Let $\pi:P\longrightarrow M$ be a $G$-principal bundle. For $p\in P$, $V_p$ denotes the space of tangent vertical vectors, that is, $V_p:=T_pP_{\pi(p)}$. The space of vertical vector fields will be denoted by $\mathfrak{X}^v(P)$ whereas $\mathfrak{g}$ stands for the Lie algebra of $G$ (i.e., the space of left invariant vector fields on $G$).

I'm trying to show::


$(i)$ There is an isomorphism of vector space $T_eG\simeq V_p$ for every $p\in P$;

$(ii)$ There is an isomorphism of Lie algebras $\mathfrak{g}\simeq \mathfrak{X}^v(P)$.


$(i)$ Define the map: \begin{align*} \displaystyle \Phi:V_p\longrightarrow T_eG, \end{align*} choosing a local trivialization $(U, \phi)$ with $\pi(p)\in U$ and setting: \begin{align*} \displaystyle \Phi(v):=d(L_{\phi_{\pi(p)}(p)}^{-1})_{\phi_{\pi(p)}(p)}(d(\phi_{\pi(p)})_p(v)). \end{align*} Above $\phi_{\pi(p)}:P_{\pi(p)}\longrightarrow G$ is the diffeomorphism induced by $\phi$, that is, $$\phi_{x}:=\textrm{pr}_2\circ \phi|_{P_x},\ \forall x\in M,$$ where $P_x:=\pi^{-1}(x)$. The inverse of $\Phi$ is the map: \begin{align*} \displaystyle \Psi:T_eG\longrightarrow V_p, \end{align*} defined by: \begin{align*} \displaystyle \Psi(v):=d(\phi_{\pi(p)}^{-1})_{\phi_{\pi(p)}(p)}(d(L_{\phi_{\pi(p)}(p)})_e(v)). \end{align*} Obs. 1 It is not difficult to show $\Phi$ and $\Psi$ don't depend on the choice of local trivializations.

$(ii)$ As to $(ii)$ I'd like to use $(i)$ to define the isomorphism. Define: $$\begin{align*} \displaystyle \Psi:\mathfrak{g}\longrightarrow \mathfrak{X}^v(P), X\longmapsto \tilde{X} \end{align*}$$ choosing a local trivialization $(U, \phi)$ with $\pi(p)\in U$ and setting: $$\begin{align*} \displaystyle \tilde{X}_p:=d(\phi_{\pi(p)}^{-1})_{\phi_{\pi(p)}(p)}(X_{\phi_{\pi(p)}(p)}). \end{align*}$$ Notice this map came from the map $\Psi$ of $(i)$ as follows: applying $X_e$ in the previous $\Psi$ and using the fact $X$ is left invariant we find this new $\Psi$.

Can anyone help me defining the inverse of $\Psi$? I thought I should use the composition $$\mathfrak{X}^v(P)\longrightarrow V_p\stackrel{\simeq}{\longrightarrow} T_eG\stackrel{\simeq}{\longrightarrow} \mathfrak{g},$$ the problem is how to choose the first map $\mathfrak{X}^v(P)\longrightarrow V_p$ for I don't have a distinguished point on $P$.


Remark. A vertical vector field on $P$ is a vector field $X\in\mathfrak{X}(P)$ such that $X_p\in V_p:=T_pP_{\pi(p)}$ for every $p\in P$.

  • $\begingroup$ Can you be more explicit about your definition of vertical vector field? I assume you want them to be $G$-invariant vector fields, tangent to the fibers of the bundle? There are quite a lot of these. $\mathfrak X^v(P)$ should be isomorphic to $\mathfrak X(M) \times \mathfrak g$, at least as vector spaces. Is that what you meant? $\endgroup$ – user98602 Dec 16 '15 at 19:23
  • $\begingroup$ That's even bigger than what I said. That's not $\mathfrak g$ - $\mathfrak g$ is finite dimensional! $\endgroup$ – user98602 Dec 16 '15 at 19:43
  • $\begingroup$ @MikeMiller I added the definition of a vertical vector field and it does not involve invariance. Are you sure of the isomorphism $\mathfrak{X}^v(P)\simeq \mathfrak{X}(M)\times \mathfrak{g}$? I know $V(P)\simeq P\times \mathfrak{g}$ as vector bundles where $V(P)$ is the vertical bundle of $P$. $\endgroup$ – PtF Dec 16 '15 at 19:43
  • $\begingroup$ @MikeMiller check this books.google.com.br/… . However the approach to show $\mathfrak{g}\simeq \mathfrak{X}^v(P)$ the author follows is different. $\endgroup$ – PtF Dec 16 '15 at 19:46
  • 1
    $\begingroup$ Since $V(P)\cong P\times\mathfrak g$, the space of vertical vector fields is isomorphic to $C^{\infty}(P,\mathfrak g)$. The fundamental vector fields (parametrized by elements of $\mathfrak g$ that you are describing are just a very special set of vertical vector fields. $\endgroup$ – Andreas Cap Dec 18 '15 at 15:13

I'd like to share my conclusions. The isomorphism $$T_eG\simeq V_p\quad (\textrm{or}\quad \mathfrak{g}\simeq V_p)$$ is correct and can also be done using the exponential map. As to my second statement $\mathfrak{g}\simeq \mathfrak{X}^v(P)$ is not right. As @Andreas Cap has observed, what really holds is $$\mathfrak{X}^v(P)\simeq C^\infty(P; \mathfrak{g}).$$ Intuitively the argument is as follows:

We have an isomorphism $\mathfrak{g}\simeq V_p$ for every $p\in P$ and since $$P\times \mathfrak{g}=\bigcup_{p\in P} \{p\}\times \mathfrak{g}\quad \textrm{and}\quad V(P)=\bigcup_{p\in P} \{p\}\times V_p,$$ one gets an isomorphism of vector bundles $$P\times \mathfrak{g}\simeq V(P),$$ where $V(P)=\{(p, v): p\in P, v\in V_p\}$ is the vertical bundle. From this it follows: $$\mathfrak{X}^v(P)=\Gamma(V(P))\simeq \Gamma(P\times \mathfrak{g})\simeq C^\infty(P; \mathfrak{g}).$$


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