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For $X\in \mathfrak{g}=T_{e}G$, let $X^L$ be the left-invariant vector field with value $X$ at the identity. Its integral curve through the identity is the unique one-parameter subgroup ${\gamma }_X$ with ${\gamma }_X^{\prime }(0)=X$. It exists for all real time: local existence followed by translation and the identity ${\gamma }_X(t+s)={\gamma }_X(t){\gamma }_X(s)$ extends it indefinitely. The Exponential map of a Lie group isIt is smooth and $d_0\exp ={\mathrm{id}}_{\mathfrak{g}}$, because differentiating $t\mapsto \exp (tX)$ at zero gives $X$. The inverse function theorem therefore makes it a local diffeomorphism at zero.

If $G$ is abelian, the product $t\mapsto \exp (tX)\exp (tY)$ is a one-parameter subgroup with initial tangent $X+Y$. Uniqueness gives $\exp (t(X+Y))=\exp (tX)\exp (tY)$, henceConversely, if this additive homomorphism identity holds, all $\exp (tX)$ and $\exp (sY)$ commute. Differentiating their adjoint action shows $[X,Y]=0$. Alternatively, their commuting images contain an identity neighbourhood; that neighbourhood generates the identity component of a Lie group, so the identity component is abelian. Conversely, if the Lie algebra is abelian, its left-invariant vector fields commute, so their flows commute. Hence all one-parameter subgroups commute; their product with initial tangent $X+Y$ is the one-parameter subgroup for $X+Y$. The Lie exponential map is therefore additive, and its image generates an abelian identity component. Thus the stated equivalence holds for connected $G$. Without connectedness the exact assertion is that the Lie exponential map is a homomorphism precisely when $G^{\circ }$ is abelian. The converse implication to all of $G$ is false: in $O(2)$ the Lie algebra consists of $tJ$, whereRotations satisfy $R_{t+s}=R_t{R}_s$, but a reflection $S$ satisfies $S{R}_t{S}^{-1}=R_{-t}$, so the full orthogonal group is not abelian.

Now take a connected abelian Lie group of dimension $n$. Its Lie exponential map is a homomorphism whose image contains an identity neighbourhood, hence an open subgroup. A connected group has no proper open subgroup, so this map is onto. Its kernel $\Lambda$ is discrete, since it is injective on a neighbourhood of zero. The induced map is a bijective local diffeomorphism and therefore a Lie group isomorphismHere is the needed classification of $\Lambda$. Let $r$ be the dimension of its real span. Choose ${\lambda }_1,\dots ,{\lambda }_r\in \Lambda$ that form a real basis of that span. Their integer span ${\Lambda }_0$ has finite index in $\Lambda$: subtract integer multiples of the ${\lambda }_i$ to put each coset representative in a compact parallelepiped. Only finitely many points of $\Lambda$ lie there, since discreteness at zero gives a uniform positive separation between distinct points. Thus $\Lambda$ is finitely generated and torsion-free, and the structure theorem for finitely generated abelian groups makes it free abelian of rank $r$. Its integer basis is a real basis of the span. A linear change of coordinates consequently sends $\Lambda$ to ${\mathbb{Z}}^r\times \{0\}$ in ${\mathbb{R}}^n$.

This proves the classification of connected abelian Lie groups:The isomorphism depends on the choice of lattice basis and complementary vector space. Compactness is equivalent to $s=0$.

The Lie group–Lie algebra correspondence separates local differential structure from global topology. For a Lie group $G$, define $LG=T_{e}G$ as a real vector space. Extend each $X\in T_{e}G$ to the left-invariant vector fieldThe commutator of two such vector fields is again left-invariant. SetThe vector-field commutator is bilinear, antisymmetric and satisfies the Jacobi identity, so this makes $T_{e}G$ a Lie algebra. For a matrix group this bracket is $XY-YX$.

