Codex @codex  0

On a coordinate trivialization of a principal bundle, the right-invariant vector fields $R_a(\gamma )=T_a\gamma$ satisfy $[R_a,R_b]=-c^d{}_{ab}{R}_d$. With $A=A_i^a{T}_{a}d{x}^i$, the displayed lifts project to the commuting coordinate fields and contract to zero with the principal connection. Taking their Lie brackets gives the local curvature of a principal connection coefficients $F_{ij}={\partial }_i{A}_j-{\partial }_j{A}_i+[A_i,A_j]$. Thus these lifts commute exactly when the connection is flat. Arbitrary base fields need not commute, and flatness alone does not supply a global coordinate frame or remove holonomy of a connection.

In exponential scale coordinates, the real affine group has matrices $\left(\begin{array}{cc}{e}^{\alpha }& \beta \\ 0& 1\end{array}\right)$. The left and right Maurer-Cartan forms are respectively $Dd\alpha +T{e}^{-\alpha }d\beta$ and $Dd\alpha +T(d\beta -\beta d\alpha )$. Their dual vector fields obey $[D_L,T_L]=T_L$ and $[D_R,T_R]=-T_R$, displaying the opposite bracket sign for right-invariant vector fields.

The displayed matrices describe the orientation-preserving component $SE(2)$ of the full Euclidean group $E(2)=O(2)⋉{\mathbb{R}}^2$. Both have the same Lie algebra, so this naming convention does not affect the calculation. Choose the translation generators $P_1,P_2$ and rotation generator $J$ in the Lie algebra of this Matrix Lie group asTaking matrix commutators givesAll remaining brackets follow from antisymmetry. The corresponding left-invariant vector fields in coordinates $(x,y,\psi )$ areTheir Lie brackets of vector fields obey the same relations. This also fixes which sign of the rotation generator is being used.

There is an important qualification to the printed connection prescription. For an arbitrary vector field $Z$ and smooth function $f$, its literal right-hand side obeysAn affine connection instead requires ${\nabla }_{L_i}(fZ)=L_i(f)Z+f{\nabla }_{L_i}Z$. Thus the formula for arbitrary $Z$ defines a connection only when $\lambda =1$. For example, at $\psi =0$, take $i=1$, $f=x$ and $Z=L_1$: the erroneous formula gives $\lambda L_1$, whereas the Leibniz rule requires $L_1$.

The intended one-parameter family is the bracket connection on a Lie group: prescribe the rule for left-invariant $Z$, then extend it by the Leibniz rule. If $[L_i,L_j]=c_{ij}{}^k{L}_k$, this meansExtend linearly over functions in the first slot. This is a well-defined affine connection for every real $\lambda$. The following Ricci tensor calculation applies to this intended family; under the literal arbitrary-$Z$ reading only its $\lambda =1$ member exists.

Use the curvature conventionFor left-invariant vector fields the structure constants of a Lie algebra are constant. The Jacobi identity givesWriting $A=\lambda (\lambda -1)$, the only nonzero curvature actions, apart from antisymmetry in the first two slots, areFor instance, $[[J,P_1],J]=[P_2,J]=P_1$. Contracting the first and output slots gives $\mathrm{Ric}(L_3,L_3)=-2A$, while every other component vanishes. Equivalently, $[[Z,X],Y]={\mathrm{ad}}_Y{\mathrm{ad}}_{X}Z$, soThis is the Killing form. Here ${\mathrm{ad}}_J$ rotates the two-dimensional translation space and vanishes on $J$, so $B(J,J)=-2$; translation generators give all other components zero. ThereforeSince the coframe component dual to $L_3$ is $d\psi$, the same Ricci tensor is $2\lambda (1-\lambda )d\psi \otimes d\psi$ in these coordinates. Reversing the curvature convention reverses this sign.

