Existence of complex line subbundle of complex bundles

Question

Suppose $M$ be a real smooth manifold, and let $E$ be a rank $r>1$-complex vector bundle over $M$.

I would like to know that if there is a complex line subbundle of $E$.

By usual obstruction theory, if it has a vanishing top Chern class, then there is an obvious line sub-bundle. For general case, it's only possible after removing some disjoint points on $M$.

Here is a question:

• Non-existence of complex line subbundle of some complex vector bundles. What would be an good example?
• Are there more sharp result than using top Chern class? Possibly, using Stiefel-Whitney's and it shows that it's possible when the top Chern class is even.

Answer 1

As discussed in the comments, if $E \to B$ is a rank $r$ complex vector bundle, then $c_r(E) \in H^{2r}(B; \mathbb{Z})$ is the first obstruction to finding a nowhere-zero section (which is equivalent to $E$ admitting a trivial line subbundle). If $\dim B \leq 2r$, the top Chern class is the only obstruction because the higher obstructions lie in $H^i(B; \pi_{i-1}(S^{2r-1}))$ for $i > 2r$. However, if $\dim B > 2r$, then the top Chern class could vanish without $E$ admitting a nowhere-zero section (see Example 3 below).

Returning to the question at hand, suppose $E \to B$ had a line subbundle $L$. Following Mariano Suárez-Álvarez's comments, the line bundle $L$ is determined up to isomorphism by $c_1(L) \in H^2(B; \mathbb{Z})$; in particular, it is trivial if and only if $c_1(L) = 0$. If $H^2(B; \mathbb{Z}) = 0$, then this is necessarily the case, so over such a space, a bundle $E \to B$ admits a line subbundle if and only if it admits a trivial line subbundle (which is equivalent to admitting a nowhere-zero section).

Example 1: A complex rank two bundle $E \to S^4$ admits a line subbundle if and only if $c_2(E) = 0$. For any $\beta \in H^4(S^4; \mathbb{Z})$, there is a complex rank two bundle $E \to S^4$ with $c_2(E) = \beta$, see this answer; choosing $\beta \neq 0$ provides examples of vector bundles which do not admit line subbundles.

More generally, we could replace $S^4$ with any four-dimensional CW complex $B$ with $H^2(B; \mathbb{Z}) = 0$.

In general, if a rank $r$ complex vector bundle $E$ admits a line subbundle $L$, then $E\cong E_0\oplus L$ for some $E_0$ with $\operatorname{rank}E_0 = \operatorname{rank}E - 1$. So $c_1(E) = c_1(E_0) + c_1(L)$, and $c_k(E) = c_{k-1}(E_0)c_1(L)$ for $k > 1$. Therefore, if $E$ admits a line subbundle, then there exists a class $\alpha \in H^2(B; \mathbb{Z})$ and classes $\eta_1, \dots, \eta_{r-1}$ with $\eta_d \in H^{2d}(B; \mathbb{Z})$ such that $c_1(E) = \eta_1 + \alpha$ and $c_k(E) = \eta_{k-1}\alpha$ for $k > 1$. These are necessary conditions for the existence of a line subbundle. In Example 1, to verify this condition, we'd need to find classes $\eta_1$ and $\alpha$ with $\eta_1 + \alpha = c_1(E) = 0$ and $\eta_1\alpha = c_2(E) = \beta$, but since $H^2(S^4; \mathbb{Z}) = 0$, we must have $\eta_1 = \alpha = 0$, so the equations cannot be solved if $\beta \neq 0$. More generally, we can obtain examples of bundles $E \to B$ which do not admit line subbundles with $H^2(B; \mathbb{Z}) \neq 0$.

Example 2: Consider a complex rank two bundle $E \to \mathbb{CP}^2$ with $c_1(E) = 0$ and $c_2(E) = x^2$ where $x \in H^2(\mathbb{CP}^2; \mathbb{Z})$ is a generator (such a bundle exists by the answer linked to in Example 1). Writing $\eta_1$ and $\alpha$ as $ax$ and $bx$ respectively for $a, b \in \mathbb{Z})$, we obtain the equations $$0 = c_1(E) = \eta_1+\alpha = ax + bx = (a + b)x \Rightarrow b = -a,$$ $$x^2 = c_2(E) = \eta_1\alpha = (ax)(bx) = abx^2 = -a^2x^2$$ which has no solutions.

As is shown in this answer, complex rank two bundles over a four-dimensional CW complex are determined by their first and second Chern classes, and every potential combination of Chern classes arises, so many different examples can be constructed this way.

Note, the necessary conditions on the Chern classes are not sufficient, as the following example illustrates.

Example 3: Consider the (unique up to isomorphism) non-trivial complex rank two bundle $E \to S^5$ described in this answer. It has $c_1(E) = 0$ and $c_2(E) = 0$ and we can find classes $\alpha$ and $\eta_1$ with $c_1(E) = \eta_1 + \alpha$ and $c_2(E) = \eta_1\alpha$, namely $\eta_1 = \alpha = 0$. If $E$ were to admit a line subbundle, then $E$ would be isomorphic to a sum of two line bundles, both of which would have to be trivial as $H^2(S^5; \mathbb{Z}) = 0$. This also provides an example of a complex vector bundle whose top Chern class vanishes, yet does not not admit a nowhere-zero section.