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The problem:

Consider the smooth quadric $Q=V(X_{0}X_{1}+X_{2}X_{3}+X_{4}^{2})\subset\mathbb{P}^{4}$ and the line $L=V(X_{0},X_{2},X_{4})$ contained in it. Prove that there exists a vector bundle $F$ on $Q$ with a section vanishing exactly at $L$.

The hardest thing is to get the transition matrices. Any help is appreciated.

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Consider on $\mathbb P^4$ the three line bundles $\mathcal L_i=\mathcal O(H_i)\; (i=0,2,4) $ associated to the divisors $H_i=\{X_i=0\}$.

They have global sections $s_i\in \Gamma(\mathbb P^4,\mathcal L_i)$ vanishing exactly on $H_i$.

The vector bundle $\mathcal L_0\oplus \mathcal L_2\oplus \mathcal L_4$ has a section vanishing exactly along $L=H_0\cap H_2\cap H_4$, namely the section $s_0\oplus s_2\oplus s_4\in \Gamma(\mathbb P^4,\mathcal L_0\oplus\mathcal L_2\oplus\mathcal L_4)$

The line bundle $F=(\mathcal L_0\oplus\mathcal L_2\oplus \mathcal L_4)|Q$ then also has a section vanishing exactly along $L$, namely the restriction to $F$ of the section $s_0\oplus s_2\oplus s_4$.

[Note that there is no need for transition matrices.]

**Edit: the transition matrices**

The transition matrices required by JSong in his comment are easy to calculate .

Since the cocycle for $\mathcal O_{P^4}(1)$ is given by $g_{ij}=\frac{X_j}{X_i}$ on $\mathbb P^4$ the transition matrices for $F$ on $Q$ are the diagonal $3\times 3$ matrices $$G_{ij}=\operatorname{diag} (\frac{z_j}{z_i},\frac{z_j}{z_i},\frac{z_j}{z_i})\quad(0\leq i,j\leq 4)$$ ($z_i$ is the restriction of $X_i$ to $Q$)

The required section $s=s_0\oplus s_2\oplus s_4\in \Gamma(Q,F)$ which vanishes exactly along $L$ has in the chart $U_i=\{z_i\neq 0\}$ the expression $$s^{(i)}(z)=\frac{ z_0}{z_i}\oplus \frac{z_2}{z_i}\oplus \frac{z_4}{z_i}\in \mathbb C^3 \quad (z\in U_i)$$

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