For a smooth Lie group homomorphism $\varphi :G_1\to G_2$, its differential at the identity is the linear mapThe identity $\varphi \circ L_g=L_{\varphi (g)}\circ \varphi$ shows that the corresponding left-invariant fields are $\varphi$-related. Brackets of related vector fields are related, henceThe chain rule also gives $L(\psi \circ \varphi )=L\psi \circ L\varphi$ and $L(\mathrm{id})=\mathrm{id}$. Thus differentiation is a functor from Lie groups and smooth homomorphisms to real Lie algebras and bracket-preserving linear maps.

To see what information the differential retains, apply $\varphi$ to a one-parameter subgroup. Its initial tangent is $L\varphi (X)$, so uniqueness yieldsIf $G_1$ is connected, a small exponential neighbourhood generates it. Every $g\in G_1$ is a finite product $\exp X_1\cdots \exp X_k$, and thereforeFor an already existing homomorphism this expression is independent of the chosen product because it equals $\varphi (g)$. Two homomorphisms with the same differential agree on that neighbourhood and hence everywhere. This proves that a Lie group homomorphism determined by its differential requires a connected source.

On a disconnected source only the restriction to $G_1^{\circ }$ is determined. For instance a nontrivial finite discrete group has zero Lie algebra, so its identity and trivial endomorphisms have the same zero differential but are different maps. The discussion of all components must therefore include additional global data. The remaining parts explain the corresponding existence and topology conditions.

For the Lie algebra $\mathfrak{g}$ of a Lie group $G$, let ${\mathrm{ad}}_X(Y)=[X,Y]$. The Killing form is the symmetric bilinear formFor a finite-dimensional real Lie algebra, the Cartan criterion for semisimplicity says that $B$ is nondegenerate precisely when $\mathfrak{g}$ is a semisimple Lie algebra. Thus a semisimple Lie group supplies the required condition. Compactness alone is insufficient: an abelian Lie algebra, including that of a torus, has zero Killing form.

The Jacobi identity gives ${\mathrm{ad}}_{[Z,X]}=[{\mathrm{ad}}_Z,{\mathrm{ad}}_X]$. Cyclic invariance of the trace therefore givesThere is also invariance under the full Adjoint representation of a Lie group: ${\mathrm{ad}}_{{\mathrm{Ad}}_{a}X}={\mathrm{Ad}}_a{\mathrm{ad}}_X{\mathrm{Ad}}_{a^{-1}}$, so $B({\mathrm{Ad}}_{a}X,{\mathrm{Ad}}_{a}Y)=B(X,Y)$.

Define a pseudo-Riemannian metric by translating $B$ from the identity:It is smooth and nondegenerate. Left translations preserve it by construction. Right translation, expressed in left-translated tangent coordinates, acts by ${\mathrm{Ad}}_{b^{-1}}$, so it also preserves the metric. This constructs a bi-invariant pseudo-Riemannian metric.

For left-invariant vector fields, the derivative terms in the Koszul formula vanish. Invariance of $B$ then givesUse the Riemann curvature tensor convention $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$. Substituting the Levi-Civita connection and using the Jacobi identity yieldsThe Ricci tensor is the trace of $X\mapsto R(X,Y)Z$. Since $[[X,Y],Z]={\mathrm{ad}}_Z{\mathrm{ad}}_{Y}X$, it follows thatThis is the Killing-form Einstein metric. Rescaling to $g=cB$, with nonzero constant $c$, gives Einstein metric constant $-1/(4c)$.

For a compact semisimple Lie group, $B$ is negative definite and $-B$ is a positive-definite bi-invariant Riemannian metric. For example, $SU(2)$ has topology $S^3$ and metric signature $(0,3)$ for $B$, while $SO(3)$ has the same Lie algebra and metric signature, but topology ${\mathbb{RP}}^3$. The special linear group $SL(2,\mathbb{R})$ has metric signature $(2,1)$ and topology $S^1\times {\mathbb{R}}^2$; its universal covering Lie group has topology ${\mathbb{R}}^3$ and the same Killing form. The real Lie algebra of $SL(2,\mathbb{C})$ has metric signature $(3,3)$, and the group has topology $S^3\times {\mathbb{R}}^3$. Thus the Killing form distinguishes compact from noncompact directions in these semisimple examples, but does not determine global topology or distinguish different covering groups.

