Confusion regarding connection form

Question

I have the following two notions of connection.

1. For a vector bundle we have a covariant derivative from sections of $E$ to sections of $E \otimes T^{*}M$ which is a $\mathbb C$-linear map and satisfies Leibniz rule.

2. We have a connection defined as splitting of vertical tangent vectors inside $TP$ where $P$ is a principal $G$ bundle [Here we care about $G=GL(n)$]

Now I have seen (and I believe I understand why) that by definition 1, a connection comes with a local 1-form and they don't glue but their exterior derivative (called curvature) gives a global 2-form.

But by definition 2, a splitting is just a $TP$-valued 1-form and since the vertical tangents at a point are isomorphic to $g$ (the Lie Algebra of $G$) we see that we get a $g$-valued 1-form.

But this latter form seems to be a global Lie Algebra valued 1-form. Why is that so? Or am I misunderstanding?

Answer 1

Too long for a comment: I learned the definition of tensorial from Greg Naber's book "Geometry, Topology and Gauge Fields, Volume 2" (Section 4.5).

Let $P$ be a principal $G$-bundle over $M$, let $V$ be a finite-dimensional vector space (in our case it will be $Lie(G)$) endowed with $\rho$ a representation of $G$ (in our case, the adjoint representation $g\cdot \xi = ad(g)\xi$, or $g\xi g^{-1}$ for matrix Lie groups).

Let $\sigma$ be a $V$-valued differential k-form over $P$. It is said to be tensorial if it satisfies:

1. $g^* \sigma = g^{-1} \cdot \sigma$, where $g^*$ is the pullback of the diffeomorphism of $P$ given by the natural $G$-action. Here $g\cdot \sigma(V_1,\dots,V_k)$ is multiplication by the $\rho$-action in the target space $V$.
2. It must be horizontal, i.e. $\sigma(V_1,\dots,V_k)$ vanishes whenever any of the $V_i$ is a vertical vector.

One can show that $k$-forms over $P$ satisfying correspond 1:1 to forms over the base $M$ with values in the associate vector bundle $P\times_\rho V$. In our case, the associated vector bundle is the adjoint bundle $ad(P)=P\times_{ad}Lie(G)$.