Let $\pi:P \rightarrow M$ be a principal $G$-bundle, and let $A \in \mathfrak{g}$, where the Lie algebra of $G$ is indicated. The *fundamental field* $A$# used to define connections is given by


$A$#$(p) := \frac{d}{dt}(\exp(At)p)|\_{t=0}$.


$A$# is well defined since $e^{At}p$ can be regarded as a vector in $\pi^{-1}(\pi(p))$. Intuitively, I try to think of $A$# as the (vertical) direction of the displacement on the fiber generated by $A$. 


By the defining properties of principal bundles (in particular, the free action of $G$), we have $\{A$# $: A \in \mathfrak{g}\} \simeq \mathfrak{g} \simeq \mathcal{V}(p)$, where $\mathcal{V}(p)$ is the vertical subspace at $p$. Something like 


$A$#$(p) = \lim\_t t^{-1}(e^{At}p - p) = A \cdot p$ 


would be (to me) a nice way of thinking about $A$#$(p)$, except that this is (at best) a formal equality. More precisely, $L\_{\exp(At)}$ is the 1-parameter group of diffeomorphisms generated by $A$#, where $L\_g$ denotes left multiplication by $g$.


**My question**: is there a better way of thinking about a fundamental field than (either the definition itself) or the notional equation above?



What is a good way to think about a fundamental field on a principal G-bundle?
If you have a homogeneous space $X$ with structure group $G$ (in your case, the fiber passing through $p$) then left multiplication give you a nice action $L: X \times G \to X$. Then for a fixed $p \in X$, $L(p) : G \to X$. The differential of this guy is a map $L(p)\_\* : TG \to TX$ which takes an element of $T\_gG$ to an element of $T\_{g \cdot p} X$. At $g = 1$, this means you get a map $L(p)\_\* : \mathfrak{g} \to T\_pX$. This is actually a representation of $\mathfrak{g}$ on the space of vector fields on $X$, and will inherit nice properties depending on how nice the action of $G$ on $X$ is. 


This map $L(p)\_\*$ is precisely the sharp map you describe: it takes an element $\xi$ of $\mathfrak{g}$ and gives a vector field on $X$ which coincides with infinitesimal multiplication by $\xi$. In your case, $G$ acts freely so $X$ looks like a copy of $G$.


If $G$ actually has a nice matrix representation then unrolling the definitions, you'll find that you really can write your proposed equation $A^\sharp(p) = A \cdot p$. For example, look at how the infinitesimal generators of $\mathfrak{so}(3)$ acts on points of the unit sphere by multiplication --- you'll easily see the vector fields whose flows are $\exp(\xi t)$.