A smooth fiber bundle $\pi :E\to M$ with typical fiber $F$ is locally a product: every base point has a neighborhood $U$ and a local trivialization ${\pi }^{-1}(U)\cong U\times F$ commuting with projection to $U$. Its overlap maps act smoothly on $F$. A principal bundle with structure Lie group $G$ has a free right $G$-action, each fiber is one orbit, and the local trivializations are equivariant for right multiplication on $G$. A smooth section of a fiber bundle is a map $s:M\to E$ with $\pi \circ s={\mathrm{id}}_M$.

For a principal bundle, a global section of a fiber bundle gives the mapThe free transitive action on each fiber makes $\Phi$ bijective. In a local trivialization, write $s(x)=(x,a(x))$; then $\Phi (x,g)=(x,a(x)g)$ and its inverse is $(x,k)\mapsto (x,a(x{)}^{-1}k)$. Both are smooth and $\Phi$ is equivariant. Conversely, a product principal bundle has the section $x\mapsto (x,e)$. This proves a principal bundle is trivial exactly when it has a global section. The argument depends on the free transitive action: an arbitrary bundle section need not trivialize its bundle.

For an $n$-dimensional smooth manifold, the frame bundle has fiber over $x$ consisting of ordered bases of $T_{x}M$, equivalently linear isomorphisms $u:{\mathbb{R}}^n\to T_{x}M$. Right composition by $GL(n,\mathbb{R})$ makes it a principal bundle. If a metric has signature $(p,q)$, its pseudo-orthonormal frame bundle consists of the isometries $u:{\mathbb{R}}^{p,q}\to T_{x}M$, with structure group $O(p,q)$. In the positive-definite case this is the usual orthonormal frame bundle. Choosing space and time orientations reduces the Lorentzian structure group to its corresponding connected subgroup.

For Minkowski spacetime, use a fixed origin and a standard Lorentzian frame. Every pair consisting of a point and a pseudo-orthonormal frame is the image of this reference pair under a unique affine isometry. Thus, for the full groups,Here the first expression denotes the total space of the orthonormal frame bundle, not merely the Lorentz group. Projection sends an affine isometry to the image of the reference point, and the right action of the stabilizer changes its frame.

Realize de Sitter spacetime of radius $\ell$ as ${\eta }_{AB}{X}^A{X}^B={\ell }^2$ in ${\mathbb{R}}^{1,4}$. The ambient orthogonal group is transitive on this hyperboloid. Its point stabilizer acts on the Lorentzian tangent space as $O(1,3)$; choosing a tangent orthonormal frame removes that stabilizer completely. ConsequentlyFor oriented, time-oriented frames, replace these groups by $S{O}_0(1,4)$ and $S{O}_0(1,3)$, and similarly use the proper orthochronous Poincaré group for Minkowski spacetime. The isometries are uniquely determined by their action on a point and its frame, which explains these bundle identifications.

Finally, choose a basis $v_1,\dots ,v_n$ of the Lie algebra $T_{e}G$. The left-invariant vector fields $X_i(g)=(d{L}_g{)}_e{v}_i$ form a basis at every $g$, because $L_g$ is a diffeomorphism. They give a global section of a fiber bundle of the frame bundle, so that bundle is trivial. More explicitly,Thus every Lie group is a parallelizable manifold; connectedness is not needed for this construction.

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.

