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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.

Suppose that $H$ is not a discrete subgroup. The identity is then an accumulation point of nonidentity subgroup elements: translate any failure of isolation to the identity. Choose $h_j\in H\setminus \{e\}$ with $h_j\to e$, and use a local exponential chart to write $h_j=\exp X_j$, where $X_j\ne 0$ and $X_j\to 0$ in the real Lie algebra $\mathfrak{g}$.

Choose a norm on this finite-dimensional vector space. By compactness of its unit sphere, pass to a subsequence with $X_j/\Vert X_j\Vert \to X$, where $\Vert X\Vert =1$. For an arbitrary real $t$, choose an integer $m_j$ nearest to $t/\Vert X_j\Vert$. Then $m_j\Vert X_j\Vert \to t$ andEvery left-hand element belongs to $H$, including for negative $m_j$. Closedness gives $\exp (tX)\in H$. Since $t$ was arbitrary and $X\ne 0$, the limit directions of a closed subgroup provide a nontrivial one-parameter subgroup in $H$. Its derivative at zero is $X$, so it cannot be the constant subgroup.

DefineIt contains zero, and $X\in \mathfrak{h}$ implies $aX\in \mathfrak{h}$ for every real scalar $a$, by replacing $t$ with $at$.

For $X,Y\in \mathfrak{h}$, we prove closure under addition using the Lie product formula in a Lie group. In the local exponential chart, let $Z(s)=\log (\exp (sX)\exp (sY))$. Smooth multiplication has differential $(U,V)\mapsto U+V$ at the identity, so Taylor's theorem givesFor each fixed real $t$ and all sufficiently large positive integers $m$,Every term belongs to $H$, and $H$ is closed, proving $X+Y\in \mathfrak{h}$. Thus $\mathfrak{h}$ is a real vector subspace. It is also closed: if $X_j\to X$ with $X_j\in \mathfrak{h}$, continuity of the Exponential map of a Lie group gives $\exp (tX)\in H$ for each $t$.

For the Lie bracket, use the Adjoint representation of a Lie group. Conjugating a one-parameter subgroup givesConsequently ${\mathrm{Ad}}_{\exp (sX)}Y\in \mathfrak{h}$. For $s\ne 0$, linearity puts $({\mathrm{Ad}}_{\exp (sX)}Y-Y)/s$ in $\mathfrak{h}$. Its limit as $s\to 0$ is ${\mathrm{ad}}_{X}Y=[X,Y]$. Closedness of $\mathfrak{h}$ therefore gives $[X,Y]\in \mathfrak{h}$, proving that the Lie algebra of a closed subgroup is a Lie subalgebra of $\mathfrak{g}$.

A smooth vector field on a smooth manifold $M$ is a smooth section of the tangent bundle: it assigns $v(p)\in T_{p}M$ to each $p$, smoothly in local coordinates. In a coordinate chart it has the form $v={\sum }_i{a}_i\partial /\partial x_i$, with smooth coefficients $a_i$. Equivalently, it acts on smooth functions as a derivation, $v(fh)=v(f)h+fv(h)$.

For a Lie group $G$, write $L_a(h)=ah$ for Left translation on a Lie group. A left-invariant vector field satisfiesThus it is determined by its value at the identity. Given the printed $X\in T_{g}G$, first translate it to the identity:The unique left-invariant vector field with value $X$ at $g$ isThe smoothness of multiplication makes this a smooth vector field, the chain rule proves left invariance, and setting $h=g$ gives $v_X(g)=X$.

We prove the completeness of left-invariant vector fields: this left-invariant vector field is a complete vector field. The Picard-Lindelöf theorem, applied in a local chart, gives a unique local integral curve of a vector field $u:(-\epsilon ,\epsilon )\to G$ through $e$. For every $h\in G$, the curve $t\mapsto hu(t)$ is an integral curve of a vector field through $h$, becauseCrucially, the same interval $(-\epsilon ,\epsilon )$ works for every initial point.

Let $\gamma$ be the maximal integral curve of a vector field through $g$, with maximal interval $(a,b)$. If $b<\infty$, choose $t_0\in (a,b)$ with $b-t_0<\epsilon /2$. The curveexists on $(t_0-\epsilon ,t_0+\epsilon )$ and agrees with $\gamma$ on the overlap by uniqueness of the local ordinary differential equation. It extends $\gamma$ past $b$, a contradiction. The same argument at a finite $a$ excludes that possibility. Thus $(a,b)=\mathbb{R}$, and uniqueness on overlapping intervals gives uniqueness on all of $\mathbb{R}$.

After establishing completeness, uniqueness also gives the one-parameter subgroup law for the global curve through $e$:Both sides, as curves in $t$, are integral curves of a vector field through $u(s)$ at $t=0$. Defining the Exponential map of a Lie group by $u(t)=\exp (t\xi )$, the answer is

The identity component $G^{\circ }$ is an open normal subgroup, and every open identity neighbourhood generates it. Let $G^{\circ }$ be the connected component of $e$. The product of connected spaces is connected, so the image of $G^{\circ }\times G^{\circ }$ under multiplication is connected and contains $e$. It is therefore contained in $G^{\circ }$. Inversion has the same property. Thus $G^{\circ }$ is a subgroup.

