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Suppose one has an $S^1$ principal bundle $p: P\rightarrow M$, and a closed 2-form $F$ on $M$. Then the pullback form $p^*F$ is closed, vanishes on vertical vectors, and is invariant under the action of $S^1$ on $P$. These are all essential characteristics of a curvature form of a connection on an $S^1$ bundle, so my question is:

Is there a connection 1-form $\alpha$ such that d$\alpha=p^*F$? As of now, I'm not even sure how to see that $p^*F$ is trivial in the cohomology of $P$.

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  • $\begingroup$ This is standard. Every connection is of the form $\omega = \psi + p^*\alpha$ for some $\alpha$, where $\psi$ is the global angular form and $\Omega = p^*F$ the curvature. Also, $\Omega = d\omega = d\psi + dp^* \alpha = -p^*e + dp^* \alpha$, where $e$ is the real Euler class which coincides with the real Chern class (which is the curvature up to a scalar). You can write this in integral coefficients and $F$ entirely determines the circle bundle with a connection. Take a look at Bott and Tu. $\endgroup$
    – user40276
    Commented Oct 12, 2016 at 5:56
  • $\begingroup$ … There's also these notes google.com.br/… that seems detailed. $\endgroup$
    – user40276
    Commented Oct 12, 2016 at 5:58
  • $\begingroup$ I believe the "notes" in the previous comment are meant to be doi.org/10.1007/978-0-8176-4959-3_2, chapter 2 of Blair's book Riemannian Geometry of Contact and Symplectic Manifolds $\endgroup$
    – David Roberts ♦
    Commented Sep 19, 2023 at 5:46

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Taking $F=0$ we satisfy those conditions, and your question is then whether there is a flat connection. But then the Chern class vanishes. Taking $F$ some area form, we need a nonzero Chern class. At the least, you need to work out whether the integral of $F$ over every 2-cycle is an integer equal to the Chern number.

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    $\begingroup$ If all of these integrals have integer values, then indeed $F$ is the curvature of some connection, but perhaps not on the circle bundle you started with. If $F$ has the same integrals as the Chern form of your circle bundle, then of course $F$ differs from the Chern form by an exact differential $d\phi$ and you add (a suitable constant multiple of) $\phi$ to your connection form to get curvature $F$. $\endgroup$
    – Ben McKay
    Commented Oct 12, 2016 at 6:46

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