# Lifting sections of a projective bundle to a vector bundle

Let $$E\to M$$ be a smooth $$\mathbb{K} = \mathbb{R}, \mathbb{C}$$ - vector bundle over a possibly non-compact connected manifold $$M$$. Denote by $$\mathbb{P}(E) \to M$$ its projectivization, which is obtained by removing the zero section of $$E$$ and fiberwise taking the projective quotient $$\sim$$ which identifies lines on each fiber:

$$\mathbb{P}(E) = \frac{E\backslash 0_M}{\sim}$$

Denote by $$\pi\colon E\backslash 0\to \mathbb{P}(E)$$ the canonical projection, which pointwise sends an element of a fiber to the class it defines in projective space. As I understand, $$\mathbb{P}(E)$$ may not admit any section. Assume it does admit a smooth section $$s\in \mathbb{P}(E)$$. I am interested in the obstruction to lift $$s\colon M\to \mathbb{P}(E)$$ to a nowhere vanishing section $$\eta\colon M\to E$$ of $$E$$ such that $$\pi(\eta) = s$$. A quick computation in $$\check{\mathrm{C}}$$ech cohomology shows that, given $$s$$, there is a unique obstruction $$c(s)\in H^1(M,\mathbb{K}^{\ast})$$ for lifting $$s$$ to a section of $$E$$ that projects to $$E$$. Now, I am not sure if this is a characteristic class of $$E$$, or if it depends on the section $$s$$ chosen (whose existence may be obstructed but I assume). I have googled the literature but I have not found this problem discussed anywhere. Notice that this is different from the problem (extensively discussed in the literature) of finding the obstruction for a projective bundle to be the projectivization of a vector bundle. Here that is taken for granted and the obstruction corresponds to lifting a section.

Thanks.

• The section defines a line bundle on $M$ which is naturally a sub-bundle of $E$ and a lift of the section to $E$ is the same thing as a section of the line bundle. The class you obtained should be the first Chern class of that line bundle, I think. – Ben Apr 9 '19 at 17:01
• Thanks for the comment @Ben, it seems very reasonable to me. Then, the obstruction class will depend on the section chosen, since different sections may determine different, non-isomorphic, sub-bundles of $E$. – Bilateral Apr 9 '19 at 17:37

This is a classical problem in (topological) obstruction theory. Moreover this will confirm Ben's guess in case $$\mathbb K = \mathbb C$$.

Assume first that $$E$$ is a complex vector bundle. Let $$\Sigma E$$ be the sphere bundle of $$E$$ (to any bundle metric) which is a strong deformation retract of $$E\setminus 0$$. This gives a circle bundle $$S^1 \to \Sigma E \to P(E)$$. Now the obtructions to lift a map $$s \colon M \to P(E)$$ to a map $$\eta\colon M \to \Sigma E$$ lie in $$H^{k+1}(M;\pi_{k}(S^1))$$, thus there is only one obstruction in $$H^2(M;\mathbb Z)$$. As Ben indicated, if there is a lift $$\eta \colon M \to \Sigma E$$ of $$s$$ then the corresponding line bundle must be topological trivial. Thus the vanishing of the first Chern class of this bundle is a necessary condition to the existence of a lift of $$s$$. And since there is only one obstruction it is also sufficient.

If $$E$$ is a real vector bundle you obtain a fiber bundle $$\mathbb Z_2 \to \Sigma E \to P(E)$$ and the only obstruction lies in $$H^1(M;\mathbb Z_2)$$ which has to be the first Stiefel-Whitney class (for the same reasons as above)

• Hi Taki, your answer makes perfect sense, of course. Let me rephrase it in a way so that we directly see the classes in $H^1(M;\mathbb K^\times)$ the OP most likely constructed using Čech-cocycles. If we don't pass to the sphere bundle but (equivalently) work with the $\mathbb K^\times$-bundle $\mathbb K^\times \to E\setminus 0\to \mathbb P(E)$ instead, ...(cont'd) – Ben Apr 10 '19 at 18:53
• (cont'd)... we see that a section $M\to \mathbb P(E)$ lifts to $E\setminus 0$ if and only if the composite with the map of the fiber sequence $\mathbb P(E)\to B\mathbb K^\times = K(\mathbb K^\times,1)$ is homotopically trivial, i. e., if and only if the associated element in $H^1(M;\mathbb K^\times)$ is trivial. – Ben Apr 10 '19 at 18:53

Assuming $$E$$ is a complex vector bundle. At the bottom I'll point out how to modify this answer in the case $$E$$ is real.

Adding on to Panagiotis's answer: $$\mathbb{P}(E)$$ comes with a tautological line bundle $$\mathcal{O}_{\mathbb{P}(E)}(-1)$$ -- if we remove the zero-section of this line bundle we get the map $$E \setminus Z \to \mathbb{P}(E)$$ discussed above (here $$Z$$ is the zero-section). In the case $$M = \mathrm{pt}$$ this is the usual tautological bundle on $$\mathbb{P}^n$$.

The obstruction to finding a nowhere vanishing section of $$\mathcal{O}_{\mathbb{P}(E)}(-1)$$ *on $$\mathbb{P}(E)$$ is the first Chern class $$c_1(\mathcal{O}_{\mathbb{P}(E)}(-1))$$ Note that a section $$\tilde{\sigma}: M \to (E \setminus Z)$$ lifting the given section $$\sigma: M \to \mathbb{P}(E)$$ is the same as a section $$\tilde{\sigma}$$ of the pulled back $$\mathbb{C}^\times$$ bundle $$\sigma^*(E \setminus Z)$$ on $$M$$.

There's an easier way to describe this bundle: a principal $$\mathbb{C}^\times$$ bundle is the same as a line bundle. Since $$\mathbb{P}(E)$$ is the family of lines in $$E$$, the section $$\sigma: M \to \mathbb{P}(E)$$ specifies a sub-line-bundle $$L \subset E$$: the fiber over $$p \in M$$ is the line in $$E_p$$ corresponding to $$\sigma(p) \in \mathbb{P}(E_p)$$. What we see here is that the obstruction to a lift $$\tilde{\sigma}$$ of $$\sigma$$ is $$c_1(L) \in H^2(M, \mathbb{Z})$$.

If $$E$$ is instead a real vector bundle, replace Chern classes with Steiffel-Whitney classes and $$\mathbb{Z}$$ coefficients with $$\mathbb{Z}/2$$ coefficients.