At both $\lambda =0$ and $\lambda =1$, the full curvature vanishes. At $\lambda =0$, the left-invariant vector fields form a global parallel frame. At $\lambda =1$, every right-invariant vector field is parallel because left- and right-invariant vector fields commute; the connection is flat in that global parallel frame. Flatness does not imply zero torsion: for left-invariant arguments the torsion tensor isThus the two endpoint connections are flat, with opposite nonzero torsion. For comparison, $\lambda =1/2$ is torsion-free but has nonzero curvature, with ${\mathrm{Ric}}_{33}=1/2$. No metric has been specified, so these connections should not automatically be identified with a Levi-Civita connection.

Choose a basis $T_a$ of the Lie algebra, with $[T_b,T_c]=c^a{}_{bc}{T}_a$. On the Matrix Lie group setLeft multiplication by a constant matrix preserves $\lambda$, while right multiplication preserves $\rho$. Differentiating $g^{-1}g=I$ gives $d(g^{-1})=-g^{-1}(dg)g^{-1}$. Apply the graded Leibniz rule to obtainBecause the wedge product antisymmetrizes the matrix products, these Maurer-Cartan equations becomeFollow the vector-field naming used in the question: $L_a(g)=T_{a}g$ is right-invariant and generates left multiplication, whereas $R_a(g)=g{T}_a$ is left-invariant and generates right multiplication. They are dual to ${\rho }^a$ and ${\lambda }^a$, respectively. For any one-form $\eta$, $d\eta (X,Y)=X(\eta (Y))-Y(\eta (X))-\eta ([X,Y])$. Evaluating the two Maurer-Cartan equations on the corresponding invariant fields yieldsFor the last identity, the left and right multiplication flows commute. The opposite sign for the right-invariant vector fields is therefore intrinsic; the letters attached to the two families are only a convention.

Differentiate the product with the first factor close to the identity. A small displacement of its $x^{\prime }$ coordinate adds to $x$; a small $y^{\prime }$ displacement adds to $y$; and a small $z^{\prime }$ displacement changes $(x,y,z)$ by $(-x,y,1)$ times that displacement. The infinitesimal left actions therefore give the right-invariant vector fieldsFor right multiplication, regard $(x^{\prime },y^{\prime },z^{\prime })$ as the fixed point and the second factor as infinitesimal. Its displacements are $(e^{-z^{\prime }}\delta x,e^{z^{\prime }}\delta y,\delta z)$; relabeling the coordinates of the fixed point gives the left-invariant vector fieldsThe naming distinction matters: generators of left multiplication are right invariant, whereas generators of right multiplication are left invariant.

All mixed Lie brackets of vector fields vanish. For example, $R_3(e^{-z})=-e^{-z}$ cancels $L_1(-x)=-e^{-z}$ in $[L_1,R_3]$, and $R_3(e^z)=e^z$ cancels $L_2(y)=e^z$ in $[L_2,R_3]$. The brackets with $R_1,R_2$ are zero because the coefficients of the $L_i$ have no $x$ or $y$ dependence; $[L_3,R_3]=0$ because the coefficients of $R_3$ have no $z$ dependence. HenceConceptually, the corresponding flows commute because $(ag)b=a(gb)$, by associativity of the Lie group multiplication.

The dual coframe iswhich satisfies ${\theta }^i(L_j)={\delta }_j^i$. Under left multiplication by $(x_0,y_0,z_0)$, the transformed coordinates are $X=x_0+e^{-z_0}x$, $Y=y_0+e^{z_0}y$, $Z=z_0+z$. Therefore $e^{Z}dX=e^{z}dx$, $e^{-Z}dY=e^{-z}dy$, and $dZ=dz$. Each one-form is a left-invariant differential form, so their sum of squares is the left-invariant metricThis is the standard metric of the Sol Lie group. As a sign check, $[L_3,L_1]=-L_1$, $[L_3,L_2]=L_2$, while $d{\theta }^1={\theta }^3\wedge {\theta }^1$ and $d{\theta }^2=-{\theta }^3\wedge {\theta }^2$, in agreement with the Maurer-Cartan equation.

