A Lie group is a group that is also a differentiable manifold, and where the product/inverse operations are smooth. We will only consider real matrix Lie groups, which are subgroups of \(GL_n(\RR)\).
The left and right translations by some element \(h \in G\) are smooth maps:
\[L_h(g) = h.g\] \[R_h(g) = g.h\]Note that \(L_g\) and \(R_h\) commute for any \(g, h\) by associativity of the group product. They are also invertible and their inverse is smooth (which makes them diffeomorphisms), with inverses given by:
\[\inv{L_h} = L_{\inv{h}}\] \[\inv{R_h} = R_{\inv{h}}\]As such, their tangent maps induce isomorphisms between tangent spaces:
\[\dd L_h(g): T_g(G) \simeq T_{hg}(G)\] \[\dd R_h(g): T_g(G) \simeq T_{gh}(G)\]where the inverses are given by:
\[\dd \inv{L_h} = \dd L_{\inv{h}}\] \[\dd \inv{R_h} = \dd R_{\inv{h}}\]In particular, the tangent space at each point \(g\) is isomorphic to the tangent space at the identity element:
\[\dd \inv{L_h}(e): T_e(G) \simeq T_g(G)\] \[\dd \inv{R_h}(e): T_e(G) \simeq T_g(G)\]These isomorphisms vary smoothly with \(g\) so that the tangent bundle \(TG\) is globally isomorphic to \(G \times T_e(G)\) (it is said trivializable): any tangent vector \(\dd g\) is uniquely determined by its coordinates in the tangent space at the identity (together with the base point \(g\)). These coordinates are called the left/right trivializations of \(\dd g\) and satisfy:
\[\dd g = \dd L_g.\db g\] \[\dd g = \dd R_g.\ds g\]where \(\db g, \ds g\) are tangent vectors at the identity element, respectively called the body-fixed and spatial coordinates in robotics. Since we are only dealing with matrix Lie groups, left/right translation derivatives are given by matrix products:
\[\dd L_h(g).\dd g = h.\dd g\] \[\dd R_h(g).\dd g = \dd g.h\]Therefore, left/right trivializations are simply:
\[\dd g = g.\db g = \ds g.g\]For reasons that will become clear later, the tangent space at the identity \(T_e(G)\) is called the Lie algebra of the group \(G\), denoted \(\alg{g}\).
The inner automorphism \(\Psi_h(g) = h.g.\inv{h}\) is obtained by composing left/right translations by \(h\):
\[\Psi_h = L_h \circ R_\inv{h} = R_\inv{h} \circ L_h\]Its differential is an isomorphism between tangent spaces, and is an automorphism when differentiated at the identity. This automorphism is called the adjoint of the group:
\[\begin{align} \Ad_h &= \dd \Psi_h(e): \alg{g} \to \alg{g}\\ &= \dd L_h.\dd R_\inv{h} = \dd R_\inv{h}.\dd L_h \\ \end{align}\]It is a group homomorphism:
\[\Ad: G \to GL(\alg{g})\] \[\Ad_{gh}=\Ad_g.\Ad_h\] \[\Ad_{\inv{h}}=\inv{\Ad_h}\]This group homomorphism provides the adjoint representation of the group over its Lie algebra. The adjoint can itself be differentiated:
\[\begin{align} \dd \Ad_g(x).\dd g &= \dd g.x.\inv{g} - g.x.\inv{g}.\dd g.\inv{g}\\ &= g.\db g.x.\inv{g} - g.x.\db g.\inv{g} \\ &= \Ad_g.\block{\db g.x - x.\db g} \\ \end{align}\]The expression \(\block{\db g.x - x.\db g}\) is called a Lie bracket, denoted by \([\db g, x]\). It is always an element of the Lie algebra. From the above, we see that the left-trivialized tangent of \(\Ad_g\) is \([\db g, .]\), which we denote by \(\ad(\db g)\). One can easily check that the right-trivialized tangent is \(\ad(\ds g)\). The Lie bracket operator is what provides the Lie algebra structure on the tangent space at the identity.
As stated above, the left/right trivializations of a tangent vector \(\dd g \in T_g G\) satisfy:
\[\dd g = g.\db g = \ds g.g\]which implies:
\[\ds g = g.\db g.\inv{g} = \Ad_g.\db g\]In other words, the group adjoint provides the conversion between body-fixed and spatial coordinates.
