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An abelian Lie group has commutative multiplication. Its Lie algebra is an abelian Lie algebra, and its Lie exponential map is a group homomorphism from the additive Lie algebra. Conversely, an additive Lie exponential map forces the identity component of a Lie group to be abelian, but does not force all disconnected components to commute.

The Exponential map of a Lie group is a group homomorphism from the additive Lie algebra exactly when its identity component of a Lie group is abelian. If it is additive, its image is an abelian subgroup containing an identity neighborhood by the local exponential chart; connectedness makes that image the entire identity component. Conversely, in an abelian identity component the product $\exp (tX)\exp (tY)$ is a one-parameter subgroup with derivative $X+Y$, giving the identity by uniqueness. The disconnected Lie group $S_3\times \mathbb{R}$ shows that the full group need not be abelian.

If a Lie group has an Adjoint representation of a Lie group element with a real eigenvector of eigenvalue whose modulus differs from one, it has no bi-invariant Riemannian metric. Conjugation is a composition of left and right translations, hence would preserve the positive identity inner product. Applying it to that eigenvector gives $\Vert v\Vert =|\lambda |\Vert v\Vert$, a contradiction. The orientation-preserving affine group of the real line supplies a connected example: conjugation by dilation of factor two multiplies the infinitesimal translation by two.

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

On a connected Lie group, a nondegenerate invariant symmetric bilinear form on a Lie algebra extends by translation to a metric preserved by both left and right translation. For a disconnected Lie group, the form must additionally be invariant under the full Adjoint representation of a Lie group. It may have indefinite signature. Its Levi-Civita connection is ${\nabla }_{X}Y=\frac{1}{2}[X,Y]$ for left-invariant vector fields, and its Ricci tensor is $-K/4$. Choosing the nondegenerate Killing form of a semisimple Lie algebra yields a canonical indefinite Einstein manifold; compact groups instead admit the positive metric $-K$.

A Riemannian metric on a Lie group is bi-invariant if both left and right translations are isometries. Its identity inner product is invariant under the Adjoint representation of a Lie group; infinitesimally,This skew-adjointness is the key additional property beyond left invariance.

Specify the displayed rule on left-invariant vector fields and extend by the Leibniz rule to an affine connection on the Lie group. In a left-invariant vector field frame $L_i$, the extension is ${\nabla }_{L_i}(z^j{L}_j)=L_i(z^j)L_j+\lambda z^j[L_i,L_j]$. Applying the bare bracket formula to arbitrary vector fields instead would violate the Leibniz rule unless $\lambda =1$. With $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$, the Jacobi identity givesfor left-invariant arguments, where $B$ is the Killing form and $\mathrm{Ric}(X,Y)=\mathrm{tr}(Z\mapsto R(Z,X)Y)$. These identities determine the tensor fields everywhere. The endpoints $\lambda =0,1$ have zero curvature, with respectively left- and right-invariant parallel frames, and opposite torsion tensors. The midpoint $\lambda =1/2$ is torsion-free but need not have zero curvature.

On a graded nilpotent Lie group whose first layer generates its Lie algebra, a horizontal curve obeys ${\dot{\gamma }}_t=(L_{{\gamma }_t}{)}_{\ast }{\dot{x}}_t$ for a control in the first layer. The displayed infimum is over horizontal curves with prescribed endpoints. The distance is left invariant and homogeneous under the graded dilations. On a free nilpotent Lie group it measures the least Euclidean length of a path realizing a prescribed truncated path signature.

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.

The Chern-Weil homomorphism turns a polynomial in curvature into a topological cohomology class. For a principal bundle with Lie group $G$, a degree-$k$ polynomial $P$ on its Lie algebra that is invariant under the adjoint action determineswhere $F$ is the curvature form of any principal connection. The Bianchi identity makes $P(F)$ a closed differential form. Changing the principal connection changes it by an exact differential form, so the de Rham cohomology class depends on the principal bundle, not that choice. Products of polynomials become cup products.^[ref]

For example, the trace-square polynomial produces $\mathrm{tr}(F\wedge F)$, underlying the first Pontryagin class. This is the bridge between the locally measured curvature and a global topological invariant.

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.

