi^{-1}(U_1)\rightarrow U_1\times\mathbb{C},\ ([z_0,z_1],b(z_0/z_1,1))\mapsto ([z_0,z_1],b)$$

The transition functions are then $f_0^1=z_0/z_1, f_1^0=z_1/z_0$.
Let $\lambda(D)$ be the line bundle determined by a divisor $D$. Now to be simple consider the case $p=[1,0]$ and $q=[0,1]$, let $f_0=z_1/z_0$ and $f_1=z_0/z_1$, then $f_0$ is well defined on $U_0$ such that $p$ is the only zero point thus the divisor $(f_0)=p=D\cap U_0$ (where $D=p+q$). Similarly we have $(f_1)=q=D\cap U_1$ so the transition function of $\lambda(D)$ is $f_0^1=f_0/f_1=z_1^2/z_0^2$, but not $z_1/z_0$ to be expected. (Dual bundles have inverse transition functions).

What's wrong with the above argument, thanks.

The hyperplane section bundle $\mathscr O(1)$ on $\mathbb P^1$ is defined by the transition functions $(g_{ij} = \frac {z_j}{z_i})_{ij}$. The pair of local functions $f_0 := \frac {z_1}{z_0}$ on $U_0$ and $f_1 := \frac {z_0}{z_1}$ on $U_1$ from your question satisfies on the intersection $U_0 \cap U_1$the compatibility rule
$$f_0 = g_{01}^2 * f_1.$$
Hence $f = (f_0, f_1)$ is a section in $\mathscr O(2)$ - not $\mathscr O(1)$. It vanishes at the two points $p=(1:0)$ and $q=(0:1).$
Note. The bundle from your question is the tautological line bundle $E = \mathscr O(-1)$.
• So $\lambda(p+q)$ is not the dual of the tautological line bundle? – Xipan Xiao Dec 31 '14 at 19:06
• $\lambda(p+q)$ is the square of the dual of the tautological line bundle. – Jo Wehler Dec 31 '14 at 22:09
• Sections of the line bundle $\mathscr O(n), n \in \mathbb N,$ are exactly the holomorphic sections with $n$ zeros, counted with multiplicities. They correspond bijectively to the homogeneous polynomials of degree $n$. – Jo Wehler Dec 31 '14 at 23:14