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Consider the Lie group $S^1$. Recall that the associated Lie algebra is $\mathbb{R}$.

Let $M$ be a manifold. Consider the second de-Rham cohomology group $H^2(M,\mathbb{R})$.

Let $\Omega\in H^2(M,\mathbb{R})$ be an integral cohomology class; that is, it is image of an element of $H^2(M,\mathbb{Z})$ under the natural map $H^2(M,\mathbb{Z})\rightarrow H^2(M,\mathbb{R})$. Then, we know that we can construct a principal $S^1$-bundle $P\rightarrow M$ over the manifold $M$, a connection $1$-form $\omega$ on the manifold $P$ such that, the associated curvature form (which is a $2$-form on $P$, but can be projected uniquely to a $2$-form on $M$) is precisely the $2$-form $\Omega$ that we have started with.

Question : How far can we go relaxing the condition that the structure group is abelian?

Question : Let $G$ be a Lie group and $\mathfrak{g}$ be its Lie algebra. Let $M$ be a manifold and $\Omega$ be a $\mathfrak{g}$-valued $2$-form on $M$. Under what conditions, can we find a principal $G$ bundle over the manifold $M$, and a connection on $P(M,G)$ whose curvature is $\Omega$?

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  • $\begingroup$ The case of integral cohomology class is called quantization problem. So, I am calling this quantization problem for non abelian groups.. Please correct me if I misunderstood the name or anything in the post above.. $\endgroup$
    – Praphulla Koushik
    Commented Apr 28, 2020 at 12:37
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    $\begingroup$ As a start, you might want to have a look at this MO question: mathoverflow.net/questions/73439/… $\endgroup$
    – Robert Bryant
    Commented Apr 30, 2020 at 9:34
  • $\begingroup$ @RobertBryant Sir, Yes, I saw that now. I did not follow all calculations but I got the idea :) Now, I am thinking of what conditions I should add to expect a positive answer... $\endgroup$
    – Praphulla Koushik
    Commented Apr 30, 2020 at 9:50

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