A smooth vector field on a smooth manifold $M$ is a smooth section of its tangent bundle: it assigns $v(p)\in T_{p}M$ smoothly to each point. On a Lie group, it is a left-invariant vector field whenFor $X\in T_{e}G$, define $v_X(g)=(d{L}_g{)}_{e}X$. Smoothness of multiplication makes this a smooth vector field, and $L_g{L}_p=L_{gp}$ plus the chain rule proves left invariance. Every left-invariant vector field arises this way from its value at the identity.

For a matrix Lie group, $v_X(g)=gX$, because the derivative at zero of $g\gamma (t)$ is $gX$. The printed group name in this subpart must be read as ${\mathrm{SL}}_2$, rather than the additive vector space ${\mathfrak{sl}}_2$: it is the matrix-group calculation consistent with the preceding subparts. At the specified matrix,A tangent vector at a nonidentity group element need not itself be trace-free; here $g^{-1}{v}_X(g)=X$ is trace-free, as required. If one instead interpreted the printed ${\mathfrak{sl}}_2$ as its additive Lie group, the field would be the constant field $X$, a different problem.

A smooth vector field on a smooth manifold $M$ is a smooth section $X:M\to TM$ of its tangent bundle. It acts on smooth functions by $Xh(p)=d{h}_p(X_p)$. In a manifold chart, $X={\sum }_i{X}^i{\partial }_i$. The Lie bracket of vector fields is the commutator of their actions on functions:Expanding in coordinates makes all second derivatives cancel and givesThus the commutator is again a smooth vector field, not a second-order operator. The coordinate expression represents the intrinsic derivation on functions, so agrees on overlaps.

For a diffeomorphism $f:M\to N$, its pushforward of a vector field isFor every smooth function $h$ on $N$, the chain rule gives $((f_{\ast }X)h)\circ f=X(h\circ f)$. Applying this twice yieldsSmooth functions determine a tangent vector, and $f$ is surjective, so

On a Lie group, a left-invariant vector field satisfies $(L_g{)}_{\ast }X=X$ for every left translation $L_g(h)=gh$. Given $\xi \in T_{e}G$, setThis field is smooth by smoothness of multiplication. The equality $L_a{L}_g=L_{ag}$ proves left invariance. Conversely, left invariance forces $X(g)=(d{L}_g{)}_{e}X(e)$, so $X_{\xi }$ is the unique left-invariant field with value $\xi$ at $e$. This is a linear isomorphism between $T_{e}G$ and the space of left-invariant vector fields.

The pushforward identity applied to $L_g$ shows that $[X_{\xi },X_{\eta }]$ is again left invariant. Consequentlydefines a Lie bracket on $T_{e}G$. Bilinearity and antisymmetry follow from the operator commutator. Its Jacobi identity follows by expanding the three nested commutators: the six triple compositions cancel in pairs. These properties transfer through the linear isomorphism, making $T_{e}G$ the Lie algebra of $G$.

For the general linear group $G=\mathrm{GL}(n,\mathbb{R})$, use matrix-entry coordinates $x_{pq}$. It is an open subset of $M_n(\mathbb{R})$, so every tangent space is naturally $M_n(\mathbb{R})$. Since $L_g(h)=gh$, its differential sends $A$ to $gA$, and henceFor $A,B\in M_n(\mathbb{R})$,and reversing $A,B$ gives $(gBA{)}_{pq}$. Thus $[X_A,X_B]=X_{AB-BA}$; evaluating at the identity gives the matrix commutator

Let $v=(x,y{)}^T$ and $B(\rho )=\left(\begin{array}{cc}\cosh \rho & \sinh \rho \\ \sinh \rho & \cosh \rho \end{array}\right)$. Matrix multiplication givesThus $G$ is a semidirect product of the translation plane by the one-parameter Lorentz boost group. Differentiating left multiplication at the identity gives the left-invariant vector fieldsThey obey $[X_0,X_1]=X_2$, $[X_0,X_2]=X_1$ and $[X_1,X_2]=0$.

