Just a question about visualization of vector bundles.
Imagine that we have a complex $3$-manifold $X$ and a vector bundle over $X$ of rank $4$.
Suppose that we express the vector bundle as a sum of $4$ line bundles, for example $O_{X}(2)\oplus O_{X}(2)\oplus O_{X}(4)\oplus O_{X}(1)$.
If we define the coordinates of the vector space as ${v_{i}}, i=1,…4$ we can define the section of each line bundle as $s_{i}:(z_{1},z_{2},z_{3})\rightarrow (z_{1},z_{2},z_{3},v_{i}=p_{i}((z_{1},z_{2},z_{3}))$
My question is: each section just must fulfill the condition of giving the identity (on $X$) when composed with the respective projection. According to that the polynomials are general in the sense that the coefficients are not completely fixed. Folllowing this the set of the $4$ sections do not define a complex manifold in $\mathbb{C}^{7}$ (as I expected).
What I am thinking wrong? Does the sections define a set of $7$-manifolds as products of $X\times M_{\alpha}$ being $X$ fixed and $M_{\alpha}$ a different $4$-manifold for each different choice $\alpha$ of the sections $s_{i}$?
Best Answer
We should think of the ratios of the $4$ sections---not all zero---as projective coordinates.
It is important not to confuse a vector bundle with the corresponding---locally free---sheaf. Let us denote by $V = L_1 \oplus L_2 \oplus L_3 \oplus L$ the total space of the vector bundle corresponding to our sheaf $\mathcal{O}_{X}(2)\oplus \mathcal{O}_{X}(2)\oplus \mathcal{O}_{X}(4)\oplus \mathcal{O}_{X}(1)$. Let us denote our section by $s: X \to V$, and the projection by $p: V \to X$. Suppose that on the open set $U$ in $X$ with coordinates $z_1$, $z_2$, $z_3$, the vector bundle $V$ is trivial, that is, $p^{-1}(U) = U \times \mathbb{C}^4$. When we give a section, in the open set $U$, our functions $p_i(z_1, z_2, z_3)$, $i = 1, 2, 3, 4$ are got fixed. Thus, the image $s(U)$ of the open set by the section $s$ is determined as the subset of $p^{-1}(U) = U \times \mathbb{C}^4$ with coordinate $(z_1, z_2, z_3, v_1, v_2, v_3, v_4)$ as the subset determined by the equations$$v_1 - p_1(z) = 0, \text{ }v_2 - p_2(z) = 0, \text{ }v_3 - p_3(z) = 0, \text{ }v_4-p_4(z) = 0.$$In this way, the image $s(X)$ of the section is a submanifold of $V$.
Sure. In fact, the locally free sheaf corresponding to a vector bundle $V$ is the sheaf of "germs of sections of $V$".