This is a very elementary question.
Fix a connection on a principal bundle, the derivative of the projection map gives an isomorphism of the horizontal subspace at a point in the bundle and the corresponding tangent space on the base. It seems possible to me that such an isomorphism may also be established by using a local section followed by horizontal projection as follows:
The image of the tangent space of a point on the base under the derivative of a local section is a linear space of the same dimension by rank-nullity but does not necessarily coincide with the horizontal subspace prescribed by the given connection. However, after the horizontal projection (i.e. taking only the horizontal component), the composite map gives an isomorphism of the tangent space of a point on the base and the corresponding horizontal subspace. In particular, horizontal projection and the derivative of a local section can be used in sequence to achieve horizontal lift of a vector field in a local trivialization.
Is this argument correct?
Thank you.
Best Answer
So your argument is that $\pi_H\circ D_xs$ is an isomorphism from $T_xM$ to $H_{s(x)}$? If so I believe this argument is correct. First note that this is injective as if $\pi_H\circ D_xs(X)$ is zero for $X\in T_xM$, then we have that $D_xs(X)\in V_{s(x)}$, however this implies that $D_{x}s(X)\in \ker D_{s(x)}\pi$, where $\pi:P\rightarrow M$ is the projection. Thus we have that: $$D_{s(x)}\pi\circ D_xs(X)=D_{x}(\pi\circ s)(X)=\text{Id}(X)=X=0$$ So by rank nullity $\pi_H\circ D_xs$ is an isomorphism.
It is mildly harder to prove however that every vector field on $M$ admits a unique horizontal vector field $X^H$ on $P$, which satisfies $D_p\pi (X^H)=X_{\pi(p)}$, and the proof of this statement only rely on local sections to prove smoothness.