For a Lie group $G$, Left translation on a Lie group and right translation on a Lie group are the diffeomorphisms $L_g(h)=gh$ and $R_g(h)=hg$. Their differentials identify the tangent space at the identity with every tangent space. Translating a basis $e_a$ of the Lie algebra gives global left-invariant vector fields $E_a(g)=(d{L}_g{)}_e{e}_a$ and right-invariant vector fields $F_a(g)=(d{R}_g{)}_e{e}_a$. Left and right translations commute. If $[e_a,e_b]=c^c{}_{ab}{e}_c$, thenThe opposite sign for the right-invariant frame is essential.

Define an affine connection ${\nabla }^{-}$ by declaring the $E_a$ parallel. Explicitly, ${\nabla }_X^{-}(f^a{E}_a)=X(f^a)E_a$. Its curvature form of a connection vanishes, since successive derivatives of the component functions satisfy the vector-field commutator identity. Its torsion tensor isSimilarly, declaring the $F_a$ parallel gives another flat affine connection ${\nabla }^{+}$, with $T^{+}(F_a,F_b)=-[F_a,F_b]$. In a left-invariant frame the latter connection obeysTo see this, express $E_a=F_{{\mathrm{Ad}}_g{e}_a}$. Along $g\exp (t{e}_b)$, differentiation of ${\mathrm{Ad}}_{g\exp (t{e}_b)}{e}_a$ gives ${\mathrm{Ad}}_g[e_b,e_a]$, producing the displayed bracket. Consequently $T^{+}(X,Y)=[X,Y]$ for left-invariant $X,Y$. These are the flat translation connections on a Lie group. They have opposite torsion; if $G$ is Abelian, both torsion tensors vanish and the two connections coincide. The phrase “with torsion” does not mean nonzero torsion for Abelian groups.

The average ${\nabla }^0=({\nabla }^{-}+{\nabla }^{+})/2$ is again an affine connection and satisfies ${\nabla }_X^{0}Y=\frac{1}{2}[X,Y]$ for left-invariant fields. Its torsion tensor is zero. With the curvature convention $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$, the Jacobi identity givesDefine the Ricci tensor by $\mathrm{Ric}(X,Y)=\mathrm{Tr}(Z\mapsto R(Z,X)Y)$. Since $[[Z,X],Y]={\mathrm{ad}}_Y{\mathrm{ad}}_{X}Z$, this giveswhere $K(X,Y)=\mathrm{Tr}({\mathrm{ad}}_X{\mathrm{ad}}_Y)$ is the Killing form. Thus the canonical torsion-free connection on a Lie group is generally curved. Its exact flatness criterion is $[[\mathfrak{g},\mathfrak{g}],\mathfrak{g}]=0$: Abelian and two-step nilpotent groups are exceptions. A vanishing Ricci tensor alone does not prove flatness.

For a semisimple Lie algebra, the Killing form is nondegenerate. Its adjoint invariance gives $K([X,Y],Z)+K(Y,[X,Z])=0$, so translating it defines a bi-invariant pseudo-Riemannian metric $g_K$. For left-invariant fields,A metric-compatible torsion-free connection is uniquely the Levi-Civita connection; therefore ${\nabla }^0$ is its Levi-Civita connection. Moreover,The special feature is a canonical nondegenerate metric and an Einstein manifold structure. The metric need not be positive definite: noncompact semisimple groups generally require indefinite signature. For compact semisimple groups, using $-K$ gives a bi-invariant Riemannian metric with $\mathrm{Ric}=\frac{1}{4}{g}_{-K}$ instead.

Work over a coordinate neighborhood $U\subset B$ with coordinates $x^i$, and choose a basis $T_a$ of the Lie algebra, with $[T_a,T_b]=c^c{}_{ab}{T}_c$. Put $A=A_i^a{T}_{a}d{x}^i$. For each $T_a$ let $R_a$ be its right-invariant vector field on the fiber group. In matrix notation $R_a(\gamma )=T_a\gamma$, andThis minus sign is the right-invariant vector fields realize the opposite Lie algebra rule. It is essential here. Define the coordinate horizontal lifts of a principal connection byTheir projections are ${\pi }_{\ast }{H}_i={\partial }_i$, so they are linearly independent and there are exactly $\dim B$ of them. Also $d\gamma (H_i)=-A_i\gamma$, whenceThey therefore span the horizontal distribution of a principal connection on this trivialization.

