Q1.
In general if $X$ is a vector field on any manifold and $\gamma$ is an integral curve, then for each fixed number $s$, the translated curve $\tilde{\gamma}(t):=\gamma(t+s)$ is also an integral curve (provided all the times are in the maximal domain of definition… which is certainly the case if the integral curve is complete). To see this, we just use the chain rule: $\widetilde{\gamma}’(t)=\gamma’(t+s)=X[\gamma(t+s)]=X(\widetilde{\gamma}(t))$, proving that $\widetilde{\gamma}$ is an integral curve of $X$.
Another general fact is that if $X$ is a left-invariant vector field on a Lie group $G$, then it is complete. Also, if $\gamma:\Bbb{R}\to $ is an integral curve of $X$, then for each $g\in G$, the mapping $\widetilde{\gamma}:=L_g\circ\gamma$ (where $L_g:G\to G$ is left-translation by $g$) satisfies
\begin{align}
\widetilde{\gamma}’(t)&=T(L_g)[\gamma’(t)]\tag{chain rule}\\
&=T(L_g)[X(\gamma(t))]\tag{$\gamma$ is integral curve}\\
&=(X\circ L_g)(\gamma(t))\tag{$X$ is left-invariant}\\
&=X(\widetilde{\gamma}(t)).
\end{align}
Hence, $\widetilde{\gamma}=L_g\circ\gamma$ is also an integral curve of $X$.
As a corollary of these two facts, we have that if $X$ is a left-invariant vector field on a Lie group $G$, then each integral curve $\gamma:\Bbb{R}\to G$ is a Lie-group homomorphism (i.e forms a 1-parameter subgroup). To prove this, fix $s\in\Bbb{R}$ and define $\gamma_1(t):=\gamma(t+s)$ and $\gamma_2(t):=\gamma(s)\cdot\gamma(t)=L_{\gamma(s)}(\gamma(t))$. By the previous two bullet points, we have that $\gamma_1$ and $\gamma_2$ are both integral curves of $X$; also it is clear that they have the same initial point of $\gamma_1(0)=\gamma_2(0)=\gamma(s)$. By uniqueness of integral curves, it follows $\gamma_1=\gamma_2$; so by arbitrariness of $s\in\Bbb{R}$, it follows that for all $s,t\in\Bbb{R}$, $\gamma(t+s)=\gamma(s)\cdot \gamma(t)$, and hence $\gamma$ is a Lie-group homomorphism.
Now, apply this reasoning to the left-invariant vector field $X_{\xi}$.
Q2.
Your equation (1) has a typo; you should write $\Phi(q,g_{\xi}(t))$ instead.
Next, given any two Lie groups $G,H$ and a Lie group homomorphism $f:G\to H$, by taking the tangent map at the identity, we get the induced map at the level of Lie algebras, $f_*\equiv Tf_e:\mathfrak{g}\equiv T_eG\to T_eH\equiv \mathfrak{h}$. We often call $f_*$ the infinitesimal version of $f$ partly because of history, because it is defined at the level of tangent spaces (and the tangent space captures ‘infinitesimal’ motions (i.e velocities of curves)).
With this in mind, note that a right action of a group on a manifold can be viewed as a group homomorphism $\widetilde{\Phi}:G\to \text{Diff}(P)$, $g\mapsto \Phi(\cdot, g)$. Glossing over the infinite-dimensional nature of the diffeomorphism group, we can differentiate this at the identity to get the induced ‘infinitesimal’ map at the level of their Lie algebras, $\widetilde{\Phi}_*:\mathfrak{g}\to \Gamma(TP)$ (since the Lie algebra of the diffeomorphism group is the space of vector fields); this is equivalent to specifying a map $\Phi’:P\times \mathfrak{g}\to TP$ such that for each $\xi\in \mathfrak{g}$, we have $\Phi’(\cdot,\xi):P\to TP$ is a vector field; this is exactly the vector field you have called $\xi_P$.
Now, why do we refer to this as a generator? Well it’s because if we’re given a complete Lie-algebra action on a manifold, then we can ‘integrate’ it to get a Lie group action. See John Lee’s Introduction to Smooth Manifolds, Theorem 20.16 for a precise statement and proof.
