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Consider a connection $d$ on a principle bundle $E\to M$. It gives gives you a curvature by the usual $$ d^2s=Fs \quad \forall s\in \Gamma(E) $$

I was wondering if the connection is completely determined by $F$.

For an abelian group it looks like it's true, but in general it seems a difficult problem.

This is actually for physics so you could use YM's equation $d\star F=0$ but I don't believe it is needed.

pushing a little further, I wanted also to know if this works for GR, is the connection completely constrained by the Riemman Tensor?

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It is not true, if the bundle is defined by the trivialization $(U_i)$, locally, the connection is defined by $d+\alpha_i$ where $d$ is the differential and $\alpha_i$ a $1$-form which takes its values in the Lie algebra${\cal G}$ of $G$ the structural group, the curvature is $d\alpha_i+\alpha_i\wedge \alpha_i$. Suppose that the Lie algebra ${\cal G}$ is commutative, and $\beta$ a closed $1$-form which takes its values in ${\cal G}$, we can define a connection by $d+\alpha_i+\beta_{\mid U_i}$ which has the same curvature.

You can also consider flat bundles (bundles whose curvature vanishes) defined for examples over a surface $M$, the classified by representations $\pi_1(M)\rightarrow G$ and are not in general non trivial, their moduli space are studied. See the paper of Goldman below:

http://www2.math.umd.edu/~wmg/SymplecticNature.pdf

Consider for example the trivial $\mathbb{R}$-bundle over $\mathbb{R}$ and $\beta=df$ where $f$ is a function defined on $\mathbb{R}$.

https://en.wikipedia.org/wiki/Connection_(vector_bundle)#Local_expression

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  • $\begingroup$ correct me if I am wrong but that doesn't change the connection. That is simply a change of coordinates. If you change the coordinates in the bundle, the gauge field changes by $\alpha\to g\alpha g^{-1} +gdg^{-1}$. In particular for an abelian field you get by doing change of coordiantes with $g=e^f$, $\alpha\to \alpha +df$ $\endgroup$ – David Jaramillo Jan 21 at 14:59
  • $\begingroup$ You can consider a closed form which is not exact. $\endgroup$ – Tsemo Aristide Jan 21 at 16:35
  • $\begingroup$ That's a local statement, you just need to define $f$ or $g$ in a small enough neighborhood. Then closed is the same as exact. $\endgroup$ – David Jaramillo Jan 21 at 16:38
  • $\begingroup$ If the bundle is commuative it will be a coordinate change relate to the first Chern class ( which classify $T^n$-bundle). $\endgroup$ – Tsemo Aristide Jan 21 at 17:02

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