Left and right translation on a Lie group ($L_g,R_g$)

For a Lie group $G$, left and right translation are the diffeomorphisms $L_g(h)=gh$ and $R_g(h)=hg$. They commute: $L_g{R}_h=R_h{L}_g$.

Declare a global left-invariant, respectively right-invariant, frame parallel to obtain two flat affine connections on a Lie group. For left-invariant vector fields, their derivatives are as displayed and their torsion tensors are $-[X,Y]$ and $+[X,Y]$. The tensors vanish on Abelian groups; two genuinely torsionful connections are not asserted there.

Declare a global frame of left-invariant vector fields parallel on a semisimple Lie group. This affine connection is flat and preserves a left-invariant metric tensor because its frame components are constant. If $[e_a,e_c]=C_a{}^b{}_c{e}_b$, its torsion tensor is $T_a{}^b{}_c=-C_a{}^b{}_c$. For fixed $c$, the matrix with entries $T_a{}^b{}_c$ is $\mathrm{ad}{e}_c$. Its trace vanishes because a semisimple Lie algebra is perfect and adjoints of brackets are commutators. The trace of the product for $c,d$ is the Killing form $K_{cd}$, proving both contractions. This flat metric connection differs from the curved torsion-free Levi-Civita connection of the Killing metric.

The average of the two flat translation connections on a Lie group is a torsion-free connection. For left-invariant vector fields, the Jacobi identity gives $R(X,Y)Z=-\frac{1}{4}[[X,Y],Z]$ with $R(X,Y)=[{\nabla }_X,{\nabla }_Y]-{\nabla }_{[X,Y]}$. Contracting gives $\mathrm{Ric}(X,Y)=-\frac{1}{4}K(X,Y)$ for the Killing form $K$. It is flat exactly when all double Lie brackets vanish, including the case of a two-step nilpotent group. For a semisimple Lie algebra, $K$ is nondegenerate and supplies a bi-invariant indefinite metric whose Levi-Civita connection is ${\nabla }^0$.

The right-invariant vector field associated with a Lie algebra element $\xi$ is obtained by right translation of its value at the identity. Its flow is $g\mapsto \exp (t\xi )g$, hence consists of left translations on a Lie group. Right invariance follows because left and right multiplication commute.

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.

For $\xi$ in the Lie algebra of $G$, the left-invariant vector field is $l_{\xi }(g)=(d{L}_g{)}_e\xi$. Its global flow is right translation ${\Phi }^t=R_{\exp (t\xi )}$.

On a hemisphere of SU(2) as the three-sphere, write $A=u_{0}I+i{u}_i{\sigma }_i$, with real coefficients satisfying $u_0^2+|\mathbf{u}{|}^2=1$. Right multiplication by $I+i{v}_j{\sigma }_j$ gives $d{u}_i=(u_0{\delta }_{ij}+u_k{ϵ}_{jki})v_j$. This right multiplication flow defines left-invariant tangent fields. With the displayed factor $i$, they obey $T_{j}A=-A{\sigma }_j$, since ${\partial }_{u_i}{u}_0=-u_i/u_0$. Applying the commutator to $A$ and using the Pauli matrix commutator identity gives $[T_i,T_j]=-2i{ϵ}_{ijk}{T}_k$. The entries of $A$ contain every coordinate, so this determines the vector-field identity, not just its action in one isolated representation.

A basis $T_a$ of the Lie algebra of a Lie group gives a global frame of a vector bundle of its tangent bundle by $X_a(g)=(d{L}_g{)}_e{T}_a$. Left translation is a diffeomorphism, so the left-invariant vector fields are independent everywhere. Every Lie group is therefore a parallelizable manifold.

On a Matrix Lie group, the left-invariant vector field associated to a tangent vector $B$ at the identity is $X_B(Q)=QB$. Its ambient derivative is $D{X}_B(Q)[H]=HB$. Therefore the Lie bracket of vector fields is $[X_{B_1},X_{B_2}](Q)=Q(B_1{B}_2-B_2{B}_1)$, and the induced Lie algebra bracket is the commutator. This fixes the sign for left invariance; right-invariant fields induce the opposite sign when evaluated with the same matrix identification.

Every smooth left-invariant vector field on a Lie group is a complete vector field. Translate a local integral curve of a vector field through the identity to every other point. The same positive local existence interval works at every initial point, so no maximal integral curve of a vector field can have a finite endpoint. Uniqueness then makes the curve through the identity a one-parameter subgroup.

A differential form $\alpha$ on a Lie group is left-invariant when $L_g^{\ast }\alpha =\alpha$ for every $g$. It is determined by its value at the identity.

Choose a nonzero element of ${\Lambda }^n{T}_e^{\ast }G$ for an $n$-dimensional Lie group and transport it by inverse left translations. This gives a smooth nowhere-zero left-invariant differential form of top degree, hence an orientation. The identity $L_{(hg{)}^{-1}}{L}_h=L_{g^{-1}}$ proves invariance. The space of invariant top forms is one-dimensional, since evaluation at the identity identifies it with ${\Lambda }^n{T}_e^{\ast }G$.

A differential form on a Lie group is bi-invariant when it is invariant under both left and right translations.

An adjoint-invariant symmetric bilinear form $B$ on a Lie algebra makes the displayed tensor alternating. If $B$ is invariant under the full group Adjoint representation, its left-translated three-form is also right-invariant. On left-invariant fields, the exterior derivative is $d\eta (W,X,Y,Z)=-2B(W,[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]])=0$ by the Jacobi identity. The Killing form supplies such a $B$ on every semisimple Lie group. An overall sign change in the lowered structure-constant convention negates the form but not its invariance or closedness.

On a connected Lie group, a left-invariant differential 1-form is closed exactly when it is bi-invariant. Connectedness is essential: on $O(2)$ every left-invariant 1-form is closed because its Lie algebra is abelian, while reflection conjugation negates every nonzero one.