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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.

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    $\begingroup$ 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. $\endgroup$ – Ben Apr 9 at 17:01
  • $\begingroup$ 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 $E$. $\endgroup$ – Bilateral Apr 9 at 17:37
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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)

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    $\begingroup$ 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) $\endgroup$ – Ben Apr 10 at 18:53
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    $\begingroup$ (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. $\endgroup$ – Ben Apr 10 at 18:53
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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.

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