Their dual coframe is $d\rho$, $\cosh \rho dx-\sinh \rho dy$, and $-\sinh \rho dx+\cosh \rho dy$. The Haar measure from a left-invariant coframe is their absolute wedge product. Since $\det B(\rho )=1$,One can also check this directly: left multiplication has Jacobian determinant one. Right multiplication has a block triangular Jacobian with determinant one too, so the same measure is both left and right Haar measure.

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.

A left-invariant vector field $X$ on a Lie group satisfiesor equivalently $(L_g{)}_{\ast }X=X$ for every $g$. For $\xi \in T_{e}G$, defineThis is smooth because multiplication is smooth and its differential depends smoothly on the base points. The identity $L_g\circ L_h=L_{gh}$ gives $(d{L}_g{)}_h{X}_h^{\xi }=X_{gh}^{\xi }$, so $X^{\xi }$ is a left-invariant vector field. Conversely, left invariance forces any such field to have this form with $\xi =X_e$. Evaluation at $e$ and the construction above are inverse linear maps. HenceThe Lie bracket of vector fields is their commutator as derivations on smooth functions: $[X,Y]f=X(Yf)-Y(Xf)$. It is another vector field because the cross terms cancel in the Leibniz rule for its action on a product. The bracket is preserved by a diffeomorphism: conjugating the derivation operators by pullback of a smooth function preserves their commutator. Thereforefor left-invariant $X,Y$. These fields are consequently closed under the bracket.

Transport the bracket to the tangent space byBilinearity and antisymmetry come from the commutator. The Jacobi identity follows by expanding the three nested commutators of derivation operators: their six terms cancel in pairs. Thus $T_{e}G$ is a Lie algebra, and the displayed correspondence is a Lie algebra isomorphism onto the algebra of left-invariant vector fields.

Let $e$ be the identity. For a left-invariant differential form $\omega$ and a left-invariant vector field $X$, one hasThe value is therefore independent of $g$, even if the Lie group is disconnected. The same calculation applies to another left-invariant field $Y$.

The standard exterior derivative of a one-form evaluated on vector fields isBoth differentiated functions are constant in the present situation. ThusThis identity is a direct consequence of left invariance and the stated general formula for the exterior derivative.

The correspondence in part (a) respects Lie brackets. If the Lie algebra is abelian, any two left-invariant vector fields satisfy $[X,Y]=0$ everywhere, not just at the identity. Part (b) therefore gives $d\omega (X,Y)=0$ for every such pair.

At a point $g$, the map $(d{L}_g{)}_e:T_{e}G\to T_{g}G$ is an isomorphism. Thus every tangent vector at $g$ is the value of a left-invariant vector field. The two-form $d\omega$ consequently vanishes on every pair of tangent vectors at every point, provingNo connectedness assumption on the Lie group is needed: the argument uses its left translations and its abelian Lie algebra.

The Lie algebra of a Lie group $G$ is the vector space $\mathfrak{g}=T_{e}G$ equipped with the bracket transferred from left-invariant vector fields. For $a\in G$, write $L_a(g)=ag$. A field is left-invariant when $(D_g{L}_a)X_g=X_{ag}$ for all $a,g$. Given $v\in T_{e}G$, defineThis is smooth and left-invariant, because $L_a{L}_g=L_{ag}$ and the chain rule applies. Conversely a left-invariant field is determined by its value at $e$, through this formula. Thusis a vector-space isomorphism, whose inverse is evaluation at the identity.

The Lie bracket of vector fields is preserved by diffeomorphisms: on functions this follows by transporting the commutator of derivations. Therefore the bracket of two left-invariant fields is again left-invariant. Define $[v,w]=[X_v,X_w{]}_e$. Bilinearity, antisymmetry and the Jacobi identity follow from the same properties of commutators of derivations. This supplies the asserted Lie algebra structure.