A smooth manifold is locally connected. In particular, a coordinate neighbourhood of $e$ can be chosen homeomorphic to an open ball, so there is a connected open neighbourhood $O$ of $e$ contained in $G^{\circ }$. Its translates $hO$, $h\in G^{\circ }$, are open and lie in $G^{\circ }$, and cover $G^{\circ }$. Therefore the identity component of a Lie group is open in $G$. Conjugation by any $a\in G$ is a homeomorphism fixing $e$, so it maps $G^{\circ }$ into itself; conjugation by $a^{-1}$ gives the reverse inclusion. Hence $G^{\circ }$ is a normal subgroup.

If $U$ is an open neighbourhood of $e$ in $G^{\circ }$, let $H=⟨U⟩$, allowing inverses in the meaning of generated subgroup. For each $h\in H$, $hU$ is open and lies in $H$, so $H={\bigcup }_{h\in H}hU$ is open in $G^{\circ }$. Every other left coset of $H$ is also open. Thus $H$ is both open and closed in the connected space $G^{\circ }$. It is nonempty, so $H=G^{\circ }$.

A quadratic form on $V={\mathbb{R}}^n$ is a function $Q(v)=B(v,v)$ for a symmetric bilinear form $B$. Equivalently, $Q(tv)=t^{2}Q(v)$ and the polarizationis a bilinear map. In coordinates there is a unique real symmetric matrix $A$ with $Q(v)=v^{T}Av$. No assumption of nondegeneracy or positive definiteness is needed.

An element $g\in {\mathrm{GL}}_n(\mathbb{R})$ stabilizes $Q$ when $Q(gv)=Q(v)$ for every $v$. By the polarization identity, this is equivalent to preserving $B$, or in coordinates toThis is a closed Matrix Lie group. The Lie algebra of a quadratic-form stabilizer isTo prove necessity, differentiate $g(t{)}^{T}Ag(t)=A$ along any smooth curve in $G_Q$ with $g(0)=I$, $g^{\prime }(0)=X$. The derivative at zero is $X^{T}A+AX$.

To prove sufficiency, suppose $X^{T}A+AX=0$ and consider the matrix exponential $g(t)=e^{tX}$. ThenIts value at zero is $A$, so $e^{tX}\in G_Q$ for every real $t$. This curve has derivative $X$ at zero, proving the claimed tangent space description even for a degenerate quadratic form.

Equivalently, the condition is $B(Xu,v)+B(u,Xv)=0$ for all $u,v$. For a nondegenerate quadratic form it is the corresponding Special orthogonal Lie algebra. To see the degenerate case explicitly, choose a basis withWriting $X=\left(\begin{array}{cc}P& R\\ S& T\end{array}\right)$, the condition becomeswhile $S,T$ are arbitrary. Thus the nullspace of the quadratic form is invariant, but arbitrary infinitesimal maps into it are allowed. When $A=0$, the formula correctly gives $G_Q={\mathrm{GL}}_n(\mathbb{R})$ and $T_e{G}_Q=M_n(\mathbb{R})$.

For a Matrix Lie group, the exponential map of a matrix Lie group is the convergent matrix exponentialIts domain is the Lie algebra $\mathfrak{g}=T_{I}G$, and $e^{tX}\in G$ for every real $t$ when $X\in \mathfrak{g}$. Because multiples of the same matrix commute, $e^{sX}{e}^{tX}=e^{(s+t)X}$ and $(e^{tX}{)}^{-1}=e^{-tX}$. Thus take $I=\mathbb{R}$ to obtain the image of a one-parameter subgroup.

Here is the precise classification of nontrivial one-parameter subgroups. For $X\ne 0$, the smooth group homomorphism ${\gamma }_X:t\mapsto e^{tX}$ has a closed kernel $K\subset \mathbb{R}$. A closed subgroup of $\mathbb{R}$ is zero, $\tau \mathbb{Z}$ for some $\tau >0$, or all of $\mathbb{R}$. Indeed, if positive elements have infimum zero, their integer multiples approximate every real number, so closedness makes the subgroup all of $\mathbb{R}$; otherwise the infimum is attained and is its least positive generator. The last possibility is excluded by ${\gamma }_X^{\prime }(0)=X\ne 0$. Give the image the quotient topology and smooth structure from $\mathbb{R}/K$. Its inclusion in $G$ is an injective immersion, since ${\gamma }_X^{\prime }(t)=e^{tX}X$ never vanishes. Consequently it is a Lie subgroup, withIn the periodic case $[0,\tau )$ is also a parameter interval covering the image, with endpoints understood modulo $\tau$.

