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:


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


  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.