Btw, if you recall, given a complete vector field $X$ on a manifold $M$, the flow $\Phi_X:M\times \Bbb{R}\to M$ is actually an action of the additive Lie group $\Bbb{R}$ on $M$. The flow is obtained from $X$ by solving a certain system of ODEs, i.e by ‘integrating’, and conversely, the vector field $X$ is obtained from the flow by differentiating at $t=0$ (which btw is the identity element of the group…so we always differentiate at the identity element of the group): $X(p):=\frac{\partial\Phi_X}{\partial t}(p,0)$, and it is often called the infinitesimal generator of the flow. So, the above terminology is just the natural generalization when we replace $\Bbb{R}$ by a more general Lie group $G$.
Ehresmann Connections.
I suggest you read my answer to Intuition behind connection 1-forms and Ehresmann connections first just to get a sense of what I’m about to say. Also, I hope you don’t mind but now I’m going to call the total space $X$ instead of $P$, and denote the group action as $m:X\times G\to X$ rather than $\Phi$ because I want to reserve $P$ for various projections, and I used $\Phi$ in the other answer to mean something else (which I’ll discuss shortly).
Ok so before we even talk about connection 1-forms, let us actually be more general and talk about a general Ehresmann connection in a fiber bundle.
Definition 1 (Ehresmann connection).
Let $(X,\pi, M)$ be a smooth fiber bundle. An Ehresmann connection on the fiber bundle is by definition a smooth subbundle $HX$ of $TX$ which is complementary to the vertical subbundle $VX:=\ker(T\pi)$, i.e $TX=HX\oplus VX$.
This is the fully general notion of a connection; in my link above, I describe the intuition for how this allows us at an infinitesimal level to ‘connect’ the different fibers. There are several equivalent ways of describing a connection; these descriptions boil down to the three ways of describing subspaces in linear algebra, namely directly, parametrically, implicitly (see the end of my link above). For now we shall discuss the direct and implicit approaches.
The definition above of an Ehresmann connection directly tells us which complement to take: by definition it gives us a subbundle $HX$ which is complementary to $VX$. Therefore, we end up with the two direct sum projections $P_H:TX\to HX$ and $P_V:TX\to VX$.
Conversely, suppose we have a smooth projection $P$ of $TX$ onto $VX$. More precisely, suppose we have a smooth fiberwise linear map $P:TX\to TX$ such that $\text{image}(P)=VX$ and $P\circ P=P$. By setting $HX:=\ker(P)$ (which will be a subbundle since $P$ has constant rank since the image is a subbundle), the condition $P\circ P=P$ implies $HX$ and $VX$ are complementary, and in fact the $P_V,P_H$ from above are respectively equal to $P$ and $I-P$.
Hence, an Ehresmann connection (i.e a choice of complementary subbundle $HX$ to $VX$) is equivalent to specifying a projection $P$. Yet another way of saying $P$ is a projection of $TX$ onto $VX$ is that $P:TX\to VX$ is a smooth fiberwise linear map such that $P|_{VX}=\text{id}_{VX}$.
We shall now use this last characterization. Ok, so having a fiberwise-linear map $P:TX\to VX$ means exactly that $P$ is a differential 1-form on $X$ with values in the vector bundle $VX$ (we call it a 1-form because it eats one tangent vector in $X$, and we say it is $VX$-valued because well… the output lands in $VX$). So, you see it is this which is the “true” connection 1-form. For completeness, let me record this as a precise definition:
Definition 1’ (Ehresmann connection 1-form).
Let $(X,\pi,M)$ be a smooth fiber bundle. An Ehresmann connection 1-form on the fiber bundle is by definition a smooth $VX$-valued differential 1-form on $X$, call it $P:TX\to VX$, such that its restriction to $VX$ is the identity, i.e $P|_{VX}=\text{id}_{VX}$.
So, by our discussion above, definitions 1 and 1’ are equivalent. So, if you want to discuss connections in general, you need to talk about complementary subbundles of $VX$, and if you want to speak in terms of connection 1-forms, then you actually need to speak of $VX$-valued 1-forms. And yes, we do need such generality, and we have to consider vector-bundle valued forms.
Principal Connections.
