Associated or induced G-connection

Now, observe that a principal bundle connection Γ in a principal bundle produces a fiber bundle connection in the associated bundles. Take the context given in the latter note. Given wE=(P×F)/G, take a representative w=[u,ξ]. We have a map

βξ:PE

being βξ(v)=[v,ξ]. Now, take as definition:

Hw:=d(βξ)u(Hu)

At this way, we also carry the parallel transport concept to the associated bundle. \cite

Particular case: vector bundle connections

In the case of a vector bundle connection with associated principal bundle the frame bundle, the induced connection is obtained, in the sense of its connection 1-form, by simply pulling back with a particular frame. Let's see in the case of the tangent bundle.

Suppose we have a vector bundle connection on the bundle TMM, induced by a principal connection on the frame bundle FMM which is described by a 1-form on FM, Θ~. Then, if we select the frame p={ei} in M, we are providing a section p of FM, and we can consider the gl(n)-valued 1-form p(Θ~) on M.
On the other hand, the frame p let us to write down a connection matrix Θ (whose entries are 1-forms) for the connection . Given a vector field X on M, the connection gives us a way to differentiate it:

Xei=jΘij(X)ej

Where Θij(X) are the components of the gl(n)-valued 1-form Θ for the given frame.

Claim:

p(Θ~)=Θ.

How does p(Θ~) acts on X?

Consider a point pM, we have

p(Θ~)(Xp)=Θ~(p(Xp)).

Now, since Θ~ is the connection induced by Θ, you can compute Θ~(V) for a vector VTf(FM), and fFM such that π(f)=p and dπf(V)=X. What Θ~(V) does is consider V as an infinitesimal curve in FM (p(t),v1(t),v2(t),) and extract the components of Xv1,Xv2,).
But in the case V=p(Xp), this curve is precisely (p(t),e1(t),e2(t),) so

Θ~(p(Xp))=Θ(Xp).

Conversely

In the case of G=GL(n) there is a converse. Any linear connection on a vector bundle E is induced from a unique principal connection on the frame bundle. See this answer in Mathstackexchange or @cap2009parabolic page 42.

Special (and important) case: when the principal bundle is the frame bundle and the associated bundle is the tangent bundle. There is an good reflection in the note connections summary.

Curvature

In the context given above. The curvature of the frame bundle is given by

Ω~=dΘ~+12[Θ~,Θ~]

and the curvature curvature of the connection on the tangent bundle , in the frame p, is

Ω=dΘΘΘ

It is clear that if Ω~=0 then Ω=0. I think that the opposite is also true:
Given two vectors V,WTfFM with dπ(V)=X,dπ(W)=Y, X,YTpM they can be interpreted as infinitesimal curves α~,β~ through f, i.e., curves α,β inside M with moving frames over them. I guess that by means of a local trivialization we can find a local section p:MFM such that restricted α,β we obtain the frames represented by V,W, that is,

p(X)=V,p(Y)=W.

And then,

Ω~(V,W)=Ω~(p(X),p(Y))=p(Ω~)(X,Y)=Ω(X,Y)=0.

Example

(xournal 197)
Consider the worked example here, where we construct the connection 1-form of a connection on the tangent bundle of R2 from a covariant derivative operator . Now, let's see how to express the induced principal connection on the frame bundle in terms of its 1-form. Consider a frame fFM at p=(x1,x2). Since FM is in this case Aff(2)=R2GL(2) it can be seen as a subgroup of GL(3) and then

f=(c11c12x1c21c22x2001)

The first components of the columns are elements of R2. If we pass from a frame f at p to a frame f at p the 1-form of the connection measures how this change fails to be constant.
Pasted image 20221030123959.png
To compute the 1-form we apply the 1-form of the original vector bundle connection (see here) to the vectors

e1:=(dx1,dx2,dc11,dc21)e2:=(dx1,dx2,dc12,dc22)

to see how every vector in the frame fails to be constant.
Pasted image 20221030124742.png
These "mistakes" are expressed like an element of VfFM. But in order to compare the mistakes, we "send them" to a common place, VeFM being e the origin with the canonical basis, or, equivalently, to gl(2). I have to think this better. Anyway, what we are doing is to express these mistakes with respect to the frame f itself, so we multiply the matrix
Pasted image 20221031093156.png
made with the mistakes by C1 obtaining ωΩ(FM)gl(2):
Pasted image 20221031093742.png

being C such that

f=(Cp01)

Remember: this 1-form tells us how much do we "fail to be constant" when we pass from the frame f to a nearby frame f, but expressing this mistake with respect to the frame f itself.