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For $X\ne 0$, the image of $t\mapsto e^{tX}$, with its intrinsic immersed Lie subgroup topology, is isomorphic to $(\mathbb{R},+)$ or the circle group. The kernel is respectively zero or $\tau \mathbb{Z}$ for a least positive period $\tau$. The case $X=0$ is the trivial group. An irrational winding in a two-dimensional torus need not be an embedded or closed subgroup, so its intrinsic topology must be distinguished from the subspace topology.

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.

Conjugation by every element of $G$ preserves a normal Lie subgroup $H$. Its differential therefore preserves $\mathfrak{h}=T_{e}H$. Differentiate that invariance along $\exp (tX)$ to obtain $\mathrm{ad}(X)\mathfrak{h}\subseteq \mathfrak{h}$, proving that $\mathfrak{h}$ is an ideal of a Lie algebra. Closed subgroups have a canonical Lie-subgroup structure; an arbitrary abstract subgroup need not.

Every finite-dimensional real Lie algebra is the Lie algebra of a connected simply connected Lie group. One construction first uses the Ado theorem to embed it into a matrix Lie algebra, integrates the corresponding left-invariant distribution by the Frobenius theorem to a connected immersed Lie subgroup, and then takes a universal covering Lie group. The immersed subgroup need not be closed; taking its ambient closure may change its Lie algebra.

A maximal torus is a connected compact abelian Lie subgroup maximal among such subgroups. Restricting a finite-dimensional unitary representation to it splits into one-dimensional weight characters. For special unitary group $\mathrm{SU}(2)$ it is the diagonal circle $\mathrm{diag}(z,z^{-1})$.

By part (i), finite-dimensional continuous unitary representations separate points of the compact group. More explicitly, if every representation sent some $g\ne e$ to the identity, all their matrix coefficients would take the same values at $g$ and $e$. Uniform density would force every continuous scalar function to do so, contradicting separation of points on a compact Hausdorff space. HenceEach representation kernel is a closed normal subgroup. We claim that the intersection can be reduced to finitely many kernels. Begin with $K_0=G$. If $K_j\ne \{e\}$, choose $g_j\in K_j\setminus \{e\}$ and a representation ${\theta }_{j+1}$ with ${\theta }_{j+1}(g_j)\ne I$. ThenAn indefinite continuation would violate the descending chain condition for closed subgroups. Thus after finitely many steps, $K_N=\{e\}$.

The direct sum $\rho ={\theta }_1\oplus \cdots \oplus {\theta }_N$ is consequently a finite-dimensional faithful representation, with continuous injective homomorphismCompactness of $G$ makes this a homeomorphism onto a compact, hence closed, image. The closed-subgroup theorem makes that image an embedded Lie subgroup of the finite-dimensional unitary group. Transporting its smooth structure to $G$ provesThis is the mechanism behind the assertion that compact groups with a descending chain condition are Lie groups: the chain condition turns an arbitrarily large separating family into one finite-dimensional faithful representation.

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.

There are two necessary qualifications to the literal formulation. The whole group $G$ is itself a closed normal subgroup, so the claim about containment in the centre must concern proper closed normal subgroups. Also, the homomorphism conclusion needs a nontrivial target $H$: for example, $SU(2)\to \{1\}$ has surjective differential onto the zero Lie algebra, but its group kernel is all of $SU(2)$ and no quotient by a finite subgroup is a one-point group. We prove the precise assertions, including these exceptional cases.

Write $\mathfrak{g}=\mathrm{Lie}(G)$. The centre $\mathfrak{z}(\mathfrak{g})$ is invariant under the Adjoint representation, so irreducibility makes it either zero or all of $\mathfrak{g}$. In the latter case $\mathfrak{g}$ is Abelian. For connected $G$, the adjoint action is then trivial: the group is generated by exponentials, and ${\mathrm{Ad}}_{\exp X}=e^{\mathrm{ad}X}=I$. A nonzero trivial representation is irreducible only in dimension one. A connected compact one-dimensional Lie group is a circle, so $G\cong U(1)$.

For every other $G$, the Lie algebra of its centre is zero. The centre is a closed compact Lie subgroup; a zero-dimensional compact Lie group is finite. Let $N$ be a closed normal subgroup. The closed-subgroup theorem identifies $\mathfrak{n}=\mathrm{Lie}(N)$ as an adjoint-invariant subspace of $\mathfrak{g}$, hence $\mathfrak{n}=0$ or $\mathfrak{n}=\mathfrak{g}$. If the latter holds, $N$ contains an identity neighbourhood and is open, so connectedness forces $N=G$. If $N$ is proper, it therefore has zero Lie algebra and is finite. For any $a\in N$, the continuous map $g\mapsto ga{g}^{-1}$ takes connected $G$ into the finite set $N$, so it is constant and equal to $a$. ThusProper closed subgroups of the circle are finite as well, and of course central.