Next, recall that if $m:X\times G\to X$ is a smooth right action, then we also get an induced right action $TX\times G\to TX$ by sending $(v_x,g)\mapsto v_x\cdot g:=T(m(\cdot,g))_x[v_x]$. So, right multiplication by $g$ maps the fiber $T_xX$ into $T_{x\cdot g}X$. I leave it to you to verify this really is a right action on $TX$; we call this the induced right action on $TX$. With this in mind, let us specialize to principal connections on a principal bundle.
Definition 2 (Principal Connection).
Let $(X,\pi,M,G)$ be a smooth principal bundle. A principal connection on $X$ is by definition an Ehresmann connection $H\equiv HX$ which is $G$-equivariant in the sense that for all $x\in M,g\in G$, we have that the fibers of $H$ are related via $H_{x\cdot g}=H_x\cdot g$.
So, the principality of the connection is a compatibility condition between the subbundle $H$ and the group action (on $X$ and the induced action on $TX$). It basically says that the subspaces $H_x$ have to vary compatibly as we vary the point $x$ within in $G$-orbit. Let me now make the corresponding definition in terms of 1-forms:
Definition 2’ (Principal connection 1-form)
Let $(X,\pi,M,G)$ be a smooth principal bundle. A principal connection 1-form on $X$ is by definition a connection 1-form on $X$ which is $G$-equivariant relative to the naturally induced right action of $G$ on $TX$ (and the obviously induced action on the subbundle $VX$).
Very explicitly, this means a smooth fiberwise-linear map $P:TX\to VX$ such that
- $P|_{VX}= \text{id}_{VX}$
- for all $v\in TX$ and $g\in G$, $P(v\cdot g)=P(v)\cdot g$ (the $\cdot$ on the LHS being the action $X\times G\to X$ and the $\cdot$ on the right being the action $TX\times G\to TX$).
Of course, the definitions (2) and (2’) are equivalent:
Theorem.
Let $(X,\pi,B,G)$ be a smooth principal bundle.
If $H$ is a principal connection in the sense of definition (2), then the corresponding direct sum projection $P=P_V:TX=H\oplus VX\to VX$ is a principal connection 1-form in the sense of definition (2’).
Conversely, if $P:TX\to VX$ is a principal connection 1-form in the sense of definition (2’), then $H:=\ker(P)$ is a principal connection in the sense of definition (2).
Note that we only need to prove the respective equivariance conditions (the other facts are true for general Ehresmann connections). This easily from the following linear algebra lemma:
Let $\mu:V_1\oplus V_2\to W_1\oplus W_2$ be linear. Then, $\mu$ restricts to linear maps $\mu_i:V_i\to W_i$ for $i\in \{1,2\}$ if and only if for $i\in \{1,2\}$ we have $P_{W_i}\circ \mu=\mu\circ P_{V_i}$.
with $\mu=T(m(\cdot,g))_x:T_{x}X=H_x\oplus V_xX\to T_{x\cdot g}=H_{x\cdot g}\oplus V_{x\cdot g}$. I leave the details to you.
In the case of a principal bundle, there is one other modification we can make to definition (2). This is due to the following fact about the vertical subbundle of a principal bundle:
Fact.
Let $(X,\pi,M,G)$ be a principal bundle, $m:X\times G\to X$ the group action map, and let $VX$ be the vertical subbundle. Then, the mapping $\Phi:X\times\mathfrak{g}\to VX$ given by $\Phi(x,\gamma):=T\left(m(x,\cdot)\right)_e[\gamma]$ is a vector bundle isomorphism; it makes the diagram below commute:
$\require{AMScd}$
\begin{CD}
X\times\mathfrak{g} @>{\Phi}>> VX \\
@V{\text{pr}_1}VV @VV{\pi|_{VX\to X}}V \\
X @>>{\text{id}_X}> X
\end{CD}
To prove this, note that $\Phi$ is obviously smooth. Next, $m$ being a principal action implies that for each $x\in X$ the mapping $m(x,\cdot):G\to X_{\pi(x)}$ is a diffeomorphism, so taking the tangent map at the identity shows that $\Phi(x,\cdot)$ is a linear isomorphism $\mathfrak{g}\to T_{m(x,e)}(X_{\pi(x)})=V_xX$. Lastly,
since the base map $\text{id}_X$ is a diffeomorphism, it follows that $\Phi$ is a vector bundle isomorphism from the trivial vector bundle $X\times\mathfrak{g}$ onto $VX$.
