Lie Derivative on Vector Bundles I (a physicist) am trying to understand more about the foundations of differential geometry. I am having some trouble disentangling the difference between the Lie derivative and the covaraint derivative.
Let $E$ be a bundle over $M$. If we wish to compare an object $W_q$ in the fiber $F_q$ over $q$ to an object $W_p$ in the fiber $F_p$ over $p$, we require an isomorphism from $E_q$ to $E_p$. This could be provided by the flow generated by a vector field $X$, such that $q = \phi_t(p)$ and $p = \phi_t^{-1}(q) = \phi_{-t}(q)$. Provided we can find
$$
(\phi_{-t})_*: \pi^{-1}[q] \to \pi^{-1}[p],
$$
then
$$
(\phi_{-t})_* W_{\phi_t(p)} \in \pi^{-1}[p]
$$
is, for small $t$, a curve in $\pi^{-1}[p]$. We can differentiate this curve at $t=0$, giving us an object in $T_p(\pi^{-1}[p]) = T_p W_p$. This is intuitvely the Lie derivative for the bundle $E$.
It sounds to me like the Lie derivative then lives in the vertical subspace tangent to a given fiber. However, when considering the connection on a vector bundle, we also use such constructions when dealing with the lifts of curves and parallel transport. Actually, the vertical subspace in this regard has the property that
$$
\pi_* V_p = 0.
$$
But since the Lie derivative of two vector fields is again another vector field, I don't see how this property can be reconciled with the Lie derivative living in the vertical subspace over a fiber. I am wondering whether the condition applies only after we define a connection and parallel transport, and isn't generally true, or whether I am mistaken about where the Lie derivative "lives". However, if we do have a connection, can the Lie derivative still be defined?
Reference used: https://cefns.nau.edu/~schulz/lieder.pdf
 A: In my opinion, there are two aspects to this question. First, the Lie derivative is not defined on general vector bundles. You need a vector bundle on which you have an action of local diffeomorphisms, since you need to form a pullback along local flows of a vector field to define a Lie derivative. An action of local diffeomorphisms is available on $TM$, $T^*M$ and more generally on all tensor bundles, but not on a general abstract vector bundle.
The second issue is the vertical tangent bundle vs. the bundle itself. In the standard theory on vector bundles, you can either avoid this as follows:  Consider a vector bundle $E\to M$, a point $x\in M$ and the fiber $E_x$ of $E$ over $x$. Then for each $y\in E_x$, the vertical tangent space $V_yE=T_yE_x\subset T_yE$ can be canonically identified with $E_x$. This works in the same way as you identify the tangent spaces to a vector space with the vector space itself. So given $z\in E_x$ you can associated a tangent vector in $V_yE$ as $\tfrac{d}{dt}|_{t=0}y+tz$ (and since the curve stays in the fiber $E_x$, this tangent vector is really vertical. Alternatively (and this is the more usual approach), you can avoid the problem by saying that you first have to evaluate the pullbacks for diffenrent times $t$ in a fixed point $x$. Then you get a curve in $E_x$ and the derivative of a curve in a vector space can be canonically viewed as an element in that vector space (just take the definition of the derivative $c'(t)$ as a limit of $\tfrac{1}{h}(c(t+h)-c(t))$ as $h$ goes to $0$). One has to prove that the resulting curve in $E_x$ is actually smooth so that one can form the derivative.
Finally if you go to fiber bundles, then Lie derivatives (which exist only if you have an action of local differeomorphisms) and covariant derivatives (which need the choice of an additional structure like a connection or a Riemannian metric) actually do have values in the vertical tangent bundle rather than in the fiber bundle itself.
