# does the curvature fixes a connection on the principal bundle?

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?

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

• 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 $g=e^f$, $\alpha\to \alpha +df$ – David Jaramillo Jan 21 at 14:59
• You can consider a closed form which is not exact. – Tsemo Aristide Jan 21 at 16:35
• 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. – David Jaramillo Jan 21 at 16:38
• If the bundle is commuative it will be a coordinate change relate to the first Chern class ( which classify $T^n$-bundle). – Tsemo Aristide Jan 21 at 17:02