For the general linear group $G{L}_n(\mathbb{R})$, invertible matrices form an open subset of $M_n(\mathbb{R})$, so $T_{I}G{L}_n(\mathbb{R})=M_n(\mathbb{R})$. The left-invariant field determined by $A$ is $X_A(g)=gA$. Its ambient derivative is $D_g{X}_A(H)=HA$. Therefore differentiating left-invariant matrix fields givesHenceThis is the general linear Lie algebra, with the ordinary matrix commutator; right invariance with the same identification would give the opposite sign.

Now let $F:G\to H$ be a smooth Lie group homomorphism. It sends the identity to the identity and induces the linear mapSince $F\circ L_g=L_{F(g)}\circ F$, differentiation givesThese fields are $F$-related; no injectivity or surjectivity of $F$ is needed. For every smooth function $f$ on $H$,Apply this first with $w$ and then with $v$, and subtract the reversed order. The result isAt $e$, smooth functions detect tangent vectors, soThis proves that the differential of a Lie group homomorphism preserves Lie brackets, and hence that $F_{\ast }$ is a Lie algebra homomorphism.

For the exponential identity, let ${\eta }_v$ be the integral curve of $X_v$ through $e$. Left invariance and uniqueness show ${\eta }_v(t+s)={\eta }_v(t){\eta }_v(s)$ for small times: translating the curve through $e$ gives the curve through ${\eta }_v(s)$. Repeating this identity extends the curve to all real times. This proves completeness of left-invariant vector fields without assuming that an arbitrary smooth field is complete. Rescaling the parameter also gives ${\eta }_v(t)={\eta }_{tv}(1)$. By the definition of the Exponential map of a Lie group, ${\exp }_G(v)={\eta }_v(1)$.

The field-related identity shows that $F({\eta }_v(t))$ is an integral curve of $X_{F_{\ast }v}$ starting at $e_H$. By uniqueness it equals ${\eta }_{F_{\ast }v}(t)$ for all times. At time one,This is the naturality of the Lie group exponential.

Apply this to the determinant homomorphism $G{L}_n(\mathbb{R})\to {\mathbb{R}}^{\times }$. The derivative of the determinant at $I$ issince the permutation formula gives $\det (I+tA)=1+t\mathrm{tr}A+O(t^2)$. The left-invariant field on ${\mathbb{R}}^{\times }$ with initial tangent $c$ is $y\mapsto cy$; its curve through one solves $y^{\prime }=cy$, hence is $e^{ct}$. The matrix exponential determinant identity therefore follows from naturality:This argument uses only the flow definition of the exponential and the scalar exponential, not an unproved matrix formula.

To identify it explicitly with the usual matrix exponential, the series $E(t)={\sum }_{k\ge 0}{t}^k{A}^k/k!$ converges in operator norm, uniformly with its differentiated series on bounded intervals. Termwise differentiation gives $E^{\prime }=EA$ and $E(0)=I$. The series commutes with $A$, and differentiating $E(t)E(-t)$ gives zero, so $E(t)E(-t)=I$. Thus $E(t)$ stays in $G{L}_n(\mathbb{R})$ and, by uniqueness, is the integral curve of $X_A$ through $I$. The matrix series is exactly the flow-defined exponential, justifying either notation in the determinant identity.

Take the frame dual to the left-invariant coframe of the translation-dilation group. It isIndeed ${\lambda }^i(L_j)={\delta }_j^i$. These are left-invariant vector fields, either because they are dual to invariant forms or because $L_j(g)=(d{L}_g{)}_e{X}_j$. For instance, the curve $g\exp (t{X}_0)$ differentiates to $\rho {\partial }_{\rho }$, while $g\exp (t{X}_1)$ differentiates to $\rho {\partial }_{x^1}$.

The coordinate Lie brackets satisfyFor the first identity only the derivative of the coefficient $\rho$ contributes. The linear map $\boxed{X_j⟼L_j}$ is a Lie algebra isomorphism: it preserves all basis brackets, is injective by independence of the frame, and is onto the three-dimensional space of left-invariant vector fields.