Two qualifications matter. If $X=0$, the image is the trivial group, a third possibility omitted by the wording “two”. Also, a general one-parameter subgroup need not be closed or embedded: an irrational winding $t\mapsto \mathrm{diag}(R(t),R(\sqrt{2}t))$ in a two-torus is an example, where $R(t)$ is a planar rotation. The classification uses the intrinsic immersed Lie subgroup structure, rather than assuming that the matrix subspace topology is the topology of $\mathbb{R}$.

For the real special linear group, differentiating $\det (I+tX)=1+t\mathrm{tr}X+O(t^2)$ at the identity shows that its Lie algebra consists of traceless matrices. Conversely $\det e^{tX}=e^{t\mathrm{tr}X}=1$ for a traceless matrix. HenceChoose the standard basis of the real sl2 Lie algebra,Their squares have the required values. Direct matrix multiplication gives the Lie bracketsWriting $[L_a,L_b]=f_{ab}{}^c{L}_c$, these specify all the nonzero Lie algebra structure constants: $f_{0+}{}^{+}=2$, $f_{0-}{}^{-}=-2$, $f_{+-}{}^0=1$, together with their negatives on reversing the lower indices.

The three requested exponentials areThus the first two images are isomorphic to the additive real group, while the last is $SO(2)$, the circle group. The PDF gives $L_{+}-L_{-}$ here; the local TeX's $L_{-}-L_{-}$ is a transcription error.

For a general generator, put $\Delta ={\alpha }_0^2+{\alpha }_{+}{\alpha }_{-}$. Direct multiplication gives $X^2=\Delta I_2$. If $\Delta >0$, the eigenvalues $\pm \sqrt{\Delta }$ give an unbounded exponential image. If $\Delta =0$ but $X\ne 0$, it is a nilpotent linear map with exponential $I_2+tX$, again unbounded. If $\Delta =-{\omega }^2<0$, the power series instead giveswith least positive period $2\pi /\omega$. Its image is a continuous image of a circle and is compact. Equivalently, $X/\omega$ is a real complex structure and is similar over $\mathbb{R}$ to the rotation generator. Therefore the exact criterion for compact one-parameter subgroups of SL2R isFor a nonzero generator only the strict inequality remains.

The transformationsform a one-parameter group with infinitesimal generatorUnder this scaling, ${\partial }_X$ has weight $-1$ and ${\partial }_T$ has weight $1$, so $u_{XT}$ has weight zero, as does $\sin u$. Thus $u_{XT}=\sin u$ is invariant and $g^s$ is a Scaling symmetry of the sine-Gordon equation.

The productis invariant under the group. A group-invariant solution therefore has the form $u(X,T)=F(z)$. Sincethe Sine-Gordon equation reduces to

Now put $w=e^{iF}$. ThenSubstitution and multiplication by $iw$ giveHenceThis is the Painlevé III equation with parameters $(\alpha ,\beta ,\gamma ,\delta )=(1/2,-1/2,0,0)$.

The one-parameter subgroup ${\gamma }_{\xi }(t)=\exp (t\xi )$ is the integral curve through the identity of $l_{\xi }$. Its defining initial-value problem is

For $\xi$ in the Lie algebra $\mathfrak{g}$, set ${\Phi }^t=R_{\exp (t\xi )}$. The one-parameter subgroup law gives the flow law, andso this is the global flow of the left-invariant vector field $l_{\xi }$.

For $f(g)=R_g^{\ast }\alpha$, every tangent vector at $g$ is the initial velocity of $g\exp (t\xi )$ for some $\xi$. Since $R_{g\exp (t\xi )}=R_{\exp (t\xi )}\circ R_g$,Thus all ${\mathcal{L}}_{l_{\xi }}\alpha$ vanish exactly when every derivative of $f$ vanishes. This is equivalent to $f$ being a locally constant function.

The matrices are rotations about the third coordinate axis. Direct multiplication givesand $g(\theta {)}^{\mathsf{T}}g(\theta )=I$. Thus they form a one-parameter subgroup of the orthogonal group $O(3)$, isomorphic to the circle group.

Differentiate the planar rotation matrix at angle zero:The even and odd powers in the matrix exponential therefore giveThis exact formula gives the whole one-parameter subgroup. Taking $X=J$ means unit angular velocity: at $t=\pi /2$, $\exp (tJ)=J$ sends $(1,0{)}^T$ to $(0,1{)}^T$, a counterclockwise quarter-turn. At $t=\pi$, it equals $-I$; at $t=2\pi$, it returns to $I$. Choosing $X=2J$ reaches the quarter-turn at $t=\pi /4$. Changing the generator changes the speed, while preserving the circular path.

For a moving vector $v(t)=R_{t}v(0)$, the ordinary differential equation $v^{\prime }(t)=Jv(t)$ says its velocity is perpendicular to its position. The beautiful compression is that one constant matrix specifies every angle of this endless family of rotations.