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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 $s=0$.

A Lie algebra homomorphism $\theta :{\mathfrak{g}}_1\to {\mathfrak{g}}_2$ always defines a local group homomorphism near the identities by the exponential charts:for sufficiently small $X$. Preservation of the BCH multiplication law makes this a local homomorphism. The issue is extending it consistently around all paths and periods in the source.

Suppose first that $G_1$ is connected and simply connected. For a piecewise smooth path $\gamma :[0,1]\to G_1$ from $e_1$ to $g$, letbe its left logarithmic velocity. Solve in $G_2$ the differential equationThe equation exists throughout the finite path interval: bounded smooth body velocity gives a uniform local existence interval at the identity, and left translation repeatedly continues the solution. Define the proposed map by the endpoint $\Phi (g)=\eta (1)$.

Bracket preservation makes this endpoint invariant under fixed-endpoint homotopy. To see the mechanism, for a two-parameter path write $A={\gamma }^{-1}{\partial }_t\gamma$ and $B={\gamma }^{-1}{\partial }_s\gamma$, using the Maurer-Cartan form rather than literal multiplication for nonmatrix groups. The Maurer-Cartan equation givesAfter applying $\theta$, the same identity holds for $a=\theta A$ and $b=\theta B$. If $c={\eta }^{-1}{\partial }_s\eta$, the transport equation impliesBoth have zero initial value because the path starts at a fixed identity. Uniqueness gives $c=b$. At the fixed endpoint, $B(s,1)=0$, hence ${\partial }_s\eta (s,1)=0$. The endpoint is therefore unchanged by the homotopy. Simple connectedness then makes it independent of the chosen path.

For paths to $g$ and $h$, concatenate the first with the left translate by $g$ of the second. Its transported endpoint is $\Phi (g)\Phi (h)$, because left logarithmic velocity is unchanged by left translation. It follows that $\Phi (gh)=\Phi (g)\Phi (h)$. Near the identity, choose $\gamma (t)=\exp (tX)$; transport yields $\Phi (\exp X)=\exp (\theta X)$. Thus $\Phi$ is smooth near the identity and, by translation, everywhere, and its differential is $\theta$. The uniqueness argument from the root solution givesThis is integration of a Lie algebra homomorphism.

For a connected source that is not simply connected, first integrate $\theta$ on its universal cover $p:{\tilde{G}}_1\to G_1$. The resulting map $\tilde{\Phi }:{\tilde{G}}_1\to G_2$ descends exactly whenIndeed two lifts of one element differ by a kernel element, so this condition is both necessary and sufficient for their images to agree. It is the period obstruction to integration of a Lie algebra homomorphism. For $G_1=G_2=S^1$, identify the Lie algebras with $\mathbb{R}$ and take $\theta (t)=ct$. The integrated map on the cover is $t\mapsto e^{ict}$, which respects the source period $2\pi$ precisely when $c\in \mathbb{Z}$. Therefore half-scaling is a valid Lie algebra homomorphism but does not integrate to a circle homomorphism. If both groups are written as simply connected covers modulo central subgroups, the equivalent descent condition is that the integrated cover map send ${\Gamma }_1$ into ${\Gamma }_2$.

For disconnected $G_1$, first solve this problem on $G_1^{\circ }$, obtaining ${\varphi }_0$. One must then assign images $a_{\gamma }$ to representatives $s_{\gamma }$ of the component group. These assignments must satisfy the conjugation and multiplication relations. Explicitly, with $s_e=e$ and $s_{\gamma }{s}_{\delta }=s_{\gamma \delta }{c}_{\gamma ,\delta }$ for $c_{\gamma ,\delta }\in G_1^{\circ }$, the conditions areWhen these can be solved, $\varphi (s_{\gamma }h)=a_{\gamma }{\varphi }_0(h)$ is a smooth global homomorphism. They may prevent existence or allow several extensions with the same differential. Thus a bracket-preserving linear map gives local data; global integration requires the source's periods and, when present, its component relations to be respected.