Every closed subgroup of a finite-dimensional Lie group is an embedded submanifold and a Lie subgroup. Closedness alone forces the compatible smooth structure; smoothness need not be assumed in advance. The proof constructs the Lie algebra of a closed subgroup from one-parameter subgroups, then uses local product coordinates for a closed subgroup.

The coadjoint action of a Lie group on the dual space of its Lie algebra is $({\mathrm{Ad}}_g^{\ast }\eta )(X)=\eta ({\mathrm{Ad}}_{g^{-1}}X)$. The inverse in this formula makes it a left group action.

The Peter-Weyl theorem makes finite-dimensional unitary representations separate elements of a compact group. Intersecting their closed kernels gives the identity subgroup. If the descending chain condition holds for closed subgroups, choose finitely many representations whose kernel intersection is already trivial: otherwise repeatedly exclude a surviving nonidentity element to obtain an infinite strict chain. Their direct sum is a faithful representation into a finite-dimensional unitary group. Compactness makes its image closed, and the closed-subgroup theorem gives the original group a compatible Lie group structure.

A compact Lie group is a Lie group whose underlying manifold is compact. Averaging any positive inner product over its Haar measure supplies an invariant inner product on each finite-dimensional continuous representation of a topological group, including the Adjoint representation.

A compact real form of a complex semisimple Lie algebra $\mathfrak{g}$ is a real Lie subalgebra $\mathfrak{u}$ whose complexification of a Lie algebra is $\mathfrak{g}$ and whose Killing form is negative definite. Its simply connected Lie group is compact. The positive internal inner product used in unitary gauge theory is therefore $-\kappa$. For ${\mathfrak{sl}}_2(\mathbb{C})$ the SU(2) Lie algebra is a compact real form, whereas ${\mathfrak{sl}}_2(\mathbb{R})$ is a different real form with indefinite Killing form.

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 complex Lie group is a complex manifold whose multiplication and inversion are holomorphic maps between complex manifolds. Its tangent space at the identity is a complex Lie algebra; the Adjoint representation of a Lie group and the Lie exponential map are holomorphic. Compatibility with a complex manifold structure is essential: an arbitrary even-dimensional real Lie group need not have holomorphic multiplication.

Complex two-by-two matrices of determinant one form a complex three-dimensional Lie group, or a real six-dimensional Lie group. Their congruence action on Hermitian two-by-two matrices gives the Lorentz spinor double cover with kernel $\{I,-I\}$. The two Weyl spinor representations are $A$ and $(A^{†}{)}^{-1}$. Treating the group as real is essential for the second, antiholomorphic representation.

The vector space ${\mathbb{R}}^3$ with the cross product as Lie bracket, $[x,y]=x\times y$. Bilinearity and $x\times x=0$ give an Alternating Lie bracket; the vector triple product identity proves the Jacobi identity because the three expressions $x\times (y\times z)=y(x\cdot z)-z(x\cdot y)$ cancel cyclically, using the symmetric dot product.

Its relation to the SO(3) Lie algebra is explicit. For a vector $x=(x_1,x_2,x_3)$, define the skew-symmetric matrixThen $[\hat{x},\hat{y}]=\widehat{x\times y}$: acting on $v$, the left side is $x\times (y\times v)-y\times (x\times v)=(x\times y)\times v$. Every real $3\times 3$ skew-symmetric matrix occurs uniquely, so this hat map is a Lie algebra isomorphism. With the standard basis vectors $e_1,e_2,e_3$,and their matrix commutators are $[X,Y]=Z$, $[Y,Z]=X$, $[Z,X]=Y$. With active rotation matrices acting on column vectors, $\exp (t\hat{x})$ rotates through angle $t\Vert x\Vert$ about the oriented axis $x$ when $x\ne 0$. In particular $\exp (tZ)e_1=(\cos t,\sin t,0)$, fixing the positive-angle convention used in the small rotation commutator.

The map $x\mapsto -(i/2)x\cdot \sigma$ is also a Lie algebra isomorphism to the compact real SU(2) Lie algebra, using the Pauli matrix commutator identity. These two identifications concern Lie algebras; they do not identify the corresponding Lie groups globally.