Now let $d\Phi$ be surjective. The submersion theorem makes $\Phi (G)$ contain an identity neighbourhood in $H$, so it is an open subgroup. Since $H$ is connected, $\Phi (G)=H$. Its group kernel $F$ is closed and normal. If $H$ is nontrivial, $F$ is proper and the preceding argument gives a finite central group kernel; this also holds for $G\cong U(1)$, whose proper closed subgroups are finite. The Lie-group first isomorphism theorem givesIf $H$ is trivial, the correct statement is instead $H=G/G$.

For the Spin groups, use the real Clifford algebra with $v^2=-\Vert v{\Vert }^2$. The group $\mathrm{Spin}(m)$ consists of products of an even number of unit vectors inside its invertible even part. Conjugation on the embedded ${\mathbb{R}}^m$ defines the standard double covering $\mathrm{Spin}(m)\to SO(m)$ with group kernel $\{\pm 1\}$. The Clifford relations show that a unit-vector reflection, with the usual twisted conjugation, is a hyperplane reflection; the even products give orientation-preserving transformations. The Cartan–Dieudonné theorem states that every orthogonal transformation is a product of reflections, proving surjectivity. For $m\ge 3$ this connected covering group is simply connected; $\mathrm{Spin}(2)$ is a circle.

Here is an explicit exterior-square construction of the requested six-dimensional map. Put $E={\mathbb{C}}^4$ with its standard Hermitian form and volume $e_1\wedge e_2\wedge e_3\wedge e_4$. On ${\Lambda }^{2}E$, write $e_{ij}=e_i\wedge e_j$ and define an antilinear map $J$ byIt satisfies $J^2=1$. Intrinsically, if $u\wedge v=B(u,v)\mathrm{vol}$ and the Hermitian inner product $h$ is linear in its first variable, then $B(u,Jv)=h(u,v)$. Because $SU(4)$ preserves both $B$ and $h$, it commutes with $J$. This is the real structure of the SU4 exterior square.

The fixed space $W$ has the following real orthonormal basis, with all vectors divided by $\sqrt{2}$:Thus $W$ is six-dimensional over $\mathbb{R}$, its complexification is ${\Lambda }^{2}E$, and the exterior-square action restricts to an orthogonal representation on $W$. Its determinant is one because $SU(4)$ is connected. We obtain $R:SU(4)\to SO(W)\cong SO(6)$.

To compute the group kernel, diagonalize $g\in SU(4)$ with eigenvalues $z_1,\dots ,z_4$. If $R(g)=I$, its complexified exterior-square action is the identity, hence $z_i{z}_j=1$ for every $i<j$. Comparing pairs shows that all $z_i$ are equal, and their common value has square one. A unitary matrix with that single eigenvalue is scalar, so $\ker R=\{\pm I_4\}$. Conversely, those two scalar matrices plainly act trivially on the exterior square.

The differential of $R$ is injective because its group kernel is finite. Both Lie algebras have dimension $15$, so the differential is an isomorphism. Its image is consequently open in connected $SO(6)$, hence all of $SO(6)$. ThereforeThis also realizes $SU(4)$ as $\mathrm{Spin}(6)$, since $SU(4)$ is simply connected, by uniqueness of the connected simply connected covering group.

Choose a real vector subspace $\mathfrak{m}$ complementary to $\mathfrak{h}$, so $\mathfrak{g}=\mathfrak{m}\oplus \mathfrak{h}$. The smooth maphas differential $(U,V)\mapsto U+V$ at $(0,0)$, an isomorphism. The inverse function theorem makes $F$ a diffeomorphism from a sufficiently small product neighborhood onto an open neighborhood of the identity in $G$.

There is a neighborhood of zero in $\mathfrak{m}$ in which $\exp U\in H$ implies $U=0$. Otherwise choose nonzero $U_j\in \mathfrak{m}$ tending to zero with $\exp U_j\in H$. A normalized subsequence tends to a unit vector $U\in \mathfrak{m}$. The limit directions of a closed subgroup from part (a) give $U\in \mathfrak{h}$, contradicting $\mathfrak{m}\cap \mathfrak{h}=0$. Shrink the product neighborhood so that this exclusion holds throughout its first factor.

If $F(U,V)\in H$, then $\exp V\in H$ by the definition of $\mathfrak{h}$, and hence $\exp U=F(U,V)\exp (-V)\in H$. The exclusion forces $U=0$. Conversely, every $F(0,V)=\exp V$ belongs to $H$. Thus the local product coordinates for a closed subgroup identify $H$ near the identity exactly with the smooth coordinate slice $U=0$.

Left translation by each element of $H$ gives the same description around that element, so $H$ is an embedded submanifold of $G$ in its subspace topology. The multiplication and inversion maps of $G$ restrict to smooth maps on this embedded submanifold, making $H$ a Lie group and an embedded Lie subgroup. Its tangent space at the identity is $\mathfrak{h}$, because $d{\exp }_0$ is the identity on this coordinate slice. It remains closed by hypothesis. This proves the closed-subgroup theorem, with no prior smoothness assumption on $H$.