For a finite-dimensional real Lie algebra $\mathfrak{h}$, the Lie third theorem asserts the existence of a connected simply connected Lie group whose Lie algebra is $\mathfrak{h}$. There are two complementary views of the construction.

Locally, the Baker--Campbell--Hausdorff formula defines multiplication on a sufficiently small neighbourhood of zero in $\mathfrak{h}$. Its leading terms are $X+Y+\frac{1}{2}[X,Y]$, its identity is zero, and inversion is $X\mapsto -X$. The convergent BCH series in finite dimensions and the Jacobi identity provide the local associative law. Continuing and globalizing this local group gives its simply connected integration; the globalization is a substantive step, rather than a claim that one exponential chart gives all of the group.

A matrix construction makes that global step more concrete. The Ado theorem gives a faithful finite-dimensional representationConsider the left-invariant distribution $D_A=A\iota (\mathfrak{h})$ on ${\mathrm{GL}}_N(\mathbb{R})$. It is involutive because $\iota (\mathfrak{h})$ is closed under commutators. The Frobenius theorem integrates it to its connected leaf through the identity. This leaf carries the intrinsic immersed-submanifold structure of the connected Lie subgroup generated by the matrix exponentials $e^{\iota (X)}$. Its tangent Lie algebra is exactly $\iota (\mathfrak{h})$; multiplication and inversion preserve the leaf and give its Lie group operations. Call this immersed group $H_0$.

Taking its universal covering Lie group givesAdo's faithfulness matters: the adjoint representation alone loses central elements. Also, the subgroup generated by the exponentials need not be closed in the ambient matrix group. For an irrational $a$, the map $t\mapsto (R_t,R_{at})$ gives a dense one-dimensional immersed subgroup of the two-dimensional torus. Its intrinsic group is $\mathbb{R}$, whereas its closure has a two-dimensional Lie algebra. Taking an ambient closure in the construction would therefore be incorrect.

Finally, any other connected Lie group with Lie algebra $\mathfrak{h}$ is obtained from this simply connected integration by a discrete central quotient, as in part (i). The simply connected integration is unique up to isomorphism, but an arbitrary corresponding group is not.

An isomorphism of Lie algebras identifies the local multiplication laws of the identity components. In exponential coordinates the local product is governed by the Baker--Campbell--Hausdorff formula,A bracket-preserving linear isomorphism carries this law to the corresponding law in the other Lie algebra. Thus the identity components are locally isomorphic. This statement is local; the exponential coordinates need not cover the whole group.

The stronger global description uses a universal covering Lie group. The universal cover of a connected Lie group acquires a group law by lifting multiplication with a chosen identity; uniqueness of lifts supplies associativity and inversion. The covering map is a local diffeomorphism and a Lie group homomorphism, with discrete kernel. That kernel is central: conjugation of one of its elements gives a continuous map from the connected covering group into a discrete set, so it is constantly that element.

By integration of the given Lie algebra isomorphism and its inverse on simply connected covers, these covers are isomorphic. Their two compositions have identity differential and therefore are identity maps by the preceding uniqueness argument. Consequently, for some connected simply connected Lie group $\tilde{G}$,where ${\Gamma }_1,{\Gamma }_2$ are discrete central subgroups. Different quotients need not be isomorphic. The additive group $\mathbb{R}$ and the circle group $\mathbb{R}/2\pi \mathbb{Z}$ have the same one-dimensional Lie algebra but different compactness and fundamental groups. Another familiar pair is the SU(2) group and the SO(3) group, with the latter the quotient of the former by $\{I,-I\}$.

Thus isomorphic Lie algebras give isomorphic connected simply connected groups, and identify arbitrary identity components up to discrete central quotients. For disconnected groups the Lie algebra also omits the group of components and its conjugation and extension data. For example $\mathbb{R}\times F$, for any finite group $F$, always has Lie algebra $\mathbb{R}$, while its component group is $F$.

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.