As an aside, note that this isomorphism does not depend on any connection; it relies solely on the principal bundle structure.
The map $\Phi$ also satisfies an equivariance condition, but it is not relative to the ‘naive’ group actions, but rather the $\text{Ad}$ map gets involved:
Lemma.
Let $(X,\pi,M,G)$ be a principal bundle and $\Phi:X\times \mathfrak{g}\to VX$ the vector bundle isomorphism introduced above. Then, $\Phi$ is $G$-equivariant relative to the right action $(x,\xi)\cdot g:=(x\cdot g, \text{Ad}_{g^{-1}}(\xi))$ of $G$ on $X\times \mathfrak{g}$, and the induced right action of $G$ on $TX$ described previously.
Explicitly, we have that for all $x\in X,\xi\in\mathfrak{g},g\in G$,
\begin{align}
\Phi(x\cdot g, \text{Ad}_{g^{-1}}(\xi))&=\Phi(x,\xi)\cdot g.
\end{align}
The proof of this is pretty easy:
\begin{align}
\Phi(x\cdot g, \text{Ad}_{g^{-1}}(\xi))
&= (x\cdot g)\cdot (g^{-1}\cdot \xi\cdot (g^{-1})^{-1})
=x\cdot \xi\cdot g
=\Phi(x,\xi)\cdot g.
\end{align}
There are quite a few things which I have simply denoted as $\cdot$, but the point is that all these actions ‘commute’, so that’s why I can freely regroup terms as I wish.
Answering your Q3.
Finally, let us address the relationship between the connection 1-form $P$ from definition (2’), and the connection 1-form $\omega$ from your definition. Note that $P:TX\to VX$ being a vector-bundle-valued 1-form is quite a complicated beast, but fortunately we have a global trivialization $\Phi:X\times \mathfrak{g}\to VX$. Hence, we can consider instead the map $\Phi^{-1}\circ P:TX\to X\times\mathfrak{g}$, and furthermore since the first component only captures the base point information, we can ignore that, and consider $\omega:=\text{pr}_2\circ\Phi^{-1}\circ P:TX\to \mathfrak{g}$.
So, the $VX$-valued $1$-form $P$ and the $\mathfrak{g}$-valued $1$-form $\omega$ contain the same amount of infomation. Finally, since $\Phi$ is a vector-bundle isomorphism, it easily follows that
- the condition $P|_{VX}=\text{id}_{VX}$ in definition (2’) is equivalent to the condition (1) you have in the definition of $\omega$.
- since $\text{pr}_2,\Phi,P$ are all $G$-equivariant (relative to the aforementioned right actions on $X\times \mathfrak{g}$ and $VX,TX$) it follows that the composition, which is $\omega$, is also $G$-equivariant. Conversely, if you give me a $G$-equivariant $\omega$, then when I reconstruct $P$, it will also be $G$-equivariant.
Thus, the definition (2’) for the connection 1-form $P$, and your definition for the connection 1-form $\omega$ are equivalent.
Finally, I hope you now understand why your remark
… because I was thinking we should take $\omega_q(v)\triangleleft g$ only.
isn’t correct. The issue is that $\omega:=\text{pr}_2\circ\Phi^{-1}\circ P$ contains the isomorphism $\Phi$ which trivializes $VX$, and it is here that the $\text{Ad}$ representation appears. Therefore, we have a choice to make between the simplicity of the object and the simplicity of its equivariance:
- we can enjoy the simple equivariance property of $P:TX\to VX$, but at the cost of it being a vector-bundle-valued $1$-form on $X$,
- Or, we can enjoy the fact that $\omega:TX\to \mathfrak{g}$ is a vector space-valued 1-form (so it’s a simpler object), at the cost of a more ‘unexpected’ equivariance condition.
So, this seems like one of the many instances of ‘conservation of difficulty’. Btw, you may also want to refer to this answer which explains the $\text{Ad}$-equivariance condition of $\omega$ at the level of vector bundles and frames.
