Let $P$ be a principal $G$ bundle, where $G$ denotes a Lie group.
Then this action defines a map $\sigma : \mathfrak{g} \to TP$ from the Lie algebra of $G$ to the vector fields on $P$, given by: $$\sigma_p(X) =\frac{d}{dt}(p \cdot e^{tX}),$$ where $X \in \mathfrak{g}$ and $p \in P$.
I'm trying to understand how this works. From my understanding if the representation of the group is a matrix representation, so some element of $Aut(M)$, then we can explicitly perform the derivative to get $$\sigma_p(X) =p \cdot X,$$ where now we have the induced Lie algebra acting on the manifold.
I don't see how this defines a vector field? Is it because, say we take any element $p \in P$ to be 2-dimensional, and $X$ a 2-dimensional representation, then $p \cdot X$ will be a 2-dimensional vector, say $p \cdot X=(a_1, a_2)$ and we are viewing this as a vector of the form $a_1 \frac{\partial}{\partial x} + a_2 \frac{\partial}{\partial y}$?
Also, do you know of any good notes for when we are not in a matrix representation?
Consider $f:\mathbb{R}\rightarrow P$ defined by $f(t)=p.e^{tX}$. It is a differentiable map, and its differential $df:\mathbb{R}\rightarrow TP$, $\sigma_p(X)=df(1)$.