The derived series of a Lie algebra is defined recursively by ${\mathfrak{g}}^{(0)}=\mathfrak{g}$ and ${\mathfrak{g}}^{(r+1)}=[{\mathfrak{g}}^{(r)},{\mathfrak{g}}^{(r)}]$, where the bracket denotes the linear span of all brackets of pairs of elements. A solvable Lie algebra is one whose derived series of a Lie algebra terminates at zero:The solvable radical $\mathrm{rad}\mathfrak{g}$ of a finite-dimensional Lie algebra is its largest solvable ideal of a Lie algebra. A semisimple Lie algebra has no nonzero solvable ideals of a Lie algebra, equivalentlyThese are definitions; identifying semisimplicity with a direct sum of simple ideals additionally uses the structure theorem in characteristic zero. The calculations and representation theorems below are over $\mathbb{C}$, as in the complex representation setting of this paper.

Use the faithful logarithm realization from the root Solution: $\log :G\to A$ identifies $L_G$ with the Lie algebra $\log G$. For $x,y\in G$, the convergent exponential identity in $A$ givesEvery Lie polynomial and its convergent sum are preserved by this realization of $L_G$. Therefore the BCH product on the underlying set of $L_G$ has exactly the same logarithm as the original product $xy$. Injectivity of the logarithm proves equality of the two products, with the same topology, identity and inverses:This proves the reverse direction of the uniform pro-p group and powerful Lie lattice correspondence directly; the forward identity alone would not imply it.

Put $U_0=G$ and $U_{j+1}=U_j^p$, where powers mean closed generated subgroups. The finite-quotient arguments of Question 3 pass to a powerful pro-p group: in particular $U_j$ is powerfully embedded, the power map is onto $U_{j+1}$, andFor the root assertion, apply the finite result to every finite quotient and use the finite intersection property on the closed sets of possible roots in the compact group. Each $U_j$ is finitely generated by the appropriate powers of a finite generating tuple, and $U_j/U_{j+1}$ is finite and elementary abelian. Thus the $U_j$ are open. They form an identity-neighbourhood basis: in any finite continuous quotient their images form a terminating lower p-series, so some $U_j$ is contained in the kernel. In particular ${\bigcap }_j{U}_j=1$. Collection also givesOne can see the extra index directly by induction from $[U_a,G]\subseteq U_{a+1}$: collect $[v,w^p]$ for $w\in U_b$. Its leading factor $[v,w{]}^p$ advances one layer, and the remaining iterated commutators, with their binomial coefficients, lie at least as deep. Products and limits preserve the inclusion. This is the usual strongly central property of the lower p-series.

A uniform pro-p group is a finitely generated powerful pro-p group for which all the indices $|U_j:U_{j+1}|$ equal $|G:G^p|$. The preliminary congruence gives surjective linear mapsIf $G$ is uniform, these maps are bijective. If $x\ne y$, write $y=xu$ and choose the unique layer $j$ with $u\in U_j\setminus U_{j+1}$. Bijectivity of ${\theta }_j$ and the power congruence imply $y^p(x^p{)}^{-1}\in U_{j+1}\setminus U_{j+2}$, so $x^p\ne y^p$. Thus the power map is injective. A finite-order element of a pro-p group has $p$-power order; repeated injectivity of the power map excludes every nonidentity such element. Hence a uniform group is a torsion-free group.

Conversely, suppose some ${\theta }_j$ has a nonzero kernel. Choose $x\in U_j\setminus U_{j+1}$ with $x^p\in U_{j+2}$. At step $r\ge j+1$, if the current $x_r$ satisfies $x_r^p\in U_{r+1}$, surjectivity of ${\theta }_r$ supplies $u_r\in U_r$ withSet $x_{r+1}=x_r{u}_r$. The power congruence gives $x_{r+1}^p\in U_{r+2}$. Since the corrections tend to $1$ in the complete profinite group, the $x_r$ converge to $z$, with $z^p=1$ and $z\equiv x(\mathrm{mod}{U}_{j+1})$. In particular $z\ne 1$. A torsion-free group cannot have such an element, so every ${\theta }_j$ is injective and all layer indices are equal. This proves the root correction criterion for uniformity andFor a uniform group the power maps are continuous bijections $U_j\to U_{j+1}$, hence homeomorphisms. Their inverses define the unique roots used below. Comparison at the first differing layer also proves the power congruences in a uniform pro-p group:

The intrinsic Lie algebra of a uniform pro-p group has the same underlying set as $G$, zero equal to the group identity, and operationsHere $x^{\lambda }$ is the limit of integral powers approaching $\lambda$. The first root is taken in $U_n$ and the second in $U_{2n}$; the commutator actually lies in $U_{2n+1}$. The following faithful algebra realization verifies convergence and the Lie algebra axioms, rather than assuming them as part of the definition.

Choose a minimal tuple $a_1,\dots ,a_d$ of topological generators. The layer isomorphisms ${\theta }_j$ show that the images of $a_i^{p^j}$ are bases of $U_j/U_{j+1}$. The ordered-coordinate argument of Question 3 therefore gives a continuous bijectionExistence follows by successive digits and compactness; uniqueness follows by the first digit at which the coordinates differ, because its images are linearly independent in the corresponding power layer. Thus this map is a homeomorphism.

Here are details of the norm needed on the group algebra. In ${\mathbb{Z}}_p[[G]]$ put $b_i=a_i-1$. Its elements have unique ordered expansions ${\sum }_{\alpha }{c}_{\alpha }{b}_1^{{\alpha }_1}\cdots b_d^{{\alpha }_d}$ with $c_{\alpha }\in {\mathbb{Z}}_p$. To verify this description, identify the ordered coordinates of $G$ with ${\mathbb{Z}}_p^d$. Elements of the completed group algebra are integral measures on this space; their values on the binomial polynomials ${\prod }_i\left(\genfrac{}{}{0ex}{}{{\lambda }_i}{{\alpha }_i}\right)$ are precisely the coefficients $c_{\alpha }$. The finite-difference identity pairs the ordered monomial $b^{\beta }$ with this binomial polynomial as ${\delta }_{\alpha \beta }$. The Mahler expansion of a continuous function therefore proves both existence and uniqueness of the measure with the specified coefficients. This also proves that the ordinary rational group algebra embeds in these expansions.

Choose $1/(p-1)<s<1$ and setThis defines the weighted group algebra of a uniform pro-p group. Its multiplication is submultiplicative in the valuation convention, $v(AB)\ge v(A)+v(B)$. Indeed, $a_i^p-1=p{b}_i+\left(\genfrac{}{}{0ex}{}{p}{2}\right)b_i^2+\cdots +b_i^p$ has valuation $1+s$; collection of ordered coordinates gives the same lower bound for $g-1$ when $g\in U_1$. Interchanging two adjacent $b_i$ in a product introduces a correction coming from $[a_i,a_j]-1$, of valuation at least $1+s>2s$. Iterating the interchange rule never lowers the total weight; the corrections have strictly increasing weight, so infinite collection converges. The same argument with $p^j$-th powers showsThe equality follows from the first nonzero coordinate digit: its leading terms are $p^j{\lambda }_i^{\prime }{b}_i$, at least one with unit ${\lambda }_i^{\prime }$, and cannot cancel in the ordered expansion. Completing the rational group algebra in this norm produces a complete normed algebra $A$ in which $G$ embeds faithfully and with its original topology.