A real Lie group is a finite-dimensional real smooth manifold with a group structure whose multiplication and inversion are smooth. Its tangent space at the identity element $e$ is the vector space $\mathfrak{g}=T_{e}G$ of velocities of smooth curves through $e$. For the general linear group ${\mathrm{GL}}_n(\mathbb{R})$ it identifies with $M_n(\mathbb{R})$. The matrix exponential and local matrix logarithm areThe exponential is globally defined, and each matrix $\exp X$ is invertible with inverse $\exp (-X)$. The logarithm here is a local map near $I$; it is not a globally defined inverse on every real invertible matrix. The differential of $\exp$ at zero is the identity, and the two power series are inverse locally.

For a general $G$, the assumed Exponential map of a Lie group with $d{\exp }_0=I$ has a smooth local inverse by the inverse function theorem, giving a local exponential chart. Define the Lie bracket from local group commutators byFor small $s,t$ the group commutator lies in that chart. More explicitly, the map $F(u,v)=\log (\exp u\exp v\exp (-u)\exp (-v))$ is smooth and has $F(u,0)=F(0,v)=0$. Its mixed second differential at $(0,0)$ is a bilinear map of $X,Y$, proving bilinearity. Swapping the two group elements inverts the commutator, and $\log (g^{-1})=-\log g$ near $e$. Thus $F(v,u)=-F(u,v)$ and $[Y,X]=-[X,Y]$. In the matrix group, expansion to the mixed term $st$ gives the familiar commutator $XY-YX$.

For $g\in G$, let $c_g(h)=gh{g}^{-1}$ and define the Adjoint representation of a Lie group by ${\mathrm{Ad}}_g=d(c_g{)}_e\in \mathrm{GL}(\mathfrak{g})$. It is smooth and satisfies ${\mathrm{Ad}}_{gh}={\mathrm{Ad}}_g{\mathrm{Ad}}_h$. Its derived representation is $\mathrm{ad}=d(\mathrm{Ad}{)}_e:\mathfrak{g}\to \mathrm{End}(\mathfrak{g})$. Naturality of the Exponential map of a Lie group under conjugation givesAt $t=0$, the derivative of the logarithm of the commutator is ${\mathrm{Ad}}_{\exp (sX)}Y-Y$: the derivative of multiplication at $(e,e)$ adds tangent vectors, and $d{\log }_e=I$. Differentiate in $s$ to concludeTo prove the Jacobi identity without assuming it in the bracket construction, first establish that the differential of a Lie group homomorphism preserves Lie brackets. For a homomorphism $f:G\to H$, the allowed exponential identity gives ${\log }_H(f({\exp }_{G}u))=d{f}_e(u)$ locally. Apply this identity to the group commutator and take the mixed derivative to obtain $d{f}_e([X,Y])=[d{f}_{e}X,d{f}_{e}Y]$. In particular $f=\mathrm{Ad}:G\to \mathrm{GL}(\mathfrak{g})$ givesApplying both sides to $Z$ yields $[[X,Y],Z]=[X,[Y,Z]]-[Y,[X,Z]]$. Rearranging with antisymmetry provesNeither injectivity of $\mathrm{ad}$ nor the existence of a global logarithm was used.

Finally let $H$ be a normal Lie subgroup of $G$, with its corresponding Lie algebra $\mathfrak{h}\subseteq \mathfrak{g}$. For each $g\in G$, conjugation restricts to $H$, so its differential preserves $\mathfrak{h}$: ${\mathrm{Ad}}_g\mathfrak{h}=\mathfrak{h}$. Differentiate along $g=\exp (tX)$ to get $[X,Y]=\mathrm{ad}(X)Y\in \mathfrak{h}$ for every $X\in \mathfrak{g}$ and $Y\in \mathfrak{h}$. Thus the Lie algebra of a normal Lie subgroup is an ideal of a Lie algebra. The assertion concerns a subgroup carrying the corresponding Lie-subgroup structure, for example any closed subgroup; no arbitrary abstract subgroup is being assigned a tangent space.

For $X\in \mathfrak{h}$, the fundamental vector field of the restricted Lie group action is $(iX{)}_M$. If $\varphi$ is the original moment map, thenThis is the defining Hamiltonian action identity, in the convention ${\iota }_{X_H}\omega =dH$.

If equivariance is included in the definition of a moment map, it is also preserved: the dual map $\pi =i^{\ast }$ intertwines the coadjoint actions of $H$, since $i({\mathrm{Ad}}_{h}X)={\mathrm{Ad}}_h(iX)$. Thusis a moment map for the restricted Hamiltonian action of the Lie subgroup $H$.

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.

Restriction to a Lie subgroup with Lie algebra inclusion $i:\mathfrak{h}\to \mathfrak{g}$ has moment map $i^{\ast }\mu$. The dual map $i^{\ast }$ composes with the original moment map.