Simliarly, the dual space of the Lie algebra \(\alg{g}^\star\) can also be used for the left/right trivialization of tangent covectors \(\tau_g^\star \in T_g^\star (G)\), via the natural pairing between tangent vectors and covectors:
\[\begin{align} \tau_g^\star.\dd g &= \tau_g^\star.\dd L_g.\db g \\ &= \tau_g^\star.\dd R_g.\ds g \\ \end{align}\]from which we identify the body-fixed coordinates \(\tau_g^{\star b} = \tau_g^\star.\dd L_g\), and spatial coordinates \(\tau_g^{\star s} = \tau_g^\star.\dd R_g\). The conversion between the two is given by:
\[\begin{align} \tau_g^{\star b} &= \tau_g^\star.\dd L_g \\ &= \tau_g^{\star s}.\inv{\dd R_g}.\dd L_g \\ &= \tau_g^{\star s}.\dd R_{\inv{g}}.\dd L_g \\ &= \tau_g^{\star s}.\Ad_g \\ \end{align}\]In other words:
\[\tau_g^{\star b} = \Ad_g^\star.\tau_g^{\star s}\]It should be noted that the mapping \(g \mapsto \Ad_g^\star\) is not a group representation since \(\Ad_{gh}^\star = \Ad_h^\star.\Ad_g^\star\). Instead, the mapping \(g \mapsto \Ad_{\inv{g}}^\star\) is used and is called the coadjoint representation of the group over its Lie coalgebra. It satisfies:
\[\tau_g^{\star s} = \Ad_{\inv{g}}^\star.\tau_g^{\star b}\]Put succintly, while the adjoint representation maps body-fixed velocities to spatial velocities, the coadjoint representation maps body-fixed forces to spatial forces.
TODO
As seen above, tangent vectors between Lie groups can be left/right trivialized over the corresponding Lie algebras. Likewise, tangent maps between tangent spaces can be left/right trivialized as linear maps between Lie algebras:
\[f: G \to H\] \[\dd f(g): T_g(G) \to T_{f(g)} H\] \[\db f(g): \alg{g} \to \alg{h} = \dd L_{\inv{f(g)}}.\dd f(g).\dd L_g\] \[\ds f(g): \alg{g} \to \alg{h} = \dd R_{\inv{f(g)}}.\dd f(g).\dd R_g\]From the above it can be checked that:
\[\db L_h = I\] \[\ds R_h = I\]Since left/right translation commute \(R_h \circ L_g = L_g \circ R_h\), we also get the following:
\[\begin{align} \ds L_h &= \Ad_h \\ \db R_h &= \Ad_{\inv{h}}\\ \end{align}\]Probably the most useful to remember in practice is \(\ds L_h = \Ad_h\), and recover the others from it. It also provides the interpretation of the group adjoint as a coordinate change.
From \(g.\inv{g} = e\), we get:
\[\dd g.\inv{g} + g.\dd\inv{g} = 0\]In other words:
\[\begin{align} \dd\inv{g} &= -\inv{g}.\dd g.\inv{g} \\ &= -\dd L_{\inv{g}}.\dd R_{\inv{g}}.\dd g \\ \end{align}\]The left/right trivializations are given by:
\[\begin{align} \db\inv{g} &= -g.\db g.\inv{g} = -\Ad_g.\db g \\ \ds\inv{g} &= -\inv{g}.\ds g.g = -\Ad_{\inv{g}}.\ds g\\ \end{align}\]Product is bilinear, hence:
\[\begin{align} \dd\block{a.b} &= \dd a.b + a.\dd b \\ &= \dd R_b.\dd a + \dd L_a.\dd b \\ \\ \db\block{a.b} &= \Ad_{\inv{b}}.\db a + \db b \\ \ds\block{a.b} &= \ds a + \Ad_a.\ds b \\ \end{align}\]Let us consider a matrix Lie group \(G\) as a subgroup of \(GL(n)\), a group element \(g \in G\) and two derivation directions \(\dd g_1, \dd g_2 \in \alg{g}\). (TODO two vector fields instead) By Schwartz’s theorem, the following holds in \(GL(n)\) taken as a vector space:
\[\dd^2 g_1.\dd g_2 = \dd^2 g_2.\dd g_1\]Now, consider two left trivializations of tangent vectors \(\dd g_1, \dd g_2\) as:
\[\begin{align} \dd g_1 & = g \omega_1 \\ \dd g_2 & = g \omega_2 \\ \end{align}\]We get:
\[\dd \omega_1.\dd g_2 = \dd \block{\inv{g}.\dd g_1}.\dd g_2 = -\inv{g}\dd g_2 \inv{g} \dd g_1 + \inv{g} \dd^2 g_1.\dd g_2\]Or, after left-trivialization:
\[\dd \omega_1.\omega_2 = -\omega_2 \omega_1 + \inv{g} \dd^2 g_1.\dd g_2\]Similarly:
\[\dd \omega_2.\omega_1 = -\omega_1 \omega_2 + \inv{g} \dd^2 g_2.\dd g_1\]The right-most terms being equal in the two equations, we obtain:
\[\dd \omega_1.\omega_2 + \omega_2 \omega_1 = \dd \omega_2.\omega_1 + \omega_1 \omega_2\]Or, equivalently:
\[\begin{align} \dd \omega_1.\omega_2 &= \dd \omega_2.\omega_1 + \omega_1 \omega_2 - \omega_2 \omega_1\\ &= \dd \omega_2.\omega_1 + [\omega_1, \omega_2]\\ \end{align}\]So the Lie bracket gives the correction to apply when switching the derivation order.