Since $v(g-1)\ge s>1/(p-1)$, the seriesconverge on the relevant balls in $A$ and are inverse there. Their convergence follows from $v_p(r!)\le (r-1)/(p-1)$ and $v_p(r)\le (r-1)/(p-1)$; formal one-variable identities prove that they are inverse. Put $X=\log x$ and $Y=\log y$. The Baker--Campbell--Hausdorff formula in $A$ now givesbecause every term of degree at least two tends to zero after this rescaling. Similarly expansion of $\exp (-p^{n}X)\exp (-p^{n}Y)\exp (p^{n}X)\exp (p^{n}Y)$ givesThe leading term is the indicated associative-algebra commutator and all terms of degree at least three tend to zero. These expressions are the logarithms of the unique roots in the intrinsic formulas. The image $\log G$ is compact and hence closed in $A$, so both limits belong to $\log G$, and the inverse logarithm takes the limits back to $G$. Also $\log (x^{\lambda })=\lambda \log x$. Consequently $\log$ identifies the intrinsic operations with addition, scalar multiplication and commutation in $A$. The Jacobi identity follows by expanding the associative products; thus $L_G$ is indeed a Lie algebra over ${\mathbb{Z}}_p$.

The commutator roots in the intrinsic formula belong to $U_1$, so their limit gives $[L_G,L_G]\subseteq \log U_1=p{L}_G$. Iterating the power congruence gives $x^{p^n}{y}^{p^n}\equiv (xy{)}^{p^n}(\mathrm{mod}{U}_{n+1})$. Taking the unique roots shows that intrinsic addition modulo $U_1$ is the original group product modulo $U_1$. Consequently $L_G/p{L}_G$ is the original elementary abelian quotient $G/U_1$, and the images of the $a_i$ form a basis. Successively removing digits expresses every element uniquely as an additive sum ${\sum }_i{\lambda }_i{a}_i$: existence uses completeness, while a nonzero relation would have a first nonzero digit contradicting the basis in $U_j/U_{j+1}$. Hence $L_G$ is a free module of rank $d$ over the p-adic integers and is a powerful Lie lattice.

Conversely a powerful Lie algebra over p-adic integers, for odd $p$, is a finite free ${\mathbb{Z}}_p$-module $L$ with a bilinear alternating Lie bracket satisfying the Jacobi identity and $[L,L]\subseteq pL$. DefineFor completeness, integrality and convergence can be verified without guessing which individual denominators are units. In ${\mathbb{Q}}_p\otimes L$ set $M=p^{-1}L$. Then $[M,M]\subseteq M$, so its integral universal enveloping algebra has an ordered basis by the Poincaré-Birkhoff-Witt theorem. Its coefficientwise $p$-adic norm is submultiplicative, and $L=pM$ consists of elements of valuation at least one. The homogeneous degree-$r$ terms in $\log (\exp X\exp Y)$ have coefficients of valuation at least $-\lfloor (r-1)/(p-1)\rfloor$: expand the logarithm, use $v_p(k!)\le (k-1)/(p-1)$ on each nonconstant exponential factor, and $v_p(\ell )\le (\ell -1)/(p-1)$ on a product of $\ell$ such factors. Therefore evaluating degree $r$ on $X,Y\in pM$ gives valuation at leastwhich tends to infinity and is at least two when $r\ge 2$. Each homogeneous term is a rational Lie polynomial by the formal Baker--Campbell--Hausdorff formula. The ordered enveloping-algebra basis shows that its intersection with ${\mathbb{Q}}_{p}M$ is exactly $M$, so the estimate places the sum in $L$, with the non-linear part in $pL$.

Associativity follows from the identity $\exp (X\ast Y)=\exp X\exp Y$ in the completed universal enveloping algebra. Its identity is $0$, its inverse is $-X$, and $X^{\ast m}=mX$, since the bracket of $X$ with itself is zero. The same coefficient estimate for the group commutator puts it in $pL$. Conjugation in the BCH group is $\exp (\mathrm{ad}X)$; each term $\mathrm{ad}(X{)}^r/r!$ preserves $p^{j}L$, because each bracket adds a factor of $p$ and $v_p(r!)<r$. Thus the submodules $p^{j}L$ are normal subgroups for $\ast$, their intersection is zero, and their finite quotients have $p$-power order. Thus $(L,\ast )$ is a powerful pro-p group, withA module basis topologically generates this group: it generates the quotient $L/pL$, and successive power digits lift that generation through all $p^{j}L$. The BCH construction therefore gives a uniform pro-p group.