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Consider the Hopf bundle $h:\mathbb{S}^3\rightarrow\mathbb{S}^2$. There is a connection $1$-form $\omega$ oh $h$ which is left $SU(2)$ invariant. In terms of the Euler angles $(\theta,\,\phi,\,\psi)$ this is given by $\omega=d\psi+\cos(\theta)d\phi$. Using $\omega$, we can form a covariant derivative on the associated bundle $\phi:\mathbb{S}^3\times_{U(1)}\mathfrak{u}(1)\rightarrow\mathbb{S}^2$. In the paper https://arxiv.org/abs/1705.02666, the author treats this as a covariant derivative on a bundle over a manifold with topology $\mathbb{C}^2$, such that the curvature is given by $d\omega$.

The only way I can think to extend $\phi$ to a bundle over $\mathbb{C}^2$ is pulling it back via $h$ to a bundle over $\mathbb{S}^3$, then extending radially over $\mathbb{C}^2\cong\mathbb{S}^3\times\mathbb{R}_{\geq 0}$. However, I am not sure how this would be well defined as the $\mathbb{R}_{\geq 0}$ component goes to $0$.

I would greatly appreciate if anyone could reconcile how to obtain such a bundle over $\mathbb{C}P^2$.

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    $\begingroup$ The bundle in question is the tautological line bundle. Its total space is $\mathbb{C}^2\setminus 0$ and the projection is the natural projection $\mathbb{C^2}\setminus \{0\}\to\mathbb{CP}^1$ that associates to a nonzero vector the line it determines $\endgroup$
    – Liviu Nicolaescu
    Commented May 9, 2020 at 9:38
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    $\begingroup$ If I understand correctly what you are asking, one thing to point out is that the bundle over $S^2$ cannot extend to a bundle over $\mathbb{C}^2$, since then the Hopf bundle would be trivial—if by 'extend' you mean along some inclusion $S^2 \hookrightarrow \mathbb{C}^2$. $\endgroup$
    – David Roberts ♦
    Commented May 9, 2020 at 10:48
  • $\begingroup$ @DavidRoberts , this makes sense, thanks you. In essence when I say `extend' what I want is the following; I have a connection from $\mathbb{S}^2$ to some some line bundle, this connection has a curvature form $F$ associated to it, I want to somehow get a bundle over $\mathbb{C}^2$ with the same curvature form (which does seem like an inclusion actually). $\endgroup$
    – Guest123412341234
    Commented May 9, 2020 at 13:18
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    $\begingroup$ If this is in fact what the author does, then it is indeed incorrect. Unless the author is using some kind of generalized covariant derivative, extending it implies extending the bundle, which is not possible here. As you say, it is indeed possible to extend it to the punctured plane but no further. $\endgroup$
    – Deane Yang
    Commented May 9, 2020 at 17:29
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    $\begingroup$ Are you sure that form extends continuously to the origin? I'm not. $\endgroup$
    – David Roberts ♦
    Commented May 10, 2020 at 2:34

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