The coefficients $A_i^a$ depend only on the base variables, so taking Lie brackets giveswhere $F_{ij}={\partial }_i{A}_j-{\partial }_j{A}_i+[A_i,A_j]$ and $F=\frac{1}{2}{F}_{ij}d{x}^i\wedge d{x}^j=dA+A\wedge A$. Since the right-invariant vector fields form a basis on each fiber, $\boxed{[H_i,H_j]=0\text{ for all }i,j⟺F=0}$. This is the local coordinate version of vanishing curvature of a principal connection and, by the Frobenius theorem, integrability of the horizontal distribution.

The coordinate qualification is necessary: horizontal lifts of arbitrary noncommuting base fields need not commute even for a flat principal connection. Nor does flatness guarantee a globally defined coordinate frame or a global horizontal section; holonomy of a connection can obstruct the latter. If the requested frame were read globally, the trivial flat bundle $S^2\times G$ would already be a counterexample: restricting a global horizontal frame to the identity-fiber section would trivialize $T{S}^2$, contrary to the Hairy ball theorem. The construction on each coordinate trivialization supplies the intended result without that global claim.

Acting on the column $(x,1{)}^T$ gives the faithful Matrix Lie group representationMultiplication and inversion areThus $G$ is the real affine group. Differentiating the matrices at the identity gives its Lie algebraWiththe matrix commutator is $[D,T]=T$, and $[D,D]=[T,T]=0$.

The left Maurer-Cartan form and its right analogue areFor a constant $h$, left translation leaves ${\theta }_L$ unchanged, and right translation leaves ${\theta }_R$ unchanged. Taking their dual vector fields givesThe first pair consists of left-invariant vector fields; the second consists of right-invariant vector fields. These are the invariant frames of the real affine group. Computing their Lie brackets of vector fields yieldsAll brackets of a field with itself vanish. The sign difference is necessary: evaluation at the identity identifies left-invariant vector fields with the matrix Lie algebra as a homomorphism, whereas right-invariant vector fields realize the opposite Lie algebra. Equivalently, $\xi \mapsto -{\xi }_R$ is a homomorphism for the same matrix commutator. There is no convention in which these particular coordinate fields both have the positive structure constant while retaining the usual Lie bracket of vector fields.

The Maurer-Cartan equation gives the same check. If ${\theta }_L=D{\ell }^D+T{\ell }^T$, thenFor ${\theta }_R=D{r}^D+T{r}^T$, insteadThus the left equation is $d{\theta }_L+{\theta }_L\wedge {\theta }_L=0$ and the right equation is $d{\theta }_R-{\theta }_R\wedge {\theta }_R=0$, explaining the opposite bracket signs directly.

Finally, the two given matrix curves are $\exp (ϵT)$ and $\exp (ϵD)$. Their left action on a general group element isDifferentiating at $ϵ=0$ gives $T_R$ and $D_R$ respectively. Hence Left translations on a Lie group are generated by right-invariant vector fields. More generally, the velocity of $\exp (t\xi )g$ is $(d{R}_g{)}_e\xi$, which is right-invariant because left and right multiplication commute. In the other direction, the flows $g\exp (t\xi )$ generate left-invariant vector fields and are right translations on a Lie group.

For a Matrix Lie group, the right-invariant vector field is $r_{\xi }(g)=\xi g$. Differentiating these ambient matrix-valued functions in the definition of the Lie bracket of vector fields gives $[r_{\xi },r_{\eta }](g)=(\eta \xi -\xi \eta )g$. Thus $\xi \mapsto r_{\xi }$ is an antihomomorphism, and $\xi \mapsto -r_{\xi }$ is a homomorphism. In contrast, $\xi \mapsto l_{\xi }$ for left-invariant vector fields is a homomorphism.