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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.

For ambient quadratic form $Q=-x_0^2-x_4^2+x_1^2+x_2^2$, associate $M=\left(\begin{array}{cc}{x}_4+x_1& x_2+x_0\\ x_2-x_0& x_4-x_1\end{array}\right)$. Its determinant is $-Q$, so the unit Anti-de Sitter spacetime hyperboloid is the special linear group of real two-by-two matrices. On its tangent space at the identity, the induced bilinear form is $\frac{1}{2}\mathrm{tr}(UV)$. The Killing form is $4\mathrm{tr}(UV)$, and left and right matrix multiplication preserve the ambient form. Consequently the induced bi-invariant pseudo-Riemannian metric is $B/8$. This model has periodic time; its causally unwrapped version is the universal covering Lie group.

Every matrix in $SU(2)$ has the unique formThe four real coordinates of $(\alpha ,\beta )$ give a smooth bijection with smooth inverse onto the unit three-sphere. Matrix multiplication identifies this model with the unit quaternions. Its round metric is invariant under left and right multiplication by unit quaternions. The actionhas kernel $\{(1,1),(-1,-1)\}$: a kernel pair must have $a=b$ from $q=1$ and must commute with all quaternions. Its derivative is injective, and both dimensions are six, givingThe full isometry group is $O(4)$. Quaternion conjugation reverses orientation and supplies the additional component. Thus an arbitrary isometry has either the displayed left-right form or that form composed with conjugation.

For the Anti-de Sitter spacetime model, writeThe hyperboloid with this quadratic form equal to $-1$ is exactly $SL(2,\mathbb{R})$. Its induced Lorentzian metric is preserved by $M\mapsto AM{B}^{-1}$ for $A,B\in SL(2,\mathbb{R})$. At the identity the metric is $g(U,V)=\frac{1}{2}\mathrm{tr}(UV)$ on traceless matrices, so it is the Killing metric divided by eight. This proves the geometric identification, including the metric rather than only the underlying manifold.

A trivial left-right action again forces $A=B$ to be a scalar commuting with all determinant-one matrices, hence $A=B=\pm I$. The differential is injective and both dimensions are six, soThis is the existing left-right double cover of SO0(2,2). The full hyperboloid isometry group is $O(2,2)$. Its other components can be obtained by also using $M\mapsto M^T$ and $M\mapsto CM{C}^{-1}$, $C=\mathrm{diag}(1,-1)$. Transposition reverses one timelike coordinate; conjugation by $C$ reverses one timelike and one spacelike coordinate, so these generate the four component classes. All isometries of these constant-curvature quadrics come from ambient orthogonal transformations: a value and an orthonormal derivative at one point determine an isometry, and the ambient group realizes every such choice.

Here ${\mathrm{AdS}}_3$ means the hyperboloid with periodic time. The physically common causally unwrapped spacetime is its universal cover, modeled by the universal covering Lie group of $SL(2,\mathbb{R})$, and is not itself $SL(2,\mathbb{R})$. Its isometries lift the above local symmetries, with the global covering and central identifications changed accordingly.

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 $F$.

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.

Use ambient quadratic form $Q=-(X^0{)}^2-(X^4{)}^2+(X^1{)}^2+(X^2{)}^2$. The printed sine coordinate has a duplicated superscript: it must be $X^2$, not another $X^1$. With that correction the embedding lies on $Q=-1$. Put $s=\sqrt{1+r^2}$. Differentiating the two timelike coordinates giveswhereas the two spatial coordinates give $(d{X}^1{)}^2+(d{X}^2{)}^2=d{r}^2+r^{2}d{\theta }^2$. Their difference is the induced Lorentzian metricThe polar coordinate $\theta$ collapses at $r=0$; this is a coordinate degeneracy, not a degeneracy of the metric.

For the matrix model of anti-de Sitter three-space, defineThen $\det M=-Q$. This linear identification of ambient space with real two-by-two matrices identifies $Q=-1$ with $SL(2,\mathbb{R})$. Its inverse recovers each $X$ by sums and differences of matrix entries, so the identification is global. The identity corresponds to $X^4=1$ and the other coordinates zero. For a traceless tangent matrix $U$, the induced squared norm isPolarization gives $g_I(U,V)=\frac{1}{2}\mathrm{tr}(UV)$.

The Killing form of ${\mathfrak{sl}}_2(\mathbb{R})$ is $B(U,V)=4\mathrm{tr}(UV)$. This normalization can be checked without assuming it: for $H=\mathrm{diag}(1,-1)$, $E=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ and $F=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$, the brackets are $[H,E]=2E$, $[H,F]=-2F$, $[E,F]=H$. Taking traces of their three-dimensional adjoint matrices gives $B(H,H)=8$, $B(E,F)=B(F,E)=4$, and all other basis pairings zero. These agree with $4\mathrm{tr}(UV)$.

Both left and right multiplication by matrices in $SL(2,\mathbb{R})$ preserve $-\det M$, hence its polarized ambient bilinear form and the induced metric. Equality at the identity therefore propagates throughout the group:In particular the Killing-form Einstein metric calculation gives $\mathrm{Ric}=-2g$, consistent with unit-radius three-dimensional Anti-de Sitter spacetime.

Every element of $SO(2,2)$ preserves the ambient quadratic form, hence the hyperboloid and its induced metric, and so acts by isometries. Define the left-right double cover of SO0(2,2) byThis is a group homomorphism, since successive actions multiply the two factors in the same order. It preserves $\det M$; as a transformation of the four-dimensional matrix space it has determinant $(\det A{)}^2(\det B^{-1}{)}^2=1$. Thus it takes values in $SO(2,2)$.

If its action is the identity, applying it to $M=I$ gives $A=B$. Commuting with every matrix then forces $A$ to be scalar. The condition $\det A=1$ leaves only $A=I$ or $A=-I$. HenceThe differential is $(a,b):M\mapsto aM-Mb$. A zero differential implies $a=b$ and that $a$ is scalar; tracelessness makes $a=b=0$. It is therefore an isomorphism onto the target Lie algebra, since both dimensions are six. The image is an open subgroup. Since $SL(2,\mathbb{R})$ is connected, the image is connected and is exactly the identity component $S{O}_0(2,2)$: an open subgroup of a connected group is also closed and must be the whole group. Thus the precise covering statement isIt is not onto the disconnected full $SO(2,2)$. For example, conjugation $M\mapsto DM{D}^{-1}$, $D=\mathrm{diag}(1,-1)$, sends $(X^0,X^4,X^1,X^2)$ to $(-X^0,X^4,X^1,-X^2)$ and belongs to the other component of $SO(2,2)$. It reverses the orientation of the negative two-plane and is not in the image of the connected source.

The Adjoint representation of a Lie group of $SL(2,\mathbb{R})$ preserves its Killing form of metric signature $(2,1)$ and gives a second double cover,Its kernel is the scalar centre; its differential is injective and both dimensions are three, so the same connected-image argument proves surjectivity onto the identity component. The full $SO(2,1)$ again has an additional component. Thus $S{O}_0(2,1)$ is locally the same Lie group geometry as $SL(2,\mathbb{R})$, with its Killing metric, and globally is its central quotient.

With the printed periodic time, the group model has topology $S^1\times {\mathbb{R}}^2$ and closed timelike curves, since varying $t$ at fixed spatial position gives a timelike circle. The causally unwrapped version of Anti-de Sitter spacetime instead has $t\in \mathbb{R}$ and is the universal covering Lie group. It must not be substituted silently for the periodic spacetime specified here.

For a connected source, first integrate on its universal covering Lie group. The resulting map descends to the original group exactly when it kills the discrete kernel of the covering map. For the circle group, a real-linear map $t\mapsto ct$ integrates to a circle endomorphism exactly when $c\in \mathbb{Z}$, since $2\pi \mathbb{Z}$ must map into itself. This is a global period condition invisible to the zero bracket of the circle's Lie algebra.