Given a Riemannian manifold $(M,g,\nabla)$ endowed with the Levi-Civita connection we define its frame bundle $\mathfrak{F}(M):= \bigsqcup_{p \in M} \mathfrak{F}(M)_p$, where $\mathfrak{F}(M)_p$ is the space of the frames at $p$. It is possible to show that is a $GL(d,\mathbb{R})$ principal bundle.
A vector space on $X \in T_u \mathfrak{F}(M)$ is vertical if it is tangent to the fiber $\mathfrak{F}(M)_{\pi u}$ and it is called horizontal if it is the tangent vector of an horizontal curve on $\mathfrak{F}(M)$, i.e. a curve $\{u_t\}$ such that for any $e \in \mathbb{R}^d$, $\{u_te\}$ is $\nabla$-parallel to $\{\pi u_t\}$.
It is possible to show that the canonical projection $\pi : \mathfrak{F}(M) \xrightarrow{} M$ induces an isomorphism on the set of horizontal vectors $H_u \mathfrak{F}(M)$.
So given a tangent vector $X_x \in T_x M$ and a frame $u_x$ at $x$ there is an unique horizontal vector $X^o_x$ such that $d\pi X^o_x=X_x$. If $X$ is a vector field in $M$, $X^o$ is an horizontal vector field on $\mathfrak{F}(M)$, called the horizontal lift of $X$ to $u$.
In the case of a Lie group i know that there is a global section of the frame bundle and any vector field is complete, so i'm expecting the horizontal lift to be complete as well (am i right?).
I'm trying to express the horizontal lift of a generic vector field on a (matrix) Lie group $G$ w.r. to its global section. This section can be trivialized as $u_h=dL_h X_i$, where $\{X_i\}$ are the generators of the Lie algebra of $G$.
I know that there is a natural bases for $H_u \mathfrak{F}(M)$ given by the vectors
$$
H_{e_i}(u_h):= \text{ the horizontal lift of } u_he \in T_{\pi u_h}M=T_h M \text{ to } u
$$
EDIT:
I notice that my question was very confusing, i apologies.
Given a local chart on a chart $(U,x^i)$ of $M$ this induces a local trivialization of the frame bundle. In particular, if $e_i$ is the standard bases of $\mathbb{R}^n$ and $u \in \pi^{-1}(U)$ we have $u e_i= e^{j}_i \frac{\partial}{\partial x^j}$. We have that $(\partial_{x^i}, \partial_{e^i_j})$ is a bases for $T_u \mathfrak{F}(M)$. In this bases the fundamental vector fields can be described as
$$
H_{e_i}(u=(x,e))=e^j_i \partial_{x^j} – e^j_i e^l_m \Gamma^k_{ij}(x) \partial_{e^k_m}
$$
My questions is:
If there is a global section $v$ there is a way to obtain a global expression for the fundamental horizontal vector fields w.r. to this section, i.e. there is an expression for $H_{e_i}(v)$ in terms of the global section of the global trivialization?
Best Answer
I am assuming that by complete you mean all integral curves of a vector field $X$ can be extended to all of $\mathbb{R}$. I also haven't thought to much about this specific case before, so if I say something incorrect, my apologies.
You are correct that the frame bundle of a Lie group $G$ has a global section. In particular, this is because a basis for the Lie algebra determines a global frame, hence if $(T_1,\dots,T_n)\in T_eG\cong \mathfrak{g}$ is a basis for the Lie algebra, then the global section is given by:
$$s(g)=(L_{g*}T_1,\dots, L_{g*}T_n)$$
as at each point $g\in G$, this is a frame for $T_gG$. This gives us a global trivialization of the frame bundle for $G$, hence:
$$\mathfrak{F}(G)\cong G\times GL_n(\mathbb{R})$$
which is itself a Lie group so every vector field is complete, including horizontal lifts.
Ok so for a general principal $G$ bundle $\pi:P\rightarrow M$ how do we obtain a horizontal lift a vector field on $M$? Well we have that any connection one form determines a horizontal sub bundle $H \subset TP$, and it follows that $D_p\pi:H_p\rightarrow T_{\pi(p)}M$ is then an isomorphism for all $p\in P$. We thus fix a connection one form on $P$. Given $X\in \mathfrak{X}(M)$ we thus define a $X^H\in\mathfrak{X}(P)$ pointwise by:
$$X^H_p=(D_p\pi)^{-1}(X_{\pi(p)})$$
We need to show that this is smooth, so let $\phi$ be a bundle chart for an open neighborhood $U$ of $\pi(p)=x$. Let $\phi(p)=(x,h)$, then $D_p\phi:T_gP\rightarrow T_xM\oplus T_gG$ is an isomorphism. Let $\pi_U:U\times G$ be the projection onto $U$, and define a smooth section of the pullback bundle $\pi_U^*TU$ by: \begin{align*} Z:(U\times G)&\longrightarrow \pi_U^*TU\\ (x,g)&\longmapsto ((x,g),X_x) \end{align*} which is smooth because $X$ is smooth, and takes values in $\pi^*_U TU\subset \pi^*_UTU\oplus \pi^*_GTG\cong T(U\times G)$, so $Z$ is actually a smooth section of $T(U\times G)$. It is then easy to see that: \begin{align*} D_{(x,g)}\pi_U(Z_{(x,g)})=X_x \end{align*} Since $\phi$ is a diffeomorphism, it follows that there exists a unique vector field $Y\in\mathfrak{X}(P_U)$ such that $\phi_*(Y)=Z$, hence for all $p\in P_U$: \begin{align*} D_p\phi(Y)=Z_{\phi(p)} \end{align*} So: \begin{align*} D_{\phi(p)}\pi_U\circ D_p\phi(Y)=X_{\pi(p)} \end{align*} however: \begin{align*} D_{\phi(p)}\pi_U\circ D_p\phi=D_p(\pi_U\circ \phi)=D_p\pi \end{align*} so: \begin{align*} D_p\pi(Y_p)=X_{\pi(p)} \end{align*} Since $X^H$ is uniquely determined, it follows that $X^H$ is locally the horizontal component of $Y$, and thus $X^H$ must be smooth. It follows that $X^H$ is the unique horizontal lift of $X$.
Regarding your question pertaining Lie groups, perhaps you can clarify what you mean specifically, because as I see it now, there's not a clear way to obtain the horizontal lift in terms of a global section. Indeed, sections determine bundle charts/local trivializations, but these bundle charts don't respect the horizontal lift (the above proof demonstrates this clearly) as that is determined by a connection one form. You could choose the flat connection, which exists in this case because the bundle is trivial, but in general it won't be a metric connection, and so can't be the Levi Civita connection.
If it helps at all, you can write the connection one form with on $G\times GL_n(\mathbb{R}) as:
$$(\phi^{-1*}\omega)_{(g,A)}=A^{-1}\pi_G^*\omega_{s(g)} A+\pi_{GL_n(\mathbb{R})}^*(A^{-1}dA)$$ where $\omega_s=s^*\omega$, $A\in GL_n(\mathbb{R})$, and $A^{-1}dA$ is the Maurer Cartan form on $GL_n(\mathbb{R})$.