How can a vector field act on a Lie Algebra element?

Question

We have the definition of a vector field as a smooth section of the tangent bundle $$X:P\longrightarrow TP,$$ where $(TP,\pi',P)$ is the tangent bundle over the total space of the principal G-bundle $(P,\pi,M)$. I.e, a vector field is an assignment of a vector ($\in TP$) to every point of a smooth manifold $P$.

I have, however, come across expressions like "$X(A)$" where $X$ is (supposedly) a vector field and $A$ is a Lie algebra element (for example at 19:31 here https://www.youtube.com/watch?v=j36o4DLLK2k).

How is the vector field defined in this case? Do we simply take for granted that the lie algebra is given a manifold structure, at which point the vector field is defined in the same way, something like $X:T_eG\longrightarrow T(T_eG)$ - assigning a vector in $T(T_eG) $ to another vector in $T_eG$?

Answer 1

This is the fundamental vector field generated by $A\in T_eG$. For $u\in P$, this is defined by $X^A(u):=\tfrac{d}{dt}|_{t=0}u\cdot exp(tA)$. Here the dot denotes the principal right action of $G$ on $P$, whose orbits are the fibers of $P$. For each $u\in P$, mapping $A$ to $X^A(u)$ induces a linear isomorphism from $T_eG$ to the vertical subspace $\ker(T_u\pi)\subset T_uP$. So these are the vectors tangent to the fibers of $P$.

Answer 2

Ok, here's what I've got so far. We have an ordinary one-form $\tilde{\theta}:P\longrightarrow T^*P$, and thus for $p\in P$ we have $\tilde{\theta}(p):T_pP\longrightarrow\mathbb{R}$ with $X_p\mapsto\tilde{\theta}(p)(X_p)$. An equivalent definition is \begin{alignat*}{2} \theta:\Gamma(TP)&\longrightarrow C^{\infty}(P) \\ X&\longmapsto \theta(X) \end{alignat*} where \begin{alignat*}{2} \theta(X):P&\longrightarrow \mathbb{R} \\ p&\longmapsto \theta(X)(p):=\tilde{\theta}(p)(X_p). \end{alignat*}

We can think of $\mathbb{R}$ as having some vector space structure, so everywhere we see it we can replace it by the Lie algebra, and produce a Lie-algebra valued one-form via $\tilde{\omega}:P\longrightarrow T^*P$, and thus for $p\in P$ we have $\tilde{\omega}(p):T_pP\longrightarrow T_eG$ with $X_p\mapsto\tilde{\omega}(p)(X_p)$. We therefore have \begin{alignat*}{2} \omega:\Gamma(TP)&\longrightarrow T_eG \\ X&\longmapsto \omega(X) \end{alignat*} where \begin{alignat*}{2} \omega(X):P&\longrightarrow T_eG \\ p&\longmapsto \omega(X)(p):=\tilde{\omega}(p)(X_p). \end{alignat*}

Thus, $\omega$ is a Lie-algebra valued one-form and $\omega(X)$ is a Lie-algebra-valued function, and can be differentiated by plugging